MA3220 Ordinary Differential Equations

MA3220 Ordinary Differential Equations Wong Yan Loi

.................................................. ............. ...... .......... ... ......... ... ....... . . . ... . . . ... . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . .................................................... . ..... ... . . . . . . . .................................. . . . . . . ........... ...... ... ... .......... . .......................... . . . . . . . . . . . . . . . . . . ...... ............................................ ........... ...... ..... .. .... .................... ............................................................. . . . . .... ........................................................................................... . . . . . . . . ... ............. .......................................................................................... ...... .... .... .......................................................................................................................................................... . . . . . . . . . .............. ....................................................................................................................................... ........ .... ................. ..................................................................................................................................... .... . . . . . . . ... ..... ..... ............ ................... ..................... ............................ ..... .. ......... .......... ................................................................................................ .... .......... ............................. ............................................................................ ............... ...... .... ... ........ ...... ........................................................................................................................... ..... .......... ............................ ............................................................................ .................... ...... .. ........ ..... ....................................................................................................................................................................................................................................... ... .... . ... . . .. .. . . .......... ............................ ........................... .................................... ................... .......... .... .... .. ........ .... ..................................... ... ............ ......................................................................................................................................................................................................................................... ........ .... ................................................................ ............................................................................................................................................................................................................................ . . . . . . ......... .......................................... .......... .................................................................................................... ...... ............................ ...... ... ............................... ....................... .......... ..................................................................................................... ... ........ ............................................................................................................................................................................................................................................................................................. ........ .... ............................................... ................ .............................................................................................................................................................................................................................................................................. ... . . . . ... ... ................ .............. ...... ................................................................................................................... .. ...................... .... .. ...... ................ ......... ................. ............................ ....................... .................. ......................................................... ... .. ... ............... ................ ................................................................................................................................................... .... ...................... ...... ......... ............... ...... ............... ........................................................................................................................................................... ... ... ... ............................... ...... .......................................................... ............................................................................................ ........................................... ...................................................... ............ ..................................................................................................................................................................................................................................................................................... ...... ......................... .. ..................... ..................... ....... ....... ................. .................................................................................................................. ..... ......... ... ... ................ ................. .................................................. ........... ... ..... ..................... ........................ ............................ ...... .. ......... ............ ..... ...................................................... ......... ........................................ .. ... ............................................................................................................................ .. . . .... ................... ......................... .............................. ........ ..................... ........... .............. .............................................. .......... .................................. .. ... ............................................ ..................................... ............... ...... .......................................................................................... ....................... ......... ... ... ... .................................. ................................... ..... ... ............................................... ......... ....................... ............ ... .......................................................................................................... .............................................. .. .. ... ................................................................................................. . . .................................................... ............... ............................... ................ ... ................... ............................................................................... ......... .................................... .. .. ... .............................. ...... .................................. . . .............................................................. ....... ............................ .. ................................................ ......... ............. ........ ..... ..... .. .. .. ...... ........ . ........................ ........................... ........ ...................... .......... ........ ......................................................................................... .... ........................................................................... .......... ....... .. .. .. ...... ...... ..................... ............... ....... ......................................... ................... ............................................................................. ................. ... ... ............. ... ... ................................................................. ... ... ............... ......................... ...... ................... ......... .............. ............... ....... ........ ................................ ... .................................................................................................................................... .... .... ............. ... ... .............................. ...... ........................... ... .............. ........................ ........ .................... ......... ... ......................................................................................................................... ... ................ ... ... ................ ............ ...... ........................... ... ... ............... .................. .......... ........ ...................... ............ ... ................................................................................................................................ .................... ... ... ...... .......... ............ ...... ........................... ... ... ............... ................. ......... ......... ....................... ......... ... ... ............................................................................................ ......................................................... ... ... ...... .......... ............ ...... ........................... ... ... ............ ... .. ............... ......... ......... ........................ ......... ... . ........................................................................................................................ ... ... ...... ......... ............... ...... ............................ ... . ... .... . . ... ......... .. ...... ...... .................. ...... .. .. ......................................................................................................................................................................................... .... ... ........ ........... ............ ...... ................................ .... ... .............. .... ... ............. ... .......... ............. ......................... ........ ............ .... . . ........................... ..................................................... .......... ..................................................... . . . . . . ... . ... . .. . . ... . ... ... ... ...... .......... ............ ...... ........................... ... . . . . . . ................... ..................... ........... .......... ...................... ....... .......... .... ................................................................................................................................................................................................................................................ . . ... . . ... ... ...... .......... ............ ...... ........................... ... . . . . . . .. ................... ... ...................................................................... ................................................................................................................... ................. ....... ......... ............... ..... ........ ... ... ... ... ...... ......... ............ ...... ............................. ... ............................................................................................................................................................................................................................................................................................................................................................................................................................................. ........... ........................... ......... ... ... ... ... ...... .......... ............ ........ ........................... ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. .. ... .. . .. . ... . . ... . .... . . . .. . .... .......... .. .... .. .... .... .. .... ... ... ... ............................ ...... ............................. .... ..... .............................. ... ........................... .................................................................................................................................................................................................................. ....... .............. .... ....... .. ... ... ... ............................ ...... ............................... ...... ..... ................................ ... .......... ............ .. .............................................................................................................................................................................................. ........... ...... ............. ...... ....... ... ... ... ... .............. ............. ...... ...................................... ....... ...................................................................................................................................... .......... .................................................................................................................................................................................................................................................................................................................................................................................................................. ............. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . . ... ... .... ....... ........... .... .................................. ......... .. .... ......... .......... ...... ... ..... .......... .......... .......... ......... ............ .... ... . .. ...... . ... . .. .. ... ... ... ................ .................... ............................................................ .................................................................................................................................................................................... ........... ..................................................................................................................................................................... ................. ............................. ......... ... ... ... ..................................... ........................................................................................................................................................................................................................... ...... ... ... ............................................................................................................................................................................................................................................................................................................................................ .............................................................................................................. ................................................................ ... ... ... ...... ........................... ............................................................................................................................................................................... ....... ... . . .................. ............................................................................................................................................. ........... ... . ... ... ........................................................................................................................................................................................................................... ........... ... ....................................................................................................................................... .............. ... ... ... ... .................................. ......... ........................................................................................................................................... ........... ... ........................................................................................................................................................... ... ... ... ............................................. ............... .............. ............... ......................................... .......... ........................ .............................................................................. ................... .... ... ... . ... .... ................. ............................... ............................... .............................................. ........... . ............................... .................................................................. . ... . .. .... ..... ...................................................................................................................................... .......... .................................................................................................... . . ... . . . . . . . . . . . . .......................................................... ..... ..... ............ ...... ......... ...................................................... ..... ....... .......... .. ... . . ........................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..... ..... .............. .................................................... ......... ........... ... ... . . ...... ...... .................... .......................................... . . .... . . . . . . ....................................... ... ...... ........ ............... ..... . . . . . . . . . . . . ....... ......... ..... ... . . . ......... .................. .......................................... . . ..... .............. ......... ............... ..... ...... ...................... ...... ...... ...... ....... ....... ....... ......... ........ . . . . . . . ............. . . ...........................................

2

Contents 1

2

3

4

First Order Differential Equations

5

1.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

1.2

Exact Equations, Integrating Factors . . . . . . . . . . . . . . . . . . . . . . . . .

8

1.3

First Order Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10

1.4

First Order Implicit Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14

1.5

Further Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

16

Linear Differential Equations

17

2.1

General Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17

2.2

Linear Equations with Constant Coefficients . . . . . . . . . . . . . . . . . . . . .

22

2.3

Operator Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

30

Second Order Linear Differential Equations

33

3.1

Exact 2nd Order Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33

3.2

The Adjoint Differential Equation and Integrating Factor . . . . . . . . . . . . . .

34

3.3

Lagrange’s Identity and Green’s Formula . . . . . . . . . . . . . . . . . . . . . .

35

3.4

Regular Boundary Value Problem . . . . . . . . . . . . . . . . . . . . . . . . . .

36

3.5

Regular Sturm-Liouville Boundary Value Problem . . . . . . . . . . . . . . . . .

38

3.6

Nonhomogeneous Boundary Value Problem . . . . . . . . . . . . . . . . . . . . .

42

3.7

Oscillations and the Sturm Separation Theorem . . . . . . . . . . . . . . . . . . .

46

3.8

Sturm Comparison Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

48

Linear Differential Systems

51

4.1

Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

51

4.2

Homogeneous Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . .

52

4.3

Non-Homogeneous Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . .

55

4.4

Homogeneous Linear Systems with Constant Coefficients . . . . . . . . . . . . . .

56

4.5

Higher Order Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . .

66

4.6

Introduction to the Phase Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . .

70

4.7

Linear Autonomous System in the Plane . . . . . . . . . . . . . . . . . . . . . . .

70

4.8

Appendix 1: Proof of Lemma 4.12 . . . . . . . . . . . . . . . . . . . . . . . . . .

77

3

4 5

6

CONTENTS Power Series Solutions 5.1 Power Series . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Series Solutions of First Order Equations . . . . . . . . . 5.3 Second Order Linear Equations and Ordinary Points . . . . 5.4 Regular Singular Points and the Method of Frobenius . . . 5.5 Bessel’s Equation . . . . . . . . . . . . . . . . . . . . . . 5.6 Appendix 2: Some Properties of the Legendre Polynomials Fundamental Theory of ODEs 6.1 Existence-Uniqueness Theorem . . . . . . . . . . 6.2 The Method of Successive Approximations . . . . 6.3 Convergence of the Successive Approximations . . 6.4 Non-local Existence of Solutions . . . . . . . . . . 6.5 Gronwall’s Inequality and Uniqueness of Solution . 6.6 Existence and Uniqueness of Solutions to Systems

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

83 . . . . . . 83 . . . . . . 86 . . . . . . 87 . . . . . . 93 . . . . . . 97 . . . . . . 102

. . . . . .

105 105 106 109 112 115 119

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

Chapter 1

First Order Differential Equations 1.1

Introduction

1. Ordinary differential equations. An ordinary differential equation (ODE for short) is a relation containing one real variable x, the real dependent variable y, and some of its derivatives y 0 , y 00 , · · · , y (n) , · · · , with respect to x. The order of an ODE is defined to be the order of the highest derivative that occurs in the equation. Thus, an n-th order ODE has the general form F (x, y, y 0 , · · · , y (n) ) = 0.

(1.1.1)

We shall always assume that (1.1.1) can be solved explicitly for y (n) in terms of the remaining n + 1 quantities as y (n) = f (x, y, y 0 , · · · , y (n−1) ), (1.1.2) where f is a known function of x, y, y 0 , · · · , y (n−1) . An n-th order ODE is linear if it can be written in the form a0 (x)y (n) + a1 (x)y (n−1) + · · · + an (x)y = r(x).

(1.1.3)

The functions aj (x), 0 ≤ j ≤ n are called coefficients of the equation. We shall always assume that a0 (x) 6≡ 0 in any interval in which the equation is defined. If r(x) ≡ 0, (1.1.3) is called a homogeneous equation. If r(x) 6≡ 0, (1.1.3) is said to be a non-homogeneous equation, and r(x) is called the non-homogeneous term. 2. Solutions. A functional relation between the dependent variable y and the independent variable x that satisfies the given ODE in some interval J is called a solution of the given ODE on J. A general solution of an n-th order ODE depends on n arbitrary constants, i.e. the solution y depends on x and n real constants c1 , · · · , cn . A first order ODE may be written as F (x, y, y 0 ) = 0. (1.1.4) 5

6

CHAPTER 1. FIRST ORDER DIFFERENTIAL EQUATIONS

In this chapter we consider only first order ODE. The function y = φ(x) is called an explicit solution of (1.1.4) in the interval J provided F (x, φ(x), φ0 (x)) = 0

for all x in J.

(1.1.5)

A relation of the form ψ(x, y) = 0 is said to be an implicit solution of (1.1.4) provided it determines one or more functions y = φ(x) which satisfy (1.1.5). The pair of equations x = x(t),

y = y(t)

(1.1.6)

is said to be a parametric solution of (1.1.4) if y(t) ˙ F x(t), y(t), = 0. x(t) ˙ Example 1.1 Consider the ODE: x + yy 0 = 0 for x ∈ (−1, 1). x2 + y 2 = 1 is an implicit solution while x = cos t, y = sin t, t ∈ (0, π) is a parametric solution.

3. Integral curves. The solutions of a first order ODE y 0 = f (x, y)

(1.1.7)

represent a one-parameter family of curves in the xy-plane. These are called integral curves. In other words, if y = y(x) is a solution to (1.1.7), then vector field F(x, y) = h1, f (x, y)i is tangent to the curve r(x) = hx, y(x)i at every point (x, y) since r0 (x) = F(x, y). 4. Elimination of constants: formation of ODE. Given a family of functions parameterized by some constants, a differential equation can be formed by eliminating the constants of this family and its derivatives. Example 1.2 The family of functions y = Aex + B sin x satisfies the ODE: y constants A and B are eliminated using the derivatives.

0000

− y = 0 when the

Exercise 1.1 Find the differential equation satisfied by the family of functions y = xc for x > 0, where c is a parameter. Ans: y 0 =

y ln y x ln x .

5. Separable equations. Typical separable equation can be written as y0 =

f (x) , g(y)

g(y)dy = f (x)dx.

or

The solution is given by Z

Z g(y)dy =

f (x)dx + c.

(1.1.8)

1.1. INTRODUCTION

7

Exercise 1.2 Solve y 0 = −2xy, y(0) = 1. 2

Ans: y = e−x . The equation y 0 = f ( xy ) can be reduced to a separable equation by letting u = f (u) = y 0 = u + xu0 , Z Z du dx = + c. f (u) − u x

y x,

i.e. y = xu. So

Exercise 1.3 Solve 2xyy 0 + x2 − y 2 = 0. Ans: x2 + y 2 = cx. 6. Homogeneous equations. A function is called homogeneous of degree n if f (tx, ty) = tn f (x, y) for all x, y, t. p For example x2 + y 2 and x + y are homogeneous of degree 1, x2 + y 2 is homogeneous of degree 2 and sin(x/y) is homogeneous of degree 0. The ODE M (x, y) + N (x, y)y 0 = 0 is said to be homogeneous of degree n if both M (x, y) and N (x, y) are homogeneous of degree n. If we write the above DE as y 0 = f (x, y), where f (x, y) = −M (x, y)/N (x, y). Then f (x, y) is homogeneous of degree 0. To solve the DE y 0 = f (x, y), where f is homogeneous of degree 0, we use the substitution y = zx. Then dy dz =z+x . dx dx Thus the DE becomes

dz = f (x, zx) = x0 f (1, z) = f (1, z). dx Consequently, the variables can be separated to yield z+x

dz dx = , f (1, z) − z x and integrating both sides will give the solution. x+y Exercise 1.4 Solve y 0 = x−y . p Ans: tan−1 (y/x) = ln x2 + y 2 + c.

Example 1.3 An equation in the form y0 =

a1 x + b1 y + c1 . a2 x + b2 y + c2

can be reduced to a homogeneous equation by a suitable substitution x = z + h, y = w + k when a1 b2 6= a2 b1 , where h and k are solutions of the system of linear equations a1 h + b1 k + c1 = 0, a2 h + b2 k + c2 = 0.

8


Exercise 1.5 Solve y 0 = Ans: tan−1

y−1 x−1

= ln

x+y−2 x−y .

p (x − 1)2 + (y − 1)2 + c.

Exercise 1.6 Solve (x + y + 1) + (2x + 2y + 1)y 0 = 0. Ans: x + 2y + ln |x + y| = c, x + y = 0.

1.2

Exact Equations, Integrating Factors

1. Exact equations. We can write a first order ODE in the following form M (x, y)dx + N (x, y)dy = 0.

(1.2.1)

(1.2.1) is called exact if there exists a function u(x, y) such that M (x, y)dx + N (x, y)dy = du =

∂u ∂u dx + dy. ∂x ∂y

Once (1.2.1) is exact, the general solution is given by u(x, y) = c, where c is an arbitrary constant. Theorem 1.1 Assume M and N together with their first partial derivatives are continuous in the rectangle S: |x − x0 | < a, |y − y0 | < b. A necessary and sufficient condition for (1.2.1) to be exact is ∂M ∂N = for all (x, y) in S. (1.2.2) ∂y ∂x When (1.2.2) is satisfied, a general solution of (1.2.1) is given by u(x, y) = c, where Z x Z y u(x, y) = M (s, y)ds + N (x0 , t)dt x0

(1.2.3)

y0

and c is an arbitrary constant. Rx Ry Proof. Let u(x, y) = x0 M (s, y)ds + y0 N (x0 , t)dt. We have to show that ∂u ∂x = M (x, y) and ∂u is immediate by the fundamental theorem of calculus. For the ∂y = N (x, y). The first equality Rx ∂ Rx ∂ second equality, we have ∂u = M (s, y)ds + N (x0 , y) = x0 ∂s N (s, y)ds + N (x0 , y) = ∂y x0 ∂y N (x, y) − N (x0 , y) + N (x0 , y) = N (x, y). Remark. In Theorem 1.1, the rectangle S can be replaced by any region which does not include any “hole”. In that case, the proof is by Green’s theorem. Exercise 1.7 Solve (x3 + 3xy 2 )dx + (3x2 y + y 3 )dy = 0. Ans: x4 + 6x2 y 2 + y 4 = c. 2. Integrating factors.

1.2. EXACT EQUATIONS, INTEGRATING FACTORS

9

A non-zero function µ(x, y) is an integrating factor of (1.2.1) if the equivalent differential equation µ(x, y)M (x, y)dx + µ(x, y)N (x, y)dy = 0

(1.2.4)

is exact. If µ is an integrating factor of (1.2.1) then (µM )y = (µN )x , i.e. N µx − M µy = µ(My − Nx ).

(1.2.5)

One may look for an integrating factor of the form µ = µ(v), where v is a known function of x and y. Plugging into (1.2.5) we find M y − Nx 1 dµ = . (1.2.6) µ dv N vx − M v y If

My −Nx N vx −M vy

is a function of v alone, say, φ(v), then µ=e

Rv

φ(v)dv

is an integrating factor of (1.2.1). Let v = x. If of (1.2.1).

My −Nx N

Let v = y. If − of (1.2.1).

is a function of x alone, say, φ1 (x), then e

My −Nx M

Rx

φ1 (x)dx

is an integrating factor

Ry

φ2 (y)dy

is an integrating factor

is a function of y alone, say, φ2 (y), then e

M −N

Let v = xy. If yNy−xMx is a function of v = xy alone, say φ3 (xy), then e factor of (1.2.1).

R xy

Exercise 1.8 Solve (x2 y + y + 1) + x(1 + x2 )y 0 = 0. Ans: xy + tan−1 x = c.

Exercise 1.9 Solve (y − y 2 ) + xy 0 = 0 Ans: y = (1 − cx)−1 , y = 0.

Exercise 1.10 Solve (xy 3 + 2x2 y 2 − y 2 ) + (x2 y 2 + 2x3 y − 2x2 )y 0 = 0 Ans: exy (1/x + 2/y) = c, y = 0. 3. Find integrating factors by inspection. The followings are some differential formulas that are often useful. x ydx − xdy d( ) = y y2

φ3 (v)dv

is an integrating

10


d(xy) = xdy + ydx d(x2 + y 2 ) = 2xdx + 2ydy d(tan−1

ydx − xdy x )= y x2 + y 2

x ydx − xdy d(ln | |) = y xy We see that the very simple ODE ydx−xdy = 0 has 1/x2 , 1/y 2 , 1/(x2 +y 2 ) and 1/xy as integrating factors. Example 1.4 Solve xdy + ydx = x cos xdx. Solution. The differential xdy + ydx can be written as d(xy). Thus the DE can be written as d(xy) = x cos xdx. Integrating, we have xy = x sin x + cos x + c. Exercise 1.11 Solve (x2 y 3 + y)dx + (x − x3 y 2 )dy = 0. Ans: ln |x/y| = 1/(2x2 y 2 ) + c and y = 0.

1.3

First Order Linear Equations

1. Homogeneous equations. A first order homogeneous linear equation is of the form y 0 + p(x)y = 0,

(1.3.1)

where p(x) is a continuous function on an interval J. Let P (x) = by eP (x) , we get d P (x) [e y] = 0, dx

Rx a

p(s)ds. Multiplying (1.3.1)

so eP (x) y = c. The general solution of (1.3.1) is given by y(x) = ce−P (x) ,

Z where P (x) =

x

p(s)ds.

(1.3.2)

a

2. Non-homogeneous equations. Now consider a first order non-homogeneous linear equation y 0 + p(x)y = q(x), where p(x) and q(x) are continuous functions on an interval J. Let P (x) = (1.3.3) by eP (x) we get d P (x) [e y] = eP (x) q(x). dx

(1.3.3) Rx a

p(s)ds. Multiplying

1.3. FIRST ORDER LINEAR EQUATIONS Thus eP (x) y(x) =

11

Z

x

eP (t) q(t)dt + c.

a

The general solution is given by −P (x)

y(x) = e Z P (x) =

Z [

x

eP (t) q(t)dt + c],

where

a

(1.3.4)

x

p(s)ds.

a

Exercise 1.12 Solve y 0 − y = e2x . Ans: y = cex + e2x . 3. The Bernoulli equation. An ODE in the form y 0 + p(x)y = q(x)y n ,

(1.3.5)

where n 6= 0, 1, is called the Bernoulli equation. The functions p(x) and q(x) are continuous functions on an interval J. Let u = y 1−n . Substituting into (1.3.5) we get u0 + (1 − n)p(x)u = (1 − n)q(x).

(1.3.6)

This is a first order linear ODE.

Exercise 1.13 Solve xy 0 + y = x4 y 3 . Ans:

1 y2

= −x4 + cx2 , or y = 0.

4. The Riccati equation. An ODE of the form y 0 = P (x) + Q(x)y + R(x)y 2

(1.3.7)

is called the Riccati equation. The functions P (x), Q(x), R(x) are continuous on an interval J. In general, the Riccati equation cannot be solved by a sequence of integrations. However, if a particular solution is known, then (1.3.7) can be reduced to a linear equation, and thus is solvable.

Theorem 1.2 Let y = y0 (x) be a particular solution of the Riccati equation (1.3.7). Set Z x H(x) = [Q(t) + 2R(t)y0 (t)]dt, x0 Z x h i Z(x) = e−H(x) c − eH(t) R(t)dt ,

(1.3.8)

x0

where c is an arbitrary constant. Then the general solution is given by y = y0 (x) +

1 . Z(x)

(1.3.9)

12


Proof. In (1.1.7) we let y = y0 (x) + u(x) to get y00 + u0 = P + Q(y0 + u) + R(y0 + u)2 . Since y0 satisfies (1.3.7), we have y00 = P + Qy0 + Ry02 . From these two equalities we get u0 = (Q + 2Ry0 )u + Ru2 .

(1.3.10)

This is a Bernoulli equation with n = 2. Set Z = u−1 and reduce (1.3.10) to Z 0 + (Q + 2Ry0 )Z = −R.

(1.3.11)

(1.3.11) is a linear equation and the solution is given by (1.3.8).

Exercise 1.14 Solve y 0 = y/x + x3 y 2 − x5 . Note y0 = x is a solution. 5

Ans: ce2x

/5

=

y−x y+x .

From (1.3.8), (1.3.9), the general solution y of the Riccati equation (1.3.7) can be written as y= where

cF (x) + G(x) , cf (x) + g(x)

(1.3.12)

f (x) = e−H(x) , g(x) = −e

−H(x)

Z

x

eH(t) R(t)dt,

x0

F (x) = y0 (x)f (x),

G(x) = y0 g(x) + 1.

Given four distinct functions p(x), q(x), r(x), s(x), we define the cross ratio by (p − q)(r − s) . (p − s)(r − q)

Property 1. The cross ratio of four distinct particular solutions of a Riccati equation is independent of x. Proof. From (1.3.12), the four solutions can be written as yj (x) =

cj F (x) + G(x) . cj f (x) + g(x)

Computations show that (y1 − y2 )(y3 − y4 ) (c1 − c2 )(c3 − c4 ) = . (y1 − y4 )(y3 − y2 ) (c1 − c4 )(c3 − c2 )

1.3. FIRST ORDER LINEAR EQUATIONS

13

The right-hand side is independent of x.

As a consequence we get Property 2. Suppose y1 , y2 , y3 are three distinct particular solutions of a Riccati equation (1.3.7). Then the general solution is given by (y1 − y2 )(y3 − y) = c, (y1 − y)(y3 − y2 )

(1.3.13)

where c is an arbitrary constant. Property 3. Suppose that y1 and y2 are two distinct particular solutions of a Riccati equation (1.3.7), then its general solution is given by Z y − y1 = [y1 (x) − y2 (x)]R(x)dx + c, ln y − y2

(1.3.14)

where c is an arbitrary constant. Proof. y and yj satisfy (1.3.7). So y 0 − yj0 = (y − yj )[Q + R(y + yj )], y 0 − yj0 = Q + R(y + yj ). y − yj Thus

y 0 − y20 y 0 − y10 − = R(y1 − y2 ). y − y1 y − y2

Integrating yields (1.3.14).

Exercise 1.15 Solve y 0 = e−x y 2 . Note y1 = ex and y2 = 0 are 2 solutions. Ans: y =

ex 1−Aex ,

y = 0.

Exercise 1.16 A natural generalization of Riccati’s equation is Abel’s equation y 0 = P (x) + Q(x)y + R(x)y 2 + S(x)y 3 , where P (x), Q(x), R(x) and S(x) are continuous functions of x on an interval J. Using the substitution z = y/x, solve the equation y0 = 3

y + xy 2 − 3y 3 . x

Ans: x/y + 3 ln |x/y − 3| = − x3 + c, y = x/3 and y = 0.

14

1.4


First Order Implicit Equations

In the above we discussed first order explicit equations, i.e. equations in the form y 0 = f (x, y). In this section we discuss solution of some first order explicit equations F (x, y, y 0 ) = 0

(1.4.1)

y = f (x, y 0 ).

(1.4.2)

which are not solvable in y 0 . 1. Method of differentiation. Consider an equations solvable in y: Let p = y 0 . Differentiating y = f (x, p) we get [fx (x, p) − p]dx + fp (x, p)dp = 0.

(1.4.3)

This is a first order explicit equation in x and p. If p = φ(x) is a solution of (1.4.3), then y = f (x, φ(x)) is a solution of (1.4.2).

Example 1.5 Clairaut’s equation is the equation of the form y = xy 0 + f (y 0 ), where f has continuous first order derivative. Let p = y 0 . We have y = xp + f (p). Differentiating we get [x + f 0 (p)]p0 = 0. When p0 = 0 we have p = c and (1.4.4) has a general solution y = cx + f (c). When x + f 0 (p) = 0 we get a solution of (1.4.4) given by parameterized equations x = −f 0 (p),

y = −pf 0 (p) + f (p).

Exercise 1.17 Solve Clairaut’s equation y = xy 0 − y 02 /4. Ans: y = cx − c2 /4, y = x2 . Exercise 1.18 Let C be the curve with parametric equation x = −f 0 (p), y = −pf 0 (p) + f (p). Show that the tangent to C at the point p = c has the equation y = cx + f (c).

(1.4.4)

1.4. FIRST ORDER IMPLICIT EQUATIONS

15

2. Method of parameterization. This method can be used to solve equations where either x or y is missing. Consider F (y, y 0 ) = 0,

(1.4.5)

where x is missing. Let p = y 0 and write (1.4.5) as F (y, p) = 0. It determines a family of curves in yp plane. Let y = g(t), p = h(t) be one of the curves, i.e. F (g(t), h(t)) = 0. Since dy dy g 0 (t)dt dx = 0 = = , y p h(t) R t 0 (t) dt + c. The solutions of (1.4.5.) are given by we have x = t0 gh(t) Z

t

x= t0

g 0 (t) dt + c, h(t)

y = g(t).

This method can also be applied to the equations F (x, y 0 ) = 0, where y is missing.

Exercise 1.19 Solve y 2 + y 02 − 1 = 0. Ans: y = cos(c − x). For the equation F (y, y 0 ) = 0 or F (x, y 0 ) = 0, if y 0 can be solved in terms of y or x, then the equation becomes explicit. Exercise 1.20 Solve y 02 − (2x + ex )y 0 + 2xex = 0. Ans: y = x2 + c1 , y = ex + c2 . 3. Reduction of order. Consider the equation F (x, y 0 , y 00 ) = 0,

(1.4.6)

where y is missing. Let p = y 0 . Then y 00 = p0 . Write (1.4.6) as F (x, p, p0 ) = 0.

(1.4.7)

It is a first order equation in x and p. If p = φ(x, c1 ) is a general solution of (1.4.7), then the general solution of (1.4.6) is Z x

y=

φ(t, c1 )dt + c2 . x0

16


Exercise 1.21 Solve xy 00 − y 0 = 3x2 . Ans: y = x3 + c1 x2 + c2 . Consider the equation F (y, y 0 , y 00 ) = 0, where x is missing. Let p = y 0 . Then y 00 =

dp dx

=

F (y, p, p

dp dy dy dx

=

(1.4.8) dp dy p.

Write (1.4.8) as

dp ) = 0. dy

(1.4.9)

It is a first order equation in y and p. If p = ψ(y, c1 ) is a general solution of (1.4.9), then we solve the equation y 0 = ψ(y, c1 ) to get a general solution of (1.4.8).

Exercise 1.22 Solve y 00 + k 2 y = 0, where k is a positive constant. Ans: y = c1 sin(kx) + c2 cos(kx).

1.5

Further Exercises

Exercise 1.23 Solve the differential equation y 2 dx + (xy − x3 )dy = 0 by using the substitution u = x−2 . Ans: 3y = 2x2 + cx2 y 3 and y = 0. Exercise 1.24 Solve the differential equation y 2 dx + (xy + tan(xy))dy = 0 by using (i) the substitution u = xy, (ii) the integrating factor cos(xy). Ans: y sin(xy) = c. Exercise 1.25 Solve the differential equation (x2 + 3 ln y)ydx − xdy = 0 (i) by using the substitution y = ez , (ii) by finding an integrating factor using the formula (1.2.6). Ans: ln y = −x2 + cx3 . ( 00

Exercise 1.26 Solve y + y =

( Ans: y =

0 if x ≤ 1 , y(0) = 0, y 0 (0) = 0. 1 − x if x > 1

0 if x ≤ 1 sin(x − 1) + 1 − x if x > 1.

Chapter 2

Linear Differential Equations 2.1

General Theory

Consider n-th order linear equation y (n) + a1 (x)y (n−1) + · · · + an−1 (x)y 0 + an (x)y = f (x),

(2.1.1)

k

d y where y (k) = dx k . Throughout this section we assume that aj (x)’s and f (x) are continuous functions defined on the interval (a, b). When f (x) 6≡ 0, (2.1.1) is called a non-homogeneous equation. The associated homogeneous equation is

y (n) + a1 (x)y (n−1) + · · · + an−1 (x)y 0 + an (x)y = 0. Let us begin with the initial value problem:   y (n) + a1 (x)y (n−1) + · · · + an (x)y = f (x),       y(x0 ) = y0 ,   y 0 (x0 ) = y1 ,     ·········      y (n−1) (x ) = y 0 n−1 .

(2.1.2)

(2.1.3)

Theorem 2.1 (Existence and Uniqueness Theorem) Assume that a1 (x), · · · , an (x) and f (x) are continuous functions defined on the interval (a, b). Then for any x0 ∈ (a, b) and for any numbers y0 , · · · , yn−1 , the initial value problem (2.1.3) has a unique solution defined on (a, b). Especially if aj (x)’s and f (x) are continuous on R then for any x0 and y0 , · · · , yn−1 , the initial value problem (2.1.3) has a unique solution defined on R. Proof of this theorem will be given in later chapter. Corollary 2.2 Let y = y(x) be a solution of the homogeneous equation (2.1.2) in an interval (a, b). Assume that there exists x0 ∈ (a, b) such that y(x0 ) = 0, y 0 (x0 ) = 0, · · · , y (n−1) (x0 ) = 0. Then y(x) ≡ 0 on (a, b). 17

(2.1.4)

18

CHAPTER 2. LINEAR DIFFERENTIAL EQUATIONS

Proof. y is a solution of the initial value problem (2.1.2), (2.1.4). From Theorem 2.1, this problem has a unique solution. Since φ(x) ≡ 0 is also a solution of the problem, we have y(x) ≡ 0 on (a, b). In the following we consider the general solutions of (2.1.1) and (2.1.2). Given continuous functions aj (x), j = 0, 1, · · · , n and f (x), define an operator L by L[y] = a0 (x)y (n) + a1 (x)y (n−1) + · · · + an (x)y.

(2.1.5)

Property 1. L[cy] = cL[y] for any constant c. Property 2. L[u + v] = L[u] + L[v]. Proof. Let’s verify Property 2. L[u + v] = a0 (x)(u + v)(n) + a1 (x)(u + v)(n−1) + · · · + an (x)(u + v) = a0 (x)u(n) + a1 (x)u(n−1) + · · · + an (x)u + a0 (x)v (n) + a1 (x)v (n−1) + · · · + an (x)v = L[u] + L[v].

Definition. An operator satisfying Properties 1 and 2 is called a linear operator. The differential operator L defined in (2.1.3) is a linear operator. Note that (2.1.1) and (2.1.2) can be written as L[y] = f (x),

(2.1.1’)

L[y] = 0.

(2.1.2’)

and

From Properties 1 and 2 we get the following conclusion. Theorem 2.3 (1) If y1 and y2 are solutions of the homogeneous equation (2.1.2) in an interval (a, b), then for any constants c1 and c2 , y = c1 y1 + c2 y2 is also a solution of (2.1.2) in the interval (a, b). (2) If yp is a solution of (2.1.1) and yh is a solution of (2.1.2) on an interval (a, b), then y = yh + yp is also a solution of (2.1.1) in the interval (a, b). Proof. (1) As L[c1 y1 + c2 y2 ] = c1 L[y1 ] + c2 L[y2 ] = 0, we see that c1 y1 + c2 y2 is also a solution of (2.1.2). (2) L[yh + yp ] = L[yh ] + L[yp ] = 0 + f (x), we see that y = yh + yp is a solution of (2.1.1).

2.1. GENERAL THEORY

19

In order to discuss structures of solutions, we need the following definition. Definition. Functions φ1 (x), · · · , φk (x) are linearly dependent on (a, b) if there exists constants c1 , · · · , ck , not all zero, such that c1 φ1 (x) + · · · + ck φk (x) = 0 for all x ∈ (a, b). A set of functions are linearly independent on (a, b) if they are not linearly dependent on (a, b).

Lemma 2.4 Functions φ1 (x), · · · , φk (x) are linearly dependent on (a, b) if and only if the following vector-valued functions     φ1 (x) φk (x)  φ0 (x)   φ0 (x)   1    (2.1.6)  , ··· ,  k   ···   ···  (n−1)

φ1

(x)

(n−1)

φk

(x)

are linearly dependent on (a, b). Proof. “⇐=” is obvious. To show “=⇒”, assume that φ1 , · · · , φk are linearly dependent on (a, b). There exists constants c1 , · · · , ck , not all zero, such that, for all x ∈ (a, b), c1 φ1 (x) + · · · + ck φk (x) = 0. Differentiating this equality successively we find that c1 φ01 (x) + · · · + ck φ0k (x) = 0, ·················· (n−1)

c1 φ 1

(n−1)

(x) + · · · + ck φk

(x) = 0.

Thus 

   φ1 (x) φk (x)  φ0 (x)   φ0 (x)      c1  1  + · · · + ck  k =0  ···   ···  (n−1) (n−1) φ1 (x) φk (x) for all x ∈ (a, b). Hence the k vector-valued functions are linearly dependent on (a, b). n Recall that, n vectors in R are linearly dependent if and only if the determinant of matrix formed by these vectors is zero. Definition. The Wronskian of n functions φ1 (x), · · · , φn (x) is defined by φ1 (x) ··· W (φ1 , · · · , φn )(x) = · · · ··· (n−1) φ (x) · · · 1

φn (x) ··· . (n−1) φn (x)

(2.1.7)

20


Note that Wronskian of φ1 , · · · , φn is the determinant of the matrix formed by the vector-valued functions given in (2.1.6). Theorem 2.5 Let y1 (x), · · · , yn (x) be n solutions of (2.1.2) on (a, b) and let W (x) be their Wronskian. (1) y1 (x), · · · , yn (x) are linearly dependent on (a, b) if and only if W (x) ≡ 0 on (a, b). (2) y1 (x), · · · , yn (x) are linearly independent on (a, b) if and only if W (x) does not vanish on (a, b). Corollary 2.6 (1) The Wronskian of n solutions of (2.1.2) is either identically zero, or nowhere zero. (2) n solutions y1 , · · · , yn of (2.1.2) are linearly independent on (a, b) if and only if the set of vectors     yn (x0 ) y1 (x0 )  y 0 (x )   y 0 (x )   n 0   1 0   , ··· ,       ··· ··· (n−1)

y1

(x0 )

(n−1)

yn

(x0 )

are linearly independent for some x0 ∈ (a, b). Proof of Theorem 2.5. Let y1 , · · · , yn be solutions of (2.1.2) on (a, b), and let W (x) be their Wronskian. Step 1. We first show that, if y1 , · · · , yn are linearly dependent on (a, b), then W (x) ≡ 0. Since these solutions are linearly dependent, from Lemma 2.4, n vector-valued functions     yn (x) y1 (x)  y 0 (x)   y 0 (x)    n  1    , ··· ,   ···   ···  (n−1)

y1

(x)

(n−1)

yn

(x)

are linearly dependent on (a, b). Thus for all x ∈ (a, b), the determinant of the matrix formed by these vectors, namely, the Wronskian of y1 , · · · , yn , is zero. Step 2. Now, assume that the Wronskian W (x) of n solutions y1 , · · · , yn vanishes at x0 ∈ (a, b). We shall show that y1 , · · · , yn are linearly dependent on (a, b). Since W (x0 ) = 0, the n vectors     y1 (x0 ) yn (x0 )  y 0 (x )   y 0 (x )   1 0   n 0   , ··· ,       ··· ··· (n−1)

y1

(x0 )

(n−1)

yn

(x0 )

are linearly dependent. Thus there exist n constants c1 , · · · , cn , not all zero, such that     y1 (x0 ) yn (x0 )  y 0 (x )   y 0 (x )  0    n 0  c1  1  + · · · + cn  =0     ··· ··· (n−1)

y1

(x0 )

(n−1)

yn

(x0 )

(2.1.8)

2.1. GENERAL THEORY

21

Define y0 (x) = c1 y1 (x) + · · · + cn yn (x). From Theorem 2.3, y0 is a solution of (2.1.2). From (2.1.8), y0 satisfies the initial conditions y(x0 ) = 0, y 0 (x0 ) = 0, · · · , y (n−1) (x0 ) = 0.

(2.1.9)

From Corollary 2.2, we have y0 ≡ 0, namely, c1 y1 (x) + · · · + cn yn (x) = 0 for all x ∈ (a, b). Thus y1 , · · · , yn are linearly dependent on (a, b).

Example 2.1 Consider the differential equation y 00 − x1 y 0 = 0 on the interval (0, ∞). Both φ1 (x) = 1 x2 1 and φ2 (x) = x2 are solutions of the differential equation. W (φ1 , φ2 )(x) = = 2x 6= 0 0 2x for x > 0. Thus φ1 and φ2 are linearly independent solutions.

Theorem 2.7 (1) Let a1 (x), · · · , an (x) and f (x) be continuous on the interval (a, b). The homogeneous equation (2.1.2) has n linearly independent solutions on (a, b). (2) Let y1 , · · · , yn be n linearly independent solutions of (2.1.2) defined on (a, b). The general solution of (2.1.2) is given by y(x) = c1 y1 (x) + · · · + cn yn (x),

(2.1.10)

where c1 , · · · , cn are arbitrary constants. Proof. (1) Fix x0 ∈ (a, b). For k = 1, 2, · · · , n, let yk be the solution of (2.1.2) satisfying the initial conditions  0 if j 6= k − 1, (j) yk (x0 ) = 1 if j = k − 1. The n vectors 

   y1 (x0 ) yn (x0 )  y 0 (x )   y 0 (x )   1 0   n 0   , ··· ,       ··· ··· (n−1) (n−1) y1 (x0 ) yn (x0 ) are lineally independent since they form the identity matrix. From Corollary 2.6, y1 , · · · , yn are linearly independent on (a, b). From Theorem 2.3, for any constants c1 , · · · , cn , y = c1 y1 + · · · + cn yn is a solution of (2.1.2). (2) Now let y1 , · · · , yn be n linearly independent solutions of (2.1.2) on (a, b). We shall show that the general solution of (2.1.2) is given by y = c1 y1 + · · · + cn yn .

(2.1.11)

22


Given a solution y˜ of (2.1.2), and fix x0 ∈ (a, b). Since y1 , · · · , yn are linearly independent on (a, b), the vectors     yn (x0 ) y1 (x0 )  y 0 (x )   y 0 (x )   n 0   1 0    , ··· ,      ··· ··· (n−1) (n−1) yn y1 (x0 ) (x0 ) are linearly independent vectors. They form a basis for Rn . Thus the vector   y˜(x0 )  y˜0 (x )    0     ··· y˜(n−1) (x0 ) can be represented as a linear combination of the n vectors, namely, there exist n constants c˜1 , · · · , cñ such that       yn (x0 ) y1 (x0 ) y˜(x0 )  y 0 (x )   y 0 (x )   y˜0 (x )   n 0   1 0    0 .  + · · · + cñ   = c˜1         ··· ··· ··· (n−1) (n−1) (n−1) y˜ (x0 ) (x0 ) yn y1 (x0 ) Let φ(x) = y˜(x) − [˜ c1 y1 (x) + · · · + cñ yn (x)]. φ(x) is a solution of (2.1.2) and satisfies the initial conditions (2.1.4) at x = x0 . By Corollary 2.2, φ(x) ≡ 0 on (a, b). Thus y˜(x) = c˜1 y1 (x) + · · · + cñ yn (x). So (2.1.11) gives a general solution of (2.1.2). Any set of n linearly independent solutions is called a fundamental set of solutions.

Now we consider the non-homogeneous equation (2.1.1). We have

Theorem 2.8 Let yp be a particular solution of (2.1.1), and y1 , · · · , yn be a fundamental set of solutions for the associated homogeneous equation (2.1.2). The general solution of (2.1.1) is given by y(x) = c1 y1 (x) + · · · + cn yn (x) + yp (x). (2.1.12) Proof. Let y be a solution of the non-homogeneous equation. Then y − yp is a solution of the homogeneous equation. Thus y(x) − yp (x) = c1 y1 (x) + · · · + cn yn (x).

2.2

Linear Equations with Constant Coefficients

Let us begin with second order linear equation with constant coefficients y 00 + ay 0 + by = 0,

(2.2.1)

2.2. LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS

23

where a and b are constants. We look for a solution of the form y = eλx . Plugging into (2.2.1) we find that, eλx is a solution of (2.2.1) if and only if λ2 + aλ + b = 0.

(2.2.2)

(2.2.2) is called the auxiliary equation or characteristic equation of (2.2.1). The roots of (2.2.2) are called characteristic values (or eigenvalues): p 1 (−a + a2 − 4b), 2 p 1 λ2 = (−a − a2 − 4b). 2

λ1 =

1. If a2 − 4b > 0, (2.2.2) has two distinct real roots λ1 , λ2 , and the general solutions of (2.2.1) is y = c1 eλ1 x + c2 eλ2 x . 2. If a2 − 4b = 0, (2.2.2) has one real root λ (we may say that (2.2.2) has two equal roots λ1 = λ2 ). The general solution of (2.2.1) is y = c1 eλx + c2 xeλx . 3. If a2 − 4b < 0, (2.2.2) has a pair of complex conjugate roots λ1 = α + iβ,

λ2 = α − iβ.

The general solution of (2.2.1) is y = c1 eαx cos(βx) + c2 eαx sin(βx).

Example 2.2 Solve y 00 + y 0 − 2y = 0, y(0) = 4, y 0 (0) = −5. Ans: λ1 = 1, λ2 = −2, y = ex + 3e−2x .

Example 2.3 Solve y 00 − 4y 0 + 4y = 0, y(0) = 3, y 0 (0) = 1. Ans: λ1 = λ2 = 2, y = (3 − 5x)e2x .

Example 2.4 Solve y 00 − 2y 0 + 10y = 0. Ans: λ1 = 1 + 3i, λ2 = 1 − 3i, y = ex (c1 cos 3x + c2 sin 3x). Now we consider n-th order homogeneous linear equations with constant coefficients y (n) + a1 y (n−1) + · · · + an−1 y 0 + an y = 0, where a1 , · · · , an are real constants.

(2.2.3)

24


y = eλx is a solution of (2.2.3) if and only if λ satisfies λn + a1 λn−1 + · · · + an−1 λ + an = 0.

(2.2.4)

The solutions of (2.2.4) are called characteristic values or eigenvalues for the equation (2.2.3). Let λ1 , · · · , λs be the distinct eigenvalues for (2.2.3). Then we can write λn + a1 λn−1 + · · · + an−1 λ + an = (λ − λ1 )m1 (λ − λ2 )m2 · · · (λ − λs )ms ,

(2.2.5)

where m1 , · · · , ms are positive integers and m1 + · · · + ms = n. We call them the multiplicity of the eigenvalues λ1 , · · · , λs respectively. Lemma 2.9 Assume λ is an eigenvalue of (2.2.3) of multiplicity m. (i) eλx is a solution of (2.2.3). (ii) If m > 1, then for any positive integer 1 ≤ k ≤ m − 1, xk eλx is a solution of (2.2.3). (iii) If λ = α + iβ, then xk eαx cos(βx), xk eαx sin(βx) are solutions of (2.2.3), where 0 ≤ k ≤ m − 1. Theorem 2.10 Let λ1 , · · · , λs be the distinct eigenvalues for (2.2.3), with multiplicity m1 , · · · , ms respectively. Then (2.2.3) has a fundamental set of solutions eλ1 x , xeλ1 x , · · · , xm1 −1 eλ1 x ; ························ ;

(2.2.6)

eλs x , xeλs x , · · · , xms −1 eλs x . Proof of Lemma 2.9 and 2.10 Consider the n-th order linear equation with constant coefficients y (n) + a1 y (n−1) + · · · + an−1 y 0 + an y = 0,

(A1)

k

d y (n) where y (k) = dx + a1 y (n−1) + · · · + an−1 y 0 + an y and p(z) = z n + a1 z n−1 + k . Let L(y) = y · · · + an−1 z + an . Note that p is a polynomial in z of degree n. Then we have

L(ezx ) = p(z)ezx 2

(A2) 2

∂ ∂ Before we begin the proof, let’s observe that ∂z∂x ezx = ∂x∂z ezx by Clairaut’s theorem because ezx is differentiable in (x, z) as a function of two variables and all the higher order partial derivatives exist and continuous. That means we can interchange the order of differentiation with respect to x d d zx and z as we wish. Therefore dz L(ezx ) = L( dz e ). For instance, one may verify directly that

d dk zx dk d zx k zx k−1 zx (e ) = xz e + kz e = ( e ). dz dxk dxk dz


25

Here one may need to use Leibniz’s rule of taking the k-th derivative of a product of two functions: (k)

(u · v)

k X k (i) (k−i) = u v . i i=0

(A3)

k

k

d d zx zx ) = L( dz ). (Strictly speaking, partial derivative notations should be More generally, dz k L(e ke used.) Now let’s prove our results.

(1) If λ is a root of p, then L(eλx ) = 0 by (A2) so that eλx is a solution of (A1). (2) If λ is a root of p of multiplicity m, then p(λ) = 0, p0 (λ) = 0, p00 (λ) = 0, . . . , p(m−1) (λ) = 0. Now for k = 1, . . . , m − 1, differentiating (A2) k times with respect to z, we have L(xk ezx ) = L(

k X dk zx dk dk k (i) zx zx e ) = L(e ) = (p(z)e ) = p (z)xk−i ezx . dz k dz k dz k i i=0

Thus L(xk eλx ) = 0 and xk eλx is solution of (A1). (3) Let λ1 , · · · , λs be the distinct roots of p, with multiplicity m1 , · · · , ms respectively. Then we wish to prove that eλ1 x , xeλ1 x , · · · , xm1 −1 eλ1 x ; ························ ; e

λs x

, xe

λs x

, ··· , x

ms −1 λs x

e

(A4) .

are linearly independent over R. To prove this, suppose for all x in R c11 eλ1 x + c12 xeλ1 x + · · · + c1m1 xm1 −1 eλ1 x + ····································+ cs1 eλs x + cs2 xeλs x + · · · + csms xms −1 eλs x = 0. Let’s write this as P1 (x)eλ1 x + P2 (x)eλ2 x + · · · + Ps (x)eλs x = 0, for all x in R, where Pi (x) = ci1 + ci2 x + · · · + cimi xmi −1 . We need to prove Pi (x) ≡ 0 for all i. By discarding those Pi ’s which are identically zero, we may assume all Pi ’s are not identically zero. Dividing the above equation by eλ1 x , we have P1 (x) + P2 (x)e(λ2 −λ1 )x + · · · + Ps (x)e(λs −λ1 )x = 0, for all x in R. Upon differentiating this equation sufficiently many times (at most m1 times since P1 (x) is a polynomial of degree m1 − 1), we can reduce P1 (x) to 0. Note that in this process, the degree of the resulting polynomial multiplied by e(λi −λ1 )x remains unchanged. Therefore, we get Q2 (x)e(λ2 −λ1 )x + · · · + Qs (x)e(λs −λ1 )x = 0, where deg Qi = deg Pi . Canceling the term eλ1 x we have Q2 (x)eλ2 x + · · · + Qs (x)eλs x = 0. For instance, if deg Pi = 0, then Pi is a nonzero constant. The resulting Qi is equal to (λi − λ1 )α Pi for some positive integer α. Since λi 6= λ1 , Qi is also a nonzero constant. Thus deg Qi = 0.

26


Repeating this procedure, we arrive at Rs (x)eλs x = 0, where deg Rs = deg Ps . Hence Rs (x) ≡ 0 which is a contradiction. Thus all the Pi (x)’s are identically zero. That means all cij ’s are zero and the functions in (A4) are linearly independent. ¯ = α − iβ is also an eigenvalue. Remark. If (2.2.3) has a complex eigenvalue λ = α + iβ, then λ k (α+iβ)x k (α−iβ)x Thus both x e and x e appear in (2.2.6), where 0 ≤ k ≤ m − 1. In order to obtain a fundamental set of real solutions, the pair of solutions xk e(α+iβ)x and xk e(α−iβ)x in (2.2.6) should be replaced by xk eαx cos(βx) and xk eαx sin(βx). In the following we discuss solution of the non-homogeneous equation. y 00 + P (x)y 0 + Q(x)y = f (x),

(2.2.7)

The associated homogeneous equation is y 00 + P (x)y 0 + Q(x)y = 0. The method applies to higher order equations. 1. Methods of variation of parameters. Let y1 and y2 be two linearly independent solutions of the associated homogeneous equation y 00 + P (x)y 0 + Q(x)y = 0 and W (x) be their Wronskian. We look for a particular solution of (2.2.7) in the form yp = u1 y1 + u2 y2 , where u1 and u2 are functions to be determined. Suppose u01 y1 + u02 y2 = 0. Differentiating this equation once, we get u001 y1 + u002 y2 = −u01 y10 − u02 y20 . Plugging yp into (2.2.7) we get u01 y10 + u02 y20 = f. Hence u01 and u02 satisfy (

u01 y1 + u02 y2 = 0, u01 y10 + u02 y20 = f.

Solving it, we find that u01 = − Integrating yields

y2 f, W

u02 =

(2.2.8)

y1 f. W

Z x y2 (t) u1 (x) = − f (t)dt, W (t) x0 Z x y1 (t) u2 (x) = f (t)dt. W (t) x0

(2.2.9)


27

Example 2.5 Solve the differential equation y 00 + y = sec x. Solution. A basis for the solutions of the homogeneous equation consists of y1 = cos x and y2 = R sin x. Now W (y1 , y2 ) = cos x cos x − (− sin x) sin x = 1. Thus u1 = − sin x sec x dx = R ln | cos x| + c1 and u2 = cos x sec x dx = x + c2 . From this, a particular solution is given by yp = cos x ln | cos x| + x sin x. Therefore, the general solution is y = c1 cos x + c2 sin x + cos x ln | cos x| + x sin x. The method of variation of parameters can also be used to find another solution of a second order homogeneous linear differential equation when one solution is given. Suppose z is a known solution of the equation y 00 + P (x)y 0 + Q(x)y = 0. We assume y = vz is a solution so that 0 = (vz)00 + P (vz)0 + Q(vz) = (v 00 z + 2v 0 z 0 + vz 00 ) + P (v 0 z + vz 0 ) + Qvz = (v 00 z + 2v 0 z 0 + P v 0 z) + v(z 00 + P z 0 + Qz) = v 00 z + v 0 (2z 0 + P z). That is

v 00 z0 = −2 − P. v0 z R R R An integration gives v 0 = z −2 e− P dx and v = z −2 e− P dx dx. We leave it as an exercise to show that z and vz are linearly independent solutions by computing their Wronskian.

Example 2.6 Given y1 = x is a solution of x2 y 00 + xy 0 − y = 0, find another solution. Solution. Let’s write the DE in the form y 00 + x1 y 0 − x12 y = 0. Then P (x) = 1/x. Thus a second linearly independent solution is given y = vx, where Z Z R 1 −2 − 1/x dx v= x e dx = x−2 x−1 dx = − 2 . 2x Therefore the second solution is y = − 12 x−1 and the general solution is y = c1 x + c2 x−1 . 2. Method of undetermined coefficients. Consider the equation y 00 + ay 0 + by = f (x), where a and b are real constants. Case 1. f (x) = Pn (x)eαx , where Pn (x) is a polynomial of degree n ≥ 0. We look for a particular solution in the form y = Q(x)eαx , where Q(x) is a polynomial. Plugging it into y 00 + ay 0 + by = f (x) we find Q00 + (2α + a)Q0 + (α2 + aα + b)Q = Pn (x).

(2.2.10)

Subcase 1.1. If α2 + aα + b 6= 0, namely, α is not a root of the characteristic equation, we choose

28


Q = Rn , a polynomial of degree n, and y = Rn (x)eαx . The coefficients of Rn can be determined by comparing the terms of same power in the two sides of (2.2.10). Note that in this case both sides of (2.2.10) are polynomials of degree n. Subcase 1.2. If α2 + aα + b = 0 but 2α + a 6= 0, namely, α is a simple root of the characteristic equation, then (2.2.10) is reduced to Q00 + (2α + a)Q0 = Pn .

(2.2.11)

We choose Q to be a polynomial of degree n + 1. Since the constant term of Q does not appear in (2.2.11), we may choose Q(x) = xRn (x), where Rn (x) is a polynomial of degree n. y = xRn (x)eαx . Subcase 1.3 If α2 + aα + b = 0 and 2α + a = 0, namely, α is a root of the characteristic equation with multiplicity 2, then (2.2.10) is reduced to Q00 = Pn .

(2.2.12)

We choose Q(x) = x2 Rn (x), where Rn (x) is a polynomial of degree n. y = x2 Rn (x)eαx .

Example 2.7 Find the general solution of y 00 − y 0 − 2y = 4x2 . Ans: y = c1 e2x + c2 e−x − 3 + 2x − 2x2 . Example 2.8 Find a particular solution of y 000 + 2y 00 − y 0 = 3x2 − 2x + 1. Ans: y = −27x − 5x2 − x3 . Example 2.9 Solve y 00 − 2y 0 + y = xex . Ans: y = c1 ex + c2 xex + 61 x3 ex . Case 2. f (x) = Pn (x)eαx cos(βx) or f (x) = Pn (x)eαx sin(βx), where Pn (x) is a polynomial of degree n ≥ 0. We first look for a solution of y 00 + ay 0 + by = Pn (x)e(α+iβ)x . Using the method in Case 1 we obtain a complex-valued solution z(x) = u(x) + iv(x),

(2.2.13)


29

where u(x) = <(z(x)), v(x) = =(z(x)). Substituting z(x) = u(x) +iv(x) into (2.2.13) and taking the real and imaginary parts, we can show that u(x) = <(z(x)) is a solution of y 00 + ay 0 + by = Pn (x)eαx cos(βx),

(2.2.14)

and v(x) = =(z(x)) is a solution of y 00 + ay 0 + by = Pn (x)eαx sin(βx).

(2.2.15)

Example 2.10 Solve y 00 − 2y 0 + 2y = ex cos x. Ans: y = c1 ex cos x + c2 ex sin x + 21 xex sin x. Alternatively to solve (2.2.14) or (2.2.15), one can try a solution of the form Qn (x)eαx cos(βx) + Rn (x)eαx sin(βx) if α + iβ is not a root of λ2 + aλ + b = 0, and xQn (x)eαx cos(βx) + xRn (x)eαx sin(βx) if α + iβ is a root of λ2 + aλ + b = 0, where Qn and Rn are polynomials of degree n The following conclusions will be useful. Theorem 2.11 Let y1 and y2 be particular solutions of the equations y 00 + ay 0 + by = f1 (x) and y 00 + ay 0 + by = f2 (x) respectively, then yp = y1 + y2 is a particular solution of y 00 + ay 0 + by = f1 (x) + f2 (x). Proof. Exercise.

Example 2.11 Solve y 00 − y = ex + sin x. Solution. A particular solution for y 00 − y = ex is given by y1 = 21 xex . Also a particular solution for y 00 − y = sin x is given by y2 = − 12 sin x. Thus 21 (xex − sin x) is a particular solution of the given differential equation. The general solution of the corresponding homogeneous differential equation is given by c1 e−x + c2 ex . Hence the general solution of the given differential equation is c1 e−x + c2 ex + 12 (xex − sin x).

30

2.3


Operator Methods

Let x denote independent variable, and y dependent variable. Introduce Dy =

d y, dx

Dn y =

dn y = y (n) . dxn

Pn We define D0 y = y. Given a polynomial L(x) = j=0 aj xj , where aj ’s are constants, we define a differential operator L(D) by n X L(D)y = aj Dj y. j=0

Then the equation n X

aj y (j) = f (x)

(2.3.1)

j=0

can be written as L(D)y = f (x).

(2.3.2)

Let L(D)−1 f denote any solution of (2.3.2). We have D−1 D = DD−1 = D0 , L(D)−1 L(D) = L(D)L(D)−1 = D0 . However, L(D)−1 f is not unique. R To see the above properties, first recall that D−1 f means a solution of y 0 = f . Thus D−1 f = f . Hence it follows that D−1 D = DD−1 = identity operator D0 . For the second equality, note that a solution of L(D)y = L(D)f is simply f . Thus by definition of L(D)−1 , we have L(D)−1 (L(D)f ) = f . This means L(D)−1 L(D) = D0 . Lastly, since L(D)−1 f is a solution of L(D)y = f (x), it is clear that L(D)(L(D)−1 f ) = f . In other words, L(D)L(D)−1 = D0 . More generally, we have: −1

Z

1.

D

f (x) =

f (x)dx + C,

2.

(D − a)−1 f (x) = Ceax + eax

3.

L(D)(eax f (x)) = eax L(D + a)f (x),

4.

L(D)−1 (eax f (x)) = eax L(D + a)−1 f (x).

Z

e−ax f (x)dx,

(2.3.3)

Proof. Property 2 is just the solution of the first order linear ODE. To prove Property 3, first observe that (D − r)(eax f (x)) = eax D(f (x)) + aeax f (x) − reax f (x) = eax (D + a − r)(f (x)). Thus (D − s)(D − r)(eax f (x)) = (D − s)[eax (D + a − r)(f (x))] = eax (D + a − s)(D + a − r)(f (x)). Now we may write L(D) = (D − r1 ) · · · (D − rn ). Then L(D)(eax f (x)) = eax L(D + a)f (x).

2.3. OPERATOR METHODS

31

This says that we can move the factor eax to the left of the operator L(D) if we replace L(D) by L(D + a). To prove Property 4, apply L(D) to the right hand side. We have L(D)[eax L(D + a)−1 f (x)] = eax [L(D + a)(L(D + a)−1 f (x))] = eax f (x). Thus L(D)−1 (eax f (x)) = eax L(D + a)−1 f (x).

Let L(x) = (x − r1 ) · · · (x − rn ). The solution of (2.3.2) is given by y = L(D)−1 f (x) = (D − r1 )−1 · · · (D − rn )−1 f (x).

(2.3.4)

Then we obtain the solution by successive integration. Moreover, if rj0 s are distinct, we can write A1 An 1 = + ··· + , L(x) x − r1 x − rn where A0j s can be found by the method of partial fractions. Then the solution is given by y = [A1 (D − r1 )−1 + · · · + An (D − rn )−1 ]f (x).

(2.3.5)

Next consider the case of repeated roots. Let the multiple root be equal to m and the equation to be solved is (D − m)n y = f (x) (2.3.6) To solve this equation, let us assume a solution of the form y = emx v(x), where v(x) is a function of x to be determined. One can easily verify that (D − m)n emx v = emx Dn v. Thus equation (2.3.6) reduces to Dn v = e−mx f (x) (2.3.7) If we integrate (2.3.7) n times, we obtain Z Z Z Z v= ··· e−mx f (x) dx · · · dx + c0 + c1 x + · · · + cn−1 xn−1

(2.3.8)

Thus we see that (D−m)−n f (x) = emx

ZZ

ZZ ···

e−mx f (x)dx · · · dx + c0 + c1 x + · · · + cn−1 xn−1

Example 2.12 Solve (D2 − 3D + 2)y = xex . Solution. First

1 D 2 −3D+2

=

1 D−2

−

1 D−1 .

Therefore

y = (D2 − 3D + 2)−1 (xex ) = (D − 2)−1 (xex ) − (D − 1)−1 (xex ) = e2x D−1 (e−2x xex ) − ex D−1 (e−x xex ) = e2x D−1 (e−x x) − ex D−1 (x) = e2x (−xe−x − e−x + c1 ) − ex ( 21 x2 + c2 ) = −ex ( 21 x2 + x + 1) + c1 e2x + c2 ex .

(2.3.9)

32


Example 2.13 Solve (D3 − 3D2 + 3D − 1)y = ex . Solution. The DE is equivalent to (D − 1)3 y = ex . Therefore,

y = (D − 1)−3 ex = ex

Z Z Z

1 e−x ex dx + c0 + c1 x + c2 x2 = ex x3 + c0 + c1 x + c2 x2 . 6

If f (x) is a polynomial in x, then (1 − D)(1 + D + D2 + D3 + · · · )f = f . Thus (1 − D)−1 (f ) = (1 + D + D2 + D3 + · · · )f . Therefore, if f is a polynomial, we may formally expand (D − r)−1 into power series in D and apply it to f . If the degree of f is n, then it is only necessary to expand (D − r)−1 up to Dn .

Example 2.14 Solve (D4 − 2D3 + D2 )y = x3 . Solution. We have y = (D4 − 2D3 + D2 )−1 f =

1 3 D 2 (1−D)2 x 3 4

= D−2 (1 + 2D + 3D2 + 4D + 5D + 6D5 )x3 = D−2 (x3 + 6x2 + 18x + 24) 4 = D−1 ( x4 + 2x3 + 9x2 + 24x) 5 4 = x20 + x2 + 3x3 + 12x2 . Therefore, the general solution is y = (c1 + c2 x)ex + (c3 + c4 x) +

x5 20

+

Exercise 2.1 For all real x, the real-valued function y = f (x) satisfies y 00 − 2y 0 + y = 2ex . (a) If f (x) > 0 for all real x, must f 0 (x) > 0 for all real x? Explain. (b) If f 0 (x) > 0 for all real x, must f (x) > 0 for all real x? Explain. Exercise 2.2 Find the general solution of y 00 − 2y 0 − 3y = x(ex + e−x ). Ans: y = c1 e−x + c2 e3x − x4 ex −

x2 −x 8 e

−

x −x . 16 e

x4 2

+ 3x3 + 12x2 .

Chapter 3

Second Order Linear Differential Equations 3.1

Exact 2nd Order Equations

The general 2nd order linear differential equation is of the form p0 (x)y 00 + p1 (x)y 0 + p2 (x)y = f (x)

(3.1.1)

(p0 y 0 − p00 y)0 + (p1 y)0 + (p000 − p01 + p2 )y = f (x)

(3.1.2)

p000 − p01 + p2 ≡ 0.

(3.1.3)

The equation can be written as

It is said to be exact if In the event that the equation is exact, a first integral to (3.1.1) is Z p0 (x)y 0 − p00 (x)y + p1 (x)y = f (x) dx + C1 .

Example 3.1 Find the general solution of the DE 1 00 1 2 1 2 y + ( − 2 )y 0 − ( 2 − 3 )y = ex . x x x x x Solution. Condition (3.1.3) is fulfilled. The first integral is 1 1 2 1 0 y + 2 y + ( − 2 )y = ex + C1 . x x x x That is

1 )y = xex + C1 x. x From the last equation, the general solution is found to be y 0 + (1 −

y=

1 x xe + C1 x + C2 xe−x . 2 33

34

3.2

CHAPTER 3. SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS

The Adjoint Differential Equation and Integrating Factor

If (3.1.1) is multiplied by a function v(x) so that the resulting equation is exact, then v(x) is called an integrating factor of (3.1.1). That is (p0 v)00 − (p1 v)0 + p2 v = 0.

(3.2.4)

This is a differential equation for v, which is, more explicitly, p0 (x)v 00 + (2p00 (x) − p1 (x))v 0 + (p000 (x) − p01 (x) + p2 (x))v = 0.

(3.2.5)

Equation (3.2.5) is called the adjoint of the given differential equation (3.1.1). A function v(x) is thus an integrating factor for a given differential equation, if and only if it is a solution of the adjoint equation. Note that the adjoint of (3.2.5) is in turn found to be the associated homogeneous equation of (3.1.1), thus each is the adjoint of the other. In this case, a first integral to (3.1.1) is Z 0 0 v(x)p0 (x)y − (v(x)p0 (x)) y + v(x)p1 (x)y = v(x)f (x) dx + C1 .

Example 3.2 Find the general solution of the DE (x2 − x)y 00 + (2x2 + 4x − 3)y 0 + 8xy = 1. Solution. The adjoint of this equation is (x2 − x)v 00 − (2x2 − 1)v 0 + (4x − 2)v = 0. By the trial of xm , this equation is found to have x2 as a solution. Thus x2 is an integrating factor of the given differential equation. Multiplying the original equation by x2 , we obtain (x4 − x3 )y 00 + (2x4 + 4x3 − 3x2 )y 0 + 8x3 y = x2 . Thus a first integral to it is 4

3

0

3

2

4

3

2

(x − x )y − (4x − 3x )y + (2x + 4x − 3x )y =

Z

x2 dx + C.

After simplification, we have y0 +

2x 1 C y= + 3 . x−1 3(x − 1) x (x − 1)

An integrating factor for this first order linear equation is e2x (x − 1)2 . Thus the above equation becomes Z Z 2x 1 e (x − 1) 2x 2 2x e (x − 1) y = (x − 1)e dx + C dx + C2 . 3 x3 That is 1 x 3 2x e2x 2x 2 e (x − 1) y = − e + C 2 + C2 . 3 2 4 2x Thus the general solution is 1 x 1 C1 −2x y= − + 2 + C2 e . (x − 1)2 6 4 x

3.3. LAGRANGE’S IDENTITY AND GREEN’S FORMULA

35

Exercise 3.1 Solve the following differential equation by finding an integrating factor of it. y 00 + Ans: y =

C1 2x−1

+

4x 8x − 8 y = 0. y0 + 2x − 1 (2x − 1)2

C2 x −2x . 2x−1 e

Exercise 3.2 The equation (p(x)y 0 )0 + q(x)y + λr(x)y = 0

(3.2.6)

is called a Sturm-Liouville equation, where λ is a real parameter. Show that the adjoint of a SturmLiouville equation is itself.

3.3

Lagrange’s Identity and Green’s Formula

Let L be the differential operator given by the left hand side of (3.1.1), that is L[y] = p0 (x)y 00 + p1 (x)y 0 +p2 (x)y. The formal adjoint of L is the differential operator defined by L+ [y] = (p0 (x)y)00 − (p1 (x)y)0 +p2 (x)y, where p000 , p01 and p2 are continuous on an interval [a, b]. Let u and v be functions having continuous second derivatives on the interval [a, b]. Direct simplification gives the following identity relating L and L+ . Theorem 3.1 (Lagrange’s identity) L[u]v − uL+ [v] =

d [P (u, v)], dx

where P (u, v) = up1 v − u(p0 v)0 + u0 p0 v. Proof. Expanding both sides, we get p0 u00 v + p1 u0 v − p0 uv 00 − 2p00 uv 0 − p000 uv + p01 uv + p1 uv 0 . Integrating Lagrange’s identity leads to Green’s formula. Corollary 3.2 (Green’s formula) b Z b + L[u]v − uL [v] dx = P (u, v)(x) . a

a

Let’s define an inner product for continuous real-valued functions on [a, b] by Z b (f, g) = f (x)g(x) dx. a

With this notation, Green’s formula becomes b (L[u], v) = (u, L [v]) + P (u, v)(x) . +

(3.3.7)

a

Though the operators L and L+ are meaningful when applied to any function y having a continuous second derivative on [a, b], it is usually more advantageous to further restrict the domains they act. In general, it is evidently more useful to restrict L and L+ so that (L[u], v) = (u, L+ [v])

(3.3.8)

36


holds for all u in the domain of L and all v in the domain of L+ . This is the case when we try to solve boundary value problems in which L and L+ are restricted to those functions that satisfy the given boundary conditions. When (3.3.8) holds, L+ is called the adjoint operator of L (i.e. without the adjective “formal”). For the special case when L+ = L and (3.3.8) holds, we say that L is self-adjoint. As an example, let L be the differential operator given by the first two terms on the left hand side of the Sturm-Liouville equation (3.2.6), that is L[y] = (p(x)y 0 )0 + q(x)y. By the exercise for the equation in (3.2.6), it follows that L+ = L. Then Lagrange’s identity and Green’s formula reduce to the followings. Theorem 3.3 (Lagrange’s identity for L[y] = (p(x)y 0 )0 + q(x)y) d [pW (u, v)], dx where W (u, v) = uv 0 − vu0 is the Wronskian of u and v. L[u]v − uL[v] = −

Corollary 3.4 (Green’s formula for L[y] = (p(x)y 0 )0 + q(x)y) b Z b (L[u], v) − (u, L[v]) = (L[u]v − uL[v]) dx = −pW (u, v)(x) . a

(3.3.9)

a

Exercise 3.3 Let L[y] = p0 (x)y 00 + p1 (x)y 0 + p2 (x)y. Prove that L+ = L if and only if p1 (x) = p00 (x). Exercise 3.4 Prove theorem 3.3. Exercise 3.5 Prove corollary 3.4. Exercise 3.6 Prove that if u and v are solutions of (p(x)y 0 )0 + q(x)y = 0, then pW (u, v)(x) is a constant for all x ∈ [a, b]

3.4

Regular Boundary Value Problem

The problem of finding a solution to a second order linear differential equation y 00 + p(x)y 0 + q(x)y = f (x), a < x < b;

(3.4.10)

satisfying the boundary conditions a11 y(a) + a12 y 0 (a) + b11 y(b) + b12 y 0 (b) = d1 , a21 y(a) + a22 y 0 (a) + b21 y(b) + b22 y 0 (b) = d2 ,

(3.4.11)

is called a two-point boundary value problem. When d1 = d2 = 0, the boundary conditions are said to be homogeneous; otherwise we refer them as nonhomogeneous. For the homogeneous equation with homogeneous boundary condition y 00 + p(x)y 0 + q(x)y = 0, a < x < b; a11 y(a) + a12 y 0 (a) + b11 y(b) + b12 y 0 (b) = 0, a21 y(a) + a22 y 0 (a) + b21 y(b) + b22 y 0 (b) = 0, one can verify the following properties:

(3.4.12)

3.4. REGULAR BOUNDARY VALUE PROBLEM

37

1. If φ(x) is a nontrivial solution to (3.4.12), then so is cφ(x) for any constant c. Thus (3.4.12) has a one-parameter family of solutions. 2. If (3.4.12) has two linearly independent solutions φ1 and φ2 , then c1 φ1 +c2 φ2 is also a solution of (3.4.12) for any constants c1 and c2 . Thus (3.4.12) has a two-parameter family of solutions. 3. The remaining possibility is that φ(x) ≡ 0 is the unique solution to (3.4.12). For nonhomogeneous equation (3.4.10), there is an additional possibility that it has no solution. The following examples illustrate some of these cases. Example 3.3 Find all solutions to the boundary value problem y 00 + 2y 0 + 5y = 0; y(0) = 2, y(π/4) = e−π/4 . Solution. The general solution to the equation y 00 +2y 0 +5y = 0 is y = c1 e−x cos 2x+c2 e−x sin 2x. Substituting the boundary conditions, we have y(0) = c1 = 2 and y(π/4) = c2 e−π/4 = e−π/4 . Thus c1 = 2, c2 = 1, and so the boundary value problem has the unique solution y = 2e−x cos 2x + e−x sin 2x. Example 3.4 Find all solutions to the boundary value problem y 00 + y = cos 2x; y 0 (0) = 0, y 0 (π) = 0. Solution. The general solution to the equation y 00 + y = cos 2x is y = c1 cos x + c2 sin x − (1/3) cos 2x. Thus y 0 = −c1 sin x + c2 cos x + (2/3) sin 2x. Substituting the boundary conditions, we have y 0 (0) = c2 = 0 and y 0 (π) = −c2 = 0. Thus the boundary value problem has a oneparameter family of solutions y = c1 cos x − (1/3) cos 2x, where c1 is any real number. Example 3.5 Find all solutions to the boundary value problem y 00 + 4y = 0; y(−π) = y(π), y 0 (−π) = y 0 (π). Solution. The general solution to the equation y 00 +4y = 0 is y = c1 cos 2x+c2 sin 2x. Since cos 2x and sin 2x are periodic functions of period π, y = c1 cos 2x + c2 sin 2x and y 0 = −2c1 sin 2x + 2c2 cos 2x automatically satisfy the boundary conditions y(−π) = y(π), y 0 (−π) = y 0 (π). Hence the boundary value problem has a two-parameter family of solutions y = c1 cos 2x + c2 sin 2x, where c1 and c2 are any real numbers. Example 3.6 Find all solutions to the boundary value problem y 00 + 4y = 4x; y(−π) = y(π), y 0 (−π) = y 0 (π). Solution. The general solution to the equation y 00 + 4y = 4x is y = x + c1 cos 2x + c2 sin 2x. Since y(−π) = −π + c1 and y(π) = π + c1 , there are no solutions satisfying y(−π) = y(π). Hence the boundary value problem has no solution. Exercise 3.7 Find all solutions to the boundary value problem y 00 + y = ex ; y(0) = 0, y(π) + y 0 (π) = 0. Ans. y = ( 21 + eπ ) sin x −

1 2

cos x + 12 ex .

38

3.5


Regular Sturm-Liouville Boundary Value Problem

Let L[y] = (p(x)y 0 )0 + q(x)y. Consider the regular Sturm-Liouville boundary value problem: L[y] + λr(x)y = 0, a < x < b; a1 y(a) + a2 y 0 (a) = 0, b1 y(b) + b2 y 0 (b) = 0,

(3.5.13)

where p(x), p0 (x), q(x) and r(x) are continuous functions on [a, b] and p(x) > 0 and r(x) > 0 on [a, b]. We only consider separated boundary conditions. Also we exclude the cases when a1 = a2 = 0 or b1 = b2 = 0. Let u and v be functions that have continuous second derivatives on [a, b] and satisfy the boundary conditions in (3.5.13). The boundary conditions in (3.5.13) imply that W (u, v)(b) = W (u, v)(a) = 0. This is because the system of equations a1 u(a) + a2 u0 (a) = 0, a1 v(a) + a2 v 0 (a) = 0 has nontrivial solutions in a1 and a2 since a1 and a2 are not both zero. Therefore the determinant of the system W (u, v)(a) must be zero. Similarly W (u, v)(b) = 0. Thus by Green’s formula (3.3.9), (L[u], v) = (u, L[v]). Therefore L is a self-adjoint operator with domain equal to the set of functions that have continuous second derivatives on [a, b] and satisfy the boundary conditions in (3.5.13). Self-adjoint operators are like symmetric matrices in that their eigenvalues are always real. Remark. Here we only require u and v satisfy the boundary conditions in (3.5.13) but not necessarily the differential equation L[y] + λr(x)y = 0. However, if u and v satisfy the differential equation, then L[u]v = uL[v] = −λruv and hence (L[u], v) = (u, L[v]) too. The regular Sturm-Liouville boundary value problem in (3.5.13) involves a parameter λ. The objective is to determine for which values of λ, the equation in (3.5.13) has nontrivial solutions satisfying the given boundary conditions. Such problems are called eigenvalue problems. The nontrivial solutions are called eigenfunctions, and the corresponding number λ an eigenvalue. If all the eigenfunctions associated with a particular eigenvalue are just scalar multiple of each other, then the eigenvalue is called simple. Theorem 3.5 All the eigenvalues of the regular Sturm-Liouville boundary value problem (3.5.13) are (i) real, (ii) have real-valued eigenfunctions and (iii) simple. Proof. (i) Let λ be a possibly complex eigenvalue with eigenfunction u for (3.5.13). Also u may be complex-valued. Thus u 6≡ 0, L[u] = −λru, and u satisfies the boundary conditions in (3.5.13). Then L[u] = L[u] = −λru = −λru. Also u satisfies the boundary conditions in (3.5.13) as a1 , a2 , b1 , b2 are real. Thus λ is an eigenvalue with eigenfunction u for (3.5.13). We have Z b (L[u], u) = (−λru, u) = −λ(ru, u) = −λ r|u|2 dx. a

By Green’s formula, Z (L[u], u) = (u, L[u]) = (u, −λru) = −λ(u, ru) = −λ a

b

r|u|2 dx.

3.5. REGULAR STURM-LIOUVILLE BOUNDARY VALUE PROBLEM Since r > 0, u 6≡ 0 and |u| ≥ 0, we have

Rb a

39

r|u|2 dx > 0. Therefore λ = λ and λ is real.

(ii) If u(x) = u1 (x) + iu2 (x) is a complex-valued eigenfunction with eigenvalue λ. By (i), λ is real. Then −λr(u1 + iu2 ) = L[u1 + iu2 ] = L[u1 ] + iL[u2 ]. Equating real and imaginary parts, we have L[u1 ] = −λru1 and L[u2 ] = −λru2 . Furthermore both u1 and u2 satisfy the boundary conditions in (3.5.13). Therefore we can take either u1 or u2 as the real-valued eigenfunction associated with λ. (iii) Suppose u1 and u2 are 2 eigenfunctions corresponding to the eigenvalue λ and both are solutions to (3.5.13). Thus a1 u1 (a) + a2 u01 (a) = 0, and a1 u2 (a) + a2 u02 (a) = 0. Since a1 and a2 are not both zero, this system of linear equations (in a1 and a2 ) has zero determinant. That is W (u1 , u2 )(a) = 0. Similarly W (u1 , u2 )(b) = 0. Suppose u1 and u2 are linearly independent on (a, b). Then W (u1 , u2 )(x) 6= 0 for all x ∈ (a, b). Since u1 and u2 are solutions of (3.5.13), we have by exercise 3.6 p(x)W (u1 , u2 )(x) = C, for all x ∈ [a, b], where C is a nonzero constant as p(x) > 0 for all x ∈ [a, b]. But this contradicts C = p(a)W (u1 , u2 )(a) = 0. Consequently u1 and u2 are linearly dependent. Therefore one must be a multiple of the other. That is λ is a simple eigenvalue. Two real-valued functions f and g defined on [a, b] are said to be orthogonal with respect to a positive weight function r(x) on the interval [a, b] if Z b f (x)g(x)r(x) dx = 0. a

Theorem 3.6 Eigenfunctions that correspond to distinct eigenvalues of the regular Sturm-Liouville boundary value problem in (3.5.13) are orthogonal with respect to the weight function r(x) on [a, b]. Proof. Let λ and µ be distinct eigenvalues with corresponding real-valued eigenfunctions u and v for (3.5.13) respectively. That is L[u] = −λru and L[v] = −µrv. Then −µ(u, rv) = (u, −µrv) = (u, L[v]) = (L[u], v) = (−λru, v) = −λ(ru, v). Since (u, rv) = (ru, v) and µ 6= λ, we have Rb (u, rv) = (ru, v) = 0. That is a ruv dx = 0 which means u and v are orthogonal with respect to the weight function r(x) on [a, b]. Theorem 3.7 The eigenvalues of the regular Sturm-Liouville boundary value problem in (3.5.13) form a countable, increasing sequence λ1 < λ2 < λ3 < · · · , with limn→∞ λn = +∞. For a proof of this result, see [4], page 333.

Example 3.7 Consider the regular Sturm-Liouville boundary value problem y 00 + λy = 0, y(0) = y(π) = 0.

(3.5.14)

40


When λ ≤ 0, the boundary value problem (3.5.14) has only the trivial solution y = 0. Thus for λ ≤ 0, it is not an eigenvalue. Let’s consider λ > 0. The general solution to the equation √ √ y 00 + λy = 0 is given by y = A cos( λ x) + B sin( λ x). Now y(0) = 0 implies that A = 0, √ and y(π) = 0 implies that B sin( λ π) = 0. Since we are looking for nontrivial solutions, we √ must have B 6= 0 so that sin( λ π) = 0. Therefore λ = n2 , n = 1, 2, 3, . . . with corresponding eigenfunctions φn (x) = Bn sin(nx). One can easily verify that (φm , φn ) = 0 for m 6= n. Thus, associated to a regular Sturm-Liouville boundary value problem, there is a sequence of orthogonal eigenfunctions {φn } defined on [a, b]. We can use these eigenfunctions to form an orthonormal system with respect to r(x) simply by normalizing each eigenfunction φn so that Z b φ2n (x)r(x) dx = 1. a

Now suppose {φn } is orthonormal with respect to a positive weight function r(x) on [a, b], that is ( Z b 0, m 6= n, φn (x)φm (x)r(x) dx = 1, m = n. a Then with any piecewise continuous1 function f on [a, b], we can identify an orthogonal expansion f (x) ∼

∞ X

cn φn (x),

n=1

where Z cn =

b

f (x)φn (x)r(x) dx. a

For instance, the eigenfunctions φn (x) = sin nx for (3.5.14) gives rise to the Fourier sine series expansion. Exercise 3.8 Consider the regular Sturm-Liouville boundary value problem y 00 + λy = 0, y(0) = 0, y 0 (π) = 0.

(3.5.15)

Show that the eigenvalues of (3.5.15) are λn = (2n − 1)2 /4, n = 1, 2, 3, . . ., with corresponding eigenfunctions 2n − 1 φn (x) = an sin x . 2 Example 3.8 Construct an orthonormal system of eigenfunctions corresponding to the regular SturmLiouville boundary value problem (3.5.15). 2n − 1 Solution. By Exercise 3.8 and Theorem 3.6, the eigenfunctions φn (x) = an sin x are 2 orthogonal with respect to r(x) = 1 on [0, π]. Direct computation gives Z π 2n − 1 2 2 an sin x dx = a2n π/2. 2 0 1A

function f (x) is piecewise continuous on a finite interval [a, b] if f (x) is continuous at every point in [a, b], except possibly for a finite number of points at which f (x) has a jump discontinuity.

3.5. REGULAR STURM-LIOUVILLE BOUNDARY VALUE PROBLEM Thus, if we take an =

41

p 2/π, then

p 2/π sin

2n − 1 x 2

is an orthonormal system of eigenfunctions. Like the theory of Fourier series, the eigenfunction expansion of f converges uniformly to f on [a, b] under suitable conditions. Theorem 3.8 Let {φn } be an orthonormal system of eigenfunctions for the regular Sturm-Liouville boundary value problem in (3.5.13). Let f be a continuous function on [a, b] such that f 0 is piecewise continuous on [a, b] and f satisfies the boundary conditions in (3.5.13). Then f (x) =

∞ X

a ≤ x ≤ b,

cn φn (x),

(3.5.16)

n=1

where

b

Z cn =

f (x)φn (x)r(x) dx. a

Furthermore, the eigenfunction expansion in (3.5.16) converges uniformly on [a, b]. For a proof of this result, see [2]. Example 3.9 Express ( f (x) =

2x/π, 0 ≤ x ≤ π/2, 1, π/2 ≤ x ≤ π,

in an eigenfunction expansion using the orthonormal system of eigenfunctions p 2n − 1 2/π sin x . 2 Solution. Referring to the problem (3.5.15), we compute Z π nπ π p 2n − 1 27/2 cn = f (x) 2/π sin x dx = 3/2 sin − . 2 2 4 π (2n − 1)2 0 Thus f (x) =

∞ X n=1

cn

p

2/π sin

2n − 1 x . 2

(3.5.17)

Since f (0) = 0, f 0 (π) = 0, the function f satisfies the boundary conditions in (3.5.15). Moreover, f is continuous and f 0 is piecewise continuous on [0, π]. Thus by Theorem 3.8, the series in (3.5.17) converges uniformly to f on [0, π]. The Sturm-Liouville boundary value problem arises from solving partial differential equations such as the heat equation in mathematical physics. The eigenfunction expansion can be used to solve the nonhomogeneous regular Sturm-Liouville boundary value problem. For further discussion on the Sturm-Liouville boundary value problem, see [4].

42

3.6


Nonhomogeneous Boundary Value Problem

We only consider the nonhomogeneous regular Sturm-Liouville boundary value problem with homogeneous boundary conditions. Let L[y] = f (x), a < x < b; a1 y(a) + a2 y 0 (a) = 0, b1 y(b) + b2 y 0 (b) = 0,

(3.6.18)

where L[y] = (p(x)y 0 )0 + q(x)y, p(x), p0 (x) and q(x) are continuous functions on [a, b] and p(x) > 0 on [a, b], and f (x) is continuous on [a, b]. Let L[y] = 0, a < x < b; a1 y(a) + a2 y 0 (a) = 0, b1 y(b) + b2 y 0 (b) = 0,

(3.6.19)

be the corresponding homogeneous regular Sturm-Liouville boundary value problem. Theorem 3.9 The nonhomogeneous problem (3.6.18) has a unique solution if and only if the homogeneous problem (3.6.19) has only the trivial solution. Proof. To prove the necessity, let z be the unique solution to (3.6.18). Suppose (3.6.19) has a nontrivial solution u. Thus u 6≡ 0 on [a, b]. Then z + u is also a solution to (3.6.18). But z + u 6≡ z on [a, b], contradicting the fact that z is the unique solution to (3.6.18). Thus (3.6.19) has only the trivial solution. To prove the sufficiency, we have to prove both the existence and uniqueness of the solution for (3.6.18). The uniqueness is easy. Suppose z1 and z2 are two solutions to (3.6.18). Then z1 − z2 is a solution of (3.6.19). Since we assume (3.6.19) has only the trivial solution, we must have z1 −z2 ≡ 0 so that z1 ≡ z2 on [a, b]. To prove the existence of solution to (3.6.18), we use the method of variation of parameters. First let y1 and y2 be nontrivial solutions to the equation L[y] = 0 satisfying respectively a1 y1 (a) + a2 y10 (a) = 0,

(3.6.20)

b1 y2 (b) + b2 y20 (b) = 0.

(3.6.21)

and

The existence of y1 and y2 follows from the existence and uniqueness theorem for initial value problems. We need to prove y1 and y2 are linearly independent. Suppose not, we may assume y1 = cy2 for some nonzero constant c. Then y1 also satisfies (3.6.21). Thus y1 is a solution of the homogeneous problem (3.6.19). Since (3.6.19) has only the trivial solution, we have y1 ≡ 0 which is a contradiction. Therefore, y1 and y2 are linearly independent. Let’s write L[y] = f in the standard form y 00 +

p0 0 q f y + y= . p p p

3.6. NONHOMOGENEOUS BOUNDARY VALUE PROBLEM

43

By the method of variation of parameters, we have y = u1 y1 + u2 y2 , where

Z

x

u1 = − a

y2 (t) f (t) dt, and W (t) p(t)

Z u2 = a

x

y1 (t) f (t) dt. W (t) p(t)

Here W (t) is the Wronskian of y1 and y2 . Since we are free to pick the antiderivative for u1 and u2 , it turns out to be convenient to choose Z b y2 (t) f (t) dt, (3.6.22) u1 = x W (t) p(t) and

Z u2 = a

x

y1 (t) f (t) dt. W (t) p(t)

(3.6.23)

Thus we obtain the following solution to L[y] = f . y(x)

Z b Z x y2 (t) f (t) y1 (t) f (t) = y1 (x) dt + y2 (x) dt. x W (t) p(t) a W (t) p(t) Z b = G(x, t)f (t) dt,

(3.6.24)

a

where

 y1 (t)y2 (x)   ,  W (t)p(t) G(x, t) = y1 (x)y2 (t)   ,  W (t)p(t)

a≤t≤x (3.6.25) x ≤ t ≤ b.

Using the fact that y1 and y2 satisfy the equation L[y] = 0, it follows from Lagrange’s identity that W (x)p(x) = C, where C is a constant. Note that C 6= 0 since p(x) > 0 and W (x) 6= 0 for all x ∈ [a, b]. Thus G(x, t) has the simpler form ( G(x, t) =

y1 (t)y2 (x)/C, a ≤ t ≤ x y1 (x)y2 (t)/C, x ≤ t ≤ b.

(3.6.26)

The function G(x, t) is called the Green’s function for the problem (3.6.18). Next we verify that Z b y(x) = G(x, t)f (t) dt, is a solution to (3.6.18). a

y 0 (x)

= = =

! Z x y2 (t)f (t) d y1 (t)f (t) y1 (x) dt + y2 (x) dt C dx C x a Z b Z x y2 (t)f (t) y1 (x)y2 (x)f (x) y1 (t)f (t) y1 (x)y2 (x)f (x) y10 (x) dt − + y20 (x) dt + C C C C a Zxb Z x y (t)f (t) y (t)f (t) 2 1 y10 (x) dt + y20 (x) dt C C x a (3.6.27) Z

d dx

b

Thus Z b y2 (t)f (t) y2 (t)f (t) 0 a1 y(a) + a2 y (a) = a1 y1 (a) dt + a2 y1 (a) dt C C a a Z b y2 (t)f (t) = (a1 y1 (a) + a2 y10 (a)) dt = 0, C a 0

Z

b

(3.6.28)

44


since y1 satisfies (3.6.20). Similarly using (3.6.21), one can verify that b1 y(b) + b2 y 0 (b)

=

(b1 y2 (b) + b2 y20 (b))

Z

b

a

y1 (t)f (t) dt = 0. C

(3.6.29)

Example 3.10 Find the Green’s function G(x, t) for the boundary value problem y 00 = f, y(0) = 0, y(π) = 0. Use the Green’s function to obtain the solution when f (x) = −6x. Solution. A general solution to y 00 = 0 is y = Ax+B. Thus y1 = x is a solution satisfying y(0) = 0, and y = π − x a solution satisfying y(π) = 0. Furthermore y1 and y2 are linearly independent with W (x) = −x − (π − x) = −π. We now compute C = p(x)W (x) = (1)(−π) = −π. Therefore the Green’s function is ( ( y1 (t)y2 (x)/C, 0 ≤ t ≤ x t(x − π)/π, 0 ≤ t ≤ x G(x, t) = = (3.6.30) y1 (x)y2 (t)/C, x ≤ t ≤ π x(t − π)/π, x ≤ t ≤ π. Hence for f (x) = −6x, the solution is given by Z π Z x Z y(x) = G(x, t)f (t) dt = (t(x − π)/π) (−6t) dt + 0

= x(π 2 − x2 ).

0

π

(x(t − π)/π) (−6t) dt

x

Since the homogeneous problem y 00 = 0, y(0) = 0, y(π) = 0 has only the trivial solution, the above solution to the nonhomogeneous problem y 00 = −6x, y(0) = 0, y(π) = 0 is unique. Theorem 3.9 can be strengthened to include other cases. Theorem 3.10 (Fredholm) The nonhomogeneous problem (3.6.18) has a solution if and only if for every solution y(x) of the homogeneous problem (3.6.19), Z

b

f (t)y(t) dt = 0. a

Proof. To prove the necessity, suppose z is a solution of (3.6.18) so that L[z] = f . Let y be any solution of (3.6.19). Thus L[y] = 0. Since L is self-adjoint, we have Z

b

f (t)y(t) dt = (f, y) = (L[z], y) = (z, L[y]) = (z, 0) = 0. a

To prove the sufficiency, let’s consider 2 cases. (Case 1) The problem (3.6.19) has only the trivial solution. By Theorem 3.9, the problem (3.6.18) has a unique solution. Rb (Case 2) The problem (3.6.19) has a nontrivial solution y1 . By hypothesis, a y1 (t)f (t) dt = 0. Also y1 satisfies both boundary conditions (3.6.20) and (3.6.21). Let y2 be any solution of L(y) = 0 such that y1 and y2 are linearly independent. Following the proof of Theorem 3.9, we can verify Z b that y(x) = G(x, t)f (t) dt is a solution to (3.6.18). For the boundary conditions, we have a

3.6. NONHOMOGENEOUS BOUNDARY VALUE PROBLEM

a1 y(a) + a2 y 0 (a)

(a1 y1 (a) + a2 y10 (a))

=

b

Z a

y2 (t)f (t) dt = 0, C

45

(3.6.31)

because y1 satisfies (3.6.20), and b1 y(b) + b2 y 0 (b)

(b1 y2 (b) + b2 y20 (b))

=

Z a

because

Rb a

b

y1 (t)f (t) dt = 0, C

(3.6.32)

y1 (t)f (t) dt = 0.

Remark. In case 2, if z is another solution of (3.6.19), then both y1 and z satisfy the boundary conditions of (3.6.19). In particular at the point a, we have a1 y1 (a) + a2 y10 (a) = 0 and a1 z(a) + a2 z 0 (a) = 0. Since a1 6= 0 and a2 6= 0, we have W (y1 , z)(a) = 0. This implies that y1 and z are linearly dependent. Thus z is just a multiple of y1 , and the problem (3.6.19) has a one-parameter family of solutions. Example 3.11 Show that the boundary value problem 5 2π y 00 + y 0 + y = f ; y(0) = 0, y( ) = 0 2 3 has a solution if and only if Z

2π 3

x

e 2 sin ( 0

3x )f (x) dx = 0. 2

Solution. Consider the homogeneous boundary value problem y 00 +y 0 + 25 y = 0; y(0) = 0, y( 2π 3 )= 3x 5 00 0 − 21 x (A cos ( 2 ) + B sin ( 3x 0. The general solution of the equation y + y + 2 y = 0 is y = e 2 )). 3x − 21 x sin ( 2 ). This also satisfies y( 2π ) = 0. As y(0) = 0, we deduce that A = 0. Thus y = Be 3 Thus the homogeneous boundary value problem has a one parameter family of solutions. Note that the given equation is not self-adjoint. However it can be converted into a self-adjoint equation by multiplying throughout by a factor ex . Thus we may write the problem as 5 2π (ex y 0 )0 + ex y = ex f (x); y(0) = 0, y( ) = 0. 2 3 By Theorem 3.10, this problem has a solution if and only if Z

2π 3

1

(Be− 2 x sin (

0

3x ))(ex f (x)) dx = 0. 2

This is equivalent to the condition Z

2π 3

x

e 2 sin ( 0

3x )f (x) dx = 0. 2

Exercise 3.9 Using Green’s function, solve the boundary value problem (xy 0 )0 − (4/x)y = −x; y(1) = 0, y(2) = 0. Ans: y =

1 2 60 (16x

ln 2 − 16x−2 ln 2 − 15x2 ln x).

46


Exercise 3.10 Show that any equation p0 (x)y 00 + p1 (x)y 0 + p2 (x)y = 0 can be made self-adjoint R by multiplying throughout by p10 e p1 /p0 dx . Exercise 3.11 Consider the regular Sturm-Liouville boundary value problem y 00 − xy + λy = 0, y(0) = 0, y(π) = 0. Let λ be an eigenvalue and φ a corresponding eigenfunction. Show that λ > 0.

3.7

Oscillations and the Sturm Separation Theorem

Consider the homogeneous second order linear differential equation. y 00 + P (x)y 0 + Q(x)y = 0

(3.7.33)

It is rarely possible to solve this equation in general. However by studying the properties of the coefficient functions, it is sometimes possible to describe the behavior of the solutions. One of the essential characteristics that is of primary interest to mathematicians is the number of zeros of a solution to (3.7.33). If a function has an infinite number of zeros in an interval [a, ∞), we say that the function is oscillatory. Therefore, studying the oscillatory behavior of a function means investigating the number and locations of its zeros. The familiar equation y 00 + y = 0 has two linearly independent solutions s(x) = sin(x) satisfying y(0) = 0, y 0 (0) = 1, and c(x) = cos(x) satisfying y(0) = 1, y 0 (0) = 0 respectively. The positive zeros of s(x) and c(x) are π, 2π, 3π, . . . , and π/2, 3π/2, 5π/2, . . ., respectively. Note that the zeros of s(x) and c(x) interlace each other in the sense that between two successive zeros of s(x), there is a zero of c(x), and vice versa. This property is described by the Sturm separation theorem. Theorem 3.11 (Sturm Separation Theorem) If y1 (x) and y2 (x) are two linearly independent solutions of y 00 + P (x)y 0 + Q(x)y = 0, then the zeros of these functions are distinct and occur alternatively in the sense that y1 (x) vanishes exactly once between any two successive zeros of y2 (x), and conversely. Proof. Let x1 and x2 be two consecutive zeros of y1 with x1 < x2 . Since y1 and y2 are linearly independent, their Wronskian W (y1 , y2 )(x) = y1 (x)y20 (x) − y10 (x)y2 (x) never vanish. It follows that y2 (x1 ) 6= 0 and y2 (x2 ) 6= 0, otherwise their Wronskian is zero at these points. Assume that the conclusion is false, that is y2 (x) 6= 0 for all x ∈ [x1 , x2 ]. Then the function f (x) = y1 (x)/y2 (x) is well-defined and continuous on [x1 , x2 ], and continuously differentiable on (x1 , x2 ). Since y1 (x1 ) = y1 (x2 ) = 0, we have f (x1 ) = f (x2 ) = 0. By Rolle’s theorem, there exists z ∈ (x1 , x2 ) such that f 0 (z) = 0. Computing f 0 (z), we find that 0 = f 0 (z) =

W (y1 , y2 )(z) y10 (z)y2 (z) − y1 (z)y20 (z) =− . 2 [y2 (z)] [y2 (z)]2

3.7. OSCILLATIONS AND THE STURM SEPARATION THEOREM

47

But this means W (y1 , y2 )(z) = 0, which contradicts the fact that y1 and y2 are linearly independent. Therefore y2 must have a zero in (x1 , x2 ). By interchanging the roles of y1 and y2 , we see that between any two consecutive zeros of y2 , there is a zero of y1 . Consequently, there cannot be more than one zero of y2 between two consecutive zeros of y1 .

... ............... .... ... ..... ... ...... ... ... .... y1.................. y.2....... ... .. ..... . ..... ...... ...... . ... . ... ... ..... ........ .... ... ... ... .. .. . . ... ... ..... ... . . ... ... . ... .. ... ... .. ..... .. ... ... . . ... ... ... .. .. ... . . ... ... . . ... ... . . . .. . . . ...............................................................• .......................................................................................• ........................................................................................... ... ... .x . ... ... . . . x2 .... . .. ... ... ... .. 1 . . . . . ... ... ... ... . .. ... ... .... ... ... ... .. ....... ... ... ... ... .... . .... .. ... ... .. ............. ..... . .. . ... ..... . . . ........... . . ............... .. ... ... ... .... ... .... ... .................... .... ...... ......... .....

Figure 3.1: Interlacing of zeros of two linearly independent solutions

Corollary 3.12 Suppose one nontrivial solution to y 00 + P (x)y 0 + Q(x)y = 0 is oscillatory on [a, ∞). Then all solutions are oscillatory.

Proof. Let y1 be a nontrivial solution that has an infinite number of zeros on [a, ∞). By the Sturm separation theorem, if y2 is any other solution such that y1 and y2 are linearly independent, then y2 must have a zero between each consecutive pair of zeros of y1 , and hence it must also have an infinite number of zeros on [a, ∞). If y1 and y2 are linearly dependent, then y2 = λy1 since y1 is nontrivial. Thus y2 also has an infinite number of zeros on [a, ∞).

Theorem 3.13 Let y be a nontrivial solution of y 00 + P (x)y 0 + Q(x)y = 0 on a closed interval [a, b]. Then y has at most a finite number of zeros in this interval.

Proof. We assume the contrary that y has an infinite number of zeros in [a, b]. By the BolzanoWeierstrass theorem, there exist in [a, b] a point x0 , and a sequence of zeros xn not equal to x0 such that lim xn = x0 . Since y is continuous and differentiable at x0 , we have n→∞

y(x0 ) = lim y(xn ) = 0 n→∞

and y 0 (x0 ) = lim

n→∞

y(xn ) − y(x0 ) = 0. xn − x0

By the existence and uniqueness theorem (Theorem 2.1), y must be the trivial solution which is a contradiction. Therefore, y has at most a finite number of zeros in [a, b].

48

3.8


Sturm Comparison Theorem

The equation y 00 + P (x)y 0 + Q(x)y = 0

(3.8.34)

u00 + q(x)u = 0

(3.8.35)

can be written in the form − 21

R

P dx and q(x) = Q(x) − 41 P (x)2 − 12 P 0 (x). It is customary to by putting y = uv, where v = e refer to (3.8.34) as the standard form, and to (3.8.35) as the normal form, of a homogeneous second order linear equation. Since v(x) > 0 for all x, the above transformation of (3.8.34) into (3.8.35) has no effect on the zeros of the solutions, and therefore leaves unaltered the oscillation behavior of the solutions. Next we shall restrict our discussion on (3.8.35).

Exercise 3.12 Using the substitution y = e−

x3 3

u , find the general solution of the equation

y 00 + 2x2 y 0 + (x4 + 2x + 1)y = 0. Show that the distance between two consecutive zeros of any nontrivial solution is π. Ans: y = Ae−

x3 3

sin(x − θ).

The Sturm Separation Theorem compares the zeros of two solutions to the same equation. For solutions of two different equations, it may still be possible to relate their zeros. For example, consider the equations y 00 + m2 y = 0 and y 00 + n2 y = 0. The first has a general solution of the form y1 (x) = A1 sin(m(x − θ1 )); and the second, y2 (x) = A2 sin(n(x − θ2 )). The distance between consecutive zeros of y1 is π/m, and of y2 is π/n. Therefore, for n > m, the distance between two zeros of y1 is greater than the distance between two zeros of y2 . Thus for n2 > m2 (equivalent to n > m), between two consecutive zeros of y1 , there is a zero of y2 . A similar result holds when the constants m2 and n2 are replaced by functions of x. Theorem 3.14 (Sturm comparison theorem) Let y1 be a nontrivial solution to y 00 + q1 (x)y = 0, a < x < b,

(3.8.36)

and let y2 be a nontrivial solution to y 00 + q2 (x)y = 0, a < x < b.

(3.8.37)

Assume q2 (x) ≥ q1 (x) for all x ∈ (a, b). If x1 and x2 are two consecutive zeros of y1 on (a, b) with x1 < x2 , then there exists a zero of y2 in (x1 , x2 ), unless q2 (x) = q1 (x) on [x1 , x2 ] in which case y1 and y2 are linearly dependent on [x1 , x2 ]. Proof. Suppose y2 (x) 6= 0 in (x1 , x2 ). We wish to show that q1 (x) = q2 (x) and y1 , y2 are linearly dependent on [x1 , x2 ]. Without loss of generality, we may assume y1 (x) > 0 and y2 (x) > 0 in (x1 , x2 ). First we have

3.8. STURM COMPARISON THEOREM d dx

49

d (W (y2 , y1 ))= dx (y2 y10 − y20 y1 ) = y20 y10 + y2 y100 − y200 y1 − y20 y10 = y2 y100 − y200 y1 = y2 (−q1 y1 ) − (−q2 y2 )y1 = y1 y2 (q2 − q1 ) ≥ 0,

for all x in (x1 , x2 ). Hence, W (y2 , y1 ) is non-decreasing on (x1 , x2 ). However, since y1 (x1 ) = y1 (x2 ) = 0 and y1 is positive on (x1 , x2 ), we must have y10 (x1 ) ≥ 0 and y10 (x2 ) ≤ 0. Therefore, W (y2 , y1 )(x1 ) = y2 (x1 )y10 (x1 ) ≥ 0 and W (y2 , y1 )(x2 ) = y2 (x2 )y10 (x2 ) ≤ 0. Since W (y2 , y1 )(x) is non-decreasing, the only way for it to be nonnegative at x1 and nonpositive d W (y2 , y1 )(x) = 0 in [x1 , x2 ]. at x2 is W (y2 , y1 )(x) = 0 for all x in [x1 , x2 ]. This also implies dx That means q1 (x) = q2 (x) in [x1 , x2 ]. Then y1 and y2 satisfy the same equation on [x1 , x2 ], and their Wronskian vanishes on this interval. Hence, y1 and y2 are linearly dependent. Remark. The Sturm comparison theorem 3.14 asserts that either y2 has a zero between x1 and x2 or y2 (x1 ) = y2 (x2 ) = 0 (since y1 and y2 are linearly dependent in the later case). Corollary 3.15 Suppose q(x) ≤ 0 for all x ∈ [a, b]. If y is a nontrivial solution of y 00 + q(x)y = 0 on [a, b], then y has at most one zero on [a, b]. Proof. Since u(x) ≡ 1 is a solution to the equation y 00 + 0 · y = 0 and q(x) ≤ 0, it follows from 3.14 that if y has two or more zeros in [a, b], then u must have a zero between them. Since u is never zero, y can have at most 1 zero in [a, b]. Example 3.12 The equation x2 y 00 + xy 0 + (x2 − p2 )y = 0, x > 0

(3.8.38)

is called Bessel’s equation of order p. For a > 0, let’s discuss the number of zeros in the interval 1 [a, a + π). The substitution y = ux− 2 transforms equation (3.8.38) into the following form. p2 − 41 d2 u + 1− u = 0, x > 0. (3.8.39) dx2 x2 1

Since y = ux− 2 , the distribution of zeros of a solution y to (3.8.39) is the same as the corresponding solution u to (3.8.39). Let’s compare the solutions of (3.8.39) with those of u00 + u = 0. Observe that u(x) = A sin(x − a) is a solution of u00 + u = 0 and has zeros at a and a + π. p2 − 1

Case 1. (p > 1/2). In this case, 4p2 − 1 > 0 so that 1 − x2 4 < 1 for x in [a, a + π). By the Sturm comparison theorem 3.14, a solution to (3.8.39) cannot have more than one zero in [a, a + π) because u(x) = A sin(x − a) does not have a zero in (a, a + π). p2 − 1

Case 2. (0 ≤ p < 1/2). In this case, 4p2 − 1 < 0 so that 1 − x2 4 > 1 for x in [a, a + π). By the Sturm comparison theorem 3.14, a solution to (3.8.39) must have a zero in (a, a + π), since a and a + π are consecutive zeros of u(x) = A sin(x − a). Case 3. (p = 1/2). In this case, (3.8.39) reduces to u00 + u = 0, which has the general solution u(x) = A sin(x − a). Consequently, it has exactly one zero in [a, a + π). For further discussion of Bessel’s equation, see section 5.5.

50


Exercise 3.13 Show that any nontrivial solution to y 00 + (x2 − 1)y = 0 has at most one zero in [−1, 1] but has infinitely many zeros in (−∞, −1) and in (1, ∞).

Exercise 3.14 Let y(x) be a nontrivial solution of y 00 + q(x)y = 0, x > 0. Suppose q(x) > 1/x2 for x > 0. Show that y(x) has an infinite number of positive zeros.

Exercise 3.15 Let q(x) be a continuous increasing function on [0, ∞), and φ(x) a nontrivial solution to y 00 + q(x)y = 0, 0 < x < ∞. Prove that if xt−1 , xt and xt+1 are three consecutive positive zeros of φ with xt−1 < xt < xt+1 , then xt+1 − xt < xt − xt−1 .

Exercise 3.16 Show that if w(x) and q(x) are continuous on [a, b] and q(x) < 0 on [a, b], then a nontrivial solution φ of the equation y 00 + w(x)y 0 + q(x)y = 0 can have at most one zero in [a, b].

Exercise 3.17 Show that the solution to the boundary value problem 5 2π y 00 + y 0 + y = f ; y(0) = 0, y( ) = 0, 2 3 1

1

where f (x) = e− 2 x sin 3x, is y = Be− 2 x sin (

1 3x 4 ) − e− 2 x sin 3x. 2 27

Chapter 4

Linear Differential Systems 4.1

Linear Systems

The following system is called a linear differential system of first order in normal form:   x0 = a11 (t)x1 + · · · + a1n (t)xn + g1 (t),   1 ······························    0 xn = an1 (t)x1 + · · · + ann (t)xn + gn (t), where aij (t) and gj (t) are continuous functions of t and 0 denotes differentiation with respect to t. Denote       a11 (t) · · · a1n (t) g1 (t) x1 (t)       x(t) =  · · ·  , g(t) =  · · ·  , A(t) =  · · · ··· ··· . an1 (t) · · · ann (t) xn (t) gn (t) We call g(t) a continuous vector field, or a continuous vector-valued function, if all its components are continuous functions. We call A(t) a continuous matrix, or a continuous matrix-valued function, if all its entries are continuous functions. Define    R  x01 (t) x1 (t)dt Z     x0 =  · · ·  , x(t)dt =  ··· . R 0 xn (t) xn (t)dt Then the linear system can be written in the matrix form: x0 = A(t)x + g(t).

(4.1.1)

x0 = A(t)x.

(4.1.2)

When g ≡ 0, (4.1.1) is reduced to

(4.1.2) is called a homogeneous differential system, and (4.1.1) is called a non-homogeneous system if g(t) 6≡ 0. We shall also call (4.1.2) the homogeneous system associated with (4.1.1), or the associated homogeneous system. 51

52

CHAPTER 4. LINEAR DIFFERENTIAL SYSTEMS

Example 4.1 x01 x02

= 2x1 + 3x2 + 3t, = −x1 + x2 − 7 sin t

is equivalent to x1 x2

!0 =

2 −1

3 1

!

x1 x2

! +

3t −7 sin t

! .

Example 4.2 Given a second order system  2 d x    2 = x + 2y + 3t, dt 2  d   y = 4x + 5y + 6t, dt2 it can be expressed into an equivalent first order differential system by introducing more variables. For this example, let u = x0 and v = y 0 . Then we have x0 u0 y0 v0

= = = =

u x + 2y + 3t . v 4x + 5y + 6t

Next, we begin with the initial value problem ( 0 x = A(t)x + g(t), x(t0 ) = x0 ,

(4.1.3)

where x0 is a constant vector. Similar to Theorem 2.1 we can show the following theorem. Theorem 4.1 Assume that A(t) and g(t) are continuous on an open interval a < t < b containing t0 . Then, for any constant vector x0 , (4.1.3) has a solution x(t) defined on this interval. This solution is unique. Especially, if A(t) and g(t) are continuous on R, then for any t0 ∈ R and x0 ∈ Rn , (4.1.3) has a unique solution x(t) defined on R.

4.2

Homogeneous Linear Systems

In this section we assume A = aij (t) is a continuous n by n matrix-valued function defined on the interval (a, b). We shall discuss the structure of the set of all solutions of (4.1.2). Lemma 4.2 Let x(t) and y(t) be two solutions of (4.1.2) on (a, b). Then for any numbers c1 , c2 , z(t) = c1 x(t) + c2 y(t) is also a solution of (4.1.2) on (a, b).

4.2. HOMOGENEOUS LINEAR SYSTEMS

53

Definition 4.1 x1 (t), · · · , xr (t) are linearly dependent in (a, b), if there exist numbers c1 , · · · , cr , not all zero, such that c1 x1 (t) + · · · + cr xr (t) = 0

for all

t ∈ (a, b).

x1 (t), · · · , xr (t) are linearly independent on (a, b) if they are not linearly dependent.

Lemma 4.3 A set of solutions x1 (t), · · · , xr (t) of (4.1.2) are linearly dependent on (a, b) if and only if x1 (t0 ), · · · , xr (t0 ) are linearly dependent vectors for any fixed t0 ∈ (a, b). Proof. Obviously “=⇒” is true. We show “⇐=”. Suppose that, for some t0 ∈ (a, b), x1 (t0 ), · · · , xr (t0 ) are linearly dependent. Then there exist constants c1 , · · · , cr , not all zero, such that c1 x1 (t0 ) + · · · + cr xr (t0 ) = 0. Let y(t) = c1 x1 (t) + · · · + cr xr (t). Then y(t) is the solution of the initial value problem x0 = A(t)x,

x(t0 ) = 0.

Since x(t) = 0 is also a solution of the initial value problem, by the uniqueness we have y(t) ≡ 0 on (a, b), i.e. c1 x1 (t) + · · · + cr xr (t) ≡ 0 on (a, b). Since cj ’s are not all zero, x1 (t), · · · , xr (t) are linearly dependent on (a, b).

Theorem 4.4 (i) The differential system in (4.1.2) has n linearly independent solutions. (ii) Let x1 , · · · , xn be any set of n linearly independent solutions of (4.1.2) on (a, b). Then the general solution of (4.1.2) is given by x(t) = c1 x1 (t) + · · · + cn xn (t),

(4.2.1)

where cj ’s are arbitrary constants. Proof. (i) Let e1 , · · · , en be a set of linearly independent vectors in Rn . Fix t0 ∈ (a, b). For each j from 1 to n, consider the initial value problem x0 = A(t)x,

x(t0 ) = ej .

From Theorem 4.1, there exists a unique solution xj (t) defined on (a, b). From Lemma 4.3, x1 (t), · · · , xn (t) are linearly independent on (a, b). (ii) Now let x1 (t), · · · , xn (t) be any set of n linearly independent solutions of (4.1.2) on (a, b). ˜ (t) Fix t0 ∈ (a, b). From Lemma 4.3, x1 (t0 ), · · · , xn (t0 ) are linearly independent vectors. Let x ˜ (t0 ) can be represented by a linear combination of x1 (t0 ), · · · , be any solution of (4.1.2). Then x xn (t0 ), namely, there exist n constants c˜1 , · · · , cñ such that ˜ (t0 ) = c˜1 x1 (t0 ) + · · · + cñ xn (t0 ). x

54


As in the proof of Lemma 4.3, we can show that ˜ (t) = c˜1 x1 (t) + · · · + cñ xn (t). x Thus c1 x1 (t) + · · · + cn xn (t) is the general solution of (4.1.2).

Recall that, n vectors 

 a11   a1 =  · · ·  , an1



······ ,

are linearly dependent if and only if the determinant a11 · · · a1n ··· ··· ··· an1 · · · ann

 a1n   an =  · · ·  ann

= 0.

In order to check whether n solutions are linearly independent, we need the following notation. Definition 4.2 The Wronskian of n vector-valued functions     x11 (t) x1n (t)     x1 (t) =  · · ·  , · · · , xn (t) =  · · ·  xn1 (t) xnn (t) is the determinant x11 (t) x12 (t) · · · x (t) x (t) · · · 21 22 W (t) ≡ W (x1 , · · · , xn )(t) = ··· ··· ··· xn1 (t) xn2 (t) · · ·

x1n (t) x2n (t) ··· xnn (t)

.

Using Lemma 4.3 we can show that Theorem 4.5 (i) The Wronskian of n solutions of (4.1.2) is either identically zero or nowhere zero in (a, b). (ii) n solutions of (4.1.2) are linearly dependent in (a, b) if and only if their Wronskian is identically zero in (a, b).

Definition 4.3 A set of n linearly independent solutions of (4.1.2) is called a fundamental set of solutions, or a basis of solutions. Let     x11 (t) x1n (t)     x1 (t) =  · · ·  , · · · , xn (t) =  · · ·  xn1 (t) xnn (t)

4.3. NON-HOMOGENEOUS LINEAR SYSTEMS

55

be a fundamental set of solutions of (4.1.2) on (a, b). The matrix-valued function    Φ(t) =  

x11 (t) x12 (t) · · · x21 (t) x22 (t) · · · ··· ··· ··· xn1 (t) xn2 (t) · · ·

x1n (t) x2n (t) ··· xnn (t)

    

is called a fundamental matrix of (4.1.2) on (a, b). Remark. (i) From Theorem 4.5, a fundamental matrix is non-singular for all t ∈ (a, b). (ii) A fundamental matrix Φ(t) satisfies the following matrix equation: Φ0 = A(t)Φ.

(4.2.2)

(iii) Let Φ(t) and Ψ(t) are two fundamental matrices defined on (a, b). Then there exists a constant, non-singular matrix C such that Ψ(t) = Φ(t)C. Theorem 4.6 Let Φ(t) be a fundamental matrix of (4.1.2) on (a, b). Then the general solution of (4.1.2) is given by x(t) = Φ(t)c, 

(4.2.3)



c1   where c =  · · ·  is an arbitrary constant vector. cn

4.3

Non-Homogeneous Linear Systems

In this section we consider the solutions of the non-homogeneous system (4.1.1), where A = aij (t) is a continuous n by n matrix-valued function and g(t) is a continuous vector-valued function, both defined on the interval (a, b). Theorem 4.7 Let xp (t) be a particular solution of (4.1.1), and Φ(t) be a fundamental matrix of the associated homogeneous system (4.1.2). Then the general solution of (4.1.1) is given by x(t) = Φ(t)c + xp (t),

(4.3.1)

where c is an arbitrary constant vector. Proof. For any constant vector c, x(t) = Φ(t)c + xp (t) is a solution of (4.1.1). On the other hand, let x(t) be a solution of (4.1.1) and set y(t) = x(t) − xp (t). Then y0 = A(t)y. From (4.2.3), there ˜ such that y(t) = Φ(t)˜ exists a constant vector c c. So x(t) = Φ(t)˜ c + xp (t). Thus (4.3.1) gives a general solution of (4.1.1).

56


Method of variation of parameters. Let Φ be a fundamental matrix of (4.1.2). We look for a particular solution of (4.1.1) in the form  u1 (t)   u(t) =  · · ·  . un (t) 

x(t) = Φ(t)u(t),

Plugging into (4.1.1) we get Φ0 u + Φu0 = AΦu + g. From (4.2.2), Φ0 = AΦ. So Φu0 = g, and thus u0 = Φ−1 g. Z

t

u(t) =

Φ−1 (s)g(s)ds.

(4.3.2)

t0

So we obtain the following: Theorem 4.8 The general solution of (4.1.1) is given by Z

t

x(t) = Φ(t)c + Φ(t)

Φ−1 (s)g(s)ds,

(4.3.3)

t0

where Φ(t) is a fundamental matrix of the associated homogeneous system (4.1.2).

( Exercise 4.1 Solve

x01 x02

Ans: x1 = c1 et + c2 e2t −

4.4

= 3x1 − x2 + t . = 2x1 + t t 2

− 14 , x2 = 2c1 et + c2 e2t −

t 2

− 14 .

Homogeneous Linear Systems with Constant Coefficients

Consider a homogeneous linear system x0 = Ax,

(4.4.1)

where A = (aij ) is a constant n by n matrix. Let us try to find a solution of (4.4.1) in the form x(t) = eλt k, where k is a constant vector, k 6= 0. Plugging it into (4.4.1) we find Ak = λk.

(4.4.2)

Definition 4.4 Assume that a number λ and a vector k 6= 0 satisfy (4.4.2), then we call λ an eigenvalue of A, and k an eigenvector associated with λ.

4.4. HOMOGENEOUS LINEAR SYSTEMS WITH CONSTANT COEFFICIENTS

57

Lemma 4.9 λ is an eigenvalue of A if and only if det (A − λI) = 0

(4.4.3)

(where I is the n × n unit matrix), namely, a11 − λ a 21 ··· an1

a12 a22 − λ ··· an2

··· ··· ··· ···

a1n a2n ··· ann − λ

= 0.

Remark. Let A be an n by n matrix and λ1 , λ2 , · · · , λk be the distinct roots of (4.4.3). Then there exist positive integers m1 , m2 , · · · , mk , such that det(A − λI) = (−1)n (λ − λ1 )m1 (λ − λ2 )m2 · · · (λ − λk )mk , and m1 + m2 + · · · + mk = n. mj is called the algebraic multiplicity (or simply multiplicity) of the eigenvalue λj . The number of linearly independent eigenvectors of A associated with λj is called the geometric multiplicity of the eigenvalue λj and is denoted by µ(λj ). We always have µ(λj ) ≤ mj . If µ(λj ) = mj then we say that the eigenvalue λj is quasi-simple. Especially if mj = 1 we say that λj is a simple eigenvalue. Note that in this case λj is a simple root of (4.4.3). Theorem 4.10 If λ is an eigenvalue of A and k is an associated eigenvector, then x(t) = eλt k is a solution of (4.4.1). Let A be a real matrix. If λ is a complex eigenvalue of A, and k is an eigenvector associated with λ, then x1 = <(eλt k),

x2 = =(eλt k)

are two linearly independent real solutions of (4.4.1).

In the following we always assume that A is a real matrix. Theorem 4.11 If A has n linearly independent eigenvectors k1 , · · · , kn associated with eigenvalues λ1 , · · · , λn respectively, then Φ(t) = (eλ1 t k1 , · · · , eλn t kn )

58


is a fundamental matrix of (4.4.1), and the general solution is given by x(t) = Φ(t)c = c1 eλ1 t k1 + · · · + cn eλn t kn , 

(4.4.4)



c1   where c =  · · ·  is an arbitrary constant vector. cn Proof. We only need to show det Φ(t) 6= 0. Since k1 , · · · , kn are linearly independent, so det Φ(0) 6= 0. From Theorem 4.5 we see that det Φ(t) 6= 0 for any t. Hence Φ(t) is a fundamental matrix. If one of the λ’s is complex, then replace eλt k and eλt k by
Example 4.3 Solve the system x = −3 1

1 −3

−3 1 !

1 −3

! x.

has eigenvalues λ1 = −2, and λ2 = −4. ! 1 For λ1 = −2 we find an eigenvector k1 = . 1 ! 1 For λ2 = −4 we find an eigenvector k2 = . −1 The general solution is given by ! ! 1 1 −2t −4t x(t) = c1 e + c2 e . 1 −1

Solution. The matrix A =

Example 4.4 Solve the system 0

x =

−3 1

1 −3

! x+e

−6 2

−2t

! .

Solution. We first solve the associated homogeneous system ! −3 1 0 x = x 1 −3 and find two linearly independent solutions x1 (t) =

e−2t e−2t

! , x2 (t) =

damental matrix is Φ = (x1 (t), x2 (t)) =

e−2t e−2t

e−4t −e−4t

! .

e−4t −e−4t

! . The fun-

4.4. HOMOGENEOUS LINEAR SYSTEMS WITH CONSTANT COEFFICIENTS −e−4t −e−2t

1 Φ = −2e−6t ! −6 . Then 2 −1

Let g(t) = e−2t

Φ

−1

−e−4t e−2t

!

−2 −4e2t

(t)g(t) =

t

Z u(t) =

Φ

−1

(s)g(s)ds =

Φ(t)u(t) = 2e

! ,

! ,

!

−t − 1 −t + 1

e2t −e4t

−2t −2e2t + 2

0

−2t

e2t e4t

1 = 2

59

! ,

1 −1

−4t

+ 2e

! .

The general solution is x = c1 e

1 1

−2t

! + c2 e

0 −4

0

Example 4.5 Solve x =

1 −1

−4t

1 1

−2t

− 2te

! + 2e

−2t

−1 1

! .

!

1 0

0 −4

!

x.

1 0

!

has eigenvalues ±2i. ! 1 . For λ = 2i, we find an eigenvector k = 2i Solution. The matrix A =

e2it

1 2i

! =(cos 2t + i sin 2t)

1 2i

! =

cos 2t −2 sin 2t

! +i

sin 2t 2 cos 2t

! .

The general solution is given by cos 2t −2 sin 2t

x(t) = c1

! + c2

sin 2t 2 cos 2t

! .

Example 4.6 Solve   x0 = −3x + 4y − 2z,   y 0 = x + z,    0 z = 6x − 6y + 5z. 

−3  Solution. The matrix A =  1 6

4 0 −6

 −2  1  has eigenvalues λ1 = 2, λ2 = 1, λ3 = −1. 5

60


 0   For λ1 = 2 we find an eigenvector k1 =  1  . 2   1   For λ2 = 1 we find an eigenvector k2 =  1  . 0   1   For λ3 = −1 we find an eigenvector k3 =  0  . −1 The general solution is given by 



       x 0 1 1      2t  t −t   y  = c1 e  1  + c2 e  1  + c3 e  0  , z 2 0 −1 namely x(t) = c2 et + c3 e−t , y(t) = c1 e2t + c2 et , z(t) = 2c1 e2t − c3 e−t . Example 4.7 Solve x0 = Ax, where 

2  A= 1 −1

 1 0  3 −1  . 2 3

Solution. The matrix A has eigenvalues λ 1 = 2, λ2 = λ3 = 3 ± i. 1   For λ1 = 2, we find an eigenvector k1 =  0  . 1   1   For λ2 = 3 + i, we find an eigenvector k2 =  1 + i  . 2−i We have   cos t + i sin t   e(3+i)t k2 = e3t  cos t − sin t + i(cos t + sin t)  , 2 cos t + sin t + i(2 sin t − cos t)   cos t   <(e(3+i)t k2 ) = e3t  cos t − sin t  2 cos t + sin t   sin t   =(e(3+i)t k2 ) = e3t  cos t + sin t  . 2 sin t − cos t


61

The general solution is 

     sin t cos t 1       x(t) = c1 e2t  0  + c2 e3t  cos t − sin t  + c3 e3t  cos t + sin t  . 2 sin t − cos t 2 cos t + sin t 1

Example 4.8 Solve x0 = Ax, where 

1  A =  −2 2

−2 1 2

 2  2 . 1

Solution. We have det(A − λI) = −(λ − 3)2 (λ + 3). A has eigenvalues λ1 = λ2 = 3, λ3 = −3 (We may say that, λ = 3 is an eigenvalue of algebraic multiplicity 2, and λ = −3 is a simple eigenvalue). For λ = 3 we solve the equation Ak = 3k, namely    −2 −2 2 k1     −2 −2 2   k2  = 0. 2 2 −2 k3   v−u    The solution is k =  u  . So we find two eigenvectors k1 =  v   1   For λ3 = −3 we find an eigenvector k3 =  1  . −1 The general solution is given by      1 −1      x(t) = c1 e3t  0  + c2 e3t  1  + c3 e−3t  1 0 

   −1 1    0  and k2 =  1  . 0 1

 1  1 . −1

Now we consider the solutions of (4.4.1) associated with a multiple eigenvalue λ, with geometric multiplicity µ(λ) less than the algebraic multiplicity. The following lemma is proved in appendix 1. Lemma 4.12 Assume λ is an eigenvalue of A with algebraic multiplicity m > 1. Then the following system (A − λI)m v = 0 (4.4.5) has exactly m linearly independent solutions.

By direct computations we can prove the following theorem.

62


Theorem 4.13 Assume that λ is an eigenvalue of A with algebraic multiplicity m > 1. Let v0 6= 0 be a solution of (4.4.5). Define vl = (A − λI)vl−1 , and let

l = 1, 2, · · · , m − 1,

tm−1 t2 vm−1 . x(t) = eλt v0 + tv1 + v2 + · · · + 2 (m − 1)!

(4.4.6)

(4.4.7)

Then x(t) is a solution of (4.4.1). (1) (m) Let v0 , · · · , v0 be m linearly independent solutions of (4.4.5). They generate m linearly independent solutions of (4.4.1) via (4.4.6) and (4.4.7).

Remark. In (4.4.6), we always have (A − λI)vm−1 = 0. If vm−1 6= 0 then vm−1 is an eigenvector of A associated with the eigenvalue λ. In practice, to find the solutions of (4.4.1) associated with an eigenvalue λ of multiplicity m, we first solve (4.4.5) and find m linearly independent solutions (1)

v0 ,

(2)

v0 ,

(m)

··· ,

v0

.

(k)

For each of these vectors, say v0 , we compute the iteration sequence (k)

vl

(k)

= (A − λI)vl−1 ,

l = 1, 2, · · · (k)

There is an integer 0 ≤ j ≤ m − 1 (j depends on the choice of v0 ) such that (k)

vj

6= 0,

(k)

(A − λI)vj

= 0.

Thus vj is an eigenvector of A associated with the eigenvalue λ. Then the iteration stops and yields a solution tj (k) t2 (k) (k) (k) (4.4.8) x(k) (t) = eλt v0 + tv1 + v2 + · · · + vj . 2 j! Example 4.9 Solve x0 = Ax, where 

−1  A= 0 1

1 −1 0

 0  4 . −4

Solution. From det(A − λI) = −λ(λ + 3)2 = 0, we find eigenvalues λ1 = −3 with multiplicity 2, and λ2 = 0 simple. For the double eigenvalue λ1 = −3, we solve   4 4 4   (A + 3I)2 v =  4 4 4  v = 0, 1 1 1

4.4. HOMOGENEOUS LINEAR SYSTEMS WITH CONSTANT COEFFICIENTS  1   =  0 , −1

(1)

and find two linearly independent solutions v0

 0   (1) =  1  . Plugging v0 , −1 



(2)

v0

(2)

v0 into (4.4.6), (4.4.7) we get  2   =  −4  , 2  

(1)

(1)

v1 = (A + 3I)v0

x(1)

   2 1     (1) (1) = e−3t (v0 + tv1 ) = e−3t  0  + t  −4  , 2 −1 

(2)

(2)

v1 = (A + 3I)v0

 1   =  −2  , 1 

   0 1     (2) (2) x(2) = e−3t (v0 + tv1 ) = e−3t  1  + t  −2  . −1 1   4   For the simple eigenvalue λ2 = 0 we find an eigenvector k3 =  4  . 1 So the general solution is x(t) = c1 x(1) + c2 x(2) + c3 k3           4 1 0 2 1           = c1 e−3t  0  + t  −4  + c2 e−3t  1  + t  −2  + c3  4  . 1 1 2 −1 −1

Example 4.10 Solve the system   x0 = 2x + y + 2z,   y 0 = −x + 4y + 2z,    0 z = 3z. Solution. Let 

2  A =  −1 0

63

 1 2  4 2 . 0 3

The eigenvalue of A is 3 with multiplicity 3. Solving the linear system:   0 0 0   (A − 3I)3 v =  0 0 0  v = 0, 0 0 0

64


we obtain 3 obvious linearly independent solutions      0 0 1       (3) (2) =  0  , v0 =  1  , v0 =  0  . 1 0 0 

(1)

v0

(j)

Plugging v0 into (4.4.6), (4.4.7) we get   (1) (1) v1 = (A − 3I)v0 =    (1) (1) v2 = (A − 3I)v1 = 

x(1)

 −1  −1  , 0  0  0 , 0 

   1 −1     (1) (1) = e3t (v0 + tv1 ) = e3t  0  + t  −1  ; 0 0



(2)

(2)

(2)

(2)

v1 = (A − 3I)v0

v2 = (A − 3I)v1

x(2)

 1   =  1 , 0   0   =  0 , 0 

   0 1     (2) (2) = e3t (v0 + tv1 ) = e3t  1  + t  1  ; 0 0



(3)

(3)

(3)

(3)

v1 = (A − 3I)v0

v2 = (A − 3I)v1

x(3)

 2   =  2 , 0   0   =  0 , 0 

   0 2     (3) (3) = e3t (v0 + tv1 ) = e3t  0  + t  2  . 1 0


65

The general solution is x(t) = c1 x(1) + c2 x(2) + c3 x(3)         1 0 −1 1       3t  3t  = c1 e  0  + t  −1  + c2 e  1  + t  1  0 0 0 0     2 0     + c3 e3t  0  + t  2  . 0 1 Remark. It is possible to reduce the number of constant vectors in the general solution of x0 = Ax by using a basis for the Jordan canonical form of A. For details of the Jordan canonical form, see appendix 1. However the following algorithm usually works well if the size of A is small. Consider an eigenvalue λ of A with algebraic multiplicity m. Start with r = m. Let v0 be a vector such that (A − λI)r v0 = 0 but (A − λI)r−1 v0 6= 0. v0 is called a generalized eigenvector of rank r associated with the eigenvalue λ. If no such v0 exists, reduce r by 1. Then v0 , v1 = (A − λI)v0 , · · · , vr−1 = (A − λI)r−1 v0 , form a chain of linearly independent solutions of (4.4.5) with vr−1 being the base eigenvector associated with the eigenvalue λ. This gives r independent solutions of x0 = Ax. x1 (t)

= eλt (v0 + tv1 + · · · +

tr−1 (r−1)! vr−1 ).

· · · xr−1 (t) = eλt (vr−2 + tvr−1 ), xr (t) = eλt vr−1 . Repeat this procedure by finding another v which is not in the span of the previous chains of vectors and the resulting base eigenvectors are linearly independent. Do this for each eigenvalue of A. Results of linear algebra shows that (1) Any chain of generalized eigenvectors constitutes a linearly independent set of vectors. (2) If two chains of generalized eigenvectors are based on linearly independent eigenvectors, then the union of these vectors is a linearly independent set of vectors (whether the two base eigenvectors are associated with different eigenvalues or with the same eigenvalue).

   Example 4.11 Solve x0 = Ax, where A =  

3 0 0 0

0 3 1 0

0 0 3 1

0 0 0 3

   . 

66


of multiplicity 4. Direct calculation gives (A − 3I) =  0 0 0 0 0 0    , (A − 3I)3 = 0, and (A − 3I)4 = 0. 0 0 0  1 0 0 0     0 1  0   0      It can be seen that v1 =   are two linearly independent eigenvectors of A.  and v4 =   0   0  1 0     0 0  1   0      Together with v2 =   and v3 =  , they form a basis of {v | (A − 3I)4 v = 0} = R4 .  0   1 

Solution.  0 0  0 0    0 1 0 0

A 0 0 0 1

has an eigenvalue λ =   0  0     , (A − 3I)2 =   0 

3 0 0 0 0

0 0 Note that (A−3I)v2 = v3 , and (A−3I)v3 = v4 . Hence {v2 , v3 , v4 } forms a chain of generalized eigenvectors associated with the eigenvalue 3. {v1 } alone is another chain. Therefore the general solution is t2 3t x(t) = e c1 v1 + c2 (v2 + tv3 + v4 ) + c3 (v3 + tv4 ) + c4 v4 . 2 That is    x(t) =  

c1 e3t c2 e3t (c2 t + c3 )e3t 2 ( c22t + c3 t + c4 )e3t

   . 

Exercise 4.2 Solve x0 = Ax, where    A= 

   Ans: x(t) = c1 e−t  

4.5

1 0 2 −1





     + c2 et   

7 0 12 −4 1 −2 0 2

5 1 10 −4 

−3 0 −5 2 

     + c3 e t   

2 0 4 −1



−1 0 −2 0



  . 



     + c4 et   

−t − 1 1 −2t 0

   . 

Higher Order Linear Equations

Consider the n-th order linear equation y (n) + a1 (t)y (n−1) + · · · + an−1 (t)y 0 + an (t)y = f (t),

(4.5.1)

4.5. HIGHER ORDER LINEAR EQUATIONS

67

k

where y (k) = ddtky . We assume that aj (t)’s and f (t) are continuous functions defined on the interval (a, b). The associated homogeneous equation is y (n) + a1 (t)y (n−1) + · · · + an−1 (t)y 0 + an (t)y = 0.

(4.5.2)

The general theory of solutions of (4.5.1) and (4.5.2) can be established by applying the results in the previous sections to the equivalent system. Recall the initial value problem  (n) (n−1)  + · · · + an (t)y = f (t),  y + a1 (t)y      y(t ) = y0 ,   0 (4.5.3) y 0 (t0 ) = y1 ,     ·········      y (n−1) (t ) = y 0 n−1 . If we let x1 = y, x2 = y 0 , x3 = y 00 ,..., xn−1 = y (n−2) , xn = y (n−1) , then (4.5.3) is equivalent to the following linear system: x01 x02 x03

...

x0n−1 x0n

= x2 = x3 = x4

... ...

= xn = −an (t)x1 − an−1 (t)x2 · · · − a1 (t)xn + f (t)

That is        

x1 x2 x3 .. . xn

0



      =      

0 0 ··· 0 −an (t)

1 0 ··· 0 −an−1 (t)

··· ··· ··· ··· ···

0 0 ··· 0 −a2 (t)

0 0 ··· 1 −a1 (t)

       

x1 x2 x3 .. . xn





      +      

0 0 0 .. . f (t)

    .   

The initial values can be expressed as        

x1 (t0 ) x2 (t0 ) x3 (t0 ) .. . xn (t0 )





      =      

y0 y1 y2 .. .

    .   

yn−1

Thus the first component x1 of each solution of this system corresponds to a solution of the original nth order linear ODE (4.5.1). In fact a solution of this system gives rise to a vector-valued function where the ith component is the (i − 1)th derivative of a solution of (4.5.1). That is

68


       

x1 x2 x3 .. . xn





      =      

y y0 y 00 .. .

    .   

y (n−1)

Remarks. 1. The existence and uniqueness theorem for first order differential systems implies the existence and uniqueness theorem for higher order linear ODEs. 2. The linear structure of the solution space of the homogeneous linear system translates to the linear structure of the solution space of the homogeneous higher order linear ODE.     yn (t) y1 (t)      yn0 (t)   y10 (t)   are n linearly independent solutions    3. If x1 (t) =  .. ..   , . . . , xn (t) =  . .     (n−1) (n−1) (t) yn y1 (t) of the equivalent linear system of (4.5.2), then the first components y1 (t), . . . , yn (t) of these vectorvalued functions form n linearly independent solutions of (4.5.2). 4. In particular the Wronskian of n solutions of (4.5.2) is the same as the Wronskian of the corresponding n solutions of the equivalent linear system. That is W (y1 , . . . , yn ) = W (x1 , . . . , xn ). Variation of parameters. From (4.3.3) we can derive the variation of parameter formula for higher order equations. Consider a second order equation y 00 + p(t)y 0 + q(t)y = f (t). (4.5.7) ! x1 Let x1 = y, x2 = y 0 , x = . Then (4.5.7) is written as x2 ! ! 0 1 0 x0 = x+ (4.5.8) −q −p f Assume y1 (t) and y2 (t) are two linearly independent solutions of the associated homogeneous equation y 00 + p(x)y 0 + q(x)y = 0. Eventually, we look for a solution of (4.5.7) in the form y = u1 y1 + u2 y2 . ! y1 y2 Choose a fundamental matrix Φ(t) = . By the method of variation of parameters (the y10 y20 matrix version), the corresponding solution of (4.5.8) is in the form x = Φu. That is ! ! ! ! y y1 y2 u1 y1 u1 + y2 u2 x= = = (4.5.9) y0 y10 y20 u2 y10 u1 + y20 u2

4.5. HIGHER ORDER LINEAR EQUATIONS

69

Now −1

Φ(t)

y20 −y10

1 = W (t)

−y2 y1

! ,

where W (t) is the Wronskian of y1 (t) and y2 (t). Using (4.3.3), we have u1 u2 Thus

!

Z =

Z u1 (t) = −

1 W (t)

y20 −y10

y2 (t) f (t)dt, W (t)

−y2 y1

!

Z u2 (t) =

0 f

! dt.

y1 (t) f (t)dt. W (t)

(4.5.10)

Note that, (4.5.9) implies y10 u1 + y20 u2 = y 0 = y10 u1 + y20 u2 + y1 u01 + y2 u02 . Hence y1 u01 + y2 u02 = 0. Plugging y 0 = y10 u1 + y20 u2 into (4.5.7) we find y10 u01 + y20 u02 = f. Solving these two equations, we find that u01 = −

y2 f, W

u02 =

y1 f. W

Again we get (4.5.10). Linear equations with constant coefficients. Now we consider linear equations with constant coefficients y (n) + a1 y (n−1) + · · · + an−1 y 0 + an y = f (t),

(4.5.11)

and the associated homogeneous equation y (n) + a1 y (n−1) + · · · + an−1 y 0 + an y = 0,

(4.5.12)

where a1 , · · · , an are real constants. Recall that (4.5.12) is equivalent to a system x0 = Ax, where

    A=   

0 0 ··· 0 −an

1 0 ··· 0 −an−1

··· ··· ··· ··· ···

0 0 ··· 0 −a2

0 0 ··· 1 −a1

    .   

The eigenvalues of A are roots of the equation det(λI − A) = λn + a1 λn−1 + · · · + an−1 λ + an = 0,

(4.5.13)

which is the same as the characteristic equation of (4.5.12). Thus the eigenvalues of A are the same as the characteristic values of (4.5.12). Then the following result can be deduced using the theory of linear differential system with constant coefficients.

70


Theorem 4.14 Let λ1 , · · · , λs be the distinct eigenvalues for (4.5.12), with multiplicity m1 , · · · , ms respectively. Then (4.5.12) has a fundamental set of solutions eλ1 t , teλ1 t , · · · , tm1 −1 eλ1 t ; ························ ;

(4.5.14)

eλs t , teλs t , · · · , tms −1 eλs t .

4.6

Introduction to the Phase Plane

Consider a system of two first order equations x0 (t) = f (x, y) y 0 (t) = g(x, y).

(4.6.1)

It is called an autonomous system since f and g are independent of t. This property implies that if the pair (x(t), y(t)) is a solution of (4.6.1), then the pair (x(t + c), y(t + c)) is also a solution of (4.6.1) for any constant c. Let the pair (x(t), y(t)) be a solution to (4.6.1). If we plot the points (x(t), y(t)) on the xy-plane, the resulting curve is called an integral curve of the system (4.6.1), and the xy-plane is called the phase plane. Let F(x, y) = hf (x, y), g(x, y)i be the vector field on the xy-plane defined by f and g. If we plot the unit vectors defined by those nonzero F(x, y) at various points of the phase plane, we get the so called direction field of (4.6.1). For any point p(t) = hx(t), y(t)i on an integral curve of (4.6.1), we have p0 (t) = hx0 (t), y 0 (t)i = hf (x, y), g(x, y)i = F(x, y). Thus F(x, y) is everywhere tangent to the integral curve p(t). See figure 4.1. By eliminating t in (4.6.1), it leads to the equation dy g(x, y) = . dx f (x, y) It is called the phase plane equation. Thus any integral curve of (4.6.1) satisfies the phase plane equation. Exercise 4.3 Solve the system x0 (t) = 1, y 0 (t) = 3x2 + x − 1. Sketch a few integral curves of the system.

4.7

Linear Autonomous System in the Plane

A linear autonomous system in the plane has the form x0 (t) = a11 x + a12 y + b1 y 0 (t) = a21 x + a22 y + b2

4.7. LINEAR AUTONOMOUS SYSTEM IN THE PLANE

71

y

.... ........ .... .... . . . . .... ... .. ............. ... ...... . .......... ......... .... . ........ ...... . . .. .. .... .... ...... . ........... .... ...... ............. ... ..... ... ............ ...... ....... ..... ...... ... .. .......... ........... ... .... .. .. ......... ...................... . . ..... .. . ...... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............. ... ..... . ........ ........ ..... ..... ..... .. ........ ....... ... ........ .. .... .. ........ .... . . ...... .. .. ........... ....... ... ........... .... .... ............. ..... ..... ... ... ........ ............ ... ..... .. .. . ......... ...................... . . .. . . . ... . .. . . .. ........ ....... . ........ . . . . . . . . . . . . . . . . . . . ..... .... ... ......... ... ... .... . ...... ...... ... ........ ........ ... .... ... .......... ... ........ .. .. .......... ........... ........... .... .... ............. ..... ..... ... .. ........ ........... ... ..... .. .. . ....... ..................... . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...... ... ...... ............ ... ..... ...... ... ..... .. ........ . . ....... . .. .. ...... ... .... ... ........... .... ......... ... ... ........... .................................... .... .... ............. ..... ..... ... .. ....... ............ ... ..... .... .. . ......... .................... . . .. .. ...... . .. ........ ........... . ..... . .. .. .. .. . . . . . . . . ...... ..... ... ......... . ..... . ........ ..... .. ....... ......... .. . ........ ... .... .. .......... .... ........ ... ... ............ ....... ... ........... ....... .... .............. ..... ..... ... .. ........ ........... ... ...... .. .. . ....... .................. . ... . . . . . . . . . . . . . . . ..... . .. .. .... .. ... .. .... .... ....... . .... . ............. ..... ...... . .......... ......... .... . ....... ...... . .. .. ..... .... ...... . ........... .... .......... ............. ... ..... ... ............ ...... ....... ..... ...... ..... ... .......... ........... ... .... .. .. ......... ...................... . ... ...... .. ..... .. ...... ... .. . ... .. .. .. . ..... . . . . . . . . .................................................................................................................................................................................................................................... x . . .. . . .... . .. ... .. . . . . . .. ..... ... .... .. ....... .... . . ... .. .. ... ........... ....... ... ........... .... .... .............. ..... ..... ... ... ........................... ... ..... .. .. . ......... ...................... . .. . . . ....... . ........ .......... . ... .. ... . .. . . .... ... .... . . . . . . . . . . ..... .... ... ......... ... ... .... . ...... ...... ... ....... ........ ... .... ... .......... ... ........ .. .. .......... ........... ........... .... .... .............. ..... ..... ... . ........ ........... ... ..... .. .. . ....... ..................... . . ... ......... ....... . ......... . .. .... .. ... .. . .. . . . . . . . . . . . . ..... .... ..... .............. ... ..... . ......... . . . ...... . . . . . .. . ... .... ... ........... .... ......... .. .. ........... ...... . ........... .... .... ........... ..... ..... ... .. ....... ............ ... ..... .. .. . ......... .................... . . .. . . .. ... .. .. . .. ....... . ......... ........... . . . . . . . . . . ...... .... ... ......... . .... . ........ ..... .. ....... ....... .. ........ ... .... .. .......... .... ........ ... ... ............ ....... ... ........... .... .... .............. ..... ..... ... .. ........ ........... ... ..... .. .. . ....... .................. . .. . ..... . . .. . .......... .. ... . . ... ...... . ........ .. .. . . ... ..... ... ........ . . ............. ... ...... . .......... . ..... . . .. . .. .. .... .... ...... . ........... .... ...... ............. ... ..... ... ............. ...... . ........ . ..... .. ...... .... .... ......................... ... .... . .. ......... ...................... . . ..... ....... .. .. . .. . ...

Figure 4.1: Direction field and an integral curve of

dx dt

= f (x, y), dy dt = g(x, y)

where aij ’s and bj ’s are constants. We assume a11 a22 − a21 a12 6= 0. By a simple translation the system can always be put in form x0 (t) = ax + by (4.7.2) y 0 (t) = cx + dy, where ad − bc 6= 0. The point (0, 0) is called a critical point of (4.7.2) since both ax + by and cx + dy are zero at x = 0, y = 0. (In general a critical point of (4.6.1) is a solution of the system of equations f (x, y) = 0, g(x, y) = 0.) Corresponding to the critical point (0, 0) of (4.7.2), we have x(t) = 0, y(t) = 0 for all t is a constant solution of (4.7.2). Our interest is to investigate the behavior of the integral curves near the critical point (0, 0). The behavior of the solutions to (4.7.2) is linked to the nature of the roots r1 and r2 of the characteristic equation of (4.7.2). That is r1 and r2 are the roots of the quadratic equation X 2 − (a + d)X + (ad − bc) = 0.

(4.7.3)

Case 1. If r1 , r2 are real, distinct and of the same sign, then the critical point (0, 0) is a node. Suppose r1 < r2 < 0. The general solution to (4.7.2) has the form x(t) = c1 A1 er1 t + c2 A2 er2 t y(t) = c1 B1 er1 t + c2 B2 er2 t ,

(4.7.4)

where A’s and B’s are fixed constants such that B1 /A1 6= B2 /A2 and c1 , c2 are arbitrary constants. When c2 = 0, we have the solution x(t) = c1 A1 er1 t , y(t) = c1 B1 er1 t . For any c1 > 0, it gives the parametric equation of a half-line A1 y = B1 x with slope B1 /A1 . As t → ∞, the point on this half-line approaches the origin (0, 0). For any c1 < 0, it represents the other half of the line A1 y = B1 x. As t → ∞, the point on this half-line also approaches the origin (0, 0). When c1 = 0, we have the solution x(t) = c2 A2 er2 t , y(t) = c2 B2 er2 t . In exactly the same way, it represents two half-lines lying on the line A2 y = B2 x with slope B2 /A2 . The two half-lines approach (0, 0) as t → ∞.

72


The two lines A1 y = B1 x and A2 y = B2 x are called the transformed axes, usually denoted by x ˆ and yˆ on the phase plane. If c1 6= 0 and c2 6= 0, the general solution (4.7.4) are parametric equations of some curves. Since r1 < 0 and r2 < 0, these curves approach (0, 0) as t → ∞. Furthermore, y (c1 B1 /c2 )e(r1 −r2 )t + B2 c1 B1 er1 t + c2 B2 er2 t = = . r t r t 1 2 x c1 A1 e + c2 A2 e (c1 A1 /c2 )e(r1 −r2 )t + A2 B2 2 As r1 − r2 < 0, we see that xy → B A2 as t → ∞ so that all the curves enter (0, 0) with slope A2 . A critical point is called a node if it is approached and entered (with a well-defined tangent line) by each integral curve as t → ∞ or t → −∞. A critical point is said to be stable if for each R > 0, there exists a positive r ≤ R such that every integral curve which is inside the circle x2 + y 2 = r2 for some t = t0 remains inside the circle x2 + y 2 = R2 for all t > t0 . Roughly speaking, a critical point is stable if all integral curves that get sufficiently close to the point stay close to the point. If the critical point is not stable, it is called unstable. A critical point is said to be asymptotically stable if it is stable and there exists a circle x2 + y 2 = r02 such that every integral curve which is inside this circle for some t = t0 approaches the critical point as t → ∞. A node is said to be proper if every direction through the node defines an integral curve, otherwise it is said to be improper.

In our situation, we have (0, 0) is an asymptotically stable improper node. See figure 4.2(i). If r1 > r2 > 0, then the situation is exactly the same except that all curves now approach and enter (0, 0) as t → −∞. So all the arrow showing the directions are reversed. See figure 4.2(ii). The point (0, 0) is a unstable improper node. Case 2. If r1 , r2 are real, distinct and of opposite sign, then the critical point (0, 0) is a saddle point. Let’s suppose r1 < 0 < r2 . The general solution is still represented by (4.7.4). The two half-line solutions x(t) = c1 A1 er1 t , y(t) = c1 B1 er1 t (for c1 > 0 and c1 < 0) still approach and enter (0, 0) as t → ∞, but the other two half-line solutions x(t) = c2 A2 er2 t , y(t) = c2 B2 er2 t approach and enter (0, 0) as t → −∞. If c1 6= 0 and c2 6= 0, the general solution (4.7.4) defines integral curves of (4.7.2), but since r1 < 0 < r2 , none of these curves approaches (0, 0) as t → ∞ or t → −∞. So (0, 0) is not a node. Instead, as t → ∞, each of these curves is asymptotic to one of the half-lines of the line A2 y = B2 x; whereas as t → −∞, each of these curves is asymptotic to one of the halflines of the line A1 y = B1 x. In this case, the critical point is called a saddle point and is certainly unstable. See figure 4.2(iii). Case 3. If r1 , r2 are real and equal, then the critical point (0, 0) is a node. Let r1 = r2 = r. In this case, the general solution may or may not involve a factor of t times ert . Let’s consider first the subcase where t is not present. Then the general solution is x(t) = c1 ert , y(t) = c2 ert , where c1 and c2 are arbitrary constants. The integral curves lie on the lines c1 y = c2 x. When r > 0, the integral curves move away from (0, 0), so (0, 0) is unstable. See figure 4.3(i). When r < 0, the integral curves approach (0, 0) and (0, 0) is asymptotically stable. See figure 4.3(ii). In either situation all integral curves lie on lines passing through (0, 0). Because every direction through (0, 0) defines an integral curve, the point (0, 0) is a proper node.


73

Now let’s discuss the case where a factor of t times ert is present. Suppose r < 0. The general solution can be written in the form x(t) = c1 Aert + c2 (A1 + At)ert y(t) = c1 Bert + c2 (B1 + Bt)ert ,

(4.7.5)

where A’s and B’s are definite constants and c1 , c2 are arbitrary constants. When c2 = 0, we obtain the solutions x(t) = c1 Aert , y(t) = c1 Bert . These are solutions representing the two half-lines (for c1 > 0 and c1 < 0) lying on the line yˆ with equation Ay = Bx and slope B/A; and since r < 0, both approach (0, 0) as t → ∞. Also since y/x = B/A, it is clear that both of these half lines enter (0, 0) with slope B/A. If c2 6= 0, the solutions (4.7.5) are curves. As r < 0, these curves approach (0, 0) as t → ∞. Furthermore, y c1 Bert + c2 (B1 + Bt)ert c1 B/c2 + B1 + Bt = = x c1 Aert + c2 (A1 + At)ert c1 A/c2 + A1 + At approaches B/A as t → ∞; so these curves all enter (0, 0) with slope B/A. We also have y/x → B/A as t → −∞. Therefore each of these curves is tangent to yˆ as t → ±∞. Consequently (0, 0) is a node that is asymptotically stable. See figure 4.2(iv). If r > 0, the situation is unchanged except that the directions of the curves are reversed and the critical point is unstable. Exercise 4.4 Find and classify the critical point of the system x0 = 2x + y + 3, y 0 = −3x − 2y − 4. Case 4. If r1 , r2 are conjugate complex but not purely imaginary, then the critical point (0, 0) is a spiral. Let r1 = α + iβ and r2 = α − iβ. First observe that the discriminant of (4.7.3) is negative. That is (a + d)2 − 4(ad − bc) = (a − d)2 + 4bc < 0.

(4.7.6)

The general solution of (4.7.2) is given by x(t) = eαt [c1 (A1 cos βt − A2 sin βt) + c2 (A1 sin βt + A2 cos βt)], y(t) = eαt [c1 (B1 cos βt − B2 sin βt) + c2 (B1 sin βt + B2 cos βt)],

(4.7.7)

where A’s and B’s are definite constants and c’s are arbitrary constants. Suppose α < 0. Then from (4.7.7) we see that x → 0 and y → 0 as t → ∞. That means all integral curves approach (0, 0) as t → ∞. Next we shall show that the integral curves do not enter (0, 0) as t → ∞. Instead they wind around like a spiral towards (0, 0). To do so, we shall show that the angular coordinate θ = tan−1 (y/x) is always strictly increasing or always strictly decreasing. That is dθ/dt > 0 for all t > 0, or dθ/dt < 0 for all t > 0. Differentiating θ = tan−1 (y/x), we have xdy/dt − ydx/dt dθ = . dt x2 + y 2 Using (4.7.2), we get dθ cx2 + (d − a)xy − by 2 = . dt x2 + y 2

(4.7.8)

74


Since we are interested in solutions that represent integral curves, we assume x2 + y 2 6= 0. Now (4.7.6) implies that b and c have opposite signs. Let’s consider the case b < 0 and c > 0. When y = 0, (4.7.8) gives dθ/dt = c > 0. If y 6= 0, dθ/dt cannot be 0. If it were, then by (4.7.8) we have cx2 + (d − a)xy − by 2 = 0 or c(x/y)2 + (d − a)(x/y) − b = 0, for some real number x/y. But this contradicts the fact that its discriminant also given by (4.7.6) is negative. Thus we conclude that dθ/dt > 0 for all t > 0 when c > 0. Similarly in case b > 0 and c < 0, dθ/dt < 0 for all t > 0. Since by (4.7.7), x and y change sign infinitely often as t → ∞, all integral curves must spiral in to (0, 0) (counterclockwise or clockwise according to c > 0 or c < 0). The critical point in this case is a spiral, and is asymptotically stable. See figure 4.3(iii). If α > 0, the situation is the same except that the integral curves approach (0, 0) as t → −∞ and the critical point is unstable. See figure 4.3(iv). Exercise 4.5 Classify the critical point at the origin and sketch the phase plane diagram for the system x0 = x − 4y, y 0 = 4x + y. See figure 4.3(v). Case 5. If r1 , r2 are purely imaginary, then the critical point (0, 0) is a centre. The general solution is given by (4.7.7) without the exponential term. Thus x(t) and y(t) are periodic functions of period 2π so that each integral curve is a closed path surrounding the origin. These curves can be shown to be ellipses by solving the phase plane equation dy cx + dy = . dx ax + by See figure 4.3(vi). Exercise 4.6 Show that integral curves of the system x0 = −βy, y 0 = βx are circles. The following result summarizes the above discussion. Theorem 4.15 Assume (0, 0) is an isolated critical point of the linear system x0 = ax + by, y 0 = cx + dy, where a, b, c, d are real and ad − bc 6= 0. Let r1 , r2 be the roots of the characteristic equation X 2 − (a + d)X + (ad − bc) = 0. The stability of the origin and the classification of the origin as a critical point depends on the roots r1 , r2 as follows. Roots

Type of critical point

Stability

distinct, positive distinct, negative opposite signs equal, positive equal, negative complex value with positive real part complex value with negative real part purely imaginary

improper node improper node saddle point proper or improper node proper or improper node spiral point spiral point centre

unstable asymptotically stable unstable unstable asymptotically stable unstable asymptotically stable stable


75

y

.. .... ... ........ ... .. ... . ... . .. ... ... yˆ. .... ... ... .... .. .. ... ..... .. . .... .... . . .. . . ..... . . ..... . . . . . . .. . ... ..... .. .... ....... ....... ........ . . . .. . ... ... . . . .. x ˆ .. ...... ........... ......... ... ........ ... ... ..... .... ... .... ..... .... ........... .. .... . . . .. ....... ..... ........ ..... ................................................... ... .... ........ ...... ......... ....... ......... . .. ... .. ........................... ..... ........ ... . .. .. ... . ......... ................................................................... ....... ......... . . . . . . . . . ....... ..... .. ............. ................................................................................................................................................................................................................................................................................................ x . .... .. . . ..... ........ .... ........................................................................ ......... .. . . . . . . . . . . . . . . . . .... ... .... .... .................. ... ......................... ......... .... . .................................................. ........ .... .... .. . ........... .... ... ... .. . ...... ..... .... ....... .... .. .. . . . . . . . . . . . . . . . ..... ...... ... .. ....... ..... ..... .... ... .... ..... ........ ..... .. ..... .... . ... ... .. . ... ..... ...... . . . . . .. . . ...... .. .... .. ..... ..... .... .. .... . .. . ..... . . .. . . . .. . . . . . . . . .. . .... . . . . . . . .... .. ... . . . . . . . .. . . . ... ...

(i) Asymptotically stable improper node: roots are real, distinct and negative

y

... .... ... ......... ... ... ... . .... . .. .. yˆ. .. ... ..... .. ... ..... ... .. .. ..... ... ... .. ..... . . ... . . . ...... . .. . .... ...... .... .... .. ..... ..... .. ............ ........ ......... x ˆ .. ....... ...... ........ .... .... . .... .. ..... .......... ....... ........ ......... .... ..... .. .... ... ........... . . . . . . . .................................................. ... ... ...... ...... ... .............. ................................. ......... . .. .. .... .......... ...... ... . . . .......... .......... ....................................................... ......... ....... ........... ........................ ..................................................................................................................................................................................................................................................................................................................... x .. ........ .. .......... ...... ....... .......... ..... ..................................................................... . . . . . . . . . . . . . ... . . . ........ . ... ......... .......... .......... ......... .... .... ................................................................................... .......... ..... .. . . . ......... ... .... ... .... ..... ..... ........... ....... .......... . . . . . . . . . .. .. ..... ..... ...... ........... .. ..... ....... .. ..... ....... ... ... .. ..... . ... ... ...... ......... .. . . . . . . . .. . ... ... . ...... . . . .. . ... .. . . . . . . . . . . ... .. . . . . . . . . .. . .. . . . . . .. . . .. . . . ....

(ii) Unstable improper node: roots are real, distinct and positive

y

..... .. ....... ... .... .. ... .... .. ... ... ... ... ... ... ... ... ... .... ... ...... ... ... ... ...... ... ... .. ... .. .. .... ... ... ... .... . .... ... ... ... ... . ... .. ... ... ... ... . ... ...... .... ... ... ... .... .... . ........ .... ....................... ... ...... ....... .. ....... ................... .. ..... .... .... ....... . . ...... ... ... ................. . . . . . . . . ..... ..................... ...... . . . ... ...... .... ..... . . . . . . ... ... .. ........ .......................... ....................... ......... ...... ... . ........... ....... ...... ... ..... .... ... ....... ...... ...... ... . .. ....... ...... ..... .. .. ........ . . ....... ...... ...... . . . . . .......... .... ....... . .......... .......... . . . . . . . . . . . .... .... ... ... ... ...... ......... ......... ..... .... .... ................................................................................................................................................................................................................................................................... x .. .. .... .... .. ... ..... ............. ............. ........ ..... ........ . . . . . . . ....... ...... ..... ..... ...... ... .... ..... ...... ........ ... ... ... ..... ...... ....... ........ .. .. .. . ...... ...... ....... .. .... .... .... ...... .......................................................... . . . . . . . . ... ..... ...... ...... .......... ... .... .... ... .... ............................... ............ . ... .. ..... .. . .. ... ... .... ... ................ ... ....... ...... ... ........................... . ....... . . . . . . . . . . . . . . . . . . . . . ..... .. ....... . . . . . . . . . ... ..... ...... ... . ... ... ... .. ... .. ... ... ... .... .... .... ........ ... ... .. ... ..... ... ... ... ... ... ... .. ...... ... ... ... ...... .... .... .... .... ... .. .. .. .. .. ... .... ... .. ... .. ...

(iii) Unstable saddle point: roots are real and of opposite sign

y

ˆ ... y .......... .. . ... ... .. .... ... .. .. .... .... ... .... .. ... .... . .. . ... ... . ..... . .. ..... ... ... .. ... ..... .. . .. . ... ... . . . . . ..... ... ... .. ... ..... ... ... ........ ... ..... ... .. ..... ... .... .......... ..... ... . . . . . . . . . . . . . .. .. ............................. ... .... .............. ... ... ..... ... ... ... ......... ... .... ...... .................. .. ... .... ..... ........... ..... ..... . . . . . . . . . . . . . . . .......................................................................................................................................................................................................................................................................... x .. . . .. ... ... ... ...... ..... .... .... .... ... ........ ....... ..... ... .... .. ................ . ...... . . . . . ........... . ... ...... .. . .. . . . . . . . . . . . . . .. . .... .. ... ... .... .................. .... ... ...... .......... ... .... .. ..... . ... ..... ......... .... ... .. ..... . . . . . . . . . . . . .. . .. ... ... ..... .. .. .. ..... ... ... .. ..... .. .. .. .... ..... .. .. .... .... . . . . .... . . . . ... . ... ... .... .. .... ... ... .. .... ... ..

(iv) Equal roots r with a factor of t, r < 0 stable. Trajectories are tangent to yˆ at the origin, and parallel to yˆ when t approaches ±∞.

Figure 4.2: Improper node and saddle point

76


y

. .... ... ... ... .. ... .. ... ... ... ... ... ... ... ... ... ... . .. ... .. ... ... ....... .... ... ... ....... ....... ........ ... ....... ........ ....... . ......... ....... . ... ....... . . . . ....... ....... . ... . . . . . . . . .. ...... ... .. ... ............. ........... ....... ..... .... ..... ............. ....... .. .. .. ....... ....... ........ ....... . . . . . . . . . . . . . . . . . .................................................................................................................................................................................................. ..... . ... ..... ....... ... .. ... ....... ....... .... ..... ..... ............... .......... .............. ... .... ..... ........... . . ....... . . . . . . . .... ... ....... ... ....... ....... ......... ....... ........ ........ . . ....... . . . . . . . . ... .... ... ....... . . . . ....... . . ... . .. . . . . . ... .. ... . . . ... . . . . . . . ... .. ... . . . ... ... .... ... .. ... ... .

y

x

(i) Proper node: equal positive roots with no factor of t, unstable ............................... ............... ......... ............................. .............. . . ....... . ...... ................. ................... ................ .......... . . . . ........... ...... .... . . . .. .. ..... ....... ............. . ..... .......... ....................................................... ............ ......... ...... . . . .... ..... ..... ....... ............................ ....... ..... .... ... ... ....... .............................................................................. ......... ....... ..... .... . . . . .. . ... ... ........... ... .. .. . . . ... .... .... ....... .................................................................... ....... .... ... .... .... ... ... ... ... .......................................................................................... .... .... .... ... ... ... ... .... .... ....................................................................................... ... ... ... ... .... .... .... .... .... .... ........................................................................................................................................... ..... ..... ..... ..... .... .. ... ... .. .. .. .. .. ................................................................................ .. ... .. .. .. .. .. ... .. .. ... ... ... ... ............................................................................................................ .... ... ... .... ... ... ... ... ... ... ... ... ... ... ... .. ......................................• ............................... .. .. ... ... .. ... ... ... ..... .... .... .... .... .... .... .... ............................................................................................................................................. .... .... .... .... .... .... ... ... ... ... ... ... ... ................................................................................................. ... .. .. .. .. .. ... ... ... ... ... ... .... ....................................................................................... ... .... ... ... ... ... ... ... ... ... ... .... .................................................................................... ... ... ... ... ... ... ... .... .... .... .................................................................. ..... .... .... .... ... ... ... ... ... ...... ............................................................. ...... ... ... .. ... .... ... ...... ...... .......................................... ..... ..... .... .... ... ... ...... ...... .......... ............ .......... ....... ...... .... .. ... ..... ...... ......... ..................... ......... ...... ...... ...... .... ..... ...... ................................ ........ ...... ..... ..... ...... .......... . . .. ... ..... ....... ........... ....... ..... ...... .......... .................... ....................... ........... ... . ....... .. ......... ....................... ................ ......................................

(iii) Spiral curve corresponding to complex roots, real part of the root α < 0, asymptotically stable, c > 0 .................y ..... .......................... ....... ... ....... ... ...... . ...... .................... ..... . ..... ..... ..... .... ................ ..... ...... ... ..... ...... ... ... ...... . ... . . . . ..... ................. ... ..... .. ........... ... .... ....... ... ... ... ...... ... ... ... ..... ... ..... ... ... ... ..... ... ................ ... ... . . . ... ........... ... ... ... ..... . ... ... ... ... ..... ... ... ... ... ... ... .. ... ... . . ... ... . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . ... ... ... ..... .................................. . . . . .. ... ... . .... ............................................ . . . ... . . . .. .. .. .. ... .... ...... .................................. . . . . . .. .. .. .. .... ... .... ................................ ... .. ... .. . . . . . ........................................................................................................................................................................................................ x .. .. .. .. ............................... .. ... ... ... .... .... .... ..... .......................................... ....... .. .. ... ........ ... ... .... ..... .. .. .. ... ... ... ... ... ............. ............... ............... ... . . . . . . . . . . . . . . . . . . . ... ... ..... .......... ... .. ... ... ........ .... . ...... ........ ...... ... ...... ..... ......... ....... .... ....... ..................... ...... ..... ..... . ....... ..... . . ..... ............................................. . . ...... .... ........ ...... ........ .............. ...........................

(v) Integral curves for x0 = x − 4y, y 0 = 4x + y

... ... ... ... . ... ... ... ... ... ... ... ... ... ... ... ... . ... . ....... . .... ... ... .......... ....... .. ......... ...... ....... ... ... ... ....... ....... ... ... . ....... . ........ . . . ... . ........ . .......... . . . . . . . . . . . . . . ... .. .. ........... .... . . .. ....... ........ ....... ..... ... .... ............ ....... ... ... .. ....... ....... .. ............ . . . . . . . . . . . . . . .................................................................................................................................................................................... .......... ..... ....... .. .. ... ....... ....... ... .. ... ............ ....... ....... .... .... .... . . . . . .......... ............... ... .... ..... .......... ... .. ........ ......... . ... ....... ... ... ....... . . . . . . . ....... . . . . . ....... ... ...... . .... ........... ....... ...... . . . . . . . ... ...... ... . . ... . . . . . . . ... . ... .... ... ... ... ... . .. ... ... .

x

(ii) Proper node: equal negative roots with no factor of t, stable ............................... ............... ......... ............................. .............. . . ....... . ...... ................. ................... ................ .......... . . . . ........... ...... .... . . . .. .. ..... ....... ............. . ..... .......... ....................................................... ............ ......... ...... . . . .... ..... ..... ....... ............................ ....... ..... .... ... ... ....... .............................................................................. ......... ....... ..... .... . . . . .. . ... ... ........... ... .. .. . . . ... .... .... ....... .................................................................... ....... .... ... .... .... ... ... ... ... .......................................................................................... .... .... .... ... ... ... ... .... .... ....................................................................................... ... ... ... ... .... .... .... .... .... .... ........................................................................................................................................... ..... ..... ..... ..... .... .. ... ... .. .. .. .. .. ................................................................................ .. ... .. .. .. .. .. ... .. .. ... ... ... ... ............................................................................................................ .... ... ... .... ... ... ... ... ... ... ... ... ... ... ... .. ......................................• ............................... .. .. ... ... .. ... ... ... ..... .... .... .... .... .... .... .... ............................................................................................................................................. .... .... .... .... .... .... ... ... ... ... ... ... ... ................................................................................................. ... .. .. .. .. .. ... ... ... ... ... ... .... ....................................................................................... ... .... ... ... ... ... ... ... ... ... ... .... .................................................................................... ... ... ... ... ... ... ... .... .... .... .................................................................. ..... .... .... .... ... ... ... ... ... ...... ............................................................. ...... ... ... .. ... .... ... ...... ...... .......................................... ..... ..... .... .... ... ... ...... ...... .......... ............ .......... ....... ...... .... .. ... ..... ..... ......... ..................... ......... ...... ...... ...... .... ..... ....... ................................ ........ ........ ..... ..... ...... .......... .. .. . ... ..... ....... ........... ......... ..... ...... .......... .................... ................ ........... ... . ....... .. ......... ....................... ................ ......................................

(iv) Spiral curve corresponding to complex roots, real part of the root α > 0, unstable, c < 0 y

.... ........ .... ... ... ... ... ... ... ... ... ... ... ......................... ....... . ..... ..... ........... . . . . ... . . ... ..... ... . . . . ... . ..................... .... . . . . . . . . ..... ... ....... . . .... . . . . . . . . . ... ... ..... ... . . .............. . . . . ... . . .... . . .. . . . . . . . .. .. ..................... ........ . . . . . . . . . . . . . .. . . ... ... .. .. . . . . . . . . . . . . . ........ ... .. .. .. ... ... ........................................................................................................• ............................................................................................................... x . .. ... .... ........ ... ... ..... .. ... . . . . . . . . . .... .... . . .... .. ... ..................... ... .. .... ... .. ... ... .......... ..... .... ... ..... ............ ..... ... ..... . . . . . . . . . . . . . . .................... .. ... ... .. .......... ... .... ............. ...... ........... ........................ .......... .. ... ..

(vi) A stable center corresponding to pure imaginary roots

Figure 4.3: Stable and unstable nodes

4.8. APPENDIX 1: PROOF OF LEMMA 4.12

4.8

77

Appendix 1: Proof of Lemma 4.12

Lemma 4.12 Let A be an n×n complex matrix and λ an eigenvalue of A with algebraic multiplicity m. Then dim {x ∈ Cn | (λI − A)m x = 0} = m. Proof. The proof consists of several steps. Let T = {x ∈ Cn | (λI − A)m x = 0}. The space T is called the generalized eigenspace corresponding to the eigenvalue λ. Step 1. T is a subspace of Cn . This is just direct verification. Step 2. T is invariant under A meaning A[T ] ⊆ T . This is because if we take a vector x in T , then (λI − A)m x = 0 so that A(λI − A)m x = 0, which is the same as (λI − A)m (Ax) = 0. Therefore, Ax is also in T . Step 3. By Step 2 which says that A[T ] ⊆ T , we may consider A as a linear transformation on the subspace T . In other words, A : T −→ T . Let µ be an eigenvalue of A : T −→ T . That is Av = µv. Then v ∈ T implies that (λI − A)m v = 0. Since Av = µv, this simplifies to (λ − µ)m v = 0. Being an eigenvector, v 6= 0 so that µ = λ. Therefore, all eigenvalues of A : T −→ T are equal to λ. Step 4. Let dim T = r. Certainly r ≤ n. Then by Step 3, the characteristic polynomial of A : T −→ T is (λ − z)r . Since T is an invariant subspace of A : Cn −→ Cn , one can choose a basis of T and then extend it to a basis of Cn so that with respect to this new basis of Cn , the matrix A is similar to a matrix where the upper left hand r × r submatrix represents A on T and the lower left hand (n − r) × r submatrix is the zero matrix. From this, we see that (λ − z)r is a factor of the characteristic polynomial of A : Cn −→ Cn . Hence r ≤ m. We also need the Cayley-Hamilton Theorem in the last step of the proof. Cayley-Hamilton Theorem Let p(z) be the characteristic polynomial of an n × n matrix A. Then p(A) = 0. Step 5. Let p(z) = (λ1 − z)m1 · · · (λk − z)mk be the characteristic polynomial of A, where λ1 , λ2 , . . . , λk are the distinct eigenvalues of A : Cn −→ Cn . For each i from 1 to k, let Ti = {x ∈ Cn | (λi I − A)mi x = 0}. By Step 4, we have dimTi ≤ mi . Let fi (z) = (λi − z)mi and gi (z) = f1 (z) · · · fî (z) · · · fk (z), where fî (z) means the polynomial fi (z) is omitted. Note that fi (z)gi (z) = p(z) for all i. 1 into partial fractions, we have the identity Resolving (λ1 −z)m1 ···(λ m k −z) k (λ1 −

z)m1

1 h1 (z) h2 (z) hk (z) ≡ + + ··· + , m m m 1 2 k · · · (λk − z) (λ1 − z) (λ2 − z) (λk − z)mk

where h1 (z), . . . , hk (z) are polynomials in z. Finding the common denominator of the right hand side and equate the numerators on both sides, we have 1 ≡ g1 (z)h1 (z) + · · · + gk (z)hk (z). Substituting the matrix A into this polynomial identity, we have g1 (A)h1 (A) + · · · + gk (A)hk (A) = I,

78


where I is the identity n × n matrix. Now for any x ∈ Cn , we have g1 (A)h1 (A)x + · · · + gk (A)hk (A)x = x. Note that each gi (A)hi (A)x is in Ti because fi (A)[gi (A)hi (A)x] = p(A)hi (A)x = 0 by the Cayley-Hamilton Theorem. This shows that any vector in Cn can be expressed as a sum of vectors where the i-summand is in Ti . In other words, Cn = T1 + T2 + · · · + Tk . Consequently, m1 + · · · mk = n ≤ dimT1 + · · · + dimTk ≤ m1 + · · · + mk so that dimTi = mi . Remarks. 1. In fact Cn = T1 ⊕ · · · ⊕ Tk , which is the primary decomposition theorem. To prove this, we know from the dimension theorem that dimT1 +dim(T2 +· · ·+Tk ) = dimCn +dimT1 ∩(T2 +· · ·+Tk ). Thus, n = m1 +(m2 +· · ·+ mk ) ≥ m1 +dim(T2 +· · ·+Tk ) = n+dimT1 ∩(T2 +· · ·+Tk ) so that dimT1 ∩(T2 +· · ·+Tk ) = 0. That is T1 ∩ (T2 + · · · + Tk ) = {0}. Similarly, Ti ∩ (T1 + · · · + Tî + · · · + Tk ) = {0}, where the summand Tî is omitted for i = 1, . . . , k. This proves that T1 + · · · + Tk is a direct sum. 2. If A is a real matrix and λ is a real eigenvalue of A of algebraic multiplicity m, then T is a real vector space and the real dimension of T is m. This is because for any set of real vectors in Rn , it is linearly independent over R if and only if it is linearly independent over C. 3. If A is a real matrix and λ is a complex eigenvalue of A of algebraic multiplicity m, then λ is also an eigenvalue of A of algebraic multiplicity m. In this case, if {v1 , . . . , vm } is a basis over C of Tλ , where Tλ is the generalized eigenspace corresponding to λ, then {v1 , . . . , vm } is a basis over C of Tλ . It can be shown that the 2m real vectors

79

We begin by finding a “cyclic vector” for each Jordan block which generates the Jordan basis for that block. Let ti = dimKer(λI − A)i , i = 1, . . . , m. Note that ti+1 − ti is the number of Jordan blocks of size greater i, and thus the number of Jordan blocks of size exactly i is si = (ti −ti−1 )−(ti+1 −ti ) for 0 < i < k. Here t0 is taken as 0. Start with the largest k such that tk−1 < tk . Note that 1 ≤ k ≤ m.

1.

Choose a basis {vk1 , . . . , vksk } of T = Ker(λI − A)k mod Ker(λI − A)k−1 . These are the cyclic vectors for the Jordan blocks of size k. There are sk = tk − tk−1 of them.

2.

If k = 1, stop. Otherwise, apply λI − A to the cyclic vectors from the previous step to obtain (λI − A)vki in Ker(λI − A)k−1 . The important point is that {(λI − A)vk1 , . . . , (λI − A)vksk } is linearly independent mod Ker(λI − A)k−2 . If k = 2, stop. Otherwise, extend to a basis of Ker(λI − A)k−1 mod Ker(λI − A)k−2 by sk−1 1 adding {vk−1 , . . . , vk−1 }, which are then the cyclic vectors for the Jordan blocks of size k − 1. Since we have extended a list of size sk to reach a length tk−1 − tk−2 , there are sk−1 = (tk−1 − tk−2 ) − (tk − tk−1 ) of these new cyclic vectors.

3.

Repeat step 2, with k replaced by k − 1.

Once the algorithm terminates, we may arrange the basis vectors as follows: { vk1 , . . . , (λI − A)k−1 vk1 , . . . , vksk , . . . , (λI − A)k−1 vksk , ············································· 1 vi , . . . , (λI − A)i−1 vi1 , . . . , visi , . . . , (λI − A)i−1 visi , ···································· v21 , (λI − A)v21 , . . . , v2s2 , (λI − A)v2s2 , v11 , . . . , v1s1 }

This puts the Jordan form of A on T into blocks of decreasing size down the diagonal.

      Example 4.12 Let A =      

2 0 0 0 0 0 0

0 2 0 0 0 0 0

0 −1 2 0 0 0 0

−1 0 0 2 0 0 0

0 0 −1 −1 2 0 0

0 0 −1 0 −1 2 0

0 0 0 −1 0 0 2

      . Find a Jordan basis for A.     

80

CHAPTER 4. LINEAR DIFFERENTIAL SYSTEMS 

     Solution. 2I − A =             2 (2I − A) =       

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 1 0 0 0 0 0

1 0 0 0 0 0 0

0 0 1 1 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

1 1 0 0 0 0 0

0 1 1 1 0 0 0

1 0 0 0 0 0 0

0 0 0 0 0 1  0 0 0 0 0 1    0 0 0 0 0 0  (2I − A)3 =   0 0 0 0 0 0   0 0 0 0 0 0   0 0 0 0 0 0 0 0 0 0 0 0 4 (2I − A) is the zero matrix so that t4

0 0 1 0 1 0 0 

0 0 0 1 0 0 0

       has rank 5, so that t1 = 2.     

      has rank 3, so that t2 = 4.     

 0 0    0   0   has rank 1, so that t3 = 6.  0   0  0 = 7, and so t5 = t6 = t7 = 7. Thus k = 4.

As (t4 − t3 ) − (t5 − t4 ) = 1, there is 1 Jordan block of size 4. As (t3 − t2 ) − (t4 − t3 ) = 1, there is 1 Jordan block of size 3. As (t2 − t1 ) − (t3 − t2 ) = 0, there is no Jordan block of size 2. As (t1 − t0 ) − (t2 − t1 ) = 0, there is no Jordan block of size 1. Let ei , i = 1, . . . , 7 be the canonical basis vectors. From the matrix 2I − A , we know that e1 , e2 are eigenvectors. Let’s follow the algorithm to find a Jordan basis for A. Start with k = 4. (k = 4) A basis for Ker(2I − A)4 mod Ker(2I − A)3 is {e6 }. Then (2I − A)(e6 ) = e3 + e5 , which is in Ker(2I − A)3 . Extending it to a basis mod Ker(2I − A)2 , we get {e3 + e5 , e7 }. Here e7 is the added vector. (k = 3) (2I − A)(e3 + e5 ) = e2 + e3 + e4 and (2I − A)(e7 ) = e4 , which are in Ker(2I − A)2 . Extending it to a basis mod Ker(2I −A), we get itself {e2 +e3 +e4 , e4 }. That is no more extension is necessary. (k = 2) (2I − A)(e2 + e3 + e4 ) = e1 + e2 , (2I − A)(e4 ) = e1 . So we choose {e1 + e2 , e1 } to be the basis for the eigenspace Ker(2I − A). (k = 1) Stop.


81

Therefore, a Jordan basis is {e6 , e3 + e5 , e2 + e3 + e4 , e1 + e2 , e7 , e4 , e1 }.       Consequently, the Jordan form of A is      

      Example 4.13 Let A =      

2 0 0 0 0 0 0

0 2 0 0 0 0 0

0 −1 2 0 0 0 0

2 0 0 0 0 0 0

−1 0 0 2 0 0 0

1 2 0 0 0 0 0

0 1 2 0 0 0 0

0 0 −1 −1 2 0 0

0 0 1 2 0 0 0

0 0 −1 0 −1 2 0

0 0 0 0 2 0 0

0 0 0 0 1 2 0

0 0 0 −1 0 0 2

0 0 0 0 0 1 2

      .     

      . Solve the system x0 = Ax.     

Solution. A has an eigenvalue 2 of algebraic multiplicity 7. We use the Jordan basis {e6 , e3 + e5 , e2 + e3 + e4 , e1 + e2 , e7 , e4 , e1 } for A. Thus 2

3

x(t) = e2t [ c1 (e6 + t(e3 + e5 ) + t2 (e2 + e3 + e4 ) + t6 (e1 + e2 )) 2 +c2 ((e3 + e5 ) + t(e2 + e3 + e4 ) + t2 (e1 + e2 )) +c3 ((e2 + e3 + e4 ) + t(e1 + e2 )) +c4 (e1 + e2 ) 2 +c5 (e7 + te4 + t2 e1 ) +c6 (e4 + te1 ) +c7 e1 ] .

    Exercise 4.7 Find a Jordan basis of the matrix A =    

3 0 0 3 0 0 0 0 0 0

−1 0 3 0 0

0 −1 0 3 0

−1 0 −1 0 3

    . Hence solve the   

system x0 = Ax.

Ans: Thereis 1 block of size 3 and 1 block of size 2. A Jordan basis is {e5 , e1 + e3 , e1 , e4 , e2 }. 2 x(t) = e3t c1 (e5 + t(e1 + e3 ) + t2 e1 ) + c2 ((e1 + e3 ) + te1 ) + c3 e1 + c4 (e4 + te2 ) + c5 e2 .

82


Chapter 5

Power Series Solutions 5.1

Power Series

An infinite series of the form ∞ X an (x − x0 )n = a0 + a1 (x − x0 ) + a2 (x − x0 )2 + · · ·

(5.1.1)

n=0

is a power series in x − x0 . In what follows, we will be focusing mostly at the point x0 = 0. That is ∞ X

an xn = a0 + a1 x + a2 x2 + · · ·

(5.1.2)

n=0

(5.1.2) is said to converge at a point x if the limit lim

m→∞

m X

an xn exists, and in this case the sum of

n=0

the series is the value of this limit. It is obvious that (5.1.2) always converges at x = 0. It can be showed that each power series like (5.1.2) corresponds to a positive real number R, called the radius of convergence, with the property that the series converges if |x| < R and diverges if |x| > R. It is customary to put R equal to 0 when the series converges only at x = 0, and equal to ∞ when it converges for all x. In many important cases, R can be found by the ratio test as follow. If each an 6= 0 in (5.1.2), and if for a fixed point x 6= 0 we have an+1 xn+1 = lim an+1 |x| = L, lim n n→∞ n→∞ an x an then (5.1.2) converges for L < 1 and diverges if L > 1. It follows from this that an R = lim n→∞ an+1 if this limit exists (we put R = ∞, if |an /an+1 | −→ ∞) The interval (−R, R) is called the interval of convergence in the sense that inside the interval the series converges and outside the interval the series diverges. Consider the following power series ∞ X

n!xn = 1 + x + 2!x2 + 3!x3 + · · ·

n=0

83

(5.1.3)

84

CHAPTER 5. POWER SERIES SOLUTIONS ∞ X x2 x3 xn =1+x+ + + ··· n! 2! 3! n=0 ∞ X

xn = 1 + x + x2 + x3 + · · ·

(5.1.4)

(5.1.5)

n=0

It is easy to verify that (5.1.3) converges only at x = 0. Thus R = 0. For (5.1.4), it converges for all x so that R = ∞. For (5.1.5), the power series converges for |x| < 1 and R = 1. Suppose that (5.1.2) converges for |x| < R with R > 0, and denote its sum by f (x). That is ∞ X

f (x) =

an xn = a0 + a1 x + a2 x2 + · · ·

(5.1.6)

n=0

Then one can prove that f is continuous and has derivatives of all orders for |x| < R. Also the series can be differentiated termwise in the sense that f 0 (x) =

∞ X

nan xn−1 = a1 + 2a2 x + 3a3 x2 + · · · ,

n=1

f 00 (x) =

∞ X

n(n − 1)an xn−2 = 2a2 + 3 · 2a3 x + · · · ,

n=2

and so on. Furthermore, the resulting series are still convergent for |x| < R. These successive differentiated series yield the following basic formula relating an to with f (x) and its derivatives. an =

f n (0) n!

(5.1.7)

Moreover, (5.1.6) can be integrated termwise provided the limits of integration lie inside the interval of convergence. If g(x) =

∞ X

bn xn = b0 + b1 x + b2 x2 + · · ·

(5.1.8)

n=0

is another power series with interval of convergence |x| < R, then (5.1.6) and (5.1.8) can be added or subtracted termwise: f (x) ± g(x) =

∞ X

(an ± bn )xn = (a0 ± b0 ) + (a1 ± b1 )x + (a2 ± b2 )x2 + · · ·

(5.1.9)

n=0

They can also be multiplied like polynomials, in the sense that f (x)g(x) =

∞ X

cn x n ,

n=0

where cn = a0 bn + a1 bn−1 + · · · + an b0 . Suppose two power series (5.1.6) and (5.1.8) converge to the same function so that f (x) = g(x) for |x| < R, then (5.1.7) implies that they have the same coefficients, an = bn for all n. In particular, if f (x) = 0 for all |x| < R, then an = 0, for all n.

5.1. POWER SERIES

85

Let f (x) be a continuous function that has derivatives of all orders for |x| < R. Can it be represented by a power series? If we use (5.1.7), it is natural to expect f (x) =

∞ X f 00 (0) 2 f (n) (0) n x = f (0) + f 0 (0)x + x + ··· n! 2! n=0

(5.1.10)

to hold for all |x| < R. Unfortunately, this is not always true. Instead, one can use Taylor’s expansion for f (x): n X f (k) (0) k f (x) = x + Rn (x), k! k=0

where the remainder Rn (x) is given by Rn (x) =

f (n+1) (x) n+1 x (n + 1)!

for some point x between 0 and x. To verify (5.1.6), it suffices to show that Rn (x) −→ 0 as n −→ ∞. Example 5.1 The following familiar expansions are valid for all x. ex =

sin x =

∞ X xn x2 x3 =1+x+ + + ··· n! 2! 3! n=0

∞ X

(−1)n

n=0 ∞ X

cos x =

x3 x5 x2n+1 =x− + − ··· (2n + 1)! 3! 5!

(−1)n

n=0

x2n x2 x4 =1− + − ··· (2n)! 2! 4!

A function f (x) with the property that a power series expansion of the form f (x) =

∞ X

an (x − x0 )n

(5.1.11)

n=0

is valid in some interval containing the point x0 is said to be analytic at x0 . In this case, an is necessarily given by f (n) (x0 ) an = , n! and (5.1.11) is called the Taylor series of f (x) at x0 . Thus ex , sin x, cos x are analytic at all points. Concerning analytic functions, we have the following basic results. 1. Polynomials, ex , sin x, cos x are analytic at all points. 2. If f (x) and g(x) are analytic at x0 , then f (x) ± g(x), f (x)g(x), and f (x)/g(x) [provided g(x0 ) 6= 0] are also analytic at x0 . 3. If f (x) is analytic at x0 , and f −1 (x) is a continuous inverse, then f −1 (x) is analytic at f (x0 ) if f 0 (x0 ) 6= 0. 4. If g(x) is analytic at x0 and f (x) is analytic at g(x0 ), then f (g(x)) is analytic at x0 . 5. The sum of a power series is analytic at all points inside the interval of convergence.

86

5.2

CHAPTER 5. POWER SERIES SOLUTIONS

Series Solutions of First Order Equations

A first order differential equation y 0 = f (x, y) can be solved by assuming that it has a power series solution. Let’s illustrate this with two familiar examples. Example 5.2 Consider the differential equation y 0 = y. We assume it has a power series solution of the form y = a0 + a1 x + a2 x2 + · · · + an xn + · · · (5.2.1) that converges for |x| < R. That is the equation y 0 = y has a solution which is analytic at the origin. Then y 0 = a1 + 2a2 x + · · · + nan xn−1 + · · · (5.2.2) has the same interval of convergence. Since y 0 = y, the series (5.2.1) and (5.2.2) have the same coefficients. That is (n + 1)an+1 = an , Thus an =

1 n an−1

=

1 n(n−1) an−2

all for n = 0, 1, 2, . . .

= ··· =

1 n! a0 .

Therefore

x3 xn x2 + + ··· + + ··· , y = a0 1 + x + 2! 3! n! where a0 is an arbitrary constant. In this case, we recognize this as the power series of ex . Thus the general solution is y = a0 ex . Example 5.3 The function y = (1+x)p , where p is a real constant satisfies the differential equation (1 + x)y 0 = py, y(0) = 1. As before, we assume it has a power series solution of the form y = a0 + a1 x + a2 x2 + · · · + an xn + · · · with positive radius of convergence. Then y0 xy 0 py

= a1 + = = pa0 +

2a2 x + 3a3 x2 + · · · + (n + 1)an+1 xn + · · · , a1 x + 2a2 x2 + · · · + nan xn + · · · , pa1 x + pa2 x2 + · · · + pan xn + · · · ,

Using (5.2.3) and equating coefficients, we have (n + 1)an+1 + nan = pan , That is an+1 =

for all n = 0, 1, 2, . . . p−n an , n+1

so that a1 = p, a2 =

p(p − 1) p(p − 1)(p − 2) p(p − 1) · · · (p − n + 1) , a3 = , ..., an = . 2 2·3 n!

(5.2.3)

5.3. SECOND ORDER LINEAR EQUATIONS AND ORDINARY POINTS

87

In other words, y = 1 + px +

p(p − 1) 2 p(p − 1)(p − 2) 3 p(p − 1) · · · (p − n + 1) n x + x + ··· + x + ··· . 2 2·3 n!

By ratio test, this series converges for |x| < 1. Since (5.2.3) has a unique solution, we conclude that (1 + x)p = 1 + px +

p(p − 1) 2 p(p − 1)(p − 2) 3 p(p − 1) · · · (p − n + 1) n x + x +···+ x +··· , 2 2·3 n!

for |x| < 1, This is just the binomial series of (1 + x)p .

5.3

Second Order Linear Equations and Ordinary Points

Consider the homogeneous second order linear differential equation y 00 + P (x)y 0 + Q(x)y = 0

(5.3.1)

Definition 5.1 The point x0 is said to be an ordinary point of (5.3.1) if P (x) and Q(x) are analytic at x0 . If at x = x0 , P (x) and/or Q(x) are not analytic, then x0 is said to be a singular point of (5.3.1). A singular point x0 at which the functions (x − x0 )P (x) and (x − x0 )2 Q(x) are analytic is called a regular singular point of (5.3.1). If a singular point x0 is not a regular singular point, then it is called an irregular singular point. Example 5.4 If P (x) and Q(x) are constant, then every point is an ordinary point of (5.3.1). Example 5.5 Consider the equation y 00 + xy = 0. Since the function Q(x) = x is analytic at every point, every point is an ordinary point. Example 5.6 In the Cauchy-Euler equation y 00 + ax1 y 0 + xa22 y = 0, where a1 and a2 are constants, the point x = 0 is a singular point, but every other point is an ordinary point. Example 5.7 Consider the differential equation y 00 +

1 8 y0 + y = 0. (x − 1)2 x(x − 1)

The singular points are 0 and 1. At the point 0, xP (x) = x(1 − x)−2 and x2 Q(x) = −8x(1 − x)−1 , which are analytic at x = 0, and hence the point 0 is a regular singular point. At the point 1, we have (x − 1)P (x) = 1/(x − 1) which is not analytic at x = 1, and hence the point 1 is an irregular singular point. To discuss the behavior of the singularities at infinity, we use the transformation x = 1/t, which converts the problem to the behavior of the transformed equation near the origin. Using the substitution x = 1/t, (5.3.1) becomes d2 y 2 1 1 dy 1 1 + − P ( ) + 4 Q( )y = 0 (5.3.2) dt2 t t2 t dt t t We define the point at infinity to be an ordinary point, a regular singular point, or an irregular singular point of (5.3.1) according as the origin of (5.3.2) is an ordinary point, a regular singular point, or an irregular singular point.

88


Example 5.8 Consider the differential equation 1 1 1 dy d2 y 1 + + y = 0. + dx2 2 x2 x dx 2x3 The substitution x = 1/t transforms the equation into d2 y 3 − t dy 1 + + y = 0. dt2 2t dt 2t Hence the point at infinity is a regular singular point of the original differential equation. Exercise 5.1 Show that the hypergeometric equation x(1 − x)y 00 + [c − (a + b + 1)x]y 0 − aby = 0, where a, b, c are constants has precisely 3 regular singular points at 0, 1, ∞.

Theorem 5.1 Let x0 be an ordinary point of the differential equation y 00 + P (x)y 0 + Q(x)y = 0, and let a0 and a1 be arbitrary constants. Then there exists a unique function y(x) that is analytic at x0 , is a solution of the differential equation in an interval containing x0 , and satisfies the initial conditions y(x0 ) = a0 , y 0 (x0 ) = a1 . Furthermore, if the power series expansions of P (x) and Q(x) are valid on an interval |x − x0 | < R, R > 0, then the power series expansion of this solution is also valid on the same interval. Proof. Let x0 be an ordinary point of (5.3.1). That is P (x) and Q(x) are analytic at x0 . For ∞ ∞ X X simplicity, we take x0 = 0. Thus P (x) = pn xn and Q(x) = qn xn , for |x| < R, with n=0

n=0

R > 0. We look for a solution of (5.3.1) in form of a power series: y =

∞ X

an xn with radius of

n=0

convergence at least R. Here a0 and a1 are the two given numbers for the initial value problem. 0

y = y 00 =

∞ X

(n + 1)an+1 xn = a1 + 2a2 x + 3a3 x2 + · · ·

n=0 ∞ X

(n + 1)(n + 2)an+2 xn = 2a2 + (2)(3)a3 x + (3)(4)a4 x2 + · · ·

n=0

0

Thus P (x)y = =

∞ X

pn x n=0 " ∞ n X X

∞ X

# (n + 1)an+1 x #

n

n=0

pn−k (k + 1)ak+1 xn .

n=0

Q(x)y =

!" n

∞ X n=0

k=0

! qn x

n

∞ X n=0

! n

an x

=

" n ∞ X X n=0

k=0

# qn−k ak xn .


89

Substituting these into (5.3.1), we have ∞ X

" (n + 1)(n + 2)an+2 +

n=0

n X

pn−k (k + 1)ak+1 +

k=0

n X

# qn−k ak xn = 0.

k=0

Equating the coefficient of xn to 0, we get the following recursion formula for an . (n + 1)(n + 2)an+2 = −

n X

[(k + 1)pn−k ak+1 + qn−k ak ]

(5.3.12)

k=0

For example, 2a2 = −(p0 a1 + q0 a0 ), (2)(3)a3 = −(p1 a1 + 2p0 a2 + q1 a0 + q0 a1 ), (3)(4)a4 = −(p2 a1 + 2p1 a2 + 3p0 a3 + q2 a0 + q1 a1 + q0 a2 ), and so on ... The formula (5.3.12) determines all a2 , a3 , . . . in terms of a0 and a1 so that the power series ∞ X y= an xn formally satisfies the DE and the given initial conditions. It remains to prove the n=0 ∞ X

series

an xn , with an defined by (5.3.12) converges for |x| < R. Recall R is the radius of

n=0

convergence for P (x) =

∞ X

p n xn

and Q(x) =

n=0

∞ X

qn xn .

(5.3.13)

n=0

Let r be a positive number such that r < R. The two series (5.3.13) converge at x = r so that the terms approach zero and are therefore bounded. Thus there exists a constant M > 0 such that |pn |rn ≤ M

and

|qn |rn ≤ M for all n.

Therefore n M X [(k + 1)|ak+1 | + |ak |] rk rn k=0 n M X ≤ n [(k + 1)|ak+1 | + |ak |] rk + M |an+1 |r. r

(n + 1)(n + 2)|an+2 | ≤

k=0

Let’s consider a sequence {bn }∞ n=0 defined as follow: b0 = |a0 |, b1 = |a1 |, and bn+2 given by (n + 1)(n + 2)bn+2 =

n M X [(k + 1)bk+1 + bk ] rk + M bn+1 r rn

(5.3.14)

k=0

Observe that 0 ≤ |an | ≤ bn for all n. Now we try to prove that Then by comparison test,

∞ X n=0

∞ X

bn xn converges for |x| < r.

n=0

an xn also converges for |x| < r. Since r is an arbitrary positive

90


number less than R, we conclude that show

∞ X

∞ X

an xn converges for |x| < R. We shall apply ratio test to

n=0

bn xn converges. From (5.3.14), replacing n by n − 1 and then by n − 2, we obtain

n=0

n(n + 1)bn+1 = (n − 1)nbn =

n−1 M X

rn−1

M rn−2

[(k + 1)bk+1 + bk ] rk + M bn r, and

k=0 n−2 X

[(k + 1)bk+1 + bk ] rk + M bn−1 r.

k=0

Now multiplying the first equation by r and by the second equation, we obtain rn(n + 1)bn+1 = =

n−1 M X

rn−2

k=0

n−2 M X

rn−2

[(k + 1)bk+1 + bk ] rk + M bn r2 [(k + 1)bk+1 + bk ] rk + rM (nbn + bn−1 )

k=0

+M bn r2 = (n − 1)nbn − M bn−1 r + rM (nbn + bn−1 ) + M bn r2 = [(n − 1)n + rM n + M r2 ]bn . Therefore

bn+1 (n − 1)n + rM n + M r2 . = bn rn(n + 1)

∞ X bn+1 xn+1 = lim bn+1 |x| = |x| , and the series Thus lim bn xn converges for |x| < r. n→∞ bn xn n→∞ bn r n=0 ∞ X Consequently, by comparison test the series an xn converges for |x| < r. n=0

Exercise 5.2 Find two linearly independent solutions of y 00 − xy 0 − x2 y = 0 Ans: y1 (x) = 1 +

1 4 12 x

+

1 6 90 x

+

3 8 1120 x

+ · · · and y2 (x) = x + 61 x3 +

3 5 40 x

+

13 7 1008 x

+ ···.

Exercise 5.3 Using power series method, solve the initial value problem (1+x2 )y 00 +2xy 0 −2y = 0, y(0) = 0, y 0 (0) = 1. Ans: y = x. Legendre’s equation (1 − x2 )y 00 − 2xy 0 + p(p + 1)y = 0, where p is a constant called the order of Legendre’s equation. p(p+1) 2x That is P (x) = − 1−x 2 and Q(x) = 1−x2 . The origin is an ordinary point, and we expect a P solution of the form y = an xn . Thus the left hand side of the equation becomes (1 − x2 )

∞ X

(n + 1)(n + 2)an+2 xn − 2x

n=0

∞ X

(n + 1)an+1 xn + p(p + 1)

n=0

∞ X n=0

an xn ,


91

or ∞ X

(n + 1)(n + 2)an+2 xn −

n=0

∞ X

(n − 1)nan xn −

n=2

∞ X

2nan xn +

n=1

∞ X

p(p + 1)an xn .

n=0 n

The sum of these series is required to be zero, so the coefficient of x must be zero for every n. This gives (n + 1)(n + 2)an+2 − (n − 1)nan − 2nan + p(p + 1)an = 0, for n = 2, 3, . . . In other words, an+2 = −

(p − n)(p + n + 1) an . (n + 1)(n + 2)

This recursion formula enables us to express an in terms of a0 or a1 according as n is even or odd. In fact, for m > 0, we have a2m = (−1)m

p(p − 2)(p − 4) · · · (p − 2m + 2)(p + 1)(p + 3) · · · (p + 2m − 1) a0 , (2m)!

a2m+1 = (−1)m

(p − 1)(p − 3) · · · (p − 2m + 1)(p + 2)(p + 4) · · · (p + 2m) a1 . (2m + 1)!

With that, we get two linearly independent solutions y1 (x) =

∞ X m=0

a2m x

2m

∞ X

and y2 (x) =

a2m+1 x2m+1 ,

m=0

and the general solution is given by p(p + 1) 2 p(p − 2)(p + 1)(p + 3) 4 y = a0 1 − x + x 2! 4! p(p − 2)(p − 4)(p + 1)(p + 3)(p + 5) 6 − x + ··· 6!

(p − 1)(p + 2) 3 (p − 1)(p − 3)(p + 2)(p + 4) 5 +a1 x − x + x 3! 5! (p − 1)(p − 3)(p − 5)(p + 2)(p + 4)(p + 6) 7 − x + ··· . 7! When p is not an integer, the series representing y1 and y2 have radius of convergence R = 1. For example, a2n+2 x2n+2 (p − 2n)(p + 2n + 1) 2 = − |x | −→ |x|2 a2n x2n (2n + 1)(2n + 2) as n −→ ∞, and similarly for the second series. In fact, by Theorem 4.1, and the familiar expansion 1 = 1 + x2 + x4 + · · · , 1 − x2

|x| < 1,

shows that R = 1 for both P (x) and Q(x). Thus, we know any solution of the form y = must be valid at least for |x| < 1.

P

an xn

92


The functions defined in the series solution of Legendre’s equation are called Legendre functions. When p is a nonnegative integer, one of these series terminates and becomes a polynomial in x. For instance, if p = n is an even positive integer, the series representing y1 terminates and y1 is a polynomial of degree n. If p = n is odd, y2 again is a polynomial of degree n. These are called Legendre polynomials Pn (x) and they give particular solutions to Legendre’s equation (1 − x2 )y 00 − 2xy 0 + n(n + 1)y = 0, where n is a nonnegative integer. It is customary to choose the arbitrary constants a0 or a1 so that the coefficient of xn in Pn (x) is (2n)!/[2n (n!)2 ]. This implies Pn (1) = 1. Then bn/2c

Pn (x) =

X k=0

(−1)k (2n − 2k)! xn−2k . − k)!(n − 2k)!

2n k!(n

The six Legendre polynomials are P0 = 1, P2 (x) = 21 (3x2 − 1), P4 (x) = 18 (35x4 − 30x2 + 3),

P1 (x) = x P3 (x) = 21 (5x3 − 3x) P5 (x) = 81 (63x5 − 70x3 + 15x)

There is also a Rodrigues’ formula for the Legendre polynomial given by Pn (x) =

1 dn 2 (x − 1)n . n!2n dxn

Hermite’s equation y 00 − 2xy 0 + 2py = 0, where p is a constant. The general solution of Hermite’s equation is y(x) = a0 y1 (x) + a1 y2 (x), where 2p 22 p(p − 2) 4 23 p(p − 2)(p − 4) 6 y1 (x) = 1 − x2 + x − x + ··· , 2! 4! 6! 2(p − 1) 3 22 (p − 1)(p − 3) 5 23 (p − 1)(p − 3)(p − 5) 7 x + x − x + ··· . y2 (x) = x − 3! 5! 7! By Theorem 4.1, both series for y1 and y2 converge for all x. Note that y1 is a polynomial if p is an even integer, whereas y2 is a polynomial if p is an odd integer. The Hermite polynomial of degree n denoted by Hn (x) is the nth-degree polynomial solution of Hermite’s equation, multiplied by a suitable constant so that the coefficient of xn is 2n . The first six Hermite’s polynomials are H0 (x) = 1, H2 (x) = 4x2 − 2, H4 (x) = 16x4 − 48x2 + 12,

H1 (x) = 2x, H3 (x) = 8x3 − 12x, H5 (x) = 32x5 − 160x3 + 120x

A general formula for the Hermite polynomials is Hn = (−1)n ex

2

dn −x2 e . dxn

5.4. REGULAR SINGULAR POINTS AND THE METHOD OF FROBENIUS

5.4

93

Regular Singular Points and the Method of Frobenius

Consider the second order linear homogeneous differential equation x2 y 00 + xp(x)y 0 + q(x)y = 0,

(5.4.1)

where p(x) and q(x) are analytic at x = 0. In other words, 0 is a regular singular point of (5.4.1). Let p(x) = p0 + p1 x + p2 x2 + p3 x3 + · · · , and q(x) = q0 + q1 x + q2 x2 + q3 x3 + · · · . Suppose (5.4.1) has a series solution of the form y = xr

∞ X n=0

an xn =

∞ X

an xn+r

(5.4.2)

n=0

An infinite series of the form (5.4.2) is called a Frobenius series, and the method that we are going to describe is called the method of Frobenius. We may assume a0 6= 0 because the series must have a first nonzero term. Termwise differentiation gives y0 =

∞ X

an (n + r)xn+r−1 ,

(5.4.3)

an (n + r)(n + r − 1)xn+r−2 .

(5.4.4)

n=0

and y 00 =

∞ X n=0

Substituting the series of y, y 0 and y 00 into (5.4.1) yields [r(r − 1)a0 xr + (r + 1)ra1 xr+1 + · · · ] + [p0 x + p1 x2 + · · · ] · [ra0 xr−1 + (r + 1)a1 xr + · · · ] +[q0 + q1 x + · · · ] · [a0 xr + a1 xr+1 + · · · ] = 0. (5.4.5) r The lowest power of x in (5.4.5) is x . If (5.4.5) is to be satisfied identically, the coefficient r(r − 1)a0 + p0 ra0 + q0 a0 of xr must vanish. As a0 6= 0, it follows that r satisfies the quadratic equation r(r − 1) + p0 r + q0 = 0.

(5.4.6)

This is the same equation obtained with the Cauchy-Euler equation. Equation (5.4.6) is called the indicial equation of (5.4.1) and its two roots (possibly equal) are the exponents of the differential equation at the regular singular point x = 0. Let r1 and r2 be the roots of the indicial equation. If r1 6= r2 , then there are two possible Frobenius solutions and they are linearly independent. Whereas r1 = r2 , there is only one possible Frobenius series solution. The second one cannot be a Frobenius series and can only be found by other means. Example 5.9 Find the exponents in the possible Frobenius series solutions of the equation 2x2 (1 + x)y 00 + 3x(1 + x)3 y 0 − (1 − x2 )y = 0. Solution. Clearly x = 0 is a regular singular point since p(x) = 23 (1 + x)2 and q(x) = − 12 (1 − x) are polynomials. Rewrite the equation in the standard form: y 00 +

3 2 (1

+ 2x + x2 ) 0 − 21 (1 − x) y + y = 0. x x2

94


We see that p0 =

3 2

and q0 = − 12 . Hence the indicial equation is 1 1 1 1 3 r(r − 1) + r − = r2 + r − = (r + 1)(r − ) = 0, 2 2 2 2 2

with roots r1 =

1 2

and r2 = −1. The two possible Frobenius series solutions are of the forms 1

y1 (x) = x 2

∞ X

an xn and y2 (x) = x−1

n=0

∞ X

an xn .

n=0

Once the exponents r1 and r2 are known, the coefficients in a Frobenius series solution can be found by substituting the series (5.4.2),(5.4.3) and (5.4.4) into the differential equation (5.4.1). If r1 and r2 are complex conjugates, we always get two linearly independent solutions. We shall restrict our attention for real solutions of the indicial equation and seek solutions only for x > 0. The solutions on the interval x < 0 can be studied by changing the variable to t = −x and solving the resulting equation for t > 0. Let’s work out the recursion relations for the coefficients. By (5.4.3), we have !" ∞ # ∞ X X 1 1 0 pn xn an (n + r)xn+r−1 x p(x)y = x n=0 n=0 !" ∞ # ∞ X X r−2 n n =x pn x an (n + r)x n=0 n=0 " # ∞ n X X = xr−2 pn−k ak (r + k) xn n=0 "k=0 # ∞ n−1 X X r−2 =x pn−k ak (r + k) + p0 an (r + n) xn . n=0

k=0

Also we have 1 1 x2 q(x)y = x2

∞ X

!

!

an xr+n n=0 ! n=0∞ ! ∞ X X r−2 n n =x qn x an x n=0 n=0 ! ∞ n X X = xr−2 qn−k ak xn n=0 k=0 ! ∞ n−1 X X r−2 =x qn−k ak + q0 an xn . n=0

qn xn

∞ X

k=0

Substituting these into the differential equation (5.4.1) and canceling the term xr−2 , we have ) ( ∞ n−1 X X an [(r + n)(r + n − 1) + (r + n)p0 + q0 ] + ak [(r + k)pn−k + qn−k ] xn = 0. n=0

k=0

5.4. REGULAR SINGULAR POINTS AND THE METHOD OF FROBENIUS

95

Thus, equating the coefficients to zero, we have for n ≥ 0, an [(r + n)(r + n − 1) + (r + n)p0 + q0 ] +

n−1 X

ak [(r + k)pn−k + qn−k ] = 0.

(5.4.7)

k=0

When n = 0, we get r(r − 1) + rp0 + q0 = 0, which is true because r is a root of the indicial equation. Then an can be determined by (5.4.7) recursively provided (r + n)(r + n − 1) + (r + n)p0 + q0 6= 0. This would be the case if the two roots of the indicial equation do not differ by an integer. Suppose r1 > r2 are the two roots of the indicial equation with r1 = r2 + N for some positive integer N . If we start with the Frobenius series with the smaller exponent r2 , then at the N -th step the process breaks off because the coefficient aN in (5.4.7) is zero. In this case, only the Frobenius series solution with the larger exponent is guaranteed to exist. The other solution may not be a Frobenius series or does not exist. The proof of the following theorem is similar to theorem 4.1, see [4]. Theorem 5.2 Assume that x = 0 is a regular singular point of the differential equation (5.4.1) and that the power series expansions of p(x) and q(x) are valid on an interval |x| < R with R > 0. Let the indicial equation (5.4.6) have real roots r1 and r2 with r1 ≥ r2 . Then (5.4.1) has at least one solution ∞ X y 1 = x r1 an xn , (a0 6= 0) (5.4.8) n=0

on the interval 0 < x < R, where an are determined in terms of a0 by the recursion formula (5.4.7) P∞ with r replaced by r1 , and the series n=0 an xn converges for |x| < R. Furthermore, if r1 − r2 is not zero or a positive integer, then equation (5.4.1) has a second independent solution y 1 = x r2

∞ X

an xn ,

(a0 6= 0)

(5.4.9)

n=0

on the same interval, where an are determined in terms of a0 by the recursion formula (5.4.7) with P∞ r replaced by r2 , and again the series n=0 an xn converges for |x| < R. Remark. (1) If r1 = r2 , then there cannot be a second Frobenius series solution. (2) If r1 − r2 = n is a positive integer and the summation of (5.4.7) is nonzero, then there cannot be a second Frobenius series solution. (3) If r1 − r2 = n is a positive integer and the summation of (5.4.7) is zero, then an is unrestricted and can be assigned any value whatever. In particular, we can put an = 0 and continue to compute the coefficients without difficulties. Hence, in this case, there does exist a second Frobenius series solution. In many cases of (1) and (2), it is possible to determine a second solution by the method of variation of parameters. For instance a second solution for the CauchyEuler equation for the case where its indicial equation has equal roots is given by xr ln x. Exercise 5.4 Find two linearly independent Frobenius series solutions of the differential equation 2x2 y 00 + x(2x + 1)y 0 − y = 0. 1

4 2 x + · · · ), y2 = x− 2 (1 − x + 21 x2 + · · · ). Ans: y1 = x(1 − 52 x + 35 P∞ (−1)n 6·4n (n+1)! n 1 More precisely, y1 = x n=0 x and y2 = x− 2 e−x . (2n+3)!

96


Example 5.10 Find the Frobenius series solutions of xy 00 + 2y 0 + xy = 0. Solution. Rewrite the equation in the standard form x2 y 00 + 2xy 0 + x2 y = 0. We see that p(x) = 2 and q(x) = x2 . Thus p0 = 2 and q0 = 0 and the indicial equation is r(r − 1) + 2r = r(r + 1) = 0 so that the exponents of the equation are r1 = 0 and r2 = −1. In this case, r1 − r2 is an integer and we may not have two Frobenius series solutions. We know there is a Frobenius series solution corresponding to r1 = 0. Let’s consider the possibility of the solution corresponding to the smaller ∞ ∞ X X exponent r2 = −1. Let’s begin with y = x−1 cn xn = cn xn−1 . Substituting this into the n=0

n=0

given equation, we obtain ∞ X

(n − 1)(n − 2)cn xn−2 + 2

n=0

∞ X

(n − 1)cn xn−2 +

n=0

∞ X

cn xn = 0,

n=0

or equivalently ∞ X

n(n − 1)cn x

n−2

n=0

or

∞ X

∞ X

+

cn xn = 0,

n=0

n(n − 1)cn xn−2 +

n=0

∞ X

cn−2 xn−2 = 0.

n=2

The cases n = 0 and n = 1 reduce to 0 · c0 = 0 and 0 · c1 = 0. Thus c0 and c1 are arbitrary and we can expect to get two linearly independent Frobenius series solutions. Equating coefficients, we obtain the recurrence relation cn−2 , for n ≥ 2. cn = − n(n − 1) It follows from this that for n ≥ 1. c2n =

(−1)n c0 (−1)n c1 and c2n+1 = . (2n)! (2n + 1)!

Therefore, we have y = x−1

∞ X

cn x n =

n=0

∞ ∞ c0 X (−1)n 2n c1 X (−1)n 2n+1 x + x . x n=0 (2n)! x n=0 (2n + 1)!

We recognize this general solution as 1 (c0 cos x + c1 sin x). x If we begin with the larger exponent, we will get the solution (sin x)/x. y=

Exercise 5.5 The equation xy 00 + (1 − x)y 0 + αy = 0, where α is a constant is called the Laguerre equation. Find a Frobenius series solution of the Laguerre equation. Show that if α = k is a nonnegative integer, the Laguerre equation has the monic polynomial solution Lk (x) = (k!)2

Ans: y = 1 +

k X

(−1)n xk−n . 2 n![(k − n)!] n=0

∞ X (0 − α)(1 − α) · · · (n − 1 − α) n x . (n!)2 n=1

5.5. BESSEL’S EQUATION

5.5

97

Bessel’s Equation

The second order linear homogeneous differential equation x2 y 00 + xy 0 + (x2 − p2 )y = 0,

(5.5.1)

where p is a constant is called Bessel’s equation. Its general solution is of the form y = c1 Jp (x) + c2 Yp (x).

(5.5.2)

The function Jp (x) is called the Bessel function of order p of the first kind and the Yp (x) is the Bessel function of order p of the second kind. These functions have been tabulated and behave √ somewhat like the trigonometric functions of damped amplitude. If we let y = u/ x, we obtain p2 − 41 d2 u + 1 − u = 0. dx2 x2

(5.5.3)

In the special case in which p = ± 21 , this equation becomes d2 u + u = 0. dx2 Hence u = c1 sin x + c2 cos x and sin x cos x y = c1 √ + c2 √ . x x

(5.5.4)

Also we see that as x → ∞ in (5.5.3), and p is finite, we would expect the solution of (5.5.1) to behave as (5.5.4). For the distribution of zeros of the solutions of Bessel’e equation, see section 3.8. It is easy to see that x = 0 is a regular singular point of Bessel’s equation. Here p(x) = 1 and q(x) = −p2 + x2 . Thus the indicial equation is r(r − 1) + r − p2 = r2 − p2 = 0. Therefore, the ∞ X exponents are ±p. Let r be either −p or p. If we substitute y = cm xm+r into Bessel’s equation, we find in the usual manner that c1 = 0 and that for m ≥ 2,

m=0

[(m + r)2 − p2 ]cm + cm−2 = 0

(5.5.5)

The case r = p ≥ 0. If we use r = p and write am in place of cm , then (5.5.5) yields the recursion formula am−2 am = − (5.5.6) m(2p + m) As a1 = 0, it follows that am = 0 for all odd values of m. The first few coefficients am for m even are a0 a0 =− 2 , a2 = − 2(2p + 2) 2 (p + 1) a2 a0 a4 = − = 4 , 4(2p + 4) 2 · 2(p + 1)(p + 2) a4 a0 a6 = − =− 6 . 6(2p + 6) 2 · 2 · 3(p + 1)(p + 2)(p + 3)

98


In general, one can show that a2m =

22m m!(p

(−1)m a0 . + 1)(p + 2) · · · (p + m)

Thus we have a solution associated with the larger exponent p y1 = a0

∞ X

(−1)m x2m+p . 2m m!(p + 1)(p + 2) · · · (p + m) 2 m=0

If p = 0, this is the only Frobenius series solution. In this case, if we choose a0 = 1, we get a solution of Bessel’s equation of order 0 given by J0 (x) =

∞ X (−1)m x2m x4 x6 x2 + − + ··· . = 1 − 22m (m!)2 4 64 2304 m=0

This special function J0 (x) is called the Bessel function of order zero of the first kind. A second linearly independent solution can be obtained by other means, but it is not a Frobenius series. The case r = −p < 0. Our theorem does not guarantee the existence of a Frobenius solution associated with the smaller exponent. However, as we shall see, it does have a second Frobenius series solution so long as p is not an integer. Let’s write bm in place of cm in (5.5.5). Thus we have b1 = 0 and for m ≥ 2, m(m − 2p)bm + bm−2 = 0

(5.5.7)

Note that there is a potential problem if it happens that 2p is a positive integer, or equivalently if p is a positive integer or an odd integral multiple of 12 . Suppose p = k/2 where k is an odd positive integer. Then for m ≥ 2, (5.5.7) becomes m(m − k)bm = −bm−2

(5.5.8)

Recall b1 = 0 so that b3 = 0, b5 = 0, · · · , bk−2 = 0 by (5.5.8). Now in order to satisfy (5.5.8) for m = k, we can simply choose bk = 0. Subsequently all bm = 0 for all odd values of m. [If we let bk to be arbitrary and non-zero, the subsequent solution so obtained is just bk y1 (x) if we also take b0 = 0. Thus no new solution arises in this situation.] So we only have to work out bm in terms of b0 for even values of m. In view of (5.5.8), it is possible to solve bm in terms of bm−2 since m(m − k) 6= 0 as m is always even while k is odd. The result is the same as before except we should replace p by −p. Thus in this case, we have a second solution y 2 = b0

∞ X m=0

22m m!(−p

(−1)m x2m−p . + 1)(−p + 2) · · · (−p + m)

Since p(x) = 1 and q(x) = x2 −p2 are just polynomials. The series representing y1 and y2 converge for all x > 0. If p > 0, then the first term in y1 is a0 xp , whereas the first term in y2 is b0 x−p . Hence y1 (0) = 0, but y2 (x) −→ ±∞ as x −→ 0, so that y1 and y2 are linearly independent. So we have two linearly independent solutions as long as p is not an integer.


99

If p = n is an nonnegative integer and we take a0 = Jn =

1 2n n! ,

the solution y1 becomes

∞ X

(−1)m x 2m+n . m!(m + n)! 2 m=0

Jn is called the Bessel function of the first kind of integral order n. Remarks. 1. For Bessel’s equation of order p, if p is not an integer, the factorials in Jp can be replaced by the so called Gamma functions and the general solution is Y = c1 Jp + c2 J−p . Here J±p =

∞ X

x 2m±p (−1)m . m!Γ(m ± n + 1) 2 m=0

If p is an integer, (5.5.7) can still be used to get a solution J−p , but it turns out it is just (−1)p Jp , so there is only one Frobenius series solution. A second solution can be obtained by considering the function Jp (x) cos pπ − J−p (x) Yp (x) = . sin pπ If p is not an integer, Yp is a solution of Bessel’s equation of order p as it is a linear combination of Jp and J−p . If p approaches to an integer, the expression of Yp gives an indeterminate form as both the numerator and denominator approach zero. To get a second solution when p = n is an integer, we take limit as p tends to n to get a solution Yn . Jp (x) cos pπ − J−p (x) 1 ∂Jp ∂J−p Yn (x) = lim = lim − (−1)n . p→n p→n π sin pπ ∂p ∂p This limit was first evaluated using L’Hôpital’s rule by Hankel in 1868. Yn is called a Bessel function of the second kind, and it follows that y = c1 Jp + c2 Yp is the general solution of Bessel’s equation in all cases, whether p is an integer or not. For a general form of Yp , see [2]. 2. The case r1 = r2 . Let L(y) = x2 y 00 + xp(x)y 0 + q(x)y. We are solving L(y) = 0 by taking P∞ a series solution of the form y(x) = xr n=0 an xn . If we treat r as a variable, then an ’s are P∞ functions of r. That is y(x, r) = xr n=0 an (r)xn . Substituting this into L(y) and requires it to be a solution, we get (5.4.7), which can be used to determine an (r) recursively provided (r + n)(r + n − 1) + (r + n)p0 + q0 6= 0. When r is near the double root r1 = r2 , this expression is nonzero so that all an can be determined from (5.4.7). This means L(y(x, r)) = a0 (r − r1 )2 xr . So if a0 6= 0, we take r = r1 , we get one Frobenius series solution y1 (x). Now let’s differentiate the above equation with respect to r. We get L(

∂ ∂y )= L(y) = a0 [(r − r1 )2 xr ln x + 2(r − r1 )xr ]. ∂r ∂r

Evaluating at r = r1 , we obtain L(

∂y ∂ ) = L(y) = 0. ∂r r=r1 ∂r r=r1

100


Consequently, we have the second solution

y2 (x) =

∞ ∞ ∞ X X X ∂y (x, r1 ) = xr1 ln x an (r1 )xn +xr1 a0 n (r1 )xn = y1 (x) ln x+xr1 a0 n (r1 )xn . ∂r n=0 n=0 n=1

Note that the sum in the last expression starts at n = 1 because a0 is a constant and a00 = 0. If we apply this method to Bessel’s equation of order p = 0, we get by choosing a0 = 1 the solutions y1 (x) =

∞ X (−1)n x 2n , (n!)2 2 n=0

y2 (x) = y1 (x) ln x − where H(n) =

and

∞ X (−1)n H(n) x 2n , (n!)2 2 n=1

Pn

1 k=1 k .

3. The case r1 − r2 is a nonnegative integer Consider x2 y 00 + xp(x)y 0 + q(x)y = 0, x > 0, P∞ P∞ where p(x) = n=0 pn xn and q(x) = n=0 qn xn . Let r1 and r2 be the roots (exponents) with r1 ≥ r2 of the indicial equation r(r − 1) + p0 r + q0 = 0. P∞ Write r1 = r2 + m, where m is a nonnegative integer. Let y1 = xr1 n=0 an xn be a Frobenius series solution corresponding to the larger exponent r1 . For simplicity, we take a0 = 1. P∞ Let u = xr2 n=0 bn xn and make a change of variable: y(x) = u(x) − bm y1 (x) ln x. We get x2 u00 + xp(x)u0 + q(x)u = bm [2xy10 + (p(x) − 1)y1 ]. P∞ Now let’s substitute u = xr2 n=0 bn xn to see if we can determine the bn ’s. Note that the first term in the power series expansion of bm [2xy10 + (p(x) − 1)y1 ] is mbm , with m ≥ 0. Hence after substituting the power series of u into the above equation, we have

(r2 (r2 − 1) + p0 r2 + q0 )b0 xr2 + A1 xr2 +1 + · · · + Am xr2 +m + · · · = mbm xr1 + · · · . (5.5.9) The first term on the left hand side is 0 as r2 is a root of the indicial equation. This means b0 can be arbitrary. The coefficients A1 , A2 , . . . are given by the main recurrence relation (5.4.7). Thus by equating A1 , . . . , Am−1 to 0, one can determine b1 , . . . bm−1 . The next term on the left hand side of (5.5.9) is the coefficient Am of xr1 . In the expression of Am given by (5.4.7), the coefficient of bm is 0. Previously, this forbids the determination of bm and possibly runs into a contradiction. Now on the right hand side of (5.5.9), if m > 0, then one can determine bm by equating the coefficients of xr1 on both sides. From then on, all the subsequent bn ’s can be determined and we get a solution of the form y(x) = u(x) − bm y1 (x) ln x. Note that if bm = 0 in this determination, then a second Frobenius series solution in fact can be obtained with the smaller exponent r2 .


101

Example 5.11 Consider x2 y 00 + xy = 0. Here p(x) = 0, q(x) = x. The exponents are 0 and 1. Hence m = 1. Corresponding to the exponent 1, the recurrence relation is n(n + 1)an + an−1 = 0 for n ≥ 0. We have the solution y1 =

∞ X (−1)n−1 n n 1 1 x = x − x2 + x3 − · · · . 2 (n!) 2 12 n=1

1 3 Now b1 [2xy10 + (p(x) − 1)y1 ] = b1 (2x(1 − x + 14 x2 − · · · ) − (x − 12 x2 + 12 x − · · · )] = 3 2 5 3 b1 [x − 2 x + 12 x − · · · ]. P∞ Substituting u = x0 n=0 bn xn into x2 u00 + xu = b1 [2xy10 + (p(x) − 1)y1 ], we get

5 3 0·(0−1)b0 +[(1)(0)b1 +b0 ]x+[(2)(1)b2 +b1 ]x2 +[(3)(2)b3 +b2 ]x3 +· · · = b1 [x− x2 + x3 −· · · ]. 2 12 5 Comparing coefficients, we have b0 = b1 , 2b2 +b1 = − 32 b1 and 6b3 +b2 = 12 b1 , · · · . Thus b1 = b0 , 5 5 5 2 5 3 b2 = − 4 b0 , b3 = 18 b0 , . . .. Therefore u = b0 (1 + x − 4 x + 18 x − · · · ). By taking b0 = 1, we 5 3 x − · · · ) − y1 (x) ln x. get the solution y = (1 + x − 54 x2 + 18

If m = 0, then r1 = r2 and the first terms on both sides of (5.5.9) are 0. Thus we can continue to determine the rest of bn ’s. In this case, the ln term is definitely present. Exercise 5.6 Find the general solution of the differential equation x2 (1 + x2 )y 00 − x(1 + 2x + 3x2 )y 0 + (x + 5x2 )y = 0. Ans: y1 = x2 (1 + x + 21 x2 + · · · ), y2 = (1 + x + 2x2 + 38 x3 + · · · ) − 2y1 ln x. The general solution is a linear combination of y1 and y2 .

102

5.6


Appendix 2: Some Properties of the Legendre Polynomials

The Legendre polynomial Pn (x) is a polynomial of degree n satisfying Legendre’s equation (1 − x2 )y 00 − 2xy 0 + n(n + 1)y = 0, where n is a nonnegative integer. It is normalized so that the coefficient of xn is (2n)!/[2n (n!)2 ]. Explicitly it is given by bn/2c

Pn (x) =

X k=0

(−1)k (2n − 2k)! xn−2k . − k)!(n − 2k)!

2n k!(n

There is also a Rodrigues’ formula for the Legendre polynomial given by Pn (x) =

1 dn 2 (x − 1)n . n!2n dxn

Note that in Rodrigues’ formula, the coefficient of xn is (2n)!/[2n (n!)2 ]. We can use Rodrigues’ formula to show that Pn (1) = 1. By this formula, we have 2n Pn (1) is the coefficient of (x − 1)n in the Taylor polynomial expansion of (x2 − 1)n at x = 1. As (x2 − 1)n = (x − 1)n (x − 1 + 2)n = (x − 1)n [(x − 1)n + n(x − 1)n−1 2 + · · · + 2n ], it is clear that the coefficient of (x − 1)n is 2n . Thus Pn (1) = 1. 1

The Legendre polynomial Pn (x) has the generating function φ(Z) = (1 − 2xZ + Z 2 )− 2 = (1 + 1 Z 2 − 2xZ)− 2 . That is Pn (x) is the coefficient of Z n in the expansion of φ. To see this, let’s expand φ as a power series in Z. It can be shown under the condition that |x| ≤ 1, this power series converges to φ(Z) for |Z| < 1. That is φ(Z) =

∞ X

An Z n , for |Z| < 1, |x| ≤ 1.

(A.1)

n=0

Using Binomial expansion, (− 21 )(− 21 − 1) 2 1 1 (1 + Z 2 − 2xZ)− 2 = 1 − (Z 2 − 2xZ) + (Z − 2xZ)2 + · · · , 2 2! we see that An is a polynomial in x of degree n. If we let x = 1, we obtain 1

φ(Z)|x=1 = (1 − 2Z + Z 2 )− 2 = (1 − Z)−1 = 1 + Z + Z 2 + Z 3 + · · · ,

|Z| < 1.

Hence An (1) = 1 for all n. Now, if we can show that An satisfies Legendre’s equation, it will be identical with Pn (x) as the An ’s are the only polynomials of degree n that satisfy the equation and have the value 1 when x = 1. Differentiating φ with respect to Z and x, we obtain (1 − 2Zx + Z 2 )

∂φ = (x − Z)φ, ∂Z

∂φ ∂φ = (x − Z) . ∂Z ∂x Substituting (A.1) into (A.2) and equating the coefficients of Z n−1 , we obtain Z

nAn − (2n − 1)xAn−1 + (n − 1)An−2 = 0

(A.2) (A.3)

(A.4)

5.6. APPENDIX 2: SOME PROPERTIES OF THE LEGENDRE POLYNOMIALS

103

Also substituting (A.1) into (A.3) and equating the coefficients of Z n−1 , we obtain x

dAn−1 dAn−2 − = (n − 1)An−1 dx dx

(A.5)

In (A.5), replace n by n + 1 to get x

dAn dAn−1 − = nAn dx dx

(A.6)

Now differentiate (A.4) with respect to x and eliminate dAn−2 /dx by (A.5), we have dAn−1 dAn −x = nAn−1 dx dx

(A.7)

We now multiply (A.6) by −x and add it to (A.7) and obtain (1 − x2 )

dAn = n(An−1 − xAn ) dx

(A.8)

Differentiating (A.8) with respect to x and simplifying the result by (A.6), we finally obtain (1 − x2 )

d2 An dAn − 2x + n(n + 1)An = 0 dx2 dx

(A.9)

This shows that An is a solution of Legendre’s equation. Using this generating function and Legendre’s equation, it can be shown that Pn (x) satisfy the following orthogonal relations. Z

(

1

Pm (x)Pn (x) dx = −1

0 2 2n+1

if m 6= n . if m = n

(A.10)

y

.. ........ ... ... ... .... P0 (x) 1 ..................................................................................................................................................................................................................................................................................................... .. ...... ......... ...... ... ..... ...... ..... ...... ...... ... ..... ............. . . ... ... . . . . ... .... ... .... .... P1 (x)............ ..................... ... ... ... ... ... ... .. ..... ... ... ... ..... .. ............ . . . . . . . . ... ... ..... ... ......... .... ... ... .... ... ... ...... ... ... .... ..... ..... ... ... ... ... .... ........ ..... . . . . ... . ... P (x) . .. ..... . . . 3 ....... .. ....... ... ........ .............. ... ..... ... ... ...... ... ................ ........ ..... ..... ........ ... .... ....... ..... ........ .... ............ ......................... ... ... ... ... .... . . . . . . . . . . . .. .. ......... .... ..... ..... .... ... ..... ... ... ..... ... ... ..... .... ... ... ... ... ......... ... .. ... .. .. .. ...... ..... .. ..... .. .. ... ...... ... .. .... ... .. ... ............. ... ..... P (x) . . ...... .... . . . 5 . . . . . . . ...... ... .. ... ..... ... ... . .. .. . ... .. ... ..... ... ... ... .... ... ... .. ... ... ... ... .. .. ..... .. .. .. ... ... ..... ... ..... .... ... ....... ... ........... .. .... .... ... .. . ... ... . . . . . . . ......... ..... . ...................................................................................................................................................................................................................................................................................................................................................................... ... ...... ... .. .. ... .. .. ... .. ............ ..... .. ... ...... ....... .. ... ... ... ... ... .. ... ...... .... ... ..... .... .......... .... ..... . ...... .. . .... ....... ... .... .... .... . ... ... ... ... ... ... ....... . . . . . . . ... ....... . .. .. . . ... ... .. . .... .. ...... ... ...... .... .... .. .. ... ......... .... ..... .. .... ... ... ... ... ... P4 (x)..... ......... ......... ........ ... ..... ......... ..... ......... .... .... ........ ....... . . . .... .... ..... . . . . . . ... .. ... . .. .......... . ...... .. ...... . .. .. ... .. .. ..... ..... .. .... ...... .......... ........... ........ ...... .... ..... ........... ... ... ... ..... ...... ........................ ........... ..... ....... . ..... . . . . . . . . . . . .... .... . . ............................... ... . . . ... ... . . . ..... .. .. ..... P2 (x) ..... ... ... ..... ..... ... ... .... .... . . . .... .... ... . ..... .. .. ... ..... . . . ........ ... . .. ...... ......... ... ...... ...... ... ...... ........ ... . ........ −1.......... .. ... .... .

Figure 5.1: The Legendre polynomials

x

104


Chapter 6

Fundamental Theory of ODEs 6.1

Existence-Uniqueness Theorem

We consider the initial value problem dx = f (t, x), dt

x(t0 ) = x0 ,

(6.1.1)

Definition 6.1 Let G be a subset in R2 . f (t, x) : G → R is said to satisfy a Lipschitz condition with respect to x in G if there exists a constant L > 0 such that, for any (t, x1 ), (t, x2 ) ∈ G, |f (t, x1 ) − f (t, x2 )| ≤ L|x1 − x2 |. L is called a Lipschitz constant. Theorem 6.1 (Picard) Let f (t, x) be continuous on the rectangle R : |t − t0 | ≤ a, |x − x0 | ≤ b (a, b > 0), and let |f (t, x)| ≤ M for all (t, x) ∈ R. Furthermore, assume f satisfies a Lipschitz condition with constant L in R. Then there is a unique solution to the initial value problem dx = f (t, x), x(t0 ) = x0 dt on the interval I = [t0 − α, t0 + α], where α = min{a, b/M }. Proof of the existence of solution will be given in section 6.2 and 6.3. The uniqueness of solution will be proved in section 6.5. 2

Example 6.1 Let f (t, x) = x2 e−t sin t be defined on G = {(t, x) ∈ R2 : 0 ≤ x ≤ 2}. Let (t, x1 ), (t, x2 ) ∈ G. 105

106

CHAPTER 6. FUNDAMENTAL THEORY OF ODES |f (t, x1 ) − f (t, x2 )| 2 2 = |x21 e−t sin t − x22 e−t sin t| 2 = |e−t sin t||x1 + x2 ||x1 − x2 | ≤ (1)(4)|x1 − x2 |

Thus we may take L = 4 and f satisfies a Lipschitz condition in G with Lipschitz constant 4. √ Example 6.2 Let f (t, x) = t x be defined on G = {(t, x) ∈ R2 : 0 ≤ t ≤ 1, 0 ≤ x ≤ 1}. √ Consider the two points (1, x), (1, 0) ∈ G. We have |f (1, x) − f (1, 0)| = x = √1x |x − 0|. However, as x → 0+ , √1x → +∞, so that f cannot satisfy a Lipschitz condition with any finite constant L > 0 on G. Proposition 6.1.1 Suppose f (t, x) has a continuous partial derivative fx (t, x) on a rectangle R = {(t, x) ∈ R2 : a1 ≤ t ≤ a2 , b1 ≤ x ≤ b2 } in the tx-plane. Then f satisfies a Lipschitz condition on R. Proof. Since |fx (t, x)| is continuous on R, it attains its maximum value in R by the extreme value theorem. Let K be the maximum value of |fx (t, x)| on R. By mean value theorem, we have |f (t, x1 ) − f (t, x2 )| = |fx (t, c)||x1 − x2 |, for some c between x1 and x2 . Therefore, |f (t, x1 ) − f (t, x2 )| ≤ K|x1 − x2 | for all (t, x1 ), (t, x2 ) ∈ R. Thus, f satisfies a Lipschitz condition in R with Lipschitz constant K. Example 6.3 Let f (t, x) = x2 be defined on G = {(t, x) ∈ R2 : 0 ≤ t ≤ 1}. First |f (t, x1 ) − f (t, x2 )| = |x21 − x22 | = |x1 + x2 ||x1 − x2 |. Since x1 and x2 can be arbitrarily large, f cannot satisfy a Lipschitz condition on G. If we replace G by any closed and bounded region, then f will satisfy a Lipschitz condition.

6.2

The Method of Successive Approximations

We will give the proof of Theorem 6.1 in several steps. Let’s fix f (t, x) to be a continuous function defined on the rectangle R : |t − t0 | ≤ a, |x − x0 | ≤ b (a, b > 0).

6.2. THE METHOD OF SUCCESSIVE APPROXIMATIONS

107

The objective is to show that on some interval I containing t0 , there is a solution φ to (6.1.1). The first step will be to show that the initial value problem (6.1.1) is equivalent to an integral equation, namely Z t

x(t) = x0 +

f (s, x(s)) ds.

(6.2.1)

t0

By a solution of this equation on I is meant a continuous function φ on I such that (t, φ(t)) is in R for all t ∈ I, and Z t

φ(t) = x0 +

f (s, φ(s)) ds. t0

Theorem 6.2 A function φ is a solution of the initial value problem (6.1.1) on an interval I if and only if it is a solution of the integral equation (6.1.2) on I. Proof. Suppose φ is a solution of the initial value problem on I. Then φ0 (t) = f (t, φ(t))

(6.2.2)

on I. Since φ is continuous on I, and f is continuous on R, the function f (t, φ(t)) is continuous on I. Integrating (6.2.2) from t0 to t we obtain Z t φ(t) − φ(t0 ) = f (s, φ(s)) ds. t0

Since φ(t0 ) = x0 , we see that φ is a solution of (6.2.1). Conversely, suppose φ satisfies (6.2.1). Differentiating we find, using the fundamental theorem of Calculus, that φ0 (t) = f (t, φ(t)) for all t ∈ I. Moreover, from (6.2.1), it is clear that φ(t0 ) = x0 and thus φ is a solution of (6.1.1). As a first approximation to the solution of (6.2.1), we consider φ0 defined by φ0 (t) = x0 . This function satisfies the initial condition φ0 (t0 ) = x0 , but does not in general satisfy (6.2.1). However, if we compute Z Z t

φ1 (t) = x0 +

t

f (s, φ0 (s)) ds = x0 + t0

f (s, x0 ) ds, t0

we might expect φ1 is a closer approximation to a solution than φ0 . In fact, if we continue the process and define successively Z t φ0 (t) = x0 , φk+1 (t) = x0 + f (s, φk (s)) ds, k = 0, 1, 2, . . . (6.2.3) t0

we might expect, on taking the limit as k → ∞, that we would obtain φk (t) → φ(t), where φ would satisfy Z t

f (s, φ(s)) ds.

φ(t) = x0 + t0

Thus φ would be our desired solution. We call the functions φ0 , φ1 , φ2 , · · · defined by (6.2.3) successive approximations to a solution of the integral equation (6.2.1), or the initial value problem (6.1.1). Example 6.4 Consider the initial value problem x0 = tx, x(0) = 1.

108

CHAPTER 6. FUNDAMENTAL THEORY OF ODES

The integral equation corresponding to this problem is Z t x(t) = 1 + s · x(s) ds, 0

and the successive approximations are given by t

Z φ0 (t) = 1,

φk+1 (t) = 1 +

sφk (s) ds, k = 0, 1, 2, . . . . 0

Rt Thus φ1 (t) = 1 + 0 s ds = 1 + established by induction that

t2 2,

φk (t) = 1 +

φ2 (t) = 1 + t2 2

1 + 2!

Rt

t2 2

0

s(1 +

2

s2 2 ) ds

1 + ··· + k!

= 1+

t2 2

t2 2

+

t4 2·4 ,

and it may be

k . 2

We recognize φk (x) as a partial sum for the series expansion of the function φ(t) = et /2 . We know that this series converges for all t and this means that φk (t) → φ(t) as k → ∞, for all x ∈ R. Indeed φ is a solution of this initial value problem. Theorem 6.3 Suppose |f (t, x)| ≤ M for all (t, x) ∈ R. Then the successive approximations φk , defined by (6.2.3), exist as continuous functions on I : |t − t0 | ≤ α = min{a, b/M }, and (t, φk (t)) is in R for t ∈ I. Indeed, the φk ’s satisfy |φk (t) − x0 | ≤ M |t − t0 |

(6.2.4)

for all t ∈ I. Note: Since for t ∈ I, |t − t0 | ≤ b/M , the inequality (6.2.4) implies that |φk (t) − x0 | ≤ b for all t ∈ I, which shows that the points (t, φk (t)) are in R for t ∈ I. The geometric interpretation of the inequality (6.2.4) is that the graph of each φk lies in the region T in R bounded by the two lines x − x0 = M (t − t0 ), x − x0 = −M (t − t0 ), and the lines t − t0 = α, t − t0 = −α. x0 + b

............................................................................................................................................................................................................................................................................................................................................................................... .. .. .... .. ... ....... ... ... .. . .. ........... ... ... .... ......... . . . . ... ........... ... ... ... . .... . ... . . . . ... . .......... ... ... . ... ..... . . . . ... . . . . . ... . . . . . .......... ... ... ... .... . . . ... . . . . . ... . . . . ......... ... ... . . ... . . . . ... . . . . ........... ... ... .... x − x0 = M (t − t0 ) ............ . . . . ......... . . ... . . . . . . . . . ... . . . . . . . . . . . .......... ... ... . . ... ... .... . . . . . ..... ... . . . . . ... . . . . . . . ......... ... ... . . ... .... . . . . . ....... . ... . . . . ...... ... . . . . . . . ........... ... ... . .... ....... ........... . . . . . . . . . . . ...... . . . ... ... . . . . . . . . . . . . . . . . . .......... ... ... ... ........ . . . . . .k .... . . .. ... . . . . . . . . . . .......... ... ... . . ... . ..... . . . . . . ........ . . . ... . . . . ... . . . . . . . . . . .......... ... ... . ... ..... . . . . . . . . . . . . ........... . . . . . . ... . . . . . ... . . . . . . . . . . . . . . . . . . . . . . . .......... ... ... . . .... . . . . . . . ....... . . . . ... . . .. . . ... . . . . . . . . . . . . . .......... ... ... . . ..... . . . . . . .......... . . . . . ... . . .. . . ... . . . . . . . . . . . . . .......... ... ... . . ..... . . . . . . . . . . . . . ............ . . . . . . . . . . . . ... . . . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ........... ... ... . .... . .. ... ... . . . . . . . . . . . . . . . . .......... ... ............. . . . . . . . . . . ........................ . . . . . . . . . . . . . . . ... ... ... ... . . . . . . . . . . . . . . .. . ...................................... . .......................................... . . . . . . . . .... .... .. ... ................................................................................................................................................................................................................................................................................................................................................................................................................................................................... ... . . . . . . . . . .................. . . . . . ....... .. ....... . . . . . . . . . . . . . . . . . .. ... ... ...... . ... . . . . . . . ............ . . . . . . ......... ... ... . ....... . . . . . . . . . . . . . . . .. ...... . . . . . . . . . . . . . . . .. 0 0 .... . . . . . . . . . . . . ................ . . . . . . . . . . . . ............ ... ... ....... . . . . . . . . . . . . . .. ... . . . . . . ........ . . . . . . .......... ... ... ....... . . . . . . . . . . . . . .. 0 . 0 ...... . . . . . . . . . . . . . .. ... . . . . ........... . . . . . . ........ ... ... ...... . . . . . . . . . . . . ... ... ... . . . . . . . ........ . . . . . . . . . . . . ........... ... ... . ...... . . . . . . . . . . . .. ....... . . . . . . . . . . . ... ... . . . ...... . . . . . . .......... ... ... ....... . . . . . . . . . ... . ... . . ....... . . . . . . ........ ... ... . .. ........ . . . . . . . . . ... . . .. . . ... . . . ......... . . . . . . . . . . .......... ... ... . . . ...... . . . . . . . . .. ....... . .. . ... . ...... . . . . . .......... ... ... . . . . ...... . . . . . . . . ...... . . . . . . .. .. ... ... ..... . . . . . ......... ... ... . . ...... . . . . . . ... .. .. . ... ...... . . . . . . . . .......... ... ... . . ....... . . . . .. .. ...... . . . . . ... .... . . . .......... . ... ... .. x − x0 = −M (t − t0 ) ......... . . . ... ....... . . . ........ . ... ... ...... . . . ... .... ... . . . ............ ... ... ....... . . ...... . . ... ... . ......... ... ... ...... ... ... ............ ... ... ........ ... ... ..... ... ... ... ... .. ... .... .....................................................................................................................................................................................................................................................................................................................................................................................

φ

T

t0 − a

•

t −α

(t , x )

x0 − b Figure 6.1: The graph of each φk lies in the region T

t +α

t0 + a

6.3. CONVERGENCE OF THE SUCCESSIVE APPROXIMATIONS

109

Proof. We prove it by induction on k. Clearly φ0 exists on I as a continuous function, and satisfies (6.2.4) with k = 0. Now suppose the theorem has been proved for the functions φ0 , φ1 , . . . , φk , with k ≥ 0. We shall prove that it is valid for φk+1 . By induction hypothesis, the point (t, φk (t)) is in R for t ∈ I. Thus the function f (t, φk (t)) exists for t ∈ I and is continuous on I. Therefore, φk+1 , which is given by Z t φk+1 (t) = x0 + f (s, φk (s)) ds, t0

exists as a continuous function on I. Moreover, Z t |f (s, φk (s))| ds ≤ M |t − t0 |, |φk+1 (t) − x0 | ≤ t0

which shows that φk+1 satisfies (6.2.4).

6.3

Convergence of the Successive Approximations

We now prove the main existence theorem Theorem 6.4 Let f (t, x) be continuous on the rectangle R : |t − t0 | ≤ a, |x − x0 | ≤ b (a, b > 0), and let |f (t, x)| ≤ M for all (t, x) ∈ R. Furthermore, assume f satisfies a Lipschitz condition with constant L in R. Then the successive approximations Z t φ0 (t) = x0 , φk+1 (t) = x0 + f (s, φk (s)) ds, k = 0, 1, 2, . . . t0

converges uniformly on the interval I = [t0 − α, t0 + α] with α = min{a, b/M }, to a solution of dx the initial value problem = f (t, x), x(t0 ) = x0 on I. dt Proof. (a) Convergence of {φk (t)}. The key to the proof is the observation that φk may be written as φk = φ0 + (φ1 − φ0 ) + (φ2 − φ1 ) + · · · + (φk − φk−1 ), and hence φk (t) is a partial sum for the series φ0 (t) +

∞ X [φp (t) − φp−1 (t)].

(6.3.1)

p=1

Therefore to show that the sequence {φk (t)} converges uniformly is equivalent to show that the series (6.3.1) converges uniformly. By Theorem 6.3, the functions φk all exist as continuous functions on I, and (t, φp (t)) is in R for t ∈ I. Moreover, |φ1 (t) − φ0 (t)| ≤ M |t − t0 |, (6.3.2)

110


for t ∈ I. Next consider the difference of φ2 and φ1 . We have Z t φ2 (t) − φ1 (t) = [f (s, φ1 (s)) − f (s, φ0 (s))] ds. t0

Therefore

Z t |f (s, φ1 (s)) − f (s, φ0 (s))| ds , |φ2 (t) − φ1 (t)| ≤ t0

and since f satisfies a Lipschitz condition |f (t, x1 ) − f (t, x2 )| ≤ L|x1 − x2 |, we have

Z t |φ2 (t) − φ1 (t)| ≤ L |φ1 (s) − φ0 (s)| ds . t0

Using (6.3.2), we obtain Z t |s − t0 | ds . |φ2 (t) − φ1 (t)| ≤ M L t0

Thus if t ≥ t0 , Z

t

|φ2 (t) − φ1 (t)| ≤ M L

(s − t0 ) ds = t0

M L(t − t0 )2 . 2

The same result is valid in case t ≤ t0 . We shall prove by induction that |φp (t) − φp−1 (t)| ≤

M Lp−1 |t − t0 |p p!

(6.3.3)

for all t ∈ I. We have proved this for p = 1 and p = 2. Let’s assume t ≥ t0 . The proof is similar for t ≤ t0 . Assume (6.3.3) is true for p = m. Using the definition of φm+1 and φm , we have Z

t

φm+1 (t) − φm (t) =

[f (s, φm (s)) − f (s, φm−1 (s))] ds, t0

and thus

Z t |φm+1 (t) − φm (t)| ≤ |f (s, φm (s)) − f (s, φm−1 (s))| ds . t0

Using a Lipschitz condition, we get Z t |φm+1 (t) − φm (t)| ≤ L |φm (s) − φm−1 (s)| ds . t0

By induction hypothesis, we obtain Z M Lm |t − t0 |m+1 M Lm t m = |s − t | ds . |φm+1 (t) − φm (t)| ≤ 0 m! t0 (m + 1)! Thus, (6.3.3) is true for all positive integer p. Since |t − t0 | ≤ α for all t ∈ I, we can further deduce from (6.3.3) that

6.3. CONVERGENCE OF THE SUCCESSIVE APPROXIMATIONS

|φp (t) − φp−1 (t)| ≤ Since the series

∞ X M (Lα)p converges to L p! p=1

series φ0 (t) +

M Lp−1 αp M (Lα)p = . p! L p!

M Lα L (e

∞ X

111

(6.3.4)

− 1), we have by Weierstrass M-test that the

[φp (t) − φp−1 (t)]

p=1

converges absolutely and uniformly on I. Thus the sequence of partial sum which is φk (t) converges uniformly on I to a limit φ(t). Next we shall show that this limit φ is a solution of the integral equation (6.2.1). (b) Properties of the limit φ. Since each φk is continuous on I and the sequence converges uniformly to φ, the function φ is also continuous on I. Now if t1 and t2 are in I, we have Z t1 f (s, φk (s)) ds ≤ M |t1 − t2 |, |φk+1 (t1 ) − φk+1 (t2 )| = t2

which implies, by letting k → ∞, |φ(t1 ) − φ(t2 )| ≤ M |t1 − t2 |.

(6.3.5)

It also follows from (6.3.5) that the function φ is continuous on I. In fact φ is uniformly continuous on I. Letting t1 = t, t2 = t0 in (6.3.5), we see that |φ(t) − φ(t0 )| ≤ M |t − t0 | which implies that the points (t, φ(t)) are in R for all t ∈ I. (c) Estimate for |φ(t) − φk (t)|. We have φ(t) = φ0 (t) +

∞ X [φp (t) − φp−1 (t)], p=1

and φk (t) = φ0 (t) +

k X [φp (t) − φp−1 (t)]. p=1

Using (6.3.4), we have X ∞ |φ(t) − φk (t)| = [φp (t) − φp−1 (t)] p=k+1 ∞ X ≤ |φp (t) − φp−1 (t)| ≤ ≤

p=k+1 ∞ X

M (Lα)p L p! p=k+1 ∞ M (Lα)k+1 X (Lα)p L (k + 1)! k+1

≤

p=0

M (Lα) eLα . L (k + 1)!

p!

112

CHAPTER 6. FUNDAMENTAL THEORY OF ODES k+1

Lα Letting k = (Lα) . In (k+1)! , we see that k → 0 as k → ∞ as k is a general term for the series e terms of k , we may rewrite the above inequality as

|φ(t) − φk (t)| ≤

M Lα e k , L

and k → 0 as k → ∞

(6.3.6)

(d) The limit φ is a solution. To complete the proof we must show that Z

t

φ(t) = x0 +

f (s, φ(s)) ds, t0

for all t ∈ I. Note that since φ is continuous, the integrand f (s, φ(s)) of the right hand side is continuous on I. Since Z t φk+1 (t) = x0 + f (s, φk (s)) ds, t0

we get the result by taking limit on both sides as k → ∞ provided we can show Z

t

Z

t

f (s, φk (s)) ds → t0

f (s, φ(s)) ds, as k → ∞. t0

Z t Z t Z t Now f (s, φ(s)) ds − |f (s, φ(s)) − f (s, φk (s))| ds f (s, φk (s)) ds ≤ t0 t0 0 tZ t |φ(s) − φk (s)| ds ≤ L t0

≤ M eLα k |t − t0 | ≤ M αeLα k → 0 as k → ∞.

by (6.3.6)

This completes the proof of the Theorem 6.4.

Example 6.5 Consider the initial value problem x0 = (sin t)x2 , x(0) = 21 . Let f (t, x) = (sin t)x2 be defined on 1 1 R = {(t, x) : |t| ≤ 1, |x − | ≤ }. 2 2 |f (t, x)| = |(sin t)x2 | ≤ 1. Thus we may take M = 1. Therefore by Theorem 6.4 a solution exists on [−α, α] where α = min{1, 12 } = 12 . In fact x(t) = (1 + cos t)−1 is a solution defined on the maximal domain (−π, π). Exercise 6.1 Consider the initial value problem a unique solution on [− 12 , 12 ].

6.4

dx dt

= tx + x10 , x(0) =

1 10 .

Show that there exists

Non-local Existence of Solutions

Theorem 6.5 Let f (t, x) be a continuous function on the strip S = {(t, x) ∈ R2 : |t − t0 | ≤ a}, where a is a given positive number, and f satisfies a Lipschitz condition with respect to S. Then the initial value problem x0 (t) = f (t, x), x(t0 ) = x0 , where (t0 , x0 ) ∈ S has a unique solution on the entire interval [−a + t0 , a + t0 ].

6.4. NON-LOCAL EXISTENCE OF SOLUTIONS

113

Remark. If f is bounded on S, the result can be deduced from Picard’s Theorem by taking b > M a. If f is not necessarily bounded, the proof is slightly different. Proof of Theorem 6.5. First note that the given region S is not bounded above or below. Hence f (t, x) needs not be bounded in S. However, as in Theorem 5.4, we shall consider the series φ0 (t) +

∞ X

(φp (t) − φp−1 (t))

p=1

whose n-th partial sum is φn (t) and φn (t) → φ(t) giving the solution of the initial value problem. Since f (t, x) is not bounded in S, we adopt a different method of estimating different terms of the series. Let M0 = |x0 | and M1 = max |φ1 (t)|. The fact that M1 exists can be seen as follows. Since f (t, x) is continuous in S, for a fixed x0 , f (t, x0 ) is a continuous function on |t − t0 | ≤ a. Rt Thus φ1 (t) = x0 + t0 f (s, x0 ) ds is a continuous function in this interval so that |φ1 (t)| attains its maximum in this interval. We take it to be M1 and let M = M0 + M1 . Thus, |φ0 (t)| = |x0 | ≤ M and |φ1 (t) − φ0 (t)| ≤ M . If t0 ≤ t ≤ t0 + a, then we have Z t Z t |φ2 (t) − φ1 (t)| = [f (s, φ1 (s)) − f (s, φ0 (s))] ds ≤ |f (s, φ1 (s)) − f (s, φ0 (s))| ds t0 Z t0 t ≤L |φ1 (s) − φ0 (s)| ds ≤ LM (t − t0 ), where L is the Lipschitz constant. t0

Now Z t Z t |f (s, φ2 (s)) − f (s, φ1 (s))| ds |φ3 (t) − φ2 (t)| = [f (s, φ2 (s)) − f (s, φ1 (s))] ds ≤ t0 t0 Z Z t t L2 M (t − t0 )2 . ≤L |φ2 (s) − φ1 (s)| ds ≤ L2 M |(s − t0 )| ds = 2 t0 t0

Hence, in general, we can prove by induction that |φn (t) − φn−1 (t)| ≤

Ln−1 M (t − t0 )n−1 . (n − 1)!

Similar argument is true for the interval t0 − a ≤ t ≤ t0 . Hence for every t with |t − t0 | ≤ a, |φn (t) − φn−1 (t)| ≤ Thus |φ0 (t)| +

∞ X

Ln−1 M |t − t0 |n−1 Ln−1 M n−1 ≤ a . (n − 1)! (n − 1)!

|φn (t) − φn−1 (t)| ≤ M + M

n=1

∞ X (La)n−1 . (n − 1)! n=1

Hence each term on the left hand side of the above equation is less than the corresponding term of the convergent series of positive constants. Hence, by Weierstrass M -test, the series on the left converges uniformly on the whole interval |t − t0 | ≤ a and let’s denote its limit by φ(t). Next we show that φ(t) is a solution of the initial value problem. We need to show that φ(t) satisfies the integral equation Z t

φ(t) − x0 −

f (s, φ(s)) ds = 0. t0

(6.4.1)

114


We know that Z

t

φn (t) − x0 −

f (s, φn−1 (s)) ds = 0.

(6.4.2)

t0

Substituting the value of x0 in (6.4.2) into the left hand side of (6.4.1), we get Z t Z t φ(t) − x0 − f (s, φ(s)) ds = φ(t) − φn (t) − [f (s, φ(s)) − f (s, φn−1 (s))] ds. t0

t0

Thus we obtain R Rt t φ(t) − x0 − t0 f (s, φ(s)) ds ≤|φ(t) − φn (t)| + t0 |f (s, φ(s)) − f (s, φn−1 (s))| ds R t (6.4.3) ≤|φ(t) − φn (t)| + L t0 |φ(s) − φn−1 (s)| ds Since φn (t) → φ(t) uniformly for t ∈ [t0 − a, t0 + a], the right hand side of (6.4.3) tends to zero as n → ∞. Hence Z

t

φ(t) − x0 −

f (s, φ(s)) ds = 0. t0

The uniqueness of solution will be proved in section 6.5.

Corollary 6.6 Let f (t, x) be a continuous function defined on R2 . Let (t0 , x0 ) ∈ R2 . Suppose that for any a > 0, f satisfies a Lipschitz condition with respect to S = {(t, x) ∈ R2 : |t| ≤ a}. Then the initial value problem x0 (t) = f (t, x), x(t0 ) = x0 has a unique solution on the entire R. Proof. If t is any real number, there is an a > 0 such that t is contained in [t0 − a, t0 + a]. For this a, the function f satisfies the condition of Theorem 5.5 on the strip {(t, x) ∈ R2 : |t − t0 | ≤ a}, since this strip is contained in the strip {(t, x) ∈ R2 : |t| ≤ a + |t0 |}. Thus there is a unique solution φ(t) to the initial value problem for all t ∈ R.

Example 6.6 Consider the initial value problem x0 = sin(tx), x(0) = 1. Let f (t, x) = sin(tx). Let a > 0. Using the mean value theorem, we have for any t ∈ [−a, a], |f (t, x1 ) − f (t, x2 )| = | sin(tx1 ) − sin(tx2 )| = |t cos(tζ)(x1 − x2 )| ≤ |t||x1 − x2 | ≤ a|x1 − x2 |. Thus f satisfies a Lipschitz condition on the strip S = {(t, x) ∈ R2 : |t| ≤ a} for any a > 0, and by corollary 6.6 there exists a unique solution on the entire R. Exercise 6.2 Let g(t) and h(t) be continuous on the interval I = [t0 − a, t0 + a], where a > 0. Prove that the initial value problem x0 + g(t)x = h(t), x(t0 ) = x0 has a solution defined on I. Exercise 6.3 Show that the initial value problem x0 = on R.

x3 et + t2 cos x, x(0) = 1 has a solution 1 + x2

6.5. GRONWALL’S INEQUALITY AND UNIQUENESS OF SOLUTION

6.5

115

Gronwall’s Inequality and Uniqueness of Solution

Theorem 6.7 Let f , g, and h be continuous nonnegative functions defined for t ≥ t0 . If t

Z f (t) ≤ h(t) +

g(s)f (s) ds, t ≥ t0 , t0

then Z

t

f (t) ≤ h(t) +

g(s)h(s)e

Rt s

g(u) du

ds, t ≥ t0 .

t0

Proof. First we are given Z

t

f (t) ≤ h(t) +

g(s)f (s) ds

(6.5.1)

t0

Let z(t) =

Rt t0

g(s)f (s) ds. Then for t ≥ t0 , z 0 (t) = g(t)f (t)

(6.5.2)

Since g(t) ≥ 0, multiplying both sides of (6.5.1) by g(t) and using (6.5.2), we get z 0 (t) ≤ g(t)[h(t) + z(t)] which gives z 0 (t) − g(t)z(t) ≤ g(t)h(t). This is a first order differential inequality which can be solved by finding an integrating factor R − tt g(u) du 0 e . Hence the solution is Z t R R − tt g(u) du − s g(u) du 0 z(t)e ≤ g(s)h(s)e t0 ds t0

Or equivalently, Z

t

z(t) ≤

g(s)h(s)e

−

Rs t0

g(u) du

Rt

e

g(u) du

t0

Z

t

ds =

g(s)h(s)e

t0

Rt s

g(u) du

ds

(6.5.3)

t0

Substituting for z(t) in (6.5.3), we get Z

t

Z

t

g(s)f (s) ds ≤

g(s)h(s)e

t0

Rt s

g(u) du

ds

(6.5.4)

t0

From (6.5.1), we can replace the left side of (6.5.4) by the lesser inequality to obtain Z

t

f (t) − h(t) ≤

g(s)h(s)e

Rt s

g(u) du

ds.

t0

116


Theorem 6.8 (Gronwall’s Inequality) Let f and g be continuous nonnegative functions for t ≥ t0 . Let k be any nonnegative constant. If Z t g(s)f (s) ds, for t ≥ t0 , f (t) ≤ k + t0

then f (t) ≤ ke

Rt t0

g(s) ds

for t ≥ t0 .

,

Proof. The result follows by letting h(t) = k for all t ≥ t0 in 6.7.

Corollary 6.9 Let f be a continuous nonnegative function for t ≥ t0 and k a nonnegative constant. If Z t f (t) ≤ k f (s) ds t0

for all t ≥ t0 , then f (t) ≡ 0 for all t ≥ t0 .

Proof. For any > 0, we can rewrite the given hypothesis as Z

t

f (t) ≤ + k

f (s) ds, t0

for all t ≥ t0 . Hence applying Gronwall’s inequality, we have Rt

f (t) ≤ e

t0

k ds

,

for all t ≥ t0 , which gives f (t) ≤ ek(t−t0 ) , for all t ≥ t0 . Since is arbitrary, we get f (t) ≡ 0 by taking limit as → 0+ . Remark. Similar results hold for t ≤ t0 when all the integrals are integrated from t to t0 . For example, in Corollary 6.9, if Z t0 f (t) ≤ k f (s) ds t

for all t ≤ t0 , then f (t) ≡ 0 for all t ≤ t0 .

Corollary 6.10 Let f (t, x) be a continuous function which satisfies a Lipschitz condition on R with a Lipschitz constant L, where R is either a rectangle or a strip. If φ and ϕ are two solutions of x0 = f (t, x), x(t0 ) = x0 , on an interval I containing t0 , then φ(t) = ϕ(t) for all t ∈ I. Proof. Let I = [t0 − α, t0 + α]. For t ∈ [t0 , t0 + α], we have Z

t

φ(t) = x0 +

f (s, φ(s)) ds, t0

6.5. GRONWALL’S INEQUALITY AND UNIQUENESS OF SOLUTION and Z

117

t

ϕ(t) = x0 +

f (s, ϕ(s)) ds. t0

Thus Z

t

|φ(t) − ϕ(t)| ≤

Z

t

|f (s, φ(s)) − f (s, ϕ(s))| ds ≤ L t0

|φ(s) − ϕ(s)| ds. t0

By Corollary 6.9, |φ(t) − ϕ(t)| ≡ 0 for t ∈ [t0 , t0 + α]. Thus φ(t) = ϕ(t) for t ∈ [t0 , t0 + α]. Similarly, φ(t) = ϕ(t) for t ∈ [t0 − α, t0 ]. Remark. If we only assume that f (t, x) is a continuous function, we can still show that (6.1.1) has at least one solution, but the solution may not be unique. Theorem 6.11 (Peano) Assume G is an open subset of R2 containing (t0 , x0 ) and f (t, x) is continuous in G. Then there exists a > 0 such that (6.1.1) has at least one solution on the interval [t0 − a, t0 + a]. The proof of Peano’s theorem uses the Arzela-Ascoli theorem. See [1] or [2]. Example 6.7 Consider the initial value problem x0 = x2/3 , x(0) = 0. We find that x(t) = 0 and 1 3 t are both solutions. x(t) = 27 Example 6.8 Suppose φ(t) is a solution to the initial value problem x0 =

x3 − x , 1 + t 2 x2

x(0) =

1 . 2

Show that 0 < φ(t) < 1 for all t ∈ J, where φ(t) is defined on the open interval J containing 0. Solution. Let φ(t) be a solution defined on the open interval J to the initial value problem, where 0 ∈ J. Suppose there exists s ∈ J such that φ(s) ≥ 1. Without loss of generality, we may assume s > 0. Since φ(t) is continuous and φ(0) = 1/2, we have by the intermediate value theorem, that φ(s0 ) = 1 for some s0 ∈ (0, s). We may take s0 to be the least value in (0, s) such that φ(s0 ) = 1. In other words, φ(t) < 1 for all t ∈ (0, s0 ) and φ(s0 ) = 1. Now consider the initial value problem x0 =

x3 − x , 1 + t2 x2

x(s0 ) = 1.

3

x −x The function f (t, x) = 1+t 2 x2 satisfies the conditions of the existence and uniqueness theorem. Thus there is a unique solution defined on an interval I = [s0 − α, s0 + α] for some α > 0. The above function φ(t) defined on J is a solution to this initial value problem, and it has the property that φ(t) < 1 for all t < s0 . However, ϕ(t) ≡ 1 is clearly a solution to this initial value problem on I. But ϕ and φ are different solutions to the initial value problem contradicting the uniqueness of the solution. Consequently, φ(t) < 1 for all t ∈ J. Similarly, φ(t) > 0 for all t ∈ J.

Corollary 6.12 Let f (t, x) be a continuous function which is defined either on a strip R = {(t, x) ∈ R2 | |t − t0 | ≤ a},

118


or a rectangle R = {(t, x) ∈ R2 | |t − t0 | ≤ a, |x − x0 | ≤ b}. Assume f satisfies a Lipschitz condition on R with a Lipschitz constant L. Let φ and ϕ be solutions defined on I = [−a + t0 , t0 + a] of x0 = f (t, x) satisfying the initial condition x(t0 ) = x0 and x(t0 ) = x1 respectively on I, then |φ(t) − ϕ(t)| ≤ |x0 − x1 |eL|t−t0 | for all t ∈ I. Remark. In particular |φ(t) − ϕ(t)| ≤ |x0 − x1 |eLa , for all t ∈ I. Thus if the initial values x0 and x1 are close, the resulting solutions φ and ϕ are also close. This gives the continuous dependence on initial value. The proof of this corollary is by Gronwall’s inequality and is left as an exercise. Corollary 6.13 Let f (t, x) be a continuous function which is defined either on a strip R = {(t, x) ∈ R2 | |t − t0 | ≤ a}, or a rectangle R = {(t, x) ∈ R2 | |t − t0 | ≤ a, |x − x0 | ≤ b}. Assume f satisfies a Lipschitz condition on R with a Lipschitz constant L. Let g(t, x) be continuous on R and |g(t, x) − f (t, x)| ≤ , for all (t, x) ∈ R. (6.5.5) Let φ and ϕ be solutions of the initial value problems x0 = f (t, x), x(t0 ) = x0 , and x0 = g(t, x), x(t0 ) = x0 respectively. Assume both φ and ϕ are defined on I = [−a + t0 , t0 + a]. Then for all t ∈ I, |φ(t) − ϕ(t)| ≤ aeLa . In particular, as g(t, x) approaches f (t, x) uniformly on R, that is, as → 0+ in (6.5.5), the solution φ(t) approaches ϕ(t) uniformly on I.

Proof. By the integral representations for φ(t) and ϕ(t), we have Z t Z t φ(t) = x0 + f (s, φ(s)) ds, ϕ(t) = x0 + g(s, ϕ(s)) ds. t0

t0

Subtracting these equations gives Z

t

φ(t) − ϕ(t) =

[f (s, φ(s)) − g(s, ϕ(s))] ds. t0

Inserting the term 0 = −f (s, ϕ(s)) + f (s, ϕ(s)) and using triangle inequality, we have for t0 ≤ t ≤ t0 + a,

6.6. EXISTENCE AND UNIQUENESS OF SOLUTIONS TO SYSTEMS Z

t

Z

|f (s, φ(s)) − f (s, ϕ(s))| ds + tZ 0 Z t t ≤L |φ(s) − ϕ(s)| ds + ds t0 Zt0t |φ(s) − ϕ(s)| ds + a. ≤L

119

t

|φ(t) − ϕ(t)| ≤

|f (s, ϕ(s)) − g(s, ϕ(s))| ds t0

t0

By Gronwall’s inequality, we obtain |φ(t) − ϕ(t)| ≤ ae

Rt t0

L ds

≤ aeLa , for t0 ≤ t ≤ t0 + a.

The last inequality is also valid for t0 − a ≤ t ≤ t0 . Thus |φ(t) − ϕ(t)| ≤ aeLa for t ∈ I. Exercise 6.4 Prove corollary 6.12.

6.6

Existence and Uniqueness of Solutions to Systems

Consider a system of differential equations   x01 = f1 (t, x1 , · · · , xn ),      x0 = f2 (t, x1 , · · · , xn ), 2  ···············     0  xn = fn (t, x1 , · · · , xn ), where x0j =

dxj dt .

Let us introduce notations 

     x1 x01 f1 (t, x)       x = · · · , x0 = · · · , f (t, x) =  · · ·  . x0n xn fn (t, x) Then the system can be written in a vector form: x0 = f (t, x).

(6.6.1)

Differential equations of higher order can be reduced to equivalent systems. Let us consider dn−1 y dn y 0 + F (t, y, y , · · · , ) = 0. dtn dtn−1 Let x1 = y,

x2 =

dy , dt

··· ,

xn =

dn−1 y . dtn−1

(6.6.2)

120


Then (6.6.2) is equivalent to the following system  0   x1 = x2 ,     x0 = x3 , 2

 ·········      x0 = −F (t, x , x , · · · , x ). 1 2 n n It can be written in the form of (6.6.1) if we let 



x1   x = · · · , xn



 x2 ,   x3 ,   f (t, x) =  .   ··· −F (t, x1 , · · · , xn )



 x1    x2   Recall that for a vector x =   .. , its magnitude |x| is defined to be  .  xn |x| = |x1 | + |x2 | + · · · + |xn |. The following 2 properties of the magnitude can be proved easily. 1. Triangle Inequality |x + y| ≤ |x| + |y|. Pn 2. If A = (aij ) is an n × n matrix and x ∈ Rn , then |Ax| ≤ |A||x| where |A| = i,j |aij |. Definition 6.2 Let G be a subset in R1+n . f (t, x) : G → Rn is said to satisfy a Lipschitz condition with respect to x in G if there exists a constant L > 0 such that, for all (t, x), (t, y) ∈ G, |f (t, x) − f (t, y)| ≤ L|x − y|. Example 6.9 Let f : R1+2 −→ R2 be given by ! 2x2 cos t f (t, x) = , where x = x1 sin t ! ! 2x cos t 2y2 cos t 2 Then |f (t, x) − f (t, y)| = − x1 sin t y1 sin t

x1 x2

! .

= |2 cos t(x2 − y2 )| + | sin t(x1 − y1 )| ≤2|x2 − y2 | + |x1 − y1 | ≤2(|x2 − y2 | + |x1 − y1 |) = 2|x − y|. Thus f satisfies a Lipschitz condition with respect to x in R3 with Lipschitz constant 2. Theorem 6.14 Suppose f is defined on a set G ⊂ R1+n of the form |t − t0 | ≤ a, |x − x0 | ≤ b, (a, b > 0)

6.6. EXISTENCE AND UNIQUENESS OF SOLUTIONS TO SYSTEMS

121

or of the form |t − t0 | ≤ a, |x| < ∞, (a > 0). If ∂f /∂xk (k = 1, . . . , n) exists, is continuous on G, and there is a constant L > 0 such that ∂f ∂xk ≤ L, (k = 1, . . . , n), for all (t, x) ∈ G, then f satisfies a Lipschitz condition on G with Lipschitz constant L. 

 f1 (t, x)    f2 (t, x)  1+n  Proof. Let f (t, x) =  −→ R.  ..  , where each fi (t, x) : R  .  fn (t, x) Thus   ∂f1 ∂xk  ∂f2   ∂xk   

∂f =  . . ∂xk  ..  ∂fn ∂xk

Let (t, z), (t, y) ∈ G ⊆ R1+n . Define F : [0, 1] −→ Rn by F(s) = f (t, sz + (1 − s)y) = f (t, y + s(z − y)). The point sz + (1 − s)y lies on the segment joining z and y, hence the point (t, sz + (1 − s)y) is in G.  ∂f1  ∂x

k n  ∂f2  n X X   ∂f dx ∂x k  . k  (zk − yk ). = Now F0 (s) =  .  ∂xk ds  . 

k=1

k=1

∂fn ∂xk

Therefore, |F0 (s)| ≤

n n X X ∂f |zk − yk | ≤ L |zk − yk | = L|z − y|, ∂xk

k=1

k=1

for s ∈ [0, 1]. Since Z f (t, z) − f (t, y) = F(1) − F(0) =

1

F0 (s) ds

0

we have |f (t, z) − f (t, y)| ≤

R1 0

|F0 (s)| ds ≤ L|z − y|.

Theorem 6.15 (Picard) Let f (t, x) be continuous on the set R : |t − t0 | ≤ a, |x − x0 | ≤ b (a, b > 0), and let |f (t, x)| ≤ M

122


for all (t, x) ∈ R. Furthermore, assume f satisfies a Lipschitz condition with constant L in R. Then there is a unique solution to the initial value problem dx = f (t, x), x(t0 ) = x0 dt on the interval I = [t0 − α, t0 + α], where α = min{a, b/M }. Theorem 6.16 Let f (t, x) be a continuous function on the strip S = {(t, x) ∈ Rn+1 : |t − t0 | ≤ a}, where a is a given positive number, and f satisfies a Lipschitz condition with respect to S. Then the initial value problem x0 (t) = f (t, x), x(t0 ) = x0 , where (t0 , x0 ) ∈ S has a unique solution on the entire interval [−a + t0 , a + t0 ]. Corollary 6.17 Let f (t, x) be a continuous function defined on Rn+1 . Let (t0 , x0 ) ∈ Rn+1 . Suppose that for any a > 0, f satisfies a Lipschitz condition with respect to S = {(t, x) ∈ Rn+1 : |t| ≤ a}. Then the initial value problem x0 (t) = f (t, x), x(t0 ) = x0 has a unique solution on the entire R. The proofs carry over directly from those for Theorem 5.1 and 5.5 and Corollary 5.6 using the method of successive approximations. That is the successive approximations Z t φ0 (t) = x0 , φk+1 (t) = x0 + f (s, φk (s)) ds, k = 0, 1, 2, . . . t0

converge uniformly on the interval I = [t0 − α, t0 + α] with α = min{a, b/M }, to a solution dx = f (t, x), x(t0 ) = x0 on I. Uniqueness is proved by Gronwall’s of the initial value problem dt inequality as before. Example 6.10 Find the first 5 successive approximations to the initial value problem x00 = −et x, x(0) = 1, x0 (0) = 0. Solution. The initial value problem is equivalent to the following initial value problem of differential system. ! ! ! ! x(t) y(t)

0

y(t) −et x(t)

=

We start with x0 (t) y0 (t) Then x1 (t) y1 (t)

! =

1 0

!

1 0

= !

Z + 0

t

x(0) y(0)

,

=

1 0

.

! , for all t ∈ R.

0 s −e × 1

! ds =

1 1 − et

! .

6.6. EXISTENCE AND UNIQUENESS OF SOLUTIONS TO SYSTEMS

x2 (t) y2 (t) x3 (t) y3 (t) x4 (t) y4 (t)

!

=

1 0

1 0

= 1 0

=

!

!

!

Z

! +

0

t

0

Z + 0

t

t

1 − es −es

1 − es −es (2 + s − es )

+

!

Z

1 2

! ds =

2 + t − e−t 1 − et

!

− es − ses + 21 e2s −es (2 + s − es )

ds =

! ds =

1 2

123

! .

2 + t − e−t − et − tet + 21 e2t 3 4

!

+ 2t − tet + 14 e2t 1 + t − et − tet

.

! .

Example 6.11 Consider the linear differential system x0 = Ax, where A = (aij ) is an n × n constant matrix. Let f (t, x) = Ax. For any a > 0 and for all |t| < a, we have |f (t, x1 )−f (t, x2 )| = Pn Pn |A(x1 −x2 )| ≤ |A||x1 −x2 |, where |A| = i=1 j=1 |aij |, so that f satisfies a Lipschitz condition on the strip S = {(t, x) ∈ Rn+1 : |t| ≤ a}. Therefore the system has a unique solution for any initial value and is defined on the entire R. Example 6.12 Let x0 = A(t)x, where A(t) = (aij (t)) is an n × n matrix of continuous functions defined on a closed interval I. Let |aij (t)| ≤ K for all t ∈ I and all i, j = 1, . . . n. Thus if f (t, x) = A(t)x, then   a1k (t)    a2k (t)  ∂f  = .  , ∂xk  ..  ank (t) which is independent of x. Therefore, n ∂f X = |aik (t)| ≤ nK ≡ L, ∂xk i=1

for all t ∈ I and k = 1, . . . , n.

By theorem 6.14 the function f satisfies a Lipschitz condition on the strip S = {(t, x) ∈ R1+n | t ∈ I} with Lipschitz constant L. Thus by corollary 6.17, the system x0 = A(t)x has a unique solution for any initial value in S and is defined on all of I. Thus if A(t) = (aij (t)) is continuous on an open interval I or R, then x0 = A(t)x has a unique solution for any initial value x(t0 ) = x0 with t0 ∈ I or R, and is defined on I or R. (See corollary 6.6.) In particular this together with corollary 6.17 prove theorem 2.1 and theorem 4.1.

124


Bibliography [1] Ravi P. Agarwal and Ramesh C. Gupta, Essentials of ordianry differential equations, McGrawHill (1991) [2] E. Coddington and N. Levinson, Theory of ordinary differential equations, Krieger, New York (1984) [3] Polking, Boggess, Arnold, Differential Equations, 2nd edition, Pearson Prentice Hall (2006) [4] George F. Simmons, Differential equations with applications and historical notes, 2nd edition, McGraw-Hill (1991)

125

MA3220 Ordinary Differential Equations

Recommend Documents