$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$ ##[6.5. Laplace operator in the disk. II](id:sect-6.5.1) ----------------------------------- > 1. [Neumann problem](#sect-6.5.1) > 2. [Laplace operator in the sector](#sect-6.5.2) ###[Neumann problem](id:sect-6.5.1) 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](./S6.4.html#mjx-eqn-eq-6.4.10)) of the previous [Section 6.4](./S6.4.html) \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) \tag{6.4.10}\label{eq-6.4.10} \end{equation} we get \begin{equation} \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} \end{equation} and feel trouble! a. We cannot satisfy (\ref{eq-6.5.3}) unless $h(\theta)$ has "free" coefficient equal to $0$ i.e. \begin{equation} \int\_0^{2\pi} h(\theta)\,d\theta=0. \label{eq-6.5.4} \end{equation} b. 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 \begin{equation} \iint\_{\mathcal{D}} u(x,y)\,dxdy=0. \label{eq-6.5.5} \end{equation} (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 \begin{equation} u(r,\theta)=\int\_0^{2\pi}G(r,\theta,\theta')g(\theta')\,d\theta' \label{eq-6.5.6} \end{equation} with \begin{gather\*} 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{gather\*} 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[a^{-2} \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 \begin{equation} 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} \end{equation} Recall that $r\< a$. Considering outside of the disk we should use ([6.4.11](./S6.4.html#mjx-eqn-eq-6.4.11)) \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). \tag{6.4.11}\label{eq-6.4.11} \end{equation} Again we need to impose condition (\ref{eq-6.5.4}); condition (\ref{eq-6.5.5}) is now replaced by \begin{equation} \lim \_{r\to \infty}u =0. \label{eq-6.5.8} \end{equation} Then $A\_0=0$ and \begin{align\*} B\_n=-\frac{1}{\pi}\frac{1}{n}a^{n+1} \int\_0^{2\pi} h(\theta')\cos(n\theta')\,d\theta',\\\\ D\_n=-\frac{1}{\pi}\frac{1}{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 \begin{equation} 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} \end{equation} ###[Laplace in the sector](id:sect-6.5.2) 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 Neumann 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.](id:remark-6.5.1)** We can consider different b.c. on these three parts of the boundary, trouble is when Neumann is everywhere. Then separating variables as in [Section 6.4](./S6.4.html) 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](./S6.4.3.html#mjx-eqn-eq-6.4.3)) \begin{equation} r^2 R''+rR'-\lambda R=0 \tag{6.4.3}\label{eq-6.5.23-3} \end{equation} we get \begin{equation} R\_n= A\_n r^{\frac{\pi n}{\alpha}}+ B\_n r^{-\frac{\pi n}{\alpha}} \label{eq-6.5.13} \end{equation} and therefore \begin{equation} 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} \end{equation} where for sector $\\{ r\< a, \ 0\<\theta \<\alpha\\}$ we should set $\_n=0$ (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. -------------- [$\Leftarrow$](./S6.4.html)  [$\Uparrow$](../contents.html)  [$\downarrow$](./S6.P.html)  [$\Rightarrow$](../Chapter7/S7.1.html)