6.5. Laplace operator in the disc. II

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## 6.5. Laplace operator in the disc. II

### Neumann problem

Consider now Neumann problem \begin{align} & u_{rr}+ \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta} =0 && \text{as }\; r < a, \label{eq-6.5.1}\\[3pt] & u_r= h(\theta) && \text{at }\; r=a. \label{eq-6.5.2} \end{align} Plugging in (\ref{eq-6.5.2}) expression (6.4.10) of the previous Section 6.4 $$u=\frac{1}{2} A_0+ \sum_{n=1}^\infty r^n \Bigl( A_n\cos(n\theta)+ C_n\sin(n\theta)\Bigr) \tag{6.4.10}\label{eq-6.4.10}$$ we get $$\sum_{n=1}^\infty n a^{n-1} \Bigl( A_n\cos(n\theta)+ C_n\sin(n\theta)\Bigr)=h(\theta) \label{eq-6.5.3}$$ and feel trouble!

1. We cannot satisfy (\ref{eq-6.5.3}) unless $h(\theta)$ has a "free" coefficient equal to $0$ i.e. $$\int_0^{2\pi} h(\theta)\,d\theta=0. \label{eq-6.5.4}$$
2. Even if (\ref{eq-6.5.4}) holds we cannot find $A_0$ so in the end solution is defined up to a constant.

So to cure (a) assume that (\ref{eq-6.5.4}) is fulfilled and to fix (b) impose condition $$\iint_{\mathcal{D}} u(x,y)\,dxdy=0. \label{eq-6.5.5}$$ (Indeed, it is $\iint u(r,\theta)\,rdrd\theta=\pi a^2 A_0$). Then $A_0=0$ and \begin{align*} A_n = \frac{1}{\pi n}a^{1-n}\int_0^{2\pi} h(\theta')\cos(n\theta')\,d\theta', \\ C_n = \frac{1}{\pi n}a^{1-n}\int_0^{2\pi} h(\theta')\sin (n\theta')\,d\theta'. \end{align*} Plugging into (\ref{eq-6.4.10}) we have $$u(r,\theta)=\int_0^{2\pi}G(r,\theta,\theta')h(\theta')\,d\theta' \label{eq-6.5.6}$$ with \begin{align*} G(r,\theta,\theta'):= &\frac{1}{\pi} \Bigl(\sum_{n=1}^\infty \frac{1}{n} r^n a^{1-n} \bigl(\cos(n\theta)\cos(n\theta')+\sin(n\theta)\sin(n\theta)\bigr) \Bigr)=\\ &\frac{1}{\pi} \Bigl(\sum_{n=1}^\infty \frac{1}{n} r^n a^{1-n} \cos (n(\theta-\theta')) \Bigr)=\\ &\frac{a}{\pi} \Bigl(\Re \sum_{n=1}^\infty \frac{1}{n} \bigl(ra^{-1}e^{i(\theta-\theta')}\bigr)^n \Bigr)=\\ &-\frac{a}{\pi} \Re \log \bigl(1-ra^{-1}e^{i(\theta-\theta')}\bigr) \end{align*} where we used that $\sum_{n=1}^\infty \frac{1}{n}z^n=-\log (1-z)$ (indeed, if we denote it by $f(z)$ then $f'(z)= \sum_{n=1}^\infty z^{n-1}=(1-z)^{-1}$ and $f(0)=0$) and plugged $z=ra^{-1}e^{i(\theta-\theta')}$ with $|z|<1$. The last expression equals \begin{multline*} -\frac{a}{2\pi} \log \bigl[ \bigl(1-ra^{-1}e^{i(\theta-\theta')}\bigr)\bigl(1-ra^{-1}e^{-i(\theta-\theta')}\bigr)\bigr]\\ =-\frac{a}{2\pi} \log \bigl[a^{-2}\bigl(a^2-2ar\cos(\theta-\theta')+r^2\bigr)\bigr]. \end{multline*} with $$G(r,\theta,\theta')= -\frac{a}{2\pi}\log \bigl[a^{-2}\bigl(a^2-2ar\cos(\theta-\theta')+r^2\bigr)\bigr]. \label{eq-6.5.7}$$ Recall that $r< a$.

Considering outside the disc we should use (6.4.11) $$u=\frac{1}{2} A_0+ \sum_{n=1}^\infty r^{-n}\Bigl( B_n\cos(n\theta)+ D_n \sin(n\theta)\Bigr). \tag{6.4.11}\label{eq-6.4.11}$$ Again we need to impose condition (\ref{eq-6.5.4}); condition (\ref{eq-6.5.5}) is now replaced by $$\lim _{r\to \infty}u =0. \label{eq-6.5.8}$$ Then $A_0=0$ and \begin{align*} B_n=-\frac{1}{\pi n}a^{n+1} \int_0^{2\pi} h(\theta')\cos(n\theta')\,d\theta',\\ D_n=-\frac{1}{\pi n} a^{n+1} \int_0^{2\pi} h(\theta')\sin(n\theta')\,d\theta'. \end{align*} Plugging into (\ref{eq-6.4.11}) we get (\ref{eq-6.5.6}) with $$G(r,\theta,\theta')= \frac{a}{2\pi}\log \bigl[r^{-2}\bigl(a^2-2ar\cos(\theta-\theta')+r^2\bigr)\bigr]. \label{eq-6.5.9}$$

### Laplace in the sector

Consider now our equation in the sector $\{ r< a, \ 0<\theta <\alpha\}$ and impose $0$ Dirichlet boundary conditions at radial parts of the boundary and non-zero on the circular part. Consider now Dirichlet problem \begin{align} & u_{rr}+ \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta} =0 && \text{as }\; r < a,\ 0<\theta<\alpha \label{eq-6.5.10}\\[3pt] &u(r,0)=u(r,\alpha)=0 &&0< r< a,\label{eq-6.5.11}\\[3pt] & u= h(\theta) && \text{at }\; r=a,\ 0<\theta<\alpha. \label{eq-6.5.12} \end{align}

Remark 1. We can consider different boundary conditions on these three parts of the boundary, trouble is when Neumann is everywhere.

Then separating variables as in Section 6.4 we get \begin{gather*} \Theta''+\lambda \Theta=0,\\[3pt] \Theta(0)=\Theta(\alpha)=0 \end{gather*} and therefore \begin{equation*} \lambda_n = \bigl(\frac{\pi n}{\alpha}\bigr)^2,\qquad \Theta_n = \sin \bigl(\frac{\pi n\theta}{\alpha}\bigr)\qquad n=1,2,\ldots \end{equation*} and plugging into (6.4.3) $$r^2 R''+rR'-\lambda R=0 \tag{6.4.3}\label{eq-6.5.23-3}$$ we get $$R_n= A_n r^{\frac{\pi n}{\alpha}}+ B_n r^{-\frac{\pi n}{\alpha}} \label{eq-6.5.13}$$ and therefore $$u= \sum_{n=1}^\infty \bigl(A_n r^{\frac{\pi n}{\alpha}}+ B_n r^{-\frac{\pi n}{\alpha}}\bigr) \sin \bigl(\frac{\pi n\theta}{\alpha}\bigr) \label{eq-6.5.14}$$ where for sector $\{ r< a, \ 0<\theta <\alpha\}$ we should set $B_n=0$ (for domain $\{ r>a, \ 0<\theta <\alpha\}$ we should set $A_n=0$ and for domain $\{ a< r< b, \ 0<\theta <\alpha\}$ we don't nix anything). The rest is easy except we don't get nice formula like Poisson formula.

Example 1. In the half-ring ${(r,\theta)\colon 1< r < 2, 0 < \theta <\pi}$ find solution by a Fourier method \begin{align*} &\Delta u=0,\ &u|_{\theta=0}=u|_{\theta=\pi}=0,\ &u|_{r=1}= 0 ,\ &u|_{r=2}=\pi-\theta. \end{align*}

Solution. Separating variables $u(r,\theta)= R(r)\Theta(\theta)$ we get $\Theta''+\lambda \Theta=0$, $\Theta(0) =\Theta(\pi ) =0$.

Solving Sturmâ€“Liouville problem we get \begin{gather*} \lambda_n = n^2, \qquad \Theta_n = \sin (n\theta) \quad n=1,2,\ldots \implies R_n=A_n r^{n}+B_nr^{-n}\\ \end{gather*} and therefore \begin{gather*} u(r,\theta) = \sum_{n=1}^\infty \bigl[A_n r^{n}+B_n r^{-n} \bigr] \sin (n\theta) \end{gather*} Plugging into boundary conditions as $r=1$ and $r=2$ we get correspondingly \begin{gather*} \left\{\begin{aligned} &\sum_{n=1}^\infty \bigl[ A_n+B_n] \sin (n\theta)= 0\implies A_n+B_n=0,\\ &\sum_{n=1}^\infty \bigl[ 2^nA_n+2^{-n}B_n] \sin (n\theta)=\pi-\theta \end{aligned}\right. \end{gather*} and therefore \begin{gather*} 2^nA_n+2^{-n}B_n =\frac{2}{\pi}\int _{0}^\pi (\pi-\theta) \sin(n\theta)\,d\theta
= \frac{2}{n} . \end{gather*} Then \begin{gather*} A_n = -B_n= \frac{2}{(2^n -2^{-n})n } \end{gather*} and therefore \begin{gather*} u(r,\theta) = \sum_{n=1}^\infty \frac{2}{(2^n -2^{-n})n } (r^n-r^{-n}) \sin(n\theta). \end{gather*}

Example 2. In the disk with a cut ${(r,\theta)\colon r < 1,\ -\pi < \theta < \pi}$ find solution by a Fourier method \begin{align*} &\Delta u=0,\\ &u|_{\theta=-\pi}=u|_{\theta=\pi}=1-r,\\ &u|_{r=1}=0. \end{align*}

Solution We cannot apply method directly since conditions at the cut are inhomogeneous. However note that $v=1+x$ satisfies both equation and conditions at the cut, so $u=v+w$ where \begin{align*} &\Delta w=0,\\ &w|_{\theta=-\pi}=w|_{\theta=\pi}=0,\\ &w|_{r=1}=-x-1= -1-\cos(\theta). \end{align*} Then after separation of variables we get $\Theta '' +\lambda \Theta =0$, $\Theta (-\pi)=\Theta(\pi)=0$ and then $\lambda = \frac{n^2}{4}$, $\Theta_n =\sin \bigl(\frac{n}{2} (\theta +\pi)\bigr)$, $n=1,2,\ldots$.

Observe that for $\Theta_n$ are even/odd with respect to $\theta$ as $n$ is odd/even respectively and since $g(\theta)=-1-\cos(\theta)$ is even, we need to consider only odd $n$: \begin{gather*} w(r,\theta)= \sum_{n=0}^\infty A_n r^{(n+\frac{1}{2})} \cos \bigl((n+\frac{1}{2})\theta\bigr) \end{gather*} with \begin{align*} A_n =&-\frac{2}{\pi} \int_0^\pi \bigl(1+\cos(\theta)\bigr) \cos \bigl((n+\frac{1}{2})\theta\bigr)\,d\theta\\ =&-\frac{1}{\pi}\int_0^\pi \Bigl( 2\cos \bigl((n+\frac{1}{2})\theta\bigr)\ + \cos \bigl((n+\frac{3}{2})\theta\bigr)+ \cos \bigl((n-\frac{1}{2})\theta\bigr)\Bigr)\,d\theta\\ =&\frac{(-1)^{n-1}}{\pi}\Bigl[ -\frac{2}{n+\frac{1}{2}} + \frac{1}{n+\frac{3}{2}} +\frac{1}{n-\frac{1}{2}}\Bigr]=\frac{2(-1)^{n-1}}{\pi (n-\frac{1}{2}) (n+\frac{1}{2}) (n+\frac{3}{2})}. \end{align*} Then \begin{gather*} w(r,\theta)= \sum_{n=0}^\infty \frac{2(-1)^{n-1}}{\pi (n-\frac{1}{2}) (n+\frac{1}{2}) (n+\frac{3}{2})} r^{(n+\frac{1}{2})} \cos \bigl((n+\frac{1}{2})\theta\bigr) \end{gather*} and \begin{gather*} u(r,\theta)=1+r\cos(\theta)+ \sum_{n=0}^\infty \frac{2(-1)^{n-1}}{\pi (n-\frac{1}{2}) (n+\frac{1}{2}) (n+\frac{3}{2})} r^{(n+\frac{1}{2})} \cos \bigl((n+\frac{1}{2})\theta\bigr). \end{gather*}