I've been asked the following question;
If $u$ is a solution of $\nabla^2u = p(x)u$, for $x \in D$, and $\nabla u \cdot n = g(x)$, for $x \in \partial D$, show that $u$ is unique.
So, to begin, I considered two possible solutions, $u_{(1)}$ and $u_{(2)}$. Then, I have the following;
$$\nabla^2 u_{(1)} = p(x) u_{(1)}$$ $$\nabla^2 u_{(2)} = p(x) u_{(2)}$$ $$\nabla u_{(1)} \cdot n = g(x)$$ $$\nabla u_{(2)} \cdot n = g(x)$$
Then, I did the following;
$$u_{(1)} \nabla^2 u_{(2)} = u_{(1)} p(x) u_{(2)} = p(x) u_{(1)}u_{(2)}$$ $$u_{(2)} \nabla^2 u_{(1)} = u_{(2)} p(x) u_{(1)} = p(x) u_{(1)}u_{(2)}$$
Then, I took the difference of these two results, and integrated both sides over the given domain, $D$;
$$\int_D (u_{(1)} \nabla^2 u_{(2)} - u_{(2)} \nabla^2 u_{(1)}) dx = \int_D ( p(x) u_{(1)}u_{(2)} - p(x) u_{(1)}u_{(2)}) dx$$ $$\implies \int_{\partial D}(u_{(1)}\nabla u_{(2)} - u_{(2)} \nabla u_{(1)})\cdot n dS = 0$$ $$\implies \int_{\partial D} (u_{(1)}g(x) - u_{(2)}g(x))dS = 0$$ $$\implies \int_{\partial D} g(x) (u_{(1)} - u_{(2)}) dS = 0$$
Now, this was just a whole bunch of mucking around with Green's Identities and integrands, but I'm a bit stuck at this point. Is it appropriate for me to divide both sides by the $g(x)$ term?? And even then, I could say that the surface integral of $u_{(1)}$ is equal to the surface integral of $u_{(2)}$, but does that really help me??
I'm just a bit unsure as to which direction I should be taking this proof. Any input would be great. :)
By setting up $v$ as outlined in the comments, we have the following system; $\nabla^2 v = p(x)v$, $\nabla v \cdot n = 0$.
From this, it can be shown quite easily that $v = 0$ for all $x$ in and on the domain, $D$.
Then, it's fairly easy to see that $v = 0 \implies u_{(1)}-u_{(2)}=0 \implies u_{(1)} = u_{(2)}$, as required!!