MA3220 Ordinary Differential Equations...Ordinary differential equations. An ordinary differential...

MA3220 Ordinary Differential Equations

Wong Yan Loi

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Contents

1 First Order Differential Equations 51.1 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

2 Linear Differential Equations 172.1 General Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2 Linear Equations with Constant Coefficients . . . . . . . . . . . . . . . . . . . . . 22

2.3 Operator Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3 Second Order Linear Differential Equations 333.1 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

4 Linear Differential Systems 514.1 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 CONTENTS

5 Power Series Solutions 835.1 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 835.2 Series Solutions of First Order Equations . . . . . . . . . . . . . . . . . . . . . . 865.3 Second Order Linear Equations and Ordinary Points . . . . . . . . . . . . . . . . . 875.4 Regular Singular Points and the Method of Frobenius . . . . . . . . . . . . . . . . 935.5 Bessel’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 975.6 Appendix 2: Some Properties of the Legendre Polynomials . . . . . . . . . . . . . 102

6 Fundamental Theory of ODEs 1056.1 Existence-Uniqueness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 1056.2 The Method of Successive Approximations . . . . . . . . . . . . . . . . . . . . . 1066.3 Convergence of the Successive Approximations . . . . . . . . . . . . . . . . . . . 1096.4 Non-local Existence of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . 1126.5 Gronwall’s Inequality and Uniqueness of Solution . . . . . . . . . . . . . . . . . . 1156.6 Existence and Uniqueness of Solutions to Systems . . . . . . . . . . . . . . . . . 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, thereal dependent variable y, and some of its derivatives y′, y′′, · · · , 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′, · · · , 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 asy(n) = f(x, y, y′, · · · , y(n−1)), (1.1.2)

where f is a known function of x, y, y′, · · · , 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 assumethat a0(x) 6≡ 0 in any interval in which the equation is defined. If r(x) ≡ 0, (1.1.3) is called ahomogeneous equation. If r(x) 6≡ 0, (1.1.3) is said to be a non-homogeneous equation, and r(x) iscalled the non-homogeneous term.

2. Solutions.A functional relation between the dependent variable y and the independent variable x that satisfiesthe 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. (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 solutionof (1.1.4) in the interval J provided

F (x, φ(x), φ′(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 determinesone 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

F

(x(t), y(t),

y(t)

x(t)

)= 0.

Example 1.1 Consider the ODE: x+ yy′ = 0 for x ∈ (−1, 1). x2 + y2 = 1 is an implicit solutionwhile x = cos t, y = sin t, t ∈ (0, π) is a parametric solution.

3. Integral curves.The solutions of a first order ODE

y′ = 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) = 〈1, f(x, y)〉 is tangentto the curve r(x) = 〈x, y(x)〉 at every point (x, y) since r′(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 formedby eliminating the constants of this family and its derivatives.

Example 1.2 The family of functions y = Aex +B sinx satisfies the ODE: y′′′′ − y = 0 when the

constants A and B are eliminated using the derivatives.

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′ = y ln yx ln x .

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

y′ =f(x)

g(y), or g(y)dy = f(x)dx. (1.1.8)

The solution is given by ∫g(y)dy =

∫f(x)dx+ c.

1.1. INTRODUCTION 7

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

Ans: y = e−x2

.

The equation y′ = f( yx ) can be reduced to a separable equation by letting u = yx , i.e. y = xu. So

f(u) = y′ = u+ xu′, ∫du

f(u)− u=

∫dx

x+ c.

Exercise 1.3 Solve 2xyy′ + x2 − y2 = 0.

Ans: x2 + y2 = cx.

6. Homogeneous equations.A function is called homogeneous of degree n if f(tx, ty) = tnf(x, y) for all x, y, t.For example

√x2 + y2 and x+ y are homogeneous of degree 1, x2 + y2 is homogeneous of degree

2 and sin(x/y) is homogeneous of degree 0.The ODE M(x, y) + N(x, y)y′ = 0 is said to be homogeneous of degree n if both M(x, y) andN(x, y) are homogeneous of degree n.If we write the above DE as y′ = f(x, y), where f(x, y) = −M(x, y)/N(x, y). Then f(x, y) ishomogeneous of degree 0. To solve the DE

y′ = f(x, y),

where f is homogeneous of degree 0, we use the substitution y = zx. Then

dy

dx= z + x

dz

dx.

Thus the DE becomesz + x

dz

dx= f(x, zx) = x0f(1, z) = f(1, z).

Consequently, the variables can be separated to yield

dz

f(1, z)− z=dx

x,

and integrating both sides will give the solution.

Exercise 1.4 Solve y′ = x+yx−y .

Ans: tan−1(y/x) = ln√x2 + y2 + c.

Example 1.3 An equation in the form

y′ =a1x+ b1y + c1a2x+ b2y + c2

.

can be reduced to a homogeneous equation by a suitable substitution x = z + h, y = w + k whena1b2 6= a2b1, where h and k are solutions of the system of linear equations a1h + b1k + c1 =

0, a2h+ b2k + c2 = 0.


Exercise 1.5 Solve y′ = x+y−2x−y .

Ans: tan−1(y−1x−1

)= ln

√(x− 1)2 + (y − 1)2 + c.

Exercise 1.6 Solve (x+ y + 1) + (2x+ 2y + 1)y′ = 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

∂xdx+

∂u

∂ydy.

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 therectangle S: |x− x0| < a, |y − y0| < b. A necessary and sufficient condition for (1.2.1) to be exactis

∂M

∂y=∂N

∂xfor all (x, y) in S. (1.2.2)

When (1.2.2) is satisfied, a general solution of (1.2.1) is given by u(x, y) = c, where

u(x, y) =

∫ x

x0

M(s, y)ds+

∫ y

y0

N(x0, t)dt (1.2.3)

and c is an arbitrary constant.

Proof. Let u(x, y) =∫ xx0M(s, y)ds +

∫ yy0N(x0, t)dt. We have to show that ∂u∂x = M(x, y) and

∂u∂y = N(x, y). The first equality is immediate by the fundamental theorem of calculus. For thesecond equality, we have ∂u

∂y =∫ xx0

∂∂yM(s, y)ds + N(x0, y) =

∫ xx0

∂∂sN(s, y)ds + N(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 + 3xy2)dx+ (3x2y + y3)dy = 0.

Ans: x4 + 6x2y2 + y4 = 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.

Nµx −Mµy = 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 andy. Plugging into (1.2.5) we find

1

µ

dµ

dv=

My −NxNvx −Mvy

. (1.2.6)

If My−NxNvx−Mvy

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

µ = e∫ v φ(v)dv

is an integrating factor of (1.2.1).

Let v = x. If My−NxN is a function of x alone, say, φ1(x), then e

∫ x φ1(x)dx is an integrating factorof (1.2.1).

Let v = y. If −My−NxM is a function of y alone, say, φ2(y), then e

∫ y φ2(y)dy is an integrating factorof (1.2.1).

Let v = xy. If My−NxyN−xM is a function of v = xy alone, say φ3(xy), then e

∫ xy φ3(v)dv is an integratingfactor of (1.2.1).

Exercise 1.8 Solve (x2y + y + 1) + x(1 + x2)y′ = 0.

Ans: xy + tan−1 x = c.

Exercise 1.9 Solve (y − y2) + xy′ = 0

Ans: y = (1− cx)−1, y = 0.

Exercise 1.10 Solve (xy3 + 2x2y2 − y2) + (x2y2 + 2x3y − 2x2)y′ = 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.

d(x

y) =

ydx− xdyy2


d(xy) = xdy + ydx

d(x2 + y2) = 2xdx+ 2ydy

d(tan−1x

y) =

ydx− xdyx2 + y2

d(ln |xy|) =

ydx− xdyxy

We see that the very simple ODE ydx−xdy = 0 has 1/x2, 1/y2, 1/(x2+y2) and 1/xy as integratingfactors.

Example 1.4 Solve xdy + ydx = x cosxdx.

Solution. The differential xdy + ydx can be written as d(xy). Thus the DE can be written asd(xy) = x cosxdx. Integrating, we have xy = x sinx+ cosx+ c.

Exercise 1.11 Solve (x2y3 + y)dx+ (x− x3y2)dy = 0.

Ans: ln |x/y| = 1/(2x2y2) + c and y = 0.

1.3 First Order Linear Equations

1. Homogeneous equations.A first order homogeneous linear equation is of the form

y′ + p(x)y = 0, (1.3.1)

where p(x) is a continuous function on an interval J . Let P (x) =∫ xap(s)ds. Multiplying (1.3.1)

by eP (x), we getd

dx[eP (x)y] = 0,

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

y(x) = ce−P (x), where P (x) =

∫ x

a

p(s)ds. (1.3.2)

2. Non-homogeneous equations.Now consider a first order non-homogeneous linear equation

y′ + p(x)y = q(x), (1.3.3)

where p(x) and q(x) are continuous functions on an interval J . Let P (x) =∫ xap(s)ds. Multiplying

(1.3.3) by eP (x) we getd

dx[eP (x)y] = eP (x)q(x).

1.3. FIRST ORDER LINEAR EQUATIONS 11

ThuseP (x)y(x) =

∫ x

a

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

The general solution is given by

y(x) = e−P (x)[

∫ x

a

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

P (x) =

∫ x

a

p(s)ds.

(1.3.4)

Exercise 1.12 Solve y′ − y = e2x.

Ans: y = cex + e2x.

3. The Bernoulli equation.An ODE in the form

y′ + p(x)y = q(x)yn, (1.3.5)

where n 6= 0, 1, is called the Bernoulli equation. The functions p(x) and q(x) are continuousfunctions on an interval J .Let u = y1−n. Substituting into (1.3.5) we get

u′ + (1− n)p(x)u = (1− n)q(x). (1.3.6)

This is a first order linear ODE.

Exercise 1.13 Solve xy′ + y = x4y3.

Ans: 1y2 = −x4 + cx2, or y = 0.

4. The Riccati equation.An ODE of the form

y′ = P (x) +Q(x)y +R(x)y2 (1.3.7)

is called the Riccati equation. The functions P (x), Q(x), R(x) are continuous on an interval J . Ingeneral, the Riccati equation cannot be solved by a sequence of integrations. However, if a particularsolution 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

H(x) =

∫ x

x0

[Q(t) + 2R(t)y0(t)]dt,

Z(x) = e−H(x)[c−

∫ x

x0

eH(t)R(t)dt],

(1.3.8)

where c is an arbitrary constant. Then the general solution is given by

y = y0(x) +1

Z(x). (1.3.9)


Proof. In (1.1.7) we let y = y0(x) + u(x) to get

y′0 + u′ = P +Q(y0 + u) +R(y0 + u)2.

Since y0 satisfies (1.3.7), we have

y′0 = P +Qy0 +Ry20 .

From these two equalities we get

u′ = (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 ′ + (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′ = y/x+ x3y2 − x5. Note y0 = x is a solution.

Ans: ce2x5/5 = 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 =cF (x) +G(x)

cf(x) + g(x), (1.3.12)

wheref(x) = e−H(x),

g(x) = −e−H(x)

∫ x

x0

eH(t)R(t)dt,

F (x) = y0(x)f(x), G(x) = y0g(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 independentof x.Proof. From (1.3.12), the four solutions can be written as

yj(x) =cjF (x) +G(x)

cjf(x) + g(x).

Computations show that

(y1 − y2)(y3 − y4)

(y1 − y4)(y3 − y2)=

(c1 − c2)(c3 − c4)

(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)

(y1 − y)(y3 − y2)= c, (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

ln

∣∣∣∣y − y1y − y2

∣∣∣∣ =

∫[y1(x)− y2(x)]R(x)dx+ c, (1.3.14)

where c is an arbitrary constant.

Proof. y and yj satisfy (1.3.7). So

y′ − y′j = (y − yj)[Q+R(y + yj)],

y′ − y′jy − yj

= Q+R(y + yj).

Thusy′ − y′1y − y1

− y′ − y′2y − y2

= R(y1 − y2).

Integrating yields (1.3.14). �

Exercise 1.15 Solve y′ = e−xy2. 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′ = P (x) +Q(x)y +R(x)y2 + S(x)y3,

where P (x), Q(x), R(x) and S(x) are continuous functions of x on an interval J . Using the substi-tution z = y/x, solve the equation

y′ =y

x+ xy2 − 3y3.

Ans: x/y + 3 ln |x/y − 3| = −x3

3 + c, y = x/3 and y = 0.


1.4 First Order Implicit Equations

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

F (x, y, y′) = 0 (1.4.1)

which are not solvable in y′.

1. Method of differentiation.Consider an equations solvable in y:

y = f(x, y′). (1.4.2)

Let p = y′. 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′ + f(y′), (1.4.4)

where f has continuous first order derivative.

Let p = y′. We have y = xp+ f(p). Differentiating we get

[x+ f ′(p)]p′ = 0.

When p′ = 0 we have p = c and (1.4.4) has a general solution

y = cx+ f(c).

When x+ f ′(p) = 0 we get a solution of (1.4.4) given by parameterized equations

x = −f ′(p), y = −pf ′(p) + f(p).

Exercise 1.17 Solve Clairaut’s equation y = xy′ − y′2/4.

Ans: y = cx− c2/4, y = x2.

Exercise 1.18 Let C be the curve with parametric equation

x = −f ′(p), y = −pf ′(p) + f(p).

Show that the tangent to C at the point p = c has the equation y = cx+ f(c).

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, (1.4.5)

where x is missing. Let p = y′ 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

dx =dy

y′=dy

p=g′(t)dt

h(t),

we have x =∫ tt0

g′(t)h(t) dt+ c. The solutions of (1.4.5.) are given by

x =

∫ t

t0

g′(t)

h(t)dt+ c, y = g(t).

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

Exercise 1.19 Solve y2 + y′2 − 1 = 0.

Ans: y = cos(c− x).

For the equation F (y, y′) = 0 or F (x, y′) = 0, if y′ can be solved in terms of y or x, then theequation becomes explicit.

Exercise 1.20 Solve y′2 − (2x+ ex)y′ + 2xex = 0.

Ans: y = x2 + c1, y = ex + c2.

3. Reduction of order.Consider the equation

F (x, y′, y′′) = 0, (1.4.6)

where y is missing. Let p = y′. Then y′′ = p′. Write (1.4.6) as

F (x, p, p′) = 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 generalsolution of (1.4.6) is

y =

∫ x

x0

φ(t, c1)dt+ c2.


Exercise 1.21 Solve xy′′ − y′ = 3x2.

Ans: y = x3 + c1x2 + c2.

Consider the equationF (y, y′, y′′) = 0, (1.4.8)

where x is missing. Let p = y′. Then y′′ = dpdx = dp

dydydx = dp

dyp. Write (1.4.8) as

F (y, p, pdp

dy) = 0. (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 solvethe equation

y′ = ψ(y, c1)

to get a general solution of (1.4.8).

Exercise 1.22 Solve y′′ + k2y = 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 y2dx + (xy − x3)dy = 0 by using the substitutionu = x−2.

Ans: 3y = 2x2 + cx2y3 and y = 0.

Exercise 1.24 Solve the differential equation y2dx + (xy + tan(xy))dy = 0 by using (i) the sub-stitution 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 substitu-tion y = ez , (ii) by finding an integrating factor using the formula (1.2.6).

Ans: ln y = −x2 + cx3.

Exercise 1.26 Solve y′′ + y =

{0 if x ≤ 1

1− x if x > 1, y(0) = 0, y′(0) = 0.

Ans: y =

{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′ + an(x)y = f(x), (2.1.1)

where y(k) = dkydxk

. Throughout this section we assume that aj(x)’s and f(x) are continuous func-tions 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′ + an(x)y = 0. (2.1.2)

Let us begin with the initial value problem:

y(n) + a1(x)y(n−1) + · · ·+ an(x)y = f(x),

y(x0) = y0,

y′(x0) = y1,

· · · · · · · · ·

y(n−1)(x0) = yn−1.

(2.1.3)

Theorem 2.1 (Existence and Uniqueness Theorem) Assume that a1(x), · · · , an(x) and f(x) arecontinuous functions defined on the interval (a, b). Then for any x0 ∈ (a, b) and for any numbersy0, · · · , 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 initialvalue 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′(x0) = 0, · · · , y(n−1)(x0) = 0. (2.1.4)

Then y(x) ≡ 0 on (a, b).

17

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 problemhas 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’)

andL[y] = 0. (2.1.2’)

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 = c1y1 + c2y2

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[c1y1 + c2y2] = c1L[y1] + c2L[y2] = 0, we see that c1y1 + c2y2 is also a solutionof (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

c1φ1(x) + · · ·+ ckφk(x) = 0

for all x ∈ (a, b). A set of functions are linearly independent on (a, b) if they are not linearlydependent on (a, b).

Lemma 2.4 Functions φ1(x), · · · , φk(x) are linearly dependent on (a, b) if and only if the followingvector-valued functions

φ1(x)

φ′1(x)

· · ·φ(n−1)1 (x)

, · · · ,

φk(x)

φ′k(x)

· · ·φ(n−1)k (x)

(2.1.6)

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),

c1φ1(x) + · · ·+ ckφk(x) = 0.

Differentiating this equality successively we find that

c1φ′1(x) + · · ·+ ckφ

′k(x) = 0,

· · · · · · · · · · · · · · · · · ·

c1φ(n−1)1 (x) + · · ·+ ckφ

(n−1)k (x) = 0.

Thus

c1

φ1(x)

φ′1(x)

· · ·φ(n−1)1 (x)

+ · · ·+ ck

φk(x)

φ′k(x)

· · ·φ(n−1)k (x)

= 0

for all x ∈ (a, b). Hence the k vector-valued functions are linearly dependent on (a, b). �

Recall that, n vectors in Rn are linearly dependent if and only if the determinant of matrix formedby these vectors is zero.

Definition. The Wronskian of n functions φ1(x), · · · , φn(x) is defined by

W (φ1, · · · , φn)(x) =

∣∣∣∣∣∣∣φ1(x) · · · φn(x)

· · · · · · · · ·φ(n−1)1 (x) · · · φ

(n−1)n (x)

∣∣∣∣∣∣∣ . (2.1.7)


Note that Wronskian of φ1, · · · , φn is the determinant of the matrix formed by the vector-valuedfunctions 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 Wron-skian.(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 nowherezero.(2) n solutions y1, · · · , yn of (2.1.2) are linearly independent on (a, b) if and only if the set of vectors

y1(x0)

y′1(x0)

· · ·y(n−1)1 (x0)

, · · · ,

yn(x0)

y′n(x0)

· · ·y(n−1)n (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 theirWronskian.

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

y1(x)

y′1(x)

· · ·y(n−1)1 (x)

, · · · ,

yn(x)

y′n(x)

· · ·y(n−1)n (x)

are linearly dependent on (a, b). Thus for all x ∈ (a, b), the determinant of the matrix formed bythese 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)

y′1(x0)

· · ·y(n−1)1 (x0)

, · · · ,

yn(x0)

y′n(x0)

· · ·y(n−1)n (x0)

are linearly dependent. Thus there exist n constants c1, · · · , cn, not all zero, such that

c1

y1(x0)

y′1(x0)

· · ·y(n−1)1 (x0)

+ · · ·+ cn

yn(x0)

y′n(x0)

· · ·y(n−1)n (x0)

= 0 (2.1.8)

2.1. GENERAL THEORY 21

Definey0(x) = c1y1(x) + · · ·+ cnyn(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′(x0) = 0, · · · , y(n−1)(x0) = 0. (2.1.9)

From Corollary 2.2, we have y0 ≡ 0, namely,

c1y1(x) + · · ·+ cnyn(x) = 0

for all x ∈ (a, b). Thus y1, · · · , yn are linearly dependent on (a, b). �

Example 2.1 Consider the differential equation y′′− 1xy′ = 0 on the interval (0,∞). Both φ1(x) =

1 and φ2(x) = x2 are solutions of the differential equation. W (φ1, φ2)(x) =

∣∣∣∣∣ 1 x2

0 2x

∣∣∣∣∣ = 2x 6= 0

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 homo-geneous 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 generalsolution of (2.1.2) is given by

y(x) = c1y1(x) + · · ·+ cnyn(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 initialconditions

y(j)k (x0) =

0 if j 6= k − 1,

1 if j = k − 1.

The n vectors y1(x0)

y′1(x0)

· · ·y(n−1)1 (x0)

, · · · ,

yn(x0)

y′n(x0)

· · ·y(n−1)n (x0)

are lineally independent since they form the identity matrix. From Corollary 2.6, y1, · · · , yn arelinearly independent on (a, b). From Theorem 2.3, for any constants c1, · · · , cn, y = c1y1 + · · · +cnyn 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 thatthe general solution of (2.1.2) is given by

y = c1y1 + · · ·+ cnyn. (2.1.11)


Given a solution y of (2.1.2), and fix x0 ∈ (a, b). Since y1, · · · , yn are linearly independent on(a, b), the vectors

y1(x0)

y′1(x0)

· · ·y(n−1)1 (x0)

, · · · ,

yn(x0)

y′n(x0)

· · ·y(n−1)n (x0)

are linearly independent vectors. They form a basis for Rn. Thus the vector

y(x0)

y′(x0)

· · ·y(n−1)(x0)

can be represented as a linear combination of the n vectors, namely, there exist n constants c1, · · · ,cn such that

y(x0)

y′(x0)

· · ·y(n−1)(x0)

= c1

y1(x0)

y′1(x0)

· · ·y(n−1)1 (x0)

+ · · ·+ cn

yn(x0)

y′n(x0)

· · ·y(n−1)n (x0)

.

Letφ(x) = y(x)− [c1y1(x) + · · ·+ cnyn(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) = c1y1(x) + · · ·+ cnyn(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 ofsolutions for the associated homogeneous equation (2.1.2). The general solution of (2.1.1) is givenby

y(x) = c1y1(x) + · · ·+ cnyn(x) + yp(x). (2.1.12)

Proof. Let y be a solution of the non-homogeneous equation. Then y − yp is a solution of thehomogeneous equation. Thus y(x)− yp(x) = c1y1(x) + · · ·+ cnyn(x). �

2.2 Linear Equations with Constant Coefficients

Let us begin with second order linear equation with constant coefficients

y′′ + ay′ + 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) wefind 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) arecalled characteristic values (or eigenvalues):

λ1 =1

2(−a+

√a2 − 4b),

λ2 =1

2(−a−

√a2 − 4b).

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 = c1eλ1x + c2e

λ2x.

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 = c1eλx + c2xe

λx.

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 = c1eαx cos(βx) + c2e

αx sin(βx).

Example 2.2 Solve y′′ + y′ − 2y = 0, y(0) = 4, y′(0) = −5.

Ans: λ1 = 1, λ2 = −2, y = ex + 3e−2x.

Example 2.3 Solve y′′ − 4y′ + 4y = 0, y(0) = 3, y′(0) = 1.

Ans: λ1 = λ2 = 2, y = (3− 5x)e2x.

Example 2.4 Solve y′′ − 2y′ + 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) + a1y(n−1) + · · ·+ an−1y

′ + any = 0, (2.2.3)

where a1, · · · , an are real constants.


y = eλx is a solution of (2.2.3) if and only if λ satisfies

λn + a1λn−1 + · · ·+ an−1λ+ 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 + a1λn−1 + · · ·+ an−1λ+ 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, xkeλx is a solution of (2.2.3).(iii) If λ = α+ iβ, then

xkeαx cos(βx), xkeαx 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

eλ1x, xeλ1x, · · · , xm1−1eλ1x;

· · · · · · · · · · · · · · · · · · · · · · · · ;

eλsx, xeλsx, · · · , xms−1eλsx.

(2.2.6)

Proof of Lemma 2.9 and 2.10

Consider the n-th order linear equation with constant coefficients

y(n) + a1y(n−1) + · · ·+ an−1y

′ + any = 0, (A1)

where y(k) = dkydxk

. Let L(y) = y(n) + a1y(n−1) + · · ·+ an−1y

′ + any and p(z) = zn + a1zn−1 +

· · ·+ an−1z + an. Note that p is a polynomial in z of degree n. Then we have

L(ezx) = p(z)ezx (A2)

Before we begin the proof, let’s observe that ∂2

∂z∂xezx = ∂2

∂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 derivativesexist and continuous. That means we can interchange the order of differentiation with respect to xand z as we wish. Therefore d

dzL(ezx) = L( ddz ezx). For instance, one may verify directly that

d

dz

dk

dxk(ezx) = xzkezx + kzk−1ezx =

dk

dxk(d

dzezx).


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

(u · v)(k) =

k∑i=0

(k

i

)u(i)v(k−i). (A3)

More generally, dk

dzkL(ezx) = L( d

k

dzkezx). (Strictly speaking, partial derivative notations should be

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, p′(λ) = 0, p′′(λ) = 0, . . . , p(m−1)(λ) = 0.Now for k = 1, . . . ,m− 1, differentiating (A2) k times with respect to z, we have

L(xkezx) = L(dk

dzkezx) =

dk

dzkL(ezx) =

dk

dzk(p(z)ezx) =

k∑i=0

(k

i

)p(i)(z)xk−iezx.

Thus L(xkeλx) = 0 and xkeλx is solution of (A1).(3) Let λ1, · · · , λs be the distinct roots of p, with multiplicity m1, · · · , ms respectively. Then wewish to prove that

eλ1x, xeλ1x, · · · , xm1−1eλ1x;

· · · · · · · · · · · · · · · · · · · · · · · · ;

eλsx, xeλsx, · · · , xms−1eλsx.

(A4)

are linearly independent over R. To prove this, suppose for all x in R

c11eλ1x + c12xe

λ1x + · · ·+ c1m1xm1−1eλ1x

+ · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·+

cs1eλsx + cs2xe

λsx + · · ·+ csmsxms−1eλsx = 0.

Let’s write this asP1(x)eλ1x + P2(x)eλ2x + · · ·+ Ps(x)eλsx = 0,

for all x in R, where Pi(x) = ci1 + ci2x + · · · + cimixmi−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 identicallyzero. Dividing the above equation by eλ1x, 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 sinceP1(x) is a polynomial of degree m1 − 1), we can reduce P1(x) to 0. Note that in this process, thedegree 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 eλ1x we have

Q2(x)eλ2x + · · ·+Qs(x)eλsx = 0.

For instance, if deg Pi = 0, then Pi is a nonzero constant. The resulting Qi is equal to (λi−λ1)αPi

for some positive integer α. Since λi 6= λ1, Qi is also a nonzero constant. Thus deg Qi = 0.


Repeating this procedure, we arrive at

Rs(x)eλsx = 0,

where deg Rs = deg Ps. Hence Rs(x) ≡ 0 which is a contradiction. Thus all the Pi(x)’s areidentically zero. That means all cij’s are zero and the functions in (A4) are linearly independent.Remark. If (2.2.3) has a complex eigenvalue λ = α + iβ, then λ = α − iβ is also an eigenvalue.Thus both xke(α+iβ)x and xke(α−iβ)x appear in (2.2.6), where 0 ≤ k ≤ m− 1. In order to obtain afundamental set of real solutions, the pair of solutions xke(α+iβ)x and xke(α−iβ)x in (2.2.6) shouldbe replaced by xkeαx cos(βx) and xkeαx sin(βx).

In the following we discuss solution of the non-homogeneous equation.

y′′ + P (x)y′ +Q(x)y = f(x), (2.2.7)

The associated homogeneous equation is

y′′ + P (x)y′ +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′′ +P (x)y′ +Q(x)y = 0 and W (x) be their Wronskian. We look for a particular solution of (2.2.7) inthe form

yp = u1y1 + u2y2,

where u1 and u2 are functions to be determined. Suppose

u′1y1 + u′2y2 = 0.

Differentiating this equation once, we get u′′1y1 + u′′2y2 = −u′1y′1 − u′2y′2. Plugging yp into (2.2.7)we get

u′1y′1 + u′2y

′2 = f.

Hence u′1 and u′2 satisfy {u′1y1 + u′2y2 = 0,

u′1y′1 + u′2y

′2 = f.

(2.2.8)

Solving it, we find thatu′1 = − y2

Wf, u′2 =

y1Wf.

Integrating yields

u1(x) = −∫ x

x0

y2(t)

W (t)f(t)dt,

u2(x) =

∫ x

x0

y1(t)

W (t)f(t)dt.

(2.2.9)


Example 2.5 Solve the differential equation y′′ + y = secx.

Solution. A basis for the solutions of the homogeneous equation consists of y1 = cosx and y2 =

sinx. Now W (y1, y2) = cosx cosx − (− sinx) sinx = 1. Thus u1 = −∫

sinx secx dx =

ln | cosx| + c1 and u2 =∫

cosx secx dx = x + c2. From this, a particular solution is givenby yp = cosx ln | cosx| + x sinx. Therefore, the general solution is y = c1 cosx + c2 sinx +

cosx ln | cosx|+ x sinx.

The method of variation of parameters can also be used to find another solution of a second orderhomogeneous linear differential equation when one solution is given. Suppose z is a known solutionof the equation

y′′ + P (x)y′ +Q(x)y = 0.

We assume y = vz is a solution so that

0 = (vz)′′ + P (vz)′ +Q(vz) = (v′′z + 2v′z′ + vz′′) + P (v′z + vz′) +Qvz

= (v′′z + 2v′z′ + Pv′z) + v(z′′ + Pz′ +Qz) = v′′z + v′(2z′ + Pz).

That isv′′

v′= −2

z′

z− P.

An integration gives v′ = z−2e−∫P dx and v =

∫z−2e−

∫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 x2y′′ + xy′ − y = 0, find another solution.

Solution. Let’s write the DE in the form y′′ + 1xy′ − 1

x2 y = 0. Then P (x) = 1/x. Thus a secondlinearly independent solution is given y = vx, where

v =

∫x−2e−

∫1/x dx dx =

∫x−2x−1 dx = − 1

2x2.

Therefore the second solution is y = − 12x−1 and the general solution is y = c1x+ c2x

−1.

2. Method of undetermined coefficients.Consider the equation y′′ + ay′ + 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′′ + ay′ + by = f(x) we find

Q′′ + (2α+ a)Q′ + (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


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 characteristicequation, then (2.2.10) is reduced to

Q′′ + (2α+ a)Q′ = 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 equationwith multiplicity 2, then (2.2.10) is reduced to

Q′′ = Pn. (2.2.12)

We choose Q(x) = x2Rn(x), where Rn(x) is a polynomial of degree n.

y = x2Rn(x)eαx.

Example 2.7 Find the general solution of y′′ − y′ − 2y = 4x2.

Ans: y = c1e2x + c2e

−x − 3 + 2x− 2x2.

Example 2.8 Find a particular solution of y′′′ + 2y′′ − y′ = 3x2 − 2x+ 1.

Ans: y = −27x− 5x2 − x3.

Example 2.9 Solve y′′ − 2y′ + y = xex.

Ans: y = c1ex + c2xe

x + 16x

3ex.

Case 2. f(x) = Pn(x)eαx cos(βx) or f(x) = Pn(x)eαx sin(βx), where Pn(x) is a polynomial ofdegree n ≥ 0.We first look for a solution of

y′′ + ay′ + by = Pn(x)e(α+iβ)x. (2.2.13)

Using the method in Case 1 we obtain a complex-valued solution

z(x) = u(x) + iv(x),


where u(x) = <(z(x)), v(x) = =(z(x)). Substituting z(x) = u(x)+iv(x) into (2.2.13) and takingthe real and imaginary parts, we can show that u(x) = <(z(x)) is a solution of

y′′ + ay′ + by = Pn(x)eαx cos(βx), (2.2.14)

and v(x) = =(z(x)) is a solution of

y′′ + ay′ + by = Pn(x)eαx sin(βx). (2.2.15)

Example 2.10 Solve y′′ − 2y′ + 2y = ex cosx.

Ans: y = c1ex cosx+ c2e

x sinx+ 12xe

x sinx.

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′′ + ay′ + by = f1(x)

and

y′′ + ay′ + by = f2(x)

respectively, then yp = y1 + y2 is a particular solution of

y′′ + ay′ + by = f1(x) + f2(x).

Proof. Exercise.

Example 2.11 Solve y′′ − y = ex + sinx.

Solution. A particular solution for y′′ − y = ex is given by y1 = 12xe

x. Also a particular solutionfor y′′ − y = sinx is given by y2 = − 1

2 sinx. Thus 12 (xex − sinx) is a particular solution of

the given differential equation. The general solution of the corresponding homogeneous differentialequation is given by c1e−x + c2e

x. Hence the general solution of the given differential equation isc1e−x + c2e

x + 12 (xex − sinx).


2.3 Operator Methods

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

Dy =d

dxy, Dny =

dn

dxny = y(n).

We define D0y = y. Given a polynomial L(x) =∑nj=0 ajx

j , where aj’s are constants, we definea differential operator L(D) by

L(D)y =

n∑j=0

ajDjy.

Then the equationn∑j=0

ajy(j) = f(x) (2.3.1)

can be written asL(D)y = f(x). (2.3.2)

Let L(D)−1f denote any solution of (2.3.2). We have

D−1D = DD−1 = D0,

L(D)−1L(D) = L(D)L(D)−1 = D0.

However, L(D)−1f is not unique.

To see the above properties, first recall that D−1f means a solution of y′ = f . Thus D−1f =∫f .

Hence it follows that D−1D = 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 definitionof L(D)−1, we have L(D)−1(L(D)f) = f . This means L(D)−1L(D) = D0. Lastly, sinceL(D)−1f is a solution of L(D)y = f(x), it is clear that L(D)(L(D)−1f) = f . In other words,L(D)L(D)−1 = D0.

More generally, we have:

1. D−1f(x) =

∫f(x)dx+ C,

2. (D − a)−1f(x) = Ceax + eax∫e−axf(x)dx,

3. L(D)(eaxf(x)) = eaxL(D + a)f(x),

4. L(D)−1(eaxf(x)) = eaxL(D + a)−1f(x).

(2.3.3)

Proof. Property 2 is just the solution of the first order linear ODE. To prove Property 3, first observethat (D − r)(eaxf(x)) = eaxD(f(x)) + aeaxf(x) − reaxf(x) = eax(D + a − r)(f(x)). Thus(D− s)(D− r)(eaxf(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)(eaxf(x)) = eaxL(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) byL(D + a).

To prove Property 4, apply L(D) to the right hand side. We have

L(D)[eaxL(D + a)−1f(x)] = eax[L(D + a)(L(D + a)−1f(x))] = eaxf(x).

Thus L(D)−1(eaxf(x)) = eaxL(D + a)−1f(x). �

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

y = L(D)−1f(x) = (D − r1)−1 · · · (D − rn)−1f(x). (2.3.4)

Then we obtain the solution by successive integration. Moreover, if r′js are distinct, we can write

1

L(x)=

A1

x− r1+ · · ·+ An

x− rn,

where A′js 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 besolved is

(D −m)ny = f(x) (2.3.6)

To solve this equation, let us assume a solution of the form y = emxv(x), where v(x) is a functionof x to be determined. One can easily verify that (D−m)nemxv = emxDnv. Thus equation (2.3.6)reduces to

Dnv = e−mxf(x) (2.3.7)

If we integrate (2.3.7) n times, we obtain

v =

∫ ∫· · ·∫ ∫

e−mxf(x) dx · · · dx+ c0 + c1x+ · · ·+ cn−1xn−1 (2.3.8)

Thus we see that

(D−m)−nf(x) = emx[∫∫

· · ·∫∫

e−mxf(x)dx · · · dx+ c0 + c1x+ · · ·+ cn−1xn−1]

(2.3.9)

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

Solution. First 1D2−3D+2 = 1

D−2 −1

D−1 . Therefore

y = (D2 − 3D + 2)−1(xex)

= (D − 2)−1(xex)− (D − 1)−1(xex)

= e2xD−1(e−2xxex)− exD−1(e−xxex)

= e2xD−1(e−xx)− exD−1(x)

= e2x(−xe−x − e−x + c1)− ex( 12x

2 + c2)

= −ex( 12x

2 + x+ 1) + c1e2x + c2e

x.


Example 2.13 Solve (D3 − 3D2 + 3D − 1)y = ex.

Solution. The DE is equivalent to (D − 1)3y = ex. Therefore,

y = (D − 1)−3ex = ex[∫∫∫

e−xex dx+ c0 + c1x+ c2x2

]= ex

[1

6x3 + c0 + c1x+ c2x

2

].

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)−1f = 1D2(1−D)2x

3

= D−2(1 + 2D + 3D2 + 4D3 + 5D4 + 6D5)x3

= D−2(x3 + 6x2 + 18x+ 24)

= D−1(x4

4 + 2x3 + 9x2 + 24x)

= x5

20 + x4

2 + 3x3 + 12x2.

Therefore, the general solution is y = (c1 + c2x)ex + (c3 + c4x) + x5

20 + x4

2 + 3x3 + 12x2.

Exercise 2.1 For all real x, the real-valued function y = f(x) satisfies

y′′ − 2y′ + y = 2ex.

(a) If f(x) > 0 for all real x, must f ′(x) > 0 for all real x? Explain.

(b) If f ′(x) > 0 for all real x, must f(x) > 0 for all real x? Explain.

Exercise 2.2 Find the general solution of y′′ − 2y′ − 3y = x(ex + e−x).

Ans: y = c1e−x + c2e

3x − x4 ex − x2

8 e−x − x

16e−x.

Chapter 3

Second Order Linear DifferentialEquations

3.1 Exact 2nd Order Equations

The general 2nd order linear differential equation is of the form

p0(x)y′′ + p1(x)y′ + p2(x)y = f(x) (3.1.1)

The equation can be written as

(p0y′ − p′0y)′ + (p1y)′ + (p′′0 − p′1 + p2)y = f(x) (3.1.2)

It is said to be exact ifp′′0 − p′1 + p2 ≡ 0. (3.1.3)

In the event that the equation is exact, a first integral to (3.1.1) is

p0(x)y′ − p′0(x)y + p1(x)y =

∫f(x) dx+ C1.

Example 3.1 Find the general solution of the DE

1

xy′′ + (

1

x− 2

x2)y′ − (

1

x2− 2

x3)y = ex.

Solution. Condition (3.1.3) is fulfilled. The first integral is

1

xy′ +

1

x2y + (

1

x− 2

x2)y = ex + C1.

That isy′ + (1− 1

x)y = xex + C1x.

From the last equation, the general solution is found to be

y =1

2xex + C1x+ C2xe

−x.

33

34 CHAPTER 3. SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS

3.2 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 calledan integrating factor of (3.1.1). That is

(p0v)′′ − (p1v)′ + p2v = 0. (3.2.4)

This is a differential equation for v, which is, more explicitly,

p0(x)v′′ + (2p′0(x)− p1(x))v′ + (p′′0(x)− p′1(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) isthus an integrating factor for a given differential equation, if and only if it is a solution of the adjointequation. Note that the adjoint of (3.2.5) is in turn found to be the associated homogeneous equationof (3.1.1), thus each is the adjoint of the other.In this case, a first integral to (3.1.1) is

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′′ + (2x2 + 4x− 3)y′ + 8xy = 1.

Solution. The adjoint of this equation is

(x2 − x)v′′ − (2x2 − 1)v′ + (4x− 2)v = 0.

By the trial of xm, this equation is found to have x2 as a solution. Thus x2 is an integrating factorof the given differential equation. Multiplying the original equation by x2, we obtain

(x4 − x3)y′′ + (2x4 + 4x3 − 3x2)y′ + 8x3y = x2.

Thus a first integral to it is

(x4 − x3)y′ − (4x3 − 3x2)y + (2x4 + 4x3 − 3x2)y =

∫x2 dx+ C.

After simplification, we have

y′ +2x

x− 1y =

1

3(x− 1)+

C

x3(x− 1).

An integrating factor for this first order linear equation is e2x(x − 1)2. Thus the above equationbecomes

e2x(x− 1)2y =1

3

∫(x− 1)e2x dx+ C

∫e2x(x− 1)

x3dx+ C2.

That is

e2x(x− 1)2y =1

3

[x

2− 3

4

]e2x + C

e2x

2x2+ C2.

Thus the general solution is

y =1

(x− 1)2

(x

6− 1

4+C1

x2+ C2e

−2x).

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′′ +4x

2x− 1y′ +

8x− 8

(2x− 1)2y = 0.

Ans: y = C1

2x−1 + C2x2x−1e

−2x.

Exercise 3.2 The equation(p(x)y′)′ + 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 Sturm-Liouville 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′′ +

p1(x)y′+p2(x)y. The formal adjoint ofL is the differential operator defined byL+[y] = (p0(x)y)′′−(p1(x)y)′+p2(x)y, where p′′0 , p

′1 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 followingidentity relating L and L+.

Theorem 3.1 (Lagrange’s identity)

L[u]v − uL+[v] =d

dx[P (u, v)],

where P (u, v) = up1v − u(p0v)′ + u′p0v.

Proof. Expanding both sides, we get p0u′′v + p1u′v − p0uv′′ − 2p′0uv

′ − p′′0uv + p′1uv + p1uv′.�

Integrating Lagrange’s identity leads to Green’s formula.

Corollary 3.2 (Green’s formula)∫ b

a

(L[u]v − uL+[v]

)dx = P (u, v)(x)

∣∣∣∣ba

.

Let’s define an inner product for continuous real-valued functions on [a, b] by

(f, g) =

∫ b

a

f(x)g(x) dx.

With this notation, Green’s formula becomes

(L[u], v) = (u, L+[v]) + P (u, v)(x)

∣∣∣∣ba

. (3.3.7)

Though the operators L and L+ are meaningful when applied to any function y having a continuoussecond 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)


holds for all u in the domain of L and all v in the domain of L+. This is the case when we try tosolve boundary value problems in which L and L+ are restricted to those functions that satisfy thegiven boundary conditions. When (3.3.8) holds, L+ is called the adjoint operator of L (i.e. withoutthe adjective “formal”). For the special case when L+ = L and (3.3.8) holds, we say that L isself-adjoint.

As an example, let L be the differential operator given by the first two terms on the left hand sideof the Sturm-Liouville equation (3.2.6), that is L[y] = (p(x)y′)′ + q(x)y. By the exercise for theequation in (3.2.6), it follows that L+ = L. Then Lagrange’s identity and Green’s formula reduceto the followings.

Theorem 3.3 (Lagrange’s identity for L[y] = (p(x)y′)′ + q(x)y)

L[u]v − uL[v] = − d

dx[pW (u, v)],

where W (u, v) = uv′ − vu′ is the Wronskian of u and v.

Corollary 3.4 (Green’s formula for L[y] = (p(x)y′)′ + q(x)y)

(L[u], v)− (u, L[v]) =

∫ b

a

(L[u]v − uL[v]) dx = −pW (u, v)(x)

∣∣∣∣ba

. (3.3.9)

Exercise 3.3 Let L[y] = p0(x)y′′ + p1(x)y′ + p2(x)y. Prove that L+ = L if and only if p1(x) =

p′0(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′)′ + q(x)y = 0, then pW (u, v)(x) is aconstant 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′′ + p(x)y′ + q(x)y = f(x), a < x < b; (3.4.10)

satisfying the boundary conditions

a11y(a) + a12y′(a) + b11y(b) + b12y

′(b) = d1,

a21y(a) + a22y′(a) + b21y(b) + b22y

′(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 themas nonhomogeneous.For the homogeneous equation with homogeneous boundary condition

y′′ + p(x)y′ + q(x)y = 0, a < x < b;

a11y(a) + a12y′(a) + b11y(b) + b12y

′(b) = 0,

a21y(a) + a22y′(a) + b21y(b) + b22y

′(b) = 0,

(3.4.12)

one can verify the following properties:

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 c1φ1+c2φ2 is also a solutionof (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. Thefollowing examples illustrate some of these cases.

Example 3.3 Find all solutions to the boundary value problem

y′′ + 2y′ + 5y = 0; y(0) = 2, y(π/4) = e−π/4.

Solution. The general solution to the equation y′′+2y′+5y = 0 is y = c1e−x cos 2x+c2e

−x sin 2x.Substituting the boundary conditions, we have y(0) = c1 = 2 and y(π/4) = c2e

−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.


y′′ + y = cos 2x; y′(0) = 0, y′(π) = 0.

Solution. The general solution to the equation y′′ + y = cos 2x is y = c1 cosx + c2 sinx −(1/3) cos 2x. Thus y′ = −c1 sinx+ c2 cosx+ (2/3) sin 2x. Substituting the boundary conditions,we have y′(0) = c2 = 0 and y′(π) = −c2 = 0. Thus the boundary value problem has a one-parameter family of solutions y = c1 cosx− (1/3) cos 2x, where c1 is any real number.


y′′ + 4y = 0; y(−π) = y(π), y′(−π) = y′(π).

Solution. The general solution to the equation y′′+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′ = −2c1 sin 2x +

2c2 cos 2x automatically satisfy the boundary conditions y(−π) = y(π), y′(−π) = y′(π). Hencethe 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.


y′′ + 4y = 4x; y(−π) = y(π), y′(−π) = y′(π).

Solution. The general solution to the equation y′′+ 4y = 4x is y = x+ c1 cos 2x+ c2 sin 2x. Sincey(−π) = −π + c1 and y(π) = π + c1, there are no solutions satisfying y(−π) = y(π). Hence theboundary value problem has no solution.

Exercise 3.7 Find all solutions to the boundary value problem

y′′ + y = ex; y(0) = 0, y(π) + y′(π) = 0.

Ans. y = ( 12 + eπ) sinx− 1

2 cosx+ 12ex.


3.5 Regular Sturm-Liouville Boundary Value Problem

Let L[y] = (p(x)y′)′ + q(x)y. Consider the regular Sturm-Liouville boundary value problem:

L[y] + λr(x)y = 0, a < x < b;

a1y(a) + a2y′(a) = 0,

b1y(b) + b2y′(b) = 0,

(3.5.13)

where p(x), p′(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 boundaryconditions 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 a1u(a) + a2u′(a) = 0, a1v(a) + a2v

′(a) = 0 has non-trivial solutions in a1 and a2 since a1 and a2 are not both zero. Therefore the determinant of thesystem 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 continuoussecond derivatives on [a, b] and satisfy the boundary conditions in (3.5.13). Self-adjoint operatorsare 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 necessarilythe 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 objec-tive is to determine for which values of λ, the equation in (3.5.13) has nontrivial solutions satisfy-ing the given boundary conditions. Such problems are called eigenvalue problems. The nontrivialsolutions are called eigenfunctions, and the corresponding number λ an eigenvalue. If all the eigen-functions associated with a particular eigenvalue are just scalar multiple of each other, then theeigenvalue 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 becomplex-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) asa1, a2, b1, b2 are real. Thus λ is an eigenvalue with eigenfunction u for (3.5.13). We have

(L[u], u) = (−λru, u) = −λ(ru, u) = −λ∫ b

a

r|u|2 dx.

By Green’s formula,

(L[u], u) = (u, L[u]) = (u,−λru) = −λ(u, ru) = −λ∫ b

a

r|u|2 dx.

3.5. REGULAR STURM-LIOUVILLE BOUNDARY VALUE PROBLEM 39

Since r > 0, u 6≡ 0 and |u| ≥ 0, we have∫ bar|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 haveL[u1] = −λru1 and L[u2] = −λru2. Furthermore both u1 and u2 satisfy the boundary conditionsin (3.5.13). Therefore we can take either u1 or u2 as the real-valued eigenfunction associated withλ.

(iii) Suppose u1 and u2 are 2 eigenfunctions corresponding to the eigenvalue λ and both are solutionsto (3.5.13). Thus

a1u1(a) + a2u′1(a) = 0, and a1u2(a) + a2u

′2(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, forall x ∈ [a, b], where C is a nonzero constant as p(x) > 0 for all x ∈ [a, b]. But this contradictsC = p(a)W (u1, u2)(a) = 0. Consequently u1 and u2 are linearly dependent. Therefore one mustbe 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 apositive weight function r(x) on the interval [a, b] if∫ b

a

f(x)g(x)r(x) dx = 0.

Theorem 3.6 Eigenfunctions that correspond to distinct eigenvalues of the regular Sturm-Liouvilleboundary 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 vfor (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(u, rv) = (ru, v) = 0. That is

∫ baruv 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′′ + λy = 0, y(0) = y(π) = 0. (3.5.14)


When λ ≤ 0, the boundary value problem (3.5.14) has only the trivial solution y = 0. Thusfor λ ≤ 0, it is not an eigenvalue. Let’s consider λ > 0. The general solution to the equationy′′ + λy = 0 is given by y = A cos(

√λx) + B sin(

√λx). Now y(0) = 0 implies that A = 0,

and y(π) = 0 implies that B sin(√λπ) = 0. Since we are looking for nontrivial solutions, we

must have B 6= 0 so that sin(√λπ) = 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 orthog-onal eigenfunctions {φn} defined on [a, b]. We can use these eigenfunctions to form an orthonormalsystem with respect to r(x) simply by normalizing each eigenfunction φn so that∫ b

a

φ2n(x)r(x) dx = 1.

Now suppose {φn} is orthonormal with respect to a positive weight function r(x) on [a, b], that is∫ b

a

φn(x)φm(x)r(x) dx =

{0, m 6= n,

1, m = n.

Then with any piecewise continuous1 function f on [a, b], we can identify an orthogonal expansion

f(x) ∼∞∑n=1

cnφn(x),

where

cn =

∫ b

a

f(x)φn(x)r(x) dx.

For instance, the eigenfunctions φn(x) = sinnx for (3.5.14) gives rise to the Fourier sine seriesexpansion.

Exercise 3.8 Consider the regular Sturm-Liouville boundary value problem

y′′ + λy = 0, y(0) = 0, y′(π) = 0. (3.5.15)

Show that the eigenvalues of (3.5.15) are λn = (2n − 1)2/4, n = 1, 2, 3, . . ., with correspondingeigenfunctions

φn(x) = an sin

(2n− 1

2x

).

Example 3.8 Construct an orthonormal system of eigenfunctions corresponding to the regular Sturm-Liouville boundary value problem (3.5.15).

Solution. By Exercise 3.8 and Theorem 3.6, the eigenfunctions φn(x) = an sin

(2n− 1

2x

)are

orthogonal with respect to r(x) = 1 on [0, π]. Direct computation gives∫ π

0

a2n sin2

(2n− 1

2x

)dx = a2nπ/2.

1A function f(x) is piecewise continuous on a finite interval [a, b] if f(x) is continuous at every point in [a, b], exceptpossibly for a finite number of points at which f(x) has a jump discontinuity.

3.5. REGULAR STURM-LIOUVILLE BOUNDARY VALUE PROBLEM 41

Thus, if we take an =√

2/π, then {√2/π sin

(2n− 1

2x

)}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-Liouvilleboundary value problem in (3.5.13). Let f be a continuous function on [a, b] such that f ′ is piecewisecontinuous on [a, b] and f satisfies the boundary conditions in (3.5.13). Then

f(x) =

∞∑n=1

cnφn(x), a ≤ x ≤ b, (3.5.16)

where

cn =

∫ b

a

f(x)φn(x)r(x) dx.

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{√2/π sin

(2n− 1

2x

)}.

Solution. Referring to the problem (3.5.15), we compute

cn =

∫ π

0

f(x)√

2/π sin

(2n− 1

2x

)dx =

27/2

π3/2(2n− 1)2sin(nπ

2− π

4

).

Thus

f(x) =

∞∑n=1

cn√

2/π sin

(2n− 1

2x

). (3.5.17)

Since f(0) = 0, f ′(π) = 0, the function f satisfies the boundary conditions in (3.5.15). Moreover,f is continuous and f ′ 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 suchas the heat equation in mathematical physics. The eigenfunction expansion can be used to solve thenonhomogeneous regular Sturm-Liouville boundary value problem. For further discussion on theSturm-Liouville boundary value problem, see [4].


3.6 Nonhomogeneous Boundary Value Problem

We only consider the nonhomogeneous regular Sturm-Liouville boundary value problem with ho-mogeneous boundary conditions. Let

L[y] = f(x), a < x < b;

a1y(a) + a2y′(a) = 0,

b1y(b) + b2y′(b) = 0,

(3.6.18)

where L[y] = (p(x)y′)′+q(x)y, p(x), p′(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].

LetL[y] = 0, a < x < b;

a1y(a) + a2y′(a) = 0,

b1y(b) + b2y′(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 homo-geneous 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 anontrivial solution u. Thus u 6≡ 0 on [a, b]. Then z + u is also a solution to (3.6.18). But z + u 6≡ zon [a, b], contradicting the fact that z is the unique solution to (3.6.18). Thus (3.6.19) has only thetrivial 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 asolution 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. Firstlet y1 and y2 be nontrivial solutions to the equation L[y] = 0 satisfying respectively

a1y1(a) + a2y′1(a) = 0, (3.6.20)

andb1y2(b) + b2y

′2(b) = 0. (3.6.21)

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

y′′ +p′

py′ +

q

py =

f

p.

3.6. NONHOMOGENEOUS BOUNDARY VALUE PROBLEM 43

By the method of variation of parameters, we have

y = u1y1 + u2y2,

where

u1 = −∫ x

a

y2(t)

W (t)

f(t)

p(t)dt, and u2 =

∫ x

a

y1(t)

W (t)

f(t)

p(t)dt.

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

u1 =

∫ b

x

y2(t)

W (t)

f(t)

p(t)dt, (3.6.22)

andu2 =

∫ x

a

y1(t)

W (t)

f(t)

p(t)dt. (3.6.23)

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

y(x) = y1(x)

∫ b

x

y2(t)

W (t)

f(t)

p(t)dt+ y2(x)

∫ x

a

y1(t)

W (t)

f(t)

p(t)dt.

=

∫ b

a

G(x, t)f(t) dt,

(3.6.24)

where

G(x, t) =

y1(t)y2(x)

W (t)p(t), a ≤ t ≤ x

y1(x)y2(t)

W (t)p(t), x ≤ t ≤ b.

(3.6.25)

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

G(x, t) =

{y1(t)y2(x)/C, a ≤ t ≤ xy1(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

y(x) =

∫ b

a

G(x, t)f(t) dt, is a solution to (3.6.18).

y′(x) =d

dx

(y1(x)

∫ b

x

y2(t)f(t)

Cdt

)+

d

dx

(y2(x)

∫ x

a

y1(t)f(t)

Cdt

)= y′1(x)

∫ b

x

y2(t)f(t)

Cdt− y1(x)y2(x)f(x)

C+ y′2(x)

∫ x

a

y1(t)f(t)

Cdt+

y1(x)y2(x)f(x)

C

= y′1(x)

∫ b

x

y2(t)f(t)

Cdt+ y′2(x)

∫ x

a

y1(t)f(t)

Cdt

(3.6.27)Thus

a1y(a) + a2y′(a) = a1y1(a)

∫ b

a

y2(t)f(t)

Cdt+ a2y

′1(a)

∫ b

a

y2(t)f(t)

Cdt

= (a1y1(a) + a2y′1(a))

∫ b

a

y2(t)f(t)

Cdt = 0,

(3.6.28)


since y1 satisfies (3.6.20). Similarly using (3.6.21), one can verify that

b1y(b) + b2y′(b) = (b1y2(b) + b2y

′2(b))

∫ b

a

y1(t)f(t)

Cdt = 0. (3.6.29)

Example 3.10 Find the Green’s function G(x, t) for the boundary value problem

y′′ = 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′′ = 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 withW (x) = −x− (π − x) = −π. We now compute C = p(x)W (x) = (1)(−π) = −π. Therefore theGreen’s function is

G(x, t) =

{y1(t)y2(x)/C, 0 ≤ t ≤ xy1(x)y2(t)/C, x ≤ t ≤ π

=

{t(x− π)/π, 0 ≤ t ≤ xx(t− π)/π, x ≤ t ≤ π.

(3.6.30)

Hence for f(x) = −6x, the solution is given by

y(x) =∫ π

0

G(x, t)f(t) dt =

∫ x

0

(t(x− π)/π) (−6t) dt+

∫ π

x

(x(t− π)/π) (−6t) dt

= x(π2 − x2).

Since the homogeneous problem y′′ = 0, y(0) = 0, y(π) = 0 has only the trivial solution, theabove solution to the nonhomogeneous problem y′′ = −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 forevery solution y(x) of the homogeneous problem (3.6.19),∫ b

a

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

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

a

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

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.

(Case 2) The problem (3.6.19) has a nontrivial solution y1. By hypothesis,∫ bay1(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

that y(x) =

∫ b

a

G(x, t)f(t) dt is a solution to (3.6.18). For the boundary conditions, we have

3.6. NONHOMOGENEOUS BOUNDARY VALUE PROBLEM 45

a1y(a) + a2y′(a) = (a1y1(a) + a2y

′1(a))

∫ b

a

y2(t)f(t)

Cdt = 0, (3.6.31)

because y1 satisfies (3.6.20), and

b1y(b) + b2y′(b) = (b1y2(b) + b2y

′2(b))

∫ b

a

y1(t)f(t)

Cdt = 0, (3.6.32)

because∫ bay1(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 boundaryconditions of (3.6.19). In particular at the point a, we have a1y1(a) + a2y

′1(a) = 0 and a1z(a) +

a2z′(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-parameterfamily of solutions.

Example 3.11 Show that the boundary value problem

y′′ + y′ +5

2y = f ; y(0) = 0, y(

2π

3) = 0

has a solution if and only if ∫ 2π3

0

ex2 sin (

3x

2)f(x) dx = 0.

Solution. Consider the homogeneous boundary value problem y′′+y′+ 52y = 0; y(0) = 0, y( 2π

3 ) =

0. The general solution of the equation y′′ + y′ + 52y = 0 is y = e−

12x(A cos ( 3x

2 ) + B sin ( 3x2 )).

As y(0) = 0, we deduce that A = 0. Thus y = Be−12x sin ( 3x

2 ). This also satisfies y( 2π3 ) = 0.

Thus the homogeneous boundary value problem has a one parameter family of solutions. Note thatthe given equation is not self-adjoint. However it can be converted into a self-adjoint equation bymultiplying throughout by a factor ex. Thus we may write the problem as

(exy′)′ +5

2exy = exf(x); y(0) = 0, y(

2π

3) = 0.

By Theorem 3.10, this problem has a solution if and only if

∫ 2π3

0

(Be−12x sin (

3x

2))(exf(x)) dx = 0.

This is equivalent to the condition∫ 2π3

0

ex2 sin (

3x

2)f(x) dx = 0.

Exercise 3.9 Using Green’s function, solve the boundary value problem

(xy′)′ − (4/x)y = −x; y(1) = 0, y(2) = 0.

Ans: y = 160 (16x2 ln 2− 16x−2 ln 2− 15x2 lnx).


Exercise 3.10 Show that any equation p0(x)y′′ + p1(x)y′ + p2(x)y = 0 can be made self-adjointby multiplying throughout by 1

p0e∫p1/p0 dx.

Exercise 3.11 Consider the regular Sturm-Liouville boundary value problem

y′′ − 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′′ + P (x)y′ +Q(x)y = 0 (3.7.33)

It is rarely possible to solve this equation in general. However by studying the properties of thecoefficient functions, it is sometimes possible to describe the behavior of the solutions. One ofthe essential characteristics that is of primary interest to mathematicians is the number of zeros ofa solution to (3.7.33). If a function has an infinite number of zeros in an interval [a,∞), we saythat the function is oscillatory. Therefore, studying the oscillatory behavior of a function meansinvestigating the number and locations of its zeros.

The familiar equation y′′ + y = 0 has two linearly independent solutions s(x) = sin(x) satisfyingy(0) = 0, y′(0) = 1, and c(x) = cos(x) satisfying y(0) = 1, y′(0) = 0 respectively. The positivezeros of s(x) and c(x) are π, 2π, 3π, . . . , and π/2, 3π/2, 5π/2, . . ., respectively. Note that the zerosof s(x) and c(x) interlace each other in the sense that between two successive zeros of s(x), thereis 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 so-lutions of

y′′ + P (x)y′ +Q(x)y = 0,

then the zeros of these functions are distinct and occur alternatively in the sense that y1(x) vanishesexactly 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 linearlyindependent, their Wronskian W (y1, y2)(x) = y1(x)y′2(x) − y′1(x)y2(x) never vanish. It followsthat 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 functionf(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, thereexists z ∈ (x1, x2) such that f ′(z) = 0. Computing f ′(z), we find that

0 = f ′(z) =y′1(z)y2(z)− y1(z)y′2(z)

[y2(z)]2= −W (y1, y2)(z)

[y2(z)]2.

3.7. OSCILLATIONS AND THE STURM SEPARATION THEOREM 47

But this meansW (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, thereis a zero of y1. Consequently, there cannot be more than one zero of y2 between two consecutivezeros of y1. �

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

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

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

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

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

• •x1 x2

y1 y2

Figure 3.1: Interlacing of zeros of two linearly independent solutions

Corollary 3.12 Suppose one nontrivial solution to y′′ + P (x)y′ + 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 Sturmseparation theorem, if y2 is any other solution such that y1 and y2 are linearly independent, theny2 must have a zero between each consecutive pair of zeros of y1, and hence it must also have aninfinite number of zeros on [a,∞). If y1 and y2 are linearly dependent, then y2 = λy1 since y1 isnontrivial. Thus y2 also has an infinite number of zeros on [a,∞). �

Theorem 3.13 Let y be a nontrivial solution of y′′ + P (x)y′ + 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 Bolzano-Weierstrass theorem, there exist in [a, b] a point x0, and a sequence of zeros xn not equal to x0 suchthat lim

n→∞xn = x0. Since y is continuous and differentiable at x0, we have

y(x0) = limn→∞

y(xn) = 0

and

y′(x0) = limn→∞

y(xn)− y(x0)

xn − x0= 0.

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


3.8 Sturm Comparison Theorem

The equationy′′ + P (x)y′ +Q(x)y = 0 (3.8.34)

can be written in the formu′′ + q(x)u = 0 (3.8.35)

by putting y = uv, where v = e−12

∫P dx and q(x) = Q(x)− 1

4P (x)2 − 12P′(x). It is customary to

refer to (3.8.34) as the standard form, and to (3.8.35) as the normal form, of a homogeneous secondorder 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 ofthe 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′′ + 2x2y′ + (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. Forsolutions of two different equations, it may still be possible to relate their zeros. For example,consider the equations

y′′ +m2y = 0 and y′′ + n2y = 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. There-fore, for n > m, the distance between two zeros of y1 is greater than the distance between two zerosof y2. Thus for n2 > m2 (equivalent to n > m), between two consecutive zeros of y1, there is azero 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′′ + q1(x)y = 0, a < x < b, (3.8.36)

and let y2 be a nontrivial solution to

y′′ + 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) withx1 < x2, then there exists a zero of y2 in (x1, x2), unless q2(x) = q1(x) on [x1, x2] in which casey1 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 linearlydependent 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 49

ddx (W (y2, y1))= d

dx (y2y′1 − y′2y1) = y′2y

′1 + y2y

′′1 − y′′2 y1 − y′2y′1

= y2y′′1 − y′′2 y1 = y2(−q1y1)− (−q2y2)y1

= y1y2(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 y′1(x1) ≥ 0 and y′1(x2) ≤ 0. Therefore,W (y2, y1)(x1) = y2(x1)y′1(x1) ≥ 0 and W (y2, y1)(x2) = y2(x2)y′1(x2) ≤ 0.

Since W (y2, y1)(x) is non-decreasing, the only way for it to be nonnegative at x1 and nonpositiveat x2 is W (y2, y1)(x) = 0 for all x in [x1, x2]. This also implies d

dxW (y2, y1)(x) = 0 in [x1, x2].That means q1(x) = q2(x) in [x1, x2]. Then y1 and y2 satisfy the same equation on [x1, x2], andtheir 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 x2or 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′′ + 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′′+0 ·y = 0 and q(x) ≤ 0, it follows from 3.14that if y has two or more zeros in [a, b], then u must have a zero between them. Since u is neverzero, y can have at most 1 zero in [a, b]. �

Example 3.12 The equation

x2y′′ + xy′ + (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[a, a+ π). The substitution y = ux−

12 transforms equation (3.8.38) into the following form.

d2u

dx2+

(1−

p2 − 14

x2

)u = 0, x > 0. (3.8.39)

Since y = ux−12 , 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 u′′ + u = 0. Observethat u(x) = A sin(x− a) is a solution of u′′ + u = 0 and has zeros at a and a+ π.

Case 1. (p > 1/2). In this case, 4p2 − 1 > 0 so that 1 − p2− 14

x2 < 1 for x in [a, a + π). By theSturm 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).

Case 2. (0 ≤ p < 1/2). In this case, 4p2 − 1 < 0 so that 1− p2− 14

x2 > 1 for x in [a, a+ π). By theSturm comparison theorem 3.14, a solution to (3.8.39) must have a zero in (a, a + π), since a anda+ π are consecutive zeros of u(x) = A sin(x− a).

Case 3. (p = 1/2). In this case, (3.8.39) reduces to u′′ + u = 0, which has the general solutionu(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.


Exercise 3.13 Show that any nontrivial solution to y′′ + (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′′ + 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 solutionto

y′′ + 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 anontrivial solution φ of the equation

y′′ + w(x)y′ + 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

y′′ + y′ +5

2y = f ; y(0) = 0, y(

2π

3) = 0,

where f(x) = e−12x sin 3x, is y = Be−

12x sin (

3x

2)− 4

27e−

12x sin 3x.

Chapter 4

Linear Differential Systems

4.1 Linear Systems

The following system is called a linear differential system of first order in normal form:x′1 = a11(t)x1 + · · ·+ a1n(t)xn + g1(t),

· · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

x′n = an1(t)x1 + · · ·+ ann(t)xn + gn(t),

where aij(t) and gj(t) are continuous functions of t and ′ denotes differentiation with respect to t.Denote

x(t) =

x1(t)

· · ·xn(t)

, g(t) =

g1(t)

· · ·gn(t)

, A(t) =

a11(t) · · · a1n(t)

· · · · · · · · ·an1(t) · · · ann(t)

.

We call g(t) a continuous vector field, or a continuous vector-valued function, if all its componentsare continuous functions. We call A(t) a continuous matrix, or a continuous matrix-valued function,if all its entries are continuous functions. Define

x′ =

x′1(t)

· · ·x′n(t)

,

∫x(t)dt =

∫x1(t)dt

· · ·∫xn(t)dt

.

Then the linear system can be written in the matrix form:

x′ = A(t)x + g(t). (4.1.1)

When g ≡ 0, (4.1.1) is reduced tox′ = A(t)x. (4.1.2)

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

51

52 CHAPTER 4. LINEAR DIFFERENTIAL SYSTEMS

Example 4.1x′1 = 2x1 + 3x2 + 3t,

x′2 = −x1 + x2 − 7 sin t

is equivalent to (x1

x2

)′=

(2 3

−1 1

)(x1

x2

)+

(3t

−7 sin t

).

Example 4.2 Given a second order systemd2x

dt2= x+ 2y + 3t,

d2y

dt2= 4x+ 5y + 6t,

it can be expressed into an equivalent first order differential system by introducing more variables.For this example, let u = x′ and v = y′. Then we have

x′ = u

u′ = x+ 2y + 3t

y′ = v

v′ = 4x+ 5y + 6t

.

Next, we begin with the initial value problem{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 containingt0. Then, for any constant vector x0, (4.1.3) has a solution x(t) defined on this interval. Thissolution 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 aunique 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) = c1x(t) + c2y(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

c1x1(t) + · · ·+ crxr(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 andonly 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

c1x1(t0) + · · ·+ crxr(t0) = 0.

Let y(t) = c1x1(t) + · · ·+ crxr(t). Then y(t) is the solution of the initial value problem

x′ = 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.c1x1(t) + · · ·+ crxr(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 thegeneral solution of (4.1.2) is given by

x(t) = c1x1(t) + · · ·+ cnxn(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 eachj from 1 to n, consider the initial value problem

x′ = 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).Fix t0 ∈ (a, b). From Lemma 4.3, x1(t0), · · · , xn(t0) are linearly independent vectors. Let x(t)

be any solution of (4.1.2). Then x(t0) can be represented by a linear combination of x1(t0), · · · ,xn(t0), namely, there exist n constants c1, · · · , cn such that

x(t0) = c1x1(t0) + · · ·+ cnxn(t0).


As in the proof of Lemma 4.3, we can show that

x(t) = c1x1(t) + · · ·+ cnxn(t).

Thus c1x1(t) + · · ·+ cnxn(t) is the general solution of (4.1.2). �

Recall that, n vectors

a1 =

a11

· · ·an1

, · · · · · · , an =

a1n

· · ·ann

are linearly dependent if and only if the determinant∣∣∣∣∣∣∣

a11 · · · a1n

· · · · · · · · ·an1 · · · 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

x1(t) =

x11(t)

· · ·xn1(t)

, · · · , xn(t) =

x1n(t)

· · ·xnn(t)

is the determinant

W (t) ≡W (x1, · · · ,xn)(t) =

∣∣∣∣∣∣∣∣∣x11(t) x12(t) · · · x1n(t)

x21(t) x22(t) · · · x2n(t)

· · · · · · · · · · · ·xn1(t) xn2(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 zeroin (a, b).(ii) n solutions of (4.1.2) are linearly dependent in (a, b) if and only if their Wronskian is identicallyzero in (a, b).

Definition 4.3 A set of n linearly independent solutions of (4.1.2) is called a fundamental set ofsolutions, or a basis of solutions. Let

x1(t) =

x11(t)

· · ·xn1(t)

, · · · , xn(t) =

x1n(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) · · · x1n(t)

x21(t) x22(t) · · · x2n(t)

· · · · · · · · · · · ·xn1(t) xn2(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:

Φ′ = 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)

where c =

c1

· · ·cn

is an arbitrary constant vector.

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 theassociated 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 y′ = A(t)y. From (4.2.3), thereexists a constant vector c such that y(t) = Φ(t)c. So x(t) = Φ(t)c + xp(t). Thus (4.3.1) gives ageneral solution of (4.1.1). �


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

x(t) = Φ(t)u(t), u(t) =

u1(t)

· · ·un(t)

.

Plugging into (4.1.1) we get

Φ′u + Φu′ = AΦu + g.

From (4.2.2), Φ′ = AΦ. So Φu′ = g, and thus u′ = Φ−1g.

u(t) =

∫ t

t0

Φ−1(s)g(s)ds. (4.3.2)

So we obtain the following:

Theorem 4.8 The general solution of (4.1.1) is given by

x(t) = Φ(t)c + Φ(t)

∫ t

t0

Φ−1(s)g(s)ds, (4.3.3)

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

Exercise 4.1 Solve

{x′1 = 3x1 − x2 + t

x′2 = 2x1 + t.

Ans: x1 = c1et + c2e

2t − t2 −

14 , x2 = 2c1e

t + c2e2t − t

2 −14 .

4.4 Homogeneous Linear Systems with Constant Coefficients

Consider a homogeneous linear system

x′ = 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) = eλtk, 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 aneigenvalue 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 a12 · · · a1n

a21 a22 − λ · · · a2n

· · · · · · · · · · · ·an1 an2 · · · 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 thereexist 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 oflinearly independent eigenvectors of A associated with λj is called the geometric multiplicity of theeigenvalue 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 thatλj 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) = eλtk

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 withλ, then

x1 = <(eλtk), x2 = =(eλtk)

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 eigenval-ues λ1, · · · , λn respectively, then

Φ(t) = (eλ1tk1, · · · , eλntkn)


is a fundamental matrix of (4.4.1), and the general solution is given by

x(t) = Φ(t)c = c1eλ1tk1 + · · ·+ cne

λntkn, (4.4.4)

where c =

c1

· · ·cn

is an arbitrary constant vector.

Proof. We only need to show det Φ(t) 6= 0. Since k1, · · · , kn are linearly independent, sodet Φ(0) 6= 0. From Theorem 4.5 we see that det Φ(t) 6= 0 for any t. Hence Φ(t) is a funda-mental matrix. If one of the λ’s is complex, then replace eλtk and eλtk by <eλtk and =eλtk. Theresulting set of solutions is a linearly independent set over R. �

Remark. Under the conditions of Theorem 4.11, the eigenvalues λ1, · · · , λn of A need not to bedistinct. In fact we only assume that all the eigenvalues of A are quasi-simple.If A has n distinct eigenvalues λ1, · · · , λn, and let k1, · · · , kn be the associated eigenvectors, thenthey are linearly independent. Hence the general solution is given by (4.4.4).

Example 4.3 Solve the system x′ =

(−3 1

1 −3

)x.

Solution. The matrix A =

(−3 1

1 −3

)has eigenvalues λ1 = −2, and λ2 = −4.

For λ1 = −2 we find an eigenvector k1 =

(1

1

).


(1

−1

).


x(t) = c1e−2t

(1

1

)+ c2e

−4t

(1

−1

).

Example 4.4 Solve the system

x′ =

(−3 1

1 −3

)x + e−2t

(−6

2

).

Solution. We first solve the associated homogeneous system

x′ =

(−3 1

1 −3

)x

and find two linearly independent solutions x1(t) =

(e−2t

e−2t

), x2(t) =

(e−4t

−e−4t

). The fun-

damental matrix is

Φ = (x1(t),x2(t)) =

(e−2t e−4t

e−2t −e−4t

).


Φ−1 =1

−2e−6t

(−e−4t −e−4t

−e−2t e−2t

)=

1

2

(e2t e2t

e4t −e4t

),

Let g(t) = e−2t

(−6

2

). Then

Φ−1(t)g(t) =

(−2

−4e2t

),

u(t) =

∫ t

0

Φ−1(s)g(s)ds =

(−2t

−2e2t + 2

),

Φ(t)u(t) = 2e−2t

(−t− 1

−t+ 1

)+ 2e−4t

(1

−1

).

The general solution is

x = c1e−2t

(1

1

)+ c2e

−4t

(1

−1

)− 2te−2t

(1

1

)+ 2e−2t

(−1

1

).

Example 4.5 Solve x′ =

(0 1

−4 0

)x.


(0 1

−4 0

)has eigenvalues ±2i.

For λ = 2i, we find an eigenvector k =

(1

2i

).

e2it

(1

2i

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

(1

2i

)=

(cos 2t

−2 sin 2t

)+ i

(sin 2t

2 cos 2t

).


x(t) = c1

(cos 2t

−2 sin 2t

)+ c2

(sin 2t

2 cos 2t

).

Example 4.6 Solve x′ = −3x+ 4y − 2z,

y′ = x+ z,

z′ = 6x− 6y + 5z.


−3 4 −2

1 0 1

6 −6 5

has eigenvalues λ1 = 2, λ2 = 1, λ3 = −1.


For λ1 = 2 we find an eigenvector k1 =

0

1

2

.

For λ2 = 1 we find an eigenvector k2 =

1

1

0

.


1

0

−1

.

The general solution is given by x

y

z

= c1e2t

0

1

2

+ c2et

1

1

0

+ c3e−t

1

0

−1

,

namelyx(t) = c2e

t + c3e−t,

y(t) = c1e2t + c2e

t,

z(t) = 2c1e2t − c3e−t.

Example 4.7 Solve x′ = Ax, where

A =

2 1 0

1 3 −1

−1 2 3

.

Solution. The matrix A has eigenvalues λ1 = 2, λ2 = λ3 = 3± i.

For λ1 = 2, we find an eigenvector k1 =

1

0

1

.

For λ2 = 3 + i, we find an eigenvector k2 =

1

1 + i

2− i

.

We have

e(3+i)tk2 = e3t

cos t+ i sin t

cos t− sin t+ i(cos t+ sin t)

2 cos t+ sin t+ i(2 sin t− cos t)

,

<(e(3+i)tk2) = e3t

cos t

cos t− sin t

2 cos t+ sin t

=(e(3+i)tk2) = e3t

sin t

cos t+ sin t

2 sin t− cos t

.



x(t) = c1e2t

1

0

1

+ c2e3t

cos t

cos t− sin t

2 cos t+ sin t

+ c3e3t

sin t

cos t+ sin t

2 sin t− cos t

.


A =

1 −2 2

−2 1 2

2 2 1

.

Solution. We havedet(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 algebraicmultiplicity 2, and λ = −3 is a simple eigenvalue).For λ = 3 we solve the equation Ak = 3k, namely −2 −2 2

−2 −2 2

2 2 −2

k1

k2

k3

= 0.

The solution is k =

v − uu

v

. So we find two eigenvectors k1 =

1

0

1

and k2 =

−1

1

0

.


1

1

−1

.


x(t) = c1e3t

1

0

1

+ c2e3t

−1

1

0

+ c3e−3t

1

1

−1

.

Now we consider the solutions of (4.4.1) associated with a multiple eigenvalue λ, with geometricmultiplicity 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 multiplicitym > 1. Then the followingsystem

(A− λI)mv = 0 (4.4.5)

has exactly m linearly independent solutions.

By direct computations we can prove the following theorem.


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, l = 1, 2, · · · ,m− 1, (4.4.6)

and let

x(t) = eλt[v0 + tv1 +

t2

2v2 + · · ·+ tm−1

(m− 1)!vm−1

]. (4.4.7)

Then x(t) is a solution of (4.4.1).Let v

(1)0 , · · · , v

(m)0 be m linearly independent solutions of (4.4.5). They generate m linearly inde-

pendent 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 multiplicitym, we firstsolve (4.4.5) and find m linearly independent solutions

v(1)0 , v

(2)0 , · · · , v

(m)0 .

For each of these vectors, say v(k)0 , we compute the iteration sequence

v(k)l = (A− λI)v

(k)l−1, l = 1, 2, · · ·

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

v(k)j 6= 0, (A− λI)v

(k)j = 0.

Thus vj is an eigenvector of A associated with the eigenvalue λ. Then the iteration stops and yieldsa solution

x(k)(t) = eλt[v(k)0 + tv

(k)1 +

t2

2v(k)2 + · · ·+ tj

j!v(k)j

]. (4.4.8)


A =

−1 1 0

0 −1 4

1 0 −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

(A + 3I)2v =

4 4 4

4 4 4

1 1 1

v = 0,


and find two linearly independent solutions v(1)0 =

1

0

−1

, v(2)0 =

0

1

−1

. Plugging v(1)0 ,

v(2)0 into (4.4.6), (4.4.7) we get

v(1)1 = (A + 3I)v

(1)0 =

2

−4

2

,

x(1) = e−3t(v(1)0 + tv

(1)1 ) = e−3t

1

0

−1

+ t

2

−4

2

,

v(2)1 = (A + 3I)v

(2)0 =

1

−2

1

,

x(2) = e−3t(v(2)0 + tv

(2)1 ) = e−3t

0

1

−1

+ t

1

−2

1

.

For the simple eigenvalue λ2 = 0 we find an eigenvector k3 =

4

4

1

.

So the general solution is

x(t) = c1x(1) + c2x

(2) + c3k3

= c1e−3t

1

0

−1

+ t

2

−4

2

+ c2e

−3t

0

1

−1

+ t

1

−2

1

+ c3

4

4

1

.

Example 4.10 Solve the system x′ = 2x+ y + 2z,

y′ = −x+ 4y + 2z,

z′ = 3z.

Solution. Let

A =

2 1 2

−1 4 2

0 0 3

.

The eigenvalue of A is 3 with multiplicity 3. Solving the linear system:

(A− 3I)3v =

0 0 0

0 0 0

0 0 0

v = 0,


we obtain 3 obvious linearly independent solutions

v(1)0 =

1

0

0

, v(2)0 =

0

1

0

, v(3)0 =

0

0

1

.

Plugging v(j)0 into (4.4.6), (4.4.7) we get

v(1)1 = (A− 3I)v

(1)0 =

−1

−1

0

,

v(1)2 = (A− 3I)v

(1)1 =

0

0

0

,

x(1) = e3t(v(1)0 + tv

(1)1 ) = e3t

1

0

0

+ t

−1

−1

0

;

v(2)1 = (A− 3I)v

(2)0 =

1

1

0

,

v(2)2 = (A− 3I)v

(2)1 =

0

0

0

,

x(2) = e3t(v(2)0 + tv

(2)1 ) = e3t

0

1

0

+ t

1

1

0

;

v(3)1 = (A− 3I)v

(3)0 =

2

2

0

,

v(3)2 = (A− 3I)v

(3)1 =

0

0

0

,

x(3) = e3t(v(3)0 + tv

(3)1 ) = e3t

0

0

1

+ t

2

2

0

.



x(t) = c1x(1) + c2x

(2) + c3x(3)

= c1e3t

1

0

0

+ t

−1

−1

0

+ c2e

3t

0

1

0

+ t

1

1

0

+ c3e3t

0

0

1

+ t

2

2

0

.

Remark. It is possible to reduce the number of constant vectors in the general solution of x′ = Ax

by using a basis for the Jordan canonical form of A. For details of the Jordan canonical form, seeappendix 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)rv0 = 0 but (A − λI)r−1v0 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−1v0,

form a chain of linearly independent solutions of (4.4.5) with vr−1 being the base eigenvector asso-ciated with the eigenvalue λ. This gives r independent solutions of x′ = Ax.

x1(t) = eλt(v0 + tv1 + · · ·+ tr−1

(r−1)!vr−1).

···

xr−1(t) = eλt(vr−2 + tvr−1),

xr(t) = eλtvr−1.

Repeat this procedure by finding another v which is not in the span of the previous chains of vectorsand 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, thenthe union of these vectors is a linearly independent set of vectors (whether the two base eigenvectorsare associated with different eigenvalues or with the same eigenvalue).

Example 4.11 Solve x′ = Ax, where A =

3 0 0 0

0 3 0 0

0 1 3 0

0 0 1 3

.


Solution. A has an eigenvalue λ = 3 of multiplicity 4. Direct calculation gives (A − 3I) =0 0 0 0

0 0 0 0

0 1 0 0

0 0 1 0

, (A− 3I)2 =

0 0 0 0

0 0 0 0

0 0 0 0

0 1 0 0

, (A− 3I)3 = 0, and (A− 3I)4 = 0.

It can be seen that v1 =

1

0

0

0

and v4 =

0

0

0

1

are two linearly independent eigenvectors of A.

Together with v2 =

0

1

0

0

and v3 =

0

0

1

0

, they form a basis of {v | (A− 3I)4v = 0} = R4.

Note that (A−3I)v2 = v3, and (A−3I)v3 = v4. Hence {v2,v3,v4} forms a chain of generalizedeigenvectors associated with the eigenvalue 3. {v1} alone is another chain. Therefore the generalsolution is

x(t) = e3t(c1v1 + c2(v2 + tv3 +

t2

2v4) + c3(v3 + tv4) + c4v4

).

That is

x(t) =

c1e

3t

c2e3t

(c2t+ c3)e3t

( c2t2

2 + c3t+ c4)e3t

.

Exercise 4.2 Solve x′ = Ax, where

A =

7 5 −3 2

0 1 0 0

12 10 −5 4

−4 −4 2 −1

.

Ans: x(t) = c1e−t

1

0

2

−1

+ c2et

1

−2

0

2

+ c3et

−1

0

−2

0

+ c4et

−t− 1

1

−2t

0

.

4.5 Higher Order Linear Equations

Consider the n-th order linear equation

y(n) + a1(t)y(n−1) + · · ·+ an−1(t)y′ + an(t)y = f(t), (4.5.1)

4.5. HIGHER ORDER LINEAR EQUATIONS 67

where y(k) = dkydtk

. 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′ + 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 inthe previous sections to the equivalent system.Recall the initial value problem

y(n) + a1(t)y(n−1) + · · ·+ an(t)y = f(t),

y(t0) = y0,

y′(t0) = y1,

· · · · · · · · ·

y(n−1)(t0) = yn−1.

(4.5.3)

If we let x1 = y, x2 = y′, x3 = y′′,..., xn−1 = y(n−2), xn = y(n−1), then (4.5.3) is equivalent tothe following linear system:

x′1 = x2

x′2 = x3

x′3 = x4... ... ...

x′n−1 = xn

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

That isx1

x2

x3...xn

′

=

0 1 · · · 0 0

0 0 · · · 0 0

· · · · · · · · · · · · · · ·0 0 · · · 0 1

−an(t) −an−1(t) · · · −a2(t) −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 originalnth order linear ODE (4.5.1). In fact a solution of this system gives rise to a vector-valued functionwhere the ith component is the (i− 1)th derivative of a solution of (4.5.1). That is


x1

x2

x3...xn

=

y

y′

y′′

...y(n−1)

.

Remarks.1. The existence and uniqueness theorem for first order differential systems implies the existenceand uniqueness theorem for higher order linear ODEs.2. The linear structure of the solution space of the homogeneous linear system translates to the linearstructure of the solution space of the homogeneous higher order linear ODE.

3. If x1(t) =

y1(t)

y′1(t)...

y(n−1)1 (t)

, . . . ,xn(t) =

yn(t)

y′n(t)...

y(n−1)n (t)

are n linearly independent solutions

of the equivalent linear system of (4.5.2), then the first components y1(t), . . . , yn(t) of these vector-valued 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 corre-sponding 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. Considera second order equation

y′′ + p(t)y′ + q(t)y = f(t). (4.5.7)

Let x1 = y, x2 = y′, x =

(x1

x2

). Then (4.5.7) is written as

x′ =

(0 1

−q −p

)x +

(0

f

)(4.5.8)

Assume y1(t) and y2(t) are two linearly independent solutions of the associated homogeneous equa-tion

y′′ + p(x)y′ + q(x)y = 0.

Eventually, we look for a solution of (4.5.7) in the form

y = u1y1 + u2y2.

Choose a fundamental matrix Φ(t) =

(y1 y2

y′1 y′2

). By the method of variation of parameters (the

matrix version), the corresponding solution of (4.5.8) is in the form x = Φu. That is

x =

(y

y′

)=

(y1 y2

y′1 y′2

)(u1

u2

)=

(y1u1 + y2u2

y′1u1 + y′2u2

)(4.5.9)

4.5. HIGHER ORDER LINEAR EQUATIONS 69

Now

Φ(t)−1 =1

W (t)

(y′2 −y2−y′1 y1

),

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

u2

)=

∫1

W (t)

(y′2 −y2−y′1 y1

)(0

f

)dt.

Thus

u1(t) = −∫

y2(t)

W (t)f(t)dt, u2(t) =

∫y1(t)

W (t)f(t)dt. (4.5.10)

Note that, (4.5.9) implies

y′1u1 + y′2u2 = y′ = y′1u1 + y′2u2 + y1u′1 + y2u

′2.

Hence y1u′1 + y2u′2 = 0. Plugging y′ = y′1u1 + y′2u2 into (4.5.7) we find y′1u

′1 + y′2u

′2 = f. Solving

these two equations, we find that

u′1 = − y2Wf, u′2 =

y1Wf.

Again we get (4.5.10).

Linear equations with constant coefficients.Now we consider linear equations with constant coefficients

y(n) + a1y(n−1) + · · ·+ an−1y

′ + any = f(t), (4.5.11)

and the associated homogeneous equation

y(n) + a1y(n−1) + · · ·+ an−1y

′ + any = 0, (4.5.12)

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

x′ = Ax,

where

A =

0 1 · · · 0 0

0 0 · · · 0 0

· · · · · · · · · · · · · · ·0 0 · · · 0 1

−an −an−1 · · · −a2 −a1

.

The eigenvalues of A are roots of the equation

det(λI−A) = λn + a1λn−1 + · · ·+ an−1λ+ 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 constantcoefficients.


Theorem 4.14 Let λ1, · · · , λs be the distinct eigenvalues for (4.5.12), with multiplicitym1, · · · , ms

respectively. Then (4.5.12) has a fundamental set of solutions

eλ1t, teλ1t, · · · , tm1−1eλ1t;

· · · · · · · · · · · · · · · · · · · · · · · · ;

eλst, teλst, · · · , tms−1eλst.

(4.5.14)

4.6 Introduction to the Phase Plane

Consider a system of two first order equations

x′(t) = f(x, y)

y′(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 ifthe 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 thephase plane.

Let F(x, y) = 〈f(x, y), g(x, y)〉 be the vector field on the xy-plane defined by f and g. If we plotthe unit vectors defined by those nonzero F(x, y) at various points of the phase plane, we get the socalled direction field of (4.6.1). For any point p(t) = 〈x(t), y(t)〉 on an integral curve of (4.6.1), wehave

p′(t) = 〈x′(t), y′(t)〉 = 〈f(x, y), g(x, y)〉 = 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

dx=g(x, y)

f(x, y).

It is called the phase plane equation. Thus any integral curve of (4.6.1) satisfies the phase planeequation.

Exercise 4.3 Solve the system x′(t) = 1, y′(t) = 3x2 + x− 1. Sketch a few integral curves of thesystem.

4.7 Linear Autonomous System in the Plane

A linear autonomous system in the plane has the form

x′(t) = a11x+ a12y + b1

y′(t) = a21x+ a22y + b2

4.7. LINEAR AUTONOMOUS SYSTEM IN THE PLANE 71

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Figure 4.1: Direction field and an integral curve of dxdt = f(x, y), dydt = g(x, y)

where aij’s and bj’s are constants. We assume a11a22 − a21a12 6= 0. By a simple translation thesystem can always be put in form

x′(t) = ax+ by

y′(t) = cx+ dy,(4.7.2)

where ad − bc 6= 0. The point (0, 0) is called a critical point of (4.7.2) since both ax + by andcx + dy are zero at x = 0, y = 0. (In general a critical point of (4.6.1) is a solution of the systemof equations f(x, y) = 0, g(x, y) = 0.) Corresponding to the critical point (0, 0) of (4.7.2), wehave x(t) = 0, y(t) = 0 for all t is a constant solution of (4.7.2). Our interest is to investigate thebehavior 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 andr2 are the roots of the quadratic equation

X2 − (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) = c1A1er1t + c2A2e

r2t

y(t) = c1B1er1t + c2B2e

r2t,(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) = c1A1e

r1t, y(t) = c1B1er1t. For any c1 > 0, it gives

the parametric equation of a half-line A1y = B1x with slope B1/A1. As t → ∞, the point onthis half-line approaches the origin (0, 0). For any c1 < 0, it represents the other half of the lineA1y = B1x. As t→∞, the point on this half-line also approaches the origin (0, 0).When c1 = 0, we have the solution x(t) = c2A2e

r2t, y(t) = c2B2er2t. In exactly the same way,

it represents two half-lines lying on the line A2y = B2x with slope B2/A2. The two half-linesapproach (0, 0) as t→∞.


The two lines A1y = B1x and A2y = B2x are called the transformed axes, usually denoted by xand y on the phase plane.If c1 6= 0 and c2 6= 0, the general solution (4.7.4) are parametric equations of some curves. Sincer1 < 0 and r2 < 0, these curves approach (0, 0) as t→∞. Furthermore,

y

x=c1B1e

r1t + c2B2er2t

c1A1er1t + c2A2er2t=

(c1B1/c2)e(r1−r2)t +B2

(c1A1/c2)e(r1−r2)t +A2.

As r1 − r2 < 0, we see that yx →B2

A2as t→∞ so that all the curves enter (0, 0) with slope B2

A2.

A critical point is called a node if it is approached and entered (with a well-defined tangent line) byeach 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 + y2 = r2

for some t = t0 remains inside the circle x2 + y2 = R2 for all t > t0. Roughly speaking, a criticalpoint is stable if all integral curves that get sufficiently close to the point stay close to the point. Ifthe critical point is not stable, it is called unstable. A critical point is said to be asymptotically stableif it is stable and there exists a circle x2 + y2 = r20 such that every integral curve which is inside thiscircle for some t = t0 approaches the critical point as t → ∞. A node is said to be proper if everydirection 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). Thepoint (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-linesolutions x(t) = c1A1e

r1t, y(t) = c1B1er1t (for c1 > 0 and c1 < 0) still approach and enter (0, 0)

as t → ∞, but the other two half-line solutions x(t) = c2A2er2t, y(t) = c2B2e

r2t approach andenter (0, 0) as t → −∞. If c1 6= 0 and c2 6= 0, the general solution (4.7.4) defines integral curvesof (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-linesof the line A2y = B2x; whereas as t→ −∞, each of these curves is asymptotic to one of the half-lines of the line A1y = B1x. In this case, the critical point is called a saddle point and is certainlyunstable. 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) = c1ert, y(t) = c2e

rt,

where c1 and c2 are arbitrary constants. The integral curves lie on the lines c1y = c2x. 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 eithersituation 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.


Now let’s discuss the case where a factor of t times ert is present. Suppose r < 0. The generalsolution can be written in the form

x(t) = c1Aert + c2(A1 +At)ert

y(t) = c1Bert + 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 obtainthe solutions x(t) = c1Ae

rt, y(t) = c1Bert. These are solutions representing the two half-lines (for

c1 > 0 and c1 < 0) lying on the line y 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

x=c1Be

rt + c2(B1 +Bt)ert

c1Aert + c2(A1 +At)ert=c1B/c2 +B1 +Bt

c1A/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 y 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 thecritical point is unstable.

Exercise 4.4 Find and classify the critical point of the system x′ = 2x+ y+ 3, y′ = −3x− 2y− 4.

Case 4. If r1, r2 are conjugate complex but not purely imaginary, then the critical point (0, 0) is aspiral.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 cosβt−A2 sinβt) + c2(A1 sinβt+A2 cosβt)],

y(t) = eαt[c1(B1 cosβt−B2 sinβt) + c2(B1 sinβt+B2 cosβt)],(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 integralcurves 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 theangular coordinate θ = tan−1(y/x) is always strictly increasing or always strictly decreasing. Thatis dθ/dt > 0 for all t > 0, or dθ/dt < 0 for all t > 0. Differentiating θ = tan−1(y/x), we have

dθ

dt=xdy/dt− ydx/dt

x2 + y2.

Using (4.7.2), we getdθ

dt=cx2 + (d− a)xy − by2

x2 + y2. (4.7.8)


Since we are interested in solutions that represent integral curves, we assume x2 + y2 6= 0. Now(4.7.6) implies that b and c have opposite signs. Let’s consider the case b < 0 and c > 0. Wheny = 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 havecx2 + (d − a)xy − by2 = 0 or c(x/y)2 + (d − a)(x/y) − b = 0, for some real number x/y. Butthis contradicts the fact that its discriminant also given by (4.7.6) is negative. Thus we conclude thatdθ/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 isa 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 → −∞ andthe 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 thesystem x′ = x− 4y, y′ = 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 periodicfunctions of period 2π so that each integral curve is a closed path surrounding the origin. Thesecurves can be shown to be ellipses by solving the phase plane equation

dy

dx=cx+ dy

ax+ by.

See figure 4.3(vi).

Exercise 4.6 Show that integral curves of the system x′ = −βy, y′ = 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 x′ = ax + by, y′ =

cx + dy, where a, b, c, d are real and ad − bc 6= 0. Let r1, r2 be the roots of the characteristicequation X2 − (a + d)X + (ad − bc) = 0. The stability of the origin and the classification of theorigin as a critical point depends on the roots r1, r2 as follows.

Roots Type of critical point Stabilitydistinct, positive improper node unstabledistinct, negative improper node asymptotically stableopposite signs saddle point unstableequal, positive proper or improper node unstableequal, negative proper or improper node asymptotically stablecomplex value with positive real part spiral point unstablecomplex value with negative real part spiral point asymptotically stablepurely imaginary centre stable


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

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

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

(iv) Equal roots r with a factor of t, r < 0 stable. Tra-jectories are tangent to y at the origin, and parallel to ywhen t approaches±∞.

Figure 4.2: Improper node and saddle point


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

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

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(iii) Spiral curve corresponding to complex roots, realpart of the root α < 0, asymptotically stable, c > 0

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

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(v) Integral curves for x′ = x− 4y, y′ = 4x+ y(vi) A stable center corresponding to pure imaginaryroots

Figure 4.3: Stable and unstable nodes

4.8. APPENDIX 1: PROOF OF LEMMA 4.12 77

4.8 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 multiplicitym. Then

dim {x ∈ Cn | (λI−A)mx = 0} = m.

Proof. The proof consists of several steps. Let T = {x ∈ Cn | (λI−A)mx = 0}. The space T iscalled 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)mx = 0 so that A(λI − A)mx = 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 onthe subspace T . In other words, A : T −→ T . Let µ be an eigenvalue of A : T −→ T . Thatis Av = µv. Then v ∈ T implies that (λI − A)mv = 0. Since Av = µv, this simplifiesto (λ − µ)mv = 0. Being an eigenvector, v 6= 0 so that µ = λ. Therefore, all eigenvalues ofA : T −→ T are equal to λ.Step 4. Let dimT = r. Certainly r ≤ n. Then by Step 3, the characteristic polynomial ofA : T −→ T is (λ − z)r. Since T is an invariant subspace of A : Cn −→ Cn, one can choose abasis of T and then extend it to a basis of Cn so that with respect to this new basis of Cn, the matrixA is similar to a matrix where the upper left hand r × r submatrix represents A on T and the lowerleft hand (n− r)× r submatrix is the zero matrix. From this, we see that (λ− z)r is a factor of thecharacteristic 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. Thenp(A) = 0.Step 5. Let p(z) = (λ1 − z)m1 · · · (λk − z)mk be the characteristic polynomial of A, whereλ1, λ2, . . . , λk are the distinct eigenvalues of A : Cn −→ Cn. For each i from 1 to k, let

Ti = {x ∈ Cn | (λiI−A)mix = 0}.

By Step 4, we have dimTi ≤ mi.Let fi(z) = (λi − z)mi and gi(z) = f1(z) · · · fi(z) · · · fk(z), where fi(z) means the polynomialfi(z) is omitted. Note that fi(z)gi(z) = p(z) for all i.Resolving 1

(λ1−z)m1 ···(λk−z)mk into partial fractions, we have the identity

1

(λ1 − z)m1 · · · (λk − z)mk≡ h1(z)

(λ1 − z)m1+

h2(z)

(λ2 − z)m2+ · · ·+ hk(z)

(λk − z)mk,

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

g1(A)h1(A) + · · ·+ gk(A)hk(A) = I,


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 theCayley-Hamilton Theorem. This shows that any vector in Cn can be expressed as a sum of vectorswhere 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 theoremthat 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 + · · ·+ Ti + · · ·+ Tk) = {0}, where thesummand Ti 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 realvector space and the real dimension of T is m. This is because for any set of real vectors in Rn, it islinearly 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 isalso an eigenvalue of A of algebraic multiplicity m. In this case, if {v1, . . . ,vm} is a basis overC of Tλ, where Tλ is the generalized eigenspace corresponding to λ, then {v1, . . . ,vm} is a basisover C of Tλ. It can be shown that the 2m real vectors <v1, . . . ,<vm,=v1, . . . ,=vm are linearlyindependent over R and form a basis of (Tλ ⊕ Tλ) ∩ Rn.

Jordan canonical form.

Consider an eigenvalue λ of A with algebraic multiplicitym. Let T = {x ∈ Cn | (λI−A)mx = 0}.We give an algorithm for finding a Jordan basis for T such that the matrix representing A on T isin Jordan form. There are many possible choices of the Jordan basis, though the Jordan form willalways consist of the same number of blocks of each size. For instance, the matrix I is in Jordanform in any basis.

Definition. Let W be a linear subspace of V . We say that {v1, . . . ,vl} in V is linearly independentmod W if

∑i αivi ∈ W implies αi = 0 for all i. We say that {v1, . . . ,vl} is a basis of V mod W

if it is linearly independent mod W and V = W + span{v1, . . . ,vl}.

To find a basis of V mod W , we can simply choose a basis {v1, . . . ,vk} for W and extend it to abasis {v1, . . . ,vk,v1, . . . ,vl} for V . Then {v1, . . . ,vl} is a basis of V mod W .


We begin by finding a “cyclic vector” for each Jordan block which generates the Jordan basis for thatblock. Let ti = dimKer(λI−A)i, i = 1, . . . ,m. Note that ti+1− ti is the number of Jordan blocksof 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 that1 ≤ k ≤ m.

1. Choose a basis {v1k, . . . ,v

skk } 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)vik in Ker(λI−A)k−1. The important point is that

{(λI−A)v1k, . . . , (λI−A)vskk }

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 byadding {v1

k−1, . . . ,vsk−1

k−1 }, which are then the cyclic vectors for the Jordan blocks of sizek − 1. Since we have extended a list of size sk to reach a length tk−1 − tk−2, there aresk−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:

{ v1k, . . . , (λI−A)k−1v1

k, . . . ,vskk , . . . , (λI−A)k−1vskk ,

· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·v1i , . . . , (λI−A)i−1v1

i , . . . ,vsii , . . . , (λI−A)i−1vsii ,

· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·v12, (λI−A)v1

2, . . . ,vs22 , (λI−A)vs22 ,

v11, . . . ,v

s11 }

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 −1 0 0 0

0 2 −1 0 0 0 0

0 0 2 0 −1 −1 0

0 0 0 2 −1 0 −1

0 0 0 0 2 −1 0

0 0 0 0 0 2 0

0 0 0 0 0 0 2

. Find a Jordan basis for A.


Solution. 2I−A =

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 1 0 1

0 0 0 0 0 1 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

has rank 5, so that t1 = 2.

(2I−A)2 =

0 0 0 0 1 0 1

0 0 0 0 1 1 0

0 0 0 0 0 1 0

0 0 0 0 0 1 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0


(2I−A)3 =

0 0 0 0 0 1 0

0 0 0 0 0 1 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

0 0 0 0 0 0 0


(2I−A)4 is the zero matrix so that t4 = 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}. Heree7 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 extensionis necessary.

(k = 2) (2I−A)(e2 + e3 + e4) = e1 + e2, (2I−A)(e4) = e1. So we choose {e1 + e2, e1} to bethe basis for the eigenspace Ker(2I−A).

(k = 1) Stop.


Therefore, a Jordan basis is

{e6, e3 + e5, e2 + e3 + e4, e1 + e2, e7, e4, e1}.

Consequently, the Jordan form of A is

2 1 0 0 0 0 0

0 2 1 0 0 0 0

0 0 2 1 0 0 0

0 0 0 2 0 0 0

0 0 0 0 2 1 0

0 0 0 0 0 2 1

0 0 0 0 0 0 2

.

Example 4.13 Let A =

2 0 0 −1 0 0 0

0 2 −1 0 0 0 0

0 0 2 0 −1 −1 0

0 0 0 2 −1 0 −1

0 0 0 0 2 −1 0

0 0 0 0 0 2 0

0 0 0 0 0 0 2

. Solve the system x′ = 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

x(t) = e2t [ c1(e6 + t(e3 + e5) + t2

2 (e2 + e3 + e4) + t3

6 (e1 + e2))

+c2((e3 + e5) + t(e2 + e3 + e4) + t2

2 (e1 + e2))

+c3((e2 + e3 + e4) + t(e1 + e2))

+c4(e1 + e2)

+c5(e7 + te4 + t2

2 e1)

+c6(e4 + te1)

+c7e1] .

Exercise 4.7 Find a Jordan basis of the matrix A =

3 0 −1 0 −1

0 3 0 −1 0

0 0 3 0 −1

0 0 0 3 0

0 0 0 0 3

. Hence solve the

system x′ = Ax.

Ans: There is 1 block of size 3 and 1 block of size 2. A Jordan basis is {e5, e1 + e3, e1, e4, e2}.x(t) = e3t

(c1(e5 + t(e1 + e3) + t2

2 e1) + c2((e1 + e3) + te1) + c3e1 + c4(e4 + te2) + c5e2

).

Chapter 5

Power Series Solutions

5.1 Power Series

An infinite series of the form∞∑n=0

an(x− x0)n = a0 + a1(x− x0) + a2(x− x0)2 + · · · (5.1.1)

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

anxn = a0 + a1x+ a2x

2 + · · · (5.1.2)

(5.1.2) is said to converge at a point x if the limit limm→∞

m∑n=0

anxn exists, and in this case the sum of

the series is the value of this limit. It is obvious that (5.1.2) always converges at x = 0. It can beshowed that each power series like (5.1.2) corresponds to a positive real numberR, called the radiusof convergence, with the property that the series converges if |x| < R and diverges if |x| > R. Itis customary to put R equal to 0 when the series converges only at x = 0, and equal to∞ when itconverges 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

limn→∞

∣∣∣∣an+1xn+1

anxn

∣∣∣∣ = limn→∞

∣∣∣∣an+1

an

∣∣∣∣ |x| = L,

then (5.1.2) converges for L < 1 and diverges if L > 1. It follows from this that

R = limn→∞

∣∣∣∣ anan+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 theseries converges and outside the interval the series diverges.

Consider the following power series∞∑n=0

n!xn = 1 + x+ 2!x2 + 3!x3 + · · · (5.1.3)

83

84 CHAPTER 5. POWER SERIES SOLUTIONS

∞∑n=0

xn

n!= 1 + x+

x2

2!+x3

3!+ · · · (5.1.4)

∞∑n=0

xn = 1 + x+ x2 + x3 + · · · (5.1.5)

It is easy to verify that (5.1.3) converges only at x = 0. Thus R = 0. For (5.1.4), it converges for allx 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

f(x) =∞∑n=0

anxn = a0 + a1x+ a2x

2 + · · · (5.1.6)

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

f ′(x) =

∞∑n=1

nanxn−1 = a1 + 2a2x+ 3a3x

2 + · · · ,

f ′′(x) =

∞∑n=2

n(n− 1)anxn−2 = 2a2 + 3 · 2a3x+ · · · ,

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

an =fn(0)

n!(5.1.7)

Moreover, (5.1.6) can be integrated termwise provided the limits of integration lie inside the intervalof convergence.

If

g(x) =

∞∑n=0

bnxn = b0 + b1x+ b2x

2 + · · · (5.1.8)

is another power series with interval of convergence |x| < R, then (5.1.6) and (5.1.8) can be addedor subtracted termwise:

f(x)± g(x) =

∞∑n=0

(an ± bn)xn = (a0 ± b0) + (a1 ± b1)x+ (a2 ± b2)x2 + · · · (5.1.9)

They can also be multiplied like polynomials, in the sense that

f(x)g(x) =

∞∑n=0

cnxn,

where cn = a0bn + a1bn−1 + · · ·+ anb0.

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, iff(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 representedby a power series? If we use (5.1.7), it is natural to expect

f(x) =

∞∑n=0

f (n)(0)

n!xn = f(0) + f ′(0)x+

f ′′(0)

2!x2 + · · · (5.1.10)

to hold for all |x| < R. Unfortunately, this is not always true. Instead, one can use Taylor’sexpansion for f(x):

f(x) =

n∑k=0

f (k)(0)

k!xk + Rn(x),

where the remainder Rn(x) is given by

Rn(x) =f (n+1)(x)

(n+ 1)!xn+1

for some point x between 0 and x. To verify (5.1.6), it suffices to show that Rn(x) −→ 0 asn −→∞.

Example 5.1 The following familiar expansions are valid for all x.

ex =

∞∑n=0

xn

n!= 1 + x+

x2

2!+x3

3!+ · · ·

sinx =

∞∑n=0

(−1)nx2n+1

(2n+ 1)!= x− x3

3!+x5

5!− · · ·

cosx =

∞∑n=0

(−1)nx2n

(2n)!= 1− x2

2!+x4

4!− · · ·

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

f(x) =

∞∑n=0

an(x− x0)n (5.1.11)

is valid in some interval containing the point x0 is said to be analytic at x0. In this case, an isnecessarily given by

an =f (n)(x0)

n!,

and (5.1.11) is called the Taylor series of f(x) at x0.Thus ex, sinx, cosx are analytic at all points. Concerning analytic functions, we have the followingbasic results.

1. Polynomials, ex, sinx, cosx 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) [providedg(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) iff ′(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.


5.2 Series Solutions of First Order Equations

A first order differential equation y′ = f(x, y) can be solved by assuming that it has a power seriessolution. Let’s illustrate this with two familiar examples.

Example 5.2 Consider the differential equation y′ = y. We assume it has a power series solutionof the form

y = a0 + a1x+ a2x2 + · · ·+ anx

n + · · · (5.2.1)

that converges for |x| < R. That is the equation y′ = y has a solution which is analytic at the origin.Then

y′ = a1 + 2a2x+ · · ·+ nanxn−1 + · · · (5.2.2)

has the same interval of convergence. Since y′ = y, the series (5.2.1) and (5.2.2) have the samecoefficients. That is

(n+ 1)an+1 = an, all for n = 0, 1, 2, . . .

Thus an = 1nan−1 = 1

n(n−1)an−2 = · · · = 1n!a0. Therefore

y = a0

(1 + x+

x2

2!+x3

3!+ · · ·+ xn

n!+ · · ·

),

where a0 is an arbitrary constant. In this case, we recognize this as the power series of ex. Thus thegeneral solution is y = a0e

x.

Example 5.3 The function y = (1+x)p, where p is a real constant satisfies the differential equation

(1 + x)y′ = py, y(0) = 1. (5.2.3)

As before, we assume it has a power series solution of the form

y = a0 + a1x+ a2x2 + · · ·+ anx

n + · · ·

with positive radius of convergence. Then

y′ = a1+ 2a2x+ 3a3x2 + · · ·+ (n+ 1)an+1x

n + · · · ,xy′ = a1x+ 2a2x

2 + · · ·+ nanxn + · · · ,

py = pa0+ pa1x+ pa2x2 + · · ·+ panx

n + · · · ,

Using (5.2.3) and equating coefficients, we have

(n+ 1)an+1 + nan = pan, for all n = 0, 1, 2, . . .

That isan+1 =

p− nn+ 1

an,

so that

a1 = p, a2 =p(p− 1)

2, a3 =

p(p− 1)(p− 2)

2 · 3, ..., an =

p(p− 1) · · · (p− n+ 1)

n!.

5.3. SECOND ORDER LINEAR EQUATIONS AND ORDINARY POINTS 87

In other words,

y = 1 + px+p(p− 1)

2x2 +

p(p− 1)(p− 2)

2 · 3x3 + · · ·+ p(p− 1) · · · (p− n+ 1)

n!xn + · · · .

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)

2x2 +

p(p− 1)(p− 2)

2 · 3x3 + · · ·+ p(p− 1) · · · (p− n+ 1)

n!xn + · · · ,

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′′ + P (x)y′ +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 analyticat 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)2Q(x) are analytic iscalled a regular singular point of (5.3.1). If a singular point x0 is not a regular singular point, thenit 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′′ + xy = 0. Since the function Q(x) = x is analytic at everypoint, every point is an ordinary point.

Example 5.6 In the Cauchy-Euler equation y′′ + a1x y′ + a2

x2 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′′ +1

(x− 1)2y′ +

8

x(x− 1)y = 0.

The singular points are 0 and 1. At the point 0, xP (x) = x(1−x)−2 and x2Q(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, wehave (x− 1)P (x) = 1/(x− 1) which is not analytic at x = 1, and hence the point 1 is an irregularsingular point.

To discuss the behavior of the singularities at infinity, we use the transformation x = 1/t, which con-verts the problem to the behavior of the transformed equation near the origin. Using the substitutionx = 1/t, (5.3.1) becomes

d2y

dt2+

(2

t− 1

t2P (

1

t)

)dy

dt+

1

t4Q(

1

t)y = 0 (5.3.2)

We define the point at infinity to be an ordinary point, a regular singular point, or an irregular singularpoint of (5.3.1) according as the origin of (5.3.2) is an ordinary point, a regular singular point, or anirregular singular point.


Example 5.8 Consider the differential equation

d2y

dx2+

1

2

(1

x2+

1

x

)dy

dx+

1

2x3y = 0.

The substitution x = 1/t transforms the equation into

d2y

dt2+

(3− t

2t

)dy

dt+

1

2ty = 0.

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′′ + [c− (a+ b+ 1)x]y′ − 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′′ + P (x)y′ +Q(x)y = 0,

and let a0 and a1 be arbitrary constants. Then there exists a unique function y(x) that is analyticat x0, is a solution of the differential equation in an interval containing x0, and satisfies the initialconditions y(x0) = a0, y

′(x0) = a1. Furthermore, if the power series expansions of P (x) andQ(x) are valid on an interval |x−x0| < R, R > 0, then the power series expansion of this solutionis 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

simplicity, we take x0 = 0. Thus P (x) =

∞∑n=0

pnxn and Q(x) =

∞∑n=0

qnxn, for |x| < R, with

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

∞∑n=0

anxn with radius of

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

y′ =

∞∑n=0

(n+ 1)an+1xn = a1 + 2a2x+ 3a3x

2 + · · ·

y′′ =

∞∑n=0

(n+ 1)(n+ 2)an+2xn = 2a2 + (2)(3)a3x+ (3)(4)a4x

2 + · · ·

Thus P (x)y′ =

( ∞∑n=0

pnxn

)[ ∞∑n=0

(n+ 1)an+1xn

]

=∞∑n=0

[n∑k=0

pn−k(k + 1)ak+1

]xn.

Q(x)y =

( ∞∑n=0

qnxn

)( ∞∑n=0

anxn

)=

∞∑n=0

[n∑k=0

qn−kak

]xn.


Substituting these into (5.3.1), we have

∞∑n=0

[(n+ 1)(n+ 2)an+2 +

n∑k=0

pn−k(k + 1)ak+1 +

n∑k=0

qn−kak

]xn = 0.

Equating the coefficient of xn to 0, we get the following recursion formula for an.

(n+ 1)(n+ 2)an+2 = −n∑k=0

[(k + 1)pn−kak+1 + qn−kak] (5.3.12)

For example,

2a2 = −(p0a1 + q0a0),(2)(3)a3 = −(p1a1 + 2p0a2 + q1a0 + q0a1),(3)(4)a4 = −(p2a1 + 2p1a2 + 3p0a3 + q2a0 + q1a1 + q0a2),

and so on ...

The formula (5.3.12) determines all a2, a3, . . . in terms of a0 and a1 so that the power series

y =

∞∑n=0

anxn formally satisfies the DE and the given initial conditions. It remains to prove the

series∞∑n=0

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

convergence for

P (x) =

∞∑n=0

pnxn and Q(x) =

∞∑n=0

qnxn. (5.3.13)

Let r be a positive number such that r < R. The two series (5.3.13) converge at x = r so that theterms 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+ 1)(n+ 2)|an+2| ≤M

rn

n∑k=0

[(k + 1)|ak+1|+ |ak|] rk

≤ M

rn

n∑k=0

[(k + 1)|ak+1|+ |ak|] rk +M |an+1|r.

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 =M

rn

n∑k=0

[(k + 1)bk+1 + bk] rk +Mbn+1r (5.3.14)

Observe that 0 ≤ |an| ≤ bn for all n. Now we try to prove that∞∑n=0

bnxn converges for |x| < r.

Then by comparison test,∞∑n=0

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


number less than R, we conclude that∞∑n=0

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

show∞∑n=0

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

n(n+ 1)bn+1 =M

rn−1

n−1∑k=0

[(k + 1)bk+1 + bk] rk +Mbnr, and

(n− 1)nbn =M

rn−2

n−2∑k=0

[(k + 1)bk+1 + bk] rk +Mbn−1r.

Now multiplying the first equation by r and by the second equation, we obtain

rn(n+ 1)bn+1 =M

rn−2

n−1∑k=0

[(k + 1)bk+1 + bk] rk +Mbnr2

=M

rn−2

n−2∑k=0

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

+Mbnr2

= (n− 1)nbn −Mbn−1r + rM(nbn + bn−1) +Mbnr2

= [(n− 1)n+ rMn+Mr2]bn.

Thereforebn+1

bn=

(n− 1)n+ rMn+Mr2

rn(n+ 1).

Thus limn→∞

∣∣∣∣bn+1xn+1

bnxn

∣∣∣∣ = limn→∞

bn+1

bn|x| = |x|

r, and the series

∞∑n=0

bnxn converges for |x| < r.

Consequently, by comparison test the series∞∑n=0

anxn converges for |x| < r. �

Exercise 5.2 Find two linearly independent solutions of y′′ − xy′ − x2y = 0

Ans: y1(x) = 1 + 112x

4 + 190x

6 + 31120x

8 + · · · and y2(x) = x+ 16x

3 + 340x

5 + 131008x

7 + · · · .

Exercise 5.3 Using power series method, solve the initial value problem (1+x2)y′′+2xy′−2y = 0,y(0) = 0, y′(0) = 1.

Ans: y = x.

Legendre’s equation(1− x2)y′′ − 2xy′ + p(p+ 1)y = 0,

where p is a constant called the order of Legendre’s equation.That is P (x) = − 2x

1−x2 and Q(x) = p(p+1)1−x2 . The origin is an ordinary point, and we expect a

solution of the form y =∑anx

n. Thus the left hand side of the equation becomes

(1− x2)

∞∑n=0

(n+ 1)(n+ 2)an+2xn − 2x

∞∑n=0

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

∞∑n=0

anxn,


or∞∑n=0

(n+ 1)(n+ 2)an+2xn −

∞∑n=2

(n− 1)nanxn −

∞∑n=1

2nanxn +

∞∑n=0

p(p+ 1)anxn.

The sum of these series is required to be zero, so the coefficient of xn must be zero for every n. Thisgives

(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)

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

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)mp(p− 2)(p− 4) · · · (p− 2m+ 2)(p+ 1)(p+ 3) · · · (p+ 2m− 1)

(2m)!a0,

a2m+1 = (−1)m(p− 1)(p− 3) · · · (p− 2m+ 1)(p+ 2)(p+ 4) · · · (p+ 2m)

(2m+ 1)!a1.

With that, we get two linearly independent solutions

y1(x) =

∞∑m=0

a2mx2m and y2(x) =

∞∑m=0

a2m+1x2m+1,

and the general solution is given by

y = a0

[1− p(p+ 1)

2!x2 +

p(p− 2)(p+ 1)(p+ 3)

4!x4

−p(p− 2)(p− 4)(p+ 1)(p+ 3)(p+ 5)

6!x6 + · · ·

]

+a1

[x− (p− 1)(p+ 2)

3!x3 +

(p− 1)(p− 3)(p+ 2)(p+ 4)

5!x5

− (p− 1)(p− 3)(p− 5)(p+ 2)(p+ 4)(p+ 6)

7!x7 + · · ·

].

When p is not an integer, the series representing y1 and y2 have radius of convergence R = 1. Forexample, ∣∣∣∣a2n+2x

2n+2

a2nx2n

∣∣∣∣ =

∣∣∣∣− (p− 2n)(p+ 2n+ 1)

(2n+ 1)(2n+ 2)

∣∣∣∣ |x2| −→ |x|2as n −→∞, and similarly for the second series. In fact, by Theorem 4.1, and the familiar expansion

1

1− x2= 1 + x2 + x4 + · · · , |x| < 1,

shows that R = 1 for both P (x) and Q(x). Thus, we know any solution of the form y =∑anx

n

must be valid at least for |x| < 1.


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 apolynomial of degree n. If p = n is odd, y2 again is a polynomial of degree n. These are calledLegendre polynomials Pn(x) and they give particular solutions to Legendre’s equation

(1− x2)y′′ − 2xy′ + n(n+ 1)y = 0,

where n is a nonnegative integer. It is customary to choose the arbitrary constants a0 or a1 so thatthe coefficient of xn in Pn(x) is (2n)!/[2n(n!)2]. This implies Pn(1) = 1. Then

Pn(x) =

bn/2c∑k=0

(−1)k(2n− 2k)!

2nk!(n− k)!(n− 2k)!xn−2k.

The six Legendre polynomials are

P0 = 1, P1(x) = x

P2(x) = 12 (3x2 − 1), P3(x) = 1

2 (5x3 − 3x)

P4(x) = 18 (35x4 − 30x2 + 3), P5(x) = 1

8 (63x5 − 70x3 + 15x)

There is also a Rodrigues’ formula for the Legendre polynomial given by

Pn(x) =1

n!2ndn

dxn(x2 − 1)n.

Hermite’s equationy′′ − 2xy′ + 2py = 0,

where p is a constant. The general solution of Hermite’s equation is y(x) = a0y1(x) + a1y2(x),where

y1(x) = 1− 2p

2!x2 +

22p(p− 2)

4!x4 − 23p(p− 2)(p− 4)

6!x6 + · · · ,

y2(x) = x− 2(p− 1)

3!x3 +

22(p− 1)(p− 3)

5!x5 − 23(p− 1)(p− 3)(p− 5)

7!x7 + · · · .

By Theorem 4.1, both series for y1 and y2 converge for all x. Note that y1 is a polynomial if p is aneven 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 ofHermite’s equation, multiplied by a suitable constant so that the coefficient of xn is 2n. The first sixHermite’s polynomials are

H0(x) = 1, H1(x) = 2x,H2(x) = 4x2 − 2, H3(x) = 8x3 − 12x,H4(x) = 16x4 − 48x2 + 12, H5(x) = 32x5 − 160x3 + 120x

A general formula for the Hermite polynomials is

Hn = (−1)nex2 dn

dxn

(e−x

2).

5.4. REGULAR SINGULAR POINTS AND THE METHOD OF FROBENIUS 93

5.4 Regular Singular Points and the Method of Frobenius

Consider the second order linear homogeneous differential equation

x2y′′ + xp(x)y′ + 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 + p1x + p2x

2 + p3x3 + · · · , and q(x) = q0 + q1x + q2x

2 + q3x3 + · · · . Suppose

(5.4.1) has a series solution of the form

y = xr∞∑n=0

anxn =

∞∑n=0

anxn+r (5.4.2)

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

y′ =

∞∑n=0

an(n+ r)xn+r−1, (5.4.3)

and

y′′ =

∞∑n=0

an(n+ r)(n+ r − 1)xn+r−2. (5.4.4)

Substituting the series of y, y′ and y′′ into (5.4.1) yields

[r(r − 1)a0xr + (r + 1)ra1x

r+1 + · · · ] + [p0x+ p1x2 + · · · ] · [ra0xr−1 + (r + 1)a1x

r + · · · ]+[q0 + q1x+ · · · ] · [a0xr + a1x

r+1 + · · · ] = 0.(5.4.5)

The lowest power of x in (5.4.5) is xr. If (5.4.5) is to be satisfied identically, the coefficient r(r −1)a0 + p0ra0 + q0a0 of xr must vanish. As a0 6= 0, it follows that r satisfies the quadratic equation

r(r − 1) + p0r + q0 = 0. (5.4.6)

This is the same equation obtained with the Cauchy-Euler equation. Equation (5.4.6) is called theindicial equation of (5.4.1) and its two roots (possibly equal) are the exponents of the differentialequation 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 Frobeniussolutions and they are linearly independent. Whereas r1 = r2, there is only one possible Frobeniusseries 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′′ + 3x(1 + x)3y′ − (1− x2)y = 0.

Solution. Clearly x = 0 is a regular singular point since p(x) = 32 (1 + x)2 and q(x) = − 1

2 (1− x)

are polynomials. Rewrite the equation in the standard form:

y′′ +32 (1 + 2x+ x2)

xy′ +

− 12 (1− x)

x2y = 0.


We see that p0 = 32 and q0 = − 1

2 . Hence the indicial equation is

r(r − 1) +3

2r − 1

2= r2 +

1

2r − 1

2= (r + 1)(r − 1

2) = 0,

with roots r1 = 12 and r2 = −1. The two possible Frobenius series solutions are of the forms

y1(x) = x12

∞∑n=0

anxn and y2(x) = x−1

∞∑n=0

anxn.

Once the exponents r1 and r2 are known, the coefficients in a Frobenius series solution can be foundby substituting the series (5.4.2),(5.4.3) and (5.4.4) into the differential equation (5.4.1). If r1 andr2 are complex conjugates, we always get two linearly independent solutions. We shall restrict ourattention for real solutions of the indicial equation and seek solutions only for x > 0. The solutionson the interval x < 0 can be studied by changing the variable to t = −x and solving the resultingequation for t > 0.

Let’s work out the recursion relations for the coefficients. By (5.4.3), we have

1xp(x)y′ =

1

x

( ∞∑n=0

pnxn

)[ ∞∑n=0

an(n+ r)xn+r−1

]

= xr−2( ∞∑n=0

pnxn

)[ ∞∑n=0

an(n+ r)xn

]

= xr−2∞∑n=0

[n∑k=0

pn−kak(r + k)

]xn

= xr−2∞∑n=0

[n−1∑k=0

pn−kak(r + k) + p0an(r + n)

]xn.

Also we have

1x2 q(x)y =

1

x2

( ∞∑n=0

qnxn

)( ∞∑n=0

anxr+n

)

= xr−2( ∞∑n=0

qnxn

)( ∞∑n=0

anxn

)

= xr−2∞∑n=0

(n∑k=0

qn−kak

)xn

= xr−2∞∑n=0

(n−1∑k=0

qn−kak + q0an

)xn.

Substituting these into the differential equation (5.4.1) and canceling the term xr−2, we have

∞∑n=0

{an[(r + n)(r + n− 1) + (r + n)p0 + q0] +

n−1∑k=0

ak[(r + k)pn−k + qn−k]

}xn = 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∑k=0

ak[(r + k)pn−k + qn−k] = 0. (5.4.7)

When n = 0, we get r(r − 1) + rp0 + q0 = 0, which is true because r is a root of the indicialequation. 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. Supposer1 > 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 processbreaks off because the coefficient aN in (5.4.7) is zero. In this case, only the Frobenius seriessolution with the larger exponent is guaranteed to exist. The other solution may not be a Frobeniusseries 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) andthat the power series expansions of p(x) and q(x) are valid on an interval |x| < R with R > 0. Letthe indicial equation (5.4.6) have real roots r1 and r2 with r1 ≥ r2. Then (5.4.1) has at least onesolution

y1 = xr1∞∑n=0

anxn, (a0 6= 0) (5.4.8)

on the interval 0 < x < R, where an are determined in terms of a0 by the recursion formula (5.4.7)with r replaced by r1, and the series

∑∞n=0 anx

n converges for |x| < R. Furthermore, if r1 − r2 isnot zero or a positive integer, then equation (5.4.1) has a second independent solution

y1 = xr2∞∑n=0

anxn, (a0 6= 0) (5.4.9)

on the same interval, where an are determined in terms of a0 by the recursion formula (5.4.7) withr replaced by r2, and again the series

∑∞n=0 anx

n 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 Frobeniusseries solution. (3) If r1 − r2 = n is a positive integer and the summation of (5.4.7) is zero, thenan is unrestricted and can be assigned any value whatever. In particular, we can put an = 0 andcontinue to compute the coefficients without difficulties. Hence, in this case, there does exist asecond Frobenius series solution. In many cases of (1) and (2), it is possible to determine a secondsolution by the method of variation of parameters. For instance a second solution for the Cauchy-Euler equation for the case where its indicial equation has equal roots is given by xr lnx.

Exercise 5.4 Find two linearly independent Frobenius series solutions of the differential equation2x2y′′ + x(2x+ 1)y′ − y = 0.

Ans: y1 = x(1− 25x+ 4

35x2 + · · · ), y2 = x−

12 (1− x+ 1

2x2 + · · · ).

More precisely, y1 = x∑∞n=0

(−1)n6·4n(n+1)!(2n+3)! xn and y2 = x−

12 e−x.


Example 5.10 Find the Frobenius series solutions of xy′′ + 2y′ + xy = 0.

Solution. Rewrite the equation in the standard form x2y′′ + 2xy′ + x2y = 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 integerand we may not have two Frobenius series solutions. We know there is a Frobenius series solutioncorresponding to r1 = 0. Let’s consider the possibility of the solution corresponding to the smaller

exponent r2 = −1. Let’s begin with y = x−1∞∑n=0

cnxn =

∞∑n=0

cnxn−1. Substituting this into the

given equation, we obtain∞∑n=0

(n− 1)(n− 2)cnxn−2 + 2

∞∑n=0

(n− 1)cnxn−2 +

∞∑n=0

cnxn = 0,

or equivalently

∞∑n=0

n(n− 1)cnxn−2 +

∞∑n=0

cnxn = 0,

or∞∑n=0

n(n− 1)cnxn−2 +

∞∑n=2

cn−2xn−2 = 0.

The cases n = 0 and n = 1 reduce to 0 · c0 = 0 and 0 · c1 = 0. Thus c0 and c1 are arbitrary andwe can expect to get two linearly independent Frobenius series solutions. Equating coefficients, weobtain the recurrence relation

cn = − cn−2n(n− 1)

, for n ≥ 2.

It follows from this that for n ≥ 1.

c2n =(−1)nc0

(2n)!and c2n+1 =

(−1)nc1(2n+ 1)!

.

Therefore, we have

y = x−1∞∑n=0

cnxn =

c0x

∞∑n=0

(−1)n

(2n)!x2n +

c1x

∞∑n=0

(−1)n

(2n+ 1)!x2n+1.

We recognize this general solution as

y =1

x(c0 cosx+ c1 sinx).

If we begin with the larger exponent, we will get the solution (sinx)/x.

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

Lk(x) = (k!)2k∑

n=0

(−1)n

n![(k − n)!]2xk−n.

Ans: y = 1 +

∞∑n=1

(0− α)(1− α) · · · (n− 1− α)

(n!)2xn.

5.5. BESSEL’S EQUATION 97

5.5 Bessel’s Equation

The second order linear homogeneous differential equation

x2y′′ + xy′ + (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 = c1Jp(x) + c2Yp(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 theBessel function of order p of the second kind. These functions have been tabulated and behavesomewhat like the trigonometric functions of damped amplitude. If we let y = u/

√x, we obtain

d2u

dx2+

(1−

p2 − 14

x2

)u = 0. (5.5.3)

In the special case in which p = ± 12 , this equation becomes

d2u

dx2+ u = 0.

Hence u = c1 sinx+ c2 cosx and

y = c1sinx√x

+ c2cosx√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) tobehave 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 andq(x) = −p2 + x2. Thus the indicial equation is r(r − 1) + r − p2 = r2 − p2 = 0. Therefore, the

exponents are±p. Let r be either−p or p. If we substitute y =

∞∑m=0

cmxm+r into Bessel’s equation,

we find in the usual manner that c1 = 0 and that for m ≥ 2,

[(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 recursionformula

am = − am−2m(2p+m)

(5.5.6)

As a1 = 0, it follows that am = 0 for all odd values of m. The first few coefficients am for m evenare

a2 = − a02(2p+ 2)

= − a022(p+ 1)

,

a4 = − a24(2p+ 4)

=a0

24 · 2(p+ 1)(p+ 2),

a6 = − a46(2p+ 6)

= − a026 · 2 · 3(p+ 1)(p+ 2)(p+ 3)

.


In general, one can show that

a2m =(−1)ma0

22mm!(p+ 1)(p+ 2) · · · (p+m).

Thus we have a solution associated with the larger exponent p

y1 = a0

∞∑m=0

(−1)m

22mm!(p+ 1)(p+ 2) · · · (p+m)x2m+p.

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

J0(x) =

∞∑m=0

(−1)mx2m

22m(m!)2= 1− x2

4+x4

64− x6

2304+ · · · .

This special function J0(x) is called the Bessel function of order zero of the first kind. A secondlinearly 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 solutionassociated with the smaller exponent. However, as we shall see, it does have a second Frobeniusseries solution so long as p is not an integer. Let’s write bm in place of cm in (5.5.5). Thus we haveb1 = 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 pis a positive integer or an odd integral multiple of 1

2 . Suppose p = k/2 where k is an odd positiveinteger. 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) form = k, we can simply choose bk = 0. Subsequently all bm = 0 for all odd values of m. [If we letbk to be arbitrary and non-zero, the subsequent solution so obtained is just bky1(x) if we also takeb0 = 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 possibleto solve bm in terms of bm−2 since m(m− k) 6= 0 as m is always even while k is odd. The result isthe same as before except we should replace p by −p. Thus in this case, we have a second solution

y2 = b0

∞∑m=0

(−1)m

22mm!(−p+ 1)(−p+ 2) · · · (−p+m)x2m−p.

Since p(x) = 1 and q(x) = x2−p2 are just polynomials. The series representing y1 and y2 convergefor all x > 0. If p > 0, then the first term in y1 is a0xp, whereas the first term in y2 is b0x−p. Hencey1(0) = 0, but y2(x) −→ ±∞ as x −→ 0, so that y1 and y2 are linearly independent. So we havetwo linearly independent solutions as long as p is not an integer.


If p = n is an nonnegative integer and we take a0 = 12nn! , the solution y1 becomes

Jn =

∞∑m=0

(−1)m

m!(m+ n)!

(x2

)2m+n

.

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 theso called Gamma functions and the general solution is Y = c1Jp + c2J−p. Here

J±p =∞∑m=0

(−1)m

m!Γ(m± n+ 1)

(x2

)2m±p.

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)pJp,so there is only one Frobenius series solution. A second solution can be obtained by considering thefunction

Yp(x) =Jp(x) cos pπ − J−p(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 ofJp and J−p. If p approaches to an integer, the expression of Yp gives an indeterminate form as boththe 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.

Yn(x) = limp→n

Jp(x) cos pπ − J−p(x)

sin pπ= limp→n

1

π

[∂Jp∂p− (−1)n

∂J−p∂p

].

This limit was first evaluated using L’Hopital’s rule by Hankel in 1868. Yn is called a Bessel functionof the second kind, and it follows that y = c1Jp + c2Yp is the general solution of Bessel’s equationin 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) = x2y′′ + xp(x)y′ + q(x)y. We are solving L(y) = 0 by takinga series solution of the form y(x) = xr

∑∞n=0 anx

n. If we treat r as a variable, then an’s arefunctions 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 determinedfrom (5.4.7). This means

L(y(x, r)) = a0(r − r1)2xr.

So if a0 6= 0, we take r = r1, we get one Frobenius series solution y1(x). Now let’s differentiatethe above equation with respect to r. We get

L(∂y

∂r) =

∂

∂rL(y) = a0[(r − r1)2xr lnx+ 2(r − r1)xr].

Evaluating at r = r1, we obtain

L(∂y

∂r

∣∣∣∣r=r1

) =∂

∂rL(y)

∣∣∣∣r=r1

= 0.


Consequently, we have the second solution

y2(x) =∂y

∂r(x, r1) = xr1 lnx

∞∑n=0

an(r1)xn+xr1∞∑n=0

a′n(r1)xn = y1(x) lnx+xr1∞∑n=1

a′n(r1)xn.

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

y1(x) =

∞∑n=0

(−1)n

(n!)2

(x2

)2n, and

y2(x) = y1(x) lnx−∞∑n=1

(−1)nH(n)

(n!)2

(x2

)2n,

where H(n) =∑nk=1

1k .

3. The case r1 − r2 is a nonnegative integerConsider

x2y′′ + xp(x)y′ + q(x)y = 0, x > 0,

where p(x) =∑∞n=0 pnx

n and q(x) =∑∞n=0 qnx

n. Let r1 and r2 be the roots (exponents) withr1 ≥ r2 of the indicial equation r(r − 1) + p0r + q0 = 0.Write r1 = r2 + m, where m is a nonnegative integer. Let y1 = xr1

∑∞n=0 anx

n be a Frobeniusseries solution corresponding to the larger exponent r1. For simplicity, we take a0 = 1.Let u = xr2

∑∞n=0 bnx

n and make a change of variable:

y(x) = u(x)− bmy1(x) lnx.

We getx2u′′ + xp(x)u′ + q(x)u = bm[2xy′1 + (p(x)− 1)y1].

Now let’s substitute u = xr2∑∞n=0 bnx

n to see if we can determine the bn’s. Note that the first termin the power series expansion of bm[2xy′1 + (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) + p0r2 + q0)b0xr2 +A1x

r2+1 + · · ·+Amxr2+m + · · · = mbmx

r1 + · · · . (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 canbe arbitrary. The coefficients A1, A2, . . . are given by the main recurrence relation (5.4.7). Thus byequating 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 bmis 0. Previously, this forbids the determination of bm and possibly runs into a contradiction. Nowon the right hand side of (5.5.9), if m > 0, then one can determine bm by equating the coefficientsof xr1 on both sides. From then on, all the subsequent bn’s can be determined and we get a solutionof the form y(x) = u(x) − bmy1(x) lnx. Note that if bm = 0 in this determination, then a secondFrobenius series solution in fact can be obtained with the smaller exponent r2.


Example 5.11 Consider x2y′′ + 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 =

∞∑n=1

(−1)n−1n

(n!)2xn = x− 1

2x2 +

1

12x3 − · · · .

Now b1[2xy′1 + (p(x) − 1)y1] = b1(2x(1 − x + 14x

2 − · · · ) − (x − 12x

2 + 112x

3 − · · · )] =

b1[x− 32x

2 + 512x

3 − · · · ].

Substituting u = x0∑∞n=0 bnx

n into x2u′′ + xu = b1[2xy′1 + (p(x)− 1)y1], we get

0·(0−1)b0+[(1)(0)b1+b0]x+[(2)(1)b2+b1]x2+[(3)(2)b3+b2]x3+· · · = b1[x−3

2x2+

5

12x3−· · · ].

Comparing coefficients, we have b0 = b1, 2b2+b1 = − 32b1 and 6b3+b2 = 5

12b1, · · · . Thus b1 = b0,b2 = − 5

4b0, b3 = 518b0, . . .. Therefore u = b0(1 + x − 5

4x2 + 5

18x3 − · · · ). By taking b0 = 1, we

get the solution y = (1 + x− 54x

2 + 518x

3 − · · · )− y1(x) lnx.

If m = 0, then r1 = r2 and the first terms on both sides of (5.5.9) are 0. Thus we can continue todetermine 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′′ − x(1 + 2x+ 3x2)y′ + (x+ 5x2)y = 0.

Ans: y1 = x2(1+x+ 12x

2 + · · · ), y2 = (1+x+2x2 + 83x

3 + · · · )−2y1 lnx. The general solutionis a linear combination of y1 and y2.


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′′ − 2xy′ + 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

Pn(x) =

bn/2c∑k=0

(−1)k(2n− 2k)!

2nk!(n− k)!(n− 2k)!xn−2k.

There is also a Rodrigues’ formula for the Legendre polynomial given by

Pn(x) =1

n!2ndn

dxn(x2 − 1)n.

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 2nPn(1) is the coefficient of (x− 1)n inthe 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−12 + · · ·+ 2n], it is clear that the coefficient of (x−1)n is 2n. ThusPn(1) = 1.

The Legendre polynomial Pn(x) has the generating function φ(Z) = (1 − 2xZ + Z2)−12 = (1 +

Z2 − 2xZ)−12 . That is Pn(x) is the coefficient of Zn 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 seriesconverges to φ(Z) for |Z| < 1. That is

φ(Z) =

∞∑n=0

AnZn, for |Z| < 1, |x| ≤ 1. (A.1)

Using Binomial expansion,

(1 + Z2 − 2xZ)−12 = 1− 1

2(Z2 − 2xZ) +

(− 12 )(− 1

2 − 1)

2!(Z2 − 2xZ)2 + · · · ,

we see that An is a polynomial in x of degree n. If we let x = 1, we obtain

φ(Z)|x=1 = (1− 2Z + Z2)−12 = (1− Z)−1 = 1 + Z + Z2 + Z3 + · · · , |Z| < 1.

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

(1− 2Zx+ Z2)∂φ

∂Z= (x− Z)φ, (A.2)

Z∂φ

∂Z= (x− Z)

∂φ

∂x. (A.3)

Substituting (A.1) into (A.2) and equating the coefficients of Zn−1, we obtain

nAn − (2n− 1)xAn−1 + (n− 1)An−2 = 0 (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 Zn−1, we obtain

xdAn−1dx

− dAn−2dx

= (n− 1)An−1 (A.5)

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

xdAndx− dAn−1

dx= nAn (A.6)

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

dAndx− xdAn−1

dx= nAn−1 (A.7)

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

(1− x2)dAndx

= n(An−1 − xAn) (A.8)

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

(1− x2)d2Andx2

− 2xdAndx

+ n(n+ 1)An = 0 (A.9)

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

∫ 1

−1Pm(x)Pn(x) dx =

{0 if m 6= n

22n+1 if m = n

. (A.10)

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P0(x)

P1(x)

P2(x)

P3(x)

P4(x)

P5(x)

Figure 5.1: The Legendre polynomials

1

−1.........

x

y

Chapter 6

Fundamental Theory of ODEs

6.1 Existence-Uniqueness Theorem

We consider the initial value problem

dx

dt= f(t, x), 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 conditionwith 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. Thenthere is a unique solution to the initial value problem

dx

dt= f(t, x), x(t0) = x0

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 solutionwill be proved in section 6.5.

Example 6.1 Let f(t, x) = x2e−t2

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)|= |x21e−t

2

sin t− x22e−t2

sin t|= |e−t2 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 = 1√

x|x− 0|.

However, as x → 0+, 1√x→ +∞, 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 onR.

Proof. Since |fx(t, x)| is continuous on R, it attains its maximum value in R by the extreme valuetheorem. 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 replaceG 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 functiondefined 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). Thefirst step will be to show that the initial value problem (6.1.1) is equivalent to an integral equation,namely

x(t) = x0 +

∫ t

t0

f(s, x(s)) ds. (6.2.1)

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

φ(t) = x0 +

∫ t

t0

f(s, φ(s)) ds.

Theorem 6.2 A function φ is a solution of the initial value problem (6.1.1) on an interval I if andonly 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

φ′(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 onI . Integrating (6.2.2) from t0 to t we obtain

φ(t)− φ(t0) =

∫ t

t0

f(s, φ(s)) ds.

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 ofCalculus, that φ′(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. Thisfunction satisfies the initial condition φ0(t0) = x0, but does not in general satisfy (6.2.1). However,if we compute

φ1(t) = x0 +

∫ t

t0

f(s, φ0(s)) ds = x0 +

∫ t

t0

f(s, x0) ds,

we might expect φ1 is a closer approximation to a solution than φ0. In fact, if we continue theprocess and define successively

φ0(t) = x0, φk+1(t) = x0 +

∫ t

t0

f(s, φk(s)) ds, k = 0, 1, 2, . . . (6.2.3)

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

φ(t) = x0 +

∫ t

t0

f(s, φ(s)) ds.

Thus φ would be our desired solution.

We call the functions φ0, φ1, φ2, · · · defined by (6.2.3) successive approximations to a solution ofthe integral equation (6.2.1), or the initial value problem (6.1.1).

Example 6.4 Consider the initial value problem x′ = tx, x(0) = 1.


The integral equation corresponding to this problem is

x(t) = 1 +

∫ t

0

s · x(s) ds,

and the successive approximations are given by

φ0(t) = 1, φk+1(t) = 1 +

∫ t

0

sφk(s) ds, k = 0, 1, 2, . . . .

Thus φ1(t) = 1 +∫ t0s ds = 1 + t2

2 , φ2(t) = 1 +∫ t0s(1 + s2

2 ) ds = 1 + t2

2 + t4

2·4 , and it may beestablished by induction that

φk(t) = 1 +

(t2

2

)+

1

2!

(t2

2

)2

+ · · ·+ 1

k!

(t2

2

)k.

We recognize φk(x) as a partial sum for the series expansion of the function φ(t) = et2/2. 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 allt ∈ 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 regionT in R bounded by the two lines x − x0 = M(t − t0), x − x0 = −M(t − t0), and the linest− t0 = α, t− t0 = −α.

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(t0, x0)t0 + αt0 − α

x0 + b

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.. .

T

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

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

x− x0 = M(t− t0)

x− x0 = −M(t− t0)

φk

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..................................................................................... ..................................................................................... t0 + at0 − a

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

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

φk+1(t) = x0 +

∫ t

t0

f(s, φk(s)) ds,

exists as a continuous function on I . Moreover,

|φk+1(t)− x0| ≤∣∣∣∣∫ t

t0

|f(s, φk(s))| ds∣∣∣∣ ≤M |t− 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),


for all (t, x) ∈ R. Furthermore, assume f satisfies a Lipschitz condition with constant L in R. Thenthe successive approximations

φ0(t) = x0, φk+1(t) = x0 +

∫ t

t0

f(s, φk(s)) ds, k = 0, 1, 2, . . .

converges uniformly on the interval I = [t0 − α, t0 + α] with α = min{a, b/M}, to a solution of

the initial value problemdx

dt= f(t, x), x(t0) = x0 on I .

Proof. (a) Convergence of {φk(t)}. The key to the proof is the observation that φk may be writtenas

φk = φ0 + (φ1 − φ0) + (φ2 − φ1) + · · ·+ (φk − φk−1),

and hence φk(t) is a partial sum for the series

φ0(t) +

∞∑p=1

[φp(t)− φp−1(t)]. (6.3.1)

Therefore to show that the sequence {φk(t)} converges uniformly is equivalent to show that theseries (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 fort ∈ I . Moreover,

|φ1(t)− φ0(t)| ≤M |t− t0|, (6.3.2)

6.3. CONVERGENCE OF THE SUCCESSIVE APPROXIMATIONS 111

|φp(t)− φp−1(t)| ≤ MLp−1αp

p!=M

L

(Lα)p

p!. (6.3.4)

Since the series∞∑p=1

M

L

(Lα)p

p!converges to M

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

series

φ0(t) +

∞∑p=1

[φp(t)− φp−1(t)]

converges absolutely and uniformly on I . Thus the sequence of partial sum which is φk(t) convergesuniformly on I to a limit φ(t). Next we shall show that this limit φ is a solution of the integralequation (6.2.1).

(b) Properties of the limit φ. Since each φk is continuous on I and the sequence converges uniformlyto φ, the function φ is also continuous on I . Now if t1 and t2 are in I , we have

|φk+1(t1)− φk+1(t2)| =∣∣∣∣∫ t1

t2

f(s, φk(s)) ds

∣∣∣∣ ≤M |t1 − 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 continuouson 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) +

∞∑p=1

[φp(t)− φp−1(t)],

and

φk(t) = φ0(t) +

k∑p=1

[φp(t)− φp−1(t)].

Using (6.3.4), we have

|φ(t)− φk(t)| =

∣∣∣∣∣∣∞∑

p=k+1

[φp(t)− φp−1(t)]

∣∣∣∣∣∣≤

∞∑p=k+1

|φp(t)− φp−1(t)|

≤∞∑

p=k+1

M

L

(Lα)p

p!

≤ML

(Lα)k+1

(k + 1)!

∞∑p=0

(Lα)p

p!

≤ML

(Lα)k+1

(k + 1)!eLα.


Letting εk = (Lα)k+1

(k+1)! , we see that εk → 0 as k → ∞ as εk is a general term for the series eLα. Interms of εk, we may rewrite the above inequality as

|φ(t)− φk(t)| ≤ M

LeLαεk, and εk → 0 as k →∞ (6.3.6)

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

φ(t) = x0 +

∫ t

t0

f(s, φ(s)) ds,

for all t ∈ I . Note that since φ is continuous, the integrand f(s, φ(s)) of the right hand side iscontinuous on I . Since

φk+1(t) = x0 +

∫ t

t0

f(s, φk(s)) ds,

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

t0

f(s, φk(s)) ds→∫ t

t0

f(s, φ(s)) ds, as k →∞.

Now∣∣∣∣∫ t

t0

f(s, φ(s)) ds−∫ t

t0

f(s, φk(s)) ds

∣∣∣∣ ≤ ∣∣∣∣∫ t

t0

|f(s, φ(s))− f(s, φk(s))| ds∣∣∣∣

≤ L∣∣∣∣∫ t

t0

|φ(s)− φk(s)| ds∣∣∣∣

≤MeLαεk|t− t0| by (6.3.6)≤MαeLαεk → 0 as k →∞.

This completes the proof of the Theorem 6.4. �

Example 6.5 Consider the initial value problem x′ = (sin t)x2, x(0) = 12 . Let f(t, x) = (sin t)x2

be defined onR = {(t, x) : |t| ≤ 1, |x− 1

2| ≤ 1

2}.

|f(t, x)| = |(sin t)x2| ≤ 1. Thus we may take M = 1. Therefore by Theorem 6.4 a solution existson [−α, α] where α = min{1, 12} = 1

2 . In fact x(t) = (1 + cos t)−1 is a solution defined on themaximal domain (−π, π).

Exercise 6.1 Consider the initial value problem dxdt = tx+ x10, x(0) = 1

10 . Show that there existsa unique solution on [− 1

2 ,12 ].

6.4 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 theinitial value problem

x′(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 > Ma.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. Hencef(t, x) needs not be bounded in S. However, as in Theorem 5.4, we shall consider the series

φ0(t) +

∞∑p=1

(φp(t)− φp−1(t))

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 theseries. Let M0 = |x0| and M1 = max |φ1(t)|. The fact that M1 exists can be seen as follows. Sincef(t, x) is continuous in S, for a fixed x0, f(t, x0) is a continuous function on |t− t0| ≤ a.Thus φ1(t) = x0 +

∫ tt0f(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

|φ2(t)− φ1(t)| =∣∣∣∣∫ t

t0

[f(s, φ1(s))− f(s, φ0(s))] ds

∣∣∣∣ ≤ ∫ t

t0

|f(s, φ1(s))− f(s, φ0(s))| ds

≤ L∫ t

t0

|φ1(s)− φ0(s)| ds ≤ LM(t− t0), where L is the Lipschitz constant.

Now

|φ3(t)− φ2(t)| =∣∣∣∣∫ t

t0

[f(s, φ2(s))− f(s, φ1(s))] ds

∣∣∣∣ ≤ ∫ t

t0

|f(s, φ2(s))− f(s, φ1(s))| ds

≤ L∫ t

t0

|φ2(s)− φ1(s)| ds ≤ L2M

∫ t

t0

|(s− t0)| ds =L2M

2(t− t0)2.

Hence, in general, we can prove by induction that

|φn(t)− φn−1(t)| ≤ Ln−1M(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)| ≤ Ln−1M |t− t0|n−1

(n− 1)!≤ Ln−1M

(n− 1)!an−1.

Thus

|φ0(t)|+∞∑n=1

|φn(t)− φn−1(t)| ≤M +M

∞∑n=1

(La)n−1

(n− 1)!.

Hence each term on the left hand side of the above equation is less than the corresponding termof the convergent series of positive constants. Hence, by Weierstrass M -test, the series on the leftconverges 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) satisfiesthe integral equation

φ(t)− x0 −∫ t

t0

f(s, φ(s)) ds = 0. (6.4.1)


We know that

φn(t)− x0 −∫ t

t0

f(s, φn−1(s)) ds = 0. (6.4.2)

Substituting the value of x0 in (6.4.2) into the left hand side of (6.4.1), we get

φ(t)− x0 −∫ t

t0

f(s, φ(s)) ds = φ(t)− φn(t)−∫ t

t0

[f(s, φ(s))− f(s, φn−1(s))] ds.

Thus we obtain∣∣∣φ(t)− x0 −∫ tt0f(s, φ(s)) ds

∣∣∣≤|φ(t)− φn(t)|+∣∣∣∫ tt0 |f(s, φ(s))− f(s, φn−1(s))| ds

∣∣∣≤|φ(t)− φn(t)|+ L

∣∣∣∫ tt0 |φ(s)− φn−1(s)| ds∣∣∣ (6.4.3)

Since φn(t)→ φ(t) uniformly for t ∈ [t0 − a, t0 + a], the right hand side of (6.4.3) tends to zero asn→∞. Hence

φ(t)− x0 −∫ t

t0

f(s, φ(s)) ds = 0.

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 thatfor any a > 0, f satisfies a Lipschitz condition with respect to S = {(t, x) ∈ R2 : |t| ≤ a}. Thenthe initial value problem

x′(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 thisa, 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 x′ = 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, andby 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 x′ + g(t)x = h(t), x(t0) = x0 has a solution defined on I .

Exercise 6.3 Show that the initial value problem x′ =x3et

1 + x2+ t2 cosx, x(0) = 1 has a solution

on R.

6.5. GRONWALL’S INEQUALITY AND UNIQUENESS OF SOLUTION 115

6.5 Gronwall’s Inequality and Uniqueness of Solution

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

f(t) ≤ h(t) +

∫ t

t0

g(s)f(s) ds, t ≥ t0,

then

f(t) ≤ h(t) +

∫ t

t0

g(s)h(s)e∫ tsg(u) du ds, t ≥ t0.

Proof. First we are given

f(t) ≤ h(t) +

∫ t

t0

g(s)f(s) ds (6.5.1)

Let z(t) =∫ tt0g(s)f(s) ds. Then for t ≥ t0,

z′(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′(t) ≤ g(t)[h(t) + z(t)]

which gives

z′(t)− g(t)z(t) ≤ g(t)h(t).

This is a first order differential inequality which can be solved by finding an integrating factore−

∫ tt0g(u) du. Hence the solution is

z(t)e−

∫ tt0g(u) du ≤

∫ t

t0

g(s)h(s)e−

∫ st0g(u) du

ds

Or equivalently,

z(t) ≤∫ t

t0

g(s)h(s)e−

∫ st0g(u) du

e∫ tt0g(u) du

ds =

∫ t

t0

g(s)h(s)e∫ tsg(u) du ds (6.5.3)

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

t0

g(s)f(s) ds ≤∫ t

t0

g(s)h(s)e∫ tsg(u) du ds (6.5.4)

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

f(t)− h(t) ≤∫ t

t0

g(s)h(s)e∫ tsg(u) du ds.

�


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

f(t) ≤ k +

∫ t

t0

g(s)f(s) ds, for t ≥ t0,

thenf(t) ≤ ke

∫ tt0g(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

f(t) ≤ k∫ t

t0

f(s) ds

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

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

f(t) ≤ ε+ k

∫ t

t0

f(s) ds,

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

f(t) ≤ εe∫ tt0k ds

,

for all t ≥ t0, which gives f(t) ≤ εek(t−t0), for all t ≥ t0. Since ε is arbitrary, we get f(t) ≡ 0 bytaking limit as ε→ 0+. �

Remark. Similar results hold for t ≤ t0 when all the integrals are integrated from t to t0. Forexample, in Corollary 6.9, if

f(t) ≤ k∫ t0

t

f(s) ds

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 witha Lipschitz constant L, where R is either a rectangle or a strip. If φ and ϕ are two solutions of

x′ = 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

φ(t) = x0 +

∫ t

t0

f(s, φ(s)) ds,

6.5. GRONWALL’S INEQUALITY AND UNIQUENESS OF SOLUTION 117

and

ϕ(t) = x0 +

∫ t

t0

f(s, ϕ(s)) ds.

Thus

|φ(t)− ϕ(t)| ≤∫ t

t0

|f(s, φ(s))− f(s, ϕ(s))| ds ≤ L∫ t

t0

|φ(s)− ϕ(s)| ds.

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) hasat 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 con-tinuous 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 x′ = x2/3, x(0) = 0. We find that x(t) = 0 andx(t) = 1

27 t3 are both solutions.

Example 6.8 Suppose φ(t) is a solution to the initial value problem

x′ =x3 − x

1 + t2x2, 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, where0 ∈ J . Suppose there exists s ∈ J such that φ(s) ≥ 1. Without loss of generality, we may assumes > 0. Since φ(t) is continuous and φ(0) = 1/2, we have by the intermediate value theorem, thatφ(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

x′ =x3 − x

1 + t2x2, x(s0) = 1.

The function f(t, x) = x3−x1+t2x2 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. Theabove function φ(t) defined on J is a solution to this initial value problem, and it has the propertythat φ(t) < 1 for all t < s0. However, ϕ(t) ≡ 1 is clearly a solution to this initial value problem onI . But ϕ and φ are different solutions to the initial value problem contradicting the uniqueness ofthe 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},


or a rectangleR = {(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 solutionsdefined on I = [−a + t0, t0 + a] of x′ = f(t, x) satisfying the initial condition x(t0) = x0 andx(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 alsoclose. 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 rectangleR = {(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 continuouson R and

|g(t, x)− f(t, x)| ≤ ε, for all (t, x) ∈ R. (6.5.5)

Let φ and ϕ be solutions of the initial value problems x′ = f(t, x), x(t0) = x0, and x′ = g(t, x),x(t0) = x0 respectively. Assume both φ and ϕ are defined on I = [−a + t0, t0 + a]. Then for allt ∈ I ,

|φ(t)− ϕ(t)| ≤ εaeLa.

In particular, as g(t, x) approaches f(t, x) uniformly onR, that is, as ε→ 0+ in (6.5.5), the solutionφ(t) approaches ϕ(t) uniformly on I .

Proof. By the integral representations for φ(t) and ϕ(t), we have

φ(t) = x0 +

∫ t

t0

f(s, φ(s)) ds, ϕ(t) = x0 +

∫ t

t0

g(s, ϕ(s)) ds.

Subtracting these equations gives

φ(t)− ϕ(t) =

∫ t

t0

[f(s, φ(s))− g(s, ϕ(s))] ds.

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 119

|φ(t)− ϕ(t)| ≤∫ t

t0

|f(s, φ(s))− f(s, ϕ(s))| ds+

∫ t

t0

|f(s, ϕ(s))− g(s, ϕ(s))| ds

≤ L∫ t

t0

|φ(s)− ϕ(s)| ds+

∫ t

t0

ε ds

≤ L∫ t

t0

|φ(s)− ϕ(s)| ds+ εa.

By Gronwall’s inequality, we obtain

|φ(t)− ϕ(t)| ≤ εae∫ tt0Lds ≤ εaeLa, for t0 ≤ t ≤ t0 + a.

The last inequality is also valid for t0 − a ≤ t ≤ t0. Thus

|φ(t)− ϕ(t)| ≤ εaeLa 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

x′1 = f1(t, x1, · · · , xn),

x′2 = f2(t, x1, · · · , xn),

· · · · · · · · · · · · · · ·

x′n = fn(t, x1, · · · , xn),

where x′j =dxjdt . Let us introduce notations

x =

x1· · ·xn

, x′ =

x′1· · ·x′n

, f(t,x) =

f1(t,x)

· · ·fn(t,x)

.

Then the system can be written in a vector form:

x′ = f(t,x). (6.6.1)

Differential equations of higher order can be reduced to equivalent systems. Let us consider

dny

dtn+ F (t, y, y′, · · · , d

n−1y

dtn−1) = 0. (6.6.2)

Let

x1 = y, x2 =dy

dt, · · · , xn =

dn−1y

dtn−1.


Then (6.6.2) is equivalent to the following system

x′1 = x2,

x′2 = x3,

· · · · · · · · ·

x′n = −F (t, x1, x2, · · · , xn).

It can be written in the form of (6.6.1) if we let

x =

x1· · ·xn

, f(t,x) =

x2,

x3,

· · ·−F (t, x1, · · · , xn)

.

Recall that for a vector x =

x1

x2...xn

, its magnitude |x| is defined to be

|x| = |x1|+ |x2|+ · · ·+ |xn|.

The following 2 properties of the magnitude can be proved easily.1. Triangle Inequality |x + y| ≤ |x|+ |y|.2. If A = (aij) is an n× n matrix and x ∈ Rn, then |Ax| ≤ |A||x| where |A| =

∑ni,j |aij |.

Definition 6.2 Let G be a subset in R1+n. f(t,x) : G→ Rn is said to satisfy a Lipschitz conditionwith 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

f(t,x) =

(2x2 cos t

x1 sin t

), where x =

(x1

x2

).

Then |f(t,x)− f(t,y)| =

∣∣∣∣∣(

2x2 cos t

x1 sin t

)−

(2y2 cos t

y1 sin t

)∣∣∣∣∣= |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)


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.

Proof. Let f(t,x) =

f1(t,x)

f2(t,x)...

fn(t,x)

, where each fi(t,x) : R1+n −→ R.

Thus

∂f

∂xk=

∂f1∂xk∂f2∂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) isin G.

Now F′(s) =

n∑k=1

∂f

∂xk

dxkds

=n∑k=1

∂f1∂xk∂f2∂xk

...∂fn∂xk

(zk − yk).

Therefore,

|F′(s)| ≤n∑k=1

∣∣∣∣ ∂f

∂xk

∣∣∣∣ |zk − yk| ≤ L n∑k=1

|zk − yk| = L|z− y|,

for s ∈ [0, 1].Since

f(t, z)− f(t,y) = F(1)− F(0) =

∫ 1

0

F′(s) ds

we have |f(t, z)− f(t,y)| ≤∫ 1

0|F′(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),



for all (t,x) ∈ R. Furthermore, assume f satisfies a Lipschitz condition with constant L in R. Thenthere is a unique solution to the initial value problem

dx

dt= f(t,x), x(t0) = x0

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 theinitial value problem

x′(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. Sup-pose that for any a > 0, f satisfies a Lipschitz condition with respect to

S = {(t,x) ∈ Rn+1 : |t| ≤ a}.

Then the initial value problemx′(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 themethod of successive approximations. That is the successive approximations

φ0(t) = x0, φk+1(t) = x0 +

∫ t

t0

f(s, φk(s)) ds, k = 0, 1, 2, . . .

converge uniformly on the interval I = [t0 − α, t0 + α] with α = min{a, b/M}, to a solution

of the initial value problemdx

dt= f(t,x), x(t0) = x0 on I . Uniqueness is proved by Gronwall’s

inequality as before.

Example 6.10 Find the first 5 successive approximations to the initial value problem

x′′ = −etx, x(0) = 1, x′(0) = 0.

Solution. The initial value problem is equivalent to the following initial value problem of differentialsystem. (

x(t)

y(t)

)′=

(y(t)

−etx(t)

),

(x(0)

y(0)

)=

(1

0

).

We start with (x0(t)

y0(t)

)=

(1

0

), for all t ∈ R.

Then (x1(t)

y1(t)

)=

(1

0

)+

∫ t

0

(0

−es × 1

)ds =

(1

1− et

).


(x2(t)

y2(t)

)=

(1

0

)+

∫ t

0

(1− es

−es

)ds =

(2 + t− e−t

1− et

).

(x3(t)

y3(t)

)=

(1

0

)+

∫ t

0

(1− es

−es(2 + s− es)

)ds =

(2 + t− e−t

12 − e

t − tet + 12e

2t

).

(x4(t)

y4(t)

)=

(1

0

)+

∫ t

0

(12 − e

s − ses + 12e

2s

−es(2 + s− es)

)ds =

(34 + t

2 − tet + 1

4e2t

1 + t− et − tet

).

Example 6.11 Consider the linear differential system x′ = 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)| =|A(x1−x2)| ≤ |A||x1−x2|, where |A| =

∑ni=1

∑nj=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 anyinitial value and is defined on the entire R.

Example 6.12 Let x′ = A(t)x, where A(t) = (aij(t)) is an n× n matrix of continuous functionsdefined 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

∂f

∂xk=

a1k(t)

a2k(t)...

ank(t)

,

which is independent of x.Therefore, ∣∣∣∣ ∂f

∂xk

∣∣∣∣ =

n∑i=1

|aik(t)| ≤ nK ≡ L, 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 x′ = A(t)x has a unique solution forany initial value in S and is defined on all of I . Thus if A(t) = (aij(t)) is continuous on an openinterval I or R, then x′ = A(t)x has a unique solution for any initial value x(t0) = x0 with t0 ∈ Ior R, and is defined on I or R. (See corollary 6.6.) In particular this together with corollary 6.17prove theorem 2.1 and theorem 4.1.

Bibliography

[1] Ravi P. Agarwal and Ramesh C. Gupta, Essentials of ordianry differential equations, McGraw-Hill (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)

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