Laplace's equation, Robin Boundary Problem

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courses.maths.ox.ac.uk/pluginfile.php/21926/mod_folder/content/0/lectures22.pdf#page=62
Corollary 105 (Robin (or Mixed) Boundary Problem) Let $R \subseteq \mathbb{R}^{3}$ be a path-connected region with piecewise smooth boundary $\partial R$ and let $f$ be a continuous function defined on $\partial R$. Further let $\beta$ be a real constant. Suppose that $\phi_{1}$ and $\phi_{2}$ are such that
\begin{align*}
\nabla^{2} \phi_{1} &=0=\nabla^{2} \phi_{2} \text { in } R \\
\beta \phi_{1}+\frac{\partial \phi_{1}}{\partial n} &=f=\beta \phi_{2}+\frac{\partial \phi_{2}}{\partial n} \text { on } \partial R .
\end{align*}
(i) If $\beta>0$ then $\phi_{1}=\phi_{2}$ in $R$.
(ii) If $\beta=0$ then $\phi_{1}-\phi_{2}$ is constant in $R$.
(iii) If $\beta<0$ then the solution need not be unique, even up to a constant.
Proof. Let $\psi=\phi_{1}-\phi_{2}$, so that
$$
\nabla^{2} \psi=0 \text { in } R, \quad \beta \psi+\frac{\partial \psi}{\partial n}=\beta \psi+\nabla \psi \cdot \mathbf{n}=0 \text { on } \partial R \text {. }
$$
If we consider the function $\psi^{2}$ then, by the Divergence Theorem and as $\nabla^{2}=\nabla \cdot \nabla$,
$$
\iint_{\partial R} \nabla\left(\psi^{2}\right) \cdot \mathrm{d} \mathbf{S}=\iiint_{R} \nabla^{2}\left(\psi^{2}\right) \mathrm{d} V
$$
Looking at the LHS and noting $\beta \psi+\nabla \psi \cdot \mathbf{n}=0$ on $\partial R$ we have
$$
\iint_{\partial R} \nabla\left(\psi^{2}\right) \cdot \mathrm{d} \mathbf{S}=\iint_{\partial R} 2 \psi(\nabla \psi \cdot \mathbf{n}) \mathrm{d} S=\iint_{\partial R} 2 \psi(-\beta \psi) \mathrm{d} S=-2 \beta \iint_{\partial R} \psi^{2} \mathrm{~d} S
$$
On the RHS we have as before
$$
\nabla^{2}\left(\psi^{2}\right)=2 \psi \nabla^{2} \psi+2 \nabla \psi \cdot \nabla \psi=2|\nabla \psi|^{2} \quad \text { as } \nabla^{2} \psi=0 \text { in } R
$$
Hence
$$
-2 \beta \iint_{\partial R} \psi^{2} \mathrm{~d} S=\iint_{R} 2|\nabla \psi|^{2} \mathrm{~d} V .
$$
If $\beta>0$ then the LHS is non-positive and the RHS is non-negative — consequently both are zero and $\nabla \psi=\mathbf{0}$ in $R$ and $\psi=0$ on $\partial R$. Hence $\psi=0$ throughout $R$.

If $\beta=0$ then the Robin Boundary Problem is just the Neumann Problem, with which we have already dealt.
The following example shows that solutions need not be unique if $\beta<0$.
Example 106 Let $R$ be the interior of the cylinder $x^{2}+y^{2}=1$. Find two solutions to the Robin Boundary Problem
$$
\nabla^{2} \psi=0 \text { in } R, \quad \psi=\frac{\partial \psi}{\partial n} \text { on } \partial R .
$$
Solution. Laplace's equation in cylindrical polars reads
$$
\nabla^{2} \psi=\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial \psi}{\partial r}\right)+\frac{1}{r^{2}} \frac{\partial^{2} \psi}{\partial \theta^{2}}+\frac{\partial^{2} \psi}{\partial z^{2}}=0 .
$$
We know for $\psi=r^{n} \sin n \theta$, where $n$ is a non-negative integer, that $\nabla^{2} \psi=0$ and on the boundary $r=1$ we have
$$
\frac{\partial \psi}{\partial n}=\frac{\partial \psi}{\partial r}=n r^{n-1} \sin n \theta=n \sin n \theta ; \quad \psi=\sin n \theta
$$
Hence when $n=1$, we see $r \sin \theta$ satisfies the given boundary problem. Other solutions are $0,r \cos \theta$ and $r \sin (\theta-\alpha)$ for arbitrary $\alpha$ so we see this solution is far from unique.

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