Originally Posted by

**Opalg** That boundary condition doesn't make sense to me. The term on the right should be a constant, not a function. Is bC supposed to be bC(x,0)?

If so, then the solution for the y problem will be of the form $\displaystyle Y = A\cos(\lambda y)+B\sin(\lambda y)$, and $\displaystyle Y' = B\lambda\cos(\lambda y) - A\lambda\sin(\lambda y)$. The boundary condition Y'(1) = 0 tells you that $\displaystyle B\cos\lambda = A\sin\lambda$. So take $\displaystyle A = \cos\lambda,\ B = \sin\lambda$. Then $\displaystyle Y' = \lambda(\sin\lambda\cos(\lambda y) - \cos\lambda\sin(\lambda y))$.

Now the boundary condition Y'(0) = bY(0) (if that is what it should be), tells you that $\displaystyle \lambda\sin\lambda = b\cos\lambda$. So the eigenvalue equation for $\displaystyle \lambda$ is $\displaystyle \tan\lambda = b/\lambda$. That has infinitely many solutions, but unfortunately you can't solve them explicitly.