Let $\Omega \subset \mathbb{R}^3$ be a bounded domain. Using an energy argument, show that the IBVP
\begin{align}
u_t &= \Delta u ~~~~~~~~~~x \in \Omega, ~t>0\\
\frac{\partial u}{\partial \nu} + \alpha u &= h(x) ~~~~~~~~x \in \partial\Omega, ~t>0\\
u(x,0)&=g(x) ~~~~~~~~~x \in \Omega
\end{align}
where $\nu$ is the exterior unit normal and $\alpha$ is a constant has at most one solution. Treat the cases $\alpha \geq 0$ and $\alpha<0$ separately. Use logarithmic convexity for the second case.
My attempted solution: By contradiction, suppose that there are two solutions $u_1$ and $u_2$ and define $v = u_1 – u_2$. Then $v$ satisfies
\begin{align}
v_t &= \Delta v ~~~~~~~~~~x \in \Omega, ~t>0\\
\frac{\partial v}{\partial \nu} + \alpha v &= 0 ~~~~~~~~~~~~~x \in \partial\Omega, ~t>0\\
v(x,0)&=0 ~~~~~~~~~~~~~x \in \Omega
\end{align}
Define the energy functional to be
$$E(t)= \frac{1}{2}\int_\Omega v^2 \,dx.$$
The case $\alpha \geq 0$ is trivial. I just showed that
$$\frac{dE}{dt}=\int_\Omega v \, v_t \,dx \leq 0$$
using Green's first identity and the conditions on $v$. Then since $E(0)=0$ we must have $E(t)=0$, and hence $v=0$.
For the $\alpha<0$ case I want to show that
$$E\frac{d^2E}{d^2t} – \left( \frac{dE}{dt} \right)^2 \geq 0\,.$$
Since $E \geq 0$ then by logarithmic convexity we would have $E=0$.
However, I'm running into some problems. I take
\begin{align}
\frac{d^2E}{dt^2}=\int_\Omega v_t^2 \,dx + \int_\Omega v \, v_{tt} \,dx\,.
\end{align}
Then, for the second term I write
\begin{align}
\int_\Omega v \, v_{tt} \,dx &=\int_\Omega v \, \Delta v_t \, dx\\
&= \int_\Omega v_t \, \Delta v \, dx \\
&= \int_\Omega (v_t)^2 \, dx
\end{align}
where I used Green's second identity and the boundary term vanished due to the boundary condition on $v$. Explicitly:
\begin{align}
\int_{\partial \Omega} v\frac{\partial v_t}{\partial \nu} – v_t \frac{\partial v}{\partial \nu} dS= \alpha \int_{\partial \Omega} -v v_t + v_t v \, dS = 0
\end{align}
by the homogeneous Robin condition.
So I get
$$E\frac{d^2E}{d^2t} – \left( \frac{dE}{dt} \right)^2 = \frac{1}{2}\int_\Omega v^2 dx \cdot 2 \int_\Omega v_t^2 dx – \left( \int_\Omega v v_t \, dx \right)^2 \geq 0$$
by the Cauchy-Schwarz inequality.
So I did the proof without even using the assumption $\alpha < 0$, which seems very strange. Did I make a mistake somewhere?
EDIT: Looks like my proof is actually correct. I guess the wording of the question just had me thinking there was an issue.
Best Answer
My original proof is actually correct. I was just confused by the wording of the problem and thought I had made a mistake. Thanks to Hans and Pavel for checking it.