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Note: 2 lectures, §9.7 in [EP], §10.8 in [BD]

A more natural setting for the Laplace equation is the circle rather than the square. On the other hand, what makes the problem somewhat more difficult is that we need polar coordinates.

Recall that the polar coordinates for the -plane are :

where and . So is distance from the origin at angle .

Now that we know our coordinates, let us give the problem we wish to solve. We have a circular region of radius 1, and we are interested in the Dirichlet problem for the Laplace equation for this region. Let denote the temperature at the point in polar coordinates. We have the problem:

(4.25) |

The first issue we face is that we do not know what the Laplacian is in polar coordinates. Normally we would find and in terms of the derivatives in and . We would need to solve for and in terms of and . While this is certainly possible, it happens to be more convenient to work in reverse. Let us instead compute derivatives in and in terms of derivatives in and and then solve. The computations are easier this way. First

Next by chain rule we obtain

Similarly for the derivative. Note that we have to use product rule for the second derivative. Let us now try to solve for . We start with to get rid of those pesky . If we add and use the fact that , we getWe’re not quite there yet, but all we are lacking is . Adding it we obtain the Laplacian in polar coordinates:

Notice that the Laplacian in polar coordinates no longer has constant coefficients.

Let us separate variables as usual. That is let us try . Then

Let us put on one side and on the other and conclude that both sides must be constant.

We get two equations: Let us first focus on . We know that ought to be -periodic in , that is, . Therefore, the solution to must be -periodic. We conclude that for a nonnegative integer . The equation becomes . When the equation is just , so we have the general solution . As is periodic, . For convenience let us write this solution asfor some constant . For positive , the solution to is

for some constants and .

Next, we consider the equation for ,

This equation has appeared in exercises before—we solved it in Exercise 2.1.6 and Exercise 2.1.7. The idea is to try a solution and if that does not work out try a solution of the form . When we obtain

and if , we get

The function must be finite at the origin, that is, when . Therefore, in both cases. Let us set in both cases as well, the constants in will pick up the slack so we do not lose anything. Therefore let

Hence our building block solutions are

Putting everything together our solution is:We look at the boundary condition in (4.25),

Therefore, the solution to (4.25) is to expand , which is a -periodic function as a Fourier series, and then the coordinate is multiplied by . In other words, to compute and from the formula we can, as usual, compute

Example 4.10.1: Suppose we wish to solve

The solution is

See the plot in Figure 4.23. The thing to notice in this example is that the effect of a high frequency is mostly felt at the boundary. In the middle of the disc, the solution is very close to zero. That is because rather small when is close to 0.

Example 4.10.2: Let us solve a more difficult problem. Suppose we have a long rod with circular cross section of radius 1 and we wish to solve the steady state heat problem. If the rod is long enough we simply need to solve the Laplace equation in two dimensions. Let us put the center of the rod at the origin and we have exactly the region we are currently studying—a circle of radius 1. For the boundary conditions, suppose in Cartesian coordinates and , the temperature is fixed at 0 when and at when .

We set the problem up. As , then on the circle of radius 1 we have . So

We must now compute the Fourier series for the boundary condition. By now the reader has plentiful experience in computing Fourier series and so we simply state that

Exercise 4.10.1: Compute the series for and verify that it really is what we have just claimed. Hint: Be careful, make sure not to divide by zero.

We now simply write the solution (see Figure 4.24) by multiplying by in the right places.

There is another way to solve the Dirichlet problem with the help of an integral kernel. That is, we will find a function called the Poisson kernel^{7} such that

While the integral will generally not be solvable analytically, it can be evaluated numerically. In fact, unless the boundary data is given as a Fourier series already, it will be much easier to numerically evaluate this formula as there is only one integral to evaluate.

The formula also has theoretical applications. For instance, as will have infinitely many derivatives, then via differentiating under the integral we find that the solution has infinitely many derivatives, at least when inside the circle, . By infinitely many derivatives what you should think of is that has “no corners” and all of its partial derivatives exist too and also have “no corners”.

We will compute the formula for from the series solution, and this idea can be applied anytime you have a convenient series solution where the coefficients are obtained via integration. Hence you can apply this reasoning to obtain such integral kernels for other equations, such as the heat equation. The computation is long and tedious, but not overly difficult. Since the ideas are often applied in similar contexts, it is good to understand how this computation works.

What we do is start with the series solution and replace the coefficients with the integrals that compute them. Then we try to write everything as a single integral. We must use a different dummy variable for the integration and hence we use instead of .

OK, so we have what we wanted, the expression in the parentheses is the Poisson kernel, . However, we can do a lot better. It is still given as a series, and we would really like to have a nice simple expression for it. We must work a little harder. The trick is to rewrite everything in terms of complex exponentials. Let us work just on the kernel.

In the above expression we recognize the geometric series. That is, recall from calculus that as long as , then

Note that starts at and that is why we have the in the numerator. It is the standard geometric series multiplied by . Let us continue with the computation.

Now that’s a forumla we can live with. The solution to the Dirichlet problem using the Poisson kernel is

Sometimes the formula for the Poisson kernel is given together with the constant , in which case we should of course not leave it in front of the integral. Also, often the limits of the integral are given as 0 to ; everything inside is -periodic in , so this does not change the integral.

Let us not leave the Poisson kernel without explaining its geometric meaning. Let be the distance from to . You may recall from calculus that this distance in polar coordinates is given precisely by the square root of . That is, the Poisson kernel is really the formula

One final note we make about the formula is to note that it is really a weighted average of the boundary values. First let us look at what happens at the origin, that is when .

So is precisely the average value of and therefore the average value of on the boundary. This is a general feature of harmonic functions, the value at some point is equal to the average of the values on a circle centered at .

What the formula says is that the value of the solution at any point in the circle is a weighted average of the boundary data . The kernel is bigger when is closer to . Therefore when computing we give more weight to the values when is closer to and less weight to the values when far from .

Exercise 4.10.4: Using the Poisson kernel, give the solution to , where is zero for outside the interval and is 1 for on the interval .

Exercise 4.10.5: a) Draw a graph for the Poisson kernel as a function of when and . b) Describe what happens to the graph when you make bigger (as it approaches 1). c) Knowing that the solution is the weighted average of with Poisson kernel as the weight, explain what your answer to part b means.

Exercise 4.10.6: Take the function to be the function on the boundary. Use the series solution to find a solution to the Dirichlet problem , . Now convert the solution to Cartesian coordinates and . Is this solution surprising? Hint: use your trig identities.

Exercise 4.10.7: Carry out the computation we needed in the separation of variables and solve , for .

Exercise 4.10.8 (challenging): Derive the series solution to the Dirichlet problem if the region is a circle of radius rather than 1. That is, solve , .

Exercise 4.10.102: Using the series solution find the solution to , . Express the solution in Cartesian coordinates (that is, using and ).

Exercise 4.10.103: a) Try and guess a solution to , . Hint: try a solution that only depends on . Also first, don’t worry about the boundary condition. b) Now solve , using superposition.

Exercise 4.10.104 (challenging): Derive the Poisson kernel solution if the region is a circle of radius rather than 1. That is, solve , .

^{7}Named for the French mathematician Siméon Denis Poisson (1781 – 1840).