Chapter 5

2D Linear Second-Order Equations

In this chapter, we delve into the realm of linear partial diﬀerential equations in 2D domains.

Our focus lies on rectangular and disk-shape d domains, as they provide us with the opportu-

nity to employ the spatial variable separation technique. Within this framework, we explore

a range of partial diﬀerential equations, namely: 1) Laplace equations, 2) Poisson equations,

3) heat equations, and 4) wave equations.

Let Ω be an open bounded set in R

2

with a smooth or piecewise smooth boundary bnd(Ω).

We begin by considering the Laplace e quation, which takes the g en eral form ∆u(x; y)=0 for

all (x; y)2Ω. This equation is deﬁned for all smooth function s u that satisfy a prescribed

boundary condition, typi cally of the form αu+β

@u

@n

=γ on the boundary bnd(Ω). Here, α; β,

and γ are constants.

The Laplace equation is a special case of the Poisson equation, which has the general form

¡∆u= f, where f is a function that is independent of u and depends only on the spatial

variables (x; y). In this equation, we seek a solution u that satisﬁes the prescribed boundary

conditions.

Moving on, we explore the steady state of a heat equation, which has the form u_

t=k∆u+ f. Here, u

t

represents the partial derivative of u with respect to time, k is a c o nstant

representing thermal diﬀusivity, ∆u is the Laplacian of u, and f represents an e xternal source

of heat.

Finally, we consider the wave equation or damped wave equation, given by

u

tt

+2ξu

t

=c

2

∆u + f. In this equation, u

tt

represents the second partial derivative of u

with respect to time, ξ is a non-negative constant known as the damping factor, c rep-

resents th e wave speed, ∆u is the Laplacian of u, and f represents an external forcing term.

Throughout this chapter, we aim to study the properties of these equations, investigate

their solutions, and u nderstand their physical interpretations in the context of 2D domains.

5.1 Overv iew of the e igenfunction expa nsion method

The eigenfunction expansion method for problems deﬁned on bounded domains in the plane

R

2

is an ex tension of the approach used for 1D problems on bounded domains in R. To

demonstrate this method, let's consider the heat prob lem:

(

u

t

=k∆u + h on Ω; t > 0

αu + β

@u

@n

= 0 on bnd(Ω); t > 0

;

1

where Ω is a bounded domain in R

2

with a piecewise smooth boundary bnd(Ω), and h is

a heat source term on Ω. Here,

@

@n

represents the directional derivative along the o utward

normal vector n on the boundary of Ω. To s o lve this problem, we ﬁrst form the associated

eigenvalue problem:

(

∆φ = ¡λφ on Ω

αφ + β

@φ

@n

= 0 on bnd(Ω)

: (5.1)

In this problem, λ is referred to as the eigenvalue, and φ represents the correspo nding

eigenfunction. The following theorem describes the properties of the eigenvalue problem:

Theorem 5.1. The eigenvalue problem possesses the following properties:

a) all eigenvalues are real, and they increases positively unbounded.

b) Eigenfunctions associated with diﬀerent eigenvalues are orthogonal in the following

sense:

hφ

n;m

; φ

n

0

;m

0

i:=

ZZ

Ω

φ

n;m

φ

n

0

;m

0

dS = 0;

if (n; m) =/ (n

0

; m

0

), where dS is the diﬀerential area in Ω.

c) The set of eigenfunctions fφ

n;m

g is an orthogonal basis for smooth functions deﬁned

in Ω. In other words, a continuously diﬀerentiable function f deﬁned on Ω can be

represented as:

f =

X

n;m

hf ; φ

n;m

i

kφ

n;m

k

2

φ

n;m

:

Problem 5. 1 . Prove that if β =/ 0 in the eigenvalue problem (5.1), the eigenvalues are greater than or

equal to

α

β

. If β =0, then all eigenvalues ar e strictly positive.

Problem 5.2. Let φ

n;m

and φ

n

0

;m

0

be eigenfunctions associated with λ

n;m

=/ λ

n

0

;m

0

. Show the following

orthogonality property

hφ

n;m

; φ

n

0

;m

0

i:=

ZZ

Ω

φ

n;m

φ

n

0

;m

0

dS = 0:

Hint:To prove, write down

(

∆φ

n;m

= ¡λ

n;m

φ

n;m

∆φ

n

0

;m

0

= ¡λ

n

0

;m

0

φ

n

0

;m

0

:

Multiply the ﬁrst equation and integrate over Ω:

ZZ

Ω

∆φ

n;m

φ

n

0

;m

0

dS = ¡λ

n;m

ZZ

Ω

φ

n;m

φ

n

0

;m

0

dS

Show that ∆ is symmetric over functions satisfying the given boundary condition, i.e.,

ZZ

Ω

∆φ

n;m

φ

n

0

;m

0

dS =

ZZ

Ω

∆φ

n

0

;m

0

φ

n;m

dS:

Since the set of eigenfunctions fφ

n;m

g forms a basis for continuously diﬀerentiable func-

tions on Ω, we can express the desired solution to the given heat problem as the series:

u =

X

n;m

U

n;m

(t) φ

n;m

; (5.2)

2 2D Linear Second-Order Equations

where U

n;m

(t) are undetermined coeﬃcient functions that need to be determined to e nsure

that the series is a valid solution to the given problem. Substituting this series into the heat

equation yields:

X

n;m

U

n;m

0

(t) φ

n;m

(x; y) =

X

n;m

kU

n;m

(t)∆ [φ

n;m

(x; y)] + h

Using the equality ∆[φ

n;m

] = ¡λ

n;m

φ

n;m

, we arrive at the following equation:

X

n;m

[U

n;m

0

(t) + kλ

n;m

U

n;m

(t)] φ

n;m

(x; y) = h:

To proceed and determine U

n;m

, we express the source term h in terms of the e igenf unction

basis fφ

n;m

g a s well:

h =

X

n;m

H

n;m

(t) φ

n;m

;

where the coeﬃcients H

n;m

(t) are determined by the inner production as:

H

n;m

=

hh; φ

n;m

i

kφ

n;m

k

2

:

Substituting this into the series equation leads to the following ordinary diﬀerential eq uation

for U

n;m

(t):

U

n;m

0

+ kλ

n;m

U

n;m

= H

n;m

(t):

The solution to this equation determines U

n;m

(t) and consequently, the solution u(x; y; t)

in the series form (5.2). It is important to note that this method provides us with a series

solution to the problem, and in most c a ses, this will be an inﬁnite series similar to the 1D

problems studied in the previous chapter.

In particular, if the heat equation is homogeneous (h is identically zero), the coe ﬃcient

functions U

n;m

will be exponential functions:

U

n;m

(t) = C

n;m

e

¡kλ

n;m

t

;

which leads to the general series solution for the homog en eous heat e quation:

u =

X

n;m

C

n;m

e

¡kλ

n;m

t

φ

n;m

(p);

where p 2 Ω, and C

n;m

are constants that can be determined through the initial condition

u(p; 0)= f. The series solution must converge to the given initial condition as t !0, which

results in:

f =

X

n;m

C

n;m

φ

n;m

(p):

Then, the constants C

n;m

for the homogeneous equation (h=0) can be determined as:

C

n;m

=

hf ; φ

n;m

i

kφ

n;m

k

2

:

5.1 Overview of the eigenfunction expansion method 3

The outlined method for solving the heat equation can be ex tende d to homogeneous or

non-homogeneous general second-order partial diﬀerential equations. Consider the following

equation:

(

a(t)u

tt

+ b(t) u

t

= k∆u + h on Ω

αu + β

@u

@n

= 0 on bnd(Ω)

:

This equation is general enough to encompass heat, wave, damped wave, and Poisson equa-

tions. To ﬁnd the solution, we represent the desired solutio n u as a series in terms of the

eigenfunctions fφ

n;m

g:

u =

X

n;m

U

n;m

(t) φ

n;m

;

where U

n;m

(t) are to be determined to satisfy the given partial diﬀerential equation. By

substituting the series into the equation, we obtain:

X

n;m

[a(t) U

n;m

00

(t) + b(t) U

n;m

0

(t) + λ

n;m

U

n;m

(t)]φ

n;m

= h:

Expanding the source term h in terms of the basis fφ

n;m

g, we arrive at the following equation

for U

n;m

:

a(t) U

n;m

00

(t) + b(t) U

n;m

0

(t) + λ

n;m

U

n;m

(t) = H

n;m

(t);

where H

n;m

(t) represents the c oeﬃcients of the expansion of h in terms of fφ

n;m

g:

h =

X

n;m

H

n;m

(t) φ

n;m

:

Solving this resulting second-order ordinary diﬀerential equation provides the general series

solution for the given partial diﬀerential equation.

5.2 Rectangular domains and Cartesian coordina te

5.2.1 Eigenfu nctions of ∆ in rectangles

We will now apply the outlined method to solve the eigenvalue problem in the Cartesian

coordinate system (x; y), where Ω is a rectangular domain deﬁned as Ω: (x₀; x₁)×(y₀; y₁).

The eigenvalue problem c a n be stated as follows:

(

∆φ(x; y) = ¡λφ(x; y) on Ω

αφ + β

@φ

@n

= 0 on bnd(Ω)

:

The rectangular geometry of the domain allows us to use the method of separation of vari-

ables. We assume th a t the eigenfunction φ(x; y) can be expressed as the product of two

separate functions, X(x) and Y (y), i.e., φ(x; y) =X(x)Y (y). Substituting this into the

eigenvalue problem, we obtain the following equation:

X

00

X

+

Y

00

Y

= ¡λ:

4 2D Linear Second-Order Equations

It is evident that this equation holds only if

X

00

(x)

X(x)

and

Y

00

(y)

Y (y)

are constants. Let us denote

X

00

(x)

X(x)

= ¡η and

Y

00

(y)

Y (y)

= ¡µ. We proceed to determine the assoc iated bo undary conditi ons

for the functions X(x) and Y (y).

The gene ral boundary condition αφ+ β

@φ

@n

=0, where n is the unit normal vector on the

boundary of Ω, le a ds to the following boundary conditions for X(x):

α

1

X(x

0

) ¡ β

1

X

0

(x

0

) = 0

α

2

X(x

1

) + β

2

X

0

(x

1

) = 0

:

Similarly, for Y (y), the boundary conditions are given by:

α

3

Y (y

0

) ¡ β

3

Y

0

(y

0

) = 0

α

4

Y (y

1

) + β

4

Y

0

(y

1

) = 0

:

In these equations, the values o f α and β depend on the speciﬁc problem and the boundary

conditions impo sed on Ω. Note that we also made use of the following relations:

@X

@n

x=x

0

= ¡X

0

(x

0

);

@X

@n

x=x

1

= X

0

(x

1

);

and similar ones for Y . The ﬁgure below depicts these relations:

y

Ω

@Y

@n

= ¡Y

0

x

@Y

@n

= Y

0

@X

@n

= X

0

@X

@n

= ¡X

0

By applying the sepa ration of variables technique, we can split the original eigenvalue

problem for the Laplacian operator ∆ into two separate eigenva lue problems: For the X(x)

function:

8

<

:

X

00

= ¡ηX

α

1

X(x

0

) ¡ β

1

X

0

(x

0

) = 0

α

2

X(x

1

) + β

2

X

0

(x

1

) = 0

;

and for the Y ( y) function:

8

<

:

Y

00

= ¡µY

α

3

Y (y

0

) ¡ β

3

Y

0

(y

0

) = 0

α

4

Y (y

1

) + β

4

Y

0

(y

1

) = 0

:

Consequently, the original eigenvalue λ can be determined by the relation λ= η+ µ, which

arises from the separation of variables approach.

5.2 Rectangular domains and Cartesian coordinate 5

Example 5.1. Below are the eigenfunctions and eigenvalues of the Laplace operator ∆ for

some well-known boundary conditions on the domain Ω: (0; a ) ×(0; b):

a) Dirichlet boundary condition:

φ(0; y) = φ(a; y) = φ(x; 0) = φ(x; b) = 0

Eigenfunctions:

φ

n;m

(x; y) = sin

nπ

a

x

sin

mπ

b

y

:

Eigenvalues:

λ

n;m

=

n

2

a

2

π

2

+

m

2

b

2

π

2

for n = 1; 2; ···, m = 1; 2; ···.

b) Neumann boundary condition:

φ

x

(0; y) = φ

x

(a; y) = φ

y

(x; 0) = φ

y

(x; b) = 0:

Eigenfunctions:

φ

n;m

(x; y) = cos

nπ

a

x

cos

mπ

b

y

:

Eigenvalues:

λ

n;m

=

n

2

a

2

π

2

+

m

2

b

2

π

2

;

for n = 0; 1; 2; ···, m = 0; 1; 2; ···.

c) Dirichlet in x-direction, and Neumann in y-direction

φ(0; y) = φ(a; y) = φ

y

(x; 0) = φ

y

(x; b) = 0:

Eigenfunctions:

φ

n;m

(x; y) = sin

nπ

a

x

cos

mπ

b

y

:

Eigenvalues:

λ

n;m

=

n

2

a

2

π

2

+

m

2

b

2

π

2

;

for n = 1; 2; ···, m = 0; 1; 2; ···.

d) Neumann in x-direction and Dirichlet in x-direction:

φ

x

(0; y) = φ

x

(a; y) = φ(x; 0) = φ(x; b) = 0:

Eigenfunctions:

φ

n;m

(x; y) = cos

nπ

a

x

sin

mπ

b

y

:

Eigenvalues:

λ

n;m

=

n

2

a

2

π

2

+

m

2

b

2

π

2

;

for n = 0; 1; 2; ···, m = 1; 2; ···.

e) Mixed in x and y-directions

φ(0; y) = φ

x

(a; y) = φ(x; 0) = φ

y

(x; b) = 0:

Eigenfunctions:

φ

n;m

(x; y) = sin

(2n ¡1)π

2a

x

sin

(2m ¡1)π

2b

y

:

6 2D Linear Second-Order Equations

Eigenvalues:

λ

n;m

=

(2n ¡1)

2

4a

2

π

2

+

(2m ¡1)

2

4b

2

π

2

;

for n = 1; 2; ···, m = 1; 2; ···.

f) Mixed in x and y-directions:

φ

x

(0; y) = φ(a; y) = φ

y

(x; 0) = φ(x; b) = 0:

Eigenfunctions:

φ

n;m

(x; y) = cos

(2n ¡1)π

2a

x

cos

(2m ¡1)π

2b

y

:

Eigenvalues:

λ

n;m

=

(2n ¡1)

2

4a

2

π

2

+

(2m ¡1)

2

4b

2

π

2

;

for n = 1; 2; ···, m = 1; 2; ···.

Example 5.2. Consider a smooth function f(x; y) deﬁned on (0; 1) ×(0; 1). Let's ﬁx y for

the function f (x; y) and consider f as a pure function of x. We can express this function

using the basis fsin(nπx)g for n = 1; 2; ··· as:

f(x; y) =

X

n=1

1

F

n

(y) sin(nπx);

where F

n

(y) is given by:

F

n

(y) = 2

Z

0

1

f(x; y) sin(nπx) dx:

Each function F

n

(y) can be represented in the basis fsin(mπy)g for m = 1; 2; ··· as:

F

n

(y) =

X

m=1

1

c

n;m

sin(mπy);

where c

n;m

is:

c

n;m

= 2

Z

0

1

F

n

(y) sin(mπy) d y:

Substituting the series for F

n

(y) into the series for f(x; y), we obtain:

f(x; y) =

X

n=1

1

X

m=1

1

c

n;m

sin(mπy) sin(nπx):

It i s important to note that the set fsin(nπx)s in(mπy)g for n =1; 2; ··· and m =1; 2; ··· forms

an orthogona l basis for functions deﬁned on the rectangle Ω=(0; 1) × (0; 1). Additionally,

the functions in this set satisfy the homogeneous Dirichlet boundary condition on bnd(Ω).

As a numerical example, let's consider f (x; y)=xy deﬁned on the rectangle (0; 1) ×(0;

1). The series representation of the function in the b a sis fsin(nπx) sin(mπy)g is:

xy =

X

n;m=1

1

4(¡1)

n+m

nmπ

2

sin(mπy) sin(nπx):

To observe the convergence of the double series to f(x; y) for (x; y) in the open rectangle,

let's consider the p o int x =

1

2

; y =

1

2

. The ﬁgure below depicts the convergence of the series

in terms of the number of iterations:

5.2 Rectangular domains and Cartesian coordinate 7

0 100 200 300 400 500

number of summations

0.23

0.24

0.25

0.26

0.27

0.28

0.29

As we observe, the double series converges to the true value very slowly. The reason for

this become s apparent. If we represent the function as a single series:

xy =

X

n=1

1

¡2(¡1)

n

nπ

y sin(nπx);

then the convergence at y =

1

2

requires fewer summations. The single series converges with

an error of less than 0.0033 a f ter 50 summations. Achieving the same accuracy for each

function F _n(y)=

¡2(¡1)

n

nπ

y requires a similar numbe r of iterations. Consequently, the double

summation needs a signiﬁcantly larger number of iterations to provide the same level of

convergence accuracy.

Exercise 5.1. Let Ω be the square (¡π; π) ×

¡

¡

π

2

;

π

2

in the xy-plane. Find the eigenvalues and

eigenfunctions of the eigenvalue problem ∆φ = ¡λφ subject to the following eigenvalue problem:

8

>

>

>

>

>

>

>

>

<

>

>

>

>

>

>

>

>

:

φ(¡π; y) = φ(π; y)

φ

x

(¡π; y) = φ

x

(π; y)

φ

¡

x; ¡

π

2

= φ

¡

x;

π

2

φ

y

¡

x; ¡

π

2

= φ

y

¡

x;

π

2

:

Use these eigenfunctions to re present the function f (x; y) = y sin(2x) deﬁned on Ω. What is the geometry

of the domain with the given boundary conditions of the eigenfunctions φ

n;m

?

Exercise 5.2. Let Ω be the unit square (¡1 ; 1) ×(0; 1).

a) Find the eig enfunctions and eigenvalues of the eigenvalue problem ∆φ = ¡λφ on Ω with the

following boundary conditions

8

<

:

u(¡1; y) = u(1; y)

u

x

(¡1; y) = u

x

(1; y )

u(x; 0) = u(x; 1) = 0

:

b) Show that these eigenfunctions are orthogonal with respect to the weight function σ = 1. Use

these eigenfunctions to approximate the function f(x; y) = x sin(πy). How many terms do you

need to use to obtain an approximati on with accuracy of 10

¡3

?

Exercise 5.3. Let Ω be the square (0; π) × (0; π). Find the eigenfunctions and eigenvalues of the

eigenvalue problem ∆φ = ¡λφ satisfying the following boundary conditions

u(0; y) = 0; u(1; y) + u

x

(1; y ) = 0

u(x; 0) = u

y

(x; 1) = 0

:

8 2D Linear Second-Order Equations