Seanie Lee

I love mathematics and statistics

Dirichlet Problem on the Unit Disc

5 minute read

Preliminaries

Suppose one has an infinite plate $(\mathbb{R}^2)$ with an initial heat distribution. Let $u(x,y)$ denote the temperature of the place at position $(x,y)\in\mathbb{R}^2$. We are mainly concerned with the steady-state heat equation

\[\begin{align*} \Delta u := \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2u}{\partial y^2}=0. \end{align*}\]

The Dirichlet problem ($D$ some open domain set, and $\partial D$ its boundary):

\[\begin{align*} \begin{cases} \Delta u = 0 \text{ on } D \\ u=f \text{ on } \partial D. \end{cases} \end{align*}\]

In polar coordinates, the condition $\Delta u=0$ becomes

\[\begin{align*} \Delta u = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} =0. \end{align*}\]

<Proof>

Let $x=r\cos\theta$ and $y=r\sin\theta$. Then

\[\begin{align*} \frac{\partial x}{\partial r} &= \cos\theta, \quad \frac{\partial x}{\partial \theta}=-r\sin\theta \\ \frac{\partial y}{\partial r}&= \sin\theta, \quad \frac{\partial y}{\partial \theta} = r\cos\theta. \end{align*}\]

The first-order partial derivative of $u$ with respect to $r$,

\[\begin{align*} \frac{\partial u}{\partial r} &= \frac{\partial u }{\partial x} \frac{\partial x}{\partial r} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial r} \\ &=\cos\theta \frac{\partial u}{\partial x} + \sin\theta \frac{\partial u}{\partial y}. \end{align*}\]

For the second-order derivative,

\[\begin{align*} \frac{\partial^2 u}{\partial r^2} &= \cos\theta \frac{\partial }{\partial r}\left(\frac{\partial u}{\partial x}\right) + \sin\theta \frac{\partial}{\partial r}\left( \frac{\partial u}{\partial y} \right) \\ &= \cos\theta \left(\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}\right)\frac{\partial x}{\partial r} + \frac{\partial}{\partial y}\left(\frac{\partial u}{\partial x}\right)\frac{\partial y}{\partial r} \right) \\ &+\sin\theta\left(\frac{\partial u}{\partial x}\left(\frac{\partial u}{\partial y} \right)\frac{\partial x}{\partial r} + \frac{\partial}{\partial y}\left( \frac{\partial u}{\partial y}\right)\frac{\partial y}{\partial r} \right) \\ &= \cos^2\theta \frac{\partial^2 u}{\partial x^2} + 2\sin\theta\cos\theta \frac{\partial^2 u}{\partial x \partial y} + \sin^2\theta \frac{\partial^2 u}{\partial y^2} . \end{align*}\]

Similarly, we get the first order derivative of $u$ with respect to $\theta$,

\[\begin{align*} \frac{\partial u}{\partial \theta} &= \frac{\partial u}{\partial x} \frac{\partial x}{\partial \theta}+ \frac{\partial u}{\partial y}\frac{\partial u}{\partial \theta} \\ &=-r\cos\theta\cdot \frac{\partial u}{\partial x} + r\cos\theta \frac{\partial u}{\partial y}. \end{align*}\]

For the second-order derivative,

\[\begin{align*} \frac{\partial^2 u}{\partial \theta^2} &= \frac{\partial}{\partial\theta}\left(-r\sin\theta \frac{\partial u}{\partial x}\right) + \frac{\partial}{\partial \theta}\left(r\cos\theta \frac{\partial u}{\partial y}\right) \\ &= -r\cos\theta \frac{\partial u}{\partial x} -r\sin\theta \frac{\partial^2}{\partial \theta \partial x}-r\sin\theta \frac{\partial u}{\partial y} + r\cos\theta \frac{\partial^2 u}{\partial \theta \partial y} \\ &= -r\cos\theta \frac{\partial u}{\partial x}-r\sin\theta \left(\frac{\partial}{\partial x}\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta}+ \frac{\partial}{\partial y}\frac{\partial u}{\partial x}\frac{\partial y}{\partial \theta} \right) - r\sin\theta \frac{\partial u}{\partial y} \\ &+r\cos\theta \left( \frac{\partial }{\partial x}\frac{\partial y}{\partial y}\frac{\partial x}{\partial \theta} + \frac{\partial }{\partial y}\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta} \right) \\ &=-r\cos\theta \frac{\partial u}{\partial x} - r\sin\theta\left(\frac{\partial^2 u}{\partial x^2}(-r\sin\theta)+\frac{\partial}{\partial x \partial y}r\cos\theta \right) -r\sin\theta \frac{\partial u}{\partial y} + r\cos\theta \left( \frac{\partial^2 u}{\partial x\partial y}(-r\sin\theta) + \frac{\partial^2 u}{\partial y^2}r\cos\theta\right) \\ &=-r\frac{\partial u}{\partial r} + r^2\sin^2\theta \frac{\partial^2 u}{\partial x^2} - 2r^2\sin\theta\cos\theta \frac{\partial^2 u}{\partial x \partial y} + r^2\cos^2\theta \frac{\partial^2 u}{\partial x \partial y}. \end{align*}\]

Thus,

\[\begin{align*} \frac{\partial^2u}{\partial r^2} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}&=\cos^2\theta \frac{\partial^2 u}{\partial x^2} + 2\sin\theta\cos\theta \frac{\partial^2 u}{\partial x \partial y} +\sin^2\theta \frac{\partial^2 u}{\partial y^2} \\ &-\frac{1}{r}\frac{\partial u}{\partial r} + \sin^2\theta \frac{\partial^2 u}{\partial x^2}-2\sin\theta\cos\theta \frac{\partial^2 u}{\partial x \partial y}+ \cos^2\theta \frac{\partial^2 u}{\partial y^2} \\ &= \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} - \frac{1}{r}\frac{\partial u}{\partial r}. \end{align*}\]

By rearranging the terms, we get

\[\begin{align*} \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}. \end{align*}\] \[\tag*{$\square$}\]

Theorem 1.1

Let $f$ be an integrable function on the unit circle. Then the Poisson integral One has,

\[\begin{align*} u(r,\theta) := (f*P_r)(\theta) \end{align*}\]

solves the heat equation on the disk. That is $u\in\mathscr{C}^2(D)$ and satisfies

i) $\Delta u =0$.

ii) If $f$ is continuous at $\theta$, then $\lim_{r\to1^-}u(r,\theta)=f(\theta)$ and if $f$ is continuous everywhere then the convergence is uniform.

iii) If $f$ is continuous, then $u(r,\theta)$ is the unique solution to the steady heat equation on the disc satisfying (i) and (ii).

<Proof>

i) Since $f$ is integrable, there is a $B>0$ such that $\lvert f(\theta)\rvert \leq B$ for all $\theta \in[-\pi, \pi]$. Thus,

\[\begin{align*} \lvert \hat{f}(m) r^{\lvert m\rvert} e^{im\theta}\rvert &= \lvert \hat{f}(\theta) r^{\lvert m\rvert}\rvert \\ &=r^{\lvert m\rvert}\left \lvert \frac{1}{2\pi}\int_{-\pi}^\pi f(\theta)e^{-im\theta}d\theta\right\rvert \\ &\leq r^{\lvert m\rvert} \cdot \frac{1}{2\pi}\int_{-\pi}^\pi \lvert f(\theta) \rvert d\theta \\ &\leq r^{\lvert m\rvert} B. \end{align*}\]

It implies that $u(r,\theta)=\sum_{m=-\infty}^\infty \hat{f}(m)r^{\lvert m\rvert}e^{im\theta}$ converges absolutely and uniformly on any disc of radius $0\leq r <1$. As a consequence of the uniform convergence, using the theorem, $u$ is differentiable term by term. Now, we want to show that $\Delta u=0$. Using the polar form of the Laplacian

\[\begin{align*} \Delta u = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}. \end{align*}\] \[\begin{align*} \Delta u(r,\theta) &= \sum_{m=-\infty}^\infty \lvert m \rvert (\lvert m \rvert -1) \hat{f}(m)r^{\lvert m\rvert-2}e^{im\theta} \\ &+ \lvert m \rvert \hat{f}(m)r^{\lvert m\rvert-2}e^{im\theta} \\ &- m^2\hat{f}(m)r^{\lvert m\rvert-2}e^{im\theta} \\ &=0. \end{align*}\]

So the Poisson integral is indeed harmonic.

ii) Restatement of the previous theorem.

iii) Suppose that $v(r,\theta)$ is another solution. Since $v$ is twice continuously differentiable, we know that $v(r,\cdot)$ has a uniform convergent Fourier series by Theorem

\[\begin{align*} \sum_{n=-\infty}^\infty a_n(r) e^{in\theta}, \quad \text{where } a_n(r) = \frac{1}{2\pi}\int_{-\pi}^\pi v(r,\theta) e^{-in\theta}d\theta. \end{align*}\]

Since

\[\begin{align*} \Delta v(r,\theta) = \sum_{n=-\infty}^\infty \left( a^{\prime\prime}_n(r) + \frac{1}{r}a^\prime_n(r) - \frac{n^2}{r^2}a_n(r)\right) e^{in\theta} =0, \end{align*}\] \[\begin{align*} a^{\prime\prime}_n(r) + \frac{1}{r}a^\prime_n(r) - \frac{n^2}{r^2}a_n(r)=0 \end{align*}\]

for all $n\in\mathbb{Z}$. By the Exercise 11 from chapter 1,

\[\begin{align*} a_n(r) = \begin{cases} A_n r^n + B_n r^{-n} \quad &(\text{if } n\neq 0) \\ A_n + B_n\log r \quad &(\text{if } n =0) \end{cases} \end{align*}\]

for some $A_n, B_n$. Since $v(r, \cdot)$ is continuous on $[-\pi, \pi]$, $v(r,\cdot)$ is bounded. Thus, $a_n(r)$ is bounded uniformly in $r\in (0,1)$ since

\[\begin{align*} \lvert a_n(r) \rvert &= \left \lvert\frac{1}{2\pi}\int_{-\pi}^\pi v(r,\theta) e^{in\theta} d\theta \right \rvert \\ &\leq \frac{1}{2\pi}\int_{-\pi}^\pi\lvert v(r,\theta) \rvert d\theta \\ &\leq M. \end{align*}\]

For $n\geq 1, B_n=0$. Otherwise $\lim_{r\to0}B_nr^{-n}=\infty$. Then

\[\begin{align*} A_n &= \lim_{r\to 1^-} a_n(r) \\ &=\lim_{r\to1^-} \frac{1}{2\pi}\int_{-\pi}^\pi v(r,\theta) e^{-in\theta}d\theta \\ &= \frac{1}{2\pi}\int_{-\pi}^\pi \lim_{r\to1^-} v(r,\theta) e^{-in\theta}d\theta \\ &= \frac{1}{2\pi}\int_{-\pi}^\pi f(\theta)e^{-in\theta}d\theta \\ &=\hat{f}(n). \end{align*}\]

Similarly, $A_n$ should be $0$ for $n\leq -1$. Otherwise $\lim_{r\to0}A_n r^n=\infty$. Then we get $B_n=\hat{f}(n)$. Lastly, $B_n$ should be $0$ for $n=0$ since $\lim_{r\to0}B_n\log r=-\infty$, which leads to $A_n=\hat{f}(n)$. Combining all these, we get

\[a_n(r) = \begin{cases} \hat{f}(n) r^n \quad &(n\geq 1) \\ \hat{f}(n) r^{-n} \quad &(n \leq -1) \\ \hat{f}(n) \quad &(n=0). \end{cases}\]

Thus, we conclude that

\[\begin{align*} v(r,\theta) = \sum_{m=-\infty}^\infty \hat{f}(m)r^{\lvert m\rvert}e^{im\theta} = u(r,\theta). \end{align*}\] \[\tag*{$\square$}\]

Theorem (Cauchy Schwarz Inequality)

Let $V$ be vector space over $\mathbb{C}$.

\[\begin{align*} \lvert \langle \mathbf{x},\mathbf{y}\rangle \rvert \leq \lVert \mathbf{x}\rVert \cdot \lVert \mathbf{y} \rVert \end{align*}\]

for all $\mathbf{x,y}\in V$.

<Proof>

First, suppose that $\lVert \mathbf{y} \rVert=0$. Since $\langle \mathbf{y},\mathbf{y}\rangle \geq 0$, $\lVert \mathbf{y} \rVert=0$ does not imply $\mathbf{y}=\mathbf{0}$. For all $t\in\mathbb{R}$,

\[\begin{align*} 0 &\leq \lVert \mathbf{x}+t\mathbf{y}\rVert^2 \\ &=\langle \mathbf{x}+t\mathbf{y}, \mathbf{x}+t\mathbf{y}\rangle \\ &= \lVert \mathbf{x}\rVert^2 + 2t \mathcal{R}(\langle \mathbf{x},\mathbf{y}\rangle ) + \lVert \mathbf{y}\rVert^2 \\ &=\lVert \mathbf{x}\rVert^2 + 2t \mathcal{R}(\langle \mathbf{x},\mathbf{y}\rangle ) \\ &\leq \lVert \mathbf{x}\rVert^2 + 2t\lvert \langle \mathbf{x},\mathbf{y}\rangle \rvert. \end{align*}\]

If $\lvert \langle \mathbf{x},\mathbf{y} \rangle \rvert \neq 0$, then taking $t\ll0$ gives a contradiction. Thus, $\lvert \langle \mathbf{x},\mathbf{y} \rangle \rvert=0$.

Second, suppose that $\lVert \mathbf{y} \rVert\neq0$. Let

\[\begin{align*} c = \frac{\langle \mathbf{x}, \mathbf{y}\rangle}{\langle \mathbf{y},\mathbf{y}\rangle}. \end{align*}\]

Then $\langle \mathbf{x}-c\mathbf{y}, \mathbf{y}\rangle=0$. So,

\[\begin{align*} \lVert \mathbf{x}\rVert^2 &= \lVert \mathbf{x}-c\mathbf{y}\rVert^2 + \lVert c\mathbf{y}\rVert^2 \\ &\geq \lvert c\rvert^2 \lVert \mathbf{y}\rVert^2 \end{align*}\]

Then,

\[\begin{align*} \lVert \mathbf{x}\rVert^2 &\geq \frac{\lvert \langle \mathbf{x},\mathbf{y}\rangle \rvert^2}{(\lVert \mathbf{y}\rVert^2)^2} \lVert \mathbf{y} \rVert^2 \Rightarrow \lvert \langle \mathbf{x},\mathbf{y}\rangle\rvert \leq \lVert \mathbf{x}\rVert \lVert \mathbf{y} \rVert \end{align*}\] \[\tag*{$\square$}\]

Reference

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