Symmetric Boundary Conditions/Eigenvalues (PDEs) Consider the following eigenvalue problem for the Laplacian
$-\Delta u = \lambda u$ in $U$
$u + a \left(\frac{\partial u}{\partial v}\right)$ on $\partial U$
where $v$ is the outward unit normal to the boundary. Suppose that $a(x) \geq  0$ for any $x \in \partial U$. Show that the boundary condition given in this system is symmetric.
 A: Recall that a symmetric boundary condition is one that satisfies 
$$<u, \Delta v > = <\Delta u, v >$$
where $< \cdot, \cdot >$ is the $L_2$ inner product. Observe, by applying Green's identity: 
$$<u, \Delta v> - <\Delta u, v> = \int_{U} (v \Delta u - u \Delta v) d\sigma = \int_{\partial U} (v \frac{\partial u}{\partial \nu} - u \frac{\partial v}{\partial \nu} ) d\sigma $$
Now we break up the boundary into cases, 
$$\int_{\partial U} (v \frac{\partial u}{\partial \nu} - u \frac{\partial v}{\partial \nu} ) d\sigma = \int_{\partial U \text{and } a(x) \not = 0} (v \frac{\partial u}{\partial \nu} - u \frac{\partial v}{\partial \nu} ) d\sigma + \int_{\partial U \text{and } a(x)= 0} (v \frac{\partial u}{\partial \nu} - u \frac{\partial v}{\partial \nu} ) d\sigma$$
By the round-robin condition, if $$a(x) = 0 \implies u = 0 = v \implies \int_{\partial U \text{and } a(x)= 0} (v \frac{\partial u}{\partial \nu} - u \frac{\partial v}{\partial \nu} ) d\sigma = 0$$
So then consider: 
$$\int_{\partial U} (v \frac{\partial u}{\partial \nu} - u \frac{\partial v}{\partial \nu} ) d\sigma = \int_{\partial U \text{and } a(x) \not = 0} (v \frac{\partial u}{\partial \nu} - u \frac{\partial v}{\partial \nu} ) d\sigma = \int_{\partial U} \frac{-vu}{a} + \frac{uv}{a} d\sigma = 0$$
Therefore, 
$$<u, \Delta v> - <\Delta u, v> = 0 \implies <u, \Delta v> = <\Delta u, v>$$
Which proves the boundary condition is symmetric!
