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.

QED