How can I solve a Laplace equation in a cube with mixed boundary conditions?

[CptXray]

Homework Statement

There's a metal cunducting cube with edge length $a$. Three of its walls: $x=y=z=0$ are grounded and the other three walls: $x=y=z=a$ are held at a constant potential $\phi_{0}$ . Find potential inside the cube.

Homework Equations

The potential must satisfy Laplace equation $$\Delta \phi = 0$$

The Attempt at a Solution

First, I postulate that a potential is a product:
$$\phi(x,y,z) = X(x)Y(y)Z(z)$$.
Plugging it into a Laplace equation(further I'm going to write functions of $X(x)$, $Y(y)$, $Z(z)$ as $X$, $Y$, $Z$ and derivatives as $X'$ and so on):

$\Delta\phi = X''YZ + Y''XZ +Z''XY = XYZ\bigg(\frac{X''}{X} + \frac{Y''}{Y} + \frac{Z''}{Z} \bigg) = 0$

Dividing both sides by $XYZ$ I get:

$$\frac{X''}{X} + \frac{Y''}{Y} + \frac{Z''}{Z} = 0$$
So, every term of the sum above has to be a constant: $\frac{X''}{X} = -\alpha^2$, $\frac{Y''}{Y} = -\beta^2$, $\frac{Z''}{Z} = \gamma^2$ it gives relation between $\alpha$, $\beta$ and $\gamma$: $\gamma^2 = \alpha^2 + \beta^2$

I know that general solutions are:
\begin{cases}
X = Ae^{i\alpha x} + Be^{-i\alpha x}
\\
Y = Ce^{i\beta y} + De^{-i\beta y}
\\
Z = Ee^{\gamma z} + Fe^{-\gamma z}
\end{cases}

Applying boundary conditions for grounded walls $\phi(x=y=z=0) = 0$
$X(0) = A + B = 0⇒X = \tilde{X}\sin{\alpha x}$
$Y(0) = C + D = 0⇒Y = \tilde{Y}\sin{\beta y}$
$X(0) = E + F = 0⇒Z = \tilde{Z}\sinh{\gamma z}$
where $\tilde{X}$, $\tilde{Y}$, $\tilde{Z}$ $∈$ $ℝ$ and are constants.

And at this point I have a problem, because when I try to find $\alpha$ and $\beta$ I don't get quantized solutions to build a series solution. Most likely it's my attempted solution that is wrong, but I'd be really happy for some guidlines or hints from you guys.

 

[phyzguy]

One question to ask yourself is this. Since the problem is completely symmetric in X,Y,Z, why should it have oscillatory solutions in X and Y, but an exponentially decaying solution in Z? How would it decide which of the three axes has the exponentially decaying solution, when all three are the same? You are on the right track, but is there any combination of α, β, and γ that is symmetric in X,Y,Z?

 

[CptXray]

One question to ask yourself is this. Since the problem is completely symmetric in X,Y,Z, why should it have oscillatory solutions in X and Y, but an exponentially decaying solution in Z?

Well, I guess the oscillatory solutions come from the conditions $X(0) = Y(0) = 0$ and hyperbolic solution from equation $Z'' = \gamma^2 Z$. I know that from symmetry there's no reson to have solution for Z in different form than the rest two functions, but that's what, again, $Z'' = \gamma^2 Z$ yields.

 

[phyzguy]

Well, I guess the oscillatory solutions come from the conditions $X(0) = Y(0) = 0$ and hyperbolic solution from equation $Z'' = \gamma^2 Z$. I know that from symmetry there's no reson to have solution for Z in different form than the rest two functions, but that's what, again, $Z'' = \gamma^2 Z$ yields.

So, is there any combination of α, β, and γ that is symmetric in X,Y,Z?

 

[CptXray]

So, is there any combination of α, β, and γ that is symmetric in X,Y,Z?

Yeah, but I thought that $\alpha, \beta, \gamma$ don't depend on $x,y,z$ but rather on some parameters $n,l,m∈ℤ$ in the final solution that should be a series.

 

[PeroK]

Yeah, but I thought that $\alpha, \beta, \gamma$ don't depend on $x,y,z$ but rather on some parameters $n,l,m∈ℤ$ in the final solution that should be a series.

Maybe you are overthinking this. Is there a simpler solution?

 

[CptXray]

Maybe you are overthinking this. Is there a simpler solution?

I'm sure there is, but that's the only version of this cube problem that I can't solve for some time. When only one wall in held at some potential it's easier because boundary conditions give really nice oscillatory solutions to work with. Here I'm stuck at the beginning.

 

[PeroK]

I'm sure there is, but that's the only version of this cube problem that I can't solve for some time. When only one wall in held at some potential it's easier because boundary conditions give really nice oscillatory solutions to work with. Here I'm stuck at the beginning.

By symmetry the three factors must be equal.

 

[CptXray]

So, the symmetry argument yields $\alpha=\beta=\gamma=0$ and this means that functions are:
$$
\begin{cases}
X(x) = X_{0} + X_{1}x
\\
Y(Y) = Y_{0} + Y_{1}y
\\
Z(z)=Z_{0} + Z_{1}z
\end{cases}
$$
conditions $X(0)=Y(0)=Z(0)=0$ eliminate $X_{0}, Y_{0}, Z_{0}$.
And general solution should be:
$$\phi(x,y,z)=\sum_{l,m,n}X_{l}Y_{m}Z_{n}xyz$$,
where $X_{l}, Y_{m}, Z_{n}$ are some constants.
Applying boundary conditions for $\phi(a,y,z)=\phi(x,a,z)=\phi(x,y,a)=\phi_{0}$ should do the trick, right?

 

[PeroK]

So, the symmetry argument yields $\alpha=\beta=\gamma=0$ and this means that functions are:
$$
\begin{cases}
X(x) = X_{0} + X_{1}x
\\
Y(Y) = Y_{0} + Y_{1}y
\\
Z(z)=Z_{0} + Z_{1}z
\end{cases}
$$
conditions $X(0)=Y(0)=Z(0)=0$ eliminate $X_{0}, Y_{0}, Z_{0}$.
And general solution should be:
$$\phi(x,y,z)=\sum_{l,m,n}X_{l}Y_{m}Z_{n}xyz$$,
where $X_{l}, Y_{m}, Z_{n}$ are some constants.
Applying boundary conditions for $\phi(a,y,z)=\phi(x,a,z)=\phi(x,y,a)=\phi_{0}$ should do the trick, right?

Perhaps you'd better check that your solution is a solution.

 

[phyzguy]

What if you try solving a problem where one side has φ=φ0, and the other 5 sides have φ=0. If you can solve this, you can superpose three of these to give your solution.