I'm trying to solve the following steady-state problem:
$$\nabla^2 u = 0 , \ \ \ (x,y,z) \in [0,\ 2]^3$$
Insulated in the faces with $x=0,y=2$. Kept at $0º$C on the faces with $x=2,z=0$. And kept at $T_0$ when $y=0$. It also satisfies Newtons Law of Cooling on the other face ($z=2$) that is
$$-ku_z(x,y,z) = hu(x,y,z) \ \ \ \ h>0, \ \ k>0$$
We now perform separation of variables: $u(x,y,z) = X(x)Y(y)Z(z)$
So our PDE becomes $$YZX^" + XZY^" + XYZ^" = 0$$ or
$$ \frac{X^"}{X} = -\frac{Y^"}{Y} - \frac{Z^"}{Z} = - \mu $$
Then $$ \frac{Y^"}{Y} + \frac{Z^"}{Z} = \mu \iff \frac{Y^"}{Y} = \mu - \frac{Z^"}{Z} = - \lambda $$
From here, the following problem arises: $$Y"(y) + \lambda Y = 0; \ \ Y(0) = T_0 ; \ \ Y´(2) = 0$$
I've tried using the Rayleigh quotient to prove that the eigenvalues are positive, but it relies on the sign of $Y'(0)$ which we don't know.
By performing the usual discussing on the values of $\lambda$ ($0, \ <0, \ >0)$ I get that $\lambda$ is an eigenvalue for all those cases, which doesn't really make sense since this eigenvalue problem comes up after separation of variables in solving Laplace's equation in a cube. So there should be only one case (or maybe two, with the case $\lambda=0$).
The general solution of $Y'' + \lambda Y = 0$ for $\lambda < 0$ is $c_1 \exp(\sqrt{-\lambda} y) + c_2 \exp(-\sqrt{-\lambda} y)$. The condition $Y'(2) = 0$ corresponds to $c_2 = c_1 \exp(4 \sqrt{-\lambda})$, and then $Y(0) = c_1 (1 + \exp(4 \sqrt{-\lambda}))$. Thus if $\lambda < 0$, for every $T_0$ there is a unique solution to your boundary value problem.