6.5. Laplace operator in the disk. II


## 6.5. 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-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 "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 of the disk 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}\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 $$G(r,\theta,\theta')= \frac{a}{2\pi}\log \bigl[r^{-2}\bigl(a^2-2ar\cos(\theta-\theta')+r^2\bigr)\bigr]. \label{eq-6.5.9}$$

### 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 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. 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 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.