boundary-value problems in rectangular...

Boundary-value Problems in Rectangular

Coordinates

Y. K. Goh

2009

Y. K. Goh

Boundary-value Problems in Rectangular Coordinates

Outline

I Separation of Variables: Heat Equation on a Slab

I Separation of Variables: Vibrating String

I Separation of Variables: Laplace Equation

I Review on Boundary Conditions

I Dirichlet’s Problems

I Neumann’s Problems

I Robin’s Problems(Optional)

I 2D Heat Equation

I 2D Wave Equation

Y. K. Goh


Separation of Variables

Separation of Variables

Y. K. Goh


One Dimensional Heat Equation (Heat Conduction

on a Slab)

Problem:Consider a unifrom slab (or rod or bar) of length L withinsulated lateral surface. Let the internal temperaturedistribution on the slab be u(x, t) at point x and time t. Giventhat at time t = the temperature distribution of the slab isf(x), and given that both ends on the slab are held at zeroconstant temperature.

Find the subsequent temperature distribution u(x, t) for0 < x < L and t > 0 on the slab. (Of course, from ourexperience we know that u→ 0 as t→∞, and our solutionfor u(x, t) should also capture this observation.)

Y. K. Goh


Boundary-value Problem

1D Heat Equation:

∂u

∂t= c2

∂2u

∂x2, 0 < x < L, t > 0; (1)

Boundary Conditions (Zero temperature at both ends):

u(0, t) = 0 and u(L, t) = 0, ∀t > 0; (2)

Initial condition (Initial temperature distribution f(x)):

u(x, 0) = f(x), 0 < x < L. (3)

Y. K. Goh


Separation of variables

Let u(x, t) = X(x)T (t), differentiate & subsitute into Eq.(1):

T

c2T=X ′′

X= k,

Which gives a set of two ODEs:

X ′′ − kX = 0, (4)

T − kc2T = 0. (5)

Now, the boundary condition becomes

u(0, t) = X(0)T (t) = 0 =⇒ X(0) = 0 ∀t > 0, (6)

u(L, t) = X(L)T (t) = 0 =⇒ X(L) = 0 ∀t > 0. (7)

Y. K. Goh


There are three possible cases for the eigenvalue k:

I k = b2 > 0I X(x) = c1 cosh bx+ c2 sinh bx,I X(0) = X(L) = 0 =⇒ c1 = c2 = 0I trivial solution

I k = 0I X(x) = c1x+ c2I X(0) = X(L) = 0 =⇒ c1 = c2 = 0I trivial solution

I k = −β2 < 0I X(x) = c1 cosβx+ c2 sinβxI X(0) = 0 =⇒ c1 = 0I X(L) = 0 =⇒ β = βn = nπ

L , n = 1, 2, . . .I c2 is arbitrary, and set c2 to 1.

Y. K. Goh


Thus, a non-trivial solution for Eq.(4) is

I X(x) = Xn(x) = sinnπ

Lx, n = 1, 2, . . . .

By using k = −β2, solve Eq.(5) T (t),

I T (t) = Tn(t) = bne−λ2

nt, n = 1, 2, . . . ,

I where λn =cnπ

L.

Combining Xn(x) and Tn(t), we get a solution for Eq.(1) andthe solution satisfies the Boundary Conditions Eq.(2)

un(x, t) = Xn(x)Tn(t) = bne−λ2

nt sinnπ

Lx, n = 1, 2, . . . , (8)

Y. K. Goh


Principle of Superposition

TheoremIf φ and ψ are solutions to a linear differential equation andsatisfy a linear boundary condition, then the linearcombination u = c1φ+ c2ψ is also a solution and satisfies thesame boundary condition. Here c1 and c2 are constants.

Since u1(x, t), u2(x, t), . . . are satisfying the 1D Heat equationand the zero temperature boundary conditions. Thus, ageneral solution is the superposition of all these un(x, t):

u(x, t) =∞∑n=1

bne−λ2

nt sinnπ

Lx. (9)

Y. K. Goh


Half-range Fourier Series

I Applying initial condition

u(x, 0) =∞∑k=1

bn sinnπ

Lx = f(x), 0 < x < L,

=⇒ bn =2

L

∫ L

0

f(x) sinnπ

Lx dx, n = 1, 2, . . .

Y. K. Goh


Summary

I Boundary-value problems

∂u

∂t= c2

∂2u

∂t2, 0 < x < L, t > 0;

u(0, t) = 0 and u(L, t) = 0, ∀t > 0;u(x, 0) = f(x), 0 < x < L.

I General solution

u(x, t) =∞∑n=1

bne−λ2

nt sinnπ

Lx, λn =

cnπ

L;

bn =2L

∫ L

0f(x) sin

nπx

Ldx.

Y. K. Goh


Sumarry for Method of Separation of Variables

1. Decompose u into products of functions of one variable.

2. Decompose the PDE into a set of ODEs.

3. Identify boundary conditions and the correspondingSturm-Liouville problems.

4. Solve the Sturm-Liouville problems and obtain thecorresponding eigenvalues.

5. Apply principle of superposition to obtain theeigenfucntion expansion of a general solution.

6. Use the initial conditions to obtain generalized Fouriercoefficients of the eigenfunction expansion.

Y. K. Goh


Example

I Given that a homogeneous rod of lenght L = π isproperly insulated except at both ends. Suppose thatboth ends of the rod are kept at a constant temperatureof zero degree Celsius. Find the temperature distributionin the rod u(x, t) for t > 0 if given that the entire rod isinitially at a temperature of 100 degree Celsius.

Y. K. Goh


One Dimensional Wave Equation (Vibrating String)

Problem:Consider a stretched string of length L with both endsfastened on the x-axis. Suppose that the string is pluckedfrom its equilibrium position and release at time t = 0.Assuming that the amplitude of the vibration at time t ≥ 0and position x is u(x, t). Suppose that the initial shape of thestring is u(x, 0) = f(x) and the initial velocity at each pointon the string is du

dx(x, 0) = g(x).

Find the subsequent motion of the string u(x, t) for0 < x < L and t > 0.

Y. K. Goh


Boundary-value problems

1D Wave Equation:

∂2u

∂t2= c2

∂2u

∂t2, 0 < x < L, t > 0; (10)

Boundary Conditions (fixed end points):

u(0, t) = 0 and u(L, t) = 0, ∀t > 0; (11)

Initial Conditions (initial displacement and initial velocity):

u(x, 0) = f(x) and∂u

∂t(x, 0) = g(x), 0 < x < L. (12)

Y. K. Goh


Separation of variables

Let u(x, t) = X(x)T (t), differentiate and subsitute intoEq. 10:

T

c2T=X ′′

X= k,

Which gives a set of two ODEs:

X ′′ − kX = 0, (13)

T − kc2T = 0. (14)

Now, the boundary condition becomes

u(0, t) = X(0)T (t) = 0 =⇒ X(0) = 0 ∀t > 0, (15)

u(L, t) = X(L)T (t) = 0 =⇒ X(L) = 0 ∀t > 0.(16)

Y. K. Goh


There are three possible cases for the constant k:I k = b2 > 0 (trivial solution)

X(x) = c1 cosh bx+ c2 sinh bx,

X(0) = X(L) = 0 =⇒ c1 = c2 = 0.

I k = 0 (trivial solution)

X(x) = c1x+ c2,

X(0) = X(L) = 0 =⇒ c1 = c2 = 0.

I k = −β2 < 0 (non-trivial solution)

X(x) = c1 cos βx+ c2 sin βx,

X(0) = 0 =⇒ c1 = 0

X(L) = 0 =⇒ β = βn =nπ

L, n = 1, 2, . . .

Y. K. Goh


A non-trivial solution for Eq. (13) is

I X(x) = Xn(x) = c2 sinnπ

Lx, n = 1, 2, . . . , with c2 is

arbitrary, and let say set c2 = 1.

By using k = −β2, solve Eq. (14) and get

I T (t) = Tn(t) = bn cosλnt+ b∗n sinλnt, n = 1, 2, . . . ,

I Here λn =cnπ

L.

Combining Xn(x) and Tn(t), we get a solution for Eq. (10)and also satisfies the Boundary Conditions Eqs. (11)

un(x, t) = Xn(x)Tn(t) = sinnπ

L(bn cosλnt+b

∗n sinλnt), n = 1, 2, . . . .

(17)

Y. K. Goh


Principle of Superposition

TheoremIf φ and ψ are solutions to a linear differential equation andsatisfy a linear boundary condition, then the linearcombination u = c1φ+ c2ψ is also a solution and satisfies thesame boundary condition. Here c1 and c2 are constants.

Since u1(x, t), u2(x, t), . . . are satisfying the 1D wave equationand the fixed ends boundary conditions. Thus, a generalsolution is the superposition of all these un(x, t):

u(x, t) =∞∑n=1

sinnπ

L(bn cosλnt+ b∗n sinλnt). (18)

Y. K. Goh


Half-range Fourier Series

I Applying initial condition

u(x, 0) =∞∑k=1

bn sinnπ

Lx = f(x), 0 < x < L,

=⇒ bn =2

L

∫ L

0

f(x) sinnπ

Lx dx, n = 1, 2, . . .

I Applying initial condition∂u

∂t(x, 0) =

∞∑k=1

λnb∗n sin

nπ

Lx = g(x), 0 < x < L,

=⇒ λnb∗n =

2

L

∫ L

0

g(x) sinnπ

Lx dx, n = 1, 2, . . .

Y. K. Goh


Summary


∂2u

∂t2= c2

∂2u

∂t2, 0 < x < L, t > 0;

u(0, t) = 0 and u(L, t) = 0, ∀t > 0;


∂t(x, 0) = g(x), 0 < x < L.

I General solution

u(x, t) =∞∑n=1

sinnπ

L(bn cosλnt+ b∗n sinλnt), λn =

cnπ

L;

bn =2L

∫ L

0f(x) sin

nπx

Ldx, b∗n =

2λnL

∫ L

0g(x) sin

nπx

Ldx.

Y. K. Goh


Sumarry for Method of Separation of Variables

1. Decompose u into products of functions of one variable.

2. Decompose the PDE into a set of ODEs.


4. Solve the Sturm-Liouville problems and obtain thecorresponding eigenvalues.

5. Apply principle of superposition to obtain theeigenfucntion expansion of a general solution.

6. Use the initial conditions to obtain generalized Fouriercoefficients of the eigenfunction expansion.

Y. K. Goh


Examples

1. Let say a string with length L = 1 is fixed at two ends.

The initial displacement of the string is f(x) = sinmπ

Lx

and with zero initial velocity. Find u(x, t).

2. Same as the previous example, but

f(x) =

{310x, 0 ≤ x ≤ 1

3;

3(1−x)10

, 13≤ x ≤ 1.

3. Now, assume that the initial displacement is 0, but thethe initial velocity is g(x) = x cosx, given that L = 1 andc = 1. Find u(x, t).

Y. K. Goh


D’Alembert’s Method (Optional)


∂2u

∂t2= c2

∂2u

∂t2, 0 < x < L, t > 0;

u(0, t) = 0 and u(L, t) = 0, ∀t;


∂t(x, 0) = g(x), 0 < x < L.

I D’Alembert solution

u(x, t) =12[f∗(x−ct)+f∗(x+ct)]+

12c

∫ x+ct

x−ctg∗(s) ds, (19)

where f∗ and g∗ are odd extension of f and g.

Y. K. Goh


Examples (Optional)

1. Let say f(x) = sin mπLx and g(x) = 0, find the solution

for the 1D wave equation.

2. Same as the previous example, but L = 1, c = 1π

,g(x) = 0, and

f(x) =

{310x, 0 ≤ x ≤ 1

3;

3(1−x)10

, 13≤ x ≤ 1.

3. Now, let say L = 1, c = 1, f(x) = 0,g(x) = x, 0 < x < 1. Find the solution.

Y. K. Goh


Two Dimensional Laplace Equation

Problem:Consider a rectangle slab of length a and width b, assumingthat it is properly insulated from top and bottom of thesurfaces. The internal temperature distributionu(x, y, t), 0 < x < a, 0 < y < b, t > 0 in the slab in now givenby 2D heat equation ∂u

∂t= c2∇2u. The four boundaries of the

slab are kept at a constant temperature of zero forx = 0, x = a and y = 0, except at y = b where thetemperature is kept at u(x, b, t) = f(x). Suppose the slab areleft for a very long time, and the temperature distribution nolonger changing with time, i.e. ∂u

∂t= 0.

Now, find the steady state temperature distribution u(x, y) ofthe slab.

Y. K. Goh


Boundary-value problems

2D Laplace Equation:

∂2u

∂x2+∂2u

∂y2= 0, 0 < x < a, 0 < y < b; (20)

Boundary Conditions:

u(0, y) = u(a, y) = 0, 0 < y < a; (21)

u(x, 0) = 0 and u(x, b) = f(x), 0 < x < a; (22)

Y. K. Goh


Solving Laplace Equation

1. Decompose u into products of functions of one variable.I u(x, y) = X(x)Y (y).

2. Decompose the PDE into a set of ODEs.I

X ′′ + kX = 0; (23)

Y ′′ − kY = 0. (24)


I X(0) = X(a) = 0, and Y (0) = 0.I The Sturm-Liouville problem isX ′′ + kX = 0, X(0) = X(a) = 0.

Y. K. Goh


4 Solve the Sturm-Liouville problems and obtain thecorresponding eigenvalues.

I Non-trivial solution is Xn(x) = sin nπa x, n = 1, 2, . . . .

I Corresponding solution for Y isYn(y) = An cosh nπ

a y +Bn sinh nπa y.

I Apply BC Y (0) = 0, thus Yn(y) = Bn sinh nπa y.

5 Apply principle of superposition to obtain theeigenfucntion expansion of a general solution.

I u(x, y) =∞∑n=1

XnY n =∞∑n=1

Bn sinhnπ

ay sin

nπ

ax.

6 Use the initial conditions to obtain generalized Fouriercoefficients of the eigenfunction expansion.

I sinhnπb

aBn =

2a

∫ ∞a

f(x) sinnπ

ax dx, n = 1, 2, . . . .

Y. K. Goh


Example

I Find the steady-state temperature distribution u(x, y) ofa 1× 2 slab, with y represent distance along the directionof the longer side of the slab. Given that one longer sideof the slab is kept at 50◦C and the other sides are kept atzero temperature.

Y. K. Goh


Review on Boundary Conditions

Review on Boundary Conditions

Y. K. Goh


One Dimensional Heat Equation Again

Recall the heat equation in our first example

∂u

∂t= c2

∂2u

∂x2, 0 < x < L, t > 0. (25)

With the initial condition,

u(x, 0) = f(x), 0 < x < L. (26)

And the boundary condition,

u(0, t) = u(L, t) = 0, ∀t > 0. (27)

Here the boundary conditions are called HomogeneousDirichlet’s Boundary Conditions.

Y. K. Goh


Homogeneous (Zero Temperature) Dirichlet’s

Boundary Condition

We already showed that the solution for the zero temperatureheat equation is

u(x, t) =∞∑n=1

bne−λ2

nt sinnπ

Lx, n = 1, 2, . . . , (28)

where bn =2

L

∫ L

0

f(x) sinnπ

Lx dx.

Now, let’s take a closer look at the B.C.

Y. K. Goh


Boundary conditions for 1D Heat Equation

Three commonly used boundary conditions are

I Dirichlet’s B.C. (the values of u are given on boundaries)

I (Homogeneous B.C.) u(0, t) = u(L, t) = 0, ∀t > 0.I (Non-homogeneous B.C.)u(0, t) = T0, u(L, t) = T1, ∀t > 0 and T0, T1 6= 0.

I Neumann’s B.C. (normal derivatives are given on boundaries)

I ∂u∂x(0, t) = ∂u

∂x(L, t) = 0, ∀t > 0.I Robin’s B.C. (αu+ ∂u

∂n are given on boundaries)

I This B.C. correspond to one end insulated and one endradiating heat.

I u(0, t) = 0, ∂u∂x(L, t) = −κu(L, t), ∀t > 0.I κ is called the convection coefficient.

Y. K. Goh


Homogeneous (Non-Zero Temperature) Dirichlet’s

Boundary Condition

∂u

∂t= c2

∂2u

∂x2, 0 < x < L, t > 0;

I.C. : u(x, 0) = f(x), 0 < x < L;

B.C. : u(0, t) = u(L, t) = T0, ∀t > 0.

By changing the variable w(x, t) = u(x, t)− T0, we couldrecovered the zero temperature boundary-value problem

∂w

∂t= c2

∂2w

∂x2, 0 < x < L, t > 0;

I.C. : w(x, 0) = f(x)− T0, 0 < x < L;

B.C. : w(0, t) = w(L, t) = 0, ∀t > 0.

Y. K. Goh


Steady-State Solutions

I The steady-state solution, or time-independent solution,is when the change of temperature distribution us(x, t)no longer depends on time t. This usually happens whent→∞.

I In the steady-state situation, ∂u∂t

= 0, and thus the heat

equation now becomes a second order ODE d2udx2 = 0.

I If the boundary conditions are homogeneousu(0, t) = u(L, t) = T0, then the steady-state solution isus(x, t) = us(x) = T0.

I If the boundary conditions are non-homogeneousu(0, t) = T0, u(L, t) = T1, then the steady-state solutionis us(x, t) = us(x) = T1−T0

Lx+ T0.

Y. K. Goh


Dirichlet’s (Non-homogeneous) B.C.

The corresponding boundary-value problem is

∂u

∂t= c2

∂2u

∂x2, 0 < x < L, t > 0; (29)

B.C.: u(0, t) = T0, u(L, t) = T1, ∀t > 0; (30)

I.C.: u(x, 0) = f(x), 0 < x < L. (31)

The strategy to solve PDE with non-homogeneous B.C. is

I Find the steady-state solution us(x) that satisfies the B.C.

I Convert the non-homogeneous problem to ahomogeneous problem by changing the variablew(x, t) = u(x, t)− us(x).

Y. K. Goh


By substituting w(x, t) = u(x, t)− us(x) into thenon-homogeneous B.C. problem, we get

∂w

∂t= c2

∂2w

∂x2, 0 < x < L, t > 0;

I.C.: w(x, 0) = f(x)− us(x), 0 < x < L;

B.C.: w(0, t) = w(L, t) = 0, ∀t > 0.

I The solution to the homogeneous BVP is

w(x, t) =∞∑n=1

bne−λ2

nt sin βnx, where

bn =2

L

∫ L

0

(f(x)− us(x)) sin βnx dx.

I Finally, the solution to the non-homogeneous problem isu(x, t) = w(x, t) + us(x).

Y. K. Goh


Example

I Solve the following non-homogeneous boundary-valueproblem:

∂u

∂t= 4

∂2u

∂x2, 0 < x < π, t > 0;

I.C.: u(x, 0) = 50, 0 < x < π;

B.C.: u(0, t) = 0, and u(π, t) = 100, ∀t > 0.

Y. K. Goh


Neumann’s Boundary Conditions


∂u

∂t= c2

∂2u

∂x2, 0 < x < L, t > 0; (32)

B.C.:∂u

∂x(0, t) =

∂u

∂x(L, t) = 0, ∀t > 0; (33)

I.C.: u(x, 0) = f(x), 0 < x < L. (34)

After separated the variables:

I X ′′ − kX = 0, X ′(0) = X ′(L) = 0;

I T − kc2T = 0.

Y. K. Goh


Consider the three cases for the eigenvalue k:

I k = b2 > 0 =⇒ X(x) = c1 cosh bx+ c2 sinh bx,I X ′(0) = X ′(L) = 0 =⇒ c1 = c2 = 0I trivial solution

I k = 0, =⇒ X(x) = c1x+ c2I X ′(0) = X ′(L) = 0 =⇒ c1 = 0, c2 arbitraryI Let choose c2 = a0/2 where a0 is a constant.

I k = −β2 < 0, =⇒ X(x) = c1 cos βx+ c2 sin βxI X ′(0) = 0 =⇒ c2 = 0I X ′(L) = 0 =⇒ β = βn = nπ

L , n = 1, 2, . . .I βc1 is arbitrary, and set βc1 to 1.

The corresponding solution for T is T (t) = Tn(t) = ane−λ2

nt.

Y. K. Goh


Combining Xn(x) and Tn(t), and applying the principle ofsuperposition, we get the general solution

u(x, t) =a0

2+∞∑n=1

ane−λ2

nt cosnπ

Lx, n = 1, 2, . . . , (35)

where

I a0 =2

L

∫ L

0

f(x) dx,

I an =2

L

∫ L

0

f(x) cosnπ

Lx dx.

Y. K. Goh


Robin’s Boundary Conditions


∂u

∂t= c2

∂2u

∂x2, 0 < x < L, t > 0; (36)

B.C.: u(0, t) = 0,∂u

∂x(L, t) = −κu(L, t), ∀t > 0;(37)

I.C.: u(x, 0) = f(x), 0 < x < L. (38)

After separated the variables:

I X ′′ − kX = 0, X(0) = 0, X ′(L) = −κX(L);

I T − kc2T = 0.

Y. K. Goh


Consider the three cases for the eigenvalue k:

I k = b2 > 0 =⇒ X(x) = c1 cosh bx+ c2 sinh bx,I X(0) = 0, X ′(L) = −κX(L) =⇒ c1 = c2 = 0I because κ, cosh bL and sinh bL are strictly positive.

I k = 0, =⇒ X(x) = c1x+ c2I X(0) = 0, X ′(L) = −κX(L) =⇒ c1 = c2 = 0I trivial solution.

I k = −β2 < 0, =⇒ X(x) = c1 cos βx+ c2 sin βxI X(0) = 0 =⇒ c1 = 0I X ′(L) = −κX(L) implies β must satisfies the non-linear

equation β cosβL+ κ sinβL = 0, which has infinitemany roots, β = βn, n = 1, 2, . . . .

I Thus, X(x) = Xn(x) ∝ sinβnx.

Y. K. Goh


I The corresponding solution for T isT (t) = Tn(t) = cne

−λ2nt, where λn = cβn.

Combining Xn(x) and Tn(t), and applying the principle ofsuperposition, we get the general solution

u(x, t) =∞∑n=1

cne−λ2

nt sin βnx, n = 1, 2, . . . , (39)

where

I cn =1∫ L

0sin2 βnx dx

∫ L

0

f(x) sin βnx dx.

Y. K. Goh


Example

I Solve the following Robin’s problem:

∂u

∂t=

∂2u

∂x2, 0 < x < 1, t > 0;

I.C.: u(x, 0) = x(1− x), 0 < x < 1;

B.C.:∂u

∂x(0, t) = 0, and

∂u

∂x(1, t) = −u(1, t), ∀t > 0.

Y. K. Goh


Two Dimensional Wave Equation

Boundary-value problem

∂2u

∂t2= c2

(∂2u

∂x2+∂2u

∂y2

), 0 < x < L, t > 0;

B.C.: u(0, y, t) = u(a, y, t) = 0, for 0 ≤ y ≤ b and t > 0;

B.C.: u(x, 0, t) = u(x, b, t) = 0, for 0 ≤ x ≤ a and t > 0;

I.C.: u(x, y, 0) = f(x, y) and∂u

∂t(x, y, 0) = g(x, y).

Separation of variables:

I u(x, y, t) = X(x)Y (y)T (t);

I u(0, y, t) = u(a, y, t) = 0 =⇒ X(0) = X(a) = 0;

I u(x, 0, t) = u(x, b, t) = 0 =⇒ Y (0) = Y (b) = 0.

Y. K. Goh


IT

c2T=X ′′

X+Y ′′

Y= −k2, gives

I T + c2k2T = 0;

IX ′′

X= −Y

′′

Y− k2 = −µ2;

I X ′′ + µ2X = 0 and Y ′′ + ν2Y = 0,I where ν = k2 − µ2.

I In summaryI X ′′ + µ2X = 0, X(0) = 0, X(a) = 0,I Y ′′ + ν2Y = 0, Y (0) = 0, Y (b) = 0,I X + c2k2T = 0, k2 = µ2 + ν2.

Y. K. Goh


I Solution of the separated equationsI X(x) = c1 sinµx+ c2 cosµx;I Y (y) = d1 sin νy + d2 cos νy;I T (t) = e1 sin ckt+ e2 cos ckt.I where ν = k2 − µ2.

I From the boundary condition for X and Y we getI c2 = 0, µ = µm = mπ

a , and c1 arbitrary (set to 1).I d2 = 0, ν = νn = nπ

b , and d1 arbitrary (set to 1).

I Thus,I X(x) = Xm(x) = sinµmx, m = 1, 2, . . . ;I Y (y) = Yn(x) = sin νny, n = 1, 2, . . . ;I T (t) = Tmn(t) = Bmn cosλmnt+B∗ sinλmnt, whereλmn = c

√µ2m + ν2

n.

Y. K. Goh


I By using the principle of superposition, the solution tothe 2D wave equation is

u(x, y, t) =∞∑n=1

∞∑m=1

(Bmn cosλt+B∗mn sinλt) sinµmx sin νny.

(40)

I The coefficients are given by

I Bmn =4ab

∫ b

0

∫ a

0f(x, y) sinµmx sin νny dx dy;

I B∗mn =4

abλmn

∫ b

0

∫ a

0g(x, y) sinµmx sin νny dx dy;

I µm =mπ

a, νn =

nπ

b, m, n = 1, 2, . . . .

Y. K. Goh


Two Dimensional Heat Equation

Boundary-value problem

I∂u

∂t= c2

(∂2u

∂x2+∂2u

∂y2

), 0 < x < L, t > 0;

I B.C.: u(0, y, t) = u(a, y, t) = 0, for 0 ≤ y ≤ b and t > 0;I B.C.: u(x, 0, t) = u(x, b, t) = 0, for 0 ≤ x ≤ a and t > 0;I I.C.: u(x, y, 0) = f(x, y), for 0 < x < a, 0 < y < b.

General solution:

I u(x, y, t) =∞∑n=1

∞∑m=1

Amn sinµmx sin νnye−λmnt;

I µm = mπa, νn = nπ

b, λmn = c

√µ2m + ν2

n;

I Amn =4

ab

∫ b

0

∫ a

0

f(x, y) sinµmx sin νny dx dy.

Y. K. Goh


boundary-value problems in rectangular...

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