6.4. Laplace operator in the disk: separation of variables

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6.4. Laplace operator in the disk: separation of variables

  1. Separation of variables
  2. Poisson formula

Separation of variables

So, consider problem \begin{align*} & \Delta u =0&& \text{as }\; x^2+y^2< a^2,\\[3pt] & u= g && \text{at }\; x^2+y^2=a^2. \end{align*} In the polar coordinates it becomes \begin{align} & u_{rr}+ \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta} =0&& \text{as }\; r < a,\label{eq-6.4.1}\\[3pt] & u= g(\theta) && \text{at }\; r=a. \label{eq-6.4.2} \end{align} First let us forget for a while about (\ref{eq-6.4.2}) and consider a simple solution $u=R(r)\Theta(\theta)$ to equation (\ref{eq-6.4.1}). Then \begin{equation*} R'' \Theta +\frac{1}{r}R'\Theta + \frac{1}{r^2}R \Theta ''=0. \end{equation*} To separate $r$ and $\theta$ we must divide by $R\Theta$ and multiply by $r^2$: \begin{equation*} \underbracket{\frac{r^2 R'' +rR'}{R}}+ \underbracket{\frac{\Theta''}{\Theta}}=0. \end{equation*} Then repeating our usual separation of variables magic spell both expressions are constant, say $\lambda$ and $-\lambda$: \begin{gather} r^2 R''+rR'-\lambda R=0,\label{eq-6.4.3}\\[3pt] \Theta''+\lambda \Theta=0. \label{eq-6.4.4} \end{gather} Now we need boundary conditions to $\Theta$ and those are periodic: \begin{equation} \Theta (2\pi)=\Theta(0),\qquad \Theta' (2\pi)=\Theta'(0). \label{eq-6.4.5} \end{equation} We already know solution to (\ref{eq-6.4.4})--(\ref{eq-6.4.5}): \begin{align} &\lambda_0=0,&& \lambda_n=n^2, &&n=1,2,\ldots.\\ &\Theta_0 =\frac{1}{2} && \Theta_{n,1}=\cos(n\theta),&& \Theta_{n,2}=\sin(n\theta). \label{eq-6.4.7} \end{align} Equation (\ref{eq-6.4.3}) is an Euler equation, we are looking for solutions $R=r^m$. Then \begin{equation} m(m-1)+m -\lambda=0\implies m^2=\lambda \label{eq-6.4.8} \end{equation} Plugging $\lambda_n=n^2$ into (\ref{eq-6.4.3}) we get $m=\pm n$ and therefore $R= A r^n+Br^{-n}$ as $n\ne 0$ and $R=A+B\log r$ as $n=0$. Therefore \begin{multline} u=\frac{1}{2}\bigl(A_0+B_0\log r\bigr)+\\ \sum_{n=1}^\infty \Bigl(\bigl(A_nr^n+B_nr^{-n}\bigr)\cos(n\theta)+ \bigl(C_nr^n+D_nr^{-n}\bigr)\sin(n\theta)\Bigr).\quad \label{eq-6.4.9} \end{multline} Here we assembled simple solutions together.

As we are looking for solutions in the disk $\{r< a\}$ we should discard terms singular as $r=0$; namely we should set $B_0=0$, $B_n=D_n=0$ for $n=1,2,\ldots$ and therefore \begin{equation} u=\frac{1}{2} A_0+ \sum_{n=1}^\infty r^n\Bigl( A_n\cos(n\theta)+ C_n \sin(n\theta)\Bigr).\quad \label{eq-6.4.10} \end{equation} If we consider equation outside of the disk (so as $r>a$) we need to impose condition $\max |u|<\infty$ and discard terms singular as $r=\infty$; namely we should sety $B_0=0$, $A_n=C_n=0$ for $n=1,2,\ldots$ and therefore \begin{equation} u=\frac{1}{2} A_0+ \sum_{n=1}^\infty r^{-n}\Bigl( B_n\cos(n\theta)+ D_n\sin(n\theta)\Bigr).\quad \label{eq-6.4.11} \end{equation} Finally, if we consider equation outside in the annulus (aka ring) $\{a< r< b\}$ we need b.c. on both circles $\{r=a\}$ and $\{r=b\}$ and we discard no terms.

Poisson formula

But we are dealing with the disk. Plugging (\ref{eq-6.4.10}) into (\ref{eq-6.4.2}) we get \begin{equation} g(\theta)=\frac{1}{2} A_0+ \sum_{n=1}^\infty a^n \Bigl( A_n\cos(n\theta)+ C_n\sin(n\theta)\Bigr)\quad \label{eq-6.4.12} \end{equation} and therefore \begin{align*} A_n=\frac{1}{\pi} a^{-n}\int_0^{2\pi} g(\theta')\cos(n\theta')\,d\theta', \\\ C_n=\frac{1}{\pi} a^{-n}\int_0^{2\pi} g(\theta')\sin(n\theta')\,d\theta'. \end{align*} Plugging into (\ref{eq-6.4.10}) we get \begin{equation} u(r,\theta)=\int_0^{2\pi}G(r,\theta,\theta')g(\theta')\,d\theta' \label{eq-6.4.13} \end{equation} with \begin{gather*} G(r,\theta,\theta'):= \frac{1}{2\pi} \Bigl(1+2\sum_{n=1}^\infty r^n a^{-n} \bigl(\cos(n\theta)\cos(n\theta')+\sin(n\theta)\sin(n\theta) \bigr) \Bigr)=\\ \frac{1}{2\pi} \Bigl(1+2\sum_{n=1}^\infty r^n a^{-n}\cos (n(\theta-\theta')) \Bigr)=\\ \frac{1}{2\pi} \Bigl(1+2\Re \sum_{n=1}^\infty \bigl(ra^{-1}e^{i(\theta-\theta')}\bigr)^n \Bigr)=\\ \frac{1}{2\pi} \Bigl(1+2\Re \frac{ra^{-1}e^{i(\theta-\theta')}}{1-ra^{-1}e^{i(\theta-\theta')}} \Bigr) \end{gather*} where we summed geometric progression with the factor $z=ra^{-1}e^{i(\theta-\theta')}$ (then $|z|=ra^{-1}<1$).

Multiplying numerator and denominator by $a^2-ra e^{-i(\theta-\theta')}$ we get $ra e^{-i(\theta-\theta')}-r^2$ and $a^2-ra [e^{-i(\theta-\theta')}+e^{i(\theta-\theta')}]+r^2a^{-2}= a^2-2ra\cos (\theta-\theta') +r^2$ and therefore we get \begin{equation*} G(r,\theta,\theta')= \frac{1}{2\pi} \Bigl(1+2 \frac{ra \cos(\theta-\theta')}{a^2-2ra\cos (\theta-\theta') +r^2} \Bigr) \end{equation*} and finally \begin{equation} G(r,\theta,\theta')= \frac{1}{2\pi}\frac{a^2-r^2}{a^2-2ra\cos (\theta-\theta') +r^2}. \label{eq-6.4.14} \end{equation} Recall that $r< a$.

Formula (\ref{eq-6.4.13})--(\ref{eq-6.4.14}) is Poisson formula.

Exercise 1. Prove that in the center of the disk (as $r=0$) \begin{equation} u(0)=\frac{1}{2\pi}\int_0^{2\pi} g(\theta')\,d\theta' \label{eq-6.4.15} \end{equation} and the right-hand expression is a mean value of $u$ over circumference $\{r=a\}$.

Exercise 2.

  1. Using (\ref{eq-6.4.11}) instead of (\ref{eq-6.4.10}) prove that for the problem \begin{align} & u_{rr}+ \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta} =0&& \text{as }\; a< r,\label{eq-6.4.16}\\[3pt] & u= g(\theta) && \text{at }\; r=a, \label{eq-6.4.17}\\[3pt] & \max |u|<\infty\label{eq-6.4.18} \end{align} solution is given (for $r>a$) by (\ref{eq-6.4.13}) but with \begin{equation} G(r,\theta,\theta')= \frac{1}{2\pi}\frac{r^2-a^2}{a^2-2ra\cos (\theta-\theta') +r^2}. \label{eq-6.4.19} \end{equation}
  2. As a corollary, prove that \begin{equation} u(\infty):=\lim _{r\to\infty}u= \frac{1}{2\pi}\int_0^{2\pi} g(\theta')\,d\theta). \label{eq-6.4.20} \end{equation}

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