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Steady-state heat equation in the unit disc

April 3, 2012

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 u(r, \theta) = A_r(f)(\theta) = \sum_{n=-\infty}^{\infty}a_nr^{|n|}e^{in\theta}. We will now show it is true, but before that, let’s note that A_r(f) converges absolutely and uniformly for integrable f when 0 \leq r < 1. Indeed, we have

|a_n| = \displaystyle\frac{1}{2\pi}\left|\int_{-\pi}^{\pi}f(x)e^{-inx}dx\right|

\leq \displaystyle\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)|dx

which is bounded since f is integrable on the circle. Thus \sum |a_mr^{|m|}e^{im\theta}| \leq \sup |f(\theta)| \sum r^{|m|} hence the uniform convergence of A_r(f) for r \in [0,1).

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

u(r, \theta) = (f*P_r)(\theta)

has the following properties:

  1. u has two continuous derivatives in the unit disc and satisfies \bigtriangleup u = 0.
  2. If \theta is any point of continuity of f, then \lim_{r\rightarrow 1} u(r,\theta) = f(\theta). If f is continuous everywhere, then this limit is uniform.
  3. If f is continuous, then u(r, \theta) is the unique solution to the steady-state heat equation in the disc which satisfies conditions 1. and 2.

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

\bigtriangleup u = \displaystyle\frac{\partial ^2u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial ^2u}{\partial \theta^2}

= \displaystyle\sum_{m=-\infty}^{\infty} |m|(|m|-1)a_mr^{|m|-2}e^{im\theta} + \sum_{m=-\infty}^{\infty}|m|a_mr^{|m|-2}e^{im\theta}

- \displaystyle\sum_{m=-\infty}^{\infty}m^2a_mr^{|m|-2}e^{im\theta}

= 0

hence 1. Since property 2 is just a restatement of the last theorem of last post, it remains only to prove 3. Suppose v(r, \theta) is another solution satisfying 1. and 2. Let’s say that for each r \in (0, 1), the function v has Fourier series

\displaystyle\sum_{n=-\infty}^{\infty}b_n(r)e^{in\theta} where b_n(r) =\displaystyle\frac{1}{2\pi}\int_{-\pi}^{\pi}v(r,\theta)e^{-in\theta}d\theta.

Since we have \bigtriangleup v = 0, consider (\dagger)

0 = \displaystyle\frac{\partial ^2v}{\partial r^2}e^{-in\theta} + \frac{1}{r}\frac{\partial v}{\partial r}e^{-in\theta} + \frac{1}{r^2}\frac{\partial ^2v}{\partial \theta^2}e^{-in\theta}

= \displaystyle\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{\partial ^2v}{\partial r^2}e^{-in\theta}d\theta + \frac{1}{2\pi r}\int_{-\pi}^{\pi}\frac{\partial v}{\partial r}e^{-in\theta}d\theta+\frac{1}{2\pi r^2}\int_{-\pi}^{\pi}\frac{\partial ^2v}{\partial \theta^2}e^{-in\theta}d\theta

Integrating by part, the third term of (\dagger) becomes

\displaystyle\frac{1}{2\pi r^2}\left(\left.\frac{\partial}{\partial \theta}v(r, \theta)e^{-in\theta}\right|_{-\pi}^{\pi} + in\int_{-\pi}^{\pi}\frac{\partial}{\partial \theta}v(r,\theta)e^{-in\theta}d\theta\right)

But \left.\frac{\partial v}{\partial \theta}(r, \theta)e^{-in\theta}\right|_{-\pi}^{\pi} = 0 since both v and e^{-in\theta} are 2\pi-periodic in \theta. Integrating by parts again gives -\frac{n^2}{2\pi r^2}\int_{-\pi}^{\pi}v(r, \theta)e^{-in\theta}d\theta so the third term of (\dagger) is equal to -n^2b_n(r)/r^2.

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

\displaystyle\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{\partial ^2v}{\partial r^2}(r, \theta)e^{-in\theta}d\theta = \frac{\partial ^2}{\partial r^2}\frac{1}{2\pi}\int_{-\pi}^{\pi}v(r, \theta)e^{-in\theta}d\theta = b''_n(r).

Similarly, the second term of (\dagger) is equal to b'_n(r)/r. Substituting all this into (\dagger), we find

b''_n(r) + \displaystyle\frac{1}{r}b'_n(r) - \frac{n^2}{r^2}b_n(r) = 0.

By exercise 3 of March 24th’s exercises entry, we know this means that we must have b_n(r) = A_nr^n + B_nr^n for certain A_n, B_n \in \textbf{R} when n \neq 0, while b_0(r) = A_0 + B_0\log(r). Since v is bounded, so are the b_n‘s so we must have B_n = 0. Moreover, since v(r, \theta) \rightarrow f(\theta) uniformly as r \rightarrow 1, (this is condition 2), we have

A_n = \displaystyle\lim_{r\rightarrow 1} b_n(r) = \lim_{r\rightarrow 1} \frac{1}{2\pi}\int_{-\pi}^{\pi} v(\theta, r)e^{-in\theta}d\theta

= \displaystyle\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\theta)e^{-in\theta}d\theta

when n \neq 0. Similarly, when n = 0, we have A_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(\theta)d\theta. In other words, the Fourier series of v is the same as the Fourier series of u. By corollary 1 of this post, this means that u = v since f is supposed continuous.



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