How to deal with this boundary condition when using separation of variables? Consider Laplace's equation $\nabla^2u = 0$ on the region $A = [0,a]\times [0,b]\times [0,c]\subset \mathbb{R}^3$ and suppose we impose the boundary conditions:
$$u(0,y,z) = \sin \frac{\pi y}{b} \sin \frac{\pi z}{c}, \quad u(a,y,z)=0 \\ u(x,0,z)=u(x,b,z)=u(x,y,0)=u(x,y,c)=0.$$
In that case if we use separation of variables we set $u(x,y,z)=X(x)Y(y)Z(z)$ and the equation reduces to
$$\frac{X''}{X}+\frac{Y''}{Y}+\frac{Z''}{Z}=0.$$
In that case reorganizing gives
$$\frac{X''}{X}+\frac{Y''}{Y}=-\frac{Z''}{Z}=\lambda$$
for some constant $\lambda$. Thus we have two new equations, one of which is
$$\frac{X''}{X} = \lambda-\frac{Y''}{Y} = \mu,$$
for some constant $\mu$. Finally we end up having three equations
$$\begin{cases}Z''+ \lambda Z &= 0, \\ Y'' + (\mu+\lambda)Y&=0, \\ X'' - \mu X &=0\end{cases}.$$
The boundary conditions which are zero on the boundaries transfer nicely to the functions $X,Y,Z$. That is, except the first condition, all the other becomes
$$X(a)=0, Y(0)=Y(b)=0, Z(0)=Z(c)=0.$$
With this the equations for $Y$ and $Z$ with those boundary conditions becomes Sturm Liouville problems. On the other hand the first condition does not separate nicely. We have
$$X(0)Y(y)Z(z) = \sin \frac{\pi y}{b} \sin \frac{\pi z}{c}.$$
This boundary condition doesn't seem to transfer nicely to $X$. So how do we deal with this boundary condition? How do we use it to solve the problem in question?
 A: $$\frac{X''}{X}+\frac{Y''}{Y}+\frac{Z''}{Z}=0.$$
$$\frac{X''}{X}+\frac{Y''}{Y}=-\frac{Z''}{Z}=\lambda$$
$$\lambda-\frac{X''}{X} = \frac{Y''}{Y} = \mu,$$
so you solve 3 auxiliary problem 2 of them are  Sturm Liouville:
$$(1)\;\begin{cases}Z''+ \lambda Z &= 0 \\ Z(0)=0=Z(c)\end{cases}
\qquad (2)\;\begin{cases} Y'' + \mu Y&=0, \\ Y(0)=0=Y(b)\end{cases}$$
The solutions are (verify!) $Z_n(z)=\sin\lambda_n z$, $Y_m(z)=\sin \mu_m z$.
The last one is a "boundary value problem" with missing condition, you need two but here you still have only one. Note that at this stage $\mu$ and $\lambda$ are  known.
$$\begin{cases}X'' - (\mu_m+\lambda_n) X &=0\\X(a)=0\end{cases}$$
The solution  is (verify!)
$$X_{mn}(x)=A_{mn}  e^{a b_{mn}} \sinh (b_{mn} (a-t)),$$ where $b_{mn}=\mu_m+\lambda_n$. 
Note that you  still have a constant ($A_{mn}$) to determine, for which you will use the last condition in the following way. The overall solution is now given by $$u(x,y,z)=\sum_{mn} A_{mn}  X_{mn}(x)Y_m(y) Z_n(z)$$
now for $x=0$ you get
$$u(0,y,z)=\sum_{mn} A_{mn}  X_{mn}(0)Y_m(y) Z_n(z) = \sin \frac{\pi y}{b} \sin \frac{\pi z}{c}$$
which you solve for $A_{mn}$  in general using orthogonality etc - you know, the formula with the integral. However, for this example matching of the coefficients should work, because you have the RHS given in the form of the series on the left side.
