Laplace Transform

Differential Equations for

Engineers

————

Lecture Notes for Math 3503

B. O. J. Tupper

Department of Mathematics and StatisticsUniversity of New Brunswick

July, 2006

Contents

1 LAPLACE TRANSFORMS 1

1.1 Introduction and definition. . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Examples of Laplace transforms. . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 The gamma function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.4 Inverse transforms and partial fractions. . . . . . . . . . . . . . . . . . . . . 10

1.5 The First Shifting Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.6 Step functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.7 Differentiation and integration of transforms. . . . . . . . . . . . . . . . . . 25

1.8 Laplace transforms of derivatives and integrals. . . . . . . . . . . . . . . . . 28

1.9 Application to ordinary differential equations. . . . . . . . . . . . . . . . . . 31

1.10 Discontinuous forcing functions. . . . . . . . . . . . . . . . . . . . . . . . . . 35

1.11 Periodic functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

1.12 Impulse functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

1.13 The convolution integral. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

2 SYSTEMS OF FIRST ORDER LINEAR EQUATIONS 53

2.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

2.2 Basic theory of systems of first order linear equations. . . . . . . . . . . . . 55

2.3 Review of eigenvalues and eigenvectors. . . . . . . . . . . . . . . . . . . . . 64

2.4 Homogeneous linear systems with constant coefficients. . . . . . . . . . . . . 71

2.5 Complex eigenvectors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

i

0 CONTENTS

2.6 Repeated eigenvalues. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

2.7 Nonhomogeneous systems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

2.8 Laplace transform method for systems. . . . . . . . . . . . . . . . . . . . . . 88

3 FOURIER SERIES 91

3.1 Orthogonal sets of functions. . . . . . . . . . . . . . . . . . . . . . . . . . . 91

3.2 Expansion of functions in series of orthogonal functions. . . . . . . . . . . . 95

3.3 Fourier series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

3.4 Cosine and sine series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

3.5 Half-range expansions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

3.6 Complex form of the Fourier series. . . . . . . . . . . . . . . . . . . . . . . . 115

3.7 Separable partial differential equations. . . . . . . . . . . . . . . . . . . . . 117

A ANSWERS TO ODD-NUMBERED PROBLEMS AND TABLES 137

A.1 Answers for Chapter 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137



A.4 Table of Laplace transforms. . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

Chapter 1

LAPLACE TRANSFORMS

1.1 Introduction and definition.

Integral transforms are powerful tools for solving linear differential equations; they

change the differential equations into algebraic equations involving the initial values of

the dependent variable and its derivatives. An integral transform changes a function f(t)

of one variable into a function F (s) of another variable by means of a definite integral of

the form

F (s) =

∫ b

aK(s, t)f(t) dt. (1.1)

The function K(s, t) is called the kernel of the transformation and F (s) is the transform of

f(t).

The Laplace transform L{f(t)} or F (s) of a function f(t), defined for t ≥ 0, is defined

by the equation

L{f(t)} = F (s) =

∫

∞

0e−stf(t) dt = lim

b→∞

∫ b

0e−stf(t) dt, (1.2)

provided that the limit exists.

The Laplace transform is a linear transform, i.e., if L{f(t)} = F (s) and L{g(t)} = G(s),

then L{αf(t) + βg(t)} = αF (s) + βG(s), where α and β are constants.

1

2 CHAPTER 1. LAPLACE TRANSFORMS

The Laplace transform L{f(t)} will exist if the following conditions are satisfied by the

function f(t):

(i) f(t) is piecewise continuous on the interval [0,∞), (1.3)

(ii) f(t) is of exponential order for t > T , i.e., there exist numbers c, M > 0, T > 0 such

that | f(t) | ≤ Mect for t > T . (1.4)

The Laplace transform then exists for s > c.

A function f(t) is piecewise continuous on [0,∞), if, in any interval 0 ≤ a ≤ t ≤ b, there

are at most a finite number of points tk, k = 1, 2, ..., n at which f has finite discontinuities

and is continuous on each open interval tk−1 < t < tk.

f(t)

t

In other words, f(t) is piecewise continuous on [0,∞) if it is continuous on the interval

except for a finite number of finite jump discontinuities. If f(t) is piecewise continuous on

a ≤ t ≤ b, where b is finite, then the integral

∫ b

af(t) dt exists.

The condition (1.4) is satisfied by many functions. For example, tn where n > 0, cos at,

sin at, eat, but is not satisfied by et2 since this grows faster than any positive linear power

of e.

1.2. EXAMPLES OF LAPLACE TRANSFORMS. 3

While the conditions (1.3) and (1.4) are sufficient to ensure the existence of the Laplace

transform, they are not necessary. For example, the function f(t) = t−12 is infinite at t = 0,

and so is not piecewise continuous there, but, as we shall see, its transform does exist.

Now we give the proof that if the conditions (1.3) and (1.4) hold, then the Laplace

transform exists.

L{f(t)} =

∫

∞

0e−stf(t) dt

=

∫ T

0e−stf(t) dt +

∫

∞

Te−stf(t) dt.

The first integral exists because f(t) is piecewise continuous. For t ≥ T , we have, by (1.4)

| e−stf(t) | ≤ e−stMect = Me−(s−c)t,

so that

∣

∣

∣

∣

∫

∞

Te−stf(t) dt

∣

∣

∣

∣

≤ M

∫

∞

Te−(s−c)tdt

= −M

[

e−(s−c)t

s − c

]

∞

T

,

which converges provided that −(s − c) is negative, i.e., s > c. This proves the result.

1.2 Examples of Laplace transforms.

Example 1. Evaluate L{1}.

L{1} =

∫

∞

0e−st1 dt =

[

−1

se−st

]

∞

0

=1

s, provided s > 0.

∴ L{1} =1

s, s > 0,

i.e., if f(t) = 1, then F (s) =1

s.


Example 2. Evaluate L{t}.

L{t} =

∫

∞

0e−stt dt =

[

t(−1

se−st)

]

∞

0

−∫

∞

01(−1

se−st) dt

= 0 −[

1

s2e−st

]

∞

0

=1

s2, provided s > 0.

∴ L{t} =1

s2, s > 0.

Example 3. Evaluate L{tn}, where n is a positive integer.

L{tn} =

∫

∞

0e−sttndt =

[

tn(−1

se−st)

]

∞

0

−∫

∞

0ntn−1(−1

se−st) dt

= 0 +n

s

∫

∞

0e−sttn−1dt , provided s > 0.

∴ L{tn} =n

sL{tn−1}.

Put n = 2. L{t2} =2

sL{t} =

2

s3.

Put n = 3. L{t3} =3

sL{t2} =

3!

s4.

Continuing this process we see that

L{tn} =n!

sn+1, s > 0.

Example 4. Evaluate L{eat}.

L{eat} =

∫

∞

0e−steatdt =

[

e−(s−a)t

−(s − a)

]

∞

0

= 0 +1

s − a=

1

s − a, s > a.

Example 5. Evaluate L{cos at} and L{sin at}.

L{cos at} =

∫

∞

0e−st cos at dt

L{sin at} =

∫

∞

0e−st sin at dt

1.2. EXAMPLES OF LAPLACE TRANSFORMS. 5

Now cos at + i sin at = eiat, so that

L{eiat} = L{cos at} + iL{sin at}

=

∫

∞

0e−steiatdt =

[

e−(s−ia)t

−(s − ia)

]

∞

0

= 0 +1

s − ia=

s + ia

s2 + a2, s > 0.

∴ L{cos at} =s

s2 + a2, L{sin at} =

a

s2 + a2.

Example 6. Evaluate L{cos2 t}.

Now cos2 t =1

2(1 + cos 2t), so

L{cos2 t} =1

2L{1} +

1

2L{cos 2t}

=1

2s+

s

2(s2 + 4).

Example 7. Evaluate L{cosh at} and L{sinh at}.

Now cosh at =1

2(eat + e−at), so

L{cosh at} =1

2L{eat} +

1

2L{e−at}

=1

2

1

s − a+

1

2

1

s + a, s > a and s > −a

=s

s2 − a2, s > | a | .

Similarly, L{sinh at} =1

2L{eat} − 1

2L{e−at}

=a

s2 − a2, s > | a | .

Example 8. Evaluate L{t2eat}

L{t2eat} =

∫

∞

0e−stt2eatdt

=

∫

∞

0t2e−(s−a)tdt


=

[

t2 · e−(s−a)t

−(s − a)

]

∞

0

−∫

∞

02t ·

[

e−(s−a)t

−(s − a)

]

dt

= 0 +2

s − a

∫

∞

0te−(s−a)tdt , s > a

=2

s − a

{[

t · e−(s−a)t

−(s − a)

]

∞

0

−∫

∞

0

e−(s−a)t

−(s − a)dt

}

= 0 +2

(s − a)2

[

e−s−a)t

−(s − a)

]

∞

0

, s > a

=2

(s − a)3[0 + 1] =

2

(s − a)3, s > a.

Example 9. Evaluate L{f(t)} where f(t) =

{

0 0 ≤ t < 21 t ≥ 2

.

L{f(t)} =

∫

∞

0e−stf(t) dt

=

∫ 2

0e−st · 0 dt +

∫

∞

2e−st · 1 dt

= 0 +

[

e−st

−s

]

∞

2

=e−2s

s, s > 0.

Example 10. Evaluate L{f(t)} where f(t) is as shown in the diagram.

f(t) =

{

kc t 0 ≤ t < c

0 t ≥ c

k

c

f(t)

t

1.3. THE GAMMA FUNCTION. 7

L{f(t)} =

∫ c

0e−st k

ct dt =

k

c

{[

t

(

e−st

−s

)]c

0

−∫ c

01e−st

−sdt

}

=k

c

{

−ce−sc

s− 0 +

[

e−st

−s2

]c

0

}

=k

c

{

−ce−sc

s− e−sc

s2+

1

s2

}

.

1.3 The gamma function.

The gamma function Γ(x) is defined by

Γ(x) =

∫

∞

0e−ttx−1dt , x > 0. (1.5)

This definite integral converges when x is positive and so defines a function of x for positive

values of x.

Now

Γ(1) =

∫

∞

0e−tdt = 1 (1.6)

and integrating by parts we see that

Γ(x + 1) =

∫

∞

0txe−tdt = x

∫

∞

0tx−1e−tdt + [−txe−t]∞0

= x

∫

∞

0tx−1e−tdt,

and hence

Γ(x + 1) = xΓ(x), (1.7)

so that when x is a positive integer

Γ(2) = 1 × Γ(1) = 1Γ(3) = 2 × Γ(2) = 2

......

...Γ(n + 1) = n!


In particular 0! = Γ(1) = 1. For this reason the gamma function is often referred to as

the generalized factorial function.

Since Γ(x) is defined for x > 0, Eq. (1.7) shows that it is defined for 0 > x > −1

and repeated use of Eq (1.7) in the form Γ(x) =1

xΓ(x + 1) shows that Γ(x) is defined in

the intervals −2 < x < −1, −3 < x < −2, etc. However, Eqs. (1.6) and (1.7) show that

Γ(0) =1

0, so the gamma function becomes infinite when x is zero or a negative integer.

In the definition (1.5) make the substitution t = u2, then

Γ(x) = 2

∫

∞

0u2x−1e−u2

du

and putting x =1

2we obtain

Γ

(

1

2

)

= 2

∫

∞

0e−u2

du. (1.8)

But, for a definite integral,

∫

∞

0e−u2

du =

∫

∞

0e−v2

dv, so that

[

Γ

(

1

2

)]2

= 4

∫

∞

0e−u2

du

∫

∞

0e−v2

dv

= 4

∫

∞

0

∫

∞

0e−(u2+v2)du dv. (1.9)

To evaluate this integral we transform to polar co-ordinates u = r cos θ, v = r sin θ and

the double integral (1.9) becomes

[

Γ

(

1

2

)]2

= 4

∫ π

2

0

∫

∞

0e−r2

r dr dθ = π.

∴ Γ

(

1

2

)

=√

π. (1.10)

Using Eq. (1.7) then shows that

Γ

(

3

2

)

=1

2Γ

(

1

2

)

=1

2

√π, (1.11)

−1

2Γ

(

−1

2

)

= Γ

(

1

2

)

⇒ Γ

(

−1

2

)

= −2√

π. (1.12)

1.3. THE GAMMA FUNCTION. 9

Example 1. Evaluate L{ta}, where a is any number satisfying a > −1.

L{ta} =

∫

∞

0e−sttadt.

Put st = u, i.e., t =u

s, dt =

du

s. Then

L{ta} =

∫

∞

0e−u ua

sa

du

s=

1

sa+1

∫

∞

0e−uuadu

=Γ(a + 1)

sa+1,

from Eq. (1.5). Note that we must have a > −1 for the integral to converge at u = 0. Thus

we have the result

L{ta} =Γ(a + 1)

sa+1for all a > −1. (1.13)

When a is a positive integer this corresponds to the result of Example 3 in Section 1.2.

Example 2. Evaluate L{t− 12 } and L{t 1

2 }.

From Eqs. (1.10), (1.12) and (1.13) we find that

L{t− 12 } =

Γ(12)

s12

=

√

π

s

and

L{t 12 } =

Γ(32)

s32

=12Γ(1

2)

s32

=1

2

√

π

s3.

Problem Set 1.3

In problems 1-14 use the definition (1.2) to find L{f(t)}.

1. f(t) =

{

−1 0 ≤ t < 11 t ≥ 1

2. f(t) =

{

4 0 ≤ t < 20 t ≥ 2

3. f(t) =

{

t 0 ≤ t < 11 t ≥ 1

4. f(t) =

{

2t + 1 0 ≤ t < 10 t ≥ 1

5. f(t) =

{

sin t 0 ≤ t < π0 t ≥ π

6. f(t) =

{

0 0 ≤ t < π2

cos t t ≥ π2


7. f(t) = et+7 8. f(t) = e−2t−5

9. f(t) = te4t 10. f(t) = t2e3t

11. f(t) = e−t sin t 12. f(t) = et cos t

13. f(t) = t cos t 14. f(t) = t sin t

In problems 15 - 40 use the results of the examples in Sections 1.2 and 1.3 to evaluate

L{f(t)}.

15. f(t) = 2t4 16. f(t) = t5 17. f(t) = 4t − 10

18. f(t) = 7t + 3 19. f(t) = t2 + 6t − 3 20. f(t) = −4t2 + 16t + 9

21. f(t) = (t + 1)3 22. f(t) = (2t − 1)3 23. f(t) = 1 + e4t

24. f(t) = t2 − e−9t + 5 25. f(t) = (1 + e2t)2 26. f(t) = (et − e−t)2

27. f(t) = 4t2 − 5 sin 3t 28. f(t) = cos 5t + sin 2t 29. f(t) = t sinh t

30. f(t) = et sinh t 31. f(t) = e−t cosh t 32. f(t) = sin 2t cos 2t

33. f(t) = sin2 t 34. f(t) = cos t cos 2t 35. f(t) = sin t sin 2t

36. f(t) = sin t cos 2t 37. f(t) = sin3 t 38. f(t) = t32

39. f(t) = t14 40. f(t) = (t

12 + 1)2

1.4 Inverse transforms and partial fractions.

The Laplace transform of a function f(t) is denoted by L{f(t)} = F (s). If we are

given F (s) and are required to find the corresponding function f(t), then f(t) is the

inverse Laplace transform of F (s) and we write

f(t) = L−1{F (s)}.

From the examples of Section 1.2 we can find the following inverse Laplace transforms:

1.4. INVERSE TRANSFORMS AND PARTIAL FRACTIONS. 11

L−1

{

1

s

}

= 1, L−1

{

n!

sn+1

}

= tn,

L−1

{

1

s − a

}

= eat, L−1

{

a

s2 + a2

}

= sin at,

L−1

{

s

s2 + a2

}

= cos at, L−1

{

a

s2 − a2

}

= sinh at,

L−1

{

s

s2 − a2

}

= cosh at.

Example 1. Evaluate L−1

{

5

s + 3

}

.

L−1

{

5

s + 3

}

= 5L−1

{

1

s + 3

}

= 5e−3t.


{

2

s2 + 16

}

.

L−1

{

2

s2 + 16

}

=1

2L−1

{

4

s2 + 16

}

=1

2sin 4t.


{

s + 1

s2 + 1

}

.

L−1

{

s + 1

s2 + 1

}

= L−1

{

s

s2 + 1

}

+ L−1

{

1

s2 + 1

}

= cos t + sin t.


{

1

s4

}

.

L−1

{

1

s4

}

=1

3!L−1

{

3!

s4

}

=1

6t3.

In order to find inverse transforms we very often have to perform a partial fraction

decomposition. Here are some typical examples:


{

s2 − 10s − 25

s3 − 25s

}

.

s2 − 10s − 25

s3 − 25s=

s2 − 10s − 25

s(s2 − 25)=

s2 − 10s − 25

s(s − 5)(s + 5).

Write the last fraction ass2 − 10s − 25

s(s − 5)(s + 5)=

A

s+

B

s − 5+

C

s + 5. We need to find A, B and

C. Multiply both sides of the equation by s(s − 5)(s + 5) and we obtain

s2 − 10s − 25 = A(s − 5)(s + 5) + Bs(s + 5) + Cs(s − 5).


The denominator is zero when s = 0, +5, −5, so put these values into the above equation.

s = 0 −25 = A(−5)(5) ∴ A = 1.

s = 5 25 − 50 − 25 = B(5)(10) ∴ B = −1.

s = −5 25 + 50 − 25 = C(−5)(−10) ∴ C = 1.

Hence

s2 − 10s − 25

s(s − 5)(s + 5)=

1

s− 1

s − 5+

1

s + 5.

∴ L−1

{

s2 − 10s − 25

s(s − 5)(s + 5)

}

= L−1

{

1

s

}

− L−1

{

1

s − 5

}

+ L−1

{

1

s + 5

}

= 1 − e5t + e−5t.


{

2s − 1

s3(s + 1)

}

.

Write2s − 1

s3(s + 1)=

A

s+

B

s2+

C

s3+

D

s + 1.

Multiply both sides of the equation by s3(s + 1) to get

2s − 1 = As2(s + 1) + Bs(s + 1) + C(s + 1) + Ds3.

Put s = 0 : −1 = C ∴ C = −1

Put s = −1 : −3 = −D ∴ D = 3,

i.e., 2s − 1 = A(s3 + s2) + B(s2 + s) − s − 1 + 3s3.

s3 terms : A + 3 = 0 ∴ A = −3

s2 terms : A + B = 0 ∴ B = 3

s terms : B − 1 = 2 ∴ B = 3.

∴

2s − 1

s3(s + 1)= −3

s+

3

s2− 1

s3+

3

s + 1.

1.5. THE FIRST SHIFTING THEOREM. 13

∴ L−1

{

2s − 1

s3(s + 1)

}

= −3L−1

{

1

s

}

+ 3L−1

{

1

s2

}

− L−1

{

1

s3

}

+ 3L−1

{

1

s + 1

}

= −3 + 3t − 1

2t2 + 3e−t.


{

s + 2

(s + 1)(s2 + 4)

}

.

Writes + 2

(s + 1)(s2 + 4)=

A

s + 1+

Bs + C

s2 + 4.

Multiply both sides by (s + 1)(s2 + 4) to get

s + 2 = A(s2 + 4) + (Bs + C)(s + 1).

Put s = −1: 1 = 5A ∴ A = 15

s2 terms 0 = A + B ∴ B = −15

s terms 1 = B + C ∴ C = 65 .

∴

s + 2

(s + 1)(s2 + 4)=

15

s + 1+

−15s + 6

5

s2 + 4.

∴ L−1

{

s + 2

(s + 1)(s2 + 4)

}

=1

5L−1

{

1

s + 1

}

− 1

5L−1

{

s

s2 + 4

}

+3

5L−1

{

2

s2 + 4

}

.

=1

5e−t − 1

5cos 2t +

3

5sin 2t.

Problem Set 1.4

Find f(t) if L{f(t)} is given by

1.s + 12

s2 + 4s2.

s − 3

s2 − 13.

3s

s2 + 2s − 8

4.2s2 + 5s − 1

s3 − s5.

s + 1

s3(s − 1)(s + 2)6.

3s2 − 2s − 1

(s − 3)(s2 + 1)

1.5 The First Shifting Theorem.

This theorem expands our ability to find Laplace transforms and their inverses.


The First Shifting Theorem: If L{f(t)} = F (s) when s > c, then

L{eatf(t)} = F (s − a), s > c + a. (1.14)

Proof:

L{eatf(t)} =

∫

∞

0e−steatf(t) dt

=

∫

∞

0e−(s−a)tf(t) dt = F (s − a).

Example 1. Evaluate L{eattn}.

L{tn} =n!

sn+1so L{eattn} =

n!

(s − a)n+1.

Example 2. Evaluate L{eat sin bt}.

L{sin bt} =b

s2 + b2so L{eat sin bt} =

b

(s − a)2 + b2.

Example 3. Evaluate L{eat cos bt}.

L{cos bt} =s

s2 + b2so L{eat cos bt} =

s − a

(s − a)2 + b2.


{

2

s2 + 2s + 2

}

.

Now s2 + 2s + 2 = (s + 1)2 + 1, so we require

L−1

{

2

(s + 1)2 + 1

}

= 2L−1

{

1

(s + 1)2 + 1

}

.

The quantity inside { } is identical with the answer to Example 2 above with a = −1, b = 1.

∴ L−1

{

2

s2 + 2s + 2

}

= 2e−t sin t.


{

3s + 9

s2 + 2s + 10

}

.

1.5. THE FIRST SHIFTING THEOREM. 15

Now

3s + 9

s2 + 2s + 10=

3(s + 1) + 6

(s + 1)2 + 9=

3(s + 1)

(s + 1)2 + 9+

6

(s + 1)2 + 9.

As in Examples 2 and 3 with a = −1, b = 3, hence

L−1

{

3s + 9

s2 + 2s + 10

}

= 3e−t cos 3t + 2e−t sin 3t.

Problem Set 1.5

Use partial fractions (if necessary) and the First Shifting Theorem to find the inverse

Laplace transforms of the following functions:

1.10 − 4s

(s − 2)22.

s2 + s − 2

(s + 1)33.

s3 − 7s2 + 14s − 9

(s − 1)2(s − 2)3

4.s2 − 6s + 7

(s2 − 4s + 5)s5.

2s − 1

s2(s + 1)36.

3!

(s − 2)4

Find the Laplace transforms of the following functions:

7. L{te8t} 8. L{t7e−5t} 9. L{e−2t cos 4t}

10. L{e3t sinh t} 11. L{

sin 2t

et

}

12. L{e2t cos2 2t}

13. L{e3t(t + 2)2} 14. L{t1/2(et + e−2t)}


1.6 Step functions.

The unit step function ua(t) is defined as follows:

ta

u (t)a

1 ua(t) =

{

0 when t < a1 when t ≥ a

(a ≥ 0). (1.15)

When a = 0 we have

u (t)

1

0

t

u0(t) =

{

0 t < 01 t ≥ 0

.

Note that the function f(t) = 1 − uc(t) is given by

tc

f(t)

1 f(t) =

{

1 0 ≤ t < c0 t ≥ c

.

The difference between two step functions, i.e.,

f(t) = ua(t) − ub(t) (b > a)

1.6. STEP FUNCTIONS. 17

has a graph of the form:

t

f(t)

a b

1

Step functions can be used to turn on or turn off portions of the graph of a function. For

example, the function f(t) = t2 when multiplied by u1(t) becomes

f(t) = t2u1(t) =

{

0 0 ≤ t < 1t2 t ≥ 1

,

so that its graph is

1

f(t)

1

t

Given the function f(t) = t2 note the graphs of the following functions:


f(t) = t2

t

f(t)

f(t)=t 2

f(t) = t2 , t ≥ 0

t

f(t)=t

f(t)

2

> 0t

f(t − 1) , t ≥ 0

t

f(t)

1

f(t- )

1

1


f(t − 1)u1(t) , t ≥ 0

1

t1

f(t)

f(t- )1

u (t)

Hence, given a function f(t), defined for t ≥ 0, the graph of the function f(t − a)ua(t)

consists of the graph of f(t) translated through a distance a to the right with the portion

from 0 to a ‘turned off’, i.e., put equal to zero.

The Laplace transform of ua(t) is

L{ua(t)} =

∫

∞

0e−stua(t) dt =

∫

∞

ae−stdt

=

[

e−st

−s

]

∞

a

=e−as

s,

i.e., L{ua(t)} =e−as

s, (s > 0). (1.16)


Example 1. Write f(t) in terms of unit step functions and find its Laplace transform

where

f(t) =

1 0 ≤ t < 10 1 ≤ t < 21 2 ≤ t < 30 t ≥ 3

.

Since f(t) can be expressed as

f(t) = u0(t) − u1(t) + u2(t) − u3(t),

we then have

L{f(t)} =1

s− e−s

s+

e−2s

s− e−3s

s=

1

s(1 − e−s)(1 + e−2s).

Example 2. Represent the function shown in the diagram below in terms of

unit step functions and find its Laplace transform.

t2 4 6 8

1

-1

-2

-3

f(t)

O

f(t) = −u0(t) − 2u2(t) + 4u4(t) − u6(t).

∴ L{f(t)} = −1

s− 2

se−2s +

4

se−4s − 1

se−6s = −1

s(1 − e−2s)(1 + 3e−2s − e−4s).


Example 3. The function shown in the diagram below is periodic with period 3. Write

the function in terms of unit step functions and find its Laplace transform.

t2 4

1

1/2

3

f(t)

1O

f(t) = u0(t) −1

2u1(t) −

1

2u2(t) + u3(t) −

1

2u4(t) −

1

2u5(t) . . .

= (u0 + u3 + u6 + . . . ) − 1

2(u1 + u4 + u7 + . . . ) − 1

2(u2 + u5 + u8 + . . . ).

L{f(t)} =1

s(1 + e−3s + e−6s + . . . ) − 1

2s(e−s + e−4s + e−7s + . . . )

− 1

2s(e−2s + e−5s + e−8s + . . . )

=1

s

1

1 − e−3s− 1

2s

e−s

1 − e−3s− 1

2s

e−2s

1 − e−3s(Geometric Series)

=2 − e−s − e−2s

2s(1 − e−3s).

The Second Shifting Theorem: If L{f(t)} = F (s), then, for a > 0,

L{f(t − a)ua(t)} = e−asF (s), (1.17)

and, conversely,

L−1{e−asF (s)} = f(t − a)ua(t). (1.18)

Proof:

L{f(t − a)ua(t)} =

∫

∞

0e−stf(t − a)ua(t) dt


=

∫

∞

ae−stf(t − a) dt.

Put v = t − a, then v = 0 when t = a, dv = dt and the integral becomes

L{f(t − a)ua(t)} =

∫

∞

0e−s(v+a)f(v) dv

= e−as

∫

∞

0e−svf(v) dv

= e−asF (s).

Example 4. Evaluate L{(t − π)uπ(t)}.

f(t) = t, so F (s) =1

s2and L{(t − π)uπ(t)} =

e−πs

s2.

Example 5. Evaluate L{tu2(t)}.

Because the step function is u2(t), i.e., has suffix 2, the function t must be written as a

function of (t − 2), i.e., t = (t − 2) + 2.

∴ L{tu2(t)} = L{(t − 2)u2(t) + 2u2(t)}

=e−2s

s2+

2e−2s

s=

(1 + 2s)

s2e−2s,

because f(t) = t in the first term.

Example 6. Evaluate L{sin(t − 3)u3(t)}.

f(t) = sin t, so L{sin(t − 3)u3(t)} =1

s2 + 1e−3s.

Example 7. Find the Laplace transform of the function

g(t) =

{

0 t < 1t2 − 2t + 2 t ≥ 1

.

Now t2 − 2t + 2 = (t − 1)2 + 1,

i.e., f(t − 1) = (t − 1)2 + 1 so f(t) = t2 + 1.


∴ L{g(t)} = L{f(t − 1)u1(t)} = e−sL{f(t)}

= e−s(2

s3+

1

s).


{

se−πs

s2 + 4

}

.

Now

L−1

{

s

s2 + 4

}

= cos 2t = f(t).

∴ L−1

{

se−πs

s2 + 4

}

= f(t − π)uπ(t) = cos 2(t − π)uπ(t) = cos 2tuπ(t).

Example 9. Evaluate L−1{e−s

s4}.

The e−s indicates that u1(t) appears in the function and so does f(t − 1).

L−1{ 1

s4} =

1

6t3 = f(t), so f(t − 1) =

1

6(t − 1)3 and

L−1{e−s

s4} =

1

6(t − 1)3u1(t).


{

1 + e−πs/2

s2 + 1

}

.

This is the sum of the inverse transforms

L−1

{

1

s2 + 1

}

and L−1

{

e−πs/2

s2 + 1

}

.

The first equals sin t; the second changes sin t to sin(t − π2 ) = − cos t and multiplies by

uπ

2(t).

∴ L−1

{

1 + e−πs/2

s2 + 1

}

= sin t − uπ

2(t) cos t.


Problem Set 1.6

Evaluate the following

1. L{(t − 1)u1(t)} 2. L{e2−tu2(t)} 3. L{(3t + 1)u3(t)}

4. L{(t − 1)3et−1u1(t)} 5. L{tet−5u5(t)} 6. L{cos t · u2π(t)}

7. L−1

{

e−2s

s3

}

8. L−1

{

(1 + e−2s)2

s + 2

}

9. L−1

{

e−πs

s2 + 1

}

10. L−1

{

se−πs/2

s2 + 4

}

11. L−1

{

e−s

s(s + 1)

}

12. L−1

{

e−2s

s2(s − 1)

}

13. L−1

{

1 − e−s

s2

}

14. L−1

{

2

s− 3e−s

s2+

5e−2s

s2

}

In Problems 15 - 20 write each function in terms of unit step functions and find the

Laplace transform of the given function.

15. f(t) =

{

2, 0 ≤ t < 3−2, t ≥ 3

16. f(t) =

1, 0 ≤ t < 40, 4 ≤ t < 51, t ≥ 5

17. f(t) =

{

0, 0 ≤ t < 1t2, t ≥ 1

18. f(t) =

{

0, 0 ≤ t < 3π2

sin t, t ≥ 3π2

19. f(t) =

{

t, 0 ≤ t < 20, t ≥ 2

20. f(t) is the staircase function,

i.e., see graph below.

t4

1

f(t)

2 3

2

3

1O

1.7. DIFFERENTIATION AND INTEGRATION OF TRANSFORMS. 25

1.7 Differentiation and integration of transforms.

Theorem: If F (s) = L{f(t)}, then

F ′(s) = L{−tf(t)}, (1.19)

and F (n)(s) = L{(−1)ntnf(t)}. (1.20)

Proof:

F (s) = L{f(t)} =

∫

∞

0e−stf(t) dt.

∴

dF

ds≡ F ′(s) =

∫

∞

0−te−stf(t) dt

=

∫

∞

0e−st[−tf(t)] dt.

∴ F ′(s) = L{−tf(t)}.

By continuing to differentiate with respect to t we see that each differentiation produces

another factor −t, so the result (1.20) follows easily.

Example 1. Evaluate L{t cos 2t}.

This is equal to (from (1.19)) −F ′(s) where F (s) = L{cos 2t} =s

s2 + 4. Hence

L{t cos t} = −(

s

s2 + 4

)

′

= −[

1(s2 + 4) − s · 2s

(s2 + 4)2

]

=s2 − 4

(s2 + 4)2.

Example 2. Evaluate L{te2t}.

f(t) = e2t∴ F (s) =

1

s − 2.

Hence

L{te2t} = −F ′(s) =1

(s − 2)2.


Example 3. Evaluate L{t2et}.

From (1.2), with n = 2, L{t2f(t)} = F ′′(s).

f(t) = et∴ F (s) =

1

s − 1∴ F ′(s) =

−1

(s − 1)2, F ′′(s) =

2

(s − 1)3.

∴ L{t2et} =2

(s − 1)3.

Example 4. Evaluate L{te−2t sin wt}.

From the First Shifting Theorem L{e−2t sin wt} =w

(s + 2)2 + w2,

i.e., if f(t) = e−2t sinwt then L{f(t)} =w

(s + 2)2 + w2= F (s).

∴ L{tf(t)} = −F ′(s) =w · 2(s + 2)

[(s + 2)2 + w2]2.

Example 5. Evaluate L−1{ 1

(s + 1)2}.

1

(s + 1)2= [− 1

s + 1]′, so if F (s) =

1

s + 1then

1

(s + 1)2= −F ′(s) and f(t) = L−1{F (s)} = e−t.

∴

1

(s + 1)2= −F ′(s) = L{te−t}.

Hence L−1

{

1

(s + 1)2

}

= te−t.


{

2s

(s2 − 4)2

}

.

Now2s

(s2 − 4)2= −

(

1

s2 − 4

)

′

= −F ′(s), where

1.7. DIFFERENTIATION AND INTEGRATION OF TRANSFORMS. 27

F (s) =1

s2 − 4= L{1

2sinh 2t}, i.e., f(t) =

1

2sinh 2t.

∴ −F ′(s) = L{t · 1

2sinh 2t},

i.e., L−1

{

2s

(s2 − 4)2

}

=1

2t sinh 2t.

Theorem: If f(t) satisfies the condition for the existence of the Laplace transform

and if limt→0+

f(t)

texists, then

L{

f(t)

t

}

=

∫

∞

sF (s) ds , (s > c) . (1.21)

Proof:

∫

∞

sF (s) ds =

∫

∞

s

[∫

∞

0e−stf(t) dt

]

ds .

Reversing the order of integration, we obtain

∫

∞

sF (s) ds =

∫

∞

0

[∫

∞

se−stf(t) ds

]

dt =

∫

∞

0f(t)

[∫

∞

se−st ds

]

dt

=

∫

∞

0f(t)

[

−1

te−st

]

∞

s

dt =

∫

∞

0e−st f(t)

tdt

= L{f(t)

t}, (s > c).

Example 7. Evaluate L−1{ln s + a

s − a}.

Now lns + a

s − a= ln(s + a) − ln(s − a),

− d

ds[ln(s + a) − ln(s − a)] = − 1

s + a+

1

s − a≡ F (s).

∴ f(t) = L−1{F (s)} = −e−at + eat = 2 sinh at.

Hence L−1

{

lns + a

s − a

}

= L−1

{∫

∞

sF (s) ds

}

=f(t)

t

= 2t−1 sinh at.


Example 8. Evaluate L−1{arc cot(s + 1)}.[

Note: If y = arc cot x, then y′ = − 1

1 + x2.

]

− d

ds[arc cot(s + 1)] =

1

1 + (s + 1)2≡ F (s)

f(t) = L−1{F (s)} = e−t sin t.

Hence L−1{arc cot(s + 1)} =f(t)

t= t−1e−t sin t.

Problem Set 1.7

Evaluate the following.

1. L{t cos 2t} 2. L{t sinh 3t} 3. L{t2 sinh t}

4. L{t2 cos t} 5. L{te2t sin 6t} 6. L{te−3t cos 3t}

7. L−1

{

s

(s2 + 1)2

}

8. L−1

{

s + 1

(s2 + 2s + 2)2

}

9. L−1

{

lns − 3

s + 1

}

10. L−1

{

lns2 + 1

s2 + 4

}

11. L−1

{

1

s− arc cot

4

s

}

12. L−1

{

arc tan1

s

}

1.8 Laplace transforms of derivatives and integrals.

Theorem: Suppose that f(t) is continuous for all t ≥ 0 and is of exponential order.

Suppose also that the derivative f ′(t) is piecewise continuous on every finite interval in the

range t ≥ 0. Then the Laplace transform of the derivative f ′(t) exists when s > c and

L{f ′(t)} = sL{f(t)} − f(0). (1.22)

Proof:

L{f ′(t)} =

∫

∞

0e−stf ′(t) dt

=[

e−stf(t)]

∞

0+ s

∫

∞

0e−stf(t) dt

= −f(0) + sL{f(t)}.

1.8. LAPLACE TRANSFORMS OF DERIVATIVES AND INTEGRALS. 29

Note that in the above proof we have assumed that f ′(t) is continuous. If f ′(t) is piecewise

continuous the proof is similar but the range of integration must be broken into parts for

which f ′(t) is continuous.

We may extend the formula (1.22) to find the Laplace transforms of higher derivatives.

For example, replacing f(t) by f ′(t) in Eq. (1.22) we obtain

L{f ′′(t)} = sL{f ′(t)} − f ′(0),

and, using Eq. (1.22) again, we obtain

L{f ′′(t)} = s2L{f(t)} − sf(0) − f ′(0). (1.23)

Similarly, we can extend this to higher derivatives to obtain

L{f (n)(t)} = snL{f} − sn−1f(0) − sn−2f ′(0) − . . . − f (n−1)(0). (1.24)

We shall use the results (1.22) and (1.23) to solve differential equations with given initial

conditions. However, we first note that (1.22) can be used to find unknown transforms.

Example 1. Evaluate L{sin2 t} using Eq. (1.22).

f(t) = sin2 t , so f(0) = 0.

f ′(t) = 2 sin t cos t = sin 2t.

∴ L{f ′(t)} = L{sin 2t} =2

s2 + 4.

But L{f ′(t)} = sL{f(t)} − f(0) = sL{sin2 t} − 0.

∴ sL{sin2 t} =2

s2 + 4

∴ L{sin2 t} =2

s(s2 + 4).

The following theorem is based on the result (1.22).

Theorem: If f(t) is piecewise continuous and of exponential order, then

L{∫ t

0f(τ) dτ

}

=1

sL{f(t)} . (1.25)


Proof: Put

∫ t

0f(τ) dt = g(t). Then f(t) = g′(t) except for points where f(t) is discon-

tinuous. Hence g′(t) is piecewise continuous on each finite interval and so, from Eq. (1.22)

L{f(t)} = L{g′(t)} = sL{g(t)} − g(0), s > c.

But g(0) = 0 and so

L{g(t)} = L{∫ t

0f(τ) dτ

}

=1

sL{f(t)}.


{

1

s2(s2 + a2)

}

.

Now L−1

{

1

s2 + a2

}

=1

asin at.

From Eq. (1.25) L−1

{

1

s(s2 + a2)

}

=1

a

∫ t

0sin aτ dτ

=1

a2(1 − cos at).

Applying Eq. (1.25) once more we obtain

L−1

{

1

s2(s2 + a2)

}

=1

a2

∫ t

0(1 − cos aτ) dτ =

1

a2

(

t − sin at

a

)

.

Problem Set 1.8

1. Use Eq. (1.22) to find L{cos2 at}.

2. Use Eq. (1.25) to find

(a) L−1

{

1

s2 + s

}

(b) L−1

{

s − 2

s(s2 + 4)

}

(c) L−1

{

1

s4(s2 + π2)

}

1.9. APPLICATION TO ORDINARY DIFFERENTIAL EQUATIONS. 31

1.9 Application to ordinary differential equations.

We now use Laplace Transform techniques to solve ordinary linear differential equations

with constant coefficients and with given initial conditions, i.e.,

y′′ + ay′ + by = f(t), y(0) = p, y′(0) = q.

We illustrate the procedure with several examples.

Example 1. Find the solution to the equation

y′′ + a2y = 0, y(0) = 0, y′(0) = 2.

Take the Laplace transform of the equation.

L{y′′} + a2L{y} = 0.

From Eq. (1.23), putting L{y} = Y (s), we obtain

s2Y (s) − sy(0) − y′(0) + a2Y (s) = 0,

i.e., (s2 + a2)Y (s) = 2 (since y(0) = 0, y′(0) = 2).

∴ Y (s) =2

s2 + a2.

Hence, y(t) =2

asin at.

Example 2. Solve 16y′′ − 9y = 0, y(0) = 3, y′(0) = 3.75.

16L{y′′} − 9L{y} = 0,

i.e., 16[s2Y (s) − sy(0) − y′(0)] − 9Y (s) = 0.


i.e., (s2 − 9

16)Y (s) = 3s + 3.75.

∴ Y (s) =3s

s2 − 916

+3.75

s2 − 916

.

∴ y(t) = 3 cosh3

4t + 5 sinh

3

4t.

Example 3. Solve y′′ + 2y′ + 17y = 0, y(0) = 0, y′(0) = 12.

L{y′′} + 2L{y′} + 17L{y} = 0

s2Y (s) − sy(0) − y′(0) + 2[sY (s) − y(0)] + 17Y (s) = 0.

(s2 + 2s + 17)Y (s) = 12.

∴ Y (s) =12

(s + 1)2 + 16.

∴ y(t) = 3e−t sin 4t.

Example 4. Solve y′′ + 4y′ + 4y = 0, y(0) = 2, y′(0) = −3.

L{y′′} + 4L{y′} + 4L{y} = 0,

i.e., s2Y (s) − sy(0) − y′(0) + 4[sY (s) − y(0)] + 4Y (s) = 0

(s2 + 4s + 4)Y (s) = 2s − 3 + 8,

i.e., Y (s) =2s + 5

(s + 2)2=

2(s + 2) + 1

(s + 2)2

=2

s + 2+

1

(s + 2)2.

∴ y(t) = 2e−2t + te−2t = (t + 2)e−2t.

Example 5. Solve y′′ − 2y′ + 10y = 0, y(0) = 3, y′(0) = 3.

L{y′′} − 2L{y′} + 10L{y} = 0,

1.9. APPLICATION TO ORDINARY DIFFERENTIAL EQUATIONS. 33

i.e., s2Y (s) − sy(0) − y′(0) − 2[sY (s) − y(0)] + 10Y (s) = 0,

i.e., (s2 − 2s + 10)Y (s) = 3s + 3 − 6 = 3(s − 1).

∴ Y (s) =3(s − 1)

(s − 1)2 + 9.

∴ y(t) = 3et cos 3t.

Example 6. Solve y′′ + y = 2, y(0) = 0, y′(0) = 3.

L{y′′} + L{y} = L{2} ⇒ s2Y (s) − sy(0) − y′(0) + Y (s) =2

s.

∴ (s2 + 1)Y (s) =2

s+ 3.

∴ Y (s) =2

s(s2 + 1)+

3

s2 + 1=

2

s− 2s

s2 + 1+

3

s2 + 1.

∴ y(t) = 2 − 2 cos t + 3 sin t.

Example 7. y′′ + 4y = 3 cos t, y(0) = 0, y′(0) = 0.

L{y′′} + 4L{y} = 3L{cos t} ⇒ s2Y (s) − sy(0) − y′(0) + 4Y (s) =3s

s2 + 1,

i.e., (s2 + 4)Y (s) =3s

s2 + 1⇒ Y (s) =

3s

(s2 + 1)(s2 + 4).

Partial fractions: Y (s) =s

s2 + 1− s

s2 + 4

∴ y(t) = cos t − cos 2t.

Example 8. y′′ − 4y = 8t2 − 4, y(0) = 5, y′(0) = 10.

L{y′′} − 4L{y} = L{8t2 − 4} ⇒ s2Y (s) − sy(0) − y′(0) − 4Y (s) =16

s3− 4

s.

(s2 − 4)Y (s) =16

s3− 4

s+ 5s + 10.

Y (s) =16

s3(s − 2)(s + 2)− 4

s(s − 2)(s + 2)+

5(s + 2)

(s − 2)(s + 2)

= −1

s− 4

s3+

12

s − 2+

12

s + 2+

1

s−

12

s − 2−

12

s + 2+

5

s − 2

= − 4

s3+

5

s − 2.


∴ y(t) = −2t2 + 5e2t.

Example 9. y′′ + 2y′ + y = e−2t, y(0) = 0, y′(0) = 0.

L{y′′} + 2L{y′} + L{y} = L{e−2t},

i.e., s2Y (s) + 2sY (s) + Y (s) =1

s + 2(because y(0) = y′(0) = 0).

∴ Y (s) =1

(s + 2)(s + 1)2=

1

s + 2− 1

s + 1+

1

(s + 1)2.

∴ y(t) = e−2t − e−t + te−t.

Example 10. Solve y′′ + 4y = 4(cos 2t − sin 2t), y(0) = 1, y′(0) = 3.

s2Y (s) − sy(0) − y′(0) + 4Y (s) = 4

[

s

s2 + 4− 2

s2 + 4

]

,

i.e., (s2 + 4)Y (s) =4s

s2 + 4− 8

s2 + 4+ s + 3.

∴ Y (s) =4s

(s2 + 4)2− 8

(s2 + 4)2+

s

s2 + 4+

3

s2 + 4.

Now4s

(s2 + 4)2= −

(

2

s2 + 4

)

′

and2

s2 + 4= L{sin 2t}.

∴ L−1

{

4s

(s2 + 4)2

}

= t sin 2t.

Also

(

s

s2 + 4

)

′

=4 − s2

(s2 + 4)2= − 1

s2 + 4+

4

(s2 + 4)2,

i.e., L−1

{

4

(s2 + 4)2

}

= −t cos 2t +1

2sin 2t.

Hence y(t) = t sin 2t + 2t cos 2t − sin 2t + cos 2t +3

2sin 2t

= t(sin 2t + 2 cos 2t) +1

2sin 2t + cos 2t

= (t +1

2)(sin 2t + 2 cos 2t).

1.10. DISCONTINUOUS FORCING FUNCTIONS. 35

Problem Set 1.9

Use Laplace transforms to solve the following initial value problems.

1. y′ − y = 1, y(0) = 0.

2. y′ + 4y = e−4t, y(0) = 2.

3. y′′ + 5y′ + 4y = 0, y(0) = 1, y′(0) = 0.

4. y′′ − 6y′ + 13y = 0, y(0) = 0, y′(0) = −3.

5. y′′ − 6y′ + 9y = t, y(0) = 0, y′(0) = 1.

6. y′′ − 4y′ + 4y = t3, y(0) = 1, y′(0) = 0.

7. y′′ − 4y′ + 4y = t3e2t, y(0) = 0, y′(0) = 0.

8. y′′ − 2y′ + 5y = 1 + t, y(0) = 0, y′(0) = 4.

9. y′′ + y = sin t, y(0) = 1, y′(0) = −1.

10. y′′ + 16y = 1, y(0) = 1, y′(0) = 2.

11. y′′ − y′ = et cos t, y(0) = 0, y′(0) = 0.

12. y′′ − 2y′ = et sinh t, y(0) = 0, y′(0) = 0.

1.10 Discontinuous forcing functions.

In some engineering problems we have to deal with differential equations in which

the forcing function, i.e., the term on the right-hand side of the differential equation, is

discontinuous. We illustrate this by a number of examples.

Example 1. Solve y′′ + 4y = g(t), y(0) = 0, y′(0) = 0,

where g(t) =

{

t 0 ≤ t < π2

π2 t ≥ π

2

,


i.e., g(t) = t − tuπ

2(t) +

π

2uπ

2(t) = t − (t − π

2)uπ

2(t).

∴ L{g(t)} = L{t} − L{(t − π

2)uπ

2(t)} =

1

s2− 1

s2e−

π

2s.

Hence, taking the Laplace transform of the differential equation we obtain

s2Y (s) + sy(0) − y′(0) + 4Y (s) =1

s2(1 − e−

π

2s),

i.e., (s2 + 4)Y (s) =1

s2(1 − e−

π

2s),

i.e., Y (s) =1

s2(s2 + 4)(1 − e−

π

2s)

=1

4

(

1

s2− 1

s2 + 4

)

(1 − e−π

2s).

Now L−1

{(

1

s2− 1

s2 + 4

)}

= t − 1

2sin 2t,

so L−1

{(

1

s2− 1

s2 + 4

)

e−π

2s

}

=

[

(

t − π

2

)

− 1

2sin 2

(

t − π

2

)

]

uπ

2(t)

= (t − π

2+

1

2sin 2t)uπ

2(t).

∴ y =1

4t − 1

8sin 2t +

1

4(t − π

2+

1

2sin 2t)uπ

2(t).

Example 2. Solve y′′ + y = f(t) , y(0) = 0 , y′(0) = 1 ,

where f(t)

{

1 0 ≤ t < π2

0 t ≥ π2

,

i.e., f(t) = u0(t) − uπ

2(t).

∴ L{f} =1

s− e−

π

2s

s.

∴ s2Y (s) − sy(0) − y′(0) + Y (s) =1

s

(

1 − e−π

2s)

.

(s2 + 1)Y (s) = 1 +1

s

(

1 − e−π

2s)

.

Y (s) =1

s2 + 1+

1

s(s2 + 1)

(

1 − e−π

2

)

=1

s2 + 1+

(

1

s− s

s2 + 1

)

(

1 − e−π

2s)

.

1.10. DISCONTINUOUS FORCING FUNCTIONS. 37

Now L−1

{

1

s− s

s2 + 1

}

= 1 − cos t.

∴ L−1

{

(1

s− s

s2 + 1)e−

π

2s

}

=[

1 − cos(t − π

2)]

uπ

2(t)

= (1 − sin t)uπ

2(t).

∴ y(t) = sin t + 1 − cos t + (1 − sin t)uπ

2(t).

Example 3. Solve y′′ + 4y = sin t − u2π(t) sin(t − 2π), y(0) = 0 , y′(0) = 0.

∴ s2Y (s) − sy(0) − y′(0) + 4Y (s) =1

s2 + 1− 1

s2 + 1e−2πs.

∴ (s2 + 4)Y (s) =1

s2 + 1(1 − e−2πs).

Y (s) =1

(s2 + 1)(s2 + 4)(1 − e−2πs)

=1

3

(

1

s2 + 1− 1

s2 + 4

)

(1 − e−2πs).

∴ y(t) =1

3

(

sin t − 1

2sin 2t

)

− 1

3

[

sin(t − 2π) − 1

2sin 2(t − 2π)

]

u2π(t)

=1

3

(

sin t − 1

2sin 2t

)

(1 − u2π(t)) .

Example 4. Solve y′′ + y′ +5

4y = g(t) , y(0) = 0 , y′(0) = 0 ,

where g(t) =

{

sin t 0 ≤ t < π0 t ≥ π

.

g(t) = (u0(t) − uπ(t)) sin t = sin t + uπ(t) sin(t − π)

(using sin(t − π) = − sin t).

∴ L{g(t)} =1

s2 + 1+

1

s2 + 1e−πs.

∴ s2Y (s) − sy(0) − y′(0) + sY (s) − y(0) +5

4Y (s) =

1

s2 + 1(1 + e−πs),

i.e., (s2 + s +5

4)Y (s) =

1

s2 + 1(1 + e−πs).


∴ Y (s) =1

(s2 + 1)(s2 + s + 54)

(1 + e−πs)

=1

17

[

−16s + 4

s2 + 1+

16s + 12

(s + 12)2 + 1

]

(1 + e−πs)

=

[

−16

17

s

s2 + 1+

4

17

1

s2 + 1+

16

17

(s + 12)

(s + 12)2 + 1

+4

17

1

(s + 12)2 + 1

]

(1 + e−πs).

∴ y(t) =−16

17cos t +

4

17sin t +

16

17e−

12t cos t +

4

17e−

12t sin t

+

[−16

17cos(t − π) +

4

17sin(t − π) +

16

17e−

12(t−π) cos(t − π)

+4

17e−

12(t−π) sin(t − π)

]

uπ(t),

i.e., y(t) =4

17(−4 cos t + sin t + 4e−

12t cos t + e−

12t sin t)

+4

17(4 cos t − sin t − 4e−

12(t−π) cos t − e−

12(t−π) sin t)uπ(t).

Example 5. Solve y′′ + 3y′ + 2y = u2(t) , y(0) = 0 , y′(0) = 1 .

s2Y (s) − sy(0) − y′(0) + 3sY (s) − 3y(0) + 2Y (s) =1

se−2s.

∴ (s2 + 3s + 2)Y (s) = 1 +1

se−2s.

∴ Y (s) =1

(s + 1)(s + 2)+

1

s(s + 1)(s + 2)e−2s

=1

s + 1− 1

s + 2+

[

12

s− 1

s + 1+

12

s + 2

]

e−2s.

∴ y(t) = e−t − e−2t +

[

1

2− e−(t−2) +

1

2e−2(t−2)

]

u2(t).

1.11. PERIODIC FUNCTIONS. 39

Problem Set 1.10

Use Laplace transforms to solve the following initial value problems

1. y′′ + 4y = f(t), where f(t) =

{

1 0 ≤ t < 10 t ≥ 1

, y(0) = 0, y′(0) = −1.

2. y′′ + 4y = u2π(t) sin t, y(0) = 1, y′(0) = 0.

3. y′′ − 5y′ + 6y = u1(t), y(0) = 0, y′(0) = 1.

4. y′′ + 4y′ + 3y = 1 − u2(t) − u4(t) + u6(t), y(0) = 0, y′(0) = 0.

5. y′′ + y = uπ(t), y(0) = 1, y′(0) = 0.

6. y(4) + 5y′′ + 4y = 1 − uπ(t), y(0) = y′(0) = y′′(0) = y′′′(0) = 0.

1.11 Periodic functions.

Let f(t) be a function which is defined for all positive t and has period T (> 0), i.e.,

f(t + T ) = f(t), all t > 0. (1.26)

Examples of periodic functions are sin t , cos t and functions such as

ta 2 3

(period )

a a

a

k

O

f(t)

and


( period = 3 )

t

f(t)

1

-1

1 2 3 4 5 6O

Theorem: Let f(t) be piecewise continuous on [0,∞) and of exponential order. If f(t) is

periodic with period T , then

L{f(t)} =1

(1 − e−sT )

∫ T

0e−stf(t) dt.

Proof:

L{f(t)} =

∫

∞

0e−stf(t) dt

=

∫ T

0e−stf(t) dt +

∫

∞

Te−stf(t) dt. (∗)

Now put t = u + T in the last integral which then becomes

∫

∞

0e−s(u+T )f(u + T ) du.

But f(u + T ) = f(u) because f is periodic so this integral is

e−sT

∫

∞

0e−suf(u) du = e−sTL{f(t)}.

Hence, (∗) becomes

L{f(t)} =

∫ T

0e−stf(t) dt + e−sTL{f(t)}.

∴ L{f(t)} =1

1 − e−sT

∫ T

0e−stf(t) dt.

Note that this can be written as

L{f(t)} =L{fT (t)}1 − e−sT

,


where

fT (t) =

{

f(t) 0 ≤ t < T0 t ≥ T

,

is called the window of length T for the function f(t).

Example 1. Evaluate L{f(t)} where

f(t) = π − t (0 ≤ t < 2π) , f(t + 2π) = f(t)

as in the diagram above.

6π5π3ππ t

π

−πO

f(t)

2π 4π

L{f(t)} =1

1 − e−2πs

∫ 2π

0e−st(π − t) dt

=[

1 − e−2πs]

−1

{

[

−(π − t)e−st

s

]2π

0

−∫ 2π

01 · 1

se−stdt

}

=[

1 − e−2πs]

−1{

πe−2πs

s+

π

s+

1

s2

[

e−st]2π

0

}

=(

1 − e−2πs)

−1{

π

s+

π

se−2πs +

1

s2e−2πs − 1

s2

}

.


f(t) = | sin at |, 0 ≤ t <π

a, f(t +

π

a) = f(t).

This is known as the rectified sine wave or the full-wave rectification of sin at. The

period isπ

awith g(t) = sin at, 0 ≤ t <

π

a, as illustrated below.


aa 4π/3π/2π/π/

f(t)

taa

1

O

L{f(t)} =1

1 − e−πs

a

∫ π

a

0e−st sin at dt

= (1 − e−πs

a )−1

[

e−st

s2 + a2(−s sin at − a cos at)

]π

a

0

=a(e−

πs

a + 1)

(1 − e−πs

a )(s2 + a2)=

a coth πs2a

(s2 + a2).


f(t) = t2 (0 ≤ t < 2π) f(t + 2π) = f(t).

L{f(t)} =(

1 − e−2πs)

−1∫ 2π

0e−stt2dt.

Now

∫ 2π

0e−stt2dt = L{[1 − u2π(t)]t2}

= L{t2 − u2π(t)[

(t − 2π)2 + 4π(t − 2π) + 4π2]

}

=2

s3− e−2πs

(

2

s3+

4π

s2+

4π2

s

)

∴ L{f(t)} =(

1 − e−2πs)

−1(

2

s3− 2

s3e−2πs − 4πe−2πs

s2− 4π2e−2πs

s

)

.

Example 4. An L − R series circuit has L = 1 henry, R = 1 ohm, E(t) given by

E(t) = t (0 ≤ t < 1) E(t + 1) = E(t)

and i(0) = 0.


The differential equation is Ldi

dt+ Ri = E(t),

i.e.,di

dt+ i = E(t).

Taking the Laplace transform of E(t) yields

L{E(t)} =1

1 − e−s

∫ 1

0te−stdt =

1

1 − e−s

[

−1

se−s − 1

s2(e−s − 1)

]

= −1

s

e−s

1 − e−s+

1

s2.

The Laplace transform of the differential equation is

sI(s) − i(0) + I(s) = L{E(t)},

i.e., I(s) =1

s2(s + 1)− 1

s(s + 1)

e−s

1 − e−s.

1

s2(s + 1)= −1

s+

1

s2+

1

s + 1= L{−1 + t + e−t}.

1

s(s + 1)=

1

s− 1

s + 1= L{1 − e−t}.

Nowe−s

1 − e−s= e−s(1 + e−s + e−2s + e−3s + · · · ). (Geometric Series)

∴

1

s(s + 1)

e−s

1 − e−s=

(

1

s− 1

s + 1

)

(

e−s + e−2s + e−3s + e−4s + · · ·)

= L{[u1(t) + u2(t) + u3(t) + u4(t) + · · · ]

−[e−(t−1)u1(t) + e−(t−2)u2(t) + e−(t−3)u3(t) + · · · ]}.

Hence, i(t) = −1 + t + e−t − [u1(t) + u2(t) + u3(t) + · · · ]

+[e−(t−1)u1(t) + e−(t−2)u2(t) + e−(t−3)u3(t) + · · · ].


Problem Set 1.11

Find the Laplace transforms of the given periodic functions.

1. f(t) =

{

1 0 ≤ t < a−1 a ≤ t < 2a

, f(t + 2a) = f(t).

2. f(t) =

{

t 0 ≤ t < 12 − t 1 ≤ t < 2

, f(t + 2) = f(t).

3. f(t) =

{

sin t 0 ≤ t < π0 π ≤ t < 2π

, f(t + 2π) = f(t).

4. f(t) = et, 0 ≤ t < 2π, f(t + 2π) = f(t).

5. Find the steady-state current in the circuit given by

R

L

E(t)

0E

432a tO a aa

E(t)

1.12 Impulse functions.

We often have to look at systems (mechanical or electrical) in which the external force

or voltage is of large magnitude but acts only for a very short time, i.e., a very high voltage

is applied to a circuit and then switched off immediately.

Consider the function defined by

δa(t − t0) =

12a t0 − a < t < t0 + a

0 t ≤ t0 − a or t ≥ t0 + a,

as shown in the first diagram on the next page. The second diagram shows how the function

changes as a gets smaller and smaller.

1.12. IMPULSE FUNCTIONS. 45

1

t

y

t +at -a t0 00

a

O

2

t

y

t0O

If a is small, then δa(t − t0) has a large constant magnitude for a short period of time

around t0. Note that the integral

I(a) =

∫ t0+a

t0−aδa(t − t0) dt =

∫ t0+a

t0−a

1

2adt = 1,

and I(a) can be written as

I(a) =

∫

∞

−∞

δa(t − t0) dt = 1,

since the function is zero outside (t0−a, t0 +a). The function δa(t− t0) is the unit impulse.

We define the “function” δ(t − t0) by

δ(t − t0) = lima→0

δa(t − t0).

The quantity δ(t− t0) is called the Dirac delta function and is an example of a generalized

function. It has the following properties:

δ(t − t0) =

{

∞ t = t00 t 6= t0

,

∫

∞

−∞

δ(t − t0) dt = 1. (1.27)

In particular, when t0 = 0 , we have

δ(t) =

{

∞ t = 00 t 6= 0

,


∫

∞

−∞

δ(t) dt = 1. (1.28)

To find the Laplace transform of δ(t−t0) we first find the Laplace transform of δa(t−t0).

L{δa(t − t0)} =

∫

∞

0e−stδa(t − t0) dt

=

∫ t0+a

t0−ae−st · 1

2adt

=

[

− 1

2ase−st

]t0+a

t0−a

=1

2ase−st0

(

eas − e−as)

=sinh as

ase−st0 .

Now let a → 0, then lima→0

sinh as

as= 1 (l’Hopital’s rule) so

L{δ(t − t0)} = lima→0

L{δa(t − t0)}

becomes

L{δ(t − t0)} = e−st0 . (1.29)

When t = t0, we have

L{δ(t)} = 1. (1.30)

Note the following result:

∫

∞

−∞

f(t)δ(t − t0) dt = f(t0), (1.31)

from which it follows that

L{f(t)δ(t − t0)} = f(t0)e−st0 . (1.32)

1.12. IMPULSE FUNCTIONS. 47

Example 1. Solve y′′ + 2y′ + 2y = δ(t − π), y(0) = 1 , y′(0) = 0.

Taking Laplace transforms

s2Y (s) − s + 2[sY (s) − 1] + 2Y (s) = e−sπ,

i.e., (s2 + 2s + 2)Y (s) = s + 2 + e−sπ.

Y (s) =s + 1

(s + 1)2 + 1+

1

(s + 1)2 + 1+

1

(s + 1)2 + 1e−sπ.

∴ y(t) = e−t cos t + e−t sin t + e−(t−π) sin(t − π)uπ(t),

i.e., y(t) = e−t cos t + e−t sin t[1 − eπuπ(t)].

Example 2. Solve y′′ + 4y = δ(t − π) − δ(t − 2π) , y(0) = y′(0) = 0.

s2Y (s) + 4Y (s) = e−sπ − e−2sπ.

Y (s) =1

s2 + 4(e−sπ − e−2sπ).

y(t) =1

2sin 2(t − π)uπ(t) − 1

2sin 2(t − 2π)u2π(t)

=1

2sin 2t(uπ(t) − u2π(t)).

Example 3. Solve y′′ + y = δ(t − π) cos t , y(0) = 0 , y′(0) = 1.

s2Y − 1 + Y = e−πs cos π = −e−πs.

∴ Y (s) =1

s2 + 1(1 − e−πs).

∴ y(t) = sin t − sin(t − π)uπ(t)

= sin t (1 + uπ(t)).


Problem Set 1.12

Solve the following initial value problems

1. y′′ + y = δ(t − 2π) , y(0) = 0 , y′(0) = 1.

2. y′′ + 2y′ + 3y = sin t + δ(t − π) , y(0) = 0 , y′(0) = 1.

3. y′′ + y = uπ

2(t) + δ(t − π) − u 3π

2(t) , y(0) = 0 , y′(0) = 0.

4. y′′ + 4y = 4δ(t − π

6) sin t , y(0) = 0 , y′(0) = 0.

5. y′′ − 2y = 1 + δ(t − 2) , y(0) = 0 , y′(0) = 1.

6. y(4) − y = δ(t − 1) , y(0) = y′(0) = y′′(0) = y′′′(0) = 0.

1.13 The convolution integral.

The inverse of the product of two Laplace transforms is not the product of the separate

inverses, i.e.,

L−1{F (s)G(s)} 6= L−1{F (s)}L−1{G(s)}.

The inverse of the product is given by the following theorem:

Theorem: If F (s) = L{f(t)} and G(s) = L{g(t)} both exist for s > a ≥ 0, then

H(s) = F (s)G(s) = L{h(t)} , s > a,

where

h(t) =

∫ t

0f(t − τ)g(τ) dτ =

∫ t

0f(τ)g(t − τ) dτ. (1.33)

The function h(t) is the convolution of f(t) and g(t) and the integrals defining h(t) are

the convolution integrals.

1.13. THE CONVOLUTION INTEGRAL. 49

We write

h(t) = (f ∗ g)(t)

so that h(t) is a “generalized product”.

Proof:

Let F (s) =

∫

∞

0e−sξf(ξ) dξ

G(s) =

∫

∞

0e−sηg(η) dη

then F (s)G(s) =

∫

∞

0e−sξf(ξ) dξ

∫

∞

0e−sηg(η) dη

=

∫

∞

0g(η) dη

∫

∞

0e−s(ξ+η)f(ξ) dξ.

Put ξ = t − η (η fixed) in second integral and let η = τ , then

F (s)G(s) =

∫

∞

0g(τ) dτ

∫

∞

τe−stf(t − τ) dt.

Reversing the order of integration, this becomes

F (s)G(s) =

∫

∞

0e−stdt

∫ t

0f(t − τ)g(τ) dτ

=

∫

∞

0e−stdt h(t) = L{h(t)}.


{

1

s2(s2 + 1)

}

.

Choose F (s) =1

s2, G(s) =

1

s2 + 1, so that

f(t) = t, g(t) = sin t.

Then h(t) =

∫ t

0(t − τ) sin τ dτ

= [−(t − τ) cos τ ]t0 −∫ t

01 cos τ dτ

= t − sin t.

Alternatively, we could have written

h(t) =

∫ t

0τ sin(t − τ) dτ,

which leads to the same result.



{

1

(s2 + a2)2

}

.

Choose F (s) = G(s) =1

s2 + a2.

Then f(t) = g(t) =1

asin at,

h(t) =1

a2

∫ t

0sin a(t − τ) sin aτ dτ.

Using the identity sinA sin B =1

2[cos(A − B) − cos(A + B)]

h(t) =1

2a2

∫ t

0[cos a(2τ − t) − cos at] dτ ( A = aτ ; B = a(t − τ) )

=1

2a2

[

1

2asin a(2τ − t) − τ cos at

]t

0

=1

2a2

[

1

2asin at − t cos at − 1

2asin(−at)

]

=1

2a3(sin at − at cos at).

Example 3. Find the Laplace transform of

f(t) =

∫ t

0(t − τ)2 cos 2τ dτ .

L{f(t)} = L{t2}L{cos 2t} =2

s3· s

s2 + 4

=2

s2(s2 + 4).

Example 4. Find L−1

{

1

s2 + 1F (s)

}

.

Let1

s2 + 1= G(s) , then g(t) = sin t.

L−1

{

1

s2 + 1F (s)

}

= h(t) =

∫ t

0sin(t − τ)f(τ) dτ,

where f(t) = L−1{F (s)}.

Example 5. Express in terms of a convolution integral, the solution of the initial value

problem

y′′ + 4y′ + 4y = g(t) , y(0) = 2 , y′(0) = −3.

1.13. THE CONVOLUTION INTEGRAL. 51

s2Y (s) − 2s + 3 + 4sY (s) − 8 + 4Y (s) = G(s).

Y (s) =2s + 5

(s + 2)2+

G(s)

(s + 2)2=

2

s + 2+

1

(s + 2)2+

G(s)

(s + 2)2.

∴ y(t) = 2e−2t + te−2t +

∫ t

0e−2(t−τ)g(τ) dτ.

Example 6. Find L−1

{

1

s4(s2 + 1)

}

.

Let G(s) =1

s2 + 1and F (s) =

1

s4. H(s) = F (s)G(s).

Then g(t) = sin t , and f(t) =1

6t3.

h(t) =

∫ t

0

1

6τ3 sin(t − τ) dτ.

Integrating by parts we obtain

h(t) =1

6t3 − t + sin t.

Problem Set 1.13

Evaluate the following using the convolution theorem.

1. L{∫ t

0(t − τ)eτdτ} 2. L{

∫ t

0sin(t − τ) cos τ dτ}

3. L{∫ t

0(t − τ) cos τ dτ} 4. L{

∫ t

0τ sin τ dτ}

5. L{t2 ∗ t3} 6. L−1

{

1

s + 5F (s)

}

7. L−1

{

s

s2 + 4F (s)

}

8. L−1

{

1

s(s + 1)

}

9. L−1

{

s

(s2 + 4)2

}

10. L−1

{

1

(s + 1)2

}

11. L−1

{

1

(s2 + 4s + 5)2

}

12. L−1

{

1

(s − 3)(s2 + 4)

}

13. Compute cos t ∗ cos t and thus show that f ∗ f is not necessarily non-negative.

14. Solve y′′ + 3y′ + 2y = cos t , y(0) = 1 , y′(0) = 0. Leave the answer in terms of a

convolution integral.


Chapter 2

SYSTEMS OF FIRST ORDER

LINEAR EQUATIONS

2.1 Introduction.

Systems of ordinary differential equations play a very important role in applied mathe-

matics. We shall illustrate the application of systems with two examples. In these examples,

and throughout this chapter, x1, x2, x3, . . . represent dependent variables which are func-

tions of the independent variable t and a prime denotes differentiation with respect to t.

Example 1. Two masses m1 and m2 are connected to two springs A and B of

negligible mass with spring constants k1 and k2, respectively. Let

x1(t) and x2(t) denote the vertical displacements of the masses

from their equilibrium positions. When the system is in motion

the spring B is subject to both an elongation and a compression;

the net elongation is x2−x1. Hence, from Hooke’s law, the springs

exert forces −k1x1 and k2(x2 − x1) on m1 and −k2(x2 − x1) on

m2. Hence, the differential equations of the system are

m

1

2

k

x

x

A

1

B

k2

m

1

2

53

54 CHAPTER 2. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS

m1d2x1

dt2= −k1x1 + k2(x2 − x1)

m2d2x2

dt2= −k2(x2 − x1).

i.e., m1x′′

1 = −(k1 + k2)x1 + k2x2

(2.1)m2x

′′

2 = k2x1 − k2x2.

Such a system may be supplemented by initial conditions, e.g., information that the

masses start from their equilibrium positions with certain velocities.

The system (2.1) is a linear system of the second order, since second derivatives of the

variables appear.

Example 2. Consider an electrical net-

work with more than one loop, as shown

in the diagram. The current i1(t) splits

into two directions at the branch point

B1. From Kirchhoff’s first law

i1(t) = i2(t) + i3(t).

A B C

L

R

i

B

R

L

1

1i i

1 1

2

3

2

2

2

1

1

C

2

2A

E

Applying Kirchhoff’s second law to each loop, the voltage drops across each part of the loop

give the following system of differential equations

E(t) = i1R1 + L1i′

2 + i2R2

E(t) = i1R1 + L2i′

3.

2.2. BASIC THEORY OF SYSTEMS OF FIRST ORDER LINEAR EQUATIONS. 55

Eliminating i1 from these equations we obtain

L1i′

2 = −(R1 + R2)i2 − R1i3 + E(t)

L2i′

3 = R1i2 + R1i3 + E(t).

This system can be supplemented by initial conditions such as i2(0) = 0 , i3(0) = 0. This

is a first order linear system.

2.2 Basic theory of systems of first order linear equations.

We first note than any nth order differential equation of the form

y(n) = F (t, y, y′, . . . , y(n−1)) (2.2)

can be reduced to a system of n first-order equations of a special form. Introduce the

variables x1, x2, . . . , xn defined by

x1 = y, x2 = y′, x3 = y′′, . . . , xn = y(n−1). (2.3)

Then Eqs. (2.2) and (2.3) can be written in the form of a system:

x′

1 = x2

x′

2 = x3

... (2.4)

x′

n−1 = xn

x′

n = F (t, x1, x2, . . . , xn).

Example 1. Consider the differential equation

y′′ + 3y′ + 2y = 0.


Put x1 = y , x2 = y′, and the equation becomes

x′

2 = −3x2 − 2x1 , with x2 = x′

1,

i.e., we have the system

x′

1 = x2

x′

2 = −2x1 − 3x2,

which can be written in vector-matrix form as

x′ = Ax,

where

x =

[

x1

x2

]

and A =

[

0 1−2 −3

]

.

The system (2.4) is a special case of the more general system

x′

1 = F1(t, x1, . . . , xn)

... (2.5)

x′

n = Fn(t, x1, . . . , xn).

However, we are interested in systems in which each of the functions F1, . . . , Fn is linear in

the variables x1, . . . , xn; such a system is said to be linear. The most general system of n

linear first order equations has the canonical form

x′

1 = a11(t)x1 + . . . + a1n(t)xn + f1(t)

... (2.6)

x′

n = an1(t)x1 + . . . + ann(t)xn + fn(t).


If the functions f1(t), . . . , fn(t) are all zero the system is said to be homogeneous; otherwise

the system is nonhomogeneous. In addition to the system of equations, there may also be

given initial conditions of the form

x1(t0) = x01 , x2(t0) = x0

2, . . . , xn(t0) = x0n. (2.7)

Theorem: If the functions aij(t) , fi(t) (i, j = 1, . . . , n) are continuous on an open interval

α < t < β, containing the point t = t0, then there exists a unique solution x1, . . . , xn of the

system of differential equations (2.6) which also satisfies the initial conditions (2.7). This

solution is valid throughout the interval α < t < β.

We write the system (2.6) in vector-matrix form, i.e.,

x′ = Ax + f(t), (2.8)

where

x =

x1...

xn

, A =

a11 . . . a1n...

...an1 . . . ann

, f =

f1(t)...

fn(t)

.

A vector x is said to be a solution of the system (2.8) if its components satisfy the system

(2.6). We assume that A and f(t) are continuous, i.e., all of their components are continuous,

on some interval α < t < β. From the last theorem this guarantees the existence of a solution

on the interval α < t < β.

We commence by considering the homogeneous system

x′ = Ax, (2.9)

i.e., system (2.8) with f(t) = 0. Specific solutions of this system will be denoted by

x(1)(t),x(2)(t), . . . ,x(k)(t).


Theorem (Principle of Superposition): If the vector functions x(1),x(2), . . . ,x(k) are

solutions of the system (2.9), then the linear combination

x = c1x(1)(t) + c2x

(2)(t) + . . . + ckx(k)(t).

is also a solution for any constants c1, . . . , ck.

Example 2. Consider the system

x′ =

[

3 21 2

]

x.

It can be shown that

x(1)(t) =

[

1−1

]

et

and

x(2)(t) =

[

21

]

e4t

are solutions of this system. From the above theorem, the vector

x = c1

[

1−1

]

et + c2

[

21

]

e4t

= c1x(1)(t) + c2x

(2)(t)

also satisfies the system. Check this!

Definition. The set of solutions x(1), . . . ,x(k) is said to be linearly dependent on some

interval α < t < β if there exist constants c1, . . . , ck such that

c1x(1) + . . . + ckx

(k) = 0,

for every t in the interval. Otherwise the vectors are said to be linearly independent.


Suppose that x(1), . . . ,x(n) are n solutions of the nth order system (2.9). We write

x(i) =

x1i

x2i...

xni

,

i.e., xmi is the mth component of the ith solution. Form a matrix by writing each solution

vector as a column of the matrix, i.e.,

Φ(t) =

x11(t) x12(t) . . . x1n(t)...

......

xn1(t) xn2(t) . . . xmn(t)

. (2.10)

Now the columns of Φ(t) are linearly independent for a given value of t if and only if

detΦ(t) 6= 0 for that value of t. This determinant is denoted by W (x(1), . . . ,x(n)) and is

called the Wronskian of the n solutions x(1), . . . ,x(n),

i.e., W = detΦ(t).

Hence, the solutions x(1), . . . ,x(n) are linearly independent at a point if and only if W 6= 0

at that point.

In fact, it can be shown that, if x(1), . . . ,x(n) are solution vectors of the system, then

either

W (x(1), . . . ,x(n)) 6= 0,

for every t in α < t < β or

W (x(1), . . . ,x(n)) = 0,

for every t in the interval. Hence, if we can show that W 6= 0 for some t0 in α < t < β,

then W 6= 0 for every t and so the solutions are linearly independent on the interval.


Example 3. Consider the system of Example 2.

The solutions

x(1) =

[

1−1

]

et , x(2) =

[

21

]

e4t

are clearly linearly independent in (−∞,∞) since neither vector is a constant multiple of

the other.

The Wronskian is given by

W (x(1),x(2)) =

∣

∣

∣

∣

et 2e4t

−et e4t

∣

∣

∣

∣

= 3e5t 6= 0,

for all real values of t.

Definition. Any set x(1), . . . ,x(n) of n linearly independent solution vectors of the ho-

mogeneous system (2.9) is said to be a fundamental set of solutions.

Definition. If x(1), . . . ,x(n) is a fundamental set of solutions of the homogeneous system

(2.9) then the general solution of the system is

x = c1x(1) + c2x

(2) + . . . + cnx(n), (2.11)

where c1, . . . , cn are arbitrary constants.

Note that the general solution (2.11) can be written in the form

x = c1

x11...

xn1

+ c2

x12...

xn2

+ . . . + cn

x1n...

xnn

=

c1x11 + c2x12 + . . . + cnx1n...

c1xn1 + c2xn2 + . . . + cnxnn


=

x11 x12 . . . x1n...

......

xn1 xn2 . . . xnn

c1...

cn

,

i.e., x = Φ(t)c, (2.12)

where c is the column vector

c =

c1...

cn

and Φ(t) is as defined in Eq. (2.10) with the solutions x(1)(t), . . . ,x(n)(t) linearly indepen-

dent. The matrix Φ(t) is said to be a fundamental matrix of the system on the interval

α < t < β. Note that, since detΦ(t) = W 6= 0, the inverse Φ−1(t) exists for every value of

t in the interval.

Example 4. For the problem of Examples 2 and 3, a fundamental matrix is

Φ(t) =

[

et 2e4t

−et e4t

]

and it follows that

Φ−1(t) =

[

13e−t −2

3e−t

13e−4t 1

3e−4t

]

.

Suppose that x(1), . . . ,x(n) are solutions of the system (2.9) which satisfy the initial

conditions

x(1)(t0) = e(1) ≡

10...0

, x(2)(t0) = e(2) =

01...0

, . . . , x(n)(t0) = e(n) ≡

00...1

,

where t0 is some point in the interval α < t < β. The fundamental matrix of this system is

a special case of Φ(t) and is usually denoted by Ψ(t). It has the property that

Ψ(t0) = I =

1 0 . . . 00 1 . . . 0...

......

0 0 . . . 1

. (2.13)


Example 5. Consider the system of Example 3. The general solution is

x = c1

[

1−1

]

et + c2

[

21

]

e4t.

Choose t0 = 0 and let the initial conditions be

x(1)(0) =

[

10

]

,x(2)(0) =

[

01

]

.

The first of these conditions leads to

[

10

]

= c1

[

1−1

]

+ c2

[

21

]

,

i.e., c1 = c2 =1

3, so that x(1)(t) =

1

3

[

1−1

]

et +1

3

[

21

]

e4t,

i.e., x(1)(t) =

[

13et + 2

3e4t

−13et + 1

3e4t

]

.

The second initial condition leads to

[

01

]

= c1

[

1−1

]

+ c2

[

21

]

,

i.e., c1 = −2

3, c2 =

1

3, so that x(2)(t) =

[

−23et + 2

3e4t

23et + 1

3e4t

]

.

Hence

Ψ(t) =

[

13et + 2

3e4t −23et + 2

3e4t

−13et + 1

3e4t 23et + 1

3e4t

]

and it is easily seen that Ψ(0) = I.


Problem Set 2.2

In problems 1 - 6 verify that the vector x is a solution of the given system.

1.dx

dt= 3x − 4y ,

dy

dt= 4x − 7y , x =

[

12

]

e−5t

2.dx

dt= −2x + 5y ,

dy

dt= −2x + 4y , x =

[

5 cos t3 cos t − sin t

]

et

3. x′ =

[

−1 14

1 −1

]

x ; x =

[

−12

]

e−32t

4. x′ =

[

2 1−1 0

]

x ; x =

[

13

]

et +

[

4−4

]

tet

5. x′ =

1 2 16 −1 0

−1 −2 −1

x ; x =

16

−13

6. x′ =

1 0 11 1 0

−2 0 −1

x ; x =

sin t−1

2 sin t − 12 cos t

− sin t + cos t

In problems 7 and 8 the given vectors are solutions of a system x′ = Ax. Determine

whether the vectors form a fundamental set on (−∞,∞).

7. x(1) =

[

1−1

]

et , x(2) =

[

26

]

et +

[

8−8

]

tet

8. x(1) =

1−2

4

+ t

122

, x(2) =

1−2

4

, x(3) =

3−612

+ t

244

9. Prove that the general solution of x′ =

0 6 01 0 11 1 0

x on the interval (−∞,∞) is

x = c1

6−1−5

e−t + c2

−311

e−2t + c3

211

e3t.


In problems 10-12 the indicated column vectors form a fundamental set of solutions for

the given system on (−∞,∞). Form a fundamental matrix Φ(t) and compute Φ−1(t).

10. x′ =

[

2 33 2

]

x ; x(1) =

[

−11

]

e−t , x(2) =

[

11

]

e5t

11. x′ =

[

4 1−9 −2

]

x ; x(1) =

[

−13

]

et , x(2) =

[

−13

]

tet +

[

01

]

et

12. x′ =

[

3 −25 −3

]

x ; x(1) =

[

2 cos t3 cos t + sin t

]

, x(2) =

[

−2 sin tcos t − 3 sin t

]

13. Find the fundamental matrix Ψ(t) satisfying Ψ(0) = I for the system in problem 10.

14. Find the fundamental matrix Ψ(t) satisfying Ψ(0) = I for the system in problem 11.

15. Find the fundamental matrix Ψ(t) satisfying Ψ[π

2

]

= I for the system in problem 12.

16. Show that the fundamental matrices Φ(t) and Ψ(t) satisfy Ψ(t) = Φ(t)Φ−1(t0).

2.3 Review of eigenvalues and eigenvectors.

Given the n×n matrix A, the equation Ax = y can be regarded as a linear transforma-

tion that maps (or transforms) a given vector x into a new vector y. In many applications

it is useful to find a vector x which is transformed into a multiple of itself by the action of

A, i.e., we need to find the solution vectors, x, of the linear system

Ax = λx, (2.14)

where λ is the proportionality factor. A vector x satisfying Eq. (2.14) is called an eigenvector

of A corresponding to the eigenvalue λ.

Eq. (2.14) can be rewritten in the form

(A − λI)x = 0, (2.15)

2.3. REVIEW OF EIGENVALUES AND EIGENVECTORS. 65

where I is the n × n identity matrix. This is a homogeneous system of linear equations

which will have non-trivial solutions if and only if

det (A − λI) = 0. (2.16)

Eq. (2.16) is a polynomial equation of degree n known as the characteristic equation.

Thus, the n × n matrix A has exactly n eigenvalues, some of which may be repeated. If a

given eigenvalue appears m times as a root of Eq. (2.16) then that eigenvalue is said to be

of multiplicity m. Each eigenvalue has at least one associated eigenvector; an eigenvalue of

multiplicity m may have q linearly independent eigenvectors where 1 ≤ q ≤ m. If all of the

eigenvalues of a matrix A are simple (i.e., of multiplicity one), then the n eigenvectors of

A are linearly independent.

Given the eigenvalues, we use Gauss-Jordan elimination to solve the system of

equations (2.15) to find the eigenvectors. These eigenvectors are determined only up to

a multiplicative constant. We can normalize the eigenvector by choosing the constant

appropriately.

We illustrate the above theory by presenting a number of examples. These examples

will cover the following possibilities:

(a) All eigenvalues real and simple.

(b) Some eigenvalues complex. (Note that for a real matrix A, complex eigenvalues must

occur in conjugate pairs).

(c) Eigenvalues of multiplicity m with m linearly independent associated eigenvectors.


(d) Eigenvalues of multiplicity m with less than m linearly independent associated eigen-

vectors.

Example 1. Find the eigenvalues and eigenvectors of the matrix A =

1 −1 43 2 −12 1 −1

.

The characteristic equation is

det (A − λI) =

∣

∣

∣

∣

∣

∣

1 − λ −1 43 2 − λ −12 1 −1 − λ

∣

∣

∣

∣

∣

∣

= 0,

i.e., − (λ3 − 2λ2 − 5λ + 6) = 0,

i.e., (λ − 1)(λ + 2)(λ − 3) = 0.

Thus A has the three simple eigenvalues λ1 = 1 , λ2 = −2 and λ3 = 3. For λ = 1,

Eq. (2.15) is

0 −1 4 03 1 −1 02 1 −2 0

⇒

0 −1 4 01 0 1 02 0 2 0

⇒

0 −1 4 01 0 1 00 0 0 0

.

Hence x1 = −x3 , x2 = 4x3 and, putting x3 = 1, the associated eigenvector is x(1) =

−141

.

For λ = −2, Eq. (2.15) is

3 −1 4 03 4 −1 02 1 1 0

⇒

3 −1 4 01 0 1 05 0 5 0

⇒

−1 −1 0 01 0 1 00 0 0 0

.

Hence x2 = −x1 , x3 = −x1 and, putting x1 = 1, the associated eigenvector is x(2) =

1−1−1

.

For λ = 3, Eq. (2.15) is

−2 −1 4 03 −1 −1 02 1 −4 0

⇒

−2 −1 4 01 0 −1 00 0 0 0

⇒

2 −1 0 01 0 −1 00 0 0 0

.

Hence x3 = x1 , x2 = 2x1 and, putting x1 = 1, the associated eigenvector is x(3) =

121

.


Note that any multiple of x(1), x(2) or x(3) is also an eigenvector.


[

3 −24 −1

]

.


det (A − λI) =3 − λ −2

4 −1 − λ= 0,

i.e., λ2 − 2λ + 5 = 0,

i.e., λ = 1 + 2i , 1 − 2i.

Thus A has the conjugate complex pair of eigenvalues λ1 = 1 + 2i , λ2 = 1 − 2i. For

λ = 1 + 2i , Eq. (2.15) is

[

2 − 2i −2 04 −2 − 2i 0

]

⇒[

1 − i −1 00 0 0

]

.

Hence x2 = (1 − i)x1 and, putting x1 = 1, the associated eigenvector is x(1) =

[

11 − i

]

.

For λ = 1 − 2i it follows that the associated eigenvector is the complex conjugate of

x(1), i.e., x(2) =

[

11 + i

]

.


0 1 11 0 11 1 0

.


det (A − λI) =−λ 1 1

1 −λ 11 1 −λ

= 0,

i.e., (λ + 1)2(λ − 2) = 0.

Thus A has the simple eigenvalue λ1 = 2 and the eigenvalue of multiplicity two, λ2 = −1.

For λ = 2, Eq. (2.15) is

−2 1 1 01 −2 1 01 1 −2 0

⇒

−2 1 1 00 1 −1 00 0 0 0

⇒

−2 0 2 00 1 −1 00 0 0 0

.


Hence x3 = x2 = x1 and the eigenvector is x(1) =

111

. For λ = −1, Eq. (2.15) is

1 1 1 01 1 1 01 1 1 0

⇒

1 1 1 00 0 0 00 0 0 0

.

Hence, we have the single equation

x1 + x2 + x3 = 0,

and thus two parameters that can be assigned arbitrary values. If we put x2 = a , x3 = b,

then x1 = −a − b and the eigenvector is of the form

−a − bab

= a

−110

+ b

−101

.

Hence, a pair of linearly independent eigenvectors associated with the repeated eigenvalue

λ2 = −1 is

x(2) =

−110

, x(3) =

−101

.

Thus, in this case, the number of linearly independent eigenvectors equals the multiplicity

of the repeated eigenvalue. Note that any linear combination of x(2) and x(3) is also an

eigenvector associated with the eigenvalue λ2 = −1.


−5 −5 −98 9 18

−2 −3 −7

.


det (A − λI) =−5 − λ −5 −9

8 9 − λ 18−2 −3 −7 − λ

= 0,

i.e., − (λ + 1)3 = 0.

Thus A has an eigenvalue, λ = −1 , of multiplicity 3. For λ = −1, Eq. (2.15) is


−4 −5 −9 08 10 18 0

−2 −3 −6 0

⇒

0 1 3 00 0 0 0

−2 0 3 0

.

Hence, x2 = −3x3 , 2x1 = 3x3 and, putting x3 = 2, we obtain the single linearly dependent

eigenvector

x(1) =

3−6

2

.


−1 −3 −90 5 180 −2 −7

.


det (A − λI) =−1 − λ −3 −9

0 5 − λ 180 −2 −7 − λ

= 0,

i.e., − (1 + λ)(λ2 + 2λ + 1) = 0,

i.e., (λ + 1)3 = 0.

Thus A has an eigenvalue, λ = −1, of multiplicity 3. For λ = −1, Eq. (2.15) is

0 −3 −9 00 6 18 00 −2 −6 0

⇒

0 1 3 00 0 0 00 0 0 0

.

Hence x1 is arbitrary, x2 = −3x3, i.e., put x1 = a, x3 = b and the eigenvector is

a−3b

b

= a

100

+ b

0−3

1

.

Thus the repeated eigenvalue λ = −1 has two linearly independent associated eigenvectors

x(1) =

100

, x(2) =

0−3

1

.



2 0 00 2 00 0 2

.


det (A − λI) =2 − λ 0 0

0 2 − λ 00 0 2 − λ

= 0,

i.e., (2 − λ)3 = 0.

Thus A has an eigenvalue, λ = 2, of multiplicity 3. For λ = 2, Eq. (2.15) is

0 0 0 00 0 0 00 0 0 0

,

i.e., x1, x2, x3 may have any values. Put x1 = a , x2 = b , x3 = c and the eigenvector is

abc

= a

100

+ b

010

+ c

001

.

Hence the repeated eigenvalue λ = 2 has three linearly independent associated eigenvectors

x(1) =

100

, x(2) =

010

, x(3) =

001

.

Problem Set 2.3

In each of the problems find the eigenvalues and eigenvectors of the given matrix.

1.

[

−1 2−7 8

]

2.

[

2 12 1

]

3.

[

−8 −116 0

]

4.

[

1 114 1

]

5.

5 −1 00 −5 95 −1 0

6.

3 0 00 2 04 0 1

7.

0 4 0−1 −4 0

0 0 −2

8.

1 6 00 2 10 1 2

9.

[

−1 2−5 1

]

10.

2 −1 05 2 40 1 2

2.4. HOMOGENEOUS LINEAR SYSTEMS WITH CONSTANT COEFFICIENTS. 71

2.4 Homogeneous linear systems with constant coefficients.

We now show how to construct the general solution of a system of homogeneous linear

equations with constant coefficients, i.e., a system of the form (2.9), x′ = Ax, where A is

a constant n× n matrix. We have seen in Section 2.2 that the systems considered there all

had solutions of the form

x = keλt, (2.17)

where k is a constant vector, so we look for solutions of this form. Since Eq. (2.17) implies

that x′ = λkeλt, substitution into Eq. (2.9) gives

λkeλt = Akeλt,

i.e., Ak = λk

and non-trivial solutions of this equation exist if and only if

det (A − λI) = 0.

This is the characteristic equation of the matrix A. Hence it follows that x = keλt is a

solution of the system (2.9) if and only if λ is an eigenvalue of A and k is the associated

eigenvector.

If the n×n matrix A possesses n distinct real eigenvalues λ1, . . . , λn, then the associated

eigenvectors k(1), . . . ,k(n) are linearly independent and

x(1) = k(1)eλ1t, x(2) = k(2)eλ2t, . . . , x(n) = k(n)eλnt (2.18)

form a fundamental set of solutions for the system and the general solution on the interval

(−∞,∞) is

x = c1k(1)eλ1t + c2k

(2)eλ2t + . . . + cnk(n)eλnt. (2.19)


Example 1. Consider the system x′ = Ax where

A =

1 −1 43 2 −12 1 −1

.

This is the matrix of Ex. 1 of Section 2.3. The eigenvalues are 1,−2, and 3 and the

associated eigenvectors are

k(1) =

−141

, k(2) =

1−1−1

, k(3) =

121

.

Thus the solutions are

x(1) =

−141

et, x(2) =

1−1−1

e−2t, x(3) =

121

e3t,

and so a fundamental matrix is

Φ(t) =

−et e−2t e3t

4et −e−2t 2e3t

et −e−2t e3t

.

The general solution is

x = c1

−141

et + c2

1−1−1

e−2t + c3

121

e3t

or, equivalently,

x = Φ(t)c =

−et e−2t e3t

4et −e−2t 2e3t

et −e−2t e3t

c1

c2

c3

.

Taking t0 = 0, the special fundamental matrix Ψ(t) is given by

Ψ(t) = Φ(t)Φ−1(t0) =

−et e−2t e3t

4et −e−2t 2e3t

et −e−2t e3t

−1 1 14 −1 21 −1 1

−1

=

−et e−2t e3t

4et −e−2t 2e3t

et −e−2t e3t

−16

13 −1

213

13 −1

12 0 −1

2

2.4. HOMOGENEOUS LINEAR SYSTEMS WITH CONSTANT COEFFICIENTS. 73

=

16et +1

3e−2t +12e3t

−23et −1

3e−2t +e3t

−16et −1

3e−2t +12e3t

−13et +1

3e−2t

43et −1

3e−2t

13et −1

3e−2t

12et −e−2t +1

2e3t

−2et +e−2t +e3t

−12et +e−2t +1

2e3t

.

It is easily seen that Ψ(0) = I.

Note that if a matrix A possesses a repeated eigenvalue of multiplicity m to which

correspond m linearly independent eigenvectors then the general solution of the system

(2.9) is of the form (2.19). In particular, if the matrix A is real and symmetric, i.e.,

A = AT , then there is always a full set of n linearly independent eigenvectors even if some

of the eigenvalues are repeated.


A =

0 1 11 0 11 1 0

.

This matrix is real and symmetric and so has real eigenvalues and three linearly inde-

pendent eigenvectors. The eigenvalues are

λ1 = 2 , λ2 = −1 , λ3 = −1,

i.e., there is an eigenvalue of multiplicity two. The associated eigenvectors are

k(1) =

111

, k(2) =

10

−1

, k(3) =

01

−1

.

(Check this!). Thus the eigenvalue of multiplicity two has two linearly independent associ-

ated eigenvectors and the general solution is

x = c1

111

e2t + c2

10

−1

e−t + c3

01

−1

e−t

or, in terms of a fundamental matrix,

x = Φ(t)c =

e2t e−t 0e2t 0 e−t

e2t −e−t −e−t

c1

c2

c3

.


2.5 Complex eigenvectors.

We assume that the coefficient matrix, A, is real but possesses a pair of conjugate

complex eigenvalues.

λ1 = α + iβ , λ2 = λ1 = α − iβ, (2.20)

where α , β are real constants. The associated eigenvectors are also complex conjugates,

i.e.,

k(1) = a + ib , k(2) = k(1)

= a − ib, (2.21)

where a and b are real vectors. Then the two linearly independent solutions of the system

which correspond to these eigenvalues are k(1)eλ1t and k(2)eλ2t = k(1)

eλ1t. However, the

solutions are complex; we need to find real solutions of the system. Since the system

x′ = Ax has real coefficients, the real and imaginary parts of the single complex solution

xc(t) = k(1)eλ1t can be shown to be two linearly independent real solutions. Thus the

general solution of the system can be written as x(t) = c1x(1)(t) + c2x

(2)(t), where

x(1)(t) = Re(xc(t))

(2.22)

x(2)(t) = Im(xc(t)).

The real and imaginary parts of k(2)eλ2t will not give rise to any additional linearly inde-

pendent real solutions.


A =

[

3 −24 −1

]

.

This is the matrix of Ex. 2 of Section 2.3. The eigenvalues are λ1 = 1 + 2i , λ2 = 1 − 2i

2.5. COMPLEX EIGENVECTORS. 75

and the associated eigenvectors are k(1) =

[

11 − i

]

, k(2) =

[

11 + i

]

.

We need use only λ1 and k(1). Thus

xc(t) =

[

11 − i

]

et(cos 2t + i sin 2t)

which expands to yield

xc(t) =

[

et cos 2t + iet sin 2tet cos 2t + et sin 2t + i(et sin 2t − et cos 2t)

]

Hence, from Eq. (2.22), the solutions are

x(1)(t) = et

[

cos 2tcos 2t + sin 2t

]

x(2)(t) = et

[

sin 2tsin 2t − cos 2t

]

,

and a fundamental matrix is

Φ(t) =

[

et cos 2t et sin 2tet cos 2t + et sin 2t et sin 2t − et cos 2t

]

.

Problem Set 2.5

In each of the following problems the given matrix A is the coefficient matrix of a linear

homogeneous system of equations x′ = Ax. Find the general solution of each system in the

form of a fundamental matrix.

1. A =

[

1 24 3

]

2. A =

[

0 28 0

]

3. A =

[

−4 2−5

2 2

]

4. A =

[

12 9

12 2

]

5. A =

[

10 −58 −12

]

6. A =

[

−6 2−3 1

]


7. A =

1 1 −10 2 00 1 −1

8. A =

−1 1 01 2 10 3 −1

9. A =

3 −1 −11 1 −11 −1 1

10. A =

1 0 10 1 01 0 1

11. A =

[

6 −15 2

]

12. A =

[

1 1−2 −1

]

13. A =

[

5 1−2 3

]

14. A =

[

4 5−2 6

]

15. A =

[

4 −55 −4

]

16. A =

[

1 −81 −3

]

17. A =

0 0 10 0 −10 1 0

18. A =

1 −1 2−1 1 0−1 0 1

19. A =

4 0 10 6 0

−4 0 4

20. A =

2 5 1−5 −6 4

0 0 2

In the following problems solve the given system subject to the indicated initial condi-

tions.

21. x′ =

[

12 01 −1

2

]

x , x(0) =

[

35

]

22. x′ =

[

6 −15 4

]

x , x(0) =

[

−28

]

2.6 Repeated eigenvalues.

We have seen that if the coefficient matrix A of the system x′ = Ax has a repeated

eigenvalue λ of multiplicity m and there are m linearly independent eigenvectors associated

with λ, then there exist m linearly independent solutions of the system corresponding to

λ. This case was covered in Section 2.4 and is essentially the same as the case of distinct

eigenvalues. In this section we consider the case of a repeated eigenvalue of multiplicity m

2.6. REPEATED EIGENVALUES. 77

with less than m associated eigenvectors.

Suppose that the coefficient matrix A has a repeated eigenvalue λ1 = λ2 of multiplicity

two and that there is only one associated eigenvector k. Then

x(1) = keλ1t (2.23)

is a solution and we need to find a second linearly independent solution. This can be

achieved by assuming a solution of the form

x(2) = mteλ1t + neλ1t, (2.24)

where m and n are constant vectors. Substituting the expression (2.24) into the equation

of the system we find that

λ1mteλ1t + meλ1t + λ1neλ1t = A(mteλ1t + neλ1t).

The coefficients of teλ1t and eλ1t are, respectively,

λ1m = Am and m + λ1n = An,

i.e., (A − λ1I)m = 0, (2.25)

(A − λ1I)n = m. (2.26)

Eq. (2.25) shows that m is the eigenvector associated with the eigenvalue λ1, i.e., m = k,

so that Eq. (2.26) becomes

(A − λ1I)n = k. (2.27)

By solving this equation for n we are able to complete the second solution given by

x(2) = kteλ1t + neλ1t. (2.28)


Example 1. Find the general solution of the system

x′ =

[

3 −41 −1

]

x = Ax.

The eigenvalues of the coefficient matrix A are given by

3 − λ −41 −1 − λ

= 0 ⇒ λ2 − 2λ + 1 = 0,

i.e., λ1 = λ2 = 1.

For λ = 1, the system is

[

2 −4 01 −2 0

]

,

i.e., x1 − 2x2 = 0.

Hence, there is a single eigenvector k =

[

21

]

and the corresponding solution of the system

is

x(1) =

[

21

]

et.

The second linearly independent solution is of the form

x(2) =

[

21

]

tet + net,

where n is a solution of the equation (A − λ1I)n = k,

i.e.,

[

2 −4 21 −2 1

]

,

i.e., n1 − 2n2 = 1. (2.29)

If we put n2 = a, where a is arbitrary, then

n =

[

2a + 1a

]

= a

[

21

]

+

[

10

]

.


Hence, x(2) =

[

21

]

tet +

[

10

]

et + a

[

21

]

et. Note that the last term is simply a multiple

of the first solution x(1), so the second linearly independent solution is

x(2) =

[

21

]

tet +

[

10

]

et

=

[

2t + 1t

]

et.

Note that the vector n could have been found by putting one of the components, n1 or n2

of n equal to zero in Eq. (2.28).

A fundamental matrix for the system is

Φ(t) =

[

2et 2tet + et

et tet

]

and the general solution is

x = c1

[

21

]

et + c2

[

2t + 1t

]

et.

Now suppose the coefficient matrix A has an eigenvalue of multiplicity three, i.e., λ1 =

λ2 = λ3 and there is only one associated eigenvector k. In this case it can be shown that

the first solution is given by Eq. (2.23), a second solution by Eq. (2.28) and a third solution

is of the form

x(3) =1

2kt2eλ1t + nteλ1t + peλ1t, (2.30)

where k is the eigenvector, n satisfies Eq. (2.27) and p satisfies

(A − λ1I)p = n. (2.31)

Example 2. Consider the system x′ = Ax where A is the matrix of Example 4 of Sec-

tion 2.3, i.e.,

x′ =

−5 −5 −98 9 18

−2 −3 −7

x.


There is a single repeated eigenvalue λ1 = λ2 = λ3 = −1 and a single eigenvector k =

3−6

2

.

Thus the first solution is

x(1) =

3−6

2

e−t.

The second solution is given by Eq. (2.28) , i.e.,

x(2) =

3−6

2

te−t + ne−t,

where n is given by (A − λ1I)n = k, i.e.,

−4 −5 −9 38 10 18 −6

−2 −3 −6 2

,

and a solution of this is n =

00

−13

, so that the second solution is

x(2) =

3−6

2

te−t +

00

−13

e−t.

The third solution is given by Eq. (2.30), i.e.,

x(3) =1

2

3−6

2

t2e−t +

00

−13

te−t + pe−t,

where p is given by (A − λ1I)p = n, i.e.,

−4 −5 −9 08 10 18 0

−2 −3 −6 −13

.

Row operations on this augmented matrix lead to

−2 0 3 53

0 1 3 23

0 0 0 0


and a solution of this is p =

−5623

0

, so that the third solution is

x(3) =1

2

3−6

2

t2e−t +

00

−13

te−t +

−5623

0

e−t.

Hence a fundamental matrix is

Φ(t) =

3e−t 3te−t (32 t2 − 5

6)e−t

−6e−t −6te−t (−3t2 + 23)e−t

2e−t (2t − 13)e−t (t2 − 1

3 t)e−t

.

Another possibility is that the coefficient matrix A has an eigenvalue of multiplicity three,

i.e., λ1 = λ2 = λ3, with two linearly independent associated eigenvectors, k(1) and k(2),

then finding the solution is a little more complicated. In this case two linearly independent

solutions are

x(1) = k(1)eλ1t , x(2) = k(2)eλ1t

and a third solution is of the form

x(3) = kteλ1t + neλ1t, (2.32)

where n is a solution of Eq. (2.27) and k is a linear combination of k(1) and k(2) chosen in

such a way that Eq. (2.27) has a solution. We demonstrate this with the following example.

Example 3. Consider the system x′ = Ax where A is the matrix of Ex. 5 of Section 2.3,

i.e.,

x′ =

−1 −3 90 5 180 −2 −7

x.

There is a single repeated eigenvalue λ1 = λ2 = λ3 = −1 and two linearly independent

associated eigenvectors

k(1) =

100

, k(2) =

0−3

1

.


Hence two linearly independent solutions are

x(1) =

100

e−t, x(2) =

0−3

1

e−t.

Eq. (2.27) is of the form

0 −3 −9 k1

0 6 18 k2

0 −2 −6 k3

,

where k =

k1

k2

k3

= ak(1) + bk(2) =

a−3bb

for some constants a and b. Thus we have

0 −3 −9 a0 6 18 −3b0 −2 −6 b

R2 + 2R1

=⇒R3 − 1

3R2

0 −3 −9 a0 0 0 2a − 3b0 0 0 0

.

For consistency we must have 2a − 3b = 0, so we take a = 3, b = 2 and so

k = 3k(1) + 2k(2) =

3−6

2

.

Thus we have the equation −3n2−9n3 = 3 and a solution of this is n1 = 0, n2 = −1, n3 = 0,

i.e., n =

0−1

0

.

Hence the third solution is

x(3) =

3−6

2

te−t +

0−1

0

e−t


Φ(t) =

e−t 0 3te−t

0 −3e−t −(6t + 1)e−t

0 e−t 2te−t

.

2.7. NONHOMOGENEOUS SYSTEMS. 83

Problem Set 2.6

In the following problems find the general solution of the given system.

1.dx

dt= 3x − y 2.

dx

dt= −6x + 5y

dy

dt= 9x − 3y

dy

dt= −5x + 4y

3.dx

dt= −x + 3y 4. x′ =

5 −4 01 0 20 2 5

x

dy

dt= −3x + 5y

5. x′ =

1 0 00 3 10 −1 1

x 6. x′ =

1 0 02 2 −10 1 0

x

7. x′ =

4 1 00 4 10 0 4

x 8. x′ =

0 4 0−1 −4 0

0 0 −2

x

2.7 Nonhomogeneous systems.

Suppose that we have a nonhomogeneous system of equations of the form

x′ = Ax + f(t). (2.33)

In solving this system, the first step is to find the solution of the homogeneous system

x′ = Ax. We know that this solution can be written in the form

x = Φ(t)c, (2.34)

where Φ(t) is a fundamental matrix of the homogeneous system and c is a column vector of

constants. In seeking a solution of the system (2.33) we replace c in Eq. (2.34) by a column


vector of functions u(t) =

u1(t)...

un(t)

so that

x = Φ(t)u(t) (2.35)

is a particular solution of the system (2.33).

Differentiating Eq. (2.35) gives

x′ = Φ′(t)u(t) + Φ(t)u′(t)

and, substituting into Eq. (2.33), we obtain

Φ′(t)u(t) + Φ(t)u′(t) = AΦ(t)u(t) + f(t). (2.36)

Now, since Φ(t) is a fundamental matrix of the homogeneous system it satisfies the equation

Φ′ = AΦ, so Eq. (2.36) becomes

Φ(t)u′(t) = f(t),

i.e., u(t) =

∫

Φ−1(t)f(t) dt.

Hence, a particular solution of the system (2.33) is

xp = Φ(t)

∫

Φ−1(t)f(t) dt, (2.37)

and the general solution of the system is

x = Φ(t)c + Φ(t)

∫

Φ−1(t)f(t) dt. (2.38)

If there is an initial condition of the form

x(t0) = x0, (2.39)


it is useful to rewrite the solution (2.38) in the form

x = Φ(t)c + Φ(t)

∫ t

t0

Φ−1(s)f(s) ds, (2.40)

so that the particular solution chosen is the specific one that is zero at t = t0. In this case

the initial condition (2.39) becomes x0 = Φ(t0)c, i.e.,

c = Φ−1(t0)x0, (2.41)

so that the solution of the initial value problem is

x = Φ(t)Φ−1(t0)x0 + Φ(t)

∫ t

t0

Φ−1(s)f(s) ds. (2.42)

If we use as fundamental matrix the matrix Ψ(t) which satisfies Ψ(t0) = I (see Section 2.2),

then this solution can be written as

x = Ψ(t)x0 + Ψ(t)

∫ t

t0

Ψ−1(s) f(s) ds. (2.43)


x′ =

[

2 −13 −2

]

x +

[

et

t

]

.

This matrix A =

[

2 −13 −2

]

has two distinct eigenvalues λ1 = 1, λ2 = −1 with associ-

ated eigenvectors k(1) =

[

11

]

, k(2) =

[

13

]

. Hence a fundamental matrix is

Φ(t) =

[

et e−t

et 3e−t

]

,

so that

Φ−1(t) =1

2

[

3e−t −e−t

−et et

]

.


Now f(t) =

[

et

t

]

and so

Φ−1(t)f(t) =1

2

[

3 − te−t

−e2t + tet

]

.

Hence∫

Φ−1(t)f(t) dt =1

2

[

3t + (t + 1)e−t

−12e2t + (t − 1)et

]

, and

xp = Φ(t)

∫

Φ−1(t)f(t) dt =

[

32 tet − 1

4et + t32 tet − 3

4et + 2t − 1

]

.


x = c1

[

11

]

et + c2

[

13

]

e−t +

[

32 tet − 1

4et + t32 tet − 3

4et + 2t − 1

]

.


x′ =

[

2 −51 −2

]

x +

[

− cos tsin t

]

.

The matrix A =

[

2 −51 −2

]

has complex eigenvalues λ1 = i, λ2 = −i with associ-

ated eigenvectors k(1) =

[

2 + i1

]

, k(2) =

[

2 − i1

]

. Hence, the a complex solution of the

homogeneous system is given by

[

2 + i1

]

eit =

[

2 + i1

]

(cos t + i sin t) =

[

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

]

,

i.e., two linearly independent real solutions are

x(1) =

[

2 cos t − sin tcos t

]

, x(2) =

[

cos t + 2 sin tsin t

]


Φ(t) =

[

2 cos t − sin t cos t + 2 sin tcos t sin t

]

.


Then Φ−1(t) is given by

Φ−1(t) =

[

− sin t cos t + 2 sin tcos t −2 cos t + sin t

]

.

Now f(t) =

[

− cos tsin t

]

and so Φ−1(t)f(t) =

[

1 − cos 2t + sin 2t− cos 2t − sin 2t

]

.

Hence

∫

Φ−1(t)f(t) dt =

[

t − 12 sin 2t − 1

2 cos 2t

−12 sin 2t + 1

2 cos 2t

]

, and

xp = Φ(t)

∫

Φ−1(t)f(t) dt =

[

2t cos t − t sin t − 32 sin t − 1

2 cos t

t cos t − 12 sin t − 1

2 cos t

]

.


x = c1

[

2 cos t − sin tcos t

]

+ c2

[

cos t + 2 sin tsin t

]

+

[

2t cos t − t sin t − 32 sin t − 1

2 cos t

t cos t − 12 sin t − 1

2 cos t

]

.

Problem Set 2.7

Find the general solution of the given system.

1. x′ =

[

3 −32 −2

]

x +

[

41

]

2. x′ =

[

1√

3√3 −1

]

x +

[

et√

3e−t

]

3. x′ =

[

1 −11 1

]

x +

[

cos tsin t

]

et 4. x′ =

[

1 14 −2

]

x +

[

e−2t

−2et

]

5. x′ =

[

4 −28 −4

]

x +

[

t−3

−t−2

]

6. x′ =

[

−4 22 −1

]

x +

[

t−1

2t−1 + 4

]

7. x′ =

[

1 14 1

]

x +

[

2−1

]

et 8. x′ =

[

2 −13 −2

]

x +

[

1−1

]

et

9. x′ =

[

−54

34

34 −5

4

]

x +

[

2tet

]

10. x′ =

[

2 −51 −2

]

x +

[

0cos t

]


2.8 Laplace transform method for systems.

If initial conditions are specified, Laplace transform methods can be used to reduce a

system of differential equations to a set of simultaneous algebraic equations and thus to find

the solution. The system need not be of the first order. We demonstrate the method with

two examples.

Example 1. Use the Laplace transform to solve the system of differential equations

dx

dt= −x + y ,

dy

dt= 2x , x(0) = 0, y(0) = 1.

Let X(s) = L{x(t)} and Y (s) = L{y(t)}. Transforming each equation of the system,

we obtain

sX(s) − x(0) = −X(s) + Y (s),

sY (s) − y(0) = 2X(s),

i.e., (s + 1)X(s) − Y (s) = 0,

−2X(s) + sY (s) = 1.

Multiplying the first equation by s and adding the second equation we obtain

(s2 + s − 2)X(s) = 1,

i.e., X(s) =1

(s + 2)(s − 1)=

13

s − 1−

13

s + 2.

∴ x(t) =1

3et − 1

3e−2t.

2.8. LAPLACE TRANSFORM METHOD FOR SYSTEMS. 89

Also

Y (s) = (s + 1)X(s) =s + 1

(s + 2)(s − 1)=

23

s − 1+

13

s + 2.

∴ y(t) =2

3et +

1

3e−2t.

Hence, the solution of the given system is

x(t) =1

3(et − e−2t) , y(t) =

1

3(2et + e−2t).

Example 2. Use the Laplace transform to solve the system of differential equations

d2x

dt2+

d2y

dt2= t2 ,

d2x

dt2− d2y

dt2= 4t,

x(0) = 8 , x′(0) = 0 , y(0) = 0 , y′(0) = 0.

Taking the Laplace transform of each equation:

s2X(s) − sx(0) − x′(0) + s2Y (s) − sy(0) − y′(0) =2

s3

s2X(s) − sx(0) − x′(0) − s2Y (s) + sy(0) + y′(0) =4

s2,

i.e., X(s) + Y (s) =8

s+

2

s5

X(s) − Y (s) =8

s+

4

s4,

∴ X(s) =8

s+

2

s4+

1

s5, Y (s) =

1

s5− 2

s4.

Hence

x(t) = 8 +1

3t3 +

1

24t4, y(t) =

1

24t4 − 1

3t3.


Problem Set 2.8

Use the Laplace transform to solve the given systems of differential equations.

1.dx

dt= x − 2y ,

dy

dt= 5x − y , x(0) = 0 , y(0) = 1.

2. 2dx

dt+

dy

dt− 2x = 1 ,

dx

dt+

dy

dt− 3x − 3y = 2 , x(0) = 0 , y(0) = 0.

3.d2x

dt2+ x − y = 0 ,

d2y

dt2+ y − x = 0 , x(0) = 0 , x′(0) = −2 , y(0) = 0 , y′(0) = 1.

4.d2x

dt2+

3dy

dt+ 3y = 0 ,

d2x

dt2+ 3y = te−t , x(0) = 0 , x′(0) = 2 , y(0) = 0.

Chapter 3

FOURIER SERIES

3.1 Orthogonal sets of functions.

Suppose we have two vectors u, v in an n-dimensional vector space V (which can be

the usual 3-space). The inner product (scalar product) is written as (u,v) or u · v and has

the following properties:

(i) (u,v) = (v,u)

(ii) (ku,v) = k(u,v) for any scalar k

(iii) (u,u) = 0 if u = 0 and (u,u) > 0 if u 6= 0

(iv) (u + v,w) = (u,w) + (v,w)

The inner product of a vector with itself is

(u,u) = u21 + u2

2 + . . . + u2n

and the non-negative square root of (u,u) is called the norm of u and is denoted by ‖u‖,

i.e.,

‖u‖ =√

(u,u), (3.1)

91

92 CHAPTER 3. FOURIER SERIES

so that (u,u) = ‖u‖2 is the squared norm of u.

Two vectors are said to be orthogonal if their inner product is zero, i.e., if (u,v) = 0. In

n-dimensional space we can find n mutually orthogonal vectors ui (i = 1, . . . , n) and these

are said to form an orthogonal set. If each vector is divided by its norm, i.e., if we form the

vectors

ei =ui

‖ui‖, (3.2)

then the unit vectors ei satisfy

(ei, ej) = δij (i, j = 1, . . . , n), (3.3)

where δij is the Kronecker delta defined by

δij =

{

0 if i 6= j1 if i = j

. (3.4)

The set of vectors ei form an orthonormal set which is denoted by {ei}. For example, the

basis vectors i, j,k of 3-space form an orthonormal set.

Every vector v in the n-dimensional space can be expressed as a linear combination of

the orthonormal vectors ei , i.e.,

v = c1e1 + c2e2 + . . . + cnen, (3.5)

where the coefficients ci are given by

(v, ei) = ci(ei, ei) = ci (i = 1, . . . , n), (3.6)

so that ci is the projection of v on ei.

Now we extend these ideas of inner product and orthogonality to functions. Suppose

that fm(x) and fn(x) are two real-valued functions defined on an interval [a, b] and which

3.1. ORTHOGONAL SETS OF FUNCTIONS. 93

are such that the integral

(fm, fn) =

∫ b

afm(x)fn(x) dx (3.7)

exists. We make the following definitions in analogy with the concepts of vector theory.

Definition 1. The inner product of two functions fm(x) and fn(x) is the number (fm, fn)

defined by Eq. (3.7).

Definition 2. Two functions fm(x) and fn(x) are said to be orthogonal on an interval

[a, b] if (fm, fn) = 0.

Definition 3. The norm of a function fm(x) is

‖fm(x)‖ =√

(fm, fm) =

√

∫ b

a[fm(x)]2 dx . (3.8)

Note that ‖fm(x)‖ ≥ 0.

Our primary interest is in infinite sets of orthogonal functions. A set of real-valued

functions f1(x), f2(x), f3(x), . . . is called an orthogonal set of functions on an interval [a, b]

if the functions are defined on [a, b] and if the condition

(fm, fn) =

∫ b

afm(x)fn(x) dx = 0 (3.9)

holds for all pairs of distinct functions in the set.

Assuming that none of the functions fm(x) has zero norm, we can form a new set of

functions {φm(x)} defined by

φm(x) =fm(x)

‖fm(x)‖ . (3.10)

The set {φm(x)} is called an orthonormal set and satisfies

∫ b

aφm(x)φn(x) dx = δmn. (3.11)


Example 1. Consider the functions fm(x) = sinmπx

ℓ, i.e.,

f1(x), f2(x), f3(x), . . . = sinπx

ℓ, sin

2πx

ℓ, sin

3πx

ℓ, . . .

These functions form an orthogonal set on [−ℓ, ℓ] since, for all m and n

(fm, fn) =

∫ ℓ

−ℓsin

mπx

ℓsin

nπx

ℓdx

=1

2

∫ ℓ

−ℓ

[

cos(m − n)πx

ℓ− cos(m + n)

πx

ℓ

]

dx

= 0 if m 6= n.

The norm of fm(x) is given by ‖fm‖ =

√

1

2

∫ ℓ

−ℓ

(

1 − cos2mπx

ℓ

)

dx =√

ℓ , so that the

corresponding orthonormal set is

{

1√ℓ

sinmπx

ℓ

}

.

This concept of orthogonal functions may be generalized in two ways. First, we say that

a set of functions {fm(x)} is orthogonal on [a, b] with respect to a weight function w(x)

where w(x) ≥ 0, if∫ b

aw(x)fm(x)fn(x) dx = 0 (m 6= n). (3.12)

The norm ‖fm(x)‖ is given by

‖fm(x)‖2 =

∫ b

aw(x)[fm(x)]2 dx (3.13)

and the set {φm(x)}, where

φm(x) =fm(x)

‖fm(x)‖ , (3.14)

forms an orthonormal set.

Note that this type of orthogonality reduces to the ordinary type by using the product

functions√

w(x)fm(x).

3.2. EXPANSION OF FUNCTIONS IN SERIES OF ORTHOGONAL FUNCTIONS. 95

Another type of orthogonality concerns complex-valued functions. A set {tm} of complex-

valued functions of a real variable x is orthogonal in the hermitian sense, on an interval [a, b]

if∫ b

atm(x)tn(x) dx = 0, m 6= n. (3.15)

The integral is the hermitian inner product (tm, tn) corresponding to the definition (3.7).

The norm of tm is real and non-negative and is given by

‖tm‖2 =

∫ b

atmtmdx =

∫ b

a([um]2 + [vm]2) dx,

where tm(x) = um(x) + ivm(x) , um(x) and vm(x) being real-valued functions of x.

Example 2. Consider the functions

tn ≡ einx = cos nx + i sinnx (n = 0,±1,±2, . . . ).

These functions form a set with hermitian orthogonality on the interval [−π, π] since

(tm, tn) =

∫ π

−π(cos mx + i sinmx)(cos nx − i sinnx) dx

=

∫ π

−π[cos(m − n)x + i sin(m − n)x] dx

= 0 (m 6= n),

(tm, tm) = ‖tm‖2 =

∫ π

−πeimxe−imxdx =

∫ π

−π1 dx = 2π,

i.e., ‖tm‖ =√

2π.

3.2 Expansion of functions in series of orthogonal functions.

Given a set {φm(x)} of functions which are orthogonal on an interval [a, b], and which

may be orthonormal but not necessarily so, we can, under certain conditions, expand a


given function f(x) in terms of a series of the functions φm(x), i.e.,

f(x) = c0φ0(x) + c1φ1(x) + c2φ2(x) + . . . ,

i.e., f(x) =∞∑

n=0

cnφn(x). (3.16)

Assuming that this expansion exists we need to determine the coefficients cn. To do this we

multiply each side of Eq. (3.16) by φm(x), where φm(x) is the mth element of the orthogonal

set to obtain

f(x)φm(x) =∞∑

n=0

cnφm(x)φn(x).

Integrating both sides of this equation over [a, b] and assuming that the integral of the

infinite sum is equivalent to the sum of the integrals, we obtain

∫ b

af(x)φm(x) dx =

∞∑

n=0

cn

∫ b

aφm(x)φn(x) dx. (3.17)

Since {φm(x)} is an orthogonal set, all the integrals on the right side of (3.17) are zero

except that one for which m = n. Hence, Eq. (3.17) reduces to

cn

∫ b

a[φn(x)]2dx =

∫ b

af(x)φn(x) dx,

i.e., cn =

∫ ba f(x)φn(x) dx

‖φn(x)‖2, (3.18)

which determines each constant cn.

The above analysis can be extended to the case when the set {φm(x)} is orthogonal with

respect to the weight function w(x). In this case we multiply Eq. (3.16) by w(x)φm(x) and

eventually obtain

cn =

∫ ba w(x)f(x)φn(x) dx

‖φn(x)‖2, (3.19)

3.2. EXPANSION OF FUNCTIONS IN SERIES OF ORTHOGONAL FUNCTIONS. 97

where now ‖φn(x)‖2 =∫ ba w(x)[φn(x)]2dx.

The series (3.16), with coefficients given by Eqs. (3.18) or (3.19), is called a generalized

Fourier series.

Note that, although we have found a formal series of the form (3.16) we have not shown

that this series actually does represent the function f(x) in (a, b) or even that it converges

in (a, b). One necessary condition for the expansion to converge to f(x) is that the set

{φm(x)} must be complete, i.e., there must be no function with positive norm which is

orthogonal to each of the functions φm(x).

In general, the functions with which we shall be concerned are sectionally continuous or

piecewise continuous. A function is said to be piecewise continuous on [a, b] if it is defined

on [a, b] and if it has only a finite number of finite discontinuities in that interval.

Problem Set 3.2

1. Define (f, g) by (f, g) =

∫ 1

0f(x)g(x) dx.

(i) Are f(x) = x and g(x) = x2 orthogonal?

(ii) Find α, β, γ such that f(x) = 1, g(x) = x+α, h(x) = x2+βx+γ are orthogonal.

(iii) Form the orthonormal set corrersponding to the above set of three vectors.

(iv) If the vectors of part (iii) are labelled φ1, φ2, φ3, express F (x) = x2 − x + 1 as a

combination F (x) = c1φ1(x) + c2φ2(x) + c3φ3(x).

2. Show that f1(x) = ex and f2(x) = (x − 1)e−x are orthogonal on the interval [0, 2].

3. Show that {fm(x)} = {sinx, sin 3x, sin 5x, . . . } is orthogonal on the interval [0, π2 ] and


find the norm of each function fm(x).

4. Show that {fm(x)} = {1, cos nπxℓ } , n = 1, 2, 3, . . . is orthogonal on the interval [0, ℓ]

and find the norm of each function fm(x).

5. Given the three functions L0(x) = 1 , L1(x) = −x + 1 , L2(x) = 12x2 − 2x + 1, verify

by direct integration that the functions are orthogonal with respect to the weight

function w(x) = e−x on the interval [0,∞).

3.3 Fourier series.

We first note that the set of functions

{

1, cosnπx

ℓ, sin

nπx

ℓ

}

, n = 1, 2, 3, . . . (3.20)

is a complete orthogonal set on the interval [−ℓ, ℓ]. This follows from the integrals

∫ ℓ

−ℓsin

nπx

ℓdx = 0 ,

∫ ℓ

−ℓcos

nπx

ℓdx = 0 (all n ≥ 1), (3.21)

∫ ℓ

−ℓsin

mπx

ℓsin

nπx

ℓdx = 0 (m 6= n), (3.22)

∫ ℓ

−ℓcos

mπx

ℓcos

nπx

ℓdx = 0 (m 6= n), (3.23)

∫ ℓ

−ℓsin

mπx

ℓcos

nπx

ℓdx = 0 (all m, n ≥ 1). (3.24)

Note also that∫ ℓ

−ℓcos2

mπx

ℓdx =

∫ ℓ

−ℓsin2 mπx

ℓdx = ℓ. (3.25)

Hence, from Section 3.2 we can, under certain circumstances, expand a given function f(x)

in terms of the functions of the set (3.20), i.e.,

f(x) =a0

2+

∞∑

n=1

(

an cosnπx

ℓ+ bn sin

nπx

ℓ

)

, (3.26)

3.3. FOURIER SERIES. 99

where the coefficients a0, an, bn are real and independent of x. Such a series may, or may

not, be convergent. If it does converge to the sum f(x), then for every integer k

f(x + 2kℓ) = f(x) (3.27)

and f(x) is a periodic function of period 2ℓ so that we need only study the series in the

interval (−ℓ, ℓ), or some other interval of length 2ℓ, such as (0, 2ℓ). A series of the form

(3.26) is called a trigonometric series.

Suppose that f(x) is a periodic function of period 2ℓ which can be represented by the

trigonometric series (3.26). We need to find the coefficients a0, an, bn; this can be done by

using the results (3.21) - (3.25). First integrate Eq. (3.26) from −ℓ to ℓ

∫ ℓ

−ℓf(x) dx =

∫ ℓ

−ℓ

a0

2dx +

∞∑

n=1

∫ ℓ

−ℓ

(

an cosnπx

ℓ+ bn sin

nπx

ℓ

)

dx

= a0ℓ + 0 + 0 ,

using Eq. (3.21). Hence

a0 =1

ℓ

∫ ℓ

−ℓf(x) dx. (3.28)

Now multiply Eq. (3.26) by cosmπx

ℓand integrate from −ℓ to ℓ.

∫ ℓ

−ℓf(x) cos

mπx

ℓdx =

∫ ℓ

−ℓ

a0

2cos

mπx

ℓdx +

∞∑

n=1

{∫ ℓ

−ℓan cos

nπx

ℓcos

mπx

ℓdx

+

∫ ℓ

−ℓbn sin

nπx

ℓcos

mπx

ℓdx

}

= 0 + amℓ + 0 ,

i.e., am =1

ℓ

∫ ℓ

−ℓf(x) cos

mπx

ℓdx. (3.29)

Similarly, multiplying by sinmπx

ℓand integrating we find

bm =1

ℓ

∫ ℓ

−ℓf(x) sin

mπx

ℓdx. (3.30)


The expressions (3.28), (3.29), (3.30) are called the Euler formulae and the coeffi-

cients a0, am, bm are called the Fourier coefficients of f(x). The series (3.26) is called the

Fourier series corresponding to f(x).

Even if the right-hand side of Eq. (3.26) does not converge to f(x) for all x in (−ℓ, ℓ) we

can still calculate the Fourier coefficients of f(x) from Eqs. (3.28) - (3.30) and then write

f(x) ∼ a0

2+

∞∑

n=1

(

an cosnπx

ℓ+ bn sin

nπx

ℓ

)

,

where the right-hand side is the Fourier series of f(x). The symbol ∼ means that f(x) is

not necessarily equal to the right-hand side which may be divergent or converge to some

function other than f(x).

In order to know whether the Fourier series does, in fact, represent the function we need

the following theorem:

Fourier’s Theorem: If a periodic function f(x), with period 2ℓ, is piecewise continuous

on −ℓ < x < ℓ and has left-hand and right-hand derivatives at each point of (−ℓ, ℓ), then

the corresponding Fourier series (3.26), with coefficients (3.28) - (3.30), is convergent to

f(x) at a point of continuity. At a point of discontinuity, the Fourier series converges to

the sum

1

2[f(x+) + f(x−)],

where f(x+) is the value of f(x) when x is approached from the right, and f(x−) is the

value of f(x) when x is approached from the left.

Summary. The Fourier series of a function f(x) defined on the interval (−ℓ, ℓ) is given by


1

2

f(x-)

O x

f(x+)

[ ] f(x+)+f(x-)

f(x) =a0

2+

∞∑

n=1

(

an cosnπx

ℓ+ bn sin

nπx

ℓ

)

, (3.26)

where a0 =1

ℓ

∫ ℓ

−ℓf(x) dx, (3.28)

an =1

ℓ

∫ ℓ

−ℓf(x) cos

nπx

ℓdx, (3.29)

bn =1

ℓ

∫ ℓ

−ℓf(x) sin

nπx

ℓdx. (3.30)

Note that when ℓ = π, i.e., when the interval is of length 2π, the expressions (3.26),

(3.28) - (3.30) take the slightly simpler form

f(x) =a0

2+

∞∑

n=1

(an cos nx + bn sinnx), (3.31)

a0 =1

π

∫ π

−πf(x) dx , an =

1

π

∫ π

−πf(x) cos nx dx , bn =

1

π

∫ π

−πf(x) sin nx dx. (3.32)

Example 1. Find the Fourier series corresponding to the function f(x) = x in the interval

(−ℓ, ℓ).

The function obviously satisfies the conditions of the theorem. Using Eqs. (3.28) - (3.30),


we obtain

a0 =1

ℓ

∫ ℓ

−ℓx dx =

1

ℓ

[

1

2x2

]ℓ

−ℓ

= 0,

an =1

ℓ

∫ ℓ

−ℓx cos

nπx

ℓdx

=1

ℓ

{

[

x · ℓ

nπsin

nπx

ℓ

]ℓ

−ℓ

− ℓ

nπ

∫ ℓ

−ℓsin

nπx

ℓdx

}

= 0 +ℓ

n2π2

[

cosnπx

ℓ

]ℓ

−ℓ= 0,

bn =1

ℓ

∫ ℓ

−ℓx sin

nπx

ℓdx

=1

ℓ

{

[

−xℓ

nπcos

nπx

ℓ

]ℓ

−ℓ

+ℓ

nπ

∫ ℓ

−ℓcos

nπx

ℓdx

}

=1

ℓ

[−ℓ2

nπcos nπ − ℓ2

nπcos(−nπ) +

ℓ2

nπ

[

sinnπx

ℓ

]ℓ

−ℓ

]

=−2ℓ

nπ(−1)n.

Hence the required series is

f(x) =∞∑

n=1

(−1)n+1 · 2ℓ

nπsin

nπx

ℓ=

2ℓ

π

[

sinπx

ℓ− 1

2sin

2πx

ℓ+

1

3sin

3πx

ℓ. . .

]

and this does converge to the function in (−ℓ, ℓ).

Example 2. Consider the step function given by

f(x) =

0 −π < x < −π2

1 −π2 < x < π

20 π

2 < x < π.

This function has only a finite number of finite discontinuities and satisfies the conditions


π

1

2−π π−π

xO

2

of the theorem. The period is 2π, i.e., ℓ = π, so we use Eq. (3.31) and (3.32)

a0 =1

π

∫ π

−πf(x) dx =

1

π

∫ π

2

−π

2

1 · dx = 1,

an =1

π

∫ π

−πf(x) cos nx dx =

1

π

∫ π

2

−π

2

cos nx dx =1

π

[

1

nsinnx

]π

2

−π

2

=2

nπsin

nπ

2.

Now sinnπ

2= 0 if n is even. If n is odd, i.e., n = 2m + 1, then

sinnπ

2= sin(mπ +

π

2) = cos mπ = (−1)m.

∴ an = 0 (n even) , a2m+1 = (−1)m 2

(2m + 1)π(m = 0, 1, 2, . . . ),

bn =1

π

∫ π

−πf(x) sin nx dx =

1

π

∫ π

2

−π

2

sin nx dx =1

π

[

− 1

ncos nx

]π

2

−π

2

= 0.

Hence the series is

f(x) =1

2+

2

πcos x − 2

3πcos 3x +

2

5πcos 5x − 2

7πcos 7x + . . . ,

i.e., f(x) =1

2+

2

π

∞∑

m=0

(−1)m

2m + 1cos(2m + 1)x.

Note that when x = −π

2and when x =

π

2, cos(2m + 1)x = 0 so at these points f(x) =

1

2,

which is the average of the two values either side of those points.


The following diagrams illustrate how the Fourier series converges to the given function.

Diagrams 1, 3, 5 and 7 show each of the individual terms of the partial sums on the same

graph. Diagrams 2, 4, 6 and 8 show the terms summed together.

–0.8

–0.6

–0.4

–0.20

0.2

0.4

0.6

0.8

1

1.2

y

–3 –2 –1 1 2 3x

–0.8

–0.6

–0.4

–0.20

0.2

0.4

0.6

0.8

1

1.2

y

–3 –2 –1 1 2 3x

DIAGRAM 11

2+

2

πcos x DIAGRAM 2

–0.8

–0.6

–0.4

–0.20

0.2

0.4

0.6

0.8

1

1.2

y

–3 –2 –1 1 2 3x

–0.8

–0.6

–0.4

–0.20

0.2

0.4

0.6

0.8

1

1.2

y

–3 –2 –1 1 2 3x

DIAGRAM 31

2+

2

πcos x − 2

3πcos 3x DIAGRAM 4


–0.8

–0.6

–0.4

–0.20

0.2

0.4

0.6

0.8

1

1.2

y

–3 –2 –1 1 2 3x

–0.8

–0.6

–0.4

–0.20

0.2

0.4

0.6

0.8

1

1.2

y

–3 –2 –1 1 2 3x

DIAGRAM 51

2+

2

πcos x − 2

3πcos 3x +

2

5πcos 5x DIAGRAM 6

–0.8

–0.6

–0.4

–0.20

0.2

0.4

0.6

0.8

1

1.2

y

–3 –2 –1 1 2 3x

–0.8

–0.6

–0.4

–0.20

0.2

0.4

0.6

0.8

1

1.2

y

–3 –2 –1 1 2 3x

DIAGRAM 71

2+

2

πcos x − 2

3πcos 3x +

2

5πcos 5x − 2

7πcos 7x DIAGRAM 8

Example 3. Find the Fourier series of the function f(x) of period 2 defined by

1

1

2

-1

-1

O

f(x) =

{

−1 −1 < x < 02x 0 ≤ x < 1

.

The period is 2 so that ℓ = 1 and hence

a0 = 1 ·∫ 1

−1f(x) dx =

∫ 0

−1(−1)dx +

∫ 1

02x dx

= [−x]0−1 +

[

x2]1

0

= −1 + 1 = 0,


an =

∫ 1

−1f(x) cos nπx dx

=

∫ 0

−1(−1) cos nπx dx +

∫ 1

02x cos nπx dx

= − 1

nπ[sinnπx]0

−1 +

[

2x1

nπsinnπx

]1

0

− 2

nπ

∫ 1

01 · sinnπx dx

= 0 + 0 +2

n2π2(cos nπ − 1)

=2

n2π2[(−1)n − 1]

=

{

0 n even

− 4n2π2 n odd, i.e., − 4

(2m + 1)2π2 (m = 0, 1, 2 . . . ),

bn =

∫ 1

−1f(x) sin nπx dx

=

∫ 0

−1(−1) sin nπx dx +

∫ 1

02x sinnπx dx

=1

nπ[cos nπx]0

−1 −[

2x1

nπcos nπx

]1

0

+2

nπ

∫ 1

0cos nπx dx

=1

nπ(1 − cos nπ) − 2

nπcos nπ +

2

n2π2[sinnπx]10

=1

nπ(1 − 3 cos nπ) + 0

=

− 2nπ (n even) = − 1

mπ (m = 1, 2, . . . )

4nπ (n odd) = 4

(2m + 1)π(m = 0, 1, 2, . . . ).

Hence f(x) =

(

− 4

π2cos πx − 4

9π2cos 3πx − . . .

)

+

(

4

πsinπx − 1

πsin 2πx +

4

3πsin 3πx − . . .

)

,

3.4. COSINE AND SINE SERIES. 107

i.e., f(x) = − 4

π2

∞∑

m=0

1

(2m + 1)2cos(2m + 1)πx +

4

π

∞∑

m=0

1

2m + 1sin(2m + 1)πx

− 1

π

∞∑

m=1

sin 2mπx.

Problem Set 3.3

In the following problems find the Fourier series of f(x) on the given interval.

1. f(x) =

{

0 −π < x < 01 0 ≤ x < π

2. f(x) =

{

−1 −π < x < 02 0 ≤ x < π

3. f(x) =

{

1 −1 < x < 0x 0 ≤ x < 1

4. f(x) =

{

0 −1 < x < 0x 0 ≤ x < 1

5. f(x) =

{

0 −π < x < 0x2 0 ≤ x < π

6. f(x) =

{

π2 −π < x < 0π2 − x2 0 ≤ x < π

7. f(x) = x + π −π < x < π 8. f(x) = 3 − 2x −π < x < π

9. f(x) =

{

0 −π < x < 0sinx 0 ≤ x < π

10. f(x) =

{

0 −π2 < x < 0

cos x 0 ≤ x < π2

11. f(x) =

0 −2 < x < 0x 0 ≤ x < 11 1 ≤ x < 2

12. f(x) =

{

2 + x −2 < x < 02 0 ≤ x < 2

13. f(x) = ex −π < x < π 14. f(x) =

{

0 −π < x < 0ex − 1 0 ≤ x < π

3.4 Cosine and sine series.

A function f(x) defined in the interval (−ℓ, ℓ) is said to be an even function of x if, for

every value of x in the interval

f(−x) = f(x). (3.33)


On the other hand, f(x) is said to be an odd function of x if, for every value of x in the

interval,

f(−x) = −f(x). (3.34)

For Eq. (3.34) to be consistent we must have f(0) = 0.

For example, f(x) = xn is an even function if n is an even integer, including zero, and

is an odd function if n is odd. Also cos x is an even function while sinx is an odd function.

The graph of an even function is symmetric with respect to the y-axis, and the graph of an

odd function is symmetric with respect to the origin.

Some properties of even and odd functions are:

(i) The product of two even functions is even.

(ii) The product of two odd functions is even.

(iii) The product of an even function and an odd function is odd.

(iv) The sum (difference) of two even functions is even.

(v) The sum (difference) of two odd functions is odd.

(vi) If f is even, then

∫ a

−af(x)dx = 2

∫ a

0f(x) dx .

(vii) If f is odd, then

∫ a

−af(x) dx = 0.

If f(x) is an even function on (−ℓ, ℓ), then from properties (i), (iii) and (vi), the Fourier

coefficients (3.28) - (3.30) become

a0 =1

ℓ

∫ ℓ

−ℓf(x) dx =

2

ℓ

∫ ℓ

0f(x) dx,

an =1

ℓ

∫ ℓ

−ℓf(x) cos

nπ

ℓx dx =

2

ℓ

∫ ℓ

0f(x) cos

nπx

ℓdx,

bn =1

ℓ

∫ ℓ

−ℓf(x) sin

nπx

ℓdx = 0.

3.4. COSINE AND SINE SERIES. 109

Hence, we have that the Fourier series of an even function f(x) on (−ℓ, ℓ) is the cosine series

f(x) =a0

2+

∞∑

n=1

an cosnπx

ℓ, (3.35)

where a0 =2

ℓ

∫ ℓ

0f(x) dx , an =

2

ℓ

∫ ℓ

0f(x) cos

nπx

ℓdx. (3.36)

The Fourier series of an odd function on (−ℓ, ℓ) is the sine series

f(x) =∞∑

n=1

bn sinnπx

ℓ, (3.37)

where bn =2

ℓ

∫ ℓ

0f(x) sin

nπx

ℓdx. (3.38)

Example 1. Find the Fourier series of the function f(x) =1

4x2 on the interval (−π, π).

The function is an even function, so the Fourier series is the cosine series given by

Eqs. (3.35), (3.36). In this case ℓ = π, so

f(x) =a0

2+

∞∑

n=1

an cos nx,

and a0 =2

π

∫ π

0f(x) dx =

1

2π

∫ π

0x2dx

=1

2π

[

1

3x3

]π

0

=1

6ππ3 =

π2

6,

an =2

π

∫ π

0f(x) cos nx dx =

1

2π

∫ π

0x2 cos nx dx

=1

2π

{[

x2 1

nsinnx

]π

0

−∫ π

02x · 1

nsinnx dx

}

=1

2π

{

0 − 2

n

([

x

(

− 1

ncos nx

)]π

0

+

∫ π

01 · 1

ncos nx dx

)}

= − 1

nπ

[

[

−x

ncos nx

]π

0+

[

1

n2sinnx

]π

0

]


= − 1

nπ

[

−π

ncos nπ + 0 + 0

]

=1

n2cos nπ =

(−1)n

n2.

∴ a1 = −1 , a2 =1

4, a3 = −1

9, a4 =

1

16, etc.

∴ f(x) =π2

12+

∞∑

n=1

(−1)n

n2cos nx

=π2

12− cos x +

1

4cos 2x − 1

9cos 3x +

1

16cos 4x − . . . .

Example 2. Find the Fourier series of the function

f(x) =

{

−x2 −π < x < 0

x2 0 ≤ x < π.

on the interval (−π, π).

This is an odd function, so the Fourier series is the sine series given by Eqs. (3.37),

(3.38).

∴ bn =2

π

∫ π

0x2 sinnx dx

=2

π

{[

x2

(

− 1

n

)

cos nx

]π

0

−∫ π

02x

(

− 1

n

)

cos nx dx

}

=2

π

{[

− 1

nx2 cos nx

]π

0

+2

n

([

x1

nsinnx

]π

0

−∫ π

011

nsinnx dx

)}

=2

π

{

− 1

nπ2 cos nπ + 0 + 0 +

2

n

[

1

n2cos nx

]π

0

}

=2

π

{

−π2

n(−1)n +

2

n3cos nπ − 2

n3

}

=2

π

{

−π2

n(−1)n +

2

n3(−1)n − 2

n3

}

.

∴ b1 =2

π

{

+π2 − 2 − 2}

=2

π(π2 − 4) ; b2 =

2

π

{

−π2

2+

2

8− 2

8

}

= −π;

3.5. HALF-RANGE EXPANSIONS. 111

b3 =2

π

{

π2

3− 2

27− 2

27

}

=2

27π(9π2 − 4) ; b4 =

2

π

{

−π2

4

}

= −π

2; etc.

∴ f(x) =2

π(π2 − 4) sin x − π sin 2x +

2

27π(9π2 − 4) sin 3x − π

2sin 4x + . . . .

3.5 Half-range expansions.

In many physical and engineering problems we need to find a Fourier series expansion

for a function f(x) which is defined on some finite interval such as (0, ℓ) and very often we

need consider only a cosine series or only a sine series. For a series of cosines only we extend

f(x) as an even function into the interval −ℓ < x < 0, and elsewhere periodically. For a

series of sines only we extend f(x) as an odd function into the interval −ℓ < x < 0, and

elsewhere periodically. These definitions of f(x) into the interval −ℓ < x < 0 are called,

respectively, the even periodic extension and the odd periodic extension of f(x).

EVEN PERIODIC EXTENSION

f(x)

x

y

l-l

f(x)

x

y

l-l

ODD PERIODIC EXTENSION

The even extension is given by

f(x) =a0

2+

∞∑

n=1

an cosnπx

ℓ, (3.39)

a0 =2

ℓ

∫ ℓ

0f(x) dx , an =

2

ℓ

∫ ℓ

0f(x) cos

nπx

ℓdx, (3.40)


and the odd extension is given by

f(x) =∞∑

n=1

bn sinnπx

ℓ, (3.41)

bn =2

ℓ

∫ ℓ

0f(x) sin

nπx

ℓdx. (3.42)

The series (3.39) and (3.41) are called the half-range expansions of the function f(x).

Example. Find both the even and the odd periodic extensions of the function

f(x) =

π

20 < x <

π

2

π − xπ

2≤ x < π

f(x)

y

xπ

π/2

π/2O

O

y

xπ

π/2

π/2−π/2

EVEN EXTENSION

−π O

y

xπ

π/2

π/2

−π/2

−π/2−π

ODD EXTENSION

The even extension is given by Eq. (3.39) with

a0 =2

π

∫ π

0f(x) dx =

2

π

∫ π

2

0

π

2dx +

2

π

∫ π

π

2

(π − x) dx

=2

π

[π

2x]π

2

0+

2

π

[

πx − 1

2x2

]π

π

2

=π

2+

2

π

[

π2 − 1

2π2 − 1

2π2 +

1

8π2

]

=3

4π,

an =2

π

∫ π

0f(x) cos nx dx =

2

π

∫ π

2

0

π

2cos nx dx +

2

π

∫ π

π

2

(π − x) cos nx dx

3.5. HALF-RANGE EXPANSIONS. 113

=2

π

[ π

2nsin nx

]π

2

0+

2

π

{

[

(π − x)1

nsin nx

]π

π

2

+

∫ π

π

2

1 · 1

nsinnx dx

}

=2

π· π

2nsin

nπ

2+

2

π

(

0 − π

2nsin

nπ

2

)

− 2

πn2[cos nx]ππ

2

= − 2

πn2

[

cos nπ − cosnπ

2

]

.

Hence

a1 = − 2

π

(

cos π − cosπ

2

)

=2

π,

a2 = − 2

4π(cos 2π − cos π) = − 1

π,

a3 = − 2

9π

(

cos 3π − cos3π

2

)

=2

9π,

a4 = − 2

16π(cos 4π − cos 2π) = 0,

a5 = − 2

25π

(

cos 5π − cos5π

2

)

=2

25π,

a6 = − 2

36π(cos 6π − cos 3π) = − 1

9π.

Hence, the even periodic extension is

f(x) =3

8π +

1

π

(

2 cos x − cos 2x +2

9cos 3x +

2

25cos 5x − 1

9cos 6x + . . .

)

.

The odd extension is given by Eq. (3.41) with

bn =2

π

∫ π

0f(x) sin nx dx =

2

π

∫ π

2

0

π

2sinnx dx +

2

π

∫ π

π

2

(π − x) sin nx dx

=

[

− 1

ncos nx

]π

2

0

+2

π

{

[

(π − x)(− 1

ncos nx)

]π

π

2

−∫ π

π

2

1

ncos nx dx

}

= − 1

ncos

nπ

2+

1

n+

2

π

[

0 +π

2ncos

nπ

2

]

− 2

πn2[sinnx]ππ

2

=1

n+

2

πn2sin

nπ

2.

Hence,


b1 = 1 +2

πsin

π

2= 1 +

2

π, b2 =

1

2+

1

2πsinπ =

1

2,

b3 =1

3+

2

9πsin

3π

2=

1

3− 2

9π, b4 =

1

4+

1

8πsin 2π =

1

4,

b5 =1

5+

2

25πsin

5π

2=

1

5+

2

25π, b6 =

1

6+

1

18πsin 3π =

1

6.

Hence, the odd periodic extension is

f(x) =

(

1 +2

π

)

sinx +1

2sin 2x +

(

1

3− 2

9π

)

sin 3x +1

4sin 4x +

(

1

5+

2

25π

)

sin 5x + . . . .

Problem Set 3.5

In problems 1 - 12 determine whether f(x) is odd or even and expand the function in

an approporiate sine or cosine series.

1. f(x) =

{

−1 −π < x < 01 0 ≤ x < π

2. f(x) =

1 −2 < x < −10 −1 < x < 11 1 < x < 2

3. f(x) = |x |, −π < x < π 4. f(x) = x, −π < x < π

5. f(x) = x2, −1 < x < 1 6. f(x) = x|x |, −1 < x < 1

7. f(x) = π2 − x2, −π < x < π 8. f(x) = x3, −π < x < π

9. f(x) =

{

x − 1 −π < x < 0x + 1 0 ≤ x < π

10. f(x) =

{

x + 1 −1 < x < 0x − 1 0 ≤ x < 1

11. f(x) = | sinx |, −π < x < π 12. f(x) = cos x, −π

2< x <

π

2

In problems 13 - 22 find the half-range cosine and sine expansions of f(x).

13. f(x) =

{

1 0 < x < 12

0 12 ≤ x < 1

14. f(x) =

{

0 0 < x < 12

1 12 ≤ x < 1

15. f(x) = cos x , 0 < x <π

216. f(x) = sin x , 0 < x < π

17. f(x) =

x 0 < x < π2

π − x π2 ≤ x < π

18. f(x) =

{

0 0 < x < π

x − π π ≤ x < 2π

19. f(x) =

{

x 0 < x < 11 1 ≤ x < 2

20. f(x) =

{

1 0 < x < 12 − x 1 ≤ x < 2

21. f(x) = x2 + x , 0 < x < 1 22. f(x) = x(2 − x) , 0 < x < 2

3.6. COMPLEX FORM OF THE FOURIER SERIES. 115

3.6 Complex form of the Fourier series.

The complex exponential function is given by

eix = cos x + i sinx , e−ix = cos x − i sinx.

Hence cos nx =1

2(einx + e−inx) and sinnx =

1

2i(einx − e−inx).

The Fourier series

f(x) =a0

2+

∞∑

n=1

(an cosnπx

ℓ+ bn sin

nπx

ℓ)

becomes

f(x) =a0

2+

∞∑

n=1

[

an1

2

(

einπx

ℓ + e−inπx

ℓ

)

+ bn · 1

2i

(

einπx

ℓ − e−inπx

ℓ

)

]

=a0

2+

∞∑

n=1

[

1

2(an +

1

ibn)e

inπx

ℓ +1

2(an − 1

ibn)e−

inπx

ℓ

]

= h0 +∞∑

n=1

(

hneinπx

ℓ + hne−inπx

ℓ

)

, (3.43)

where h0 =a0

2, hn =

1

2(an − ibn) , hn =

1

2(an + ibn).

By writing hn as h−n, Eq. (3.43) can be written as

f(x) =∞∑

n=−∞

hneinπx

ℓ . (3.44)

The Fourier coefficients are given by

h0 =a0

2=

1

2ℓ

∫ ℓ

−ℓf(x) dx, (3.45)

hn =1

2(an − ibn) =

1

2ℓ

∫ ℓ

−ℓf(x)

(

cosnπx

ℓ− i sin

nπx

ℓ

)

dx,


i.e., hn =1

2ℓ

∫ ℓ

−ℓf(x)e−

inπx

ℓ dx, (3.46)

hn =1

2(an + ibn) =

1

2ℓ

∫ ℓ

−ℓf(x)e

inπx

ℓ dx. (3.47)

Hence, the complex Fourier series is given by Eq. (3.43) with Eqs. (3.45) - (3.47). Alterna-

tively, the Fourier series is given by Eq. (3.44) with hn given by Eq. (3.46).

If f(x) is an even function, then in its Fourier expansion bn = 0. Hence hn = hn =1

2an,

so that the complex Fourier coefficients are real. Similarly, if f(x) is an odd function, then

a0 = an = 0 and hn = −ibn so that the complex Fourier coefficients are pure imaginary.

Example. Find the complex Fourier series of the function

f(x) = ex, −π < x < π, f(x + 2π) = f(x).

We use the form (3.44) so that

hn =1

2π

∫ π

−πexe−inxdx =

1

2π

[

1

1 − ine(1−in)x

]π

−π

=1

2π

1

1 − in

(

e(1−in)π − e−(1−in)π)

=1

2π

1 + in

1 + n2

(

eπe−inπ − e−πeinπ)

.

Now einπ = cos nπ + i sinnπ = (−1)n. Similarly e−inπ = (−1)n. Hence

hn =1

2π

1 + in

1 + n2(−1)n(eπ − e−π) =

1 + in

π(1 + n2)(−1)n sinhπ,

∴ f(x) =sinhπ

π

∞∑

n=−∞

1 + in

1 + n2(−1)neinx.

Separating this expression for f(x) into real and imaginary parts we obtain

f(x) =sinhπ

π

∞∑

n=−∞

(−1)n

1 + n2[(cos nx − n sinnx) + i(n cos nx + sinnx)].

3.7. SEPARABLE PARTIAL DIFFERENTIAL EQUATIONS. 117

The terms corresponding to n = −1. . . . ,−∞ can be included in a sum from n = 1 to ∞

by replacing n by −n and the n = 0 term can be listed separately to give

f(x) =sinhπ

π

{

1 +∞∑

n=1

(−1)n

1 + n2[(cos nx − n sinnx) + i(n cos nx + sinnx)]

+∞∑

n=1

(−1)n

1 + n2[(cos nx − n sinnx) + i(−n cos nx − sinnx)]

}

=sinhπ

π

{

1 + 2∞∑

n=1

(−1)n

1 + n2cos nx − 2

∞∑

n=1

(−1)nn

1 + n2sinnx

}

,

which is the corresponding real Fourier series.

Problem Set 3.6

Find the complex form of the Fourier series of the following functions f(x).

1. f(x) = x − π < x < π.

2. f(x) =

{

−1 −π < x < 01 0 < x < π

.

3. f(x) = e−x − π

2< x <

π

2.

3.7 Separable partial differential equations.

Many problems in applied mathematics can be reduced to the solution of the partial

differential equation

∇2V = L∂2V

∂t2+ M

∂V

∂t+ N, (3.48)

where V is a physical quantity depending on the three cartesian space co-ordinates x, y, z

and on time t, ∇2 is the differential operator

∇2 =

(

∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)

, (3.49)


known as the Laplacian, and L, M, N are functions of x, y, z or constants. The following

four special cases of Eq. (3.48) are of particular importance.

(a) Laplace’s equation. Here L = M = N = 0 and examples of quantitites which can

be represented by the function V are:

(i) The gravitational potential in a region devoid of attracting matter.

(ii) The electrostatic potential in a uniform dielectric.

(iii) The magnetic potential.

(iv) The velocity potential in the irrotational motion of a homogeneous fluid.

(v) The steady state temperature in a uniform solid.

(b) Poisson’s Equation. In this equation L = M = 0 and N is a given function of x, y, z.

Examples of quantities represented by V are:

(i) The gravitational potential in a region in which N is proportional to the density

of the material at a point (x, y, z) in the region.

(ii) The electrostatic potential in a region in which N is proportional to the charge

distribution.

(iii) In the two-dimensional (z absent) form of the equation in which N is a constant,

V is a measure of the shear stress entailed by twisting a long bar of specified

cross-section.

(c) Heat conduction equation. When L = N = 0 and M = k−2, where k2 is the

diffusivity of a homogeneous isotropic body, Eq. (3.48) gives the temperature V at a


point (x, y, z) of the body. In certain circumstances the same equation can be used

in diffusion problems, the quantity V then being the concentration of the diffusing

substance.

(d) The wave equation. Eq. (3.48) with M = N = 0 and L = c−2 arises in investigations

of waves propagated with velocity c independent of wave length. Typical examples of

quantities which can be represented by the function V are:

(i) Components of the displacement in vibrating systems.

(ii) The velocity potential of a gas in the theory of sound.

(iii) Components of the electric or magnetic vector in the electromagnetic theory of

light.

In the two-dimensional case, i.e., when there are two independent variables x1 and x2, the

general second-order homogeneous linear partial differential equation for a function u(x1, x2)

can be written in the form

A∂2u

∂x21

+ B∂2u

∂x1∂x2+ C

∂2u

∂x22

+ D∂u

∂x1+ E

∂u

∂x2+ Fu = 0, (3.50)

where A, B, C, D, E, F are real constants. Such an equation can be classified into one of

three types; the equation is said to be

(a) hyperbolic if B2 − 4AC < 0 ,

(b) parabolic if B2 − 4AC = 0 ,

(c) elliptic if B2 − 4AC > 0.

Three important partial differential equations which are special cases of both Eqs. (3.48)

and (3.50) are:


1. The one-dimensional heat conduction equation

k2 ∂2u

∂x2=

∂u

∂t, (3.51)

where k2 is the thermal diffusivity, and u(x, t) is the temperature. This equation

governs the temperature distribution in a straight bar of uniform cross-section and

homogeneous material. It is a parabolic equation.

2. The one-dimensional wave equation

c2 ∂2u

∂x2=

∂2u

∂t2(3.52)

describes the motion of a vibrating string, where c is the wave velocity for the string.

It is a hyperbolic equation.

3. The two-dimensional Laplace equation

∂2u

∂x2+

∂2u

∂y2= 0. (3.53)

This equation arises in problems involving time-independent potential functions. It is

an elliptic equation.

We shall first turn our attention to the one-dimensional heat conduction equation, ap-

plying it to the study of the temperature distribution in a bar of thermal diffusivity k2 and

of length ℓ with the ends of the bar denoted by x = 0 and x = ℓ. In order to solve such a

problem we require an initial condition, such as the temperature distribution in the bar at

time t = 0. Only one such condition is required since the differential equation contains only

the first derivative with respect to t. However, we also require two boundary conditions,


i.e., conditions on x, since the differential equation contains the second derivative with re-

spect to x. The boundary conditions may be that the ends of the bar are held at fixed

temperatures or that the ends are insulated or that one end is at a fixed temperature while

the other end is insulated. The initial and boundary conditions that we will use may be

stated as follows:

Initial condition:

u(x, 0) = f(x). (3.54)

This states that at time t = 0 , the temperature distribution in the bar is given by f(x).

Boundary conditions: Either

u(0, t) = 0 , u(ℓ, t) = 0 , (3.55)

which states that both ends of the bar are held at zero temperature, or

ux(0, t) = 0 , ux(ℓ, t) = 0 , (3.56)

which states that there is no heat flow at the ends of the bar, i.e., the ends are insulated.

In order to solve Eq. (3.51) we assume that u(x, t) is a product function, i.e.,

u(x, t) = X(x)T (t). (3.57)

Substituting this into Eq. (3.51) we obtain

k2X ′′T = XT , (3.58)

where the prime denotes differentiation with respect to x and the dot denotes differentiation

with respect to t. Dividing through by XT , Eq. (3.58) becomes

X ′′

X= k−2 T

T. (3.59)


The left side of this equation is a function of x only while the right side is a function of t

only. Hence, each side must be equal to a constant, i.e.,

X ′′

X= k−2 T

T= α , (3.60)

so that the single partial differential equation is replaced by the two ordinary differential

equations

X ′′ − αX = 0, (3.61)

T − αk2T = 0. (3.62)

The product of two solutions of Eq. (3.61), (3.62), respectively, for any value of α is a

solution of Eq. (3.51). However, we require solutions that satisfy the boundary conditions

(3.55) or (3.56) and this severely restricts the possible values of α.

First we consider the case of the boundary conditions (3.55). The first of these gives

u(0, t) = X(0)T (t) = 0 and, since T (t) = 0 would imply that u(x, t) = 0 for all x, the only

possibility is X(0) = 0. Similarly, the second condition of (3.55) gives u(ℓ, t) = X(ℓ)T (t) = 0

which implies that X(ℓ) = 0. Hence, the boundary conditions (3.55) imply that

X(0) = X(ℓ) = 0. (3.63)

Now, in solving Eq. (3.61) there are three cases to consider, namely α = 0, α > 0 , (i.e.,

α = λ2) , and α < 0 , (i.e., α = −λ2).

Case 1. α = 0. Eq. (3.61) becomes X ′′ = 0, i.e.,

X = ax + b,


where a, b are constants. Substituting x = 0 and x = ℓ, the boundary conditions (3.63)

lead to a = b = 0, i.e., X = 0, so that u(x, t) = 0. This is not an acceptable solution so we

discard it.

Case 2. α = λ2. Eq. (3.61) becomes X ′′ − λ2X = 0, i.e.,

X = K1eλx + K2e

−λx.

The boundary conditions (3.63) give

K1 + K2 = 0 , K1eλℓ + K2e

−λℓ = 0 ,

and the solution of this system is K1 = K2 = 0, i.e., X = 0 , so that u(x, t) = 0. Again, we

discard this solution.

Case 3. α = −λ2. Eq. (3.61) becomes X ′′ + λ2X = 0, i.e.,

X = K1 cos λx + K2 sinλx.

The boundary conditions (3.63) give

K1 = 0 , K2 sinλℓ = 0.

We discard the possibility K2 = 0 since this implies that X = 0, so the solution is given by

sinλℓ = 0, i.e.,

λℓ = nπ , (3.64)

where n is a non-zero integer. Thus there are an infinite number of solutions. The constant

α in Eq. (3.61) is given by

α = −λ2 =−n2π2

ℓ2(3.65)


and X is proportional to sin(nπx

ℓ

)

.

From Eq. (3.65), the differential equation (3.62) for T becomes

T =−n2π2

ℓ2k2T, (3.66)

so that T is proportional to e−n

2π2

k2

t

ℓ2 . Thus, neglecting constant multipliers, the functions

un(x,t) = e−n

2π2

k2

t

ℓ2 sinnπx

ℓ(n = 1, 2, . . . ) (3.67)

are each solutions of Eq. (3.51) and satisfy the boundary conditions (3.55). Note that we

consider only positive values of n because negative values of n give the same solutions. Since

the differential equation (3.51) and the boundary conditions (3.55) are linear and homo-

geneous, it follows that any linear combination of the un(x, t) also satisfies the differential

equation and boundary conditions. Consequently, the solution of the differential equation

can be written as

u(x, t) =∞∑

n=1

cnun(x, t) =∞∑

n=1

cne−n

2π2

k2

t

ℓ2 sinnπx

ℓ. (3.68)

To complete the solution we have to satisfy the initial condition (3.54). Putting t = 0

in Eq. (3.68) we obtain

u(x, 0) = f(x) =∞∑

n=1

cn sinnπx

ℓ,

so that the cn are the Fourier coefficients for the Fourier sine series corresponding to f(x),

i.e.,

cn =2

ℓ

∫ ℓ

0f(x) sin

nπx

ℓdx. (3.69)

This completes the solution for u(x, t) which is Eq. (3.68) with cn given by Eq. (3.69).


Example 1. Find the solution to the heat conduction problem

k2uxx = ut 0 < x < ℓ , t > 0,

u(0, t) = 0 , u(ℓ, t) = 0 t > 0,

u(x, 0) = x(ℓ − x).

The solution is of the form (3.68) with cn given by Eq. (3.69) with f(x) = x(ℓ− x), i.e.,

cn =2

ℓ

∫ ℓ

0x(ℓ − x) sin

nπx

ℓdx

=2

ℓ

{

[

x(ℓ − x)(− ℓ

nπcos

nπx

ℓ)

]ℓ

0

+ℓ

nπ

∫ ℓ

0(ℓ − 2x) cos

nπx

ℓdx

}

=2

ℓ[0] +

2

nπ

{

[

(ℓ − 2x)ℓ

nπsin

nπx

ℓ

]ℓ

0

+2ℓ

nπ

∫ ℓ

0sin

nπx

ℓdx

}

=2

nπ[0] +

4ℓ

n2π2

[−ℓ

nπcos

nπx

ℓ

]ℓ

0

=4ℓ2

n3π3(cos 0 − cos nπ)

=4ℓ2

n3π3[1 − (−1)n]

=

{

0 n even8ℓ2

n3π3 n odd.

Put n = 2m + 1 (m = 0, 1, 2, . . . ), then

c2m+1 =8ℓ2

(2m + 1)3π3

and the final solution is

u(x, t) =∞∑

m=0

8ℓ2

(2m + 1)3π3e

−(2m+1)2π2

k2

t

ℓ2 sin(2m + 1)πx

ℓ


=8ℓ2

π3

[

e−π

2k2

t

ℓ2 sinπx

ℓ+

1

27e

−9π2

k2

t

ℓ2 sin3πx

ℓ+

1

125e

−25π2

k2

t

ℓ2 sin5πx

ℓ+ . . .

]

.

Example 2. Find the solution to the heat conduction problem

100uxx = ut 0 < x < 1 , t > 0,

u(0, t) = 0 , u(1, t) = 0 t > 0,

u(x, 0) = sin 2πx − 2 sin 5πx 0 ≤ x ≤ 1.

In this case k2 = 100 , ℓ = 1 , f(x) = sin 2πx − 2 sin 5πx.

The solution is of the form

u(x, t) =

∞∑

n=1

cne−100n2π2t sin nπx.

When t = 0 , u(x, 0) = sin 2πx − 2 sin 5πx =∞∑

n=1

cn sinnπx.

Hence c2 = 1 , c5 = −2 , cn = 0(n 6= 2, 5) and the solution is

u(x, t) = e−400π2t sin 2πx − 2e−2500π2t sin 5πx.

Now consider the case of the boundary conditions (3.56). The first of these gives

ux(0, t) = X ′(0)T (t) = 0 which, to avoid u(x, t) = 0 for all x, requires that X ′(0) = 0.

Similarly, the second condition implies that X ′(ℓ) = 0. Hence, the boundary conditions

(3.56) imply that

X ′(0) = X ′(ℓ) = 0. (3.70)

As in the previous case we consider the three cases in which α = 0 , α = λ2 ,

α = −λ2 , respectively, in Eq. (3.61).


Case 1. α = 0.

We again obtain X = ax + b and each of the conditions (3.70) imply that a = 0 so that

X is a constant, i.e.,

X =1

2c0. (3.71)

Case 2. α = λ2.

We again obtain X = K1eλx + K2e

−λx, so that X ′ = K1λeλx − K2λe−λx and the bound-

ary conditions (3.70) lead to

K1λ − K2λ = 0 , K1λeλℓ − K2λe−λℓ = 0

and the solution of this system is K1 = K2 = 0, i.e., X = 0, so that u(x, t) = 0 and we

discard this solution.

Case 3. α = −λ2.

As before we obtain X = K1 cos λx + K2 sinλx. Then X ′ = −K1λ sin λx + K2λ cos λx

and the boundary conditions (3.70) give

K2λ = 0 , −K1λ sinλℓ = 0.

Hence K2 = 0 and, since K1 cannot also be zero, we must have sin λℓ = 0, i.e.,

λℓ = nπ, (3.72)

where n is a non-zero integer. Thus there is an infinite number of solutions of Eq. (3.51)

satisfying the boundary conditions (3.56) of the form

un(x, t) = e−n2

π2

k2

ℓ2t cos

nπx

ℓ(n = 1, 2, . . . ) (3.73)


together with the solution of Case 1. Thus the solution of the differential equation can be

written as

u(x, t) =1

2c0 +

∞∑

n=1

cne−n2

π2

k2

ℓ2t cos

nπx

ℓ. (3.74)

We now have to satisfy the initial condition (3.54). Putting t = 0 in Eq. (3.74) we obtain

u(x, 0) = f(x) =1

2c0 +

∞∑

n=1

cn cosnπx

ℓ, (3.75)

so that c0 and cn are the Fourier coefficients for the Fourier cosine series corresponding to

f(x), i.e.,

c0 =2

ℓ

∫ ℓ

0f(x) dx , (3.76)

cn =2

ℓ

∫ ℓ

0f(x) cos

nπx

ℓdx. (3.77)

Thus, in the case of the boundary conditions (3.56), the complete solution is given by

Eq. (3.74) with c0, cn given by Eq. (3.76) , (3.77).

Example 3. Consider a uniform rod of length ℓ with an initial temperature given by

sinπx

ℓ, 0 ≤ x ≤ ℓ. Assume that both ends of the bar are insulated. Find a formal series

expansion for the temperature u(x, t). What is the steady-state temperature as t → ∞?

The solution is of the form (3.74) with c0 and cn given by Eqs. (3.76) and (3.77) with

f(x) = sinπx

ℓ, i.e.,

c0 =2

ℓ

∫ ℓ

0sin

πx

ℓdx

=2

ℓ

[

− ℓ

πcos

πx

ℓ

]ℓ

0

= − 2

π(cos π − cos 0) =

4

π.


cn =2

ℓ

∫ ℓ

0sin

πx

ℓcos

nπx

ℓdx

=2

ℓ

∫ ℓ

0

1

2

[

sin(n + 1)πx

ℓ− sin

(n − 1)πx

ℓ

]

dx

=1

ℓ

[

− ℓ

(n + 1)πcos(n + 1)

πx

ℓ+

ℓ

(n − 1)πcos(n − 1)

πx

ℓ

]ℓ

0

(n 6= 1)

= − 1

(n + 1)πcos(n + 1)π +

1

(n − 1)πcos(n − 1)π +

1

(n + 1)πcos 0 − 1

(n − 1)πcos 0

=

{

0 n odd, n 6= 1−4

(n2 − 1)πn even . (3.78)

When n = 1

c1 =2

ℓ

∫ ℓ

0sin

πx

ℓcos

πx

ℓdx

=1

ℓ

∫ ℓ

0sin

2πx

ℓdx

=1

ℓ

[

− ℓ

2πcos

2πx

ℓ

]ℓ

0

= − 1

2π(cos 2π − cos 0) = 0.

Hence the solution is

u(x, t) =2

π−

∞∑

n=1

cne−n2

π2

k2

ℓ2t cos

nπx

ℓ,

where cn are given by Eq. (3.78).

The steady state temperature is the limit as t → ∞ in the expression for u(x, t), i.e.,

uss =2

π.

Now consider the one-dimensional wave equation (3.52) as applied to the vibrations of

an elastic string tightly stretched between two points at the same horizontal level, distance


ℓ apart, so that the x-axis lies along the string. These vibrations are described by Eq. (3.52)

provided that damping effects, such as air resistance, are neglected and that the amplitude

of the motion is not too large. The constant c2 appearing in Eq. (3.52) is given by

c2 = T/ρ, (3.79)

where T is the tension in the string and ρ is the mass per unit length of the string. As

we remarked earlier, c is called the wave velocity for the string, i.e., the velocity at which

waves are propagated along the string.

Since Eq. (3.52) is of second order with respect to x and also with respect to t, it follows

that we require two boundary conditions and two initial conditions. Assuming that the

ends of the string are fixed, the boundary conditions are

u(0, t) = 0 , u(ℓ, t) = 0 , (3.80)

and the initial conditions are

u(x, 0) = f(x) , 0 ≤ x ≤ ℓ, (3.81)

which describes the initial position of the string, and

ut(x, 0) = g(x) , 0 ≤ x ≤ ℓ, (3.82)

which describes the initial velocity, where f(x) and g(x) are given functions which, for the

consistency of Eqs. (3.80) to (3.82) must satisfy

f(0) = f(ℓ) = 0 , g(0) = g(ℓ) = 0. (3.83)

The string may be set in motion by plucking, i.e., by pulling the string aside and letting it

go from rest. In this case f(x) 6= 0, but g(x) = 0. Alternatively, the string may be struck


while in an initial horizontal position, in which case f(x) = 0 but g(x) 6= 0. Of course, the

actual initial conditions may be some combination of these two possibilities.

To solve Eq. (3.52) we assume a separable solution of the form

u(x, t) = X(x)T (t) (3.84)

and substitution of this into Eq. (3.52) yields

X ′′

X=

1

c2

T

T= α , (3.85)

where α is a constant. As in the case of the heat-conduction equation with boundary

conditions (3.55), a non-trivial solution is obtained only if α < 0, i.e., α = −λ2, in which

case Eq. (3.85) leads to the two ordinary differential equations

X ′′ + λ2X = 0 , (3.86)

T + c2λ2T = 0 , (3.87)

for which the solutions are

X = K1 cos λx + K2 sin λx , (3.88)

T = K3 cos cλt + K4 sin cλt , (3.89)

where K1, . . . , K4 are arbitrary constants.

The boundary conditions (3.80) become

X(0) = 0 , X(ℓ) = 0 , (3.90)

which lead to K1 = 0 , K2 sinλℓ = 0. In order to avoid the trivial solution u(x, t) = 0

everywhere we require sin λℓ = 0, i.e.,

λ =nπ

ℓ, (3.91)


so that the solution for X(x) is

X = K2 sinnπx

ℓ(n = 1, 2, 3, . . . ). (3.92)

From Eqs. (3.89), (3.91) and (3.92) we see that the solutions for u satisfying the boundary

conditions are

un = (An cosnπct

ℓ+ Bn sin

nπct

ℓ) sin

nπx

ℓ(3.93)

and the general solution is the sum of all such solutions, i.e.,

u(x, t) =∞∑

n=1

(An cosnπct

ℓ+ Bn sin

nπct

ℓ) sin

nπx

ℓ. (3.94)

Putting t = 0 in Eq. (3.94) we have

u(x, 0) = f(x) =∞∑

n=1

An sinnπx

ℓ, (3.95)

so that the An are the Fourier coefficients for the half-range expansion sine series for f(x),

i.e.,

An =2

ℓ

∫ ℓ

0f(x) sin

nπx

ℓdx. (3.96)

To determine Bn, we differentiate Eq. (3.94) with respect to t to obtain

ut(x, t) =∞∑

n=1

(−Annπc

ℓsin

nπct

ℓ+ Bn

nπc

ℓcos

nπct

ℓ) sin

nπx

ℓ,

so that

ut(x, 0) = g(x) =∞∑

n=1

Bnnπc

ℓsin

nπx

ℓ, (3.97)

and hence the coefficients Bnnπc

ℓare the Fourier coefficients for the half-range expansion

sine series for g(x), i.e.,

Bnnπc

ℓ=

2

ℓ

∫ ℓ

0g(x) sin

nπx

ℓdx,

i.e., Bn =2

nπc

∫ ℓ

0g(x) sin

nπx

ℓdx. (3.98)


Example 4. Find the solution of the wave equation for a vibrating string of length ℓ

subject to the boundary conditions

u(0, t) = 0 , u(ℓ, t) = 0,

and the initial conditions

u(x, 0) = 0 , ut(x, 0) = x(ℓ − x).

In this problem f(x) = 0 , g(x) = x(ℓ − x), so that An = 0 and

Bn =2

nπc

∫ ℓ

0x(ℓ − x) sin

nπx

ℓdx

=2

nπc

{

[−ℓ

nπx(ℓ − x) cos

nπx

ℓ

]ℓ

0

+ℓ

nπ

∫ ℓ

0(ℓ − 2x) cos

nπx

ℓdx

}

=2ℓ

n2π2c

{

[

(ℓ − 2x)ℓ

nπsin

nπx

ℓ

]ℓ

0

+2ℓ

nπ

∫ ℓ

0sin

nπx

ℓdx

}

=4ℓ2

n3π3c· ℓ

nπ

[

cosnπx

ℓ

]ℓ

0

=4ℓ3

n4π4c[1 − (−1)n].

Hence the solution for u(x, t) is

u(x, t) =4ℓ3

π4c

∞∑

n=1

1

n4[1 − (−1)n] sin

nπct

ℓsin

nπx

ℓ.

When the tension T in the string is large enough, the vibrating string will produce

a musical sound. This sound is the result of standing waves. The solution (3.94) is a

superposition of product solutions called standing waves or normal modes

u(x, t) = u1(x, t) + u2(x, t) + u3(x, t) + . . . .


The product solutions (3.93) can be written as

un(x, t) = Cn sin

(

nπct

ℓ+ αn

)

sinnπx

ℓ, (3.99)

where Cn =√

A2n + B2

n and sinαn =An

Cn, cos αn =

Bn

Cn. For n = 1, 2, 3, . . . the stand-

ing waves are essentially the graphs of sinnπx

ℓwith a time varying amplitude given by

Cn sin

(

nπct

ℓ+ αn

)

. Alternatively, we see from Eq. (3.99) that, at a fixed value of x, each

product function un(x, t) represents simple harmonic motion with amplitude Cn| sinnπx

ℓ|

and frequency fn =nc

2ℓ, i.e., each point on a standing wave vibrates with a different ampli-

tude but with the same frequency. When n = 1

u1(x, t) = C1 sin

(

πct

ℓ+ αn

)

sinπx

ℓ,

is called the first standing wave, the first normal mode, or the fundamental mode of vibration.

l0

FIRST STANDING WAVE

0 l/2 l

node

SECOND STANDING WAVE

The first three standing waves, or normal modes, are shown in the figures. The dashed

line graphs represent the standing waves at various values of time. The points in the


02l/3l/3

l

nodes

THIRD STANDING WAVE

interval (0, ℓ) for which sinnπx

ℓ= 0 correspond to points on a standing wave where there

is no motion; these points are called nodes. In general, the nth normal mode of vibration

has n − 1 nodes.

The frequency f1 =c

2ℓ=

1

2ℓ

√

T

ρof the first normal mode is called the fundamental

frequency, or first harmonic, and is directly related to the pitch produced by a stringed

instrument. The frequencies fn of the other normal modes, which are integer multiples of

the fundamental frequency, are called overtones. The second harmonic is the first overtone,

and so on.

Problem Set 3.7

1. Consider the conduction of heat in a copper rod 100 cm. in length whose ends are

maintained at 0◦C for all t > 0. Find an expression for the temperature u(x, t) if the

initial temperature distribution in the rod is given by

(a) u(x, 0) =

{

x 0 ≤ x < 50100 − x 50 ≤ x ≤ 100

.

(b) u(x, 0) =

0 0 ≤ x < 2550 25 ≤ x < 750 75 ≤ x ≤ 100

.

2. Consider a uniform rod of length ℓ with an initial temperature given by sinπx

ℓ,


0 ≤ x ≤ ℓ. Assume that both ends of the bar are insulated. Find a Fourier series ex-

pansion for the temperature u(x, t). What is the steady-state temperature as t → ∞?

3. Find the displacement u(x, t) in an elastic string that is fixed at its ends, x = 0 and

x = ℓ, and is set in motion by plucking it at its centre. The initial displacement f(x)

is defined by

f(x) =

Ax 0 ≤ x ≤ 12ℓ

A(ℓ − x) 12ℓ < x ≤ ℓ

,

where A is a constant.

4. Find the displacement u(x, t) in an elastic string of length ℓ, fixed at both ends, that

is set in motion from its straight equilibrium position with the initial velocity g(x)

defined by

g(x) =

x 0 ≤ x ≤ 14ℓ

14ℓ 1

4ℓ < x < 34ℓ

ℓ − x 34ℓ ≤ x ≤ ℓ

.

Appendix A

ANSWERS TO

ODD-NUMBERED PROBLEMS

AND TABLES

A.1 Answers for Chapter 1.

Problem Set 1.3 (p. 9)

1.1

s

(

2e−s − 1)

3.1

s2(1 − e−s) 5.

−(e−sπ + 1)

s2 + 1

7.e7

s − 19.

1

(s − 4)211.

1

(s + 1)2 + 1

13.s2 − 1

(s2 + 1)215.

48

s517.

4

s2− 10

s

19.2

s3+

6

s2− 3

s21.

6

s4+

6

s3+

3

s2+

1

s23.

1

s+

1

s − 4

25.1

s+

2

s − 2+

1

s − 427.

8

s3− 15

s2 + 929.

2s

(s2 − 1)2

31.s + 1

s(s + 2)33.

2

s(s2 + 4)35.

1

2

[

− s

s2 + 9+

s

s2 + 1

]

37.6

(s2 + 1)(s2 + 9)39.

Γ(54)

s5/4

Problem Set 1.4. (p. 13)

1. 3 − 2e−4t 3. 2e−4t + e2t 5. −5

8− 3

4t − 1

4t2 +

2

3et − 1

24e−2t

137

138 APPENDIX A. ANSWERS TO ODD-NUMBERED PROBLEMS AND TABLES


1. 2(t − 2)e2t 3. tet − 1

2t2e2t

5. −t + 5 − 3

2e−tt2 − 4e−tt − 5e−t 7.

1

(s − 8)2

9.s + 2

s2 + 4s + 2011.

2

s2 + 2s + 5

13.2

(s − 3)3+

4

(s − 3)2+

4

(s − 3)


1.1

s2e−s 3.

(

3

s2+

10

s

)

e−3s

5.

[

1

(s − 1)2+

5

s − 1

]

e−5s 7.1

2(t − 2)2u2(t)

9. − sin t uπ(t) 11. [1 − e−(t−1)]u1(t)

13. t − (t − 1)u1(t) 15.2

s(1 − 2e−3s), f(t) = 2u0 − 4u3

17.

(

2

s3+

2

s2+

1

s

)

e−s, f(t) = t2u1 19.1

s2− 1

s2e−2s − 2

se−2s, f(t) = t(u0 − u2)


1.s2 − 4

(s2 + 4)23.

2(3s2 + 1)

(s2 − 1)35.

12(s − 1)

(s2 − 4s + 40)2

7.1

2t sin t 9.

1

t(e−t − e3t) 11. 1 +

1

tsin 4t


1.s2 + 2a2

s(s2 + 4a2)

A.1. ANSWERS FOR CHAPTER 1. 139


1. et − 1 3.1

3(4e−t − e−4t)

5.1

27(2 + 3t − 2e3t + 30te3t) 7.

1

20t5e2t

9. cos t − 1

2sin t − 1

2t cos t 11.

1

2(1 + et sin t − et cos t)


1.1

4[1 + 2 sin 2t − cos 2t − u1(t) + cos 2(t − 1)u1(t)]

3.1

6[1 − 3e2(t−1) + 2e3(t−1)]u1(t) + e3t − e2t

5. cos t + (1 + cos t)uπ(t)


1.1

stanh

1

2as 3.

1

s2 + 1· 1

1 − e−πs

5.Eo

R

[(

eRa

L − 1)

ua(t) −(

e2Ra

L − 1)

u2a(t) +(

e3Ra

L − 1)

u3a(t) − . . .]


1. sin t(1 + u2π(t))

3. uπ

2(t) − u 3π

2(t) − sin t

[

uπ

2(t) + uπ(t) + u 3π

2(t)]

5.1√2

sinh√

2t +1

2cosh

√2t − 1

2+

1√2

sinh√

2(t − 2)u2(t)


1.1

s2(s − 1)3.

1

s(s2 + 1)5.

12

s7

7.

∫ t

0f(t − τ) cos 2τ dτ 9.

1

4t sin 2t

11.1

2e−2t(sin t − t cos t) 13.

1

2(t cos t + sin t) [< 0 when t = π]




7. Fundamental Set

11. Φ(t) =

[

−et −tet

3et (3t + 1)et

]

, Φ−1(t) =

[

−(3t + 1)e−t −te−t

3e−t e−t

]

13. Ψ(t) =

[ 12e−t + 1

2e5t −12e−t + 1

2e5t

−12e−t + 1

2e5t 12e−t + 1

2e5t

]

15. Ψ(t) =

[

sin t − 3 cos t 2 cos t−5 cos t sin t + 3 cos t

]


1. λ1 = 6 , λ2 = 1 , x(1) =

[

27

]

, x(2) =

[

11

]

3. λ1 = λ2 = −4 , x(1) =

[

1−4

]

5. λ1 = 0 , λ2 = 4 , λ3 = −4 , x(1) =

94525

, x(2) =

111

, x(3) =

191

7. λ1 = λ2 = λ3 = −2 , x(1) =

2−1

0

, x(2) =

001

9. λ1 = 3i , λ2 = −3i , x(1) =

[

1 − 3i5

]

, x(2) =

[

1 + 3i5

]


1. x = c1

[

12

]

e5t + c2

[

1−1

]

e−t

3. x = c1

[

21

]

e−3t + c2

[

25

]

et

5. x = c1

[

52

]

e8t + c2

[

14

]

e−10t


7. x = c1

100

et + c2

231

e2t + c3

102

e−t

9. x = c1

111

et + c2

110

e2t + c3

101

e2t

11. x = c1

[

cos t2 cos t + sin t

]

e4t + c2

[

sin t2 sin t − cos t

]

e4t

13. x = c1

[

cos t− cos t − sin t

]

e4t + c2

[

sin t− sin t + cos t

]

e4t

15. x = c1

[

5 cos 3t4 cos 3t +3 sin 3t

]

+ c2

[

5 sin 3t4 sin 3t −3 cos 3t

]

17. x = c1

100

+ c2

− cos tcos tsin t

+ c3

sin t− sin t

cos t

19. x = c1

010

e6t + c2

cos 2t0

−2 sin 2t

e4t + c3

sin 2t0

2 cos 2t

e4t

21. x = 3

[

11

]

e12t + 2

[

01

]

e−12t


1. x = c1

[

13

]

+ c2

(

[

13

]

t +

[

14

−14

])

3. x = c1

[

11

]

e2t + c2

([

11

]

t +

[

−130

))

e2t

5. x = c1

100

et + c2

01

−1

e2t + c3

01

−1

t +

010

e2t

7. x = c1

100

e4t + c2

100

t +

010

e4t + c3

100

t2

2+

010

t +

001

e4t


1. x = c1

[

11

]

+ c2

[

32

]

et +

[

−5t − 3−5t

]


3. x = c1

[

cos tsin t

]

et + c2

[

sin t− cos t

]

et +

[

cos tsin t

]

tet

5. x = c1

[

12

]

+ c2

([

12

]

t − 1

2

[

01

])

− 2

[

12

]

ln t +

[

25

]

t−1 −[

120

]

t−2

7. x = c1

[

12

]

e3t + c2

[

1−2

]

e−t +1

4

[

1−8

]

et

9. x = c1

[

11

]

e−12t + c2

[

1−1

]

e−2t +

[

5232

]

t −[

174154

]

+

[

1612

]

et


1. x = −2

3sin 3t , y = −1

3sin 3t + cos 3t

3. x = −1

2t − 3

4

√2 sin

√2t , y = −1

2t +

3

4

√2 sin

√2t



1. (i) Not orthogonal

(ii) α = −1

2, β = −1 , γ =

1

6

(iii) φ1(x) = 1 , φ2(x) = 2√

3(x − 1

2) , φ3(x) = 6

√5(x2 − x − 1

6)

(iv) F (x) =5

6φ1(x) +

1

6√

5φ3(x)

3. Norm of each function is1

2

√π


1. f(x) =1

2+

1

π

∞∑

n=1

1

n[1 − (−1)n] sin nx

3. f(x) =3

4+

∞∑

n=1

{

(−1)n − 1

n2π2cos nπx − 1

nπsin nπx

}

5. f(x) =1

6π2 + 2

∞∑

n=1

(−1)n

n2cos nx +

2

π

∞∑

n=1

(−1)n − 1

n3sin nx + π

∞∑

n=1

(−1)n+1

nsinnx


7. f(x) = π + 2

∞∑

n=1

(−1)n+1

nsinnx

9. f(x) =1

π− 1

π

∞∑

n=1

1

n2 − 1

[

1 − (−1)n+1]

cos nx +1

2sinx

11. f(x) =3

8+

2

π2

∞∑

n=1

{

− 1

n2(1 − cos

nπ

2) cos

nπx

2+

[

1

n2sin

nπ

2− (−1)nπ

2n

]

sinnπx

2

}

13. f(x) =2 sinhπ

π

[

1

2+

∞∑

n=1

(−1)n

n2 + 1(cos nx − n sinnx)

]


1. ODD f(x) =2

π

∞∑

n=1

[1 − (−1)n]

nsin nx

3. EVEN f(x) =π

2+

2

π

∞∑

n=1

[(−1)n − 1]

n2cos nx

5. EVEN f(x) =1

3+

4

π2

∞∑

n=1

(−1)n

n2cos nπx

7. EVEN f(x) =2

3π2 + 4

∞∑

n=1

(−1)n+1

n2 cos nx

9. ODD f(x) =2

π

∞∑

n=1

1 − (−1)n(π + 1)

nsinnx

11. EVEN f(x) =2

π+

2

π

∞∑

n=1

1 + (−1)n

1 − n2cos nx

13. EVEN EXTENSION f(x) =1

2+

2

π

∞∑

n=1

1

nsin

nπ

2cos nπx

ODD EXTENSION f(x) =2

π

∞∑

n=1

1

n

(

1 − cosnπ

2

)

sinnπx


π+

4

π

∞∑

n=1

(−1)n

1 − 4n2cos 2nx


π

∞∑

n=1

n

4n2 − 1sin 2nx


17. EVEN EXTENSION f(x) =π

4+

2

π

∞∑

n=1

2 cos nπ2 − 1 − (−1)n

n2cos nx


π

∞∑

n=1

1

n2sin

nπ

2sinnx


4+

4

π

∞∑

n=1

(cos nπ2 − 1)

n2cos

nπx

2

ODD EXTENSION f(x) =∞∑

n=1

[

4

n2π2sin

nπ

2− 2

nπ(−1)n

]

sinnπx

2


6+

2

π2

∞∑

n=1

[3(−1)n − 1]

n2cos nπx

ODD EXTENSION f(x) = 4∞∑

n=1

[

(−1)n+1

nπ+

(−1)n − 1

n3π2

]

sinnπx


1. f(x) = i∞∑

n=−∞

1

n(−1)n+1einx (n 6= 0)

3. f(x) =2 sinh π

2

π

∞∑

n=−∞

(−1)n (1 − 2in)

1 + 4n2e2inx


1. (a) u(x, t) =400

π2

∞∑

n=1

1

n2sin

nπ

2sin

nπx

100e

(

−nπ2

c2

t

104

)

(b) u(x, t) =100

π

∞∑

n=1

1

n

[

cosnπ

4− cos

3nπ

4

]

sinnπx

100e

(

−nπ2

c2

t

104

)

3. u(x, t) =4Aℓ

π2

∞∑

n=1

1

n2sin

nπ

2sin

nπx

ℓcos

nπct

ℓ

A.4. TABLE OF LAPLACE TRANSFORMS. 145

A.4 Table of Laplace transforms.

f(t) = L−1{F (s)} F (s) = L{f(t)}

11

s, s > 0

eat 1

s − a, s > a

sin ata

s2 + a2, s > 0

tn, n = positive integern!

sn+1, s > 0

tp, p > −1Γ(p + 1)

sp+1, s > 0

cos ats

s2 + a2, s > 0

sinh ata

s2 − a2, s > | a |

cosh ats

s2 − a2, s > | a |

eat sin btb

(s − a)2 + b2, s > a

eat cos bts − a

(s − a)2 + b2, s > a

tneat , n = positive integer n!(s − a)−n−1 , s > a

uc(t)e−cs

s, s > 0

uc(t)f(t − c) e−csF (s)

ectf(t) F (s − c)

f(ct)1

cF(s

c

)

, c > 0∫ t

0f(t − τ)g(τ) dτ F (s)G(s)

δ(t − c) e−cs

f (n)(t) snF (s) − sn−1f(0) − . . . − f (n−1)(0)

(−t)nf(t) F (n)(s)

t−1f(t)

∫

∞

sF (s) ds

∫ t

0f(u) du s−1F (s)

1

2a3sin at − t

2a2cos at

1

(s2 + a2)2

1

2at sin at

s

(s2 + a2)2

f(t + T ) = f(t)1

(1 − e−sT )

∫ T

0e−stf(t) dt

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Laplace Transform