The Two Dimensional Heat Equation Lecture 3 6 Short

The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions

The two dimensional heat equation

Ryan C. Daileda

Trinity University

Partial Differential EquationsMarch 6, 2012

Daileda The 2D heat equation


Physical motivation

Consider a thin rectangular plate made of some thermallyconductive material. Suppose the dimensions of the plate are a b.

The plate is heated in some way, and then insulated along itstop and bottom.

Our goal is to mathematically model the way thermal energymoves through the plate.



We let

u(x , y , t) = temperature of plate at position (x , y) andtime t.

For a fixed t, the height of the surface z = u(x , y , t) gives thetemperature of the plate at time t and position (x , y).

Under ideal assumptions (e.g. uniform density, uniform specificheat, perfect insulation, no internal heat sources etc.) one canshow that u satisfies the two dimensional heat equation

ut = c22u = c2(uxx + uyy) (1)

for 0 < x < a, 0 < y < b.



At the edges of the plate we impose some sort of boundaryconditions. The simplest are homogeneous Dirichletconditions:

u(0, y , t) = u(a, y , t) = 0, 0 y b, t 0,

u(x , 0, t) = u(x , b, t) = 0, 0 x a, t 0. (2)

Physically, these correspond to holding the temperature along theedges of the plate at 0.

The way the plate is heated initially is given by the initialcondition

u(x , y , 0) = f (x , y), (x , y) R , (3)

where R = [0, a] [0, b].



Solving the 2D wave equation: homogeneous Dirichlet

boundary conditions

Goal: Write down a solution to the heat equation (1) subject tothe boundary conditions (2) and initial conditions (3).

As usual, we

separate variables to produce simple solutions to (1) and(2), and then

use the principle of superposition to build up a solution thatsatisfies (3) as well.



Separation of variables

Assuming thatu(x , y , t) = X (x)Y (y)T (t),

plugging into the heat equation (1) and using the boundaryconditions (2) yields the separated equations

X BX = 0, X (0) = 0, X (a) = 0, (4)

Y CY = 0, Y (0) = 0, Y (b) = 0, (5)

T c2(B + C )T = 0. (6)



We have already seen that the solutions to (4) and (5) are

Xm(x) = sinmx , m =mpi

a, B = 2m

Yn(y) = sin ny , n =npi

b, C = 2n ,

for m, n N. Using these values in (6) gives

Tmn(t) = e2mnt ,

where

mn = c2m +

2n = cpi

m2

a2+

n2

b2.



Superposition

Assembling these results, we find that for any pair m, n 1 wehave the normal mode

umn(x , y , t) = Xm(x)Yn(y)Tmn(t) = sinmx sin ny e2mnt .

For any choice of constants Amn, by the principle of superpositionwe then have the general solution to (1) and (2)

u(x , y , t) =

m=1

n=1

Amn sinmx sin ny e2mnt .



Initial conditions

We must determine the values of the coefficients Amn so that oursolution also satisfies the initial condition (3). We need

f (x , y) = u(x , y , 0) =

n=1

m=1

Amn sinmpi

ax sin

npi

by

which is just the double Fourier series for f (x , y). We know thatif f is a C 2 function then

Amn =4

ab

a0

b0

f (x , y) sinmpi

ax sin

npi

bdy dx .



Conclusion

Theorem

Suppose that f (x , y) is a C 2 function on the rectangle[0, a] [0, b]. The solution to the heat equation (1) withhomogeneous Dirichlet boundary conditions (2) and initialconditions (3) is given by

u(x , y , t) =

m=1

n=1

Amn sinmx sin ny e2mnt ,

where m =mpi

a, n =

npi

b, mn = c

2m +

2n , and

Amn =4

ab

a0

b0

f (x , y) sinmpi

ax sin

npi

by dy dx .



Remarks:

We have not actually verified that this solution is unique, i.e.that this is the only solution to problem.

We will prove uniqueness later using the maximum principle.

Example

A 2 2 square plate with c = 1/3 is heated in such a way that thetemperature in the lower half is 50, while the temperature in the

upper half is 0. After that, it is insulated laterally, and the

temperature at its edges is held at 0. Find an expression that gives

the temperature in the plate for t > 0.



We must solve the heat equation problem (1) - (3) with

f (x , y) =

{50 if y 1,

0 if y > 1.

The coefficients in the solution are

Amn =4

2 2

20

20

f (x , y) sinmpi

2x sin

npi

2y dy dx

= 50

20

sinmpi

2x dx

10

sinnpi

2y dy

= 50

(2(1 + (1)m+1)

pim

)(2(1 cos npi2 )

pin

)

=200

pi2(1 + (1)m+1)(1 cos npi2 )

mn.



Since

mn =pi

3

m2

4+

n2

4=

pi

6

m2 + n2

the solution is

u(x , y , t) =200

pi2

m=1

n=1

((1 + (1)m+1)(1 cos npi2 )

mnsin

mpi

2x

sinnpi

2y epi

2(m2+n2)t/36).



Steady state solutions

To deal with inhomogeneous boundary conditions in heatproblems, one must study the solutions of the heat equationthat do not vary with time.

These are the steady state solutions. They satisfy

ut = 0.

In the 1D case, the heat equation for steady states becomes

uxx = 0.

The solutions are simply straight lines.



Laplaces equation

In the 2D case, we see that steady states must solve

2u = uxx + uyy = 0. (7)

This is Laplaces equation.

Solutions to Laplaces equation are called harmonicfunctions.

See assignment 1 for examples of harmonic functions.



We want to find steady state solutions to (7) that satisfy theDirichlet boundary conditions

u(x , 0) = f1(x), u(x , b) = f2(x), 0 < x < a (8)

u(0, y) = g1(x), u(a, y) = g2(y), 0 < y < b (9)

The problem of finding a solution to (7) - (9) is known as theDirichlet problem.

We will begin by assuming f1 = g1 = g2 = 0.

The general solution to the Dirichlet problem will be obtainedby superposition.



Solution of the Dirichlet problem on a rectangle

Goal: Solve the boundary value problem

2u = 0, 0 < x < a, 0 < y < b,

u(x , 0) = 0, u(x , b) = f2(x), 0 < x < a,

u(0, y) = u(a, y) = 0, 0 < y < b.

Picture:



Separation of variables

Setting u(x , y) = X (x)Y (y) leads to

X + kX = 0 , Y kY = 0,

X (0) = X (a) = 0 , Y (0) = 0.

We know the nontrivial solutions for X :

X (x) = Xn(x) = sinnx , n =npi

a, k = 2n.

for n N = {1, 2, 3, . . .}. The corresponding solutions for Y are

Y (y) = Yn(y) = Aneny + Bne

ny .



The condition Y (0) = 0 yields An = Bn. Choosing An = 1/2 wefind that

Yn(y) = sinhny .

This gives the separated solutions

un(x , y) = Xn(x)Yn(y) = sinnx sinhny ,

and superposition gives the general solution

u(x , y) =

n=1

Bn sinnx sinhny .



The general solution satisfies the Laplace equation (7) inside therectangle, as well as the three homogeneous boundary conditionson three of its sides (left, right and bottom).

We now determine the values of Bn to get the boundary conditionon the top of the rectangle. This requires

f2(x) = u(x , b) =n=1

Bn sinhnpib

asin

npi

ax ,

which is the Fourier sine series for f2(x) on 0 < x < a.

Appealing to previous results, we can now summarize our findings.



Conclusion

Theorem

If f2(x) is piecewise smooth, the solution to the Dirichlet problem

2u = 0, 0 < x < a, 0 < y < b,

u(x , 0) = 0, u(x , b) = f2(x), 0 < x < a,

u(0, y) = u(a, y) = 0, 0 < y < b.

is

u(x , y) =n=1

Bn sinnx sinhny ,

where n =npi

aand Bn =

2

a sinh npiba

a0

f2(x) sinnpi

ax dx .



Remarks: As we have noted previously:

We have not proven this solution is unique. This requires themaximum principle.

If we know the sine series expansion for f2(x) then we can usethe relationship

Bn =1

sinh npiba

(nth sine coefficient of f2(x))

to avoid integral computations.



Example

Example

Solve the Dirichlet problem on the square [0, 1] [0, 1], subject tothe boundary conditions

u(x , 0) = 0, u(x , b) = f2(x), 0 < x < a,

u(0, y) = u(a, y) = 0, 0 < y < b.

where

f2(x) =

{75x if 0 x 23 ,

150(1 x) if 23 < x 1.



We have a = b = 1 and we require the sine series of f2(x). Thegraph of f2(x) is:

According to exercise 2.4.17 (with p = 1, a = 2/3 and h = 50):

f2(x) =450

pi2

n=1

sin 2npi3n2

sin npix .



Thus,

Bn =1

sinh npi

(450

pi2sin 2npi3n2

)=

450

pi2sin 2npi3

n2 sinh npi,

and

u(x , y) =450

pi2

n=1

sin 2npi3n2 sinh npi

sin npix sinh npiy .


The 2D heat equationHomogeneous Dirichlet boundary conditionsSteady state solutions

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The Two Dimensional Heat Equation Lecture 3 6 Short