math 110, spring 2004 notesbdriver/110-s04/lecture notes/110notes... · 2004. 5. 25. · bruce k....

Bruce K. Driver

Math 110, Spring 2004 Notes

May 25, 2004 File:110notes.tex

Springer

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Contents

1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 The spectral theorem for symmetric matrices . . . . . . . . . . . . . . . 11.2 Cylindrical and Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . 5

1.2.1 Cylindrical coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2.2 Spherical coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2 PDE Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.1 The Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.1.1 d’Alembert’s solution to the 1-dimensional waveequation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2 Heat Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3 Other Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.4 Worked Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3 Linear ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.1 First order linear ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.2 Solving for etA using the Spectral Theorem . . . . . . . . . . . . . . . . . 323.3 Second Order Linear ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.4 ODE Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4 Linear Operators and Separation of Variables . . . . . . . . . . . . . . 414.1 Introduction to Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.2 Application / Separation of variables . . . . . . . . . . . . . . . . . . . . . . . 47

5 Orthogonal Function Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . 535.1 Generalities about inner products on function spaces . . . . . . . . . 535.2 Convergence of the Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . 575.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 585.4 Proof of Theorem 5.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

5.4.1 Gibb’s Phenomenon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

4 Contents

5.5 Fourier Series on Other Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . 76

6 Boundary value problem examples . . . . . . . . . . . . . . . . . . . . . . . . . 81

7 Boundary value generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 877.1 Linear Algebra of the Strurm-Liouville Eigenvalue Problem . . . 877.2 General Elliptic PDE Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

8 PDE Applications and Duhamel’s Principle . . . . . . . . . . . . . . . . 978.1 Interpretation of d’Alembert’s solution to the 1-d wave equation 978.2 Solving 1st - order equations using 2nd - order solutions . . . . . . 97

8.2.1 The Solution to the Heat Equation on R . . . . . . . . . . . . . 998.3 Duhamel’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1008.4 Application of Duhamel’s principle to 1 - d wave and heat

equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

9 Problems In other Coordinates Systems . . . . . . . . . . . . . . . . . . . 1159.1 A Heat equation in spherical coordinates . . . . . . . . . . . . . . . . . . . 1159.2 Problems with cylindrical symmetries . . . . . . . . . . . . . . . . . . . . . . 1169.3 Strum Liouville problem in cylindrical coordinates . . . . . . . . . . . 1199.4 Bessel Equation and Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1209.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

A Some Complex Variables Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

B Assigned Homework Problems: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

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1

Preliminaries

1.1 The spectral theorem for symmetric matrices

Let A be a real N ×N matrix,

A =

a11 a12 . . . a1N

a21 a22 . . . a2N

.... . .

...aN1 aN2 . . . aNN

, (1.1)

and

f =

f1

f2

...fN

∈ RN

be a given vector. As usual we will let ei denote the vector in RN with allentries being zero except for the ith which is taken to be one.

We will write

(u, v) := u · v =N∑

i=1

uivi = utrv and

|u|2 = (u, u) =N∑

i=1

u2i = utru.

Recall that viNi=1 ⊂ RN is said to be an orthonormal basis if

(vi, vj) = δij :=

1 if i = j0 if i 6= j

. (1.2)

The following proposition and its infinite dimensional analogue will be thebasis for much of this course.

2 1 Preliminaries

Proposition 1.1. If viNi=1 ⊂ RN satisfies Eq. (1.2) then viNi=1 is a basisfor RN and if u ∈ RN we have

u =N∑

i=1

(u, vi) vi. (1.3)

Proof. Suppose that u =∑N

i=1 aivi for some ai ∈ R. Then

(u, vj) =

(N∑

i=1

aivi, vj

)=

N∑i=1

ai (vi, vj) =N∑

i=1

aiδij = aj .

In particular if u = 0 we learn that aj = (u, vj) = 0 and we have shown thatviNi=1 is a linearly independent set. Since dim

(RN)

= N, it now followsthat viNi=1 is a basis for RN and hence every u ∈ RN may be written in theform u =

∑Ni=1 aivi. By what we have just proved, we must have ai = (u, vi) ,

i.e. Eq. (1.3) is valid.

Definition 1.2. A matrix A as in Eq. (1.1) is symmetric A = Atr, i.e. ifaij = aji for all i, j.

The following characterization of a symmetric matrix will be more usefulfor our purposes.

Lemma 1.3. If A is a real N ×N matrix then, for all u, v ∈ RN ,

(Au, v) =(u,Atrv

). (1.4)

Moreover A is symmetric iff

(Au, v) = (u,Av) for all u, v ∈ RN . (1.5)

Proof. Eq. (1.4) is a consequence of the following matrix manipulations

(Au, v) = (Au)tr v = utrAtrv =(u,Atrv

)which are based on the fact that (AB)tr = BtrAtr. Hence if A is symmetric,then Eq. (1.5) holds. Conversely, if Eq. (1.5) holds, by taking u = ei andv = ej in Eq. (1.5) we learn that

aji =

a1i

...aN,i

, ej

= (Aei, ej) = (ei, Aej) =

ei,

a1j

...aN,j

= aij .

Corollary 1.4. Suppose that A = Atr and v, w ∈ RN are eigenvectors of Awith eigenvalues λ and µ respectively. If µ 6= λ then v and w are orthogonal,i.e. (v, w) = 0.


1.1 The spectral theorem for symmetric matrices 3

Proof. If Av = λv and Aw = µw with λ 6= µ then

λ (v, w) = (λv,w) = (Av,w) = (v,Aw) = (v, µw) = µ (v, w)

or equivalently, (λ− µ) (v, w) = 0. Sine λ 6= µ, we must conclude that (v, w) =0.

The following important theorem from linear algebra gives us a methodfor guaranteeing that a matrix is diagonalizable. Again much of this course isbased on an infinite dimensional generalization of this theorem.

Theorem 1.5 (Spectral Theorem). If A in Eq. (1.1) is a symmetric ma-trix, then A has an orthonormal basis of eigenvectors, v1, . . . , vN and thecorresponding eigenvalues, λ1, λ2, . . . , λN are all real.

Example 1.6. Suppose that

A :=[

12 −

32

− 32

12

], (1.6)

then

p (λ) = det (A− λI) = det[

12 − λ −

32

− 32

12 − λ

]=(

12− λ)2

− 94

which we set equal to zero to learn(12− λ)2

=94

or equivalently,(λ− 1

2

)= ± 3

2 and hence A has eigenvalues,

λ1 = −1 and λ2 = 2.

Since

A+ I :=[

32 −

32

− 32

32

]∼=[

1 −10 0

]and

A− 2I =[− 3

2 −32

− 32 −

32

]∼=[

1 10 0

]we learn that

v1 =[

11

]←→ λ1 = −1

v2 =[

1−1

]←→ λ2 = 2.


4 1 Preliminaries

Notice that (v1, v2) = 0 as is guaranteed by Corollary 1.4. The normalizedeigenvectors are given by 2−1/2v1 and 2−

12 v2. Consequently if f ∈ R2, we

have

f =(2−1/2v1, f

)2−1/2v1 +

(2−1/2v2, f

)2−1/2v2

=12

(v1, f) v1 +12

(v2, f) v2. (1.7)

Remark 1.7. As above, it often happens that naturally we find a orthogonalbut not orthonormal basis viNi=1 for RN , i.e. (vi, vj) = 0 if i 6= j but(vi, vi) 6= 1. We can still easily expand in terms of these vectors. Indeed,|vi|−1

vi

N

i=1is an orthonormal basis for RN and therefore if f ∈ RN we

have

f =N∑

i=1

(f, |vi|−1

vi

)|vi|−1

vi =N∑

i=1

(f, vi)|vi|2

vi.

Example 1.8. Working as above, one shows the symmetric matrix,

A :=

1 7 −27 1 −2−2 −2 10

, (1.8)

has characteristic polynomial given by

p (λ) = det (A− λI) = −(λ3 − 12λ2 − 36λ+ 432

)= − (λ− 6) (λ− 12) (λ+ 6) .

Thus the eigenvalues of A are given by λ1 = −6, λ2 = 6 and λ3 = 12 and thecorresponding eigenvectors are

v1 :=

−110

↔ −6, v2 :=

111

↔ 6, v3 :=

−1−12

↔ 12.

Again notice that v1, v2, v3 is an orthogonal set as is guaranteed by Corollary1.4. Relative to this basis we have the expansion

f = (f, v1)v1

|v1|2+ (f, v2)

v2

|v2|2+ (f, v3)

v3

|v3|2

=12

(f, v1) v1 +13

(f, v2) v2 +16

(f, v2) v3.

For example if f = (1, 2, 3)tr , then

f =12v1 + 2v2 +

12v3. (1.9)


1.2 Cylindrical and Spherical Coordinates 5

Exercise 1.1. Verify that the vectors vi3i=1 are eigenvectors of A in Eq.(1.8) which have the stated eigenvalues. Hint: you are only asked to verifynot solve from scratch.

Exercise 1.2. Find eigenvectors vi3i=1 and corresponding eigenvalues λi3i=1

for the symmetric matrix,

A :=

−2 1 11 −2 11 1 −2

.Make sure you choose them to be orthogonal. Also express the followingvectors,

f = (1, 0, 2)tr and g = (0, 1, 2)tr and h = (−1, 1, 0)tr ,

as linear combinations of the vi3i=1 that you have found.

Exercise 1.3. Suppose that A is a N × N symmetric matrix and viNi=1 isa basis of eigenvectors of A with corresponding eigenvalues λiNi=1 . Supposef ∈ RN has been decomposed as

f =N∑

i=1

aivi.

Show:

1. Anf =∑N

i=1 aiλni vi.

2. More generally, suppose that

p (λ) = a0 + a1λ+ a2λ2 + · · ·+ anλ

n

is a polynomial in λ, then

p (A) f =N∑

i=1

aip (λi) vi

and in particular p (A) v = p (λ) v is Av = λv.

1.2 Cylindrical and Spherical Coordinates

Our goal in this section is to work out the Laplacian in cylindrical and spher-ical coordinates. We will need these results later in the course. Our methodis to make use of the following two observations:


6 1 Preliminaries

1. If ui3i=1 is any orthonormal basis for R3 then

∇f · ∇g =3∑

i=1

(∇f, ui) (∇g, ui) =3∑

i=1

∂uif∂ui

g

and2. if g has compact support in a region Ω, then by integration by parts∫

Ω

∆fgdV = −∫

Ω

∇f · ∇gdV. (1.10)

The following theorem is a far reaching generalization of Eq. (1.10).

Theorem 1.9 (Divergence Theorem). Let Ω ⊂ Rn be an open boundedregion with smooth boundary, n : ∂Ω → Rn be the unit outward pointingnormal to Ω. If Z ∈ C1(Ω,Rn), then∫

∂Ω

Z(x) · n(x)dσ(x) =∫Ω

∇ · Z(x) dx. (1.11)

Corollary 1.10 (Integration by parts). Let Ω ⊂ Rn be an open boundedregion with smooth boundary, n : ∂Ω → Rn be the unit outward pointingnormal to Ω. If Z ∈ C1(Ω,Rn) and f ∈ f ∈ C1(Ω,R), then∫

Ω

f(x)∇·Z(x) dx = −∫Ω

∇f(x) ·Z(x) dx+∫

∂Ω

f (x)Z(x) ·n(x)dσ(x). (1.12)

Also if g ∈ C2(Ω,R), then∫Ω

f(x)∆g(x) dx = −∫Ω

∇f(x)·∇g(x) dx+∫

∂Ω

f (x)∇g(x)·n(x)dσ(x). (1.13)

Proof. Eq. (1.12) follows by applying Theorem 1.9 with Z replaced by fZmaking use of the fact that

∇ · (fZ) = ∇f · Z + f∇ · Z.

Eq. (1.13) follows from Eq. (1.12) by taking Z = ∇g.

1.2.1 Cylindrical coordinates

Recall that cylindrical coordinates, see Figures 1.1, are determined by

(x, y, z) = R(ρ, θ, z) ≡ (ρ cos θ, ρ sin θ, z).

In these coordinates we have

dV = ρdρdθdz.



Fig. 1.1. Cylindrical and polar coordinates.

Proposition 1.11 (Laplacian in Cylindrical Coordinates). The Lapla-cian in cylindrical coordinates is given by

∆f =1ρ∂ρ (ρ∂ρf) +

1ρ2∂2

θf + ∂2zf. (1.14)

Proof. We further observe that

Rρ(ρ, θ, z) = (cos θ, sin θ, 0)Rθ(ρ, θ, z) = (−ρ sin θ, ρ cos θ, 0)Rz(ρ, θ, z) = (0, 0, 1)

so that Rρ(ρ, θ, z), ρ−1Rθ(ρ, θ, z),Rz(ρ, θ, z)

is an orthonormal basis for R3. Therefore,

(∇f,∇g) = (∇f,Rρ) (∇g,Rρ) +(∇f, ρ−1Rθ

) (∇g, ρ−1Rθ

)+ (∇f,Rz) (∇g,Rz)

=∂f

∂ρ

∂g

∂ρ+

1ρ2

∂f

∂θ

∂g

∂θ+∂f

∂z

∂g

∂z.

If g has compact support in a region Ω, then by integration by parts,


8 1 Preliminaries∫Ω

∆fgdV = −∫

Ω

∇f · ∇gdV

= −∫

Ω

[∂f

∂ρ

∂g

∂ρ+

1ρ2

∂f

∂θ

∂g

∂θ+∂f

∂z

∂g

∂z

]· ρdρdθdz

=∫

Ω

[∂ρ (ρ∂ρf) +

1ρ∂2

θf + ρ∂2zf

]g · dρdθdz

=∫

Ω

[1ρ∂ρ (ρ∂ρf) +

1ρ2∂2

θf + ∂2zf

]gρdρdθdz

=∫

Ω

[1ρ∂ρ (ρ∂ρf) +

1ρ2∂2

θf + ∂2zf

]gdV.

Since this formula holds for arbitrary g with small support, we conclude that

∆f =1ρ∂ρ (ρ∂ρf) +

1ρ2∂2

θf + ∂2zf.

1.2.2 Spherical coordinates

We will now work out the Laplacian in spherical coordinates by a similarmethod. Recall that spherical coordinates, see Figures 1.2, are determined by

Fig. 1.2. Defining spherical coordinates of a point in R3.

(x, y, z) = R(r, θ, ϕ) ≡ (r sinϕ cos θ, r sinϕ sin θ, r cosϕ).

In this coordinate systems we have

dV = r2 sinϕdrdθdϕ.

See Figure 1.3.



Fig. 1.3. A picture proof that dxdydz = r2 sin φdrdθdφ, where r2 sin φdrdθdφ shouldbe viewed as (r sin φdθ)(rdφ)dr.

Proposition 1.12 (Laplacian in spherical coordinates). The Laplacianin spherical coordinates is given by

∆f =1r2∂r(r2∂rf) +

1r2 sinϕ

∂ϕ(sinϕ∂ϕf) +1

r2 sin2 ϕ∂2

θf. (1.15)

A simple computation shows (as will be needed later) that

1r2∂r(r2∂rf) =

1r∂2

r (rf). (1.16)

Proof. Since

Rr(r, θ, ϕ) = (sinϕ cos θ, sinϕ sin θ, cosϕ)Rθ(r, θ, ϕ) = (−r sinϕ sin θ, r sinϕ cos θ, 0)Rϕ(r, θ, ϕ) = (r cosϕ cos θ, r cosϕ sin θ,−r sinϕ)

it is easily verified thatRr(r, θ, ϕ),

1r sinϕ

Rθ(ρ, θ, z),1rRϕ(ρ, θ, z)

is an orthonormal basis for R3. Therefore

(∇f,∇g) = (∇f,Rr) (∇g,Rr) +1

r2 sin2 ϕ(∇f,Rθ) (∇g,Rθ)

+1r2

(∇f,Rϕ) (∇g,Rϕ)

=∂f

∂r

∂g

∂r+

1r2 sin2 ϕ

∂f

∂θ

∂g

∂θ+

1r2∂f

∂ϕ

∂g

∂ϕ.

If g has compact support in a region Ω, then


10 1 Preliminaries∫Ω

∆fgdV = −∫

Ω

∇f · ∇gdV

= −∫

Ω

[∂f

∂r

∂g

∂r+

1r2 sin2 ϕ

∂f

∂θ

∂g

∂θ+

1r2∂f

∂ϕ

∂g

∂ϕ

]· r2 sinϕdrdϕdθ

= −∫

Ω

[r2 sinϕ

∂f

∂r

∂g

∂r+

1sinϕ

∂f

∂θ

∂g

∂θ+ sinϕ

∂f

∂ϕ

∂g

∂ϕ

]· drdϕdθ

=∫

Ω

[∂r

(r2∂rg

)sinϕ+

1sinϕ

∂2θf + ∂ϕ (sinϕ · ∂ϕf)

]g · drdϕdθ

=∫

Ω

[1r2∂r

(r2∂rg

)+

1r2 sin2 ϕ

∂2θf +

1r2 sinϕ

∂ϕ (sinϕ∂ϕf)]g · r2 sinϕdrdϕdθ

=∫

Ω

[1r2∂r

(r2∂rg

)+

1r2 sin2 ϕ

∂2θf +

1r2 sinϕ

∂ϕ (sinϕ∂ϕf)]g · dV.

Since this formula holds for arbitrary g we conclude that

∆f =1r2∂r(r2∂rf) +

1r2 sinϕ

∂ϕ(sinϕ∂ϕf) +1

r2 sin2 ϕ∂2

θf.

1.2.3 Exercises

In the following two exercises, I am using the conventions in the Lecture notesand not the book.

Exercise 1.4. Compute ∆f where f is given in cylindrical coordinates as:

f = ρ3 cos θ + zρ

Exercise 1.5. Compute ∆f where f is given in spherical coordinates as:

f = r−1 + cos θ sinϕ.


2

PDE Examples

2.1 The Wave Equation

Example 2.1 (Wave Equation for a String). Suppose that we have a stretchedstring supported at x = 0 and x = L and y = 0. Suppose that the string onlyundergoes vertical motion (pretty bad assumption). Let u(t, x) and T (t, x)denote the height and tension respectively of the string at (t, x), δ(x) denotethe density in equilibrium and T0 be the equilibrium string tension. Let J =

Fig. 2.1. A piece of displace string

[x, x+∆x] ⊂ [0, L], then

PJ(t) :=∫

J

ut(t, x)δ(x)dx

is the momentum of the piece of string above J. (Notice that δ(x)dx is theweight of the string above x.) Newton’s equations state

dPJ(t)dt

=∫

J

utt(t, x)δ(x)dx = Force on String.

Since the string is to only undergo vertical motion we require

12 2 PDE Examples

T (t, x+∆x) cos(αx+∆x)− T (t, x) cos(αx) = 0

for all ∆x and therefore that T (t, x) cos(αx) =: H for some constant H, i.e.the horizontal component of the tension is constant. Looking at Figure 2.2,the tension on the piece of string above J = [a, b] at the right endpoint b must

Fig. 2.2. Computing the net vertical force due to tension on the part of the stringabove [a, b].

be given by H (1, ux (t, b)) while the tension at the left endpoint, a, must begiven by H (−1,−ux (t, a)) . So the net tension force on the string above J is

H [ux(t, b)− ux(t, a)] = H

∫ b

a

uxx (t, x) dx.

Finally there may be a component due to gravity and air resistance, say

gravity = −g∫ b

a

δ(x)dx and

air resistance = −∫ b

a

k(x)ut(t, x)dx.

So Newton’s equations become∫ b

a

utt(t, x)δ(x)dx =∫ b

a

[Huxx (t, x)− gδ (x)− k(x)ut(t, x)] dx.

Differentiating this equation in b at b = x then shows

utt(t, x)δ(x) = Huxx(t, x)− gδ(x)− k(x)ut(t, x)

or equivalently that

utt(t, x) =H

δ(x)uxx(t, x)− g − k(x)

δ(x)ut(t, x). (2.1)


2.1 The Wave Equation 13

Example 2.2 (Wave equation. for a drum head). Suppose that u (t, x) repre-sents the height at time t of a drum head over a point x ∈ Ω — Ω being thebase of the drum head, see Figure 2.3. As for the string we will make the sim-

Fig. 2.3. A deformed membrane attached to a “wire” base. We are also computethe tension density on a region of the membrane above a region V in the plane.

plifying assumption that the membrane only moves vertically or equivalentlythat the horizontal component of tension/unit-length is a constant value, H.

Let V ⊂ Ω be a test region and consider the membrane which lie above Vas in Figure 2.3. Then

PV (t) :=∫

V

ut(t, x)δ(x)dx

is the momentum of the piece of string above V where δ(x)dx is the weightof the membrane above x. Newton’s equations state

dPV (t)dt

=∫

V

utt(t, x)δ(x)dx = Force on membrane.

To find the vertical force on the membrane above V, let x ∈ ∂V, then

(n (x) ,∇u (t, x) · n (x)) =d

ds|0(x+ sn (x) , u(t, x+ sn (x))

is a vector orthogonal to the boundary of the region above V and by as-sumption the tension/unit-length at x is H (n (x) ,∇u (t, x) · n (x)) . Thus thevertical component of the force on the membrane above V is given by

H

∫∂V

∇u (t, x) · n (x) d` (x) = H

∫V

∇ · ∇u (t, x) dx = H

∫V

∆u (t, x) dx.


14 2 PDE Examples

Finally there may be a component due to gravity and air resistance, say

gravity = −∫

V

gδ(x)dx and

air resistance = −∫

V

k(x)ut(t, x)dx.

So Newton’s equations become∫V

utt(t, x)δ(x)dx =∫

V

[H∆u (t, x)− gδ(x)− k(x)ut(t, x)] dx.

Since V is arbitrary, this implies

δ(x)utt(t, x) = H∆u (t, x)− gδ(x)− k(x)ut(t, x)


utt(t, x) =H

δ (x)∆u (t, x)− g − k(x)

δ(x)ut(t, x). (2.2)

Example 2.3 (Wave equation for a metal bar). Suppose that have a metal wirewhich we is going to be deformed and then released. We would like to find theequation that the displacement u (t, x) of the section of the bar originally atlocation x must solve, see Figure 2.4 below.

Fig. 2.4. The picture represents an elastic bar in its un-deformed state and then in adeformed state. The quantity y (t, x) represents the displacement of the section thatwas originally at location x in the un-deformed bar. In the above figure y (t, x) < 0which y (t, x + ∆x) > 0.



To do this will write down Newton’s equation of motion. First off, thelongitudinal force on the left face of the section which was originally betweenx and x+∆x is approximately,

−AEu (t, x+∆x)− u(t, x)∆x

,

where E is Young’s modulus of elasticity and A is the area of the bar. (Theminus represents the fact that we must pull to the left to get the currentconfiguration in the figure.) Letting ∆x→ 0, we find the force of the sectionthat was originally at x is given by −AEux (t, x) . Now suppose that ∆x isnot necessarily small. Then we have the momentum of the region of the baroriginally between x and x+∆x is given by∫ x+∆x

x

ut (t, x) δ (x) dx

where δ (x) is the linear mass density. Therefore,

Mass× acceleration =d

dt

∫ x+∆x

x

ut (t, x) δ (x) dx =∫ x+∆x

x

utt (t, x) δ (x) dx

= the net force on this section of the bar= −AEux (t, x) +AEux (t, x+∆x)

where −AEux (t, x) is the force on left end and AEux (t, x+∆x) is the forceon the right end. Hence we have∫ x+∆x

x

utt (t, x) δ (x) dx = AEux (t, x+∆x)−AEux (t, x)

which upon differentiating in ∆x at ∆x = 0 shows

utt (t, x) δ (x) = AEuxx (t, x) .

2.1.1 d’Alembert’s solution to the 1-dimensional wave equation

Here we are going to try to find solutions to the wave equation, ytt = a2yxx.Since this equation may be written as(

∂2t − a2∂2

x

)y = 0 (2.3)

and (∂2

t − a2∂2x

)= (∂t − a∂x) (∂t + a∂x)

we are lead to consider the wave equation in the new variables,

u = x+ at and v = x− at.


16 2 PDE Examples

In these variables we have

∂t =∂u

∂t∂u +

∂v

∂t∂v = a∂u − a∂v and

∂x =∂u

∂x∂u +

∂v

∂x∂v = ∂u + ∂v

from which it follows that

∂t − a∂x = −2a∂v and ∂t + a∂x = 2a∂u

and hence the wave equation in (u, v) – coordinates becomes,

0 = (∂t − a∂x) (∂t + a∂x) y = −2a∂v2a∂uy = −4a2yuv,

i.e. yuv = 0. Integrating this equation in v shows yu = F (u) and then inte-grating in u shows

y =∫F (u) du+ ψ (v) = ϕ (u) + ψ (v) .

Thus we have shown if y solves the wave equation then

y (t, x) = ϕ (x+ at) + ψ (x− at) (2.4)

for some functions ϕ and ψ.

Exercise 2.1. Show that if y (t, x) has the form given in Eq. (2.4) with ϕ andψ being twice continuously differentiable functions, then y solves the wave Eq.(2.3).

To get a unique solution to Eq. (2.3) we must introduce some initial con-ditions. For example, let us further assume that

y (0, x) = f (x) and yt (0, x) = 0.

This then implies that

f (x) = ϕ (x) + ψ (x) and0 = aϕ′ (x)− aψ′ (x) ,

The latter equation shows that ψ (x) = ϕ (x) + C and using this in the firstequation implies that

f (x) = 2ϕ (x) + C

orϕ (x) =

12

(f (x)− C) and ψ (x) =12

(f (x) + C) .

Thus we have found the solution to be given by

y (t, x) =12f (x+ at) + f (x− at) .

In the homework you are asked to generalize this result to prove the followingtheorem.



Theorem 2.4 (d’Alembert’s solution). If f (x) is twice continuously dif-ferentiable and g (x) is continuously differentiable for x ∈ R, then the uniquesolution to

ytt = a2yxx with (2.5)y (0, x) = f (x) and yt (0, x) = g (x) (2.6)

is given by

y(t, x) =12

[f(x+ at) + f(x− at)] +12a

∫ x+at

x−at

g(s)ds. (2.7)

Example 2.5. Here we wish to solve for x ≥ 0 and t ≥ 0,

∂2t y = ∂2

xy with y (0, x) = f (x) and y (0, x) = 0 with y (t, 0) = 0.

As before we know that y (t, x) = ϕ (x+ t) + ψ (x− t) . We must now imple-ment all of the boundary conditions,

f (x) = y (0, x) = ϕ (x) + ψ (x)0 = y (0, x) = ϕ′ (x)− ψ′ (x) and0 = y (t, 0) = ϕ (t) + ψ (−t) .

This suggests that we define ψ (−t) := −ϕ (t) for t > 0, and also that

ϕ (x) = ψ (x) + C

f (x) = 2ψ (x) + C

or

ψ (x) =12

(f (x)− C)

ϕ (x) =12

(f (x) + C) .

Thus our answer is given by

y (t, x) =12

[f (x+ t) + f (x− t)]

where by above,

12

(f (−x)− C) = ψ (−x) = −ϕ (x) = −12

(f (x) + C)

and thusf (−x) := −f (x) .

Thus we havey (t, x) =

12

[f (x+ t) + f (x− t)]

where f is extend to all of R to be an odd function.


18 2 PDE Examples

2.2 Heat Equations

Example 2.6 (Heat or Diffusion Equation in 1-dimension). Let us consider thetemperature in a rod Ω. We will let

1. δ (x) denote the linear density of the rod2. c (x) denote the heat capacity of the rod per unit mass at x3. κ (x) be the thermal conductivity of the rod at x. By Newton’s Law of

cooling, the heat flow from left to right in the rod at location x should beapproximately equal to

κ (x)∆

(u (x)− u (x+∆)) . (2.8)

Notice the ∆ appearing in the denominator represents the fact that thethicker the insulation in your house the less heat transfer that you have.Passing to the limit in Eq. (2.8) then gives Fourier’s law, namely the heatflow from left to right in the rod at location x is given by

− κ (x)u′ (x) . (2.9)

(In the book, it is typically assumed that δ (x) = δ, κ (x) = K andc (x) = σ are all constant.)

4. u (t, x) be the temperature of the rod at time t and location x.5. H (t, x) represent heat source at x and time t. For example we may be

passing a current through the wire and the resistance of the wire is bothspatially and time dependent. Alternatively we may be heating the wirewith an external source.

Let B = [a, b] be a sub-region of the rod, see Figure 2.5. Then

Fig. 2.5. Part of a rod with a test region B = [a, b] being examined.

E (t) =∫ b

a

u (t, x) δ (x) c (x) dx

represents the heat energy in B at time t. Hence


2.2 Heat Equations 19

E (t) =∫ b

a

ut (t, x) δ (x) c (x) dx

is the rate of change of heat energy in B. This may alternatively be computedas the rate at which heat enters the system which is given by

E (t) =∫ b

a

H (t, x) dx+ (κ (b)ux (t, b)− κ (a)ux (t, a))

=∫ b

a

[H (t, x) +

d

dx(κ (x)ux (t, x))

]dx.

Hence we conclude that∫ b

a

ut (t, x) δ (x) c (x) dx =∫ b

a

[H (t, x) +

d

dx(κ (x)ux (t, x))

]dx

for all sub-intervals Ω in the rod and therefore (again just differentiatein b) that

δ (x) c (x)ut (t, x) =d

dx(κ (x)ux (t, x)) +H (t, x) . (2.10)

This equation may be written as

ut (t, x) = Lu (t, x) + h (t, x)

where

Lf (x) :=1

p (x)d

dx

(κ (x)

d

dxf (x)

),

p (x) = δ (x) c (x) and h (t, x) :=H (t, x)p (x)

.

If we further assume that the rod in not perfectly insulated along itslength and the ambient temperature is not constant, we may end up withanother terms in computing E (t) of the form∫ b

a

Q (x) [u (t, x)− T (x)] dx

and we would then arrive at a heat equation of the form

ut (t, x) = Lu (t, x) + h (t, x)

where

Lf (x) :=1

p (x)d

dx

(κ (x)

d

dxf (x)

)+

1p (x)

q (x) f (x) (2.11)

for some function q (x) and a modified function h (t, x) .


20 2 PDE Examples

Fig. 2.6. A test volume B in Ω centered at x with outward pointing normal, n.

Example 2.7 (Heat or Diffusion Equation in d - dimensions). Suppose thatΩ ⊂ Rd is a region of space filled with a material, δ(x) is the density ofthe material at x ∈ Ω and c(x) is the heat capacity. Let u(t, x) denote thetemperature at time t ∈ [0,∞) at the spatial point x ∈ Ω. Now suppose thatB ⊂ Rd is a “little” volume in Rd, ∂B is the boundary of B, and EB(t) is theheat energy contained in the volume B at time t. Then

EB(t) =∫

B

δ(x)c(x)u(t, x)dx.

So on one hand,

EB(t) =∫

B

δ(x)c(x)u(t, x)dx (2.12)

while on the other hand,

EB(t) =∫

∂B

(κ(x)∇u(t, x) · n(x)) dσ(x), (2.13)

where κ(x) is a d × d–positive definite matrix representing the conductionproperties of the material, n(x) is the outward pointing normal to B at x ∈∂B, and dσ denotes surface measure on ∂B.

In order to see that we have the sign correct in (2.13), suppose that x ∈ ∂Band ∇u(x) ·n(x) > 0, then the temperature for points near x outside of B arehotter than those points near x inside of B and hence contribute to a increasein the heat energy inside of B. (If we get the wrong sign, then the resultingequation will have the property that heat flows from cold to hot!)

Comparing Eqs. (2.12) to (2.13) after an application of the divergencetheorem shows that∫

B

δ(x)c(x)u(t, x)dx =∫

B

∇ · (κ(·)∇u(t, ·))(x) dx. (2.14)

Since this holds for all volumes B ⊂ Ω, we conclude that the temperaturefunctions should satisfy the following partial differential equation.


2.3 Other Equations 21

δ(x)c(x)u(t, x) = ∇ · (κ(·)∇u(t, ·))(x) . (2.15)


u(t, x) =1

δ(x)c(x)∇ · (κ(x)∇u(t, x)). (2.16)

Setting gij(x) := κij(x)/(δ(x)c(x)) and

zj(x) :=d∑

i=1

∂(κij(x)/(δ(x)c(x)))/∂xi

the above equation may be written as:

u(t, x) = Lu(t, x), (2.17)

where

(Lf)(x) =∑i,j

gij(x)∂2

∂xi∂xjf(x) +

∑j

zj(x)∂

∂xjf(x). (2.18)

The operator L is a prototypical example of a second order “elliptic” differ-ential operator.

Example 2.8 (Laplace and Poisson Equations). Laplace’s Equation is of theform Lu = 0 and solutions may represent the steady state temperature distri-bution for the heat equation. Equations like ∆u = −ρ appear in electrostaticsfor example, where u is the electric potential and ρ is the charge distribution.

2.3 Other Equations

Example 2.9 (Shrodinger Equation and Quantum Mechanics).

i∂

∂tψ(t, x) = −∆

2ψ(t, x) + V (x)ψ(t, x) with ‖ψ(·, 0)‖2 = 1.

Interpretation,∫A

|ψ(t, x)|2 dt = the probability of finding the particle in A at time t.

(Notice similarities to the heat equation.)

Example 2.10 (Maxwell Equations in Free Space).

∂E∂t

= ∇×B

∂B∂t

= −∇×E

∇ ·E = ∇ ·B = 0.


22 2 PDE Examples

Notice that

∂2E∂t2

= ∇× ∂B∂t

= −∇× (∇×E) = ∆E−∇ (∇ ·E) = ∆E

and similarly, ∂2B∂t2 = ∆B so that all the components of the electromagnetic

fields satisfy the wave equation.

Example 2.11 (Traffic Equation). Consider cars travelling on a straight roadwith coordinate x ∈ R, let u(t, x) denote the density of cars on the road attime t and location x ∈ R, and v(t, x) be the velocity of the cars at (t, x).Then for J = [a, b] ⊂ R, NJ(t) :=

∫ b

au(t, x)dx is the number of cars in the set

J at time t. We must have∫ b

a

u(t, x)dx = NJ(t) = u(t, a)v(t, a)− u(t, b)v(t, b)

= −∫ b

a

∂

∂x[u(t, x)v(t, x)] dx.

Since this holds for all intervals [a, b], we must have

u(t, x) = − ∂

∂x[u(t, x)v(t, x)] .

To make life more interesting, we may imagine that v(t, x) = −F (u(t, x), ux(t, x)),in which case we get an equation of the form

∂

∂tu =

∂

∂xG(u, ux) where G(u, ux) = −u(t, x)F (u(t, x), ux(t, x)).

A simple model might be that there is a constant maximum speed, vm andmaximum density um, and the traffic interpolates linearly between 0 (whenu = um) to vm when (u = 0), i.e. v = vm(1− u/um) in which case we get

∂

∂tu = −vm

∂

∂x(u(1− u/um)) .

Example 2.12 (Burger’s Equation). Suppose we have a stream of particlestravelling on R, each of which has its own constant velocity and let u(t, x)denote the velocity of the particle at x at time t. Let x(t) denote the trajec-tory of the particle which is at x0 at time t0. We have C = x(t) = u(t, x(t)).Differentiating this equation in t at t = t0 implies

0 = [ut(t, x(t)) + ux(t, x(t))x(t)] |t=t0 = ut(t0, x0) + ux(t0, x0)u(t0, x0)

which leads to Burger’s equation

0 = ut + u ux.


2.4 Worked Examples 23

Example 2.13 (Minimal surface Equation). Let D ⊂ R2 be a bounded regionwith reasonable boundary, u0 : ∂D → R be a given function. We wish to findthe function u : D → R such that u = u0 on ∂D and the graph of u, Γ (u) hasleast area. Recall that the area of Γ (u) is given by

A(u) = Area(Γ (u)) =∫

D

√1 + |∇u|2dx.

Assuming u is a minimizer, let v ∈ C1(D) such that v = 0 on ∂D, then

0 =d

ds|0A(u+ sv) =

d

ds|0∫

D

√1 + |∇(u+ sv)|2dx

=∫

D

d

ds|0√

1 + |∇(u+ sv)|2dx

=∫

D

1√1 + |∇u|2

∇u · ∇v dx

= −∫

D

∇ ·

1√1 + |∇u|2

∇u

v dx

from which it follows that

∇ ·

1√1 + |∇u|2

∇u

= 0.

Example 2.14 (Navier – Stokes). Here u(t, x) denotes the velocity of a fluidad (t, x), p(t, x) is the pressure. The Navier – Stokes equations state,

∂u

∂t+ ∂uu = ν∆u−∇p+ f with u(0, x) = u0(x) (2.19)

∇ · u = 0 (incompressibility) (2.20)

where f are the components of a given external force and u0 is a given di-vergence free vector field, ν is the viscosity constant. The Euler equationsare found by taking ν = 0. Equation (2.19) is Newton’s law of motion again,F = ma. See http://www.claymath.org for more information on this Mil-lion Dollar problem.

2.4 Worked Examples

Example 2.15 (6.3 in book). Slab / wire along [0, c] with surface conductanceat the end being H and K being the thermal conductivity along the rod.Assume outside temperature at 0 is 0 and at c is T, Find the steady statetemperature of the rod. The relevant heat equation is:


24 2 PDE Examples

σδut = Kuxx withKu′ (0) = H (u (0)− outside temp) = H (u (0)− 0) andKu′ (c) = −H (u (c)− outside temp) = H (T − u (c)) .

Notice the signs used here. Recall that Ku′ (x) represents the heat movement(or Flux) from right to left in the bar at location x, therefore,

Ku′ (0) = rate of heat leaving bar at 0 whileH (u (0)− outside temp) = rate of heat leaving bar at 0 do surface conductance

Similarly,

Ku′ (c) = rate of heat entering bar at c whileH (u (c)− T ) = rate of heat leaving bar at 0 do surface conductance

and henceKu′ (c) = −H (u (c)− T ) = H (T − u (c)) .

So in steady state we have ut = 0 and so

u′′ (x) = 0 of u (x) = mx+ b.

Now

Km = Ku′ (0) = Hu (0) = Hb andKm = Ku′ (c) = H (T − u (c)) = H (T −mc− b)

and so if we let h := H/K, then

m = hb andm = h (T −mc− b) = hT − hmc− hb = hT − hmc−m

from which it follows that (2 + hc)m = hT or

m =hT

2 + hcand b =

m

h=

T

2 + hc.

Thus the solution is given by

u (x) =T

2 + hc[hx+ 1] .

Example 2.16 (10.1b in book.). Solve uxy = 2x for x, y > 0 and u (0, y) = 0and u (x, 0) = x2. Ans.,

ux = 2xy + C (x)

and so


2.4 Worked Examples 25

u (x, y) = x2y +K (x) + F (y)

Sinceu (x, 0) = x2 = K (x) + F (0)

and0 = u (0, y) = K (0) + F (y)

we havex2 = K (x)−K (0) and

u (x, y) = x2y +K (x)−K (0)

= x2y + x2 = x2 (y + 1) .


3

Linear ODE

3.1 First order linear ODE

We would like to solve the ordinary differential equation

u (t) = Au (t) with (3.1)u (0) = f. (3.2)

The method of separation of variables or eigenvector expansionsproposed to begin by looking for solutions of the form u (t) = T (t) v to Eq.(3.1). Here v is a fixed vector in ∈ RNand T (t) is some unknown function oft. Substituting u (t) = T (t) v into Eq. (3.1) gives

T (t) v = T (t)Av


Av =T (t)T (t)

v.

Since the left side of this equation is independent of t we must have

T (t)T (t)

= λ (3.3)

for some λ ∈ R. The solution to Eq. (3.3) is of course T (t) = etλT (0) andtherefore we have shown the following lemma.

Lemma 3.1. If u (t) = T (t) v solves Eq. (3.1), then v is an eigenvector ofA and if λ is the corresponding eigenvalue (i.e. Av = λv) then

u (t) = eλtT (0) v.

Conversely if Av = λv then u (t) = eλtv solves Eq. (3.1).

28 3 Linear ODE

Proposition 3.2 (Principle of superposition). If u (t) and v (t) solves Eq.(3.1) then so does u (t) + cv (t) for any c ∈ R.

Proof. This is a simple consequence of the fact that matrix multiplicationand differentiation are linear operations. In detail,

d

dt(u (t) + cv (t)) = u (t) + cv (t) = Au (t) + cAv (t)

= A (u (t) + cv (t)) .

Consequently if Avi = λivi for i = 1, 2, . . . , k, then

u (t) =∑

i

etλivi.

solves Eq. (3.1).

Theorem 3.3. Suppose the matrix A is diagonalizable, i.e. there exists a basisviNi=1 for RN consisting of eigenvectors A. Then to any f ∈ RN there is aunique solution, u (t) , to Eqs. (3.1) and (3.2). Moreover, if we expand f interms of the basis viNi=1 as

f =N∑

i=1

aivi,

then the unique solution to Eqs. (3.1) and (3.2) is given by

u (t) =N∑

i=1

aietλivi. (3.4)

Proof. The fact that Eq. (3.4) solves Eqs. (3.1) and (3.2) follows from theprinciple of superposition and the fact that etλi = 1 when t = 0.

Conversely, suppose that u solves Eqs. (3.1) and (3.2), then

u (t) =N∑

i=1

ai (t) vi (3.5)

for some functions ai (t) with ai (0) = ai. Now on one hand

u (t) =N∑

i=1

ai (t) vi

while on the other hand

u (t) = Au (t) = AN∑

i=1

ai (t) vi =N∑

i=1

ai (t)Avi =N∑

i=1

ai (t)λivi.


3.1 First order linear ODE 29

Subtracting these two equation shows

0 =N∑

i=1

ai (t) vi −N∑

i=1

ai (t)λivi =N∑

i=1

(ai (t)− ai (t)λi) vi.

Since viNi=1 is a basis for RN it follows, for all i, that

ai (t) = ai (t)λi with ai (0) = ai

and therefore, ai (t) = etλiai. Putting this result back into Eq. (3.5) gives Eq.(3.4).

Definition 3.4. If f ∈ RN we will write etAf for the solution, u (t) , to Eqs.(3.1) and (3.2).

Fact 3.5 Eqs. (3.1) and (3.2) have a unique solution independent as towhether A has a basis of eigenvectors or not. Moreover we may compute etA

using the matrix power series expansion,

etAf =∞∑

n=0

tn

n!Anf. (3.6)

Notice that formula in Eq. (3.6) is consistent with our previous results.For example if v ∈ RN and Av = λv, then Anv = λnv and therefore,

∞∑n=0

tn

n!Anv =

∞∑n=0

tn

n!λnv = etλv = etAv.

More generally, if u =∑k

i=1 aivi with Avi = λivi, then

etAu =k∑

i=1

aietλivi =

k∑i=1

aietAivi.

Remark 3.6. As the notation suggests, it is true that

etAesA = e(t+s)A

as you are asked to prove in Exercise 3.1 below. However, it is not generallytrue that

e(A+B) = eAeB = eBeA,

see Proposition 3.9 below.

Example 3.7. Let us find etA when

A =

1 1 00 2 20 0 3

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30 3 Linear ODE

The eigenvalues of A are given as the roots of the characteristic polyno-mial,

p (λ) = det (A− λI) = (1− λ) (2− λ) (3− λ) .

These roots are λ = 1, λ = 2, and λ = 3. As usual we find the correspond-ing eigenvectors as solutions to the equation (A− λI)u = 0. The result is,eigenvectors:

v1 :=

100

↔ 1, v2 :=

110

↔ 2, v3 :=

121

↔ 3.

Since 010

=

110

−1

00

and 0

01

=

121

− 2

110

+

100

it follows that the columns of etA are given by

etA

100

= et

100

etA

010

= etA

110

− etA

100

= e2t

110

− et

100

and

etA

001

= etA

121

− 2etA

110

+ etA

100

= e3t

121

− 2e2t

110

+ et

100

and therefore,

etA =

et −et + e2t et − 2e2t + e3t

0 e2t −2e2t + 2e3t

0 0 e3t

.Example 3.8. Let

A :=

1 7 −27 1 −2−2 −2 10

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3.1 First order linear ODE 31

Solve the equation

u (t) = Au (t) with

u (0) = v1 + 2v2 + 3v3 =

−110

+ 2

111

+ 3

−1−12

i.e.

u (0) = f =

−208

.Answer. Recall from Example 1.8 that A has eigenvectors

v1 :=

−110

↔ −6, v2 :=

111

↔ 6, v3 :=

−1−12

↔ 12.

Therefore,

(f, v1)|v1|2

= 1,(f, v2)|v2|2

=63

= 2 and(f, v3)|v3|2

=186

= 3

so thatf = v1 + 2v2 + 3v3.

Therefore,u (t) = e−6tv1 + 2e6tv2 + 3e12tv3

solves the ODE.

Proposition 3.9. Let A and B be two N × N matrices. Then the followingare equivalent:

1. 0 = [A,B] := AB −BA.2. etAB = BetA for all t ∈ R,3. etAesB = esBetA for all s, t ∈ R.

Moreover if [A,B] = 0 then e(A+B) = eAeB and in particular

etAesA = e(t+s)A for all s, t ∈ R. (3.7)

Proof. If [A,B] = 0, then

d

dtetABe−tA = etA [A,B] e−tA = 0

and therefore, etAB = BetA for all t ∈ R. It now follows that


32 3 Linear ODE

d

dsetAesB = etABesB = BetAesB

d

dsesBetA = BesBetA

and so by uniqueness of solutions to these ODE we conclude etAesB = esBetA

for all s, t ∈ R. If etAesB = esBetA for all s, t ∈ R then

AB =d

dt|0etAB =

d

dt|0d

ds|0etAesB =

d

dt|0d

ds|0esBetA =

d

dt|0BetA = BA.

For the last assertion, let T (t) := etAetB , then

d

dtT (t) = AetAetB + etABetB = AetAetB +BetAetB

= (A+B)T (t) with T (0) = I.

So again by uniqueness of solutions,

etAetB = T (t) = et(A+B).

3.2 Solving for etA using the Spectral Theorem

Example 3.10. Let

A :=[

12 −

32

− 32

12

]as in Example 1.6 with eigenvectors/eigenvalues given by

v1 =[

11

]←→ λ1 = −1

and

v2 =[

1−1

]←→ λ2 = 2.

Recall thatf =

12

(v1, f) v1 +12

(v2, f) v2

and hence

etAf =12

(v1, f) etAv1 +12

(v2, f) etAv2

=12

(v1, f) e−tv1 +12

(v2, f) e2tv2. (3.8)

Taking f = e1 and f = e2 then implies


3.2 Solving for etA using the Spectral Theorem 33

etAe1 =12etAv1 +

12etAv2 =

12e−tv1 +

12e2tv2

=12e−t

[11

]+

12e2t

[1−1

]=

12

[e−t + e2t

e−t − e2t

].

and

etAe2 =12etAv1 −

12etAv2 =

12e−tv1 −

12e2tv2

=12e−t

[11

]− 1

2e2t

[1−1

]=

12

[e−t − e2t

e−t + e2t

].

Thus we may conclude that

etA =[etAe1 e

tAe2]

=12

[e−t + e2t e−t − e2t

e−t − e2t e−t + e2t

].

Alternatively, from Eq. (3.8)

etAf =12

(v1, f) e−tv1vtr1 f +

12e2tv2v

tr2 f

and therefore

etA =12

(v1, f) e−tv1vtr1 +

12e2tv2v

tr2

=12e−t

[11

] [1 1]+

12e2t

[1−1

] [1 −1

]=

12

[e−t + e2t e−t − e2t

e−t − e2t e−t + e2t

].

Example 3.11. The matrix

A =[−2 1

1 −2

]has eigenvectors/eigenvalues given by

v1 :=[

11

]↔ −1 and v2 :=

[−11

]↔ −3

As usual, if f ∈ R2 then

f =12

(v1, f) v1 +12

(v2, f) v2.


34 3 Linear ODE

It then follows that

etAf =12

(v1, f) etAv1 +12

(v2, f) etAv2

=12

(v1, f) e−tv1 +12

(v2, f) e−3tv2.

Taking f = e1 and then f = e2 gives

etAe1 =12

(v1, e1) e−tv1 +12

(v2, e1) e−3tv2

=12e−tv1 −

12e−3tv2

=12

[e−t + e−3t

e−t − e−3t

]and similarly

etAe2 =12

[e−t − e−3t

e−t + e−3t

].

Therefore,

etA =[etAe1 e

tAe2]

=12

[e−t + e−3t

e−t − e−3te−t − e−3t

e−t + e−3t

].

Example 3.12. Continuing the notation and using the results of Example 1.8,

A :=

1 7 −27 1 −2−2 −2 10

with eigenvectors/eigenvalues given by

v1 :=

−110

↔ −6, v2 :=

111

↔ 6, v3 :=

−1−12

↔ 12.

andf =

12

(f, v1) v1 +13

(f, v2) v2 +16

(f, v2) v3.

For example if f = (1, 2, 3)tr , then

f =12v1 + 2v2 +

12v3

and hence


3.3 Second Order Linear ODE 35

etA (1, 2, 3)tr =12etAv1 + 2etAv2 +

12etAv3

=12e−6tv1 + 2e6tv2 +

12e12tv3.

A straightforward but tedious computation shows

etA =16

3e−6t + 2e6t + e12t −3e−6t + 2e6t + e12t 2e6t − 2e12t

−3e−6t + 2e6t + e12t 3e−6t + 2e6t + e12t 2e6t − 2e12t

2e6t − 2e12t 2e6t − 2e12t 2e6t + 4e12t

.This can alternatively be done using a computer algebra package, which iswhat I did.

3.3 Second Order Linear ODE

We would like to solve the ordinary differential equation

u (t) = Au (t) with (3.9)u (0) = f and u (0) = g (3.10)

for some f, g ∈ RN and A a N×N matrix. Again we might begin by trying tofind solutions to Eq. (3.9) by considering functions of the form u (t) = T (t) v.In order for u (t) = T (t) v to be a solution we must have

T (t) v = T (t)Av

and working as above we concluded that there must exists λ such that

T (t) = λT (t) and Av = λv.

The general solution to the equation

T (t) = λT (t)

isT (t) = cλ (t)T (0) + sλ (t) T (0)

where

cλ (t) :=

cos√−λt if λ ≤ 0

cosh√λt if λ ≥ 0

and

sλ (t) :=

sin

√−λt√−λ

if λ < 0t if λ = 0

sinh√

λt√λ

if λ > 0.


36 3 Linear ODE

Theorem 3.13. Suppose the matrix A is diagonalizable, i.e. there exists abasis viNi=1 for RN consisting of eigenvectors A with corresponding eigen-values λiNi=1 ⊂ R. Then for any f, g ∈ RN there is a unique solution, u (t) ,to Eqs. (3.9) and (3.10). Moreover, if we expand f and g in terms of the basisviNi=1 as

f =N∑

i=1

aivi and g =N∑

i=1

bivi

then the unique solution to Eqs. (3.9) and (3.10) is given by

u (t) =N∑

i=1

[aicλi (t) + bisλi (t)] vi. (3.11)

Proof. It is easy to check that u defined as in Eq. (3.11) solves Eqs.(3.9) and (3.10) which proves the existence of solutions. The uniqueness ofsolutions may also be proved similarly to what was done in Theorem 3.3.Indeed, suppose that

u (t) =N∑

i=1

αi (t) vi

then the equation, u = Au, is equivalent to

N∑i=1

αi (t) vi = u (t) = Au (t) =N∑

i=1

αi (t)Avi =N∑

i=1

αi (t)λivi

and since viNi=1 is a basis for RN we must have

αi (t) = λiαi (t) for all i. (3.12)

Moreover,

N∑i=1

aivi = f = u (0) =N∑

i=1

αi (0) vi and

N∑i=1

bivi = g = u (0) =N∑

i=1

αi (0) vi

implies thatαi (0) = ai and αi (0) = bi for all i. (3.13)

This completes the proof, since the unique solution to Eqs. (3.12) and (3.13)is given by

αi (t) = aicλi(t) + bisλi

(t) .


3.3 Second Order Linear ODE 37

Notation 3.14 From now on, let us agree that

cos√−λt := cosh

√λt if λ ≥ 0

sin√−λt√−λ

:=sinh√λt√λ

if λ > 0 and

sin√−λt√−λ

:= t if λ = 0.

With the above notation it is natural to write the general solution Eqs.(3.9) and (3.10) as

u (t) =(cos√−At

)f +

sin√−At√−A

g

with the understanding that

1. cos√−At and sin

√−At√−A

are linear (i.e. matrices) and2. if Av = λv then (

cos√−At

)v :=

(cos√−λt

)v and

sin√−At√−A

v :=sin√−λt√−λ

v.


A :=

1 7 −27 1 −2−2 −2 10


v1 :=

−110

↔ −6, v2 :=

111

↔ 6, v3 :=

−1−12

↔ 12.

We will solve,

u (t) = Au (t) with

u (0) = f = (1, 2, 3)tr and u (0) = g = (1,−1, 1)tr .

As above we have

f =12

(f, v1) v1 +13

(f, v2) v2 +16

(f, v2) v3

=12v1 + 2v2 +

12v3


38 3 Linear ODE

andg = 0v1 +

13v2 +

26

(f, v2) v3 =13

(2v2 + v3) .

Therefore,

cos(√−At

)f =

12

cos(√

6t)v1 + 2 cosh

(√6t)v2 +

12

cosh(√

12t)v3

andsin(√−At

)√−A

g =13

(2sinh

(√6t)

√6

v2 +sinh

(√12t)

√12

v3

)and the solution is given by

u (t) = cos(√−At

)f +

sin(√−At

)√−A

g

=12

cos(√

6t)v1

+

[2 cosh

(√6t)

+23

sinh(√

6t)

√6

]v2

+

[12

cosh(√

12t)

+13

sinh(√

12t)

√12

]v3.

3.4 ODE Exercises

Exercise 3.1. Here you are asked to give another proof of Eq. (3.7). Let Abe an N ×N, matrix, f ∈ RN and s, t ∈ R. Show

etAesAf = e(t+s)Af.

Outline: Let u (t) := etAesAf and v (t) = e(t+s)Af and show both u and vsolve the differential equation,

w (t) = Aw (t) with w (0) = esAf

and then use uniqueness of solutions of this equation (see Fact 3.5) to concludethat u (t) = v (t) .

Exercise 3.2. Let

A =(

0 1−1 0

).

Show

etA =(

cos t sin t− sin t cos t

)using the following three methods.


3.4 ODE Exercises 39

1. Showingd

dt

(cos t sin t− sin t cos t

)= A

(cos t sin t− sin t cos t

)and (

cos t sin t− sin t cos t

)∣∣∣∣t=0

= I =(

10

01

)and

2. by explicitly summing the series

etA =∞∑

n=0

tn

n!An.

3. Show d2

dt2 etA = −etA and then solve this equation using etA|t=0 = I and

ddt |0e

tA = A.

Exercise 3.3. Combine Exercises 3.1 and 3.2 to give a proof of the trigono-metric identities:

cos(s+ t) = cos s cos t− sin s sin t (3.14)and

sin (s+ t) = cos s sin t+ cos t sin s. (3.15)

Exercise 3.4. Let a, b, c ∈ R and

A =

0 a b0 0 c0 0 0

.

Show

etA =

1 at bt+ 12act

2

0 1 ct0 0 1

by summing the matrix power series. Also find et(λI+A) where λ ∈ R and I isthe 3× 3 identity matrix.

Exercise 3.5. Let

A :=

−2 1 11 −2 11 1 −2

and

f = (1, 0, 2)tr and g = (0, 1, 2)tr .

be as in Exercise 1.2. Solve the following equations

u (t) = Au (t) with u (0) = f andu (t) = Au (t) with u (0) = f and u (0) = g.


40 3 Linear ODE

Write your solutions in the form

u (t) =3∑

i=1

ai (t) vi

where the functions ai are to be determined.Hint: Recall from Exercise 1.2 (you should have shown) that

v1 =

111

↔ 0, v2 =

−101

↔ −3, v3 =

−12−1

↔ −3.

is an orthogonal basis of eigenvectors (with corresponding eigenvalues) for A.


4

Linear Operators and Separation of Variables

Definition 4.1. A linear combination of the vectors vini=1 ⊂ R3 (orvini=1 ⊂ V with V being any vector space) is a vector of the form,

c1v1 + c2v2 + · · ·+ cnvn,

with cini=1 being real (or complex) constants.

We are going to be interested in the case that the vector space V consistsof a class of functions on some domain, Ω ⊂ Rn. If u1 and u2 are functionson Ω ⊂ Rn, and c1, c2 ∈ R we write (c1u1 + c2u2) =: u for the function,u : Ω → R such that

u (x) = c1u1 (x) + c2u2 (x) for all x ∈ Ω.

For example we may consider, u1 + u2, u1 + 3u2, and 0 = 0u1 + 0u2.

Definition 4.2. A linear space of functions, V, is a class of functionswith common domain so that if u1, u2 are in the class then so is c1u1 + c2u2

for all c1, c2 ∈ R, i.e. the space of functions V is closed under taking linearcombinations.

Example 4.3.

D = f : R→ R : f is differentiable on R

orC = f : R→ R : f is continuous on R.

Consider operator, L : D → all functions on R defined by Lf = f ′. Thisoperator is linear, namely, we have

L(c1f1 + c2f2) = (c1f1 + c2f2)′ = c1f′1 + c2f

′2 = c1L(f1) + c2L(f2).

It is interesting to note that L does not map D to C. For example, let

42 4 Linear Operators and Separation of Variables

f(x) =x2 sin 1

x x 6= 00 x = 0

then

f ′(x) =

2x sin 1

x − cos 1x x 6= 0

limx→0

x2 sin 1x

x = 0 x = 0

so that ϕ ∈ D however f ′ /∈ C.

Definition 4.4. A Linear operator, is a mapping, L, of one linear space offunctions to another such that

L(c1u1 + c2u2) = c1L(u1) + c2L(u2)

for all u1, u2 in the domain function space and c1, c2 ∈ R.

An induction argument shows the linearity condition in Definition 4.4 im-plies

L

(n∑

i=1

ciui

)=

N∑i=1

ciL(ui)

for all ui in the domain function space and ci ∈ R.

Example 4.5. Let Ω be some open subset of R2, for example Ω = R2 orΩ =

x ∈ R2|x2

1 + x22 < 5

and let D denote those functions u : Ω → R such

that u and all of its partial derivatives up to order two exist and are continuous.(In the future we denote this class of functions by C2 (Ω) .) Then the followingare example of linear operators taking D to the class of continuous functionson Ω :

1. Lu = ∂2u∂x2

2. Lu = ∂2u∂x ∂y

3. (Lu) (x, y) = x∂u∂y (x, y) + y ∂u

∂x (x, y) .

Whereas, the following operator is an example of a non-linear operator;Lu = ∂

∂xu+ u2. To see this operator is not linear, notice that

L(u1 + u2) =∂

∂xu1 +

∂

∂xu2 + (u1 + u2)2

6= ∂

∂xu1 + u2

1 +∂

∂xu2 + u2

2 = Lu1 + Lu2.

For example, let u1 = 1x and u2 = 1

x+5 (also see Exercise 13.9), then

L(u1) = − 1x2

+1x2

= 0

L(u2) = − 1(x+ 5)2

+1

(x+ 5)2= 0


4 Linear Operators and Separation of Variables 43

while

L(u1 + u2) = L

(1x

+1

x+ 5

)=−1x2− 1

(x+ 5)2+(

1x

+1

x+ 5

)2

= − 1x2− 1

(x+ 5)2+

1x2

+1

(x+ 5)2+

2x+ 5)

=2

x(x+ 5)6= 0.

Definition 4.6. If L,M are two linear operators on the same class of func-tions we define L+M

(L+M)(u) := Lu+Mu.

If M is a linear operator on the range-space of L we also define LM by((LM)u) = L(Mu).

These new operators are still linear, for example,

(LM)(c1u1 + c2u2) = L(c1M(u1) + c2M(u2))= c1L(M(u1)) + c2L(M(u2)))= c1(LM)(u1) + c2(LM)(c2).

However, it is in general not true (see Exercise 13.2) that LM = ML, in factLM may be defined while ML is not defined. For example, let Lu = x2u andMu = ∂

∂xu then taking u = exy we find

LM(u) = L

(∂

∂xexy

)= L(y exy) = x2yexy

whileM(Lu) =

∂

∂x(x2exy) = 2xexy + yx2exy 6= LM (u) .

In general, in this class we will be interested in linear differential operatorsof the form

Lu = Auxx +Buxy + Cuyy +Dux + Euy + Fu

where A,B,C,D,E, F are functions of x and y. The homogeneous partialdifferential equations, Lu = 0 is shorthand notation for u solving the equation,

Auxx +Buxy + Cuyy +Dux + Euy + Fu = 0.

Lemma 4.7 (Principle of Superposition). If L is a linear differential op-erator as above and u1 and u2 solve the homogeneous partial differential equa-tions, Lu1 = 0 and Lu2 = 0, then any linear combination, c1u1 + c2u2 alsosatisfies the same equation, namely,

L (c1u1 + c2u2) = 0.



Example 4.8 (Homogeneous Wave Equation ). The wave equation utt =a2uxx, is equivalent to writing Lu = 0 where L = ∂2

∂t2 −a2 ∂2

∂x2 . So if u1, . . . , uM

are solutions to the wave equation, (i.e., Lun = 0), then any linear combina-tion,

c1u1 + · · ·+ cnuM ,

is another solution as well. (See Exercises 13.4, 13.6, 13.8 for more on this andthe issue of boundary conditions.) To be more explicit let us notice that

1. Show that un (t, x) := sin(nx) sin(ant) for n ∈ N all solve the equation,Lu = 0. Indeed,

Lun =∂2

∂t2[sin(nx) sin(ant)]− a2 ∂

2

∂x2[sin(nx) sin(ant)]

=∂

∂t(sin(nx)an cos(ant))− a2 ∂

∂x(n cos(nx) sin(ant))

= −a2n2 sin(nt) sin(ant)− a2[−n2 sin(nx) sin(ant)] = 0.

2. So by the superposition principle,

u(x, y) =N∑

n=1

cn sin(nx) sin(ant)

with c1, . . . , cM ∈ R also satisfies the wave equation. (Later we will allowfor infinite linear combinations and we will then choose the constants, ci,so that certain boundary conditions are satisfied.)

4.1 Introduction to Fourier Series

In this section, I would like to explain how certain functions like sinnx andcosnx are going to appear in our study of partial differential equations. Sup-pose L is the differential operator, L = d2

dx2 , acting on functions on Ω = [a, b] .Define the inner product,

(f, g) :=∫

Ω=[a,b]

f (x) g (x) dx,

for functions f, g : Ω → R. Two integration by parts now shows,

(Lf, g) =∫

Ω

f ′′ (x) · g (x) dx = −∫

Ω

f ′ (x) · g′ (x) dx+ f ′ (x) g (x) |ba

=∫

Ω

f (x) · g′′ (x) dx+ [f ′ (x) g (x)− f (x) · g′ (x)] |ba

= (f, Lg) + [f ′ (x) g (x)− f (x) · g′ (x)] |ba.


4.1 Introduction to Fourier Series 45

There are now a number of boundary conditions that may be imposed on fand g so that boundary terms in the previous equation are zero. For examplewe may assume f, g ∈ Dper where Dper denotes those twice continuouslydifferentiable functions such that f (b) = f (a) and f ′ (b) = f ′ (a) . Or wemight assume f, g ∈ DDirichlet or f, g ∈ DNeumann where DDirichlet (DNeumann)consists of those twice continuously differentiable functions such that f (a) =0 = f (b) (f ′ (a) = 0 = f ′ (b)). In any of these cases, we will have

(Lf, g) = (f, Lg)

and so in analogy with the Spectral Theorem 1.5 we should expect that L hasan orthonormal basis of eigenvectors. Let us find these eigenvectors in a fewexamples. Before doing this it is useful to record a few integrals.

Lemma 4.9. Let n be a positive integer, then∫ π

0

sin2 nxdx =∫ π

0

cos2 nxdx =π

2, (4.1)∫ π

−π

sin2 nxdx =∫ π

−π

cos2 nxdx = π, (4.2)

and∫ π

−π

sinnx cosnxdx = 0. (4.3)

Proof. Recall that

cos 2θ = cos2 θ − sin2 θ = 1− 2 sin2 θ = 2 cos2 θ − 1.

Therefore, taking θ = nx and integrating we find,

0 =12n

sin 2nx|π0 =∫ π

0

cos 2nxdx

=∫ π

0

[1− 2 sin2 nx

]dx =

∫ π

0

[2 cos2 nx− 1

]dx

= π − 2∫ π

0

sin2 nxdx = 2∫ π

0

cos2 nxdx− π

which gives Eq. (4.1). Similarly, replacing∫ π

0by∫ π

−πabove shows Eq. (4.2) is

valid as well. Finally,∫ π

−π

sinnx cosnxdx =12n

sin2 nx|π−π = 0− 0 = 0.

Example 4.10 (Fourier Series). Let a = −π and b = π, L = d2

dx2 and D = Dper

so that



(f, g) =∫ π

−π

f (x) g (x) dx (4.4)

and

(Lf, g) = − (f ′, g′) = −∫ π

−π

f ′ (x) g′ (x) dx = (f, Lg).

Thus if Lf = λf with f ∈ D we have

λ (f, f) = (Lf, f) = −∫ π

−π

[f ′ (x)]2 dx ≤ 0

from which it follows that λ ≤ 0. So we need only look for negative eigenvalues.If λ = 0 the eigenvalue equation becomes f ′′ = 0 and hence f (x) = Ax+ B.We will only have f ∈ D if A = 0 and therefore let f0 = 1.

We may now suppose that λ = −ω2 < 0 in which case the eigenvalueequation becomes

f ′′ = −ω2f

which hasf (x) = A cosωx+B sinωx

as the general solution. We still must enforce the boundary values. For examplef (π) = f (−π) implies

A cos (−ωπ) +B sin (−ωπ) = A cosωπ +B sinωπ

or B sinωπ = 0. Similarly, f ′ (π) = f ′ (−π) implies

−ωA sin (−ωπ) + ωB cos (−ωπ) = −ωA sinωπ + ωB cosωπ

that A sinωπ = 0. Hence we either have A = B = 0 (in which case f ≡ 0which is not allowed) or sinωπ = 0 from which it follows that ω = n ∈ Z.Hence we have

β := cosnx, sinnx : n ∈ N ∪ 1as our possible eigenvectors. The eigenvalue associated to cosnx and sinnx isλn = −n2. By Lemma 4.9, (cosnx, sinnx) = 0, so that β is an orthogonal setand moreover,

(cosnx, cosnx) = (sinnx, sinnx) = π and(1, 1) = 2π.

Thus we expect that any reasonable function f on [−π, π] may be written as

f (x) =12π

(f, 1) 1 +1π

∞∑n=1

[(f, cosn (·)) cosnx+ (f, sinn (·)) sinnx] . (4.5)

See Theorem 5.8, Theorem 5.17 and Theorem 7.2 below for more details onthis point.


4.2 Application / Separation of variables 47

Example 4.11 (Fourier Sine Series / Dirichlet boundary conditions). Supposea = 0 and b = π, L = d2

dx2 and D = DDirichlet so if f, g ∈ D then f (0) = 0 =f (π) . We now take

(f, g) =∫ π

0

f (x) g (x) dx

and working as in the previous example we find

un (x) = sinnx with λn = −n2 for n ∈ N.

By Lemma 4.9,(sinn (·) , sinn (·)) =

π

2and so by Theorem 7.2,

f (x) =2π

∞∑n=1

(f, sinn (·)) sinnx

for any “reasonable” function f on [0, π] .

Example 4.12 (Fourier Cosine Series / Neumann boundary conditions.). Sup-pose a = 0 and b = π, L = d2

dx2 and D = DNeumann so if f, g ∈ D thenf ′ (0) = 0 = f ′ (π) .

Again we take

(f, g) =∫ π

0

f (x) g (x) dx

and we find the eigenfunctions and eigenvalues to be

un (x) = cosnx with λn = −n2 for n ∈ N∪0 .

By Lemma 4.9,

(cosn (·) , cosn (·)) =π

2for n ∈ N

and (1, 1) = π,

Thus Theorem 7.2 asserts that any “reasonable” function f on [0, π] may bewritten as

f (x) =1π

(f, 1) 1 +2π

∞∑n=1

(f, cosn (·)) cosnx.

4.2 Application / Separation of variables

Example 4.13. Use “separation of variables” to solve the heat equation,



ut (t, x) = uxx (t, x) with u (t, 0) = u (t, 5) = 0 (4.6)and u (0, x) = f (x) .

The technique is to first ignore the nonhomogeneous condition u (0, x) = f (x)and look for any solutions to the Eq. (4.6) of the form

u (t, x) = T (t)X (x) .

From this we get,T (t)X (x) = T (t)X ′′ (x)

or equivalently thatT (t)T (t)

=X ′′ (x)X (x)

= λ

where λ is a constant. Thus we require that

X ′′ (x) = λX (x) with X (0) = 0 and X (5) = 0.

The solutions to this Sturm-Liouville problem are given by

Xn (x) = sinnπx

5with λ = λn = −

(nπ5

)2

.

This then forces Tn (t) = e−t(nπ5 )2

. Thus we find that

un (t, x) = e−t(nπ5 )2

sinnπx

5

are all solutions to Eq. (4.6). We then look for a general solution to ourproblem in the form

u (t, x) =∞∑

n=1

bnun (t, x)

where we wish to choose the constants, bn such that

f (x) = u (0, x) =∞∑

n=1

bnun (0, x) =∞∑

n=1

bn sinnπx

5.

Letting

(f, g) =∫ 5

0

f (x) g (x) dx

we have (sin

nπx

5, sin

mπx

5

)= δmn

52

and therefore,

bn =

(f, sin nπx

5

)(sin nπx

5 , sin nπx5

) =25

∫ 5

0

f (x) sinnπx

5dx.



Example 4.14. Use separation of variables to solve Laplace’s equation,

uxx (x, y) + uyy (x, y) = 0 for 0 ≤ x ≤ 5 and 0 ≤ y ≤ 2

with boundary conditions,

u (0, y) = u (5, y) = 0,u (x, 0) = 0 and u (x, 2) = f (x) .

To do this we will work as above and begin by ignoring the non-homogeneousboundary condition and solve the rest by separation of variables. So we writeu (x, y) = X (x)Y (x) and require that

X ′′

X+Y ′′

Y= 0 with X (0) = X (5) = 0 = Y (0) .

As before we must have X ′′ = λX and we know the solutions are given by

Xn (x) = sinnπx

5with λ = λn = −

(nπ5

)2

.

It them implies that

Y ′′ (y) =(nπ

5

)2

Y (y) with Y (0) = 0.

From this we concluded that

Yn (y) = sinhnπy

5

and we find thatun (x, y) = sin

nπx

5sinh

nπy

5in this case. So working as above we try to find a solution of the form

u (x, y) =∞∑

n=1

bnun (x, y) .

All the boundary conditions are now satisfied except for

f (x) = u (x, 2) =∞∑

n=1

bnun (x, 2) =∞∑

n=1

bn sinh2nπ5

sinnπx

5.

By the same logic as above we must have

bn sinh2nπ5

=25

∫ 5

0

f (x) sinnπx

5dx

and thus that



u (x, y) =∞∑

n=1

bn sinh2nπ5

sinnπx

5

with

bn =2

5 sinh 2nπ5

∫ 5

0

f (x) sinnπx

5dx.

Example 4.15. Solve the wave equation, for 0 ≤ x ≤ 5 and t ∈ R,

utt (t, x) = ux x (t, x) with u (t, 0) = u (t, 5) = 0and u (0, x) = f (x) and ut (0, x) = 0.

We could go through separation of variables here to answer this question, butthis is getting tedious. I will just write down the answer as

u (t, x) = cos(√−∂2

xt)f (x) .

As we have seen,

f (x) =∞∑

n=1

bn sinnπx

5

with

bn =25

∫ 5

0

f (x) sinnπx

5dx

and hence

u (t, x) =∞∑

n=1

bn cos(√−∂2

xt)

sinnπx

5

=∞∑

n=1

bn cos(nπt

5

)sin

nπx

5.

It is interesting to notice that since

sin (A+B) = cosA sinB + sinA cosB

we havesin (A+B) + sin (A−B) = 2 sinA cosB.

Thus we may write,

sinnπx

5cos(nπt

5

)=

12

[sin(nπ (x+ t)

5

)+ sin

(nπ (x− t)

5

)]and thus

u (t, x) =12

∞∑n=1

bn

[sin(nπ (x+ t)

5

)+ sin

(nπ (x− t)

5

)]=

12

[F (x+ t) + F (x− t)]



where

F (x) =∞∑

n=1

bn sin(nπx

5

)= the 5− periodic extensions of f (x) ,

i.e. if x = 5n+ y with 0 < y < 5 and n ∈ Z, then F (x) = f (y) .


5

Orthogonal Function Expansions

5.1 Generalities about inner products on function spaces

Let Ω be a region in Rd (most of the time d will be one for us) and p : Ω →(0,∞) be a positive function. For functions f, g : Ω → R define

(f, g) :=∫

Ω

f (x) g (x) p (x) dx.

This is an example of a inner product, i.e. something that behaves like thedot product on RN . For example we still have the following properties:

(f1 + cf2, g) = (f1, g) + c (f2, g)(f, g) = (g, f)

‖f‖2 := (f, f) = 0 implies f = 0.

The following computation will be used frequently in this class:

‖f + g‖2 = (f + g, f + g) = ‖f‖2 + ‖g‖2 + (f, g) + (g, f)

= ‖f‖2 + ‖g‖2 + 2(f, g). (5.1)

Definition 5.1. Two functions f, g : Ω → R are orthogonal and we writef ⊥ g iff (f, g) = 0. More generally, a collection of functions, ϕini=1 , is anorthogonal set if ϕi ⊥ ϕj (i.e. (ϕi, ϕj) = 0) for i 6= j. If we further have‖ϕi‖ = 1 then we say ϕini=1 is an orthonormal set.

Exercise 5.1. Put in some exercise on orthogonal sets from the book here.

Theorem 5.2 (Schwarz Inequality). For all f, g : Ω → R,

|(f, g)| ≤ ‖f‖‖g‖

and equality holds iff f and g are linearly dependent.

54 5 Orthogonal Function Expansions

Proof. If g = 0, the result holds trivially. So assume that g 6= 0 andobserve; if f = αg for some α ∈ C, then (f, g) = α ‖g‖2 and hence

|(f, g)| = |α| ‖g‖2 = ‖f‖‖g‖.

Now suppose that f ∈ H is arbitrary, let h := f − ‖g‖−2(f, g)g. (So z is the“orthogonal projection” of f onto g, see Figure 5.1.) Then

Fig. 5.1. The picture behind the proof of the Schwarz inequality.

0 ≤ ‖h‖2 =∥∥∥∥f − (f, g)

‖g‖2g

∥∥∥∥2

= ‖f‖2 +(f, g)2

‖g‖4‖g‖2 − 2

(f,

(f, g)‖g‖2

g

)= ‖f‖2 − (f, g)2

‖g‖2

from which it follows that 0 ≤ ‖g‖2‖f‖2 − (f, g)2 with equality iff h = 0 orequivalently iff f = ‖g‖−2(f, g)g.

Corollary 5.3 (Triangle inequality). Let f, g : Ω → R be functions anda ∈ R, then

‖f + g‖ ≤ ‖f‖+ ‖g‖ and (5.2)‖af‖ = |a| ‖f‖ . (5.3)

Proof.

‖f + g‖2 = ‖f‖2 + ‖g‖2 + 2(f, g)

≤ ‖f‖2 + ‖g‖2 + 2‖f‖‖g‖ = (‖f‖+ ‖g‖)2.

Taking the square root of this inequality shows Eq. (5.2) holds. Taking thesquare root of the identity,

‖af‖2 =∫

Ω

|a|2 |f (x)|2 dx = |a|2∫

Ω

|f (x)|2 dx = |a|2 ‖f‖2,

proves Eq. (5.3).


5.1 Generalities about inner products on function spaces 55

Proposition 5.4 (Pythagorean’s Theorem). Suppose that ϕini=1 is anorthogonal set, then ∥∥∥∥∥

n∑i=1

ϕi

∥∥∥∥∥2

=n∑

i=1

‖ϕi‖2. (5.4)

Proof. Let s :=∑n

i=1 ϕi, then

‖s‖2 = (s, s) =

(n∑

i=1

ϕi, s

)=

n∑i=1

(ϕi, s)

and

(ϕi, s) =

ϕi,n∑

j=1

ϕj

=n∑

j=1

(ϕi, ϕj) = (ϕi, ϕi) = ‖ϕi‖2 .

The last two equations proves Eq. (5.4).

Theorem 5.5 (Best Approximation Theorem). Suppose ϕini=1 is anorthonormal set and ai ∈ R, then∥∥∥∥∥f −

n∑i=1

aiϕi

∥∥∥∥∥2

=

∥∥∥∥∥f −n∑

i=1

(f, ϕi)ϕi

∥∥∥∥∥2

+n∑

i=1

|(f, ϕi)− ai|2 (5.5)

and therefore the best approximation to f by functions of the form∑n

i=1 aiϕi

occurs when ai = (f, ϕi) .

Proof. The function (vector),

h := f −n∑

i=1

(f, ϕi)ϕi,

is orthogonal to ϕini=1 since

(h, ϕj) =

(f −

n∑i=1

(f, ϕi)ϕi, ϕj

)= (f, ϕj)−

n∑i=1

(f, ϕi) (ϕi, ϕj)

= (f, ϕj)−n∑

i=1

(f, ϕi) δij = (f, ϕj)− (f, ϕj) = 0.

Since

f −n∑

i=1

aiϕi = f −n∑

i=1

(f, ϕi)ϕi +n∑

i=1

[(f, ϕi)− ai]ϕi

= h+n∑

i=1

[(f, ϕi)− ai]ϕi,



it follows by Pythagorean’s Theorem, Proposition 5.4, that∥∥∥∥∥f −n∑

i=1

aiϕi

∥∥∥∥∥2

= ‖h‖2 +n∑

i=1

‖[(f, ϕi)− ai]ϕi‖2

= ‖h‖2 +n∑

i=1

|(f, ϕi)− ai|2

=

∥∥∥∥∥f −n∑

i=1

(f, ϕi)ϕi

∥∥∥∥∥2

+n∑

i=1

|(f, ϕi)− ai|2 .

Definition 5.6. Let f be a function such that∫

Ω|f (x)|2 p (x) dx < ∞ and

ϕi∞i=1 be an orthonormal set, we will write f ∼∑∞

i=1 (f, ϕi)ϕi to mean

limn→∞

∫Ω

∣∣∣∣∣f (x)n∑

i=1

(f, ϕi)ϕi (x)

∣∣∣∣∣2

p (x) dx

= limn→∞

∥∥∥∥∥f −n∑

i=1

(f, ϕi)ϕi

∥∥∥∥∥2

= 0.

We say ϕi∞i=1 is complete (or closed in the book’s terminology) if f ∼∑∞i=1 (f, ϕi)ϕi whenever ‖f‖2 <∞.

Corollary 5.7 (Bessel’s (In)equality ). Suppose ϕini=1 is an orthonor-mal set, then

n∑i=1

|(f, ϕi)|2 ≤ ‖f‖2 for all f, (5.6)

Moreover we get equality iff f =∑n

i=1 (f, ϕi)ϕi. These statements remaintrue even when n = ∞ provided we interpret, f =

∑∞i=1 (f, ϕi)ϕi to mean

f ∼∑∞

i=1 (f, ϕi)ϕi. So we have f ∼∑∞

i=1 (f, ϕi)ϕi iff Pythagorean’s theoremholds, i.e. iff

∞∑i=1

|(f, ϕi)|2 = ‖f‖2.

Proof. Taking ai = 0 in Eq. (5.5) shows

‖f‖2 =

∥∥∥∥∥f −n∑

i=1

(f, ϕi)ϕi

∥∥∥∥∥2

+n∑

i=1

|(f, ϕi)|2

and hence that

n∑i=1

|(f, ϕi)|2 = ‖f‖2 −

∥∥∥∥∥f −n∑

i=1

(f, ϕi)ϕi

∥∥∥∥∥2

≤ ‖f‖2


5.2 Convergence of the Fourier Series 57

with equality iff f =∑n

i=1 (f, ϕi)ϕi. Letting n→∞ in the previous equationshows,

∞∑i=1

|(f, ϕi)|2 = ‖f‖2 − limn→∞

∥∥∥∥∥f −n∑

i=1

(f, ϕi)ϕi

∥∥∥∥∥2

≤ ‖f‖2

with equality iff

limn→∞

∥∥∥∥∥f −n∑

i=1

(f, ϕi)ϕi

∥∥∥∥∥2

= 0,

i.e. iff f ∼∑∞

i=1 (f, ϕi)ϕi.

5.2 Convergence of the Fourier Series

For this section it will be convenient to define

(f, g) =∫ π

−π

f (y) g (y)1πdy

Recall from Example 4.10, if f : R→ R is “reasonable” 2π - periodic function(i.e. f (x+ 2π) = f (x) for all x ∈ R), we expect by analogy with the finitedimensional spectral theorem that

f (x) =12a0 +

∞∑n=1

[an cosnx+ bn sinnx] (5.7)

where

an := (f, cosn (·)) =1π

∫ π

−π

f (y) cosny dy for n = 0, 1, 2, . . .

and

bn := (f, sinn (·)) =1π

∫ π

−π

f (y) sinny dy for n = 1, 2, . . . .

The following theorem gives a precise version of this statement.

Theorem 5.8 (Fourier Convergence Theorem). Let f : R→ R be a 2π- periodic function which is piecewise continuous on (−π, π). Then at pointsx ∈ X where f ′ (x±) exist we have

12a0 +

∞∑n=1

[an cosnx+ bn sinnx] =f (x+) + f (x−)

2.



Fact 5.9 If f : [−π, π] → R is any function such that∫ π

−π|f (x)|2 dx < ∞,

we may still define

fN (x) =12a0 +

N∑n=1

[an cosnx+ bn sinnx]

with an and bn as above. With this definition we will always have,

1π

limN→∞

∫ π

−π

|f (x)− fN (x)|2 dx = limN→∞

‖f − fN‖2 = 0,

i.e. that

f (x) ∼ 12a0 +

∞∑n=1

[an cosnx+ bn sinnx] .

5.3 Examples

Remark 5.10. We will use the following identities repeatedly.

sin (A+B) = cosA sinB + sinA cosB, (5.8)cos (A+B) = cosA cosB − sinA sinB, (5.9)

sinA cosB =12

(sin (A+B) + sin (A−B)) (5.10)

cosA cosB =12

(cos (A+B) + cos (A−B)) (5.11)

sinA sinB =12

(cos (A−B)− cos (A+B)) . (5.12)

Example 5.11. Suppose

f (x) =

1 if 0 < x < π−1 if −π < 0 < x

,

then an := (f, cosn (·)) = 0 because f is odd while

bn = (f, sinn (·)) =2π

∫ π

0

sinny dy = − 2πn

cosny|π0

=2πn

(1− cosnπ) =

0 if n is even4

πn if n is odd.

Thus we conclude that

f (x) ∼∑

n odd

4πn

sinnx =∞∑

n=1

4π (2n− 1)

sin (2n− 1)x.


5.3 Examples 59

The series converges for every x ∈ R and x ∈ (−π, π) \ 0 it converges tof (x) and at x = 0 it converges to 0. By the way, by Bessel’s equality we have

∞∑n=1

(4

π (2n− 1)

)2

= ‖f‖2 =1π

∫ π

−π

|f (x)|2 dx

=1π

∫ π

−π

1dx = 2

and hence we conclude that

1 +19

+125

+149

+ · · · =∞∑

n=1

1(2n− 1)2

=π2

8.

Here are some related graphs wherein fN (x) :=∑N

n=1:n odd4

πn sinnx.



Fig. 5.2. A plot of f5 (x) = 4π

(sin x + 1

3sin 3x + 1

5sin 5x

). which is approximating

f (x) .

Fig. 5.3. A plot of f11 (x) .


5.3 Examples 61

Fig. 5.4. A plot of f19 (x) . These last few picture illustrate what is known as theGibb’s effect; namely the convergence is not uniform. There is always the pesky littlebump appearing near 0 (and ±π).

Fig. 5.5. A plot (at t = 1/2) of 4π

(e−t sin x + 1

3e−9t sin 3x + 1

5e−25t sin 5x

)which is

approximating the solution to the heat equation with periodic boundary condtionsand with initial condition, u (0, x) = f (x) .

The following lemma will be useful in simplifying the computations insome of the examples below.

Lemma 5.12. Suppose f : [a, b] → R is a continuous function and p (x) =∑nk=0 pkx

k is a polynomial, then∫fpdx = pF1 − p′F2 + p′′F3 − · · ·+ (−1)n

p(n)Fn+1 + C

whereF1 =

∫fdx and Fk+1 =

∫Fkdx.



Proof. This is simply a matter of repeated integration by parts. Explicitly,∫fpdx =

∫F ′1pdx = F1p−

∫F1p

′dx+ C.∫F1p

′dx = F2p′ −∫F2p

′′dx+ C

and hence ∫fpdx =

∫F ′1pdx = F1p− F2p

′ +∫F2p

′′dx+ C

= F1p− F2p′ + F2p

′′dx−∫F3p

(3)dx+ C,

etc.Using this fact we may now compute the Fourier series of a number of

functions.

Example 5.13. Let (f, g) = 1π

∫ π

−πf (x) g (x) dx, then

1.

(x, 1) = 0, (x, cosnx) = 0 by symmetry and

(x, sinnx) =1π

∫ π

−π

x sinnxdx =[x

(− 1n

cosnx)− 1n2

sinnx]π

−π

=1n

[−2 cosnπ] =2n

(−1)n+1

wherein we have used

F1 =∫

sinnxdx = − 1n

cosnx and

F2 = − 1n

∫cosnxdx =

1n2

sinnx.

Thus we expect that

x =∞∑

n=1

2n

(−1)n+1 sinnx = 2(

11

sinx− 12

sin 2x+13

sin 3x− 14

sin 4x+ . . .

).


5.3 Examples 63

Plot of x and

21

sinx− 22

sin 2x+23

sin 3x− 24

sin 4x

.

2. f (x) = x2 – expansion.(x2, sinnx

)= 0 by symmetry and(

x2, cosnx)

=1π

∫ π

−π

x2 cosnxdx

=[x2

(1n

sinnx)

+ 2x1n2

cosnx− 2sinnxn3

]π

−π

=1π

2π1n2

[cosnπ + cosn (−π)] = 41n2

(−1)n

wherein we have used

F1 =∫

cosnxdx =1n

sinnx,

F2 =∫

1n

sinnxdx = − 1n2

cosnx and

F3 =∫− 1n2

cosnxdx = − 1n3

sinnx.

We also have (x2, 1

)=

1π

∫ π

−π

x2dx =x3

3|π−π =

2π2

3

Thus we expect from Eq. (4.5) that



x2 =12(x2, 1

)1 +

∞∑n=1

[(x2, cosn (·)

)cosnx+

(x2, sinn (·)

)sinnx

]=

12

2π2

31 +

∞∑n=1

41n2

(−1)n cosnx

=π2

3+ 4

∞∑n=1

(−1)n

n2cosnx

=π2

3+ 4

[− cosx+

122

cos 2x− 132

cos 3x+ . . . .

].

Plot of x2 (in red) and

π2

3+ 4

[− cosx+

122

cos 2x− 132

cos 3x+ . . .

].

3. Integrating the equation,

x2 =π2

3+ 4

∞∑n=1

(−1)n

n2cosnx,

one expects,


5.3 Examples 65

x3

3=π2

3x+ 4

∞∑n=1

(−1)n

n2

sinnxn

=π2

3

[2∞∑

n=1

(−1)n+1

nsinnx

]− 4

∞∑n=1

(−1)n+1

n3sinnx

=∞∑

n=1

[2π2

3(−1)n+1

n− 4

(−1)n+1

n3

]sinnx

= 2∞∑

n=1

(−1)n+1

[(πn)2 − 6

3n3

]sinnx

and hence that

x3 = 2∞∑

n=1

(−1)n+1

[(πn)2 − 6

n3

]sinnx

also see top of page 77 of the book4. The expansions for functions of the form f (x) = ax+ bx2 + cx3 are now

easily found. For example, if

x2 − πx =π2

3+ 4

∞∑n=1

(−1)n

n2cosnx− π

∞∑n=1

2n

(−1)n+1 sinnx.

The plots of this function and the following approximation,

S2 (x) =π2

3+ 4

(−112

cosx+122

cos 2x)− 2π

(11

sinx− 12

sin 2x),

S3 (x) =π2

3+ 4

(−112

cosx+122

cos 2x+132

cos 3x)

− 2π(

11

sinx− 12

sin 2x+13

sin 3x)

and

S4 (x) =π2

3+ 4

(−112

cosx+ · · ·+ 142

cos 4x)

− 2π(

11

sinx− 12

sin 2x+13

sin 3x− 14

sin 4x)

are given in Figure 5.6 below.

Example 5.14. Suppose that f (x) is a function defined for 0 ≤ x ≤ π. Supposewe extend f to be an odd function by setting



Fig. 5.6. Plot of x2 − πx and the approximats, S2, S3, and S4. Notice that theapproximations are not doing to well at the end points. This is because they areconvergeing to π2 at x = π and x = −π.

Fig. 5.7. The function f with its extension to an odd function on [−π, π].

F (x) :=

f (x) if 0 < x < π−f (−x) if −π < x < 0,

see Figure 5.7. Then computing the Fourier series of F, we learn

an := (F, cosn (·)) =1π

∫ π

−π

F (y) cosny dy = 0 for n = 0, 1, 2, . . .

and

bn := (F, sinn (·)) =1π

∫ π

−π

F (y) sinny dy =2π

∫ π

−π

f (y) sinny dy.

From this we learn that

f (x) =∞∑

n=1

(2π

∫ π

−π

f (y) sinny dy)

sinnx.


5.3 Examples 67

Notice that sinny : n ∈ N form an orthonormal set relative to the innerproduct,

(f, g) :=2π

∫ π

0

f (x) g (x) dx.

As an explicit example, let us consider the sin - series expansion of cosx for0 < x < π. For this we have

bn =2π

∫ π

0

cos y sinny dy

=1π

∫ π

0

(sin ((n+ 1) y) + sin ((n− 1) y)) dy

= − 1π

(cos ((n+ 1) y)

n+ 1+

cos ((n− 1) y)n− 1

)π

0

= − 1π

((−1)n+1

n+ 1+

(−1)n−1

n− 1− 1n+ 1

− 1n− 1

)

= −1n-even−2π

(1

n+ 1+

1n− 1

)= 1n-even

2π

2nn2 − 1

= 1n-even4π

n

n2 − 1

and we conclude that

cosx =4π

∑n=2,4,6,...

n

n2 − 1sinnx

=4π

∞∑n=1

2n4n2 − 1

sin 2nx.

Here is a plot of cosx along with 4π

(2

22−1 sin 2x+ 442−1 sin 4x+ 6

62−1 sin 6x)

and 4π

(2

22−1 sin 2x+ · · ·+ 12122−1 sin 12x

)Page: 67 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28


Example 5.15. Suppose that f (x) is a function defined for 0 ≤ x ≤ π. Supposewe extend f to be an even function by setting

F (x) :=f (x) if 0 < x < πf (−x) if −π < x < 0,

see Figure 5.8. Then computing the Fourier series of F, we learn

Fig. 5.8. The function f with its extension to an even function on [−π, π].

an := (F, cosn (·)) =1π

∫ π

−π

F (y) cosny dy =2π

∫ π

0

f (y) cosny dy

and

bn := (F, sinn (·)) =1π

∫ π

−π

F (y) sinny dy = 0.

From this we learn that

f (x) =a0

2+

∞∑n=1

an cosnx

wherean :=

2π

∫ π

0

f (y) cosny dy.

Notice that cosny : n ∈ N ∪

1√2

form an orthonormal set relative to the

inner product,

(f, g) :=2π

∫ π

0

f (x) g (x) dx.

As an explicit example, let us consider the cos - series expansion of sinx for0 < x < π. For this we have


5.3 Examples 69

an :=2π

∫ π

0

sin y cosny dy

=1π

∫ π

0

(sin ((n+ 1) y) + sin ((1− n) y)) dy

= − 1π

(cos ((n+ 1) y)

n+ 1+

cos ((1− n) y)1− n

)π

0

= 2 · 1n-even1π

(1

1 + n+

11− n

)=

4π

1n-even1

1− n2= − 4

π1n-even

1n2 − 1

.

So we conclude that

sinx =2π− 4π

∑n=2,4,6,...

1n2 − 1

cosnx

=2π− 4π

∞∑n=1

14n2 − 1

cos 2nx

=2π

(1− 2

22 − 1cos 2x− 2

42 − 1cos 4x− 2

62 − 1cos 6x− . . .

).

Plot of sinx and the function

2π

(1− 2

22 − 1cos 2x− 2

42 − 1cos 4x

)which consists of the first 3 terms in the cosine expansion of sinx.



5.4 Proof of Theorem 5.8

Before giving the proof of Theorem 5.8, we will need the following simpleconsequence of Bessel’s inequality in Corollary 5.7.

Lemma 5.16. Suppose f is a continuous function on [−π, π] or more gener-ally any function such that

∫ π

−π|f (x)|2 dx < ∞ and an and bn are given as

above, then

12a20 +

∞∑n=1

[|an|2 + |bn|2

]≤ 1π

∫ π

−π

|f (x)|2 dx <∞. (5.13)

Moreover we also have

limN→∞

∫ π

−π

f (y) sin((

N +12

)y

)dy = 0. (5.14)

Proof. Since

1√2, cosnx, sinnx : n ∈ N

forms an orthonormal set, it fol-

lows from Corollary 5.7 that

12a20 +

∞∑n=1

[|an|2 + |bn|2

]=(f,

1√2

)2

+∞∑

n=1

(f, cosn (·))2 + (f, sinn (·))2

≤ (f, f) =1π

∫ π

−π

|f (x)|2 dx <∞.

In particular this implies that

limn→∞

∫ π

−π

f (y) cosny dy = π limn→∞

bn = 0 and

limn→∞

∫ π

−π

f (y) sinny dy = π limn→∞

an = 0.

Since

sin((

N +12

)y

)= cos

(12y

)sin (Ny) + sin

(12y

)cos (Ny) ,

Eq. (5.14) now follows from the previous limit formulas with f replaced byf (y) cos 1

2y and f (y) sin 12y respectively.

Proof of the Fourier Convergence Theorem 5.8.. To concentrateon the basic ideas of the argument, I am only going to give the proof underthe additional assumption that f is continuously differentiable. The full proofmay be found in the book.


5.4 Proof of Theorem 5.8 71

Let

fN (x) =12a0 +

N∑n=1

[an cosnx+ bn sinnx] . (5.15)

We begin by deriving a more tractable form for the function fN (x) . To dothis notice that

an cosnx+ bn sinnx

=1π

(∫ π

−π

f (y) cosny dy)

cosnx+1π

(∫ π

−π

f (y) sinny dy)

sinnx

=1π

∫ π

−π

f (y) (cosny cosnx+ sinny sinnx) dy

=1π

∫ π

−π

f (y) cosn(x− y)dy

wherein the last equality we have used Eq. (3.14) with t = nx and s = −ny.Using this observation in Eq. (5.15) shows

fN (x) =12π

∫ π

−π

f (y) dy +1π

N∑n=1

∫ π

−π

f (y) cosn(x− y)dy

=1π

∫ π

−π

f (y)DN (x− y)dy, (5.16)

where

DN (θ) =12

+N∑

n=1

cosnθ =sin((N + 1

2

)θ)

2 sin(

12θ) , (5.17)

the second equality being Problem 14 of Section 32 of the book. (Also seeRemark 5.18 below.)

To see what DN (θ) looks like, see Figure 5.9.Making the change of variables, z = x− y in Eq. (5.16) and using the fact

that f and DN are 2π - periodic, we have

fN (x) =1π

∫ π

−π

f (y)DN (x− y)dy = − 1π

∫ x−π

x+π

f (x− z)DN (z)dz

=1π

∫ π+x

π−x

f (x− z)DN (z)dz =1π

∫ π

π

f (x− z)DN (z)dz.

Also notice that it follows from Eq. (5.17) that

1π

∫ π

−π

DN (y)dy = (1, DN )

=

(1,

12

+N∑

n=1

cosn (·)

)=

12

(1, 1) = 1.



Fig. 5.9. This is a plot D1 and D10.

Hence we may write

fN (x)− f (x) =1π

∫ π

π

[f (x− z)− f (x)]DN (z)dz

=1π

∫ π

π

f (x− z)− f (x)2 sin 1

2zsin((

N +12

)z

)dz

=1π

∫ π

π

g (z) sin((

N +12

)z

)dz (5.18)

where

g (z) :=

f(x−z)−f(x)

2 sin 12 z

if z 6= 0−f ′ (z) if z = 0.

Notice that, by l’Hopital’s rule, g is continuous and hence

1π

∫ π

π

|g (z)|2 dz <∞

and so we may let N →∞ in Eq. (5.18) with the aid of Lemma 5.16 to find

limN→∞

fN (x)− f (x) =1π

limN→∞

∫ π

π

g (z) sin((

N +12

)z

)dz = 0.

The following strengthens the convergence of the sum in Eq. (5.7) when: R→ R is piecewise C1.

Theorem 5.17 (Uniform Convergence of Fourier Series). If f : R→ Ris piecewise C1 – the convergence in Eq. (5.7) is uniform. To be explicit, ifwe let fN (x) be as in Eq. (5.15), i.e.



fN (x) =12πa0 +

1π

N∑n=1

[an cosnx+ bn sinnx] ,

thenmax

x|f (x)− fN (x)| → 0 as N →∞. (5.19)

Proof. We have

maxx|f (x)− fN (x)| = 1

πmax

x

∣∣∣∣∣∞∑

n=N+1

[an cosnx+ bn sinnx]

∣∣∣∣∣≤ 1π

maxx

∞∑n=N+1

|an cosnx+ bn sinnx|

≤ 1π

∞∑n=N+1

|an|+1π

∞∑n=N+1

|bn| . (5.20)

By the Cauchy-Schwarz inequality,

∞∑n=1

|an| =∞∑

n=1

n |an| ·1n≤

√√√√ ∞∑n=1

(n |an|)2 ·

( ∞∑n=1

1n2

). (5.21)

Finally, by an integration by parts (where the boundary terms vanish usingthe 2π - periodicity of the function f) we find

nan =1π

∫ π

−π

f (y)n cosny dy =1π

∫ π

−π

f (y)d

dycosny dy

= − 1π

∫ π

−π

f ′ (y) cosny dy = − (f ′, cos (n (·)))

Therefore by Bessel’s inequality, with f replaced by f ′, it follows that∑∞n=1 (n |an|)2 < ∞ and so by Eq. (5.21),

∑∞n=1 |an| < ∞. Similarly we

may also show that∑∞

n=1 |bn| < ∞ and hence it Eq. (5.19) follows from Eq.(5.20).

Remark 5.18 (The Dirichlet kernel for those who know complex variables).Recall Euler’s formula which states, eiθ = cos θ + i sin θ. With this notationwe have

DN (θ) =12

+N∑

n=1

cosnθ

=12

+N∑

n=1

12(einθ − e−inθ

)=

12

N∑n=−N

einθ.



Letting α = eiθ/2, we have

2DN (θ) =N∑

n=−N

α2n =α2(N+1) − α−2N

α2 − 1=α2N+1 − α−(2N+1)

α− α−1

=2i sin(N + 1

2 )θ2i sin 1

2θ=

sin(N + 12 )θ

sin 12θ

.

and therefore

DN (θ) :=12

N∑n=−N

einθ =sin(N + 1

2 )θ2 sin 1

2θ, (5.22)

with the understanding that the right side of this equation is N + 12 whenever

θ ∈ 2πZ.

5.4.1 Gibb’s Phenomenon

Problem 38.7: This problems deals with the Fourier expansion of the func-tion

f (x) =π

2(1−π<x<0 + 10<x<π) .

The expansion is given by

sN (x) = 2N∑

n=1

sin((2n− 1)x)2n− 1

= 2(

sinx1

+sin 3x

3+

sin 5x5

+sin 7x

7+

sin 9x9

+sin 11x

11. . .

)whose plots are The goal of this problem is to show how that convergence isnot uniform which is referred to as Gibb’s phenomenon in this setting.

(a) We have

2 sinx cos((2n− 1)x) = sin 2nx− sin((2n− 2)x).

So summing both sides of the above equation on n,

2 sinxN∑

n=1

cos((2n− 1)x) =N∑

n=1

[sin 2nx− sin((2n− 2)x)]

= sin 2x+ sin 4x+ · · ·+ sin((2N − 2)x) + sin 2Nx− sin 2x− sin 4x− · · · − sin((2N − 2)x)

= sin 2Nx.

Thus,

2N∑

n=1

cos((2n− 1)x) =sin 2Nx

sinx(5.23)



Fig. 5.10. Plot of s2.

Fig. 5.11. A plot of S6.

for x 6= kπ, k an integer. Since for 0 < x < π,

sN (x) = 2N∑

n=1

sin((2n− 1)x)2n− 1

,

we may then write

s′N (x) = 2N∑

n=1

cos(2n− 1)x) =sin 2Nx

sinx.

(b) To find local extrema in 0 < x < π, we of course must find thepoints in the interval at which s′N (x) = 0. This occurs when sin 2Nx = 0, at2Nx = 0,±π,±2π, . . .. So the extrema of sN (x) on the interval 0 < x < π



occur when

x =π

2N,

2π2N

,3π2N

, . . . ,(2N − 1)π

2N.

Thus the first extremum on the relevant interval occurs at x = π2N and since

s′′N (x) =(sinx)(2N cos 2Nx)− (sin 2Nx)(cosx)

sin2 x

s′′N (π

2N) =

2N cosπsin(

π2N

) = − 2Nsin(

π2N

) < 0,

we know a relative maximum occurs at this point.(c & d) Integrating Eq. (5.23) implies

sN (π/ (2N)) = 2N∑

n=1

sin((2n− 1)π/ (2N))2n− 1

= 2N∑

n=1

sin((2n− 1)x)2n− 1

|π/(2N)0

= 2N∑

n=1

∫ π/(2N)

0

cos((2n− 1)x)dx

=∫ π/(2N)

0

sin 2Nxsinx

dx

=∫ π/(2N)

0

sin 2Nxx

dx+∫ π/(2N)

0

(x− sinx)x sinx

sin 2Nxdx

=∫ π

0

sinxx

dx+∫ π/(2N)

0

(x− sinx)x sinx

sin 2Nxdx (5.24)

wherein we have used1

sinx− 1x

=(x− sinx)x sinx

.

Because ∣∣∣∣ (x− sinx)x sinx

sin 2Nx∣∣∣∣ ≤ ∣∣∣∣ (x− sinx)

x sinx

∣∣∣∣is bounded near 0, we may let N →∞ in Eq. (5.24) to find,

limN→∞

sN

( π

2N

)=∫ π

0

sinxx

dx =: σ ∼= 1.8519.

5.5 Fourier Series on Other Intervals

Suppose f (x) is defined for −c < x < c. By setting F (y) := f(

cπy), we get

a function defined for −π < y < π. This function may be expanded into aFourier series as


5.5 Fourier Series on Other Intervals 77

f( cπy)

= F (y) =12a0 +

∞∑n=1

[an cosny + bn sinny]

where

an :=1π

∫ π

−π

f( cπy)

cosny dy and

bn :=1π

∫ π

−π

f( cπy)

sinny dy.

Making the change of variables, x = cπy (or y = π

c x) in the above equationsgives

f (x) =12a0 +

∞∑n=1

[an cosn

π

cx+ bn sinn

π

cx]

where

an :=1c

∫ c

−c

f (x) cosnπ

cx dx

bn :=1c

∫ c

−c

f (x) sinnπ

cx dx.

The convergence properties of these sum are the same as those for the Fourierseries on (−c, c). Similarly if f (x) is defined on 0 < x < c then we have thesine and cosine series expansions

f (x) =∞∑

n=1

bn sinnπ

cx for 0 < x < c, and

f (x) =12a0 +

∞∑n=1

an cosnπ

cx for 0 < x < c,

where now

an :=2c

∫ c

0

f (x) cosnπ

cx dx

bn :=2c

∫ c

0

f (x) sinnπ

cx dx.

Example 5.19. We have seen

y =∞∑

n=1

2n

(−1)n+1 sinny for − π < y < π.

By letting y = πc x in the above formula, we conclude that

π

cx =

∞∑n=1

2n

(−1)n+1 sinnπ

cx for − c < x < c,



i.e.

x =∞∑

n=1

2cπn

(−1)n+1 sinnπ

cx for − c < x < c.

Suppose that we want the cosine expansion of x on the interval, 0 < x < c.In this case we will have

x =12a0 +

∞∑n=1

an cosnπ

cx

with

an =2c

∫ c

0

x cosnπ

cx dx =

2c

∫ c

0

x( c

πn

) d

dxsinn

π

cx dx

=2cx( c

πn

)sinn

π

cx|c0 −

2c

( c

πn

)∫ c

0

sinnπ

cx dx

=2c

( c

πn

)2

cosnπ

cx|c0 =

2c

( c

πn

)2

((−1)n − 1)

for n 6= 0 and

a0 =2c

∫ c

0

x dx =x2

c|c0 = c

Thus we havex =

c

2− 4cπ2

∑n=1,3,5,...

1n2

cosnπ

cx.

For example if c = 1, we have

x =12− 4π2

(cosπx+

132

cos 3πx+152

cos 5πx+ . . .

),

see Figure 5.19 below.

A plot of x and its Fourier cosine series approximation,12 −

4π2

(cosπx+ 1

32 cos 3πx+ 152 cos 5πx

).


5.5 Fourier Series on Other Intervals 79

This series is convergent for all 0 ≤ x ≤ c. Taking x = 0 in this expansion wemay conclude that

0 =12− 4π2

(1 +

132

+152

+ . . .

)or that (

1 +132

+152

+ . . .

)=π2

8

and taking x = 1 in this expansion we find

1 =12− 4π2

(−1− 1

32cos 3πx− 1

52cos 5πx+ . . .

)which again gives (

1 +132

+152

+ . . .

)=π2

8.


6

Boundary value problem examples

We will begin in this chapter by solving three problems related to the operatorL = d2

dx2 on [0,−π] with Neumann boundary conditions and the function

f (x) = x (π − x) = xπ − x2

Recall that the eigenfucntions/eigenvalues to this problem are

Xn (x) = cosnx with λn = −n2

for n ∈ N∪0 . Also we have and we have

f (x) ∼ a0

2+

∞∑n=1

anXn (x)

where

82 6 Boundary value problem examples

an =2π

∫ π

0

f (x)Xn (x) dx =2π

∫ π

0

x (π − x) cosnxdx

= − 2πn2

∫ π

0

x (π − x) d2

dx2cosnxdx

= − 2πn2

[(x (π − x) d

dxcosnx

)|π0 −

∫ π

0

(π − 2x)d

dxcosnxdx

]=

2πn2

[∫ π

0

(π − 2x)d

dxcosnxdx

]=

2πn2

[[(π − 2x) cosnx] |π0 −

∫ π

0

(−2) cosnxdx]

=2πn2

[[(π − 2x) cosnx] |π0 + 2

sinnxn|π0]

= −π 2πn2

[(−1)n + 1] = − 2n2

[(−1)n + 1]

anda0 =

2π

∫ π

0

x (π − x) dx =13π2.

Therefore we have

x (π − x) =π2

6− 2

∞∑n=1

1n2

[(−1)n + 1] cosnx

=π2

6− 4

(122

cos 2x+142

cos 4x+ . . .

)

Plot of π2

6 − 4(

122 cos 2x+ 1

42 cos 4x)

which is approximating x (π − x) .

Example 6.1.

utt = uxx with ux (t, 0) = 0 = ux (t, π) and

u(0, x) = 0 and ut (0, x) = xπ − x2.


6 Boundary value problem examples 83

The answer is given by

u (t, x) =sin√−Lt√−L

f (x) =sin√−Lt√−L

(a0

2X0 (x) +

∞∑n=1

anXn (x)

)

=

(a0

2t+

∞∑n=1

anXn (x)sinntn

)

=π2

6t− 2

∞∑n=1

1n3

[(−1)n + 1] cosnx sinnt

=π2

6t− 4

(123

cos 2x sin 2t+143

cos 4x sin 4t+ . . .

)Example 6.2.

ut = uxx withux (t, 0) = 0 = ux (t, π) and

u (0, x) = xπ − x2.

Answer is given by

u (t, x) = etLf (x) = etL

(a0

2X0 (x) +

∞∑n=1

anXn (x)

)

=

(a0

2etLX0 (x) +

∞∑n=1

anetLXn (x)

)

=

(a0

2+

∞∑n=1

ane−tn2

Xn (x)

)

=π2

6− 2

∞∑n=1

1n3

[(−1)n + 1] e−tn2cosnx

=π2

6− 4

(122e−t22

cos 2x+142e−t42

cos 4x+ . . .

)→ π2

6.

Notice that ∫ π

0

x (π − x) dx =16π3

because heat is conserved with Neumann boundary conditions.

Example 6.3. Here we take 0 < x < π and 0 < y < 3 and we wish to solve

uxx + uyy = 0,ux (0, y) = 0 = ux (π, y) ,u (x, 3) = 0 and u(x, 0) = x (π − x) .



In this case we use the separation of variables mantra and look for Y (y)such that Y (3) = 0 and so that un (x, y) = Xn (x)Y (y) solves Laplace’sequation. To do this we have

0 = ∆un = −n2Xn (x)Y (y) +Xn (x)Y ′′ (y)

from which we must demand that

Y ′′ (y)− n2Y (y) = 0

so thatY (y) = Aeny +Be−ny

and then requiring0 = Y (3) = Ae3n +Be−3n

shows thatB = −Ae6n

and thus thatY (y) = Aeny −Ae6ne−ny.

Taking A = − 12e−3n gives

Y (y) = sinhn (3− y) .

For n = 0 we have Y ′′ (y) = 0, so that Y (y) = Ay + B and to match theboundary condition we should take Y (y) = 3− y. Thus we look for a solutionof the form

u (x, y) =a0

2(3− y) +

∞∑n=1

an cosnx sinhn (3− y)

with

f (x) =π2

6− 2

∞∑n=1

1n2

[(−1)n + 1] cosnx

= u (x, 0) =a0

23 +

∞∑n=1

an cosnx sinhn3

and so

a0

23 =

π2

6or a0 =

π2

9and

an sinhn3 = −21n2

[(−1)n + 1] or an = −21

n2 sinhn3[(−1)n + 1] .

Thus the solution is given by

u (x, y) =π2

18(3− y)− 2

∞∑n=1

1n2

[(−1)n + 1] cosnxsinhn (3− y)

sinh 3n.


6 Boundary value problem examples 85

Example 6.4. Suppose now 0 ≤ x ≤ π and 0 ≤ y ≤ 1 and we wish to solveLaplace’s equation,

∆u (x, y) := uxx (x, y) + uyy (x, y) = 0 with (6.1)u (0, y) = u (π, y) = 0, u (x, 1) = 0 (6.2)

and u (x, 0) = f (x) . (6.3)

Solution: By separation of variable, one finds the functions

un (x, y) = sinnx · sinhn (1− y)

solve Eqs. (6.1) and (6.2) for all n ∈ N. By the superposition principle,

u (x, y) =∞∑

n=1

bnun (x, y) =∞∑

n=1

bn sinnx · sinhn (1− y) (6.4)

will also solve Eqs. (6.1) and (6.2). Finally we wish to chose the bn so that

f (x) =∞∑

n=1

bn sinnx · sinhn (1− y) |y=0 =∞∑

n=1

bn sinnx · sinhn.

But from the theory of the Fourier sine series, we know that

bn sinhn =2π

∫ π

0

f (x) sinnxdx

and therefore we must choose

bn =2

π sinhn

∫ π

0

f (y) sinnydy.

We thus find the solution to our problem is

u (x, y) =∞∑

n=1

(2

π sinhn

∫ π

0

f (y) sinnydy)

sinnx · sinhn (1− y) .

For example if f (x) = sin 2x− sin 5x, then we want

sin 2x− π sin 5x =∞∑

n=1

bn sinnx · sinhn

and hence that b2 sinh 2 = 1 and b5 sinh 5 = −π with all of the other bn = 0.Thus we find

u (x, y) =1

sinh 2sin 2x · sinh 2 (1− y)− π

sinh 5sin 5x · sinh 5 (1− y) .



Example 6.5 (Example 39.2 on p. 122). Here we wish to solve,

ut = kuxx with u (t, 0) = 0, u (t, π) = u0 and u(x, 0) = 0.

Notice that the boundary condition is not homogeneous. To take care of thiswe begin by finding the equilibrium solution, Φ (x) to the problem. Namely,Φ′′ (x) = 0 with Φ (0) = 0 and Φ (π) = u0. The answer is of course, Φ (x) =xπu0.

We now writeu (t, x) = U (t, x) + Φ (x)

and notice that

U (t, x) = u (t, x) = kuxx = k (Uxx (t, x) + Φ′′ (x)) = kUxx (t, x)

and U now has the boundary conditions,

U (t, 0) = 0, U (t, π) = 0 and U(x, 0) = −Φ (x) = −xπu0.

Hence the solution to our problem now is

U (t, x) = −ektLΦ (x) = −u0

πektLx.

Recalling that

x =∞∑

n=1

2n

(−1)n+1 sinnx = 2(

11

sinx− 12

sin 2x+13

sin 3x− 14

sin 4x+ . . .

)it now follows that

U (t, x) = −ektLΦ (x) = −u0

π

∞∑n=1

2n

(−1)n+1ektL sinnx

= −u0

π

∞∑n=1

2n

(−1)n+1e−ktn2

sinnx

and the full solution is given by

u (t, x) =x

πu0 −

u0

π

∞∑n=1

2n

(−1)n+1e−ktn2

sinnx

=u0

π

[x+ 2

∞∑n=1

(−1)n

ne−ktn2

sinnx

].


7

Boundary value generalities

7.1 Linear Algebra of the Strurm-Liouville EigenvalueProblem

Suppose L is a differential operator on functions on Ω = [a, b] of the form

Lf (x) :=1

p (x)d

dx

(κ (x)

d

dxf (x)

)+

1p (x)

q (x) f (x) . (7.1)

Define the inner product,

(f, g) :=∫

Ω=[a,b]

f (x) g (x) p (x) dx,

for functions f, g : Ω → R. With this notation we find

(Lf, g) =∫

Ω

[d

dx

(κ (x)

d

dxf (x)

)+ q (x) f (x)

]· g (x) dx

=∫

Ω

[−κ (x)

d

dxf (x) · d

dxg (x) + q (x) f (x) g (x)

]dx+ κ (x) f ′ (x) g (x) |ba

=∫

Ω

f (x) ·[d

dx

(κ (x)

d

dxg (x)

)+ q (x) g (x)

]dx

+ [κ (x) (f ′ (x) g (x)− f (x) g′ (x))]|ba . (7.2)

Hence we see that

(Lf, g)− (f, Lg) = [κ (x) (f ′ (x) g (x)− f (x) g′ (x))]|ba . (7.3)

Let (a1, a2) and (b1, b2) be two non-zero vectors in R2 and define

Bf (a) = a1f (a) + a2f′ (a) and

Bf (b) = b1f (b) + b2f′ (b) .

88 7 Boundary value generalities

In the sequel we will be interested on imposing the boundary conditionson Bf (a) = 0 = Bf (b) . If we assume that f and g satisfy these boundaryconditions in Eq. (7.3) then it follows that (f (a) , f ′ (a)) and (g (a) , g′ (a))line on the same line and therefore

W (f, g) (a) := f ′ (a) g (a)− f (a) g′ (a) = det[g (a) g′ (a)f (a) f ′ (a)

]= 0.

(W (f, g) is called the Wronskian of f and g.) Similar reasoning shows

f ′ (b) g (b)− f (b) g′ (b) = det[g (b) g′ (b)f (b) f ′ (b)

]= 0

and therefore it follows from Eq. (7.3) that L satisfies the symmetry condition,

(Lf, g) = (f, Lg) if Bf = 0 = Bg on ∂Ω = a, b .

It also worth noting that if κ (a) = κ (b) and f and g satisfy periodic bound-ary conditions; i.e.

f (a) = f (b) and f ′ (a) = f ′ (a) (7.4)

then we still have(Lf, g) = (f, Lg) .

As we will see we are going to be interested in the following eigenvalueproblem, namely we will be interested in finding solutions to the eigenfunctionequation,

Lf = −λf with Bf = 0 on ∂Ω.

This may be rewritten as

d

dx

[κ (x)

d

dxf (x)

]+ [q (x) + λp (x)] f (x) = 0 with Bf = 0 on ∂Ω.

This is the general form of the Sturm-Liouville eigenvalue problem as inChapter 6 of the book. There κ (x) = r (x) .

Let DB denote those functions f : Ω → R with are twice continuouslydifferentiable and satisfy the boundary conditions, Bf = 0 on ∂Ω and ifκ (a) = κ (b) let Dper denote those functions f : Ω → R with are twicecontinuously differentiable and satisfy periodic boundary conditions in Eq.(7.4).

The next result is the formal analogue of Corollary 1.4.

Proposition 7.1. If f, g ∈ DB or κ (a) = κ (b) and f, g ∈ Dper such thatLf = −λf and Lg = −µg with µ 6= λ, then (f, g) = 0.

Proof. The same proof as that given for Corollary 1.4 work here withoutchange.

The next theorem is an analogue of the spectral theorem for the operatorL.


7.1 Linear Algebra of the Strurm-Liouville Eigenvalue Problem 89

Theorem 7.2 (Strurm-Liouville Spectral Theorem). Let L be as above,assume κ > 0 on [a, b] and let D = DB or D = Dper (in which case weassume additionally that κ (b) = κ (a)), then there exists un ∈ D and λn ∈ Rsuch that:

1. −Lun = λnun for all n,2. the eigenvalues are increasing, i.e.

λ1 ≤ λ2 ≤ λ3 ≤ . . . ,

3. limn→∞ λn =∞ (in fact # n : λn ≤ a ∼ a1/2 or equivalently λn ∼ n2).4. Every “nice” function f on [a, b] may be expanded as

f (x) =∞∑

n=1

(f, un)un (x)

=∞∑

n=1

[∫ b

a

f (x)un (x) p (x) dx

]un (x) .

Example 7.3 (Problem 53.2 on p. 173). Consider the eigenfunction problem,

X ′′ + λX = 0 with X (0) = 0 and hX (1) +X ′ (1) = 0

with h > 0. Notice again that

−λ (X,X) = (X ′′, X) =∫ 1

0

X ′′ (x)X (x) dx

= −∫ 1

0

X ′ (x)X ′ (x) dx+X ′ (x)X (x) |10

= − (X ′, X ′)− hX (1)2 (7.5)

from which it follows that λ ≥ 0. Moreover if λ = 0, then X ′ = 0 and henceX ′ is constant. But since hX (1) + X ′ (1) = 0 it follows that X = 0, hencewe must have λ > 0. Therefore, X (x) = sinαx for some α. Implementing thelast boundary condition then gives,

h sinα+ α cosα = 0

or equivalently thattanα = −α/h.

Let αn denote the solutions to this equation, see Figure 7.3 below.



The x – coordinates of the points of intersection of the line with the curvesgives the values of α when h = 1.

So we have

Xn (x) = Cn sinαnx where tanαn = −αn/h with αn > 0 and λn = α2n.

Let us now choose Cn so that (Xn, Xn) = 1. To this end we have

(Xn, Xn) = C2n

∫ 1

0

[sinαnx]2dx.

Now recall that cos 2θ = 1− 2 sin2 θ, and therefore,

sin2 θ =1− cos 2θ

2.

Using this above gives∫ 1

0

[sinαnx]2dx =

12

∫ 1

0

(1− cos 2αnx) dx

=12

(1− sin 2αnx

2αn|10)

=12

(1− sin 2αn

2αn

)=

12

(1− sinαn cosαn

αn

).

Since tanαn = −αn/h, we have

sinαn = −αn

hcosαn (7.6)

and therefore,∫ 1

0

[sinαnx]2dx =

12

(1 +

cos2 αn

h

)=

12h(h+ cos2 αn

).

So we need to choose Cn so that


7.2 General Elliptic PDE Theory 91

C2n

h+ cos2 αn

2h= 1,

i.e.

Cn =√

2hh+ cos2 αn

and hence the final answer is given by

Xn (x) =√

2hh+ cos2 αn

sinαnx where tanαn = −αn/h.

Here αn > 0 and λn = α2n.

Alternative way to compute∫ 1

0[sinαnx]

2dx. From Eq. (7.5),

−α2n

∫ 1

0

[sinαnx]2dx = −

∫ 1

0

[αn cosαnx]2dx− h sin2 αn

and therefore∫ 1

0

[sinαnx]2dx =

∫ 1

0

[cosαnx]2dx+

h

α2n

sin2 αn

=∫ 1

0

[1− sin2 αnx

]dx+

h

α2n

sin2 αn.

From this equation and Eq. (7.6) it follows that

2∫ 1

0

[sinαnx]2dx = 1 +

h

α2n

sin2 αn = 1 +h

α2n

(−αn

hcosαn

)2

= 1 +cos2 αn

h=h+ cos2 αn

h

and therefore that ∫ 1

0

[sinαnx]2dx =

h+ cos2 αn

2has before.

7.2 General Elliptic PDE Theory

In this section we will state the generalization of theorem 7.2 to higher di-mensional situations. Let Ω be a “nice” open bounded region in Rd (typicallywe have in mind d = 1, 2, 3 here). Suppose that κi,j (x) and p (x) be smoothfunctions on Ω such that matrix κ (x)

κ (x) :=

κ11 (x) κ12 (x) . . . κ1d (x)κ21 (x) κ22 (x) . . . κ2d (x)

.... . .

...κd1 (x) κd2 (x) . . . κdd (x)



Fig. 7.1. A region represnting some material of some region in space.

is positive definite, i.e. κtr = κ and κv · v > 0 for all v ∈ Rd. We now form theinner product

(u, v) :=∫

Ω

u (x) v (x) p (x) dV

for function u, v on Ω and let

Lu :=d∑

i,j=1

1p∂i (κij∂ju) + γu. (7.7)

Example 7.4. Look at the formula for the Laplacian in polar, cylindrical, andspherical coordinates to find natural operators written in this form. (Thereare some singularities involved here which are artifices of these coordinatessystem. We will have to deal with them later.) The general form of the heatequation also produced such operators L as in Eq. (7.7) above.

By the divergence Theorem 1.9,

(Lu, v) =∫

Ω

d∑i,j=1

1p∂i (κij∂ju) + γu

vpdV=∫

Ω

d∑i,j=1

∂i (κij∂ju) · v + γuvp

dV=∫

Ω

− d∑i,j=1

κij∂ju∂iv + γuvp

dV +∫

∂Ω

d∑i,j=1

(κij∂ju)nivdσ

=∫

Ω

[−κ∇u · ∇v + γuv] p · dV +∫

∂Ω

(κ∇u · n) v dσ

where n (x) is the outward pointing normal and dσ is the surface measure on∂Ω. By interchanging the roles of u and v in the above formula, it follows that



(u, Lv) = −∫

Ω

[∇u · κ∇v + γuv] dV +∫

∂Ω

(κ∇v · n)u dσ

= −∫

Ω

[κ∇u · ∇v + γuv] dV +∫

∂Ω

(κ∇v · n)u dσ

and therefore,

(Lu, v)− (u, Lv) =∫

∂Ω

[(κ∇u · n) v − (κ∇v · n)u] dσ. (7.8)

Let us now further suppose λ is a given function on ∂Ω and u and v satisfythe boundary conditions,

Bu (x) := κ (x)∇u (x) · n (x) + α (x)u (x) set= 0

where we allow for α (x) = ∞ by which we mean u (x) = 0 at such points.Then using these boundary condition in Eq. (7.8) shows that

[(κ∇u · n) v − (κ∇v · n)u] = αvu− αuv = 0 on ∂Ω

and hence we have

(Lu, v) = (u, Lv) whenever Bu = Bv = 0.

The following is an analogue of the spectral theorem for matrices in thiscontext.

Theorem 7.5. Keeping the above set up, there exists an orthonormal setun∞n=1 of eigenvectors for (−L,B) , i.e. Bun = 0 on ∂Ω and −Lun = λnun.Moreover these may be chosen so that:

1. the eigenvalues are increasing, i.e.

λ1 ≤ λ2 ≤ λ3 ≤ . . . ,

2. limn→∞ λn = ∞ (in fact # n : λn ≤ a ∼ ad/2 or equivalently λn ∼n2/d).

3. Every “nice” function f on Ω may be expanded as

f (x) =∞∑

n=1

(f, un)un (x)

=∞∑

n=1

[∫Ω

f (x)un (x) p (x) dV]un (x) .

Example 7.6. Let us find the eigenfunctions u : Ω = [0, a] × [0, b] → R suchthat

∆u = λu and u = 0 on ∂Ω.



As usual, let us try to find solutions of the form

u (x, y) = X (x)Y (y) .

In order for this to work we must have

λ =u

u=∆u

u=X ′′ (x)X (x)

+Y ′′ (y)Y (y)

and therefore,

X ′′ (x)X (x)

= k andY ′′ (y)Y (y)

= λ− k with

X (0) = 0 = X (a) and Y (0) = 0 = Y (b)

for some constant k. The solutions to these one-dimensional Strum-Liouvilleproblems are already known, namely we must have

X (x) = sinmπ

ax and Y (y) = sinn

π

by

for some integers m,n. The functions,

um,n (x, y) := sinmπ

ax sinn

π

by

satisfy∆um,n = λm,num,n with um,n = 0 on ∂Ω

where

λm,n := −π2

[(ma

)2

+(nb

)2].

All of these functions are orthogonal and in fact normalized if we define

(f, g) :=2a· 2b

∫ a

0

dx

∫ b

0

dyf (x, y) g (x, y)

and we may expand f as

f =∞∑

m,n=1

Bm,num,n

where

Bm,n = (f, um,n) =2a· 2b

∫ a

0

dx

∫ b

0

dyf (x, y)um,n (x, y) .

To see this is correct, notice that

f (x, y) =∞∑

n=1

(2b

∫ b

0

f (x, y) sinnπ

bydy

)sinn

π

by (7.9)



and

f (x, y) =∞∑

m=1

(2a

∫ a

0

f (x, y) sinmπ

axdx

)sinm

π

ax. (7.10)

Formally inserting Eq. (7.10) into Eq. 7.9) shows

f (x, y) =∞∑

n=1

(2b

∫ b

0

dy

[ ∞∑m=1

(2a

∫ a

0

f (x, y) sinmπ

axdx

)sinm

π

ax

]sinn

π

by

)sinn

π

by

=∞∑

n=1

∞∑m=1

(2b

∫ b

0

dy2a

∫ a

0

dxf (x, y) sinmπ

ax sinn

π

by

)sinm

π

ax sinn

π

by

=∞∑

m,n=1

Bm,num,n (x, y) .

It also interesting to observe that if ∆u = λu with u = 0 on ∂Ω and welet

vm (y) :=2a

∫ a

0

u (x, y) sinmπ

axdx,

then

λvm (y) = λ2a

∫ a

0

u (x, y) sinmπ

axdx =

2a

∫ a

0

∆u (x, y) sinmπ

axdx

=2a

∫ a

0

∂2xu (x, y) sinm

π

axdx+

2a

∫ a

0

∂2yu (x, y) sinm

π

axdx

= −(mπ

a

)2 2a

∫ a

0

u (x, y) sinmπ

axdx+

2a∂2

y

∫ a

0

u (x, y) sinmπ

axdx

= −(mπ

a

)2

vm (y) + ∂2yvm (y)

and hence

∂2yvm (y) =

[λ+

(mπ

a

)2]vm (y) with vm (0) = vm (b) = 0.

Since vm (y) is an eigenfunction of ∂2y with Dirichlet boundary conditions, we

must have vm (y) = Cm sinnmπb y for some constant Cm and some integer nm

such thatλ+ π2

(ma

)2

= −π2(nm

b

)2

,

i.e. we must have

λ = −π2

[(ma

)2

+(nm

b

)2].

From all of this it follows that



u (x, y) =∞∑

m=1

vm (y) sinmπ

ax

=∞∑

m=1

Cm sinnmπ

by sinm

π

ax,

with only a finite number of non-zero terms in the above sum. This shows thatthe functions um,n : m,n ∈ N forms a basis for all of the eigenfunctions of∆ with Dirichlet boundary conditions on Ω.


8

PDE Applications and Duhamel’s Principle

8.1 Interpretation of d’Alembert’s solution to the 1-dwave equation

Example 8.1. We may use d’Alembert’s solution to the wave equation to for-

mally work out the meaning of cos(t√−a2∂2

x

)and

sin(

t√−a2∂2

x

)√−a2∂2

x

. To see what

we should get, let A2 = −a2∂2x and A =

√−a2∂2

x then

y (t, x) = cos(t√−a2∂2

x

)f (x) +

sin(t√−a2∂2

x

)√−a2∂2

x

g (x) for −∞ < x <∞

should solve Eqs. (2.5) and (2.6). By comparing this with Eq. (2.7),d’Alembert’s solution which I recall here,

y(t, x) =12

[f(x+ at) + f(x− at)] +12a

∫ x+at

x−at

g(s)ds,

we conclude that

cos(t√−a2∂2

x

)f(x) =

12

[f(x+ at) + f(x− at)] and (8.1)

sin(t√−a2∂2

x

)√−a2∂2

x

g(x) =12a

∫ x+at

x−at

g(s)ds. (8.2)

We will use these results and the results of the next section to allow of tosolve the forced wave equation.

8.2 Solving 1st - order equations using 2nd - ordersolutions

Lemma 8.2 (A Key Fourier Transform Formula). For all λ ∈ R andt > 0,

98 8 PDE Applications and Duhamel’s Principle∫ ∞

−∞

e−14t s2

√4πt

cos (λs) ds = e−tλ2. (8.3)

Proof. Fix t > 0 and let

g (λ) :=∫ ∞

−∞

e−14t s2

√4πt

cos (λs) ds.

Then

g′ (λ) = −∫ ∞

−∞

e−14t s2

√4πt

s sin (λs) ds

= 2t∫ ∞

−∞

d

ds

e−14t s2

√4πt

· sin (λs) ds

= −2t∫ ∞

−∞

e−14t s2

√4πt

· dds

sin (λs) ds

= −2tλ∫ ∞

−∞

e−14t s2

√4πt

· cos (λs) ds = −2tλg (λ) .

Solving this ODE for g gives,

g (λ) = e−tλ2g (0) .

This completes the proof since g (0) = 1 as we now show. Letting s =√tx,

g2 (0) =

(∫ ∞

−∞

e−14t s2

√4πt

ds

)2

=

(∫ ∞

−∞

e−14 x2

√4π

dx

)2

=∫ ∞

−∞

e−14 x2

√4π

dx

∫ ∞

−∞

e−14 y2

√4π

dy =∫∫

R2

e−14 (x2+y2)

4πdxdy

=∫ ∞

0

∫ 2π

0

e−14 r2

4πrdrdθ =

12

∫ ∞

0

e−14 r2rdr = −e− 1

4 r2|∞0 = 1.

wherein the fifth equality we have gone to polar coordinates.

Theorem 8.3 (Solving for etA via cos(√−At

)). Suppose A is a N × N

symmetric matrix with all non-positive eigenvalues. Then

etA =∫ ∞

−∞

e−14t s2

√4πt

cos(√−As

)ds for all t > 0. (8.4)

Proof. Formally we are taking λ =√−A in Eq. (8.3). To rigorously

prove Eq. (8.4), let viNi=1 be an orthonormal basis of eigenvectors for A. Byassumption we may write eigenvalue for vi as −λ2

i , i.e. Avi = −λ2i vi.

Then


8.2 Solving 1st - order equations using 2nd - order solutions 99∫ ∞

−∞

e−14t s2

√4πt

cos(√−As

)vids =

∫ ∞

−∞

e−14t s2

√4πt

cos (λis) vids

= e−tλ2i vi = etAvi

and the result follows since both sides of Eq. (8.4) are linear.

8.2.1 The Solution to the Heat Equation on R

Example 8.4 (Heat Equation). Let us try to formally use Theorem 8.3 to solvethe heat equation,

ut (t, x) = ∂2xu (t, x) with u (0, x) = f (x) .

According to theorem 8.3 and Example 8.1 with a = 1, the solution shouldbe given by

u (t, x) =(et∂2

xf)

(x) =∫ ∞

−∞

e−14t s2

√4πt

cos(√−∂2

xs)f (x) ds

=∫ ∞

−∞

e−14t s2

√4πt

12

[f(x+ s) + f(x− s)] ds

=∫ ∞

−∞f(x− s)e

− 14t s2

√4πt

ds =∫ ∞

−∞f(y)

e−14t (x−y)2

√4πt

dy.

Exercise 8.1. Suppose f is a bounded continuous function, show

u (t, x) :=∫ ∞

−∞f(y)

e−14t (x−y)2

√4πt

dy =∫ ∞

−∞f(y)p (t, x− y) dy (8.5)

solves the heat equation, ut (t, x) = ∂2xu (t, x) for t > 0 where

p (t, x) :=e−

14t x2

√4πt

. (8.6)

Hint: first show p (t, x) solves the heat equation for t > 0. Then check usolves the heat equation by differentiating past the integral, which you shouldassume to be valid here.

It is a fact that we will discuss later that

limt↓0

u (t, x) = f (x) . (8.7)

This is based on the idea that p (t, x) is approximating a “δ – function,” seeFigure 8.1

We will abbreviate all this by the suggestive formula,(et∂2

xf)

(x) =∫ ∞

−∞f(y)p (t, x− y) dy. (8.8)


100 8 PDE Applications and Duhamel’s Principle

Fig. 8.1. Plots of x → p (t, x) for t = 2, t = 14, t = 1

32and t = 1

64. Notice that p (t, x)

is being more and more concentrated near x = 0 as t ↓ 0 while always keeping thetotal area under x → p (t, x) equal to one.

8.3 Duhamel’s Principle

Theorem 8.5 (Duhamel’s Principle I). Suppose A is an N ×N, f ∈ RN

and h (t) ∈ RN be given. Then the ordinary differential equation,

u (t) = Au (t) + h (t) with (8.9)u (0) = f (8.10)

has a unique solution given by

u (t) = etAf +∫ t

0

e(t−τ)Ah (τ) dτ. (8.11)

In words, u (t) is constructed by “adding” together the solutions to a bunch ofinitial value problems where h ≡ 0. Namely, etAf is the solution to

u (t) = Au (t) with u (0) = f

while e(t−τ)Ah (τ) is the solution to

u (t) = Au (t) with u (τ) = h (τ) .

Hence we have

u (t) =(

solution at time t tou (t) = Au (t) & u (0) = f

)+∫ t

0

(solution at time t− τ to

u (t) = Au (t) & u (0) = h (τ)

)dτ.


8.3 Duhamel’s Principle 101


u (t) =(

solution at time t tou (t) = Au (t) & u (0) = f

)+∫ t

0

(solution at time t to

u (t) = Au (t) & u (τ) = h (τ)

)dτ.

Proof. Suppose u solves Eq. (8.9), then by the product rule

d

dt

[e−tAu (t)

]= −Ae−tAu (t) + e−tAu (t)

= −e−tAAu (t) + e−tA (Au (t) + h (t))

= e−tAh (t) .

Integrating this equation then shows

e−tAu (t) = u (0) +∫ t

0

e−τAh (τ) dτ = f +∫ t

0

e−τAh (τ) dτ.

Multiplying this equation by etA on the left shows that if u exists it must begiven by Eq. (8.11).

To prove existence, let u now be defined by Eq. (8.11) and notice that wemay write it as

u (t) = etA

[f +

∫ t

0

e−τAh (τ) dτ].

Thus u (0) = f and, by the product rule and the fundamental theorem ofcalculus,

u (t) = AetA

[f +

∫ t

0

e−τAh (τ) dτ]

+ etA[e−tAh (t)

]= Au (t) + h (t) .


A :=

1 7 −27 1 −2−2 −2 10


v1 :=

−110

↔ −6, v2 :=

111

↔ 6, v3 :=

−1−12

↔ 12.

We wish to solve,



u (t) = Au (t) + tf with u (0) = 0

wheref = (1, 2, 3)tr =

12v1 + 2v2 +

12v3

The solution is

u (t) =∫ t

0

τe(t−τ)Afdτ

=∫ t

0

τ

(12e−6(t−τ)v1 + 2e6(t−τ)v2 +

12e12(t−τ)v3

)dτ.

To work this out we notice that∫ t

0

τe−λτdτ = − d

dλ

∫ t

0

e−λτdτ = − d

dλ

[e−λτ

−λ|τ=tτ=0

]=

d

dλ

e−λt − 1λ

=1λ2

(1− (tλ+ 1) e−tλ

)and therefore, ∫ t

0

τeλ(t−τ)dτ = eλt 1λ2

(1− (tλ+ 1) e−tλ

)=

1λ2

(eλt − (1 + tλ)

).

Hence the answer is given by

u (t) =(

12·62

(e−6t − 1 + 6t

)v1 + 2

62

(e6t − 1− 6t

)v2

+ 12·122

(e12t − 1− 12t

)v3

).

Theorem 8.7 (Duhamel’ s Principle II). Suppose A is an N ×N, f, g ∈RN and h (t) ∈ RN be given. Then the ordinary differential equation,

u (t) = Au (t) + h (t) with (8.12)u (0) = f and u (0) = g (8.13)

has a unique solution given by

u (t) =(cos√−At

)f +

sin√−At√−A

g +∫ t

0

sin√−A (t− τ)√−A

h (τ) dτ. (8.14)

Again, in words, u (t) is constructed by “adding” the solutions to a bunch ofinitial value problems where h ≡ 0. Namely,(

cos√−At

)f +

sin√−At√−A

g



is the solution to

u (t) = Au (t) with u (0) = f and u (0) = g

while sin√−A(t−τ)√−A

h (τ) is the solution to

u (t) = Au (t) with u (τ) = 0 and u (τ) = h (τ)

so

u (t) =

solution at time t tou (t) = Au (t) withu (0) = f & u (0) = g

+∫ t

0

solution at time t tou (t) = Au (t) with

u (τ) = 0 & u (τ) = h (τ)

dτ.

Proof. The best way to understand this theorem is to reduce it to the firstversion of Duhamel’s principle in Theorem 8.5. To this end, let v (t) = u (t) ,

then the pair(u (t)v (t)

)∈ RN × RN solves the equation,

d

dt

(u (t)v (t)

)=(u (t)v (t)

)=(v (t)u (t)

)=(

v (t)Au (t) + h (t)

)=(

0 IA 0

)(u (t)v (t)

)+(

0h (t)

).

Let

B :=(

0 IA 0

)– a 2N × 2N matrix. Then by Theorem 8.5, we have(

u (t)v (t)

)= etB

(fg

)+∫ t

0

e(t−τ)B

(0

h (τ)

)dτ. (8.15)

When h ≡ 0, we know that

u (t) =(cos√−At

)f +

sin√−At√−A

g

from which we deduce

etB

(fg

)=

((cos√−At

)f + sin

√−At√−A

g

∗

). (8.16)



The second component of etB

(fg

)is easily found as well (just differentiate

the first component) but we will not need it. Because of Eq. (8.16),

e(t−τ)B

(0

h (τ)

)=

(sin

√−A(t−τ)√−A

h (τ)∗

). (8.17)

Hence taking the first component of Eq. (8.15), using Eqs. (8.16) and (8.17),gives Eq. (8.14).

We may also directly check that Eq. (8.14) works. Indeed, for simplicityassume f = 0 = g so that

u (t) =∫ t

0


h (τ) dτ.

Then u (0) = 0,

u (t) =sin√−A (t− τ)√−A

h (τ) |τ=t +∫ t

0

d

dt


h (τ) dτ

=∫ t

0

cos√−A (t− τ)h (τ) dτ

(so u (0) = 0) and similarly,

u (t) = cos√−A (t− τ)h (τ) |τ=t +

∫ t

0

d

dtcos√−A (t− τ)h (τ) dτ

= h (t) +∫ t

0

−√−A sin

√−A (t− τ)h (τ) dτ

= h (t)−(√−A)2∫ t

0


h (τ) dτ

= h (t) +Au (t)

as desired.


A :=

1 7 −27 1 −2−2 −2 10


v1 :=

−110

↔ −6, v2 :=

111

↔ 6, v3 :=

−1−12

↔ 12.



We wish to solve,

u (t) = Au (t) + h with u (0) = 0 = u (0)

whereh = (1, 2, 3)tr =

12v1 + 2v2 +

12v3.

The solution is given by

u (t) =∫ t

0


hdτ

=∫ t

0


(12v1 + 2v2 +

12v3

)dτ

=∫ t

0

sin√

6(t−τ)√6

12v1 + sinh

√6(t−τ)√6

2v2

+ sinh√

12(t−τ)√12

12v3

dτ.

Now ∫ t

0

sin a (t− τ)a

dτ =cos a (t− τ)

a2|τ=tτ=0 =

1− cos ata2

and similarly,∫ t

0

sinh a (t− τ)a

dτ = −cosh a (t− τ)a2

|τ=tτ=0 = −1− cosh at

a2

so that

u (t) =1− cos

√6t

612v1 +

cosh√

6t− 16

2v2 +cosh

√12t− 1

1212v3.

Let us do a quick check that this solution is correct. For example let us checkthat

u (t) =1− cos

√6t

6v1

solves u (t) = Au (t) + v1. This is the case since,

u (t)−Au (t) = cos√

6t · v1 −Au (t)

= cos√

6t · v1 −1− cos

√6t

6Av1

= cos√

6t · v1 +(1− cos

√6t)v1 = v1

as desired.

Exercise 8.2. Let

A :=

−2 1 11 −2 11 1 −2



and h = (−1, 1, 0)tr . Solve the following equations for u,

u (t) = Au (t) + h with u (0) = 0 = (0, 0, 0)tr andu (t) = Au (t) + h with u (0) = 0 and u (0) = 0.

Write your solutions in the form

u (t) =3∑

i=1

ai (t) vi

where the functions ai are to be determined.

8.4 Application of Duhamel’s principle to 1 - d wave andheat equations

Using the formulas for cos(t√−∂2

x

)and

sin(

t√−∂2

x

)√−∂2

x

in Eqs. (8.1) and (8.2)

respectively we are now in a position to formally apply Duhamel’s principlein order to solve the forced wave equation;

utt = uxx + h with u(0, ·) = f and ut(0, ·) = g. (8.18)

Theorem 8.9. If f ∈ C2(R,R) and g ∈ C1(R,R), and h ∈ C(R2,R) suchthat hx exists and hx ∈ C(R2,R), then Eq. (8.18) has a unique solution u(t, x)given by Eq. (8.19).

Proof. By a formal application of Theorem 8.5 with A = ∂2x suggest that

u(t, ·) = cos(t√−∂2

x)f +sin(t

√−∂2

x)√−∂2

x

g +∫ t

0

sin((t− τ)√−∂2

x)√−∂2

x

h(τ, ·)dτ.

Moreover using the formulas in Eqs. (8.1) and (8.2) then implies

u(t, x) =12

[f(x+ t) + f(x− t)]+12

∫ t

−t

g(x+s)ds+12

∫ t

0

dτ

∫ x+t−τ

x−t+τ

dy h(τ, y).

(8.19)To verify that u defined in Eq. (8.19) satisfies Eq. (8.18) it suffices (by whatwe have already done) to assume f = g = 0 so that

u(t, x) =12

∫ t

0

dτ

∫ x+t−τ

x−t+τ

dy h(τ, y).

Now


8.4 Application of Duhamel’s principle to 1 - d wave and heat equations 107

ut =12

∫ t

0

[h(τ, x+ t− τ) + h(τ, x− t+ τ)] dτ,

utt =12

∫ t

0

[hx(τ, x+ t− τ)− hx(τ, x− t+ τ)] dτ + h(t, x)

ux(t, x) =12

∫ t

0

dτ [h(τ, x+ t− τ)− h(τ, x− t+ τ)] and

uxx(t, x) =12

∫ t

0

dτ [hx(τ, x+ t− τ)− hx(τ, x− t+ τ)]

so that utt − uxx = h and u(0, x) = ut(0, x) = 0.The only thing left to prove is the uniqueness assertion. For this suppose

that v is another solution, then (u− v) solves the wave equation (8.18) withf = g = 0 and hence by the uniqueness assertion in Theorem 2.4 (with a = 1),u− v ≡ 0.

Similarly we may solve the forced heat equation as well.

Theorem 8.10 (The Forced Heat Equation). Suppose that f : R→ Rand h : R2 → R are bounded continuous functions, then the function

u (t, x) =∫ ∞

−∞p (t, x− y) f (y) dy +

∫ t

0

dτ

∫ ∞

−∞p (t− τ, x− y)h (τ, y) dy

(8.20)solves the heat equation

ut = uxx + h with limt↓0

u(t, x) = f (x) . (8.21)

Proof. Formally applying Theorem 8.5 with A = ∂2x suggests that

u (t, x) = et∂2xf (x) +

∫ t

0

[e(t−τ)∂2

xh (τ, ·)](x) dτ.

In light of Eq. (8.8), this equation then gives rise to Eq. (8.20). It is of coursepossible to directly check that Eq. (8.20) solves Eq. (8.21), however I will notstop to do it here.

Example 8.11 (Problem 40.4 on p. 128). Solve the following problem on 0 <x < π,

ut = kuxx with u (0, x) = 0 andu (t, 0) = 0 and u (t, π) = F (t) .

where F (0) = 0.To do this, let ϕ (t, x) = x

πF (t) and then let U (t, x) = u (t, x) − ϕ (t, x) .We then have


108 8 PDE Applications and Duhamel’s Principle(∂t − k∂2

x

)U =

(∂t − k∂2

x

)[u (t, x)− ϕ (t, x)] = −x

πF (t) with

U (0, x) = −ϕ (0, x) = 0

Hence the solution to the problem is

U (t, x) = −∫ t

0

e(t−τ)L x

πF (τ) dτ

= −e(t−τ)L x

πF (τ) |t0 +

∫ t

0

F (τ)d

dτe(t−τ)L x

πdτ

= −xπF (t)−

∫ t

0

F (τ)Le(t−τ)L x

πdτ

and therefore,

u (t, x) =x

πF (t)− x

πF (t)−

∫ t

0

F (τ)Le(t−τ)L x

πdτ

= −∫ t

0

F (τ)Le(t−τ)L x

πdτ.

Recalling that

x =∞∑

n=1

2n

(−1)n+1 sinnx

we have

Le(t−τ)Lx =∞∑

n=1

2n

(−1)n+1Le(t−τ)L sinnx

=∞∑

n=1

2n

(−1)n+1 (−kn2)e−(t−τ)kn2

sinnx

we get

u (t, x) =∞∑

n=1

2nπ

(−1)n (−kn2)(∫ t

0

e−(t−τ)kn2F (τ) dτ

)sinnx

=2kπ

∞∑n=1

(−1)n+1

(∫ t

0


)n sinnx.

Example 8.12 (Problem 40.4 on p. 128). Solve the following problem on 0 <x < π,

ut = kuxx with u (0, x) = 0 andu (t, 0) = 0 and u (t, π) = F (t) .

where F (0) = 0.



To do this, let ϕ (t, x) = xπF (t) and then let U (t, x) = u (t, x) − ϕ (t, x) .

We then have(∂t − k∂2

x

)U =

(∂t − k∂2

x

)[u (t, x)− ϕ (t, x)] = −x

πF (t) with

U (0, x) = −ϕ (0, x) = 0U (t, 0) = 0 = U (t, π) .


U (t, x) = −∫ t

0

e(t−τ)L x

πF (τ) dτ

= −e(t−τ)L x

πF (τ) |t0 +

∫ t

0

F (τ)d

dτe(t−τ)L x

πdτ

= −xπF (t)−

∫ t

0

F (τ)Le(t−τ)L x

πdτ

and therefore,

u (t, x) =x

πF (t)− x

πF (t)−

∫ t

0

F (τ)Le(t−τ)L x

πdτ

= −∫ t

0

F (τ)Le(t−τ)L x

πdτ.

Recalling that

x =∞∑

n=1

2n

(−1)n+1 sinnx

we have

Le(t−τ)Lx =∞∑

n=1

2n

(−1)n+1Le(t−τ)L sinnx

=∞∑

n=1

2n

(−1)n+1 (−kn2)e−(t−τ)kn2

sinnx

we get

u (t, x) =∞∑

n=1

2nπ

(−1)n (−kn2)(∫ t

0


)sinnx

=2kπ

∞∑n=1

(−1)n+1

(∫ t

0


)n sinnx.

Example 8.13 (Forced Wave Equation problem). Consider the forced waveequation,



utt = uxx + sinx

2with u (t, 0) = ux (t, π) = 0

u (0, x) = f (x) and ux (0, x) = g (x) .

Letting L = d2

dx2 acting on functions satisfying h (0) = 0 = h′ (π) , the formalsolution to our problems is given by

u (t, x) = cos(t√−L)f (x) +

sin(t√−L)

√−L

g (x)

+∫ t

0

sin((t− τ)

√−L)

√−L

(sin

x

2

)dτ.

In order to work this out we need to recall (see Problem 53.1 on p. 173) thatthe eigenfunctions

ϕn (x) = sinαnx where αn =2n− 1

2for n ∈ N

and Lϕn = −α2nϕn and∫ π

0

sin2 αnxdx =∫ π

0

1− cos 2αnx

2dx =

π

2− 1

2sin 2αnx

2αn|π0 =

π

2.

Thus we have

f (x) =∞∑

n=1

fn sinαnx and g (x) =∞∑

n=1

gn sinαnx

where

fn =2π

∫ π

0

f (x) sinαnxdx and gn =2π

∫ π

0

g (x) sinαnxdx.

The solution to our problem is thus

u (t, x) =∞∑

n=1

fn cos(t√−L)

sinαnx+∞∑

n=1

gn

sin(t√−L)

√−L

sinαnx

+∫ t

0

sin ((t− τ) /2)1/2

(sin

x

2

)dτ

=∞∑

n=1

fn cos (αnt) sinαnx+∞∑

n=1

gnsin (αnt)αn

sinαnx

− 4 cos ((t− τ) /2) |t0(sin

x

2

)=

∞∑n=1

fn cos (αnt) sinαnx+∞∑

n=1

gnsin (αnt)αn

sinαnx+ 4[1− cos

t

2

]sin

x

2.



Let us check that

v (t, x) = 4[1− cos

t

2

]sin

x

2does satisfy the correct equation. It is easily seen that v (0, x) = v (0, x) = 0and

vtt − vxx = 4(

14

cost

2sin

x

2+

14

[1− cos

t

2

]sin

x

2

)= sin

x

2as required.

Example 8.14 (Ressonance in the Wave Equation). Consider now the forcedequation with a time dependent forcing, namely

utt = uxx + sint

2sin

x

2with u (t, 0) = ux (t, π) = 0

u (0, x) = 0 and ux (0, x) = 0.

The solution is given by

u (t, x) =∫ t

0

sin((t− τ)

√−L)

√−L

(sin

τ

2sin

x

2

)dτ

=∫ t

0

sin((t− τ) 1

2

)12

(sin

τ

2sin

x

2

)dτ

= 2∫ t

0

sin (t− τ) 12

sinτ

2dτ sin

x

2

= 2(

sin12t− 1

2t cos

12t

)sin

x

2

=(

2 sin12t− t cos

12t

)sin

x

2

wherein we have computed this integral using the identity,

sinA sinB =12

(cos (A−B)− cos (A+B))

with A = (t− τ) 12 and B = τ/2 to find

2∫ t

0

sin(t− τ

2

)sin

τ

2dτ =

∫ t

0

(cos(t

2− τ)− cos

(t

2

))dτ

=(− sin

(t

2− τ)|t0 − t cos

t

2

)= 2 sin

t

2− t cos

t

2.

This is an example the phenomenon of resonance, namely, the amplitude ofu (t, x) can be arbitrarily large for most values of large t.



Proposition 8.15. The solution to

utt = kuxx with u (0, x) = 0 = ut (0, x) andu (t, 0) = 0, and u (t, π) = F (t) ,

where 0 < x < π and (0) = 0 is given by

u (t, x) =1π

∞∑n=1

(−1)n+1∫ t

0

sin ((t− τ)n)F (τ) dτ sinnx.

Moreover if F (τ) = sin τ, then

u (t, x) =1π

(12

sin t− 12t cos t

)sinnx+ (Terms remaining bounded) .

Notice that t cos t can be arbitrarily large, another manifestation of resonance.

Proof. To prove this, let ϕ (t, x) = xπF (t) and then let U (t, x) = u (t, x)−

ϕ (t, x) . We then have(∂2

t − k∂2x

)U =

(∂2

t − k∂2x

)[u (t, x)− ϕ (t, x)] = −x

πF (t) with

U (0, x) = −ϕ (0, x) = 0

Ut (0, x) = −F (0)x

πand

U (t, 0) = 0 = U (t, π) .


U (t, x) = −F (0)sin(t√−L)

√−L

x

π−∫ t

0

sin((t− τ)

√−L)

√−L

x

πF (τ) dτ.

Integrating the last term by parts shows,∫ t

0

sin((t− τ)

√−L)

√−L

x

πF (τ) dτ =

[sin((t− τ)

√−L)

√−L

x

πF (τ)

]τ=t

τ=0

−∫ t

0

d

dτ

sin((t− τ)

√−L)

√−L

x

πF (τ) dτ

= −sin(t√−L)

√−L

x

πF (0)

+∫ t

0

cos((t− τ)

√−L) xπF (τ) dτ.

Combining the last two equations then implies that

U (t, x) = −∫ t

0

cos((t− τ)




We now integrate by parts one more time to find,

U (t, x) = − cos((t− τ)

√−L) xπF (τ)

∣∣∣τ=t

τ=0+∫ t

0

d

dτcos((t− τ)

√−L) xπF (τ) dτ

= −xπF (t) +

∫ t

0

√−L sin

((t− τ)


and therefore,

u (t, x) =x

πF (t)− x

πF (t) +

∫ t

0

√−L sin

((t− τ)


=∫ t

0

√−L sin

((t− τ)


=∫ t

0

√−L sin

((t− τ)

√−L) 1π

( ∞∑n=1

2n

(−1)n+1 sinnx

)F (τ) dτ

=1π

∞∑n=1

1n

(−1)n+1∫ t

0

n sin ((t− τ)n)F (τ) dτ (sinnx)

=1π

∞∑n=1

(−1)n+1∫ t

0

sin ((t− τ)n)F (τ) dτ sinnx.

To be more explicit, let us suppose that F (τ) = sin τ, in this case∫ t

0

sin ((t− τ)n) sin τdτ =12

∫ t

0

(cos ((n+ 1) τ − nt)− cos (nt+ (1− n) τ)) dτ

=12

[sin ((n+ 1) τ − nt)

n+ 1− sin (nt+ (1− n) τ)

(1− n)

]τ=t

τ=0

for n 6= 1 and for n = 1 we have∫ t

0

sin ((t− τ)) sin τdτ =12

sin t− 12t cos t.

Therefore

u (t, x) =1π

(12

sin t− 12t cos t

)sinx+

∑(Terms remaining bounded) .


9

Problems In other Coordinates Systems

9.1 A Heat equation in spherical coordinates

Example 9.1 (Example 40.3 on p. 126 (Spherical Coordinates)). Let x ∈ R3,|x| =

√x · x and

Ω =x ∈ R3 : |x| ≤ a

.

We wish to solve

ut (t, x) = ∆u (t, x) for x ∈ Ω withu (t, x) = 0 if |x| = a and u (0, x) = f (|x|) for x ∈ Ω

We will look for a solution which only depends on r = |x| , where r is theradial variable in spherical coordinates. Recalling from Eqs. (1.15) and (1.16)that

∆g =1r∂2

r (rg) +1

r2 sinϕ∂ϕ(sinϕ∂ϕg) +

1r2 sin2 ϕ

∂2θg

our equations becomes

ut (t, r) =1r∂2

r (ru (t, r)) =1r

(ru)rr (t, r) with

u (t, a) = 0 and u (0, r) = f (r) for 0 ≤ r ≤ a.

Hence if we let v (t, r) := ru (t, r) , we find that

vt (t, r) = vrr (t, r) with v (t, a) = 0 = v (t, 0) and v (0, r) = rf (r) for 0 ≤ r ≤ a.

We can now solve this equation for v by the standard means, namely,

rf (r) =∑

bn sinnπr

a

withbn :=

2a

∫ a

0

rf (r) sinnπr dr.

116 9 Problems In other Coordinates Systems

Fig. 9.1. The region Ω.

Then we have, with L = ∂2

∂r2 with Dirichlet boundary conditions, that

v (t, r) = etL∑

bn sinnπr

a=∑

bnetL sinnπ

r

a

=∑

bn exp(t

(−n

2π2

a2

))sinnπ

r

a

and so

u (t, r) =1r

∑bn exp

(t

(−n

2π2

a2

))sinnπ

r

a.

9.2 Problems with cylindrical symmetries

Example 9.2 (Example 43.1 on p. 137 (Harmonic functions in cylindrical co-ordinates)). In this example, Ω denotes those points which may be written inpolar coordinates (ρ, θ) with 1 < ρ < b and 0 < θ < π, see Figure 9.1 below.We wish to solve

∆u = 0 with (9.1)u (ρ, 0) = 0 = u (ρ, π) for 1 < ρ < b and (9.2)u (1, θ) = f (θ) and u (b, θ) = g (θ) for 0 < θ < π. (9.3)

Recalling that ∆ in polar coordinates is given by

∆u =1ρ∂ρ (ρ∂ρu) +

1ρ2∂2

θu (9.4)

we look for some solutions to Eq. (9.1) and Eq. (9.2) of the form u (ρ, θ) =R (ρ)Θ (θ) . To short cut things, from experience we know that Θ (θ) = sinnθfor some n ∈ N and hence

un (ρ, θ) = R (ρ) sinnθ.

Plugging this into Eq. (9.1) then implies


9.2 Problems with cylindrical symmetries 117

0 =1ρ∂ρ (ρ∂ρR)− n2

ρ2R.

This is an Euler type equation which may be solved by looking for solution ofthe form R (ρ) = ρα. This happens iff

0 = α2ρα−2 − n2ρα−2

and hence we must have α2 = n2 for α = ±n. Thus we find

un (ρ, θ) =(Anρ

n +Bnρ−n)sinnθ.

So we now look for solutions of the form

u (ρ, θ) =∞∑

n=1

(Anρ

n +Bnρ−n)sinnθ

and we wish to choose An and Bn so that Eq. (9.3) is satisfied. If we write

f (θ) =∞∑

n=1

f (n) sinnθ and g (θ) =∞∑

n=1

g (n) sinnθ

with

f (n) =2π

∫ π

0

f (θ) sinnθdθ and

g (n) =2π

∫ π

0

g (θ) sinnθdθ

then we must have∞∑

n=1

f (n) sinnθ =∞∑

n=1

(An +Bn) sinnθ and

∞∑n=1

g (n) sinnθ =∞∑

n=1

(Anb

n +Bnb−n)sinnθ

which implies that

An +Bn = f (n)

Anbn +Bnb

−n = g (n) .

Solving these equations gives

Bn

(bn − b−n

)= bnf (n)− g (n)

An

(b−n − bn

)= b−nf (n)− g (n)

or



Fig. 9.2. The unit disk in R2.

Bn =bnf (n)− g (n)bn − b−n

An =b−nf (n)− g (n)

b−n − bn=g (n)− b−nf (n)

bn − b−n

and we find

u (ρ, θ) =∞∑

n=1

(g (n)− b−nf (n)

bn − b−nρn +

f (n) bn − g (n)bn − b−n

ρ−n

)sinnθ

=∞∑

n=1

1bn − b−n

(g (n)

(ρn − ρ−n

)+ f (n)

((b

ρ

)n

−(ρb

)n))

sinnθ.

Example 9.3 ( Laplace’s equation on the disk). We wish to solve Laplace’sequation,

∆u = 0 on Ω with u = f on ∂Ω

where Ω =(x, y) ∈ R2 : x2 + y2 < 1

is the unit disk as in the Figure below.

The usual separation of variables arguments shows that

un (ρ, θ) = (A cosnθ +B sinnθ)(aρ−n + bρn

)for n = 1, 2, 3, . . .

and for n = 0 we have 0 = 1ρ∂ρ (ρ∂ρR) , which implies ρ∂ρR = a and R =

a ln ρ+ b so thatu0 (ρ, θ) = 1 · (a ln ρ+ b)

all solve ∆u = 0. However, we want our solution to be continuous and hencebounded near (0, 0) which requires us to take a = 0. Thus we look for asolution of the form

u (ρ, θ) =A0

2+

∞∑n=1

(An cosnθ +Bn sinnθ) ρn.


9.3 Strum Liouville problem in cylindrical coordinates 119

To satisfy the boundary condition we must require,

f (θ) =A0

2+

∞∑n=1

(An cosnθ +Bn sinnθ) ,

i.e.An =

1π

∫ π

−π

f (θ) cosnθdθ and Bn =1π

∫ π

−π

f (θ) sinnθdθ.

You are asked to show in Problem 48.7 that the solution may be written as

u (ρ, θ) =12π

∫ π

−π

1− ρ2

1− 2ρ cos (θ − ψ) + ρ2f (ψ) dψ for ρ < 1.

This is called the Poisson integral formula.

9.3 Strum Liouville problem in cylindrical coordinates

In this section we will consider problems with symmetry about the z - axis.We will, for simplicity, restrict ourselves the region,

Ω =(x, y) ∈ R2 : x2 + y2 ≤ c2

⊂ R2

and consider problems with associated to Dirichlet boundary conditions. Thatis we wish to solve the heat and wave equations,

ut = ∆u with u = 0 on ∂Ω andutt = ∆u with u = 0 on ∂Ω.

Following the strategy used throughout this course we first will look forsolutions to the eigenfunction problem,

∆ϕ = −λϕ on Ω with ϕ = 0 on ∂Ω. (9.5)

As above we will let

(f, g) :=∫

Ω

f (x, y) g (x, y) dxdy. (9.6)

Let us recall that if ϕ is a solution to Eq. (9.5), then

−λ (ϕ,ϕ) = (−λϕ, ϕ) = (∆ϕ,ϕ) = −∫

Ω

|∇ϕ|2 dxdy

and hence it follows that λ ≥ 0. Moreover, if λ = 0, then∫

Ω|∇ϕ|2 dxdy = 0

which implies ∇ϕ = 0. This condition along with the Dirichlet boundaryconditions implies that ϕ ≡ 0.



Conclusion: When looking for solutions to Eq. (9.5) we may restrict ourattention to the case where λ > 0.

Because of the symmetry of the domain, we will view ϕ as a function ofpolar coordinate, (ρ, θ) in which case Eq. (9.5) may be written as

1ρ∂ρ (ρ∂ρϕ) +

1ρ2∂2

θϕ = −λϕ with ϕ (c, θ) = 0, (9.7)

see Eq. (1.14). Moreover in polar coordinates, Eq. (9.6) becomes

(f, g) :=∫ 2π

0

∫ c

0

f (ρ, θ) g (ρ, θ) ρdρdθ. (9.8)

The structure of Eq. (9.7), suggests we look for solutions of the form, ϕ (ρ, θ) =R (ρ)Θ (θ) . Plugging this into Eq. (9.7) implies

1ρ∂ρ (ρ∂ρR)Θ +

R

ρ2∂2

θΘ = −λRΘ

and then multiplying by ρ2

RΘ gives

ρ

R∂ρ (ρ∂ρR) + λρ2 = −∂

2θΘ

Θ= k

for some constant k. We begin by recalling the solutions to

∂2θΘ = −kΘ with Θ (0) = Θ (2π) and Θ′ (0) = Θ′ (2π)

are of the formΘn (θ) = An cosnθ +Bn sinnθ

for some n = 0, 1, 2, . . . and constants A and B. For such a Θn we have k = n2.Fixing such an n, we must now solve

ρ

R∂ρ (ρ∂ρR) + λρ2 = n2 with R (c) = 0.

Multiplying this equation through by R, we get

ρ∂ρ (ρ∂ρR) +(λρ2 − n2

)R = 0 with R (c) = 0


ρ2R′′ + ρR′ +(λρ2 − n2

)R = 0 with R (c) = 0.

9.4 Bessel Equation and Functions

Before proceeding, we will need to understand better the solutions to theordinary differential equation,

ρ∂ρ (ρ∂ρR) +(λρ2 − n2

)R = 0 with λ > 0. (9.9)


9.4 Bessel Equation and Functions 121

Lemma 9.4. If R solves Eq. (9.9), x :=√λρ and

y (x) := R (ρ) = R

(x√λ

),

then y solves Bessel’s equation of order n,

0 = xd

dx

(xd

dxy

)+(x2 − n2

)y

= x2y′′ (x) + xy′ (x) +(x2 − n2

)y (x) . (9.10)

Alternatively put, if y solves Bessel’s equation of order n, then R (ρ) :=y(√

λρ)

will solve Eq. (9.9).

Proof. By the chain rule,

ρd

dρ= ρ

dx

dρ

d

dx=

x√λ

√λd

dx= x

d

dx

and hence in this new variable, Eq. (9.9) becomes

0 = xd

dx

(xd

dxy

)+

(λ

(x√λ

)2

− n2

)y

= xd

dx

(xd

dxy

)+(x2 − n2

)y

and hence we have eliminate the λ from the equation.The previous results show that if y solves Bessel’s equation of order n and

y(√

λc)

= 0, then

u (ρ, θ) = (An cosnθ +Bn sinnθ) y(√

λρ)

(9.11)

satisfies∆u = −λu with u = 0 on ∂Ω. (9.12)

To understand this better we need to understand the solution to Eq. (9.10)better. We will do this in the next few results.

Theorem 9.5. Every solution to Bessel’s Eq. (9.10) of order n which isbounded near x = 0 is a multiple the Bessel function of the first kind oforder n defined by

Jn (x) :=∞∑

k=0

(−1)k

k! · (n+ k)!

(x2

)n+2k

. (9.13)



Fig. 9.3. A plot of J0, J3 and J10. Notice that J0 (0) = 1 whereas Jn (0) = 0 for alln ≥ 1.

Proof. This solution can be found by the usual power series method forsolving such equations, namely plugging in the series expansion of the form

y (x) =∞∑

k=0

akxk+r

into Eq. (9.10) and then finding relations that r and the coefficients ak have tosatisfy. This method produces the solution in Eq. (9.13) above. Moreover, itcan be verified by straightforward computation that Jn (x) does indeed solveEq. (9.10).

To find more solutions one may use the method of “reduction of order” toshow that the general solution to Eq. (9.13) is of the form

y (x) = A

∫1

J2n(x)

1xdx+B

for some constants A and B. Since Jn (x) ∼ xn for x near zero we learn that

u (x) ∼∫

1x2n+1

dx ∼ x−2n for x near 0 if n ≥ 1,

u (x) ∼ ln |x| for x near 0 if n ≥ 1

and therefore, we find u (x) Jn (x) ∼ x−n for x near zero if n ≥ 1 or ln |x| J0 (x)if n = 0. At any rate, Jn (x) is the only solution to Bessel’s equation which isbounded near x = 0.

Theorem 9.6. The functions Jn (x) behave as shown in Figure 9.3. Moreprecisely, there are constants cn and θn such that

Jn(x) ≈ cnx−1/2 sin(x+ θn) as x→∞. (9.14)



In particular each Bessel function Jn has an infinite number of zeros

0 < αn,1 < αn,2 < . . . αn,j < . . .

and limj→∞αn,j

j = π. (By convention, when n = 0 we let α0,1 denote the firstzero of J0 which is positive.)

Proof. We give the idea for the proof without giving the full proof whichwould take us to far afield. We begin by writing a solution to Bessel’s Eq.(9.13) in the form y (x) = xru (x) where r is to be determined so that u solvesa second order differential equation where u′ (x) does not appear. Since

y′ = rxr−1u+ xru′ and

y′′ = r (r − 1)xr−2u+ 2rxr−1u′ + xru′′

we have

0 = x2y′′ + xy′ +(x2 − n2

)y

= r (r − 1)xru+ 2rxr+1u′ + xr+2u′′

+ rxru+ xr+1u′ +(x2 − n2

)xru

= xr+2u′′ + (2r + 1)xr+1u′ +(r2 + x2 − n2

)xru.

So we choose r = − 12 and then divide the above result by xr+2 to learn u

solves,

u′′ +(

1 +1/4− n2

x2

)u = 0.

For x large this equation looks like u′′ + u = 0 (the spring “constant,” 1 +(14 − n

2)x−2 → 1 as x → ∞), and thus we expect u(x) ≈ c sin(x + θ) for

x >> 1. Thus we are lead to expect that Jn(x) = x−1/2u (x) should behaveas described in Eq. (9.14).

Corollary 9.7 (Eigenfunctions). The functions,

un,j (ρ, θ) := (An cosnθ +Bn sinnθ) Jn

(αn,j

ρ

c

), (9.15)

satisfy

∆un,j = −α2

n,j

c2un,j with un,j = 0 on ∂Ω. (9.16)

Proof. This is a consequence of Eqs. (9.11), (9.12), Lemma 9.4, Theorem9.5 and Theorem 9.6.

Theorem 9.8 (Recurrence relations for Bessel functions). Let Jn (x)be Bessel function of the first kind of order n in Eq. (9.13), then:



d

dx

[x−nJn (x)

]= −x−nJn+1 (x) or equivalently (9.17)

xJ ′n (x) = nJn (x)− xJn+1 (x) , for n = 0, 1, 2, 3, . . . , (9.18)

d

dx[xnJn (x)] = xnJn−1 (x) or equivalently (9.19)

xJ ′n (x) = −nJn (x) + xJn−1 (x) , for n = 1, 2, 3, . . . , (9.20)

andxJn+1 (x) = 2nJn (x)− xJn−1 (x) for n = 1, 2, 3, . . . (9.21)

In particular from Eq. (9.18) with n = 0,

J1 (x) = −J ′0 (x) (9.22)

and form Eq. (9.20) with n = 1, xJ ′1 (x) = −J1 (x) + xJ0 (x) which may bewritten as

xJ0 (x) =d

dx(xJ1 (x)) . (9.23)

Proof. Computing

d

dx

[x−nJn (x)

]= 2−n d

dx

∞∑k=0

(−1)k

k! · (n+ k)!

(x2

)2k

= 2−n∞∑

k=1

(−1)kk

k! · (n+ k)!

(x2

)2k−1

= 2−n∞∑

k=1

(−1)k

(k − 1)! · (n+ k)!

(x2

)2k−1

= 2−n∞∑

k=0

(−1)k+1

k! · (n+ k + 1)!

(x2

)2k+1

= −x−n∞∑

k=0

(−1)k

k! · (n+ k + 1)!

(x2

)2k+1+n

= −x−nJn+1 (x) .

This relation also implies

−Jn+1 (x) = xn d

dx

[x−nJn (x)

]= −nx−1Jn (x) + J ′n (x)

which his equivalent to Eq. (9.18). Eq. (9.20) is proved similarly, see Problems69.1 from Brown and Churchill. Subtracting Eq. (9.20) from Eq. (9.18) givesEq. (9.21).



Theorem 9.9 (Bessel function normalizations). Let αn,j denote the jth

zero of Jn (x) . Then∫ 1

0

J2n (αn,jρ) ρdρ =

12

(J ′n (αn,j))2 =

12

(Jn+1 (αn,j))2 (9.24)

More generally, if ϕn,j (ρ) = Jn (αn,jρ) where αn,j = αn,j/c is the jth – zeroof the function α→ Jn (cα) , then∫ c

0

ϕ2n,j (ρ) ρdρ =

c2

2(J ′n (αn,jc))

2 =c2

2(Jn+1 (αn,j))

2. (9.25)

When n = 0, this may be written as∫ c

0

J20 (αn,jρ) (ρ) ρdρ =

c2

2(J ′0 (αn,jc))

2 =c2

2J2

1 (αn,jc) . (9.26)

Proof. To prove Eq. (9.24) we begin by recalling that R (ρ) := Jn (αρ)where α = αn,j , solves the differential equation,

1ρ∂ρ (ρ∂ρR) +

(α2 − n2

ρ2

)R = 0


∂ρ (ρ∂ρR) +(ρα2 − n2

ρ

)R = 0.

Therefore,

0 =∫ 1

0

[∂ρ (ρ∂ρR) +

(ρα2 − n2

ρ

)R

]ρR′dρ

=12

∫ 1

0

[∂ρ (ρ∂ρR)2 +

(ρ2α2 − n2

)∂ρR

2]dρ

=12

(ρ∂ρR)2 |10 +12

∫ 1

0

(ρ2α2 − n2

)∂ρR

2dρ

=12

[(ρ∂ρR)2 +

(ρ2α2 − n2

)R2]|10 −

12

∫ 1

0

∂ρ

(ρ2α2 − n2

)R2dρ

=12

[(ρ∂ρR)2 +

(ρ2α2 − n2

)R2]|10 − α2

∫ 1

0

R2ρdρ

and we have shown,∫ 1

0

R2ρdρ =1

2α2

[(ρ∂ρR)2 +

(ρ2α2 − n2

)R2]|10.

Thus we have shown,


126 9 Problems In other Coordinates Systems∫ 1

0

J2n (αn,jρ) ρdρ =

12α2

n,j

[(ραn,jJ

′n (αn,jρ))

2 +(ρ2α2 − n2

)J2

n (αn,jρ)]|10

=1

2α2n,j

[(αn,jJ

′n (αn,j))

2 +(α2 − n2

)J2

n (αn,j1)]

=12

(J ′n (αn,j))2.

To get the second equality in Eq. (9.24), set x = αn,j in recursion relation,

xJ ′n (x) = nJn (x)− xJn+1 (x) ,

to find

αn,jJ′n (αn,j) = nJn (αn,j)− αn,jJn+1 (αn,j) = −αn,jJn+1 (αn,j) ,

which implies that [J ′n (αn,j)]2 = [Jn+1 (αn,j)]

2.

More generally, if ϕn,j (ρ) = Jn

(αn,j

ρc

), then letting x = ρ

c or ρ = cx wefind∫ c

0

ϕ2n,j (ρ) ρdρ =

∫ c

0

J2n

(αn,j

ρ

c

)ρdρ

= c2∫ 1

0

J2n (αn,jx)xdx =

c2

2(J ′n (αn,j))

2 ==c2

2J2

n+1 (αn,j) .

Notice that αn,j = αn,j/c is the jth – zero of the function α→ Jn (cα) . Withthis notation we have

ϕn,j (ρ) = Jn (αn,jρ) and∫ c

0

ϕ2n,j (ρ) ρdρ =

c2

2(J ′n (αn,jc))

2 =c2

2J2

n+1 (αn,j) .

Corollary 9.10 (Eigenfunction Expansions). Any “reasonable function”,F (ρ, θ) , may be expressed as

F (ρ, θ) =12

∞∑j=1

A0,jJ0

(α0,j

ρ

c

)+

∞∑n=1

∞∑j=1

(An cosnθ +Bn sinnθ) Jn

(αn,j

ρ

c

)(9.27)

where

An,j :=2

c2J2n+1 (αn,j)

∫ c

ρ=0

∫ π

θ=−π

F (ρ, θ) Jn

(αn,j

ρ

c

)cosnθρdρdθ (9.28)

and

Bn,j :=2

c2J2n+1 (αn,j)

∫ c

ρ=0

∫ π

θ=−π

F (ρ, θ) Jn

(αn,j

ρ

c

)sinnθρdρdθ. (9.29)



Proof. The existence of an expansion in Eq. (9.27) is to be expectedfrom our previous experience with expansions related to symmetric eigenvalueproblems. (We will have to take this one on faith.) Given such an expansionexists, the usual orthogonality arguments shows that the coefficients must begiven as in Eqs. (9.28) and (9.29).

Corollary 9.11 (Heat and Wave Eq. Solutions). The solution to the heatequation,

ut = ∆u with u = 0 on ∂Ω andu (t, ρ, θ) = F (ρ, θ)

is given by

u (t, ρ, θ) =12

∞∑j=1

A0,je−tα2

0,jJ0

(α0,j

ρ

c

)+

∞∑n=1

∞∑j=1

e−tα2n,jJn

(αn,j

ρ

c

)(An cosnθ +Bn sinnθ) .

The solution to the wave equation,

utt = ∆u with u = 0 on ∂Ω andu (t, ρ, θ) = F (ρ, θ) and ut (t, ρ, θ) = 0

is given by

u (t, ρ, θ) =12

∞∑j=1

A0,j cos (α0,jt) J0

(α0,j

ρ

c

)+

∞∑n=1

∞∑j=1

cos (αn,jt) Jn

(αn,j

ρ

c

)(An cosnθ +Bn sinnθ) .

The following result will enable us to work out some explicit examples.

Theorem 9.12 (Reduction formula, Problem 69.3).∫ x

0

snJn−1 (s) ds = xnJn (x) for n ∈ N (9.30)∫ x

0

sJ0 (s) ds = xJ1 (x) (9.31)∫ x

0

snJ0 (s) ds = xnJ1 (x) + (n− 1)xn−1J0 (x)− (n− 1)2∫ x

0

sn−2J0 (s) ds

(9.32)

The scaled version of Eq. (9.32) is


128 9 Problems In other Coordinates Systems∫ x

0

tnJ0 (αt) dt =xn

αJ1 (αx) +

n− 1α2

xn−1J0 (αx)

− (n− 1)2

α2

∫ x

0

tn−2J0 (αt) dt. (9.33)

Proof. Equation (9.30) is an easy consequence of the fundamental theoremof calculus and Eq. (9.19). The proof of Eq. (9.32) is Problem 69.3, see Solution??. Making the change of variables, s = αt, we find∫ x

0

tnJ0 (αt) dt =∫ αx

0

sn

αnJ0 (s)

ds

α= α−(n+1)

∫ αx

0

snJ0 (s) ds

=α−(n+1)

[(αx)n

J1 (αx) + (n− 1) (αx)n−1J0 (αx)

− (n− 1)2∫ αx

0sn−2J0 (s) ds

]=xn

αJ1 (αx) +

n− 1α2

xn−1J0 (αx)

− (n− 1)2

αn+1αn−1

∫ x

0

tn−2J0 (αt) dt

=xn

αJ1 (αx) +

n− 1α2

xn−1J0 (αx)− (n− 1)2

α2

∫ x

0

tn−2J0 (αt) dt.

9.5 Examples

Example 9.13. We want to expand 1 in terms of the functions ϕj (x) =J0 (αjx) where αj is the jth – root of J0 (x) . To this end we have

1 =∞∑

j=1

(1, ϕj)(ϕj , ϕj)

ϕj =∞∑

j=1

(1, ϕj)12J

21 (αj)

ϕj (x)

where

(f, g) :=∫ 1

0

f (x) g (x)xdx.

Now, using Eq. (9.23) shows

(1, ϕj) =∫ 1

0

xϕj (x) dx =∫ 1

0

xJ0 (αjx) dx

=1α2

j

∫ αj

0

sJ0 (s) ds =1α2

j

∫ αj

0

d

ds[sJ1 (s)] ds =

1αjJ1 (αj) .

Hence we find


9.5 Examples 129

1 =∞∑

j=1

(1, ϕj)(ϕj , ϕj)

ϕj =∞∑

j=1

J1 (αj)12J

21 (αj)

ϕj (x)

= 2∞∑

j=1

1αjJ1 (αj)

J0 (αjx) .

Solution to Exercise (76.1). We want to expand x in terms of the func-tions ϕj (x) = J0 (αjx) where αj is the jth – root of J0 (x) . To this end wehave

x =∞∑

j=1

(x, ϕj)(ϕj , ϕj)

ϕj =∞∑

j=1

(x, ϕj)12J

21 (αj)

ϕj (x)

where

(f, g) :=∫ 1

0

f (x) g (x)xdx.

Using the reduction formula in Eq. (9.33) we with x = 1 and t replaced by x,∫ 1

0

xnJ0 (αjx) dx =1αjJ1 (αj)+

n− 1α2

j

J0 (αj)−(n− 1)2

α2j

∫ 1

0

xn−2J0 (αjx) dx,

with n = 2, we find

(x, ϕj) =∫ 1

0

x2ϕj (x) dx =∫ 1

0

x2J0 (αjx) dx

=1αjJ1 (αj) +

1α2

j

J0 (αj)−1α2

j

∫ 1

0

J0 (αjx) dx

=1αjJ1 (αj)−

1α2

j

∫ 1

0

J0 (αjx) dx.

Therefore we find,

x =∞∑

j=1

(x, ϕj)(ϕj , ϕj)

ϕj =∞∑

j=1

1αjJ1 (αj)− 1

α2j

∫ 1

0J0 (αjx) dx

12J

21 (αj)

ϕj (x)

= 2∞∑

j=1

[1− 1

αjJ1 (αj)

∫ 1

0

J0 (αjx) dx]J0 (αjx)αjJ1 (αj)

= 2∞∑

j=1

[1− 1

α2jJ1 (αj)

∫ αj

0

J0 (s) ds

]J0 (αjx)αjJ1 (αj)

.

Solution to Exercise (76.3). We want to expand

f (x) =

1 if 0 < x < 10 if 1 < x < 2.



We will begin by first expanding f (2x) instead, and for this we have

f (2x) =∞∑

j=1

(f (2x) , ϕj)(ϕj , ϕj)

ϕj =∞∑

j=1

(f (2x) , ϕj)12J

21 (αj)

ϕj (x)

= 2∞∑

j=1

(f (2x) , ϕj)J2

1 (αj)ϕj (x)

where, using Eq. (9.31), we have,

(f (2x) , ϕj) =∫ 1

0

f (2x) J0 (αjx)xdx =∫ 1/2

0

J0 (αjx)xdx

= α−2j

∫ αj/2

0

J0 (s) sds = α−2j ·

αj

2J1

(αj

2

)=

12αj

J1

(αj

2

).

Therefore,

f (2x) = 2∞∑

j=1

12αj

J1

(αj

2

)J2

1 (αj)J0 (αjx)

=∞∑

j=1

J1

(αj

2

)αjJ2

1 (αj)J0 (αjx) .

Replacing x by x/2 shows

f (x) =∞∑

j=1

J1

(αj

2

)αjJ2

1 (αj)J0 (αjx/2)

and then letting αj = αj/2, shows

f (x) =12

∞∑j=1

J1 (αj)αjJ2

1 (2αj)J0 (αjx)

where J0 (2αj) = 0 for each j.

Solution to Exercise (77.2). In this problem we are asked to solve

∆u = 0 on Ω := (ρ, θ, z) : 0 ≤ ρ, z ≤ 1 withu (1, θ, z) = 0, uz (ρ, θ, 0) = 0 and u (ρ, θ, 1) = 1.

Given the symmetry of this problem, the solution we are looking for shouldonly depend on (ρ, z) and we should begin by looking for solutions of the form


9.5 Examples 131

u (ρ, z) = J0 (α0,jρ)Z (z) .

Since1ρ∂ρ (ρ∂ρJ0 (α0,jρ)) = −α2

0,jJ0 (α0,jρ)

we have

0 = ∆u =1ρ∂ρ (ρ∂ρu) +

1ρ2∂2

θu+ ∂2zu

= −α20,jJ0 (α0,jρ)Z (z) + J0 (α0,jρ)Z ′′ (z)

and we conclude thatZ ′′ (z) = α2

0,jZ (z) .

The solutions to this equation are given by

Z (z) = A coshα0,jz +B sinhα0,jz.

Because of the boundary condition, uz (ρ, θ, 0) = 0, we must require B = 0and we will look for a solution of the form

u (ρ, z) =∞∑

j=1

AjJ0 (α0,jρ) coshα0,jz.

To fix the constants Aj we must require that

1 =∞∑

j=1

AjJ0 (α0,jρ) coshα0,j .

Recalling from Example 9.13 that

1 = 2∞∑

j=1

1αjJ1 (αj)

J0 (αjρ)

we must require that

Aj coshα0,j = 21

αjJ1 (αj)

and hence our solution is given by

u (ρ, z) = 2∞∑

j=1

J0 (α0,jρ)αjJ1 (αj)

coshα0,jz

coshα0,j.

Solution to Exercise (77.7). Here we are to solve,

ut = ∆u− bu = (∆− b)uon Ω := (t, ρ, θ) : 0 ≤ ρ ≤ 1 and t > 0 with

u (t, 1, θ) = 0 and u (0, ρ, θ) = 1.



In this case we have

u (t, ρ) = et(∆−b)1 = e−tbet∆1

= e−tbet∆

2∞∑

j=1

1αjJ1 (αj)

J0 (αjρ)

= 2e−tb

∞∑j=1

J0 (αjρ)αjJ1 (αj)

e−tα20,j .


A

Some Complex Variables Facts

Here we suppose w (t) = c (t) + id (t) where c (t) and d (t) are two real valuedfunctions of t. An important example of such a complex valued function isfound in the next definition.

Definition A.1 (Euler’s Formula). For t ∈ R let

eit := cos t+ i sin t (A.1)

and for z = x+ iy let

ez := exeiy = ex (cos y + i sin y) . (A.2)

Notice that any complex number, z = x+ iy, may be written as z = reiθ

where (r, θ) are the polar coordinates of the point (x, y) ∈ R2.

Definition A.2. If c (t) and d (t) are differentiable, then we define

w (t) := c (t) + id (t)

and ∫ β

α

w (t) dt :=∫ β

α

c (t) dt+ i

∫ β

α

d (t) dt

Example A.3. If w (t) = et + i sin t, then

w (t) = et − i cos t and∫ π/2

0

w (t) dt =∫ π/2

0

(et + i sin t

)dt = e

12 π − 1 + i.

Example A.4. Suppose w (t) = eit, then

d

dteit =

d

dt(cos t+ i sin t) = − sin t+ i cos t

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

134 A Some Complex Variables Facts

and ∫ b

a

eitdt =∫ b

a

(cos t+ i sin t) dt =∫ b

a

cos tdt+ i

∫ b

a

sin tdt

= (sin t− i cos t) |ba =eit

i|ba.

Example A.6 below, generalizes this result.

Theorem A.5 (These definitions work just as in real variables). Ifz (t) = a (t) + ib (t) and w (t) = c (t) + id(t) and λ = u+ iv ∈ C then

1. ddt (w (t) + z (t)) = w (t) + z (t)

2. ddt [w (t) z (t)] = wz + wz

3.∫ β

α[w (t) + λz (t)] dt =

∫ β

αw (t) dt+ λ

∫ β

αz (t) dt

4.∫ β

αw(t)dt = w(β)− w(α) In particular if w = 0 then w is constant.

5. ∫ β

α

w(t)z(t)dt = −∫ β

α

w(t)z(t)dt+ w (t) z (t) |βα.

6. ∣∣∣∣∣∫ β

α

w (t) dt

∣∣∣∣∣ ≤∫ β

α

|w (t)| dt.

Proof. 1. and 4. are easy.2.

d

dt[wz] =

d

dt(ac− bd) + i

d

dt(bc+ ad)

= (ac− bd) + i(bc+ ad)

+ (ac− bd) + i(bc+ ad)= wz + wz.

3. The only interesting thing to check is that∫ β

α

λz (t) dt = λ

∫ β

α

z (t) dt.

Again we simply write out the real and imaginary parts:

∫ β

α

λz (t) dt =∫ β

α

(u+ iv) (a (t) + ib (t)) dt

=∫ β

α

(ua(t)− vb(t) + i [ub(t) + va(t)]) dt

=∫ β

α

(ua(t)− vb(t)) dt+ i

∫ β

α

[ub(t) + va(t)] dt


A Some Complex Variables Facts 135

while ∫ β

α

λz (t) dt = (u+ iv)∫ β

α

[a (t) + ib (t)] dt

= (u+ iv)

(∫ β

α

a (t) dt+ i

∫ β

α

b (t) dt

)

=∫ β

α

(ua(t)− vb(t)) dt+ i

∫ β

α

[ub(t) + va(t)] dt.

Shorter Alternative: Just check it for λ = i, this is the only new thingover the real variable theory.

5.

w (t) z (t) |βα =∫ β

α

d

dt[w (t) z (t)] dt =

∫ β

α

w(t)z(t)dt+∫ β

α

w(t)z(t)dt.

6. (Skip this one!) Let ρ ≥ 0 and θ ∈ R be chosen so that∫ β

α

w (t) dt = ρeiθ,

then ∣∣∣∣∣∫ β

α

w (t) dt

∣∣∣∣∣ = ρ = e−iθ

∫ β

α

w (t) dt =∫ β

α

e−iθw (t) dt

=∫ β

α

Re[e−iθw (t)

]dt ≤

∫ β

α

∣∣Re[e−iθw (t)

]∣∣ dt≤∫ β

α

∣∣e−iθw (t)∣∣ dt =

∫ β

α

|w (t)| dt.

Example A.6. Suppose z = x+ iy, then

ezt := exteiyt := ext cos yt+ iext sin yt

and so, again by definition,

d

dtezt =

d

dt

(ext cos yt+ iext sin yt

)= ext (x cos yt− y sin yt) + iext (x sin yt+ y cos yt)

= (x+ iy)(ext cos yt+ iext sin yt

)= zezt.

A better proof. By the product rule and Example A.4,

d

dtetz =

d

dt

[etxeity

]= xetxeity + etxiyeity = zetxeity = zetz.


136 A Some Complex Variables Facts

Using this fact and item 4. of Theorem A.5 we may conclude,∫ b

a

eztdt =ezt

z|ba.

If we write out what this means by comparing the real and imaginary partsof both sides we find∫ b

a

eztdt =∫ b

a

ext cos (yt) dt+ i

∫ b

a

ext sin (yt) dt

while

ezt

z=ext cos yt+ iext sin yt

x+ iy

x− iyx− iy

=ext

x2 + y2[x cos yt+ y sin yt+ i (x sin yt− y cos yt)]

from which we may conclude that∫ b

a

ext cos (yt) dt =ext

x2 + y2[x cos yt+ y sin yt]

∣∣∣∣ba

and∫ b

a

ext sin (yt) dt =ext

x2 + y2[x sin yt− y cos yt]

∣∣∣∣ba

.

Theorem A.7 (Addition formula for ez). The function ez defined by Eq.(A.2) satisfies

Proposition A.8. 1. e−z = 1ez and

2. ew+z = ewez.

Proof. By the previous example we know

d

dtetz = zetz with e0z = e0+i0 = 1.

Similarly, using the chain rule or by direct computation, one shows

d

dte−tz = −zetz with e0z = e0+i0 = 1.

1. By the product rule,

d

dt

[e−tzetz

]= −ze−tzetz + e−tzzetz = 0

and therefore e−tzetz is independent of t and hence e−tzetz = e−0ze0z = 1.Taking t = 1 proves 1.

2. Again by the product rule shows


A Some Complex Variables Facts 137

d

dt

[e−t(w+z)etwetz

]=[

− (w + z) e−t(w+z)etwetz

+e−t(w+z)wetwetz + e−t(w+z)etwzetz

]= 0

and so e−t(w+z)etwetz = e−t(w+z)etwetz|t=0 = 1. Taking t = 1 then showse−(w+z)ewez = 1 and then using Item 1. we get item 2.

Corollary A.9 (Addition formulas cos and sin). For α, β ∈ R we have

cos (α+ β) = cosα cosβ − sinα sinβsin (α+ β) = cosα sinβ + cosβ sinα.

Proof. These follow by comparing the real and imaginary parts of theidentity

eiαeiβ = ei(α+β) = cos (α+ β) + i sin (α+ β)

while

eiαeiβ = (cosα+ i sinα) · (cosβ + i sinβ)= cosα cosβ − sinα sinβ + i (cosα sinβ + cosβ sinα) .


B

Assigned Homework Problems:

1. Exercises from Lecture notes: 1.1, 1.2, 1.3, 2.1

Exercises from the Book: 6.1, 6.2, 7.1, 10.1, 10.4, 10.52. Exercises from Lecture notes: 3.1, 3.2, 3.3, 3.4

Exercises from the Book: 6.4, 6.5, 8.1, 10.6, 10.76.4-6.5, 8.1, 8.4, 10.6, 10.7

3. Exercises from Lecture notes: 3.5,

Exercises from the Book: 13.2, 13.4, 13.6, 13.7, 13.9

4. Exercises from the Book: 15.1, 15.2, 15.4a, 17.3, 17.4, 17.6, 22.2

5. Exercises from the Book: 22.3, 22.4, 24.3, 24.4, 24.5, 24.6, 28.1, 28.26. Exercises from the Book: 28.4, 28.5 32.1, 32.2, 32.3, 32.5, 32.6a, (32.14

optional), 33.2, 33.57. Exercises from the Book: 33.15, 40.1, 40.2, 40.3, 43.1, 43.4, 53.18. Exercises from Lecture notes: 8.2.

From book: 40.6, 40.7, 41.1, 41.2, 41.8, 43.2, 43.5, 45.7a (Read b only)9. Exercises from the Book: 45.2, 45.5, 48.1, 53.7, 69.1, 69.3, 76.4

math 110, spring 2004 notesbdriver/110-s04/lecture notes/110notes... · 2004. 5. 25. · bruce k....

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