<s>[INST] Analytical solution of Laplace's equation with robin/third boundary condition I'm new to PDE solution, and feel difficult to solve the Laplace's equation with robin boundary conditions.
The equation $\Delta u=0$ with boundary conditions shown in the picture:
picture of problem.
Does this problem has an analytical solution? if so, what is it?
 [/INST] Using separation of variables $u = X(x)Y(y)$, we obtain the ODEs
\begin{align}
X'' &= -\lambda X \\
Y'' &= \lambda Y
\end{align}
where we used $-\lambda$ for our separation constant, with associated boundary conditions
\begin{align}
u(L,y) &= 0 \implies X(L) = 0 \\
u_{x}(0,y) &= 0 \implies X'(0) = 0 \\
u_{y}(x,0) - hu(x,0) &= 0 \implies Y'(0) - hY(0) = 0
\end{align}
Solving the ODE in $X$, we find non-trivial solutions only if $\lambda > 0$ which yields
$$X = A \cos \sqrt{\lambda} x + B \sin \sqrt{\lambda} x$$
Now
\begin{align}
X'(0) &= \sqrt{\lambda} B \\
&= 0 \\
\implies B &= 0 \quad \text{(why?)} \\
\therefore X(L) &= A \cos \sqrt{\lambda} L \\
&= 0 \\
\implies \sqrt{\lambda} L &= \frac{(2n + 1) \pi}{2}, \quad n \ge 0 \quad \text{($A \ne 0$ for non-trivial solutions)} \\
\implies \lambda &= \frac{(2n + 1)^{2} \pi^{2}}{4L^{2}}, \quad n \ge 0 \quad(*)
\end{align}
and hence
$$X_{n} = A \cos \left( \frac{(2n + 1) \pi x}{2L} \right)$$
Using the eigenvalue $(*)$ and solving the ODE in $Y$ yields
$$Y = \frac{1}{h} \cdot \frac{(2n + 1) \pi }{2L} \cosh \left(\frac{(2n + 1) \pi y}{2L} \right) + \sinh \left(\frac{(2n + 1) \pi y}{2L} \right)$$
and so the general solution is given by 
$$u(x,y) = \sum_{n \ge 0} A_{n} \cos \left( \frac{(2n + 1) \pi x}{2L} \right) \left[ \frac{1}{h} \cdot \frac{(2n + 1) \pi }{2L} \cosh \left(\frac{(2n + 1) \pi y}{2L} \right) + \sinh \left(\frac{(2n + 1) \pi y}{2L} \right)\right]$$
Applying the inhomogeneous condition, we find
\begin{align}
u(x,L) &= u_{0} \\
&= \sum_{n \ge 0} A_{n} \cos \left( \frac{(2n + 1) \pi x}{2L} \right) \left[ \frac{1}{h} \cdot \frac{(2n + 1) \pi }{2L} \cosh \left(\frac{(2n + 1) \pi}{2} \right) + \sinh \left(\frac{(2n + 1) \pi}{2} \right)\right] \\
&= \sum_{n \ge 0} C_{n} \cos \left( \frac{(2n + 1) \pi x}{2L} \right)
\end{align}
where 
$$C_{n} = A_{n} \left[ \frac{1}{h} \cdot \frac{(2n + 1) \pi }{2L} \cosh \left(\frac{(2n + 1) \pi}{2} \right) + \sinh \left(\frac{(2n + 1) \pi}{2} \right)\right]$$
You can now solve for the coefficients $C_{n}$ using orthogonality relations. Note that when doing the integrals, a change of variable might help.
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