Given PDE $\Delta u= -1$ for $|x|<1,|y|<1$. With boundary conditions $u=0$ for $|x|=1$ and $\frac{\partial u}{\partial x}-\frac{\partial u}{\partial y}=0$ for $|y|=1$. Show that there is at most one solution in $|x|<1,|y|<1$.
I'm not used to seeing boundary conditions of this type. I want to do something like assume $w$ is the difference of two solutions and then $0=\int w \Delta w = \cdots$, but I'm not getting anything with this BC.
You are on the right track. Let $u_1, u_2$ be two solutions, and let $w = u_1 - u_2$. You want to show that $w$ is constant, in which case it must be $w \equiv 0$ since $u_1 = u_2 = 0$ whenever $|x| = 1$. Let $\Omega$ be the square $\{ (x,y): |x| < 1, |y| < 1 \}$. Then $$ \int_{\Omega} |Dw|^2 \, dx = - \int_{\Omega} w \Delta w \, dx + \int_{\partial \Omega} \frac{\partial w}{\partial \nu} w \, dS$$ where $dS$ is the surface measure and $\nu$ is vector normal to the surface. Because $\Delta w = 0$, the first term vanishes, and now we're left with the integral over $\partial \Omega$. Because $\partial \Omega$ is just the four sides of a square, you'll be able to explicitly calculate that each of the four integrals is zero by using the known boundary conditions. As a final hint, it might be useful to remember that $f \frac{\partial f}{\partial x} = \frac{1}{2} \frac{\partial}{\partial x}[f^2]$.
