8.6 The Dirichlet Problem for Laplace’s Equation

Back

Remark 8.6.1. We cannot hope to solve the general Dirichlet Problem for

Laplace’s Equation, but we shall treat a few simple cases.

Suppose we take a rectangle in the plane, and lift up one of the sides of the

rectangle by a function. See figure 8.5. It will be useful to think of it as a

8.6. THE DIRICHLET PROBLEM FOR LAPLACE’S EQUATION 177

wire frame. We are going to find the equation of the soap film which will be

formed when the whole thing is dipped in soap7.

Formally, we have 0 _ x _ a, 0 _ y _ b as the region U, We have that there

is some unique function f : U −! R which is unknown, but that we have:

1. f(x, 0) = 0, 8x 2 [0, a]

2. f(0, y) = 0, 8y 2 [0, b]

3. f(x, b) = 0, 8x 2 [0, a]

4. f(a, y) = h(y), 8y 2 [0, b]

5.

@2f

@x2 +

@2f

@y2 = 0

for some given function h. I have illustrated h in figure 8.5 with a nice

parabolic function, but let us keep h general at the moment. The problem

is to find f, the soap film function. I remind you that this is not being

done because we care about soap, but because very much more significant

problems can be done using the same methods, and it is useful to have clear

pictures of a simple sort in your mind.

The first thing we do is make an assumption which is not immediately justifiable

or even reasonable, but which actually works.

Separation of Variables

Suppose that the function f(x, y) can be written as a product of

functions of x and y separately.

Write

f(x, y) = p(x)q(y)

Then differentiating partially with respect to x gives

@f

@x

= q(y)

dp

dx

,

@2f

@x2 = q

d2p

dx2

7 I am simplifying here: the Partial Differential Equation for Soap films or area minimisation

is non-linear; the general problem for solving it for given boundary conditions

is known as Plateau’s Problem. The PDE is approximated well by Laplace’s Equation

provided the non-linear effects are small, which will happen if the function f is not too

different from an affine function, and solving Laplace’s Equation for a given boundary

condition is often a good start on the Plateau Problem. From now on, I shall cheerfully

talk of soap films as if they were exactly solved by Laplace’s Equation

178 CHAPTER 8. PARTIAL DIFFERENTIAL EQUATIONS

and similarly

@f

@y

= p(x)

dq

dy

,

@2f

@y2 = p

d2q

dy2

Now Laplace’s Equation gives us

q¨p + p¨q = 0

Or

¨p

p

= −

¨q

q

= c

for some constant c.

Thus we have reduced the partial differential equation to two simultaneous

Ordinary Differential Equations, ¨p = c p and ¨q = −c q, which we know how

to solve8.

The boundary conditions for these can be worked out from the boundary

values for the original PDE: we have that f(0, y) = p(0)q(y) = 0, for all the

y 2 [0, b], and since q = 0 will not be a solution at x = a, we must have

p(0) = 0. Similarly q(0) = 0, q(b) = 0. We also deduce that p(a)q(y) = h(y).

You may be coming to feel that what is going to happen is that we are going

to have that nice parabola at f(a, y) simply scaled down progressively to zero

as we reduce x to zero. This seems physically reasonable.

First we solve

¨q = −c q

a familiar old face. We recall that the general solution is

q(y) = Asin(

p

cy) + B cos(

p

cy)

if we suppose c is positive, and we just swap p, q otherwise.

Now

q(0) = 0 ) B = 0, and q(b) = 0 ) c = (

n_

b

)2, n = 1, 2, · · ·

8 I have used the notation p˙ and q¨ rather casually; we are differentiating with respect

to different variables here. I interpret the dot as ‘Differentiate with respect to the (single)

variable’. Other, sterner, folk insist that we use Newton’s dot notation only when the

variable is time. I have tried it in other notations, and it is longer and harder to read.

8.6. THE DIRICHLET PROBLEM FOR LAPLACE’S EQUATION 179

So we get solutions for q of the form

q(y) = An sin(

n_y

b

)

Going back to the equation for p, we have that

p(x) = C sinh(

p

cx) + Dcosh(

p

cx)

is the general solution and we know that p(0) = 0 and hence that D = 0.

But we now know something about c, from the boundary conditions on q,

so we can conclude that for every positive integer n, there is a solution in

waiting,

fn(x, y) = cn sinh(

n_x

b

) sin(

n_y

b

)

These ‘solutions in waiting’ as I have called them, exist for all positive integers

n, and for all constants cn, and the sum of any set of them is also a solutionin-

waiting. In order to be a real solution, we have to find a sum of them which

also satisfy the final remaining boundary condition, they have to agree with

the function h, the parabola in the figure. We cannot reasonably expect the

sum of a finite collection of sine functions to be a parabola, but we can get

closer and closer– we are just doing Fourier Theory.

This means that

f(a, y) =

X1

n=1

cn sinh(

n_x

b

) sin(

n_y

b

) = h(y)

Each Fourier coefficient is cn sinh(n_x

b ), given by

cn sinh(

n_a

b

) =

2

b

Z b

0

h(y) sin(

n_y

b

) dy

If we can do the integrals, we can calculate the coefficients as far as we like,

and in some happy cases we can get explicit solutions.

We can get a reasonable agreement with figure 8.5 if we put b = _ and

h(y) = sin(y). We therefore work through the following example:

Example 8.6.1.

Problem

Let the harmonic function f : [0, 1] × [0, _] satisfy the following boundary

conditions:

180 CHAPTER 8. PARTIAL DIFFERENTIAL EQUATIONS

Figure 8.6: Function on the boundary

1. f(x, 0) = 0, 8x 2 [0, 1]

2. f(0, y) = 0, 8y 2 [0, _]

3. f(1, y) = 0, 8y 2 [0, _]

4. f(1, y) = sin(y), 8y 2 [0, _]

Sketch the graph of f on the boundary of the rectangle, and calculate the

function f on the interior.

Solution The graph is illustrated in fig 8.6.

We suppose that the solution is separable, f(x, y) = p(x)q(y).

This tells us that

¨p

p

= −

¨q

q

= c

and we have no way of knowing whether c is positive or negative until we

investigate the boundary conditions, since we could always interchange x and

y in the problem.

Since we know that at x = 1 we have that f(1, y) = sin(y), we have

p(1)q(y) = sin(y), which tells us that p(1) = 1 and

q(y) = sin(y)

8.6. THE DIRICHLET PROBLEM FOR LAPLACE’S EQUATION 181

is a possibility, with c = 1. In which case,

¨p

p

= 1

and we have

p(x) = Acosh(x) + B sinh(x)

is a solution in waiting. It, or some (possibly infinite) sum of such solutions,

must satisfy the boundary conditions not so far used. These are straightforward:

we must have

p(0) = 0; p(1) = 1

and this immediately tells us that

A = 0,B =

1

sinh(1)

is a solution. So the final solution is

f(x, y) =

sinh(x)

sinh(1)

sin(y)

It is straightforward to verify (1) that the boundary conditions are all satisfied

and (2) that the function is harmonic (everywhere). Since we have a

uniqueness theorem, we have produced the only possible solution.

Exercise 8.6.1.

Verify that the given solution satisfies Laplace’s Equation.

Remark 8.6.2. If you have any soul in you at all, you will now stand up

and clap for half an hour at something so wonderful.

Example 8.6.2.

Problem

Let a square of side _ units be made of metal, and let three of the four sides

be kept at a temperature of 0o. Let the last side have temperature sin(x) at

distance x along from one end. Find the temperature at the centre of the

plate when the system is in equilibrium.

Solution

e_/2 − e−_/2

e_ − e−_

182 CHAPTER 8. PARTIAL DIFFERENTIAL EQUATIONS

Figure 8.7: A Little Problem

This is essentially the same as the last worked example of course. Only the

names have been changed to protect the guilty9. There is a scaled version of

the last solution because the plate is bigger. I have expanded the sinh function

out for those of you who like to see everything in terms of exponentials.

Go on, check it out, then clap! It isn’t quite as wonderful as Euler’s Formula,

ei_ = −1, but it is pretty damned smart and can be verified by experiment,

which is more than can be said for the Euler Formula10.

Exercise 8.6.2.

1. Suppose the function f of the preceding worked examples had been

modified so that it is defined on the interval

[−_/2, _/2] × [−_/2, _/2]

and is zero along two opposite sides and is a cosine function along the

other two opposite sides, as in fig 8.7. Find an explicit form for the

harmonic function from first principles.

2. You have calculated the Fourier Series for a certain number of functions

by now: choose some function where you have the Fourier Series already

worked out and use it as an alternative to my function h(y) = sin(y)

to obtain a Fourier Series solution to your very own soap film problem.

9The innocent are not in need of protection, their strength is as the strength of ten,

because their hearts are pure.

10 What is marvellous isn’t the answer, it is that somebody of the same species as you

was smart enough to figure it out, and you are smart enough to follow the argument.

If this doesn’t strike you as astonishingly wonderful, you are probably dead but haven’t

noticed yet.

8.7. LAPLACE ON DISKS 183

3. The constraints on the boundary look rather strong, and you might

wonder what you could do with a case where one edge was fixed to

have height h1(y) and the opposite edge was fixed to have height h2(y).

Deal with the case where one end of a square of side _ is made to

have height sin(y), the opposite is made to have height −sin(y) and

the other two are kept at height zero. Do this by finding solutions to

(a) the case where three sides are zero and one side is at height sin(y),

which has been done, (b) the case where the opposite side is kept at

height −sin(y) and all other sides kept at zero and then (c) adding up

the answers. After all, if two functions satisfy Laplace’s Equation, so

does their sum.(!) Of course, if the two functions h1, h2 are the same

there might be a quicker method, as in our earlier worked example.

Навигация