
### Laplace operator in the disk. 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-1}\\[3pt] & u_r= h(\theta) && \text{at }\; r=a. \label{eq-2} \end{align} Plugging in (\ref{eq-2}) expression (23.10) of the previous Lecture23 $$u=\frac{1}{2} A_0+ \sum_{n=1}^\infty r^n\Bigl( A_n\cos(n\theta)+ C_n\sin(n\theta)\Bigr) \tag{23.10}\label{eq-23-10}$$ we get $$\sum_{n=1}^\infty n a^{n-1}\Bigl( A_n\cos(n\theta)+ C_nr\sin(n\theta)\Bigr)=h(\theta) \label{eq-3}$$ and feel trouble!

a. We cannot satisfy (\ref{eq-3}) unless $h(\theta)$ has "free" coefficient equal to $0$ i.e.

$$\int_0^{2\pi} h(\theta)\,d\theta=0. \label{eq-4}$$

b. Even if (\ref{eq-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-4}) is fulfilled and to fix (b)impose condition $$\iint_{\mathcal{D}} u(x,y)\,dxdy=0. \label{eq-5}$$ (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-23-10}) we have $$u(r,\theta)=\int_0^{2\pi}G(r,\theta,\theta')g(\theta')\,d\theta' \label{eq-6}$$ 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 $$G(r,\theta,\theta')= -\frac{a}{2\pi}\log \bigl[a^{-2}\bigl(a^2-2ar\cos(\theta-\theta')+r^2\bigr)\bigr]. \label{eq-7}$$ Recall that $r<a$.

Considering outside of the disk we should use (23.11) $$u=\frac{1}{2} A_0+ \sum_{n=1}^\infty r^{-n}\Bigl( B_n\cos(n\theta)+ D_n \sin(n\theta)\Bigr). \tag{23.11}\label{eq-23-11}$$ Again we need to impose condition (\ref{eq-4}); condition (\ref{eq-5}) is now replaced by $$\lim _{r\to \infty}u =0. \label{eq-8}$$ 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-23-11}) we get (\ref{eq-6}) with $$G(r,\theta,\theta')= \frac{a}{2\pi}\log \bigl[r^{-2}\bigl(a^2-2ar\cos(\theta-\theta')+r^2\bigr)\bigr]. \label{eq-9}$$

#### 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 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-10}\\[3pt] &u(r,0)=u(r,\alpha)=0 &&0<r<a,\label{eq-11}\\[3pt] & u= h(\theta) && \text{at }\; r=a,\ 0<\theta<\alpha.\label{eq-12} \end{align} Remark. We can consider different b.c. on these three parts of the boundary, trouble is when Neumann is everywhere.

Then separating variables as in Lecture 23 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 (23.3) $$r^2 R''+rR'-\lambda R=0\tag{23.3}\label{eq-23-3}$$ we get $$R_n= A_n r^{\frac{\pi n}{\alpha}}+ B_n r^{-\frac{\pi n}{\alpha}} \label{eq-13}$$ and therefore $$u= \sum_{n=1}^\infty \bigl(A_n r^{\frac{\pi n}{\alpha}}+ B_n r^{-\frac{\pi n}{\alpha}}\bigr) \sin \bigl(\frac{\pi n\theta}{\alpha}\bigr) \label{eq-14}$$ 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.