# Steady-state heat equation in the unit disc

This is the conclusion to the last 4 posts. Recall that in a motivational problem for the study of Fourier series, we were led to believe that the solution to the steady-state heat equation on the disc was given by We will now show it is true, but before that, let’s note that converges absolutely and uniformly for integrable when . Indeed, we have

which is bounded since is integrable on the circle. Thus hence the uniform convergence of for

Theorem (Stein’s book p.57):Let be an integrable function defined on the unit circle. Then the function defined in the unit disc by the Poisson integralhas the following properties:

- has two continuous derivatives in the unit disc and satisfies
- If is any point of continuity of then If is continuous everywhere, then this limit is uniform.
- If is continuous, then is the unique solution to the steady-state heat equation in the disc which satisfies conditions 1. and 2.

**Proof:** For 1, recall that converges absolutely and uniformly for so it’s differentiable term by term. Moreover, the differentiated series is absolutely and uniformly convergent thus can be differentiated twice in the unit disc. Therefore, we get

hence 1. Since property 2 is just a restatement of the last theorem of last post, it remains only to prove 3. Suppose is another solution satisfying 1. and 2. Let’s say that for each the function has Fourier series

where

Since we have consider

Integrating by part, the third term of becomes

But since both and are -periodic in Integrating by parts again gives so the third term of is equal to

Moreover, using Leibniz’s rule for integration (which will be proved in exercise 3), we get, for the first term of

Similarly, the second term of is equal to Substituting all this into we find

By exercise 3 of March 24th’s exercises entry, we know this means that we must have for certain when , while Since is bounded, so are the ‘s so we must have Moreover, since uniformly as (this is condition 2), we have

when Similarly, when we have In other words, the Fourier series of is the same as the Fourier series of By corollary 1 of this post, this means that since is supposed continuous.

QED

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