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

In summary: I thought.By symmetry the three factors must be equal.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_{
  • #1
CptXray
23
3

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.
 
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  • #2
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?
 
  • #3
phyzguy said:
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.
 
  • #4
CptXray said:
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?
 
  • #5
phyzguy said:
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.
 
  • #6
CptXray said:
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?
 
  • #7
PeroK said:
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.
 
  • #8
CptXray said:
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.
 
  • #9
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?
 
  • #10
CptXray said:
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.
 
  • #11
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.
 
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FAQ: How can I solve a Laplace equation in a cube with mixed boundary conditions?

1. What is the Laplace equation in a cube?

The Laplace equation in a cube is a partial differential equation that describes the distribution of temperature or potential in a three-dimensional cube. It is also known as the Poisson equation and is used in various fields of science, such as physics, engineering, and mathematics.

2. How is the Laplace equation in a cube derived?

The Laplace equation in a cube is derived from the general Laplace equation, which is a second-order partial differential equation. In a cube, the equation takes into account the three spatial dimensions and is solved using boundary conditions and appropriate mathematical techniques, such as separation of variables or the method of images.

3. What are the applications of the Laplace equation in a cube?

The Laplace equation in a cube has various applications, including solving problems related to heat conduction, electrostatics, fluid flow, and diffusion. It is also used in image processing, signal processing, and numerical analysis. Additionally, the Laplace equation is a fundamental building block for more complex equations in physics and engineering.

4. What are the boundary conditions for solving the Laplace equation in a cube?

The boundary conditions for solving the Laplace equation in a cube depend on the specific problem being solved. In general, the boundary conditions specify the values of the solution at the boundaries of the cube, such as temperature or potential values. These conditions are essential for obtaining a unique solution to the equation.

5. Are there any limitations to using the Laplace equation in a cube?

While the Laplace equation in a cube is a powerful tool for solving various problems, it does have some limitations. It assumes a steady-state system, meaning that the temperature or potential does not change with time. It also assumes a linear relationship between the solution and its derivatives, which may not always hold true in real-life scenarios. Additionally, the Laplace equation may not be able to accurately describe highly complex systems with irregular boundaries or varying material properties.

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