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

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

=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

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

+2ξu

∆u + f. In this equation, u

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

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:

(

=k∆u + h on Ω; t > 0

αu + β

= 0 on bnd(Ω); t > 0

;

where Ω is a bounded domain in R

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

a heat source term on Ω. Here,

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

= 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

i:=

Ω

n;m

dS = 0;

if (n; m) =/ (n

; m

), 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 =

n;m

hf ; φ

n;m

kφ

n;m

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

equal to

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

Problem 5.2. Let φ

n;m

and φ

be eigenfunctions associated with λ

n;m

=/ λ

. Show the following

orthogonality property

hφ

n;m

; φ

i:=

Ω

n;m

dS = 0:

Hint:To prove, write down

(

∆φ

n;m

= ¡λ

n;m

∆φ

= ¡λ

Multiply the ﬁrst equation and integrate over Ω:

Ω

∆φ

n;m

dS = ¡λ

n;m

Ω

n;m

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

Ω

∆φ

n;m

dS =

Ω

∆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 =

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:

n;m

(t) φ

n;m

(x; y) =

n;m

(t)∆ [φ

n;m

(x; y)] + h

Using the equality ∆[φ

n;m

] = ¡λ

n;m

, we arrive at the following equation:

n;m

(t) + kλ

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 =

n;m

(t) φ

n;m

;

where the coeﬃcients H

n;m

(t) are determined by the inner production as:

n;m

hh; φ

n;m

kφ

n;m

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

for U

n;m

(t):

n;m

+ kλ

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:

n;m

(t) = C

n;m

¡kλ

n;m

;

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

u =

n;m

¡kλ

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 =

n;m

(p):

Then, the constants C

n;m

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

n;m

hf ; φ

n;m

kφ

n;m

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

+ b(t) u

= k∆u + h on Ω

αu + β

= 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

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:

n;m

[a(t) U

n;m

(t) + b(t) U

n;m

(t) + λ

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

(t) + b(t) U

n;m

(t) + λ

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

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

= 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:

= ¡λ:

4 2D Linear Second-Order Equations

It is evident that this equation holds only if

(x)

X(x)

and

(y)

Y (y)

are constants. Let us denote

(x)

X(x)

= ¡η and

(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

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

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



X(x

) ¡ β

) = 0

X(x

) + β

) = 0

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



Y (y

) ¡ β

) = 0

Y (y

) + β

) = 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=x

= ¡X

);



x=x

= X

);

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

Ω

= ¡Y

= Y

= X

= ¡X

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:

= ¡ηX

X(x

) ¡ β

) = 0

X(x

) + β

) = 0

;

and for the Y ( y) function:

= ¡µY

Y (y

) ¡ β

) = 0

Y (y

) + β

) = 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:

n;m

(x; y) = sin



nπ



sin



mπ



Eigenvalues:

n;m

for n = 1; 2; ···, m = 1; 2; ···.

b) Neumann boundary condition:

(0; y) = φ

(a; y) = φ

(x; 0) = φ

(x; b) = 0:

Eigenfunctions:

n;m

(x; y) = cos



nπ



cos



mπ



Eigenvalues:

n;m

;

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

(x; 0) = φ

(x; b) = 0:

Eigenfunctions:

n;m

(x; y) = sin



nπ



cos



mπ



Eigenvalues:

n;m

;

for n = 1; 2; ···, m = 0; 1; 2; ···.

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

(0; y) = φ

(a; y) = φ(x; 0) = φ(x; b) = 0:

Eigenfunctions:

n;m

(x; y) = cos



nπ



sin



mπ



Eigenvalues:

n;m

;

for n = 0; 1; 2; ···, m = 1; 2; ···.

e) Mixed in x and y-directions

φ(0; y) = φ

(a; y) = φ(x; 0) = φ

(x; b) = 0:

Eigenfunctions:

n;m

(x; y) = sin



(2n ¡1)π



sin



(2m ¡1)π



6 2D Linear Second-Order Equations

Eigenvalues:

n;m

(2n ¡1)

(2m ¡1)

;

for n = 1; 2; ···, m = 1; 2; ···.

f) Mixed in x and y-directions:

(0; y) = φ(a; y) = φ

(x; 0) = φ(x; b) = 0:

Eigenfunctions:

n;m

(x; y) = cos



(2n ¡1)π



cos



(2m ¡1)π



Eigenvalues:

n;m

(2n ¡1)

(2m ¡1)

;

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) =

n=1

(y) sin(nπx);

where F

(y) is given by:

(y) = 2

f(x; y) sin(nπx) dx:

Each function F

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

(y) =

m=1

n;m

sin(mπy);

where c

n;m

is:

n;m

= 2

(y) sin(mπy) d y:

Substituting the series for F

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

f(x; y) =

n=1

m=1

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 =

n;m=1

4(¡1)

n+m

nmπ

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 =

; y =

. 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 =

n=1

¡2(¡1)

nπ

y sin(nπx);

then the convergence at y =

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)

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) ×

;



in the xy-plane. Find the eigenvalues and

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

φ(¡π; y) = φ(π; y)

(¡π; y) = φ

(π; y)

x; ¡



= φ



x; ¡



= φ



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

u(¡1; y) = u(1; y)

(¡1; y) = u

(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

(1; y ) = 0

u(x; 0) = u

(x; 1) = 0

8 2D Linear Second-Order Equations

Here you may need to ﬁnd the roots of the trigonometric function sin(z) + z cos(z). Use an online

application to determine a few roots of the function.

5.2.2 Heat, wave an d Poisson equations on rectangles

In this section, we will utilize th e eigenfunction expansion method to express the solution

of heat, wave, and Poisson equations as series expansions using the eigenfunctions of the

operator ∆, subject to speciﬁc bo undary conditions on rectangular domains. To illustrate

this method, we will solve several examples and provide detailed explanations.

Poisson's equation

The solution to Poisson's equation ¡∆u = f deﬁned on the rectangle (x

; x

) ×(y

; y

) in the

xy-plane can be expressed as a series in terms of the eigenfunction basis fφ

n;m

(x; y)g as:

u(x; y) =

n;m

(x; y);

where U

n;m

are undete rmined constants. Substituting this series into the Poisson equation

yields:

n;m

(x; y) = f (x; y);

and then U

n;m

can be determined using the inner produc t as:

n;m

hf ; φ

n;m

kφ

n;m

;

provided that λ

n;m

=/ 0.

Example 5.3. As a numerical example, let Ω = (0; π) × (0; π), and consider the Poisson

equation:



¡∆u = f on Ω

u = 0 on bnd(Ω)

;

where f(x; y)=y sin(x). The eig enfunctions of ∆ on Ω subject to the homogeneous Dirichlet

boundary c ondition are given by:

n;m

(x; y) 2fsin(nx) sin(my)g:

Since hf ; φ

n;m

i= 0 for n =/ 1, we can seek a solution in the single series form:

u(x; y) =

m=1

sin(my) sin(x):

Substituting this series into the Poisson equation yields:

m=1

1;m

sin(my) = y;

which determines U

as: U

2(¡1)

1;m

. Therefore, the seri es solution t o the equation is

obtained as:

u(x; y) = sin(x)

m=1

¡2(¡1)

m(1 + m

)

sin(my):

5.2 Rectangular domains and Cartesian coordinate 9

Exercise 5.4. The solution obtained above is a series solution to the given Poisson problem. In some

cases, it is possible to derive a closed-form solution. For the given problem, consider the solution as

u(x; y) = Y (y) sin(x). Substitute this into the equation and show that Y (y) must be equal to:

Y (y) = y ¡

sinh(y)

sinh(1)

The series representation of Y (y) in terms of the functions fsin(mπy)g is as the derived in the solution

of the example. The ﬁgure below depicts the function Y (y) and its series truncated up to three terms:

0 /2

0.2

0.4

0.6

0.8

1.2

The surface solu tion is illustrated b elow:

Remark 5.1. Consider the Poisson equation



¡∆u = φ

i;j

on Ω

u = 0 on bnd(Ω)

;

where φ

i;j

is the (i; j)

eigenfunction of ∆ on Ω. This equation can be interpreted in the

system interpretation as shown in the following block-diagram:

i;j

u =

i; j

i;j

Poisson system

¡∆

¡1

10 2D Linear Second-Order Equations

Since φ

i;j

is the eigenfunction of ∆ with the given boundary condition, the response of

the system to this input is equal to u =

i; j

i;j

. Since the set fφ

n;m

g is a basis for functions

deﬁne don Ω, we can use the superposition principle to determine the s o lution to the equation

¡∆u = f :=

n;m

The above equation is equivalent to the following system

u =

i; j

i;j

i; j

f =

i;j

Poisson system

¡∆

¡1

Exercise 5.5. Let Ω be the square (0; π) × (0; π). Solve the Poisson equation ¡∆u = y sin



on Ω

where u satisﬁes the following boundary conditions



u(0; y) = u

(π; y) = 0

u(x; 0) = u

(x; π) = 0

Exercise 5.6. Let Ω be the square (¡1; 1) ×(¡1; 1). Solve the Poisson equation ¡∆u = (1 + x)sin(πy)

on Ω where u satisﬁes the following boundary conditions

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

(0; y ) = u

(1; y )

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

Exercise 5.7. Let Ω be the unit square (0; 1) ×(0; 1), and consider the Poisson equation ¡∆u = f where

u satisﬁes the following boundary conditions



(0; y ) = u

(1; y ) = 0

(x; 0) = u

(x; 1) = 0

a) Find the solution to the problem if f = y cos(πx).

b) Show that there is not a solution to the problem if f = y sin(πx). Explain the reason for the non-

existence of the solution.

Heat problems on rectangles

Let's solve the following heat problem on Ω: (0; 1) ×(0; 1)

= ∆u + h on Ω

u = 0 on bnd(Ω)

u(x; y; 0) = 0

Since the set φ

n;m

2 fsin(nπx) sin(mπy)g forms a basis for the functions deﬁned on Ω, we

express the solution u(x; y; t) as

u =

n;m=1

n;m

(t)φ

n;m

;

for undetermined coeﬃcients U

n;m

(t). This series is a valid solution if it satisﬁes the e quation:

n;m=1

n;m

(t) φ

n;m

(x; y) =

n;m=1

¡λ

n;m

(t) φ

n;m

(x; y) + h(x; y; t);

5.2 Rectangular domains and Cartesian coordinate 11

where λ

n;m

= (n

+ m

)π

. Moving the second summation to the left side, we can write

n;m=1

n;m

(t) + λ

n;m

(t)] φ

n;m

(x; y) = h(x; y; t):

To proceed, we use the representation of h as a series in terms of eigenfunctions fφ

n;m

g a s:

h(x; y; t) =

n;m=1

n;m

(t) φ

n;m

(x; y);

where H

n;m

(t) are determined by the inner product

n;m

(t) = 4

h(x; y; t) φ

n;m

(x; y) dxdy:

For instance, let h = φ

1;1

(x; y). In this case, we have:

n;m

(t) =



1 (n; m) = (1; 1)

0 (n; m) =/ (1; 1)

According to the given initial condition, we can write the initial value problems for U

n;m

as:

(

n;m

(t) + λ

n;m

(t) = 0

n;m

(0) = 0

; (n; m) =/ (1; 1);

and

(

n;m

(t) + λ

n;m

(t) = 1

n;m

(0) = 0

; (n; m) = (1; 1):

Solving the above system yields: U

n;m

(t) = 0 for (n; m) =/ (1; 1) and

1;1

(t) =

2π

(1 ¡e

¡2π

Finally, we arrive at:

u(x; y; t) =

2π

(1 ¡e

¡2π

) sin(πx) sin(πy):

System interpretation

Consider the equation

= k∆u + H

i; j

(t) φ

i;j

(x; y) on Ω

u = 0 on bnd(Ω)

u(x; y; 0) = 0

;

where φ

i;j

(x; y) is an eigenfunction of ∆ with the given boundary condition. From a system

perspective, the heat system is triggered by the source te rms H

i; j

(t) φ

i;j

(x; y). This can be

depicted in the following diagram:

12 2D Linear Second-Order Equations

¡k∆)

¡1

u = U

i;j

(t) φ

i;j

(t)φ

i;j

Heat system

In this diagram, the function U

i;j

(t) is the solution to the ordinary diﬀere ntial equation

i;j

+ kλ

i;j

= H

i;j

In the above example, the heat system is triggered by the source φ

1;1

(x; y), resulting in the

response:

1;1

1 ¡e

¡λ

1;1

(x; y):

Now, if the heat sy stem is triggered by the source term

h =

i; j

i;j

(t) φ

i;j

(x; y);

then by the superposition principle, the response can be written as the summation depicted

in the followi ng diagram:

u =

i;j

(t) φ

i;j

(t)φ

i;j

Heat system

¡k∆)

¡1

A heat system can be triggered by the initial condition itself. Consider the fo llowing

problem on Ω = (0; 1) ×(0; 1):

= ∆u on Ω

u = 0 on bnd(Ω)

u(x; y; 0) = φ

i;j

; (5.3)

where φ

= sin(iπx) sin(jπx). It can be shown that the respon se of the heat s ystem to this

initial condition is given by:

= e

¡λ

(x; y):

Now, let's con sider a slightly diﬀerent problem:

= ∆u + δ(t) φ

on Ω

u = 0 on bnd(Ω)

u(x; y; 0) = 0

;

where the initial condition appears as an external source multiplied by the Dirac delta

function δ(t). In this case, we seek the solution in the form:

= U(t) φ

(x; y):

Substituting this into the equation yields:

+ λ

U = δ(t):

5.2 Rectangular domains and Cartesian coordinate 13

This ordinary differential equation can be solved using the Laplace transform method,

resulting in: U(t) = e

¡λ

u(t), where u(t) is the He aviside function deﬁned as:

u(t) =



1 t > 0

0 t < 0

Thus, the solution u

(x; y; t) for t> 0 is the same as the solution to the problem (5.3), namely

¡λ

(x; y).

Exercise 5.8. What is the diﬀerence between the solution to the problem

= ∆u + φ

on Ω

u = 0 on bnd(Ω)

u(x; y; 0) = 0

;

and t he solution to the followin g one

= ∆u + u(t)φ

on Ω

u = 0 on bnd(Ω)

u(x; y; 0) = 0

;

where u(t) is the Heaviside function.

Geometrical interpretation

From a geometric perspective, we can rephrase the argument in the above remark as follows:

each eigenfunction φ

i;j

deﬁnes a direction in an inﬁnite-dimensional vector space. Along

each direction (eigenfunction), the heat partial diﬀe rential equation reduces to an ordinary

diﬀerential equation, as depicted in the following diagram:

i;j

+ kλ

i;j

= H

i;j

1;2

i;j

1;1

Exercise 5.9. Write down the series solut ion of the following heat problem deﬁned on Ω: (0; π) ×(0; π)

= ∆u

u(0; y; t) = u(π; y; t) = 0

u(x; 0; t) = u

(x; π; t) = 0

u(x; y; 0) = x sin



Exercise 5.10. Write down the series so lution of the following heat problem deﬁned on Ω:(0; π) ×(0; π)

= ∆u + e

¡t

(0; y; t) = u

(π; y; t) = 0

(x; 0; t) = u

(x; π; t) = 0

u(x; y; 0) = 0

14 2D Linear Second-Order Equations

Exercise 5.11. Consider the heat problem

= ∆u + sin(πx) sin(πy)

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

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

a) The source terms is independent of time t. For this, we can consider the solution as

u(x; y; t) = V (x; y) + w(x; y; t):

Substituting u into the equation leads to a Poisson equation for V .

b) The equation for a w will be a homogeneous equation with the general solution

w(x; y; t) =

n;m

¡λ

n;m

(x; y);

where φ

n;m

= sin(nπx) sin(mπy). Find the solution of the problem if u(x; y; 0) = 0. This solution

is the sa me the solution we obtained by employing the eigenfunction expansion method.

Exercise 5.12. Consider the heat problem

= ∆u ¡ y sin(x)

u(0; y; t) = u(π; y; t) = 0

u(x; 0; t) = u(x; π; t) = 0

u(x; y; 0) = 0

a) The source terms is independent of time t. For this, we can consider the solution as

u(x; y; t) = V (x; y) + w(x; y; t):

Substituting u into the equation leads to a Poisson equation for V . Find a closed form solution

for V ( x; y).

b) The equation f or a w will be a homogeneous equation. Choose a single series for w based on the

form of the source term. Use the initial condition for u and obtain the valued series solution of

the problem.

Wave problem

Let's solve the following wave equation deﬁned on the unit square Ω = (0; 1) ×(0; 1):

= c

∆u + h on Ω

u = 0 on bnd(Ω)

u(x; y; 0) = 0

(x; y; 0) = 0

Representing the desired solution u in terms of the eigenfunctions fφ

n;m

g, we reach the

following equation:

n;m=1

n;m

(t) + c

n;m

] φ

n;m

(x; y) = h(x; y; t):

Using the representation of h as a series in terms of φ

n;m

h =

n;m=1

n;m

(t) φ

n;m

(x; y);

5.2 Rectangular domains and Cartesian coordinate 15

we arrive at the following equation for the coeﬃcients U

n;m

+ c

n;m

= H

n;m

(t):

For example, let h = sin(iπx) sin(jπy). Since H

(t) = 1, and H

n;m

(t) = 0 for (n; m) =/ (i; j),

we obtain th e following system for U

n;m

(

+ λ

= 1

(0) = U

(0) = 0

;

(

n;m

+ λ

n;m

= 0

n;m

(0) = U

n;m

(0) = 0

; (n; m) =/ (i; j):

The solution to the second system is U

n;m

(t) = 0, and for the ﬁrst one is

(t) =



1 ¡cos



;

Therefore, the solution to the given wave problem is:

(x; y; t) =



1 ¡cos



(x; y):

Remark 5.2. The logic b eh ind the solution of the above example is clear. From a system

perspective, the response of the wave system (@

¡ c

∆)

¡1

to the input H

i;j

(t) φ

i;j

(x; y)

is given by U

i; j

(t) φ

i; j

(x; y), where U

i;j

(t) satisﬁes the second-order ordinary diﬀerential

equation:

i;j

+ c

i;j

= H

i; j

(t):

Thus, we can represent the solution as shown in the block-diagram below:

u = U

i;j

(t) φ

i;j

(t)φ

i;j

Wave system

¡c

∆)

¡1

For general source terms h =

i; j

i;j

, the response of the system will be smmation

as depicted in the diagram below:

u =

i;j

(t) φ

i;j

(t)φ

i;j

Wave system

¡c

∆)

¡1

An equivalent interpretation, which is ge o metric, is that the wave partial diﬀerential

equation re duces to a second-order ordinary diﬀerential equation along each direction deﬁned

by φ

in an inﬁn ite-dimensional vector space spanned by the set fφ

n;m

(x; y)g for n = 1;2; ···,

m = 1; 2; ··· .

Exercise 5.13. Solve the following problem

= c

∆u + sin(t) sin( πx) sin(πy) on Ω

u = 0 on bnd(Ω)

u(x; y; 0) = 0

(x; y; 0) = 0

;

16 2D Linear Second-Order Equations

where Ω is the unit square Ω = (0; 1) ×(0; 1).

Exercise 5.14. Solve the following wave problem deﬁned on Ω = (0; 1) ×(0; 1)

= ∆u ¡sin(t)y sin



u(0; y) = u

(1; y ) = 0

u(x; 0) = u

(x; 1) = 0

u(x; y; 0) = 0

(x; y; 0) = 0

Exercise 5.15. Consider the following wave equation

= ∆u ¡ y sin



u(0; y) = u

(1; y ) = 0

u(x; 0) = u

(x; 1) = 0

u(x; y; 0) = 0

(x; y; 0) = 0

a) Since the source term is indepen dent o f time, we can consider the solution u as follows

u(x; y; t) = V (x; y) + w(x; y; t):

Substituting u into the equation results to a Poisson equa tion for V . Find t he closed form solution

for this equation.

b) Write a single summation series solution for w based on the form of the source term. Use the

initial condition and det ermine the solution u.

Exercise 5.16. Consider the following wave equation Ω = (0; 1) ×(0; 1)

= ∆u ¡sin(πx) sin(πy)

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

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

u(x; y; 0) = sin(πx) sin(πy)

(x; y; 0) = 0

Exercise 5.17. Solve the following damped wave equation

+ 0.2u

= ∆u

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

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

u(x; y; 0) = sin(πx) sin(πy)

(x; y; 0) = 0

5.2.3 Laplace equation

So far, we have solved partial diﬀerential equations with homogeneous boundary conditions.

To illustrate the solution to problems with non-homogeneous boundary conditions, let's

consider the followi ng heat problem:

= ∆u on Ω

u = f on bnd(Ω)

u(x; y; 0) = 0

5.2 Rectangular domains and Cartesian coordinate 17

Since the source term in the heat equation is the time-indep en de nt boundary condition f ,

we can express the solution u as:

u(x; y; t) = V (x; y) + w(x; y; t):

By substituting u into the equatio n, we obtain the Laplace equation:



∆V = 0 on Ω

V = f on bnd(Ω)

In this section, our focus is on solving La place's equation.

Let Ω be the rectangle (0; a) ×(0; b). Consider the Laplace equation ∆u = 0 subject to

the bou ndary condition shown below

Ω

u + β

= g(x)

u + β

= f(x)

u + β

= 0

u + β

= 0

Using the separation u(x; y) = X(x ) Y (y), we arrive at the equation:

= 0:

Since u satisﬁes the homogeneous boundary condition in the x-direction as:



u + β

= 0

u + β

= 0

;

we arrive at the following eigenvalue problem for X(x):

= ¡λX

X(0) + β

(0) = 0

X(a) + β

(a) = 0

Problem 5.3. Prove that eigenvalue λ of the problem are non-negative.

Let this eigenvalue problem to be solved for eigenfunctions φ

(x) and eigenvalues λ

for

n = 1; 2; ···. The equation for Y satisﬁes the second-order ODE:

¡λ

Y = 0:

If λ

= 0 is an eigenvalue, the is solved for Y

(y) = A

+ B

y. For n ≥1, the equatio n is solved

for the following functions:

(y) = A

cosh( λ

y) + B

sinh( λ

y):

18 2D Linear Second-Order Equations

The general solution of the Laplace equation is then obtained as:

u(x; y) = (A

+ B

y) φ

(x) +

n=1

cosh( λ

y) + B

sinh( λ

y)] φ

(x):

The parameters A

; B

are determined by applying the boundary conditions at y = 0 and

y = b.

Example 5.4. Let's sol ve the following Laplace's equation on the rectangular domain Ω : (0;

1) ×(0; 1):

∆u = 0

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

u(x; 0) = sin(2πx); u(x; 1) = ¡sin(2πx)

Since the boundary condition is Dirichlet homogeneous in the x-direction, the eigenfunctions

corresponding to this condition are given by φ

(x)= s in(nπx), with associated eigenvalues

. We can express the general series solution as:

u(x; y) =

n=1

cosh(nπy) + B

sinh(nπy)] sin(nπx):

To determine the coeﬃcients A

and B

, we apply the boundary conditions at y=0 and y=1.

From u(x;0)=sin(πx), we ﬁnd that A

=1, and A

= 0 for all n =/ 2. From u(x; 1)=¡sin(2πx),

we obtain: B

= ¡

1 + cosh(2π)

sinh(2π)

, and B

= 0 for n =/ 1. Thus, we obtain

u(x; y) =

sinh(2π(1 ¡y)) ¡sinh(2πy)

sinh(2π)

sin(2πx):

The ﬁgure below illustrate the surface of u(x; y):

Example 5.5. Let's solve the Laplace equation ∆u=0 on the unit square (0; 1) ×(0; 1) with

the following boundary conditions:



u(0; y) = ¡1; u(1; y) = 1

u(x; 0) + u

(x; 0) = 0; u

(x; 1) = 0

5.2 Rectangular domains and Cartesian coordinate 19

Since the problem is homogeneous in the y-direction, we ﬁrst solve the following eigenvalue

problem for φ(y):

= ¡λφ

φ(0) + φ

(0) = 0

(1) = 0

It can be shown that the eigenva lues of the problem are strictly positive and satisfy the

equation:

sin( λ

) + cos( λ

) = 0:

Accordingly, the eigenfunctions are obtained as:

(y) = sin( λ

y) ¡ λ

cos( λ

The general series solution to the problem is:

u(x; y) =

n=1

cosh( λ

x) + B

sinh( λ

x)] φ

(y):

To determine the coeﬃcients A

and B

, we apply the boundary conditions. At x=0, we

obtain:

¡1 =

n=1

(y);

which gives

¡1

kφ

(y) dy =

¡1

kφ

At x = 1, we reach the equation

1 =

n=1

¡1

kφ

cosh( λ

) + B

sinh( λ

)

(y);

and thus:

1 + cosh( λ

)

kφ

sinh( λ

)

The ﬁnal series solution is obtained as:

u(x; y) =

n=1

¡1

kφ

cosh( λ

x) +

1 + cosh( λ

)

kφ

sinh( λ

)

(y):

Example 5.6. Consider the following Laplace e quation on Ω = (0; 1) ×(0; 1)

∆u = 0

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

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

20 2D Linear Second-Order Equations

Since the boundary conditions are non-homogeneous in both directions, we can split the

problem into two sub-problems, each homogeneous in one direction:

(1)

∆u = 0

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

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

; (2)

∆u = 0

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

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

The solution to the ﬁrst problem is given by:

(x; y) =

n=1

¡2(¡1)

nπ sinh(nπ)

sinh(nπy) sin(nπx):

And the solution to the second sub-problem is:

(x; y) =

n=1

¡2(¡1)

nπ sinh(nπ)

sinh(nπx) sin(nπy):

The superpos ition solution to the original equation is

u(x; y) =

n=1

¡2(¡1)

nπ sinh(nπ)

[sinh(nπy) sin(nπx) + sinh(nπx) sin(nπy)]:

On the other hand, it can be veriﬁed that the function u=xy satisﬁes the given Laplace

equation. According to the uniqueness theorem, this closed-form solution and the series

solution must be equal, resulting in the equation:

xy =

n=1

¡2(¡1)

nπ sinh(nπ)

[sinh(nπy) sin(nπx) + sinh(nπx) sin(nπy)]:

Exercise 5.18. Solve the Laplace equation ∆u = 0 satisfying boundary conditions given below

u(x; 0) = ¡1; u(π; y) = 0; u(x; π) = 1; u(0; y) = 0

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

u(x; 0) = ¡1; u(2π; y) = 1; u(x; π) = ¡1; u(π; y) = 1

(x; 0) = 0; u(2π; y) = 1; u

(x; π) = 0; u(π; y) = 1

Exercise 5.19. Consider the Laplace equation ∆u = 0 for function u(x; y) deﬁned on the domain Ω:

(0; 1) ×(0; 1) with the following boundar y conditions

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



πy



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



πx



a) Find a series solution to the given equation.

5.2 Rectangular domains and Cartesian coordinate 21

b) The solution derived in part a) is a series form of the closed form solution u(x; y)= sin

πx



sin

πy



ver ify that this function satisﬁes the Laplace equation and the given boundary conditions.

c) To make sure that these two solutions are the same, choose some arbitrary point or points inside

Ω and verify tha t the series and the closed form solution are the same.

Exercise 5.20. In the domain Ω = (0; 1) ×(0; 1):

a) Solve the Laplace equation

∆u = 0

u(x; 0) = 0; u(0; y) = 0

(1; y ) = y; u(x; 1) = x

b) So lve t he Poisson equation ∆u = y with the boundary conditions given in part a).

5.2.4 Problems

Problem 5.4. Let Ω be the unit square (0; 1) ×(0; 1). Consider the following Poisson equation on Ω



∆u = xy on Ω

u = 0 on bnd(Ω)

a) Find the double series solution of the pr ob lem in terms of the eigenfunctions of ∆ satisfying the

given boundary condition.

b) As we observed above, the convergence of double series is usually slower than single series. Let's

consider the solution to the problem as the following single series

u(x; y) =

n=1

(y) sin(nπx):

Substitute thi s series into the Poisson equation as obtain an ordinary diﬀerential equation of order

2 for U

(y).

c) Repeat the argument for the single series

u =

n=1

(x) sin(nπy):

Problem 5.5. Let Ω be the rectangle (0; 1) ×(0; 1). Consider th e Poisson equation

∆u = x sin(πy)

on Ω where u satisﬁes the homogeneous Dirichlet boundary condition: u = 0 on bnd(Ω).

a) To solve the equation and determine u = u(x; y), we express the solution u as the series in terms

of the eigenfunctions in the s et fsin(nπx) sin(mπy)g. Note that these functions satisfy the given

bo undary condition:

u(x; y) =

n=1

m=1

n;m

sin(nπx) sin(mπy):

Substitute the series into the diﬀerential equation and determine the coeﬃcients C

n;m

. Hint: Note

that the solution could be expressed as

u = sin(πy)

n=1

sin(nπx);

22 2D Linear Second-Order Equations

according to the source terms of the equation: x sin(πy).

b) Try a closed form solution of the form u(x; y) = U (x) sin(πy). Substitute this into the equation

and show that U (x) is of the form

U(x) =



sinh(πx)

sinh(π)

¡x



c) According to the uniqueness theorem of the problem, the series solution derived in part a) must be

equal to the closed form solution derived in part b). Chose an arbitrary p oint on Ω and compare

two solutions at the chosen point.

Problem 5.6. Consider the following Poisson equation on Ω = (0; a) ×(0; b)

(

∆u = f (x; y) on Ω

= 0 on bnd(Ω)

Show that the problem is solvable only if

Ω

f(x; y) dxdy = 0:

Problem 5.7. Consider the following Poisson equation

∆u = y sin



u(0; y) = u

(1; y ) = 0

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

Find a series solution to the problem and then ﬁnd a closed form solution to it.

Problem 5.8. Find a series or closed form solution to the following Poisson equation

∆u = sin(x)

on the domain Ω: (0; π) ×(0; π) satisfying the following boundary conditions

u(0; y) = 0; u(π; y) = 0; u

(x; 0) = 0; u

(x; π) = 0:

Problem 5.9. Let Ω be the rectangle Ω: (0; 1) ×(0; 1). Solve the following heat equ ation on Ω:

= k∆u on Ω

= 0 on bnd(Ω)

u(x; y; 0) = cos(2πx) cos(πy)

Problem 5.10. Let Ω be the rectangle (0; π) × (0; π). Cons ider the following damped wave equation

on Ω:

+ 0.2u

= c

∆u

(0; y; t) = u

(π; y; t) = 0

u(x; 0; t) = u(x; π; t) = 0

a) Write down the general series solution of the equation in terms of the appropriate eigenfunctions

of the opera tor ∆.

b) Determine the solution of the problem if the initial conditions are given by: u(x; y; 0) = 0, and

(x; y; 0) = cos(x) sin(2y).

5.2 Rectangular domains and Cartesian coordinate 23

Problem 5.11. Let Ω be the rectangle (0; π) ×(0; π). Consider the heat equation

= ∆u + sin



sin



u(0; y; t) = u

(π; y; t) = 0

u(x; 0; t) = u

(x; π; t) = 0

u(x; y; 0) = 0

The initial condition of the system is zero, and the dynamic is driven by the source term h=sin



sin



This source term is independent of time t. To solve the equation and determine the solution u(x;

y; t), we ﬁrst write u as

u(x; y; t) = V (x; y) + W (x; y; t):

a) Substitute u into equation. This leads to a Poisson equation for V (x; y). Determine V .

b) Write down the general series solution for W (x; y; t).

c) Apply the initial condition for u, and determine the solution to the given problem.

d) An alternative and more straightforward way to solve the given equa tion is as follows: the source

term is an eigenfunction of the operator ∆ with the given boundary condition. The response of

the heat system to this input is

u(x; y; t) = U (t) sin





sin





Determine U(t) and verify that the result conﬁrms the result of part (c).

Problem 5.12. Now, let us transfer the source term h to the initial condition and consid er the following

equation on Ω := (0; π) ×(0; π)

= ∆u

u(0; y; t) = u

(π; y; t) = 0

u(x; 0; t) = u

(x; π; t) = 0

u(x; y; 0) = sin



sin



a) Determine the solution to the given equation.

b) Show that the solution is equal to the solution to the following system

= ∆u + δ(t) sin



sin



u(0; y; t) = u

(π; y; t) = 0

u(x; 0; t) = u

(x; π; t) = 0

u(x; y; 0) = 0

;

where δ(t) is the Dirac delt a function.

Problem 5.13. Solve the following wave problem deﬁned on Ω = (0; π) ×(0; π)

= ∆u

(0; y; t) = u

(π; y; t) = 0

(x; 0; t) = u

(x; π; t) = 0

u(x; y; 0) = 3cos(2x) + y

Problem 5.14. Use the superpos ition principle and solve the following heat problem on Ω := (0; π) ×(0;

π)

= ∆u + sin



sin



u(0; y; t) = u

(π; y; t) = 0

u(x; 0; t) = u

(x; π; t) = 0

u(x; y; 0) = sin



sin



¡sin



sin



24 2D Linear Second-Order Equations

Problem 5.15. Consider the following heat equation on Ω := (0; 1) ×(0; 1)

= ∆u

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

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

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

a) Write the solution of the problem as the series

u(x; y; t) =

n;m=1

n;m

(t) φ

n;m

(x; y);

where φ

n;m

are the eigenfunctions of ∆ with given boundary conditions. Derive an ordinary

diﬀerential equation for undetermined coeﬃcients U

n;m

(t)

b) So lve t he obtained diﬀerential equation and writ e the general series s olution of the equation.

c) Apply the given initial condition and derive the particular series solution of the problem.

d) Verify that the solution is equivalent to the superposition of the solution of the following equations

(1)

= ∆u

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

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

u(x; y; 0) = x

; (2)

= ∆u

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

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

u(x; y; 0) = y

Problem 5.16. Let Ω be the rectangle (0; 1) ×(0; 1). Consider the following wave equation

= ∆u + x sin(πy) on Ω

u = 0 on Ω

u(x; y; 0) = 0

(x; y; 0) = 0

a) Since the external source term is independent of time t, we can express the solution u as

u(x; y; t) = V (x; y) + w(x; y; t);

where V satisﬁes the Poisson equation



¡∆V = x sin(πy) on Ω

V = 0 on bnd(Ω)

Find a closed form solution for V (x; y).

b) The function w satisﬁes the fol lowing equati on

= ∆w on Ω

w = 0 on bnd(Ω)

w(x; y; 0) = ¡V (x; y)

(x; y; 0) = 0

Find the series solution for w(x; y; t).

Problem 5.17. Let Ω be the set Ω = (0; π) ×(0; π).

a) Solve the Poisson equation



∆u = xy on Ω

u = 0 on bnd(Ω)

5.2 Rectangular domains and Cartesian coordinate 25

b) Use the result obtained in the above part and solve the following heat problem

= ∆u ¡xy on Ω

u = 0 on bnd(Ω)

u(x; y; 0) = x

Problem 5.18. Solve the following problem on Ω = (0; π) ×(0; π)

= ∆u + txy on Ω

u = 0 on bnd(Ω)

u(x; y; 0) = x:

Problem 5.19. Let Ω be the rectangle (0; 1) × (0; 1). Find a series solution for the following wave

equation

= c

∆u + e

¡t

x sin(πy) on Ω

u = 0 on Ω

u(x; y; 0) = 0

(x; y; 0) = 0

Problem 5.20. We aim to solve the following heat problem on Ω := (0; 1) ×(0; 1)

= L[u]

bnd(Ω)

u(x; y; 0) = e

¡x

sin(πx) sin(πy)

;

where L is the following operator

L[u] = u

+ 2u

+ u

a) Determine the eigenvalues and eigenfunctions of the operator L satisfying the given bou ndary

conditions. In what sense are these eigenfunctions orthogonal?

b) Write the series solution of the given problem in terms of this eigenfunctions.

Problem 5.21. We aim to solve the following wave equation on Ω = (0; 1) ×(0; 1)

= ∆u

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

(0; y; t) = u

(1; y; t) = 0

a) Find the eigenvalues and eigenfunctions of the associated eigenvalue problem, that is

∆φ = ¡λφ

φ(x; 0) = φ(x; 1) = 0

(0; y ) = φ

(1; y ) = 0

b) Use the eigenfunctions expansion method and write t he general series solution of the wave equa-

tion as follows:

u(t; x; y) =

n=0

m=1

(t) φ

(x; y):

Find diﬀerentia l equations that U

n;m

satisfy, and then solve them.

26 2D Linear Second-Order Equations

c) Assume that the initial conditions are given by u(x; y; 0) = 0, u

(x; y; 0) = y cos(πx). Show that

the solution looks like the following one

u(t; x; y) = cos(πx)

m=1

sin( λ

t) sin(mπy);

for s ome constants β

. Determine these parameters.

If you are interested in making a video to visualize your solution, the following co de can helps

%set time of running

T=5;

[x,y,t]=meshgrid(0:0.01:1,0:0.01:1,0:0.05:T);

u=zeros(size(x));

for m=1:10

u=u-2*(-1)^m*cos(pi*x).*sin(sqrt(1+m^2*pi^2)*t).*sin(m*pi*y)/

(m*pi*sqrt(1+m^2*pi^2));

end

% The fil e filename.avi is saved on your system. You can change the name if

you wish

obj=VideoWriter('filename.avi');

open(obj);

for i=1:20*T+1

surf(x(:,:,1),y(:,:,1),u(:,:,i),'edgecolor','none');

axis([0 1 0 1 -0.25 0.25]);

hold off;

f= getframe(gcf);

writeVideo(obj,f);

pause(0.1)

end

obj.close();

Problem 5.22. Consider the following heat problem on Ω = (0; 1) ×(0; 1)



= 0.1(∆u + 2u

) on Ω

u = 0 on bnd(Ω)

a) Find the eigenfunctions and eigenvalues of the following eigenvalue problem



∆φ + 2φ

= ¡λφ on Ω

φ = 0 on bnd Ω

b) Write the solution of the given heat equation as follows:

u(t; x; y) =

n=1

m=1

(t) φ

(x; y):

Find a diﬀerential equation for U

(t). Solve it and write down the general solution of the

equation.

c) Assume u(x; y; 0) = xe

¡x

sin(2πy). Find the solutio n that satisﬁes the given initial condition.

d) Run the following code and observe the change of the temperature with respect to time:

%set time of running

T=1;

[x,y,t]=meshgrid(0:0.01:1,0:0.01:1,0:0.05:T);

u=zeros(size(x));

5.2 Rectangular domains and Cartesian coordinate 27

for n=1:10

l=1+(4+n^2)*pi^2;

u=u-2*(-1)^n*sin(2*pi*y).*exp(-0.1*l*t).*sin(n*pi*x)/(n*pi);

end

% The fil e filename.avi is saved on your system. You can change the name if

you wish

obj=VideoWriter('filename.avi');

open(obj);

for i=1:20*T+1

surf(x(:,:,1),y(:,:,1),u(:,:,i),'edgecolor','none'); view(2);colormap(jet);

axis equal;

caxis([-0.2,0.2]); axis([0 1 0 1]);

hold off;

f= getframe(gcf);

writeVideo(obj,f);

pause(0.2)

end

obj.close();

Problem 5.23. In the domain Ω = (1; e) ×(0; 1):

a) Solve the eigenvalue problem

(

+xφ

+ φ

= ¡λφ on Ω

φ = 0 on bnd(Ω)

b) So lve t he wave equation

= x

+xu

+ u

+ ty on Ω

u = 0 on bnd(Ω)

u(x; y; 0) = 0

(x; y; t) = 0

Problem 5.24. Let Ω be the set Ω = (¡π; π) ×(¡π; π). Consider the following eigenvalue problem

∆φ = ¡λφ

φ(¡π; y) = φ(π; y); φ

(¡π; y) = φ

(π; y)

φ(x; ¡π) = φ(x; π); φ

(x; ¡π) = φ

(x; π)

a) Show the eigenvalues are non-negative: λ ≥0.

b) Show that eigenfunctions of the problem are

(x; y) = fcos(nx + my); sin(nx + my)g;

for n; m = 0; 1; ···.

c) Show that

¡π

(x; y) φ

(x; y) dxdy =



1 (n; m) = (n

; m

)

0 (n; m) =/ (n

; m

)

d) solve the following heat problem:

= ∆u

φ(¡π; y; t) = φ( π; y; t); φ

(¡π; y; t) = φ

(π; y; t)

φ(x; ¡π; t) = φ(x; π; t); φ

(x; ¡π; t) = φ

(x; π; t)

u(x; y; 0) = xy

28 2D Linear Second-Order Equations

Problem 5.25. Let D be the domain shown below

` = 1

Consider the following equa tion on D

= 2∆u ¡2u

;

with the homogeneous Dirichlet boundary condition. Use a transformation to transform the domain D

to an unit square. Derive the appropriate diﬀerential equation in the new coordinate. Solve the new

equation and then write down the solution of the given equation on D.

Problem 5.26. Let Ω be the unit square (0; 1) ×(0; 1). Consider the following Laplace equation

∆u = 0

u(0; y) = ¡1; u(1; y) = 1

u(x; 0) = @

u(x; 1) = 0

a) Find a series solution to the equation.

b) Verify that the the function u = 2x ¡1 solves the equation and thus it is equal to the series so lution

obtained in part a).

c) Now consider the following Poisson equation

∆u = y

u(0; y) = ¡1; u(1; y) = 1

u(x; 0) = @

u(x; 1) = 0

Consider the solution u as u = V + w, where V satisﬁes the Laplace equation given above. Find

a series solution for w, and determine u(x; y).

Problem 5.27. Solve the following Poisson equations

∆u = sin(πy)

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

u(x; 0) = ¡1; u(x; 1) = 1

∆u = sin(2x) + sin(3y)

u(0; y) = ¡1; u(π; y) = 1

u(x; 0) = ¡1; u(x; π) = 1

∆u = xy

(0; y ) = 0; u

(π; y) = 0

u(x; 0) = 0; u(x; π) = 1

Problem 5.28. We aim to solve the following equation

= ∆u + 2u

+ u + y; (x; y) 2(0; 1) ×(0; 1);

5.2 Rectangular domains and Cartesian coordinate 29

with the boundary conditions

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

and t he initial condition u(x; y; 0) = 0.

a) Show that the steady state solution V (x; y) is

V (x; y) =

n=1



sinh(nπy) +

2((¡1)

¡1)



¡x

sin(nπx):

Determine the value of B

b) Show that the transient solution has the form

w(x; y; t) =

m=1

n=1

¡(n

)π

¡x

sin(nπx) sin(mπy)

Determine the value of A

n;m

Problem 5.29. Let Ω be the rectangle (0; 1) ×(0; 1)

u = 0

Ω

u = 0

u = 0 u = sin(πy)

a) Find a closed form solution to the Laplace equation ∆u =0 with the boundary conditions given

in the ﬁgure.

b) So lve t he following wave problem

= ∆u ¡sin(t) on Ω

bnd(Ω)

: given in the ﬁgure

u(x; y; 0) =

sinh(πx)

sinhπ

sin(πy)

(x; y; 0) = 0

Problem 5.30. In the domain Ω = (0; 1) ×(0; 1):

a) Solve the eigenvalue problem

(

∆ φ + 2φ

¡2φ

= ¡λφ on Ω

φ = 0 on bnd(Ω)

;

b) Find σ(x; y) such that the obtained eigenfunctions are orthogonal with respect to σ.

c) Solve the heat equation

= ∆u + 2u

¡2u

on Ω

u = 3 on bnd(Ω)

u(x; y; 0) = xye

y ¡ x

+ 3

d) Find the steady state solution.

30 2D Linear Second-Order Equations

Problem 5.31. Solve the following heat problem

= ∆u

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

u(x; 0; t) = x; u

(x; 1; t) = 0

u(x; y; 0) = xy

Problem 5.32. Let Ω be the square (0; 1) ×(0; 1). Solve the following wave equation on Ω

= 4∆u + 4xy on Ω

u = 1 on bnd(Ω)

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

(x; y; 0) = 0

Problem 5.33. Consider the equation @

u+ @

u ¡@

u= 0 in the domain shown in the following ﬁgure

u = y ¡x

¡1

u = x ¡ y

u = x + y

u = ¡x ¡ y

Use an appropriate transformation to rewrite the equation in the normal form. Note that this

transformation will rotate the domain too. Find a solution to the new problem, and then obtain the

solution to the original problem.

5.3 Disk domain and pol ar coordinate

In this section, we focus on studying partial diﬀerential equations deﬁned on a disk or a

semi-disk. Unlike the rectangular domain, the geometry of a disk does not lend itself easily

to the separation of variable technique when using Cartesian coordinates. However, we can

overcome this challenge by introduci ng polar coo rdinates, which involves a transformation

from the rectangular domain [¡π; π] ×[0; a) to the (θ; r)-plane. By employing the separation

of variable technique in polar coordinates, we can eﬀectively solve the eigenvalue problems

associated with these domains.

5.3.1 Eigenfu nctions of ∆ in a disk

Let Ω be a disk of radius a. We consider the following eigenvalue problem for φ(r; θ) on Ω:



∆φ = ¡λφ

φ(a; θ) = 0

Recall that in polar coordinates, the Laplacian operator ∆ takes the form:

∆

(r;θ)

φ = φ

θθ

5.3 Disk domain and polar coordinate 31

By using the separation of variables φ(r; θ)=R(r)Θ(θ), the eigenvalue problem for φ reduces

to:

+ rR

= ¡λr

This equation holds true only if the ratio

is a constant, denoted as ¡α. Therefore, we

have the eigenvalue problem for Θ(θ):

= ¡αΘ:

To determine suitable boundary c o nditions for Θ(θ), we take into account the geometry of

the disk. We impose the foll owing boundary conditions:



Θ(¡π) = Θ(π)

(¡π) = Θ

(π)

These conditions ensure that the eigenfunctions of the e igenvalue problem Θ''=¡αΘ are

smoothly 2π-periodic, as required by the geometry of the disk.

The eigenpairs of the eigenvalue problem Θ

=¡αΘ for the disk are given by:

(θ) 2f1; cos(nθ); sin(nθ)g;

for n = 1; 2; ···. This set of eigenfunctions forms the Fourier orthogonal basis for functions

deﬁned on the interval [¡π; π]. Let f(θ) be a co ntinuously diﬀerentiable function deﬁned on

[¡π; π]. This function can be expressed in terms of the eigenfunctions as:

f(θ) = a

n=1

cos(nθ) +

n=1

sin(nθ);

where the coeﬃcients are given by:

2π

f(θ) dθ; a

f(θ) cos(nθ) dθ; b

f(θ) sin(nθ) dθ:

These coeﬃcients represent the contributions of the diﬀerent eigenfunctions to the function

f(θ).

Problem 5.34. Solve the eigenvalue problem for Θ(θ) with the periodic bo undary condition and show

that the eigenfunctions are 1; cos(nθ); sin(nθ) with associated eigenva lues n 2N.

For α = n

(n= 0; 1; 2; ··· ), the equation for R(r) reduces to the following ordinary

eigenvalue problem:

+ rR

¡n

R = ¡λr

The boundary condition that R(r) satisﬁes is R(a) = 0 in accord ance with the condition

φ(a; θ) = 0.

Since λ > 0, we can use the substitution x= λ

r to rewrite the above equation in the

following standard form known as a Bessel equation:

+ x

+ (x

¡n

)R = 0: (5.4)

32 2D Linear Second-Order Equations

In our book on ord inary diﬀerential equati o ns, we have studied this equation in detail using

the power series method. If we represent the solution R(x) in the form:

R(x) = x

k=0

;

substituting the series into the diﬀerential equation a nd performing some alge braic manip-

ulations, we arrive at the following series:

R(x) =

k=0

(¡1)

k! (n + k)!





2k+n

This series is known as the ﬁrst-type Bessel function and is denoted by J

(x). The ﬁgure

below depicts J

(x) for n = 0; 1; 2

5 10 15

−

(

)

(

)

(

)

For x = λ

r, the solution to the equation for R(r) is:

(r) = J

( λ

r):

The eigenvalues λ are determined by the boundary condition R(a) = 0. This condition leads

to the equation J

( λ

a) = 0. This equation can not be solved in closed form, we have to use

numerical methods to draw the zeros of the equation J

(x)= 0. As we observe from the above

graph, there are inﬁnitely many zeros for each ﬁxed n. Let us denote these zeros as z

n;p

, i.e.,

n;p

) = 0 fo r p = 1; 2; ··· for ea ch ﬁxed n. The table below includes some of zeros of J

(x):

p z

1 2.4048 3.8317 5.1356 6.3802

2 5.5201 7.0156 8.4172 9.7610

3 8.6537 10.1735 11.6198 13.0152

4 11.7915 13.3237 14.7960 16.2235

5 14.9309 16.4706 17.9598 19.4094

6 18.0711 19.6159 21.1170 22.5827

7 21.2116 22.7601 24.2701 25.7482

8 24.3525 25.9037 27.4206 28.9083

9 27.4935 29.0468 30.5692 32.0648

10 30.6346 32.1897 33.7165 35.2187

The eigenvalues λ a re determined in terms o f these zeros as:

n;p

5.3 Disk domain and polar coordinate 33

Therefore, the e igenf unction s for the eigenvalue problem for R(r) are obtained as: J

n;p



For sake of simplicity, we denote th ese functions by J

n;p

(r).

Remark 5.3. For each ﬁxed n, we obtain a set of inﬁnite eigenfunctions fJ

n;p

(r)g for p =1;

2; ···. Fo r example, for n = 0 we obtain

0;1

(r); J

0;2

(r); J

0;3

(r); ···g;

and for n = 1:

1;1

(r); J

1;2

(r); J

1;3

(r); ···g:

For each ﬁxed n, the set fJ

n;p

(r)g forms a basis for piecewise continuously diﬀerentiable

functions f(r) deﬁned on [0; a]. The ﬁgure below illustrate the set fJ

1;n

g for n = 1; ···; 4.

0 0.2 0.4 0.6 0.8 1

-0.4

-0.2

0.2

0.4

0.6

Problem 5.35. Prove that for each ﬁxed n, the eigenfunction J

n;p

(r) are orthogonal with respect to

the weight function σ = r, i.e.,

n; p

; J

n; p

n;p

(r) J

n; p

(r) rdr = 0;

for p =/ p

To understand how each set in the family of sets fJ

n;p

(r)g is a basis, let's express the

function f(r)=r

in terms of two bases: fJ

0;1

; J

0;2

; :::g and fJ

1;1

; J

1;2

); :::g. We will truncate

both approximations by summing up to ten terms. For the ﬁrst basis fJ

0;1

; J

0;2

; :::g, the

approximation of f (r) is given by:

f(r) ≈c

0;1

(r) + c

0;2

(r) + ···+ c

0;10

(r);

where c

are coeﬃci ents which are determined by the relation

hf ; J

0;p

f(r) J

0;p

(r) rdr

0;p

(r) rdr

Similarly, for the second basis fJ

1;1

; J

1;2

); :::g, the approximation of f (r) is give n by:

f(r) ≈d

1;1

(r) + d

1;2

(r) + ···+ d

1;10

(r);

34 2D Linear Second-Order Equations

where d

are coeﬃcients which are determined by the relation

hf ; J

1;p

f(r) J

1;p

(r) rdr

1;p

(r) rdr

The resulting approximations will give us an idea of how well the sets fJ

0;1

; J

0;2

; :::g and

1;1

; J

1;2

; :::g approximate the function f(r)=r

when truncated to ten terms. The ﬁgure

below illustrate the result:

0 0.5 1

-0.2

0.2

0.4

0.6

0.8

0 0.5 1

-0.2

0.2

0.4

0.6

0.8

Exercise 5.21. As we know, the set fsin(nπr)g is an orthogonal basis for functions deﬁned on [0; 1].

The ﬁgure below show sin(nπr) for n = 1; ···; 4.

0 0.2 0.4 0.6 0.8 1

-1

-0.5

0.5

Approximate th e func tion f(r) = r

in terms of fsin(nπr )g for n = 1; ···; 10 and compare t he er ror

with the approximation in terms of fJ

1;p

(r)g for p = 1; ···; n. The error is deﬁned as

err :=



jf (r ) ¡S

(r)j



;

where S

(r) is the truncated series approximation up to N terms.

Exercise 5.22. Approximate function f (r) = sin(πr) for r 2[0; 1] in terms of basis fJ

0;p

(r)g for p =1; ···;

10. Draw f(r) and its a pproximation in the same coordinate. If you wish, you can use the following

cod e in Matlab:

Z0=[2.4048 5.5201 8.6537 11.7915 14.9309 18.0711 21.2116 24.3525 27.4935 30.6346];

5.3 Disk domain and polar coordinate 35

f=@(r) sin(pi*r);

C=integral(@(r) r*f(r)*besselj(1,Z0(:)*r),0,1,'arrayvalued',true)./...

integral(@(r) r*besselj(1,Z0(:)*r).^2,0,1,'arrayvalued',true);

r=0:0.01:1;

S=besselj(1,Z0(:)*r);

fhat=C'*S;

plot(r,f(r),r,fhat)

Remark 5.4. The obtained solutions J

(x) for the equation (5.4) correspond to the ﬁrst-

type of Bessel functions. In the book on ordinary diﬀerential equations, we also encountered

the seco nd solution to the Bessel equation, denoted by Y

(x). The s eries expansion of Y

(x)

is given by:

(x) = cJ

(x) log(x) + x

¡n

k=0

;

where c and C

are co nstants. It can be observed that these solutions are singular at x=0,

which corresponds to the center of the disk (r=0). The ﬁgure below illustrates a f ew examples

of Y

(x):

5 10 15

−

(

)

(

)

(

)

For p artial diﬀerential equations deﬁned on the interior of a disk, we typically ig nore these

singular solutions (Y

) as they are non-physical. H owever, for equations deﬁned on a annulus

shape or the exterior of a disk, we need to consider bot h J

and Y

functions for the solution.

In our discussion of equations deﬁned on the interior of a disk, we will focus on the ﬁrst-

type Bessel functions (J

) and disregard the singular solutions (Y

) for their non-physical

nature.

Having Θ

(θ) and J

n;p

(r), the eig enfunctions of the eigenvalue problem ∆

(r;θ)

φ = ¡λφ

are obtained as

n;p

(r; θ) 2fJ

n;p

(r) cos(nθ); J

n;p

(r) sin(nθ)g;

with the associated ei g enva lues λ

n;p

, whe re z

n;p

are zeros of the Bessel functions J

i.e., J

n;p

) = 0 for n = 0; 1; 2; ···, and p = 1; 2; ···.

Theorem 5.2. The set of functions fJ

n;p

(r) cos(nθ); J

n;p

(r) sin(nθ)g for n = 0; 1; ···, p = 1;

2; ···, forms an orthogonal basis for continuously diﬀerentiable functions deﬁned in the region

[¡π; π] ×[0; a]. These functions are orthogonal with respect to the weight function σ = r.

36 2D Linear Second-Order Equations

This means that any c o ntinuously diﬀerentiable function f (r; θ) deﬁned in the region

[¡π; π]×[0; a] can be expressed as a linear combination of these eigenfunctions as follows:

f(r; θ) =

p=1

0;p

(r) +

n;p=1

n;p

(r) cos(nθ) +

n;p=1

n;p

(r) sin(nθ):

The orthogonality property ensures th a t the expansion coeﬃcients can be determined using

the inner product weighted by the weight function σ =r as follows:

0;p

hf ; J

0;p

¡π

f(r; θ) J

0;p

(r) rdrdθ

2π

0;p

(r) rdr

;

n;p

hf ; J

n;p

cos(nθ)i

n;p

cos(nθ)k

¡π

f(r; θ) J

n;p

(r) cos(nθ) rdrdθ

n;p

(r) rdr

;

n;p

hf ; J

n;p

sin(nθ)i

n;p

sin(nθ)k

¡π

f(r; θ) J

n;p

(r) sin(nθ) rdrdθ

n;p

(r) rdr

These formulas allow us to determine the expansion coeﬃcie nts A

0;p

; A

n;p

, and B

n;p

using

the given function f(r; θ) and the properties of the Bessel f unctions J

n;p

(r).

Example 5.7. Let u s represent the function f (r; θ) = rθ deﬁ ne d on [¡π; π] ×[0; 1] in terms

of fφ

n;p

(r; θ)g. Since θ is an odd function in [¡π; π], the series representatio n of f contains

only sin(nθ):

rθ =

n=1

p=1

n;p

0;p

(r) sin(nθ);

where

n;p

hf ; J

n;p

sin(nθ)i

n;p

sin(nθ)k

¡π

rθJ

n;p

(r) sin(nθ) rdrdθ

n; p

(r) rdr

The double inte g ral at the top can be calculated as

¡π

rθJ

n;p

(r) sin(nθ) rdrdθ =



n;p

(r) r



¡π

θ sin(nθ) dθ



¡2π(¡1)

n;p

(r) r

dr:

5.3 Disk domain and polar coordinate 37

Therefore, B

n;p

are simpliﬁed as

n;p

¡2(¡1)

n;p

(r) r

n;p

(r) rdr

Exercise 5.23. Consider the function f(r; θ) = sin(rθ) inside the unit disk.

a) Represent this function in terms of basis fJ

n;p

(r) cos(nθ); J

n;p

(r) sin(nθ )g for n = 0; 1; 2; ···. Note

the the representation holds only for points inside the disk and is not valid at the boundary r = 1.

b) As a function of θ, write the Fourier series of f in terms of the basis fcos(nθ); sin(nθ)g, for n = 0;

1; 2; ··· as

f(r; θ) =

n=0

(r) cos(nθ) +

n=0

(r) cos(nθ):

c) As a function of r, write the s eries representation o f f in terms of basis fJ

1;p

(r)g as

f(r; θ) =

n=1

(θ) J

1;p

(r):

5.3.2 Heat, wave an d Poisson equations on a disk

We utilize the eigenfunction expansion method to solve partial diﬀerential equations on a

disk. In order to demonstrate the method, we will solve a few examples.

Poisson equation on a disk

In this section, we will a ddr ess the Poisson equation on a disk Ω of radius a > 0:



¡∆u = f(r; θ)

u(a; θ) = 0

To solve this equation, we express the desired solution as a series in terms of the basis

n;p

(r) cos(nθ); J

n;p

(r) sin(nθ)g:

u(r; θ) =

n=0 ; p=1

n;p

(r) cos(nθ) +

n;p=1

n;p

(r) sin(nθ);

where A

n;p

and B

n;p

are undetermined coeﬃcients. To determine these coeﬃcients, we sub-

stitute this series into the Poisson equation, yielding:

n=0 ;p=1

n;p

n; p

(r) cos(nθ) +

n;p=1

n;p

(r) sin(nθ) = f(r; θ)

Next, we ﬁnd the coeﬃcients A

n;p

and B

n;p

by taking the inner product of both sides of the

Poi sson equation with J

n; p

(r) cos(nθ) and J

n;p

(r) sin(nθ), respectively.

38 2D Linear Second-Order Equations

Example 5.8. Let's consider the Poisson equation on a disk of radius a = 1:

(

¡∆u = r

sin(3θ )

u(1; θ) = 0

We seek a series solution of the f o rm:

u(r; θ) =

n=0;p=1

n;p

(r) cos(nθ) +

n;p=1

n;p

(r) sin(nθ):

Substituting this series into the Poisson equation, we obtain:

n=0 ;p=1

n;p

n; p

(r) cos(nθ) +

n;p=1

n;p

n; p

(r) sin(nθ) = r

sin(3θ):

The equality holds for A

n;p

= 0, for all n, and for B

n;p

= 0 for n =/ 3. For n = 3, we have

p=1

3;p

(r) sin(3θ) = r

sin(3θ);

This yields the expression for B

3;p

= ¡

; J

3;p

0;3

3;p

Finally, the solution for u(r; θ) is given as:

u(r; θ) = sin(3θ)

p=1

; J

3;p

0;3

3;p

(r):

In the example above, we obtained the solution in a series form. However, it is also

possible to ﬁnd a closed-form solution for the equation as follows. If we interpret the problem

in the system form as:

u = ¡∆

¡1

[f(r; θ)];

we can re present this mathematical equation in a block-diagram form as shown below:

U(r) sin(3θ)

∆

¡1

sin(3θ)

Since the system ∆

¡1

is triggered by the function sin(3θ), and since sin(3θ) is p art of the

eigenfunctions of ∆, we can assume that the response of this system is of the form:

u(r; θ) = R(r ) sin(3θ);

where R(r) is an undetermined function. We determine this function such that u (r; θ)

satisﬁes the Poisson equation and the bo undary condition R(1)=0. Substituting u(r; θ ) into

the Poisson equation yiel ds:

(r) sin(3θ) +

(r) sin(3θ) ¡

9R(r)

sin(3θ ) = ¡r

sin(3θ);

5.3 Disk domain and polar coordinate 39

leading to the following Cauchy-Euler equation:

(

(r) + rR

(r) ¡9R(r) = ¡r

U(1) = 0

The general solution to this Cauchy-Euler equation is:

U(r) = Ar

+ Br

¡3

Assuming that the solution remains bounded at r = 0, B must be equal to zero, and thus:

U(r) = Ar

By performing the boundary condition at r = 1, we obtain the solution as:

U(r) =

(1 ¡r);

and consequently:

u(r; θ) =

(1 ¡r) sin(3θ):

This s o luti o n is known as the closed-form solution to the given problem. B y the uniqueness

theorem, these two solutions must be the same, i.e.,

(1 ¡r) sin(3θ) = sin(3θ)

p=1

; J

3;p

0;3

3;p

(r):

Exercise 5.24. Find a series solutio n to the Poisson equation ∆u = r on the unit disk satisfying the

homogeneous Dirichlet boundary conditions. Find a closed form solution for the equation. Draw the

functions and a truncated series in the same coordinate. You ca n use the followin g code to generate the

ﬁgure

Z0=[2.4048 5.5201 8.6537 11.7915 14.9309 18.0711 21.2116 24.3525 27.4935 30.6346];

f=@(r) r;

C=-integral(@(r) r*f(r)*besselj(0,Z0(:)*r),0,1,'arrayvalued',true)./...

integral(@(r) r*besselj(0,Z0(:)*r).^2,0,1,'arrayvalued',true)./(Z0(:).^2);

r=0:0.01:1;

S=besselj(0,Z0(:)*r);

fhat=C'*S;

plot(r,(r.^3-1)/9,r,fhat)

Exercise 5.25. Find a series and closed form solution for the Poisson equation ∆u = r sin(3θ) satisfying

the boundary conditio n u(1; θ) = 0.

Heat problems on a disk

Let's solve the following heat problem

= ∆u + h(r; θ; t)

u(a; θ; t) = 0

u(r; θ; 0) = 0

40 2D Linear Second-Order Equations

Since the set of eigenfunctions fφ

n;p

(r; θ)g is a basis for functi o ns deﬁne don the unit disk,

we can express the desired solution u(r; θ; t) as:

u(r; θ; t) =

n=0 ; p=1

n;p

(t) J

n; p

(r) cos(nθ) +

n=1 ;p=1

n;p

(t) J

n;p

(r) sin(nθ);

for undetermined coeﬃcients functions U

n;p

(t) and V

n;p

(t). Substituting these series into the

heat equation, we reach the following equation:

n=0;p=1

n;p

(t) + λ

n;p

] J

n;p

(r) cos(nθ) +

n=1;p=1

n;p

(t) + λ

n;p

] J

n;p

(r) sin(nθ) = h:

To proceed, we write the function h in terms of the basis fφ

n;p

g a s

h(r; θ; t) =

n=0;p=1

n;p

(t) J

n;p

(r) cos(nθ) +

n=1 ; p=1

n;p

(t) J

n;p

(r) sin(nθ);

where α

n;p

and β

n;p

are respectively:

n; p

hh; J

n;p

(r) cos(nθ)i

n;p

(r) cos(nθ)k

; β

n;p

hh; J

n;p

(r) sin(nθ)i

n; p

(r) sin(nθ)k

From these representation, we arrive at the following equations f o r U

n;p

and V

n;p

(

n;p

+ λ

n;p

n; p

= α

n;p

(0) = 0

;

(

n;p

+ λ

n; p

n;p

= β

n;p

(0) = 0

Note that the initial condition is in accordance with th e initial condition u(r; θ; 0) = 0.

Example 5.9. Let's consider the following heat p roblem on a unit disk

= ∆u + e

¡t

3;1

(r) sin(3θ)

u(1; θ; t) = 0

u(r; θ; 0) = 0

From a system point of view, the heat system is triggered by the source term:

h=e

¡t

3;1

(r)sin(3θ). Since the term J

3;1

(r)sin(3θ) is an eigenfunction of the Laplacian oper-

ator, we seek a response of the system in the form:

u(r; θ; t) = U (t)J

3;1

(r) sin(3θ);

where U (t) is an undetermined functi o n. The blo ck-diagram below illustrates this approach:

¡∆)

¡1

u = U(t) J

3;1

(r) sin(3θ)

¡t

3;1

(r) sin(3θ)

Substituting u = U(t) J

3;1

(r) sin(3θ) into the heat equation yields:

+ λ

3;1

U = e

¡t

;

5.3 Disk domain and polar coordinate 41

which is s o lved for the general solution

U(t) = Ce

¡λ

3;1

¡t

3;1

¡1

The initial condition u(r; θ; 0) = 0 implies U(0) = 0, that in turn determines C as: C =

3;1

¡ 1

. Finally, the given equation is solved for:

u(r; θ; t) =

¡t

¡e

¡λ

3;1

¡1

3;1

(r) sin(3θ):

The ﬁgure below illustrates the change of disk in some instances of ti me:

Exercise 5.26. Consider the following heat problem:

= ∆u + e

¡t

sin(3θ)

u(1; θ; t) = 0

u(r; θ; 0) = 0

The only source terms t ha t triggers the heat system is h = e

¡t

sin(3θ). This source can be represented

as the series

h =

p=1

¡t

3;p

(r) sin(3θ):

The system block-diagram is illustrated below:

sin(3θ)

p=1

(t) J

3;p

(r)

sin(3θ)

p=1

3;p

(r)

¡∆)

¡1

42 2D Linear Second-Order Equations

Take the solution u as follows:

u(r; θ; t) =

p=1

(t) J

3;p

(r) sin(3θ):

Determine the coeﬃcients fu nctions V

(t) and write down the series solution of the problem.

Exercise 5.27. Consider the following heat problem on the unit disk

= ∆u

u(1; θ; t) = 0

u(r; θ; 0) = r cos(θ)

The only source that triggers the system is the initial condition r cos(θ ) which can be represented as

r cos(θ) = cos(θ)

p=1

1;p

(r):

Consider the solution of the equation as:

u(r; θ; t) =

p=1

(t) J

1;2

(r) cos(θ):

Determine U

(t) a s write down the series so lution of the problem.

Exercise 5.28. Consider the following heat equation

= ∆u ¡r sin(θ)

u(1; θ; t) = 0

u(r; θ; 0) = 0

a) Take u as the summation

u(r; θ; t) = V (r; θ) + w(r; θ; t):

Find a closed form solution for the Poisson equation for V .

b) Wh at diﬀerential equation does w(r; θ; t) satisfy? Find the s eries solution of this equati on and

then write the solution for the original problem.

Wave problem

Let's solve the following wave equation on the unit disk:

= c

∆u + h(r; θ; t)

u(1; θ; t) = 0

u(r; θ; 0) = 0

(r; θ; 0) = 0

Employing the ei g enfunction expansion me thod, the solution u can be expressed as:

u(r; θ; t) =

n=0 ; p=1

n;p

(t) J

n;p

(r) cos(nθ) +

n=1 ; p=1

n;p

(t) J

n;p

(r) sin(nθ):

The ordinary diﬀerential equations f o r U

n;p

and V

n;p

are:

n;p

+ c

n;p

= α

n;p

(0) = 0

n;p

(0) = 0

;

n;p

+ c

n;p

= β

n;p

(0) = 0

n;p

(0) = 0

;

5.3 Disk domain and polar coordinate 43

where α

n;p

(t) and β

n;p

(t) are respectively the coeﬃcients of the expansion of h in te rms of

n;p

(r) cos(nθ)g and fJ

n;p

(r) sin(nθ)g.

Example 5.10. Let's consider the following wave problem

= c

∆u + sin(t) sin(θ)

u(1; θ; t) = 0

u(r; θ; 0) = 0

(r; θ; 0) = 0

Since the expansion of h is:

h = sin(θ)

p=1

h1; J

1;p

sin(t) J

1;p

(r);

we seek for u as:

u(r; θ; t) =

p=1

(t) s in(θ) J

1;p

(r);

where U

satisﬁes the equation:

+ c

1;p

h1; J

1;p

sin(t)

(0) = 0

For c =/

1;p

, the above ODE is solved for

(t) =

h1; J

1;p

¡1)

sin(t) ¡

c λ

1;p

sin

c λ

1;p



Finally, the solution can be expressed as:

u(r; θ; t) = sin(θ)

p=1

h1; J

1;p

¡1)

sin(t) ¡

c λ

1;p

sin

c λ

1;p



1;p

(r):

Exercise 5.29. Find a series solution to the following wave equation

= ∆u + e

¡t

u(1; θ; t) = 0

u(r; θ; 0) = 0

(r; θ; 0) = 0

Exercise 5.30. Find a series solution to the following wave problem

= ∆u

u(1; θ; t) = 0

u(r; θ; 0) = J

1;2

(r)sin(θ)

(r; θ; 0) = J

1;3

(r) sin(θ)

44 2D Linear Second-Order Equations

Exercise 5.31. Consider the following wave equation

= ∆u ¡r sin(θ)

u(1; θ; t) = 0

u(r; θ; 0) = 0

(r; θ; 0) = 0

Since the source term is independent of time t, we can take t he solution u as

u(r; θ; t) = V (r; θ) + w(r; θ; t);

where V satisﬁes a Poisson equation.

a) Solve the Poisson equation and determine its solution V (r; θ).

b) Based on the solution V , determine the correct series solution to the wave equation for w(r; θ; t).

Exercise 5.32. Find the series solution for the following wave problem

= c

∆u

u(1; θ; t) = 0

u(r; θ; 0) = r sin(θ)

(r; θ; 0) = 0

5.3.3 Laplace equation

We co nsider the following problem on the domain Ω: [¡π; π] ×[0; a]:



∆u = 0

u(a; θ) = f(θ)

To solve this problem, we employ the method of separation of variables and assume a solution

of the form u(r; θ)=R(r)Θ(θ). Thi s leads to the e quation:

+ rR

= 0:

Noting that the ratio

should be a dimensionless constant, we e xpress it as

= ¡λ. As

mentioned before, the function Θ(θ) must satisfy periodic boun dary conditions:



Θ(¡π) = Θ(π)

(¡π) = Θ

(π)

;

This gives rise to the following eigenvalue problem:

= ¡λΘ

Θ(¡π) = Θ(π)

(¡π) = Θ

(π)

Solving this eigenvalue problem yields the eigenfunctions fcos(nθ); sin(nθ)g for n=0; 1; 2; ···:

Considering the ratio

= ¡n

, the equation for R(r) simpliﬁes to the Cauchy-Euler eq ua-

tion:

+ rR

¡n

R = 0:

5.3 Disk domain and polar coordinate 45

For n = 0, the equation is solved as

(r) = A

+ B

ln(r);

and for n ≥1, the s o lution becomes

(r) = A

+ B

¡n

Additionally, the bo undedness condition R(0) being bounded implies B

=0 for n ≥0. Con-

sequently, the separated solution can be written as:

(r; θ) = R

(r) Θ

(θ) = fr

cos(nθ); r

sin(nθ)g:

The general series solution of the Laplace equation is a linear combination o f the separated

solutions:

u(r; θ) = A

n=1

cos(nθ) + B

sin(nθ)]

The coeﬃcients are determined by the boundary conditions: u(a; θ) = f (θ):

f(θ) = A

n=1

cos(nθ) + B

sin(nθ)]:

Using the inner product and orthogonality properties of the functions fcos(nθ); sin(nθ)g, we

can determine the coeﬃcients as follows:

2π

¡π

f(θ); A

πa

¡π

f(θ) cos (nθ); B

πa

¡π

f(θ) sin (nθ):

Example 5.11. Let's solve the foll owing equation on the disk r ≤1:



∆u = 0

u(1; θ) = 1 + 3cos(2θ)

The contribution of the boundary terms 1 to the solution is the constant function u = 1. The

contribution of the second boundary term to the solution is: u = 3r

cos(3θ). Consequently,

the solution to the given Laplace equation is given as:

u(r; θ) = 1 + 3r

cos(3θ):

The ﬁgure below depicts the surface solution of this Lap lace equation:

46 2D Linear Second-Order Equations

Example 5.12. Let's solve the foll owing equation on the disk r ≤1:



∆u = 0

u(1; θ) = θ

Since θ can be expressed by the series:

θ =

n=1

¡2

(¡1)

sin(nθ);

the contribution of this term to the solution is

u(r; θ) =

n=1

¡2

(¡1)

sin(nθ):

Exercise 5.33. Solve the Laplace equation ∆u = 0 inside the unit disk with the boundary conditions

given below

a) u(1; θ) = θ

b) u(1; θ) = cos





c) u(1; θ) = sin

Exercise 5.34. Solve the following Laplace equation on the unit disk



∆u = 0

u(1; θ) + u

(1; θ) = sin(θ)

Show that the Laplace equation does not have a solution if the boundary condition is changed to the

following one:

u(1; θ) ¡u

(1; θ) = sin(θ) :

Exercise 5.35. We aim to prove that the following equation deﬁned inside the unit disk has a unique

solution



∆u = 0

u(1; θ) + u

(1; θ) = f(θ)

a) Assume that u

(r; θ), u

(r; θ) are two solu tions to the problem. The function u = u

¡u

solves

the following equation:



∆u = 0

u(1; θ) + u

(1; θ) = 0

Since u(r; θ), the solution to the equation, is smooth, we can express it as the following Fourier

series:

u(r; θ) = U

(r) +

n=1

(r) cos(nθ) + B

(r) sin(nθ):

Substitute this into the equation ∆u = 0 and conclude A

(r) = A

, B

(r) = B

for n = 0; 1; ···.

b) Finally use the boundary condition and show u(r; θ) = 0 and conclude u

(r; θ) = u

(r; θ).

Exercise 5.36. Consider the following Laplace equation inside the unit ball



∆u = 0

(1; θ) = f(θ)

;

5.3 Disk domain and polar coordinate 47

where f(θ) is a smooth function in θ.

a) Is there a unique solution to the problem?

b) Find a solution if f (θ ) = cosθ.

c) Is there any so lution if f (θ) = 1 ?

d) Show that the necessary condition for the problem to have a solution is

¡π

f(θ)dθ = 0:

Exercise 5.37. In the case of solving a Laplace equation in the exterior of a disk, that is, in the region

Ω

:= f(r; θ); r > ag, we keep only the terms R

(r) = r

¡n

and reject R

(r) = r

and ln(r). The reason

is that only the b ounded solutions of Laplace equation are meaningful physically. Solve the following

Laplace equation in the exterior of a disk of radius a = 2



∆u = 0

u(2; θ) = 1 + 2sin(2θ)

Exercise 5.38. Let Ω be the region outside of the unit disk an d interior of the disk of radius 2. SOlve

the following Laplace equation on Ω

∆u = 0

u(1; θ) = sin(θ)

u(2; θ) = sin(2θ)

5.3.4 Eigenvalue problem on an annulus

Let Ω be the region enclosed by two disks r < b and r < a, i.e.,

Ω = f(r; θ); a < r < bg:

Consider the follow ing eigenvalue problem



∆φ = ¡λφ

φ(a; θ) = φ(b; θ) = 0

For the sep a rated solution φ

(r; θ) = R

(r) Θ

(θ), the s o luti o n for R

(r) reduces to the

following one:

(r) = A

( λ

r) + B

( λ

r):

Note that in this case we should keep the second type of the Bessel functions Y

( λ

r) as the

origin is excluded and thus Y

is bounded in the annulus domain Ω. The boundary conditions

for R

(r) results to the f o llowing equations:

(

( λ

a) + B

( λ

a) = 0

( λ

b) + B

( λ

b) = 0

A straightforward algebraic manipulation leads to the following equation:

( λ

a) Y

( λ

b) ¡J

( λ

b) Y

( λ

a) = 0:

48 2D Linear Second-Order Equations

Taking x = λ

a, the above equation reduces to the following one:

(x) Y





¡J





(x) = 0:

If the zeroth of the above equation are denote d by x = s

, the eigenvalues are obtained as

. Substituting λ

into the above s ystem of e quations gives

) + B

) = 0





+ B





= 0

Now, constants A

, and B

are simply determined by the relation

= ¡

)

Hence

(r) = Y



¡J

) Y



Note that R

(r) satisfy the equation

+ r

¡n

= ¡λ

;

and therefore

; R

= 0; p =/ q:

The eigenfunctions of ∆ in Ω are

(r; θ) = fR

(r) cos(nθ); R

(r) sin(nθ)g;

for n = 0; 1; 2; ··· with associated eigenvalue λ

n;p

. These eigenparis satisfy the following

relation:

∆φ

= ¡

The table below gives some values of s

for b = 2a.

p s

1 3.1230 3.1966 3.4069

2 6.2734 6.3123 6.4278

3 9.4182 9.4445 9.5228

4 12.5614 12.5812 12.6404

5 15.7040 15.7198 15.7673

The ﬁgure b elow depicts a few of eigenfunctions R

1;p

(r) in the region Ω: f(r; θ); 1 < r <rg.

As we observed they satisfy the homogeneous boundary conditions at r = 1 and r = 2

5.3 Disk domain and polar coordinate 49

1 1.2 1.4 1.6 1.8 2

-0.2

-0.15

-0.1

-0.05

0.05

0.1

Example 5.13. Let's solve the foll owing Poisson equati o n on Ω := f(r; θ); 1 < r < 2g



∆u = r sin(θ)

u(1; θ) = u(2; θ) = 0

The series solution of the problem is expressed as

u(r; θ) = sin(θ)

p=1

1;p

(r):

Substituting this into the equation gives B

¡1

1;p

hr; R

1;p

Here, we calculate a few of coeﬃcients B

= 1.1314 B

= ¡0.143; B

= 0.13416; B

= ¡0 .0362; B

= 0.0 4 86

We also can ﬁnd a closed form solution as follows. We take the solution o f the form u(r;

θ) = R(r) sin(θ). Substituting this solution into the Poisson equation gives the following

equation for R(r):

+ rR

¡R = r

This equation is solved for

R(r) = Ar + Br

¡1

Applying the boundary con ditions gives: A =

¡5

and B =

, and consequently:

u(r; θ) =



r +

¡1



sin(θ ):

by the uniqueness, we conclude the equality

r +

¡1

p=1

1;p

(r):

50 2D Linear Second-Order Equations

The ﬁgure below depicts This fun ction and its approximation up to only three terms in the

same coordinate:

1 1.2 1.4 1.6 1.8 2

-0.2

-0.15

-0.1

-0.05

Exercise 5.39. Solve the following heat equation on the domain Ω := f(r; θ); 1 < r < 2g

= ∆u + e

¡t

sin(θ)

u(1; θ; t) = 0

u(2; θ; t) = 0

u(r; θ; 0) = 0

5.3.5 Problems

Problem 5.36. Let Ω denote a disk with radius a = 1. Consider the followi ng problem on Ω



∆u = r cosθ

u(1; θ) = 0

a) Find a double series solution to the problem in terms of the eigenfunctions fJ

n; p

(r) cos(nθ);

n;p

(r)sin(nθ)g.

b) Find a closed form solution to the problem and justify it is equal to the series solution.

Problem 5.37. Consider the Poisson equatio n ∆u = θ deﬁned on the unit disk satisfying the homog e-

neous Dirichlet boundary condition.

a) Find a series solution to the problem in terms of the basis fJ

n; p

(r) cos(nθ); J

n;p

(r) sin(nθ)g. This

is a double series solution to the problem.

b) Consider the solution u as follows

u(r; θ) =

n=1

(r) sin(nθ):

Note that θ is an odd function on [¡π; π], and its Fourier series contains only sine terms. Obtain

a diﬀerential equation for U

(r) and solve it to determine the single series solution.

Problem 5.38. Let Ω denote a disk with radius a = 1. Consider the followi ng problem on Ω



∆u = sin(θ)

(1; θ) = 0

Here t he boundary condition is the homogeneous Neumann boundary condition.

a) Find a closed form solution to the problem by take u as: u = U(r) sin(θ).

5.3 Disk domain and polar coordinate 51

b) To ﬁnd a series solution to the problem, we need to solve the eigenvalue problem



∆φ = ¡λφ

(1; θ) = 0

Show that the eigenfunctions are

n; p

(r; θ) 2fJ

(ζ

n; p

r) cos(nθ ); J

(ζ

n;p

r) sin(nθ)g;

for n = 0; 1; 2; ··· where ζ

n; p

are the roots of J

(x). Here are a few of them:

1;1

= 1.8412; ζ

1;2

= 5.3314; ζ

1;3

= 8.5363; ζ

1;4

= 11.7060; ζ

1;5

= 14.8636

Problem 5.39. Let Ω be the unit disk. Consider the Poisson problem ∆u = f(r; θ) on Ω satisfying the

homogeneous Neumann boundary condition u

(1; θ) = 0.

a) Show that the condition for solvability of the problem is:

Ω

f(r; θ) rdrdθ = 0:

Hint: Use the Gauss theorem for the integral

Ω

∆udS:

b) Show that the eigenvalue problem



∆φ = ¡λφ

φ(1; θ) = 0

;

accepts the eigenvalue λ = 0. What is the relationship of this result to the solvability condition

of the Poisson equation?

Problem 5.40. Find a single series solution to the following problem on the unit disk



∆u = θ

u(1; θ) = cosθ

Problem 5.41. Let Ω

denote the exterior domain of the unit disk in R

. Find a closed form solution

to the following problem on Ω

(

∆u =

sinθ

u(1; θ) = 0

Problem 5.42. Consider the following diﬀerential equation

(

∆u +

= r sin(θ)

u(1; θ) = 0

We do not have any inf ormation about the eigenfunctions of the operator

∆ +



. However, we can

ﬁnd a closed form solution to the problem. Take u as u(r; θ) = R (r) sin(θ). Obtain a Cauchy-Euler

equation for R(r), solve and and determine the solution u.

Problem 5.43. Co nsider the Poisson equation ∆u = sin(θ) on the region Ω: f(r; θ); 1 < r < 2g. Assume

that u satisﬁes the boundary conditions u

(1; θ) = 0 and u(2; θ) = 0.

a) Find eigenvalues and eigenfunctions of the eigenvalue problem

∆φ = ¡λφ on Ω

(1; θ) = 0

φ(2; θ) = 0

Use a numerical method to ﬁnd ﬁrst ﬁve eigenvalues. Use the results and write a series solution

for the given problem.

52 2D Linear Second-Order Equations

b) Consider the closed form solution u a s u = R(r) sin(θ). Determine R(r).

c) With the r esult in part a), solve the following heat equation on Ω

= ∆u ¡e

sin(θ) on Ω

(1; θ; t) = 0

u(2; θ; t) = 0

u(r; θ; 0) = 0

Problem 5.44. Solve the Poisson equation ∆ u = cos(θ) on the domain Ω = f(r; θ); 0 ≤r < 1; 0 ≤θ ≤πg

satisfying the boundary conditions

u(r; 0) = u(r; π ) = 0:

Problem 5.45. Let Ω be the unit disk. Consider the following heat equation on Ω

= ∆u ¡r

u(t; 1; θ) = 0

u(0; r; θ) = cosθ

Take the solution u as

u(r; θ; t) = V (r; θ) + w(r; θ; t);

where V satisﬁes the following Poisson equation



∆u = r

u(1; θ) = 0

a) Find a closed form solution to the equation for V .

b) Use the result in part a) and write downs the value series solution for w. Write down the solution

for u.

Problem 5.46. Let Ω be the unit disk. Consider the following heat problem on Ω:

= ∆u ¡r

u(1; θ; t) = cos(θ)

u(r; θ; 0) =

¡1)

Consider the solution u as

u(r; θ; t) = V (r; θ) + w(r; θ; t);

where V satisﬁes the following Poisson equation

(

∆u = r

u(1; θ) = cos(θ)

a) Find a closed form solution to the equation for V .

b) Use the result in part a) and write downs the value series solution for w. Write down the solution

for u.

Problem 5.47. Let Ω be the unit disk. Consider the following wave equation

= ∆u ¡cos θ

u(1; θ; t) = 0

u(r; θ; 0) =

cosθ

(r; θ; 0) = 0

Take the solution u as

u(r; θ; t) = V (r; θ) + w(r; θ; t);

5.3 Disk domain and polar coordinate 53

where V satisﬁes a Poisson equation.

a) Solve the Poisson equation for V

b) Use the result of part a) and write the series s olution for w.

Problem 5.48. Let C denote the unit circle of radius 1. As it is known, the Laplacian ∆ on C has the

form ∆ u = u

θθ

. Solve the following problem on C

(

= u

θθ

¡e

¡t

sinθ

u(θ; 0) = 0

Problem 5.49. Let C denote the unit circle of radius 1. Solve the following wave equation on C

= u

θθ

u(θ; 0) = f(θ)

(θ; 0) = 0

;

where f(θ) is the following function

f(θ) =

(

< θ <

0 otherwise

Problem 5.50. Let Ω be the unit disk. Write a series solution to the following problem

= ∆u ¡e

¡t

cosθ

u(1; θ; t) = cos(2θ)

u(r; θ; 0) = cos(3θ)

Problem 5.51. Let Ω be the unit disk. Wri te the series solution to the following problem on Ω in the

series form

= c

∆u + sin(t)θ

u(1; θ; t) = 0

u(r; θ; 0) = 0

(r; θ; 0) = 0

Problem 5.52. Let Ω be the domain out side of the disk r = 1 and inside the di sk r = 2. We aim to

solve the heat problem u

= ∆u on Ω subject to the boundary condition shown in the ﬁgure:

u = 1

u = ¡1

a) Solve the Laplace equation ∆V = 0 subject to the boundary condition given in the ﬁgure. Hint:

the solution is a pure function of r and independent of θ.

b) So lve t he given heat problem if u(r; θ; 0) = 0 by taking u as u = V (r) + w (r; θ; t).

Problem 5.53. Solve the following wave equation on Ω := f(r; θ ); 1 < r < 2g

= ∆u

(1; θ; t) = 0

u(2; θ; t) = 0

u(r; θ; 0) = cos

(r ¡1)



sin(θ)

(r; θ; 0) = 0

54 2D Linear Second-Order Equations

Problem 5.54. Let Ω be the domain



(r; θ); 0 ≤r < 1 ; 0 < θ <



. We aim to solve the f ollowing wave

equation for the function u(r; θ; t):

= c

∆u + r sin(3θ) on Ω

Subject to the following boundary conditions:

u(r; 0; t) = 0

; t



= 0

u(1; θ; t) = 0

a) Find the eigenvalues and eigenfunctions of the associated eigenvalue problem.

b) Find a series solution to the problem if the initial conditions are given by: u(r; θ; 0) = 0 and

(r; θ; 0) = 0.

Problem 5.55. Let Ω denote the ﬁrst quarter of a unit disk shown in the following ﬁgure

u = 0

u = 2 sin(4θ)

a) Solve the Laplace equation ∆u = 0 in Ω with the boundary conditions given in the ﬁgure.

b) Find a closed form solution to the Poisson equation



∆u = sin(2θ) on Ω

u = 0 on bnd(Ω)

c) Find a series solution to the following heat problem

= ∆u ¡sin(2θ)

u(r; 0) = u



= 0

u(1; θ) = 2 sin(4θ)

u(r; θ; t) = 2r

sin(4θ) +

lnr sin(2θ)

Problem 5.56. Consider the following ﬁgure

u = 0

u = 1

u = ¡ 1

Ω

u = 0

a) Solve the Laplace equation ∆V = 0 on the domain shown in the ﬁgure and associated boundary

condition.

b) So lve t he Poisson equation ∆u = θ onΩ. Hint: Use the single series for u as:

u(r; θ) =

n=1

(r) sin(nθ):

5.3 Disk domain and polar coordinate 55

Problem 5.57. We aim to solve the heat equation u

= ∆u inside the domain Ω := f(r; θ); 0 ≤ r < 1;

0 ≤θ ≤πg subject to the boundary conditions



u(1; θ; t) = 1:

u(r; 0; t) = u(r; π; π) = 0

a) Determine the eigenfunctions and eigenvalues of the Laplacian operator ∆ on Ω sa tisfying the

homogeneous boundary conditions:



∆φ = ¡λφ on Ω

φ = 0 on bnd(Ω)

b) So lve t he Laplace equation ∆V = 0 on Ω adher es to the prescribed boundary conditions.

c) Solve the given heat problem utilizing the separation u = V (r; θ) + w(r; θ; t).

Problem 5.58. Let Ω be the semi-disk: Ω =



(r; θ); 0 ≤r < 1;

¡π

≤θ ≤



a) Determine the ei gen fun ctions and eigenvalues of the Laplacia n operator ∆ in Ω subject to toe

homogeneous Dirichlet boundar y condition.

b) Consider the following heat problem on Ω:

= ∆u ¡r sin(2θ) on Ω

u = 0 on bnd(Ω)

u(r; θ; 0) = 0

Since the source term is independent of time, we can consider the solution u as:

u(r; θ; t) = V (r; θ) + w(r; θ; t);

where the f unct ion V satisﬁes the Po isson equitation



∆V = r sin(2θ) on Ω

V = 0 on bnd(Ω)

Take V of the form V = R(r) sin(2θ) for an undetermined function R(r). Substitute V into the

Poisson equation and derive a Cauchy-Euler equation. Solve this equation and determine V . This

is a closed form solution for the Po isson equation.

c) Write down the equation for w, and determine the correct series solution to it based on the form

of the sour ce term h = r sin(2θ). Determine the solution u(r; θ; t).

Problem 5.59. Consider the following domain:

◦

Ω

u = 1

◦

u = ¡1

u = 0

a) Determine the eigenfunctions and eigenvalues of the Laplacian operator ∆ on the domain Ω

subject to the homogeneous Dirichlet boundary condition.

b) So lve the Laplace equation ∆V = 0 o n the above domain with the given boundary conditions

shown in the ﬁgure.

c) Solve the heat equation



= ∆u

u(r; θ; t) = 0

;

where u satisﬁes the given boundary conditions in the ﬁgu re.

56 2D Linear Second-Order Equations