The background of this problem is as follows:
An infinite straight metal rod of constant thermal conductivity $k$ has cross-section that is a path-connected region S bounded by a simple closed curve $C$. The temperature $T(x, y)$ in each cross-section satisfies Poisson’s equation
$−k(\frac{∂^2T}{∂^2x}+\frac{∂^2T}{∂^2y})$=$Q(x,y)$
with Newton’s law of cooling giving the boundary condition $−k\frac{∂T}{∂n} = h (T − T_a)$ for $(x, y) \in C$, where Q is the given volumetric heat source, $h$ is the constant heat transfer coefficient, $T_a$ is the constant ambienttemperature and $\frac{∂T}{∂n}$ denotes the outward normal derivative of $T$ on $C$.
My issue is that I am asked to prove that there is 'at most' one solution when $h>0$. I understand the proof for uniqueness up to a constant of this problem, however, I don't understand how this idea can be extended to show that there is at most one solution when $h>0$. Thanks in advance for your help.
The key point is that this is not the Neumann problem, but rather a Robin-type problem. This is due to the fact that both $\partial T / \partial n$ and $T$ appear in the boundary condition. If it were Neumann, it would only involve $\partial T / \partial N$, and if it were Dirichlet it would only involve $T$. Robin is a linear combination of the Dirichlet and Neumann.
If you had two solutions $T_1, T_2$, then you could subtract one from the other to see that $T = T_1 - T_2$ solves $$ -k\left( \frac{\partial^2 T}{\partial x^2} +\frac{\partial^2 T}{\partial y^2} \right) =0 \text{ in } C $$ with the boundary condition $$ -k \frac{\partial T}{\partial n} = h T. $$ Here's a hint on how to finish: using this, you can mimic the proof you know for the Neumann case.