# Homework Help: Solve Laplace equation

1. Aug 6, 2014

### skrat

1. The problem statement, all variables and given/known data
I got to this while solving a physical problem, therefore it is hard for me to write the problem statement if there is none. I can write the whole physical problem here but it wouldn't make any sense.
So, here is how it goes.

I got to a Laplace equation $\triangle \phi =0$ and I also have boundary conditions that say:

$\phi (x=-r/2,y=0, z=d) = -U/2$

$\phi (x=r/2,y=0,z=d)= U/2$ and

$\frac{d}{dz} \phi (x,y,z=d)=0$
$\frac{d}{dz} \phi (x,y,z=0)=0$

2. Relevant equations

3. The attempt at a solution

We have actually never done this before, so I don't really know how to start. Any ideas? :/

2. Aug 6, 2014

### Staff: Mentor

$\phi=\frac{Ux}{r}$?

3. Aug 6, 2014

Ammmm. ? :D

4. Aug 6, 2014

### SteamKing

Staff Emeritus
It would probably be better to write the whole problem out. As it is, what you have provided raises as many questions as it answers.

5. Aug 6, 2014

### Orodruin

Staff Emeritus
Your boundary conditions are not sufficient to uniquely define the solution, they are all fulfilled by Chester's solution but I can add an arbitrary linear function of y to this and still have a solution fulfilling all your boundary conditions. What is the region you are trying to solve in? As it is you have just given boundary conditions in two planes (z=0 and z=d) as well as two points (which happen to also lie in the plane z = d ...).

6. Aug 6, 2014

### skrat

Problem statement:
On a thin infinite metal plate with thickness $d$ and specific conductivity $\sigma$ two electrodes are attached. The distance between the two electrodes is $r$ and the potential is $U$.
Describe magnetic field anywhere in space, due to the currents in stationary state on the metal plate. (Ignore the wires which bring the current to the electrodes)
End of problem

We already discussed this on this forum, and we found out that:

$\nabla \cdot {\bf J}=-\frac{\partial \rho}{\partial t}$ since we have stationary conditions, this is than $\nabla \cdot {\bf J}=0$.

Now because $J=\sigma E$ than $\nabla E=0$ but $E=-\nabla \phi$ so $\nabla ^2 \phi =0$.

Now the boundary conditions:
$\phi (x=-r/2,y=0, z=d) = -U/2$
$\phi (x=r/2,y=0,z=d)= U/2$

and also $\frac{d}{dz} \phi (x,y,z=d)=0$ beacuse there is no component of electric field perpendicular to the surface of the plate. It might be inside the metal plate, but I don't care about that, it is Important that there is none on the surface of the plane. And I guess the same goes for the other side of the plate, so for $z=0$ than $\frac{d}{dz} \phi (x,y,z=0)=0$

Now I have to solve this.

7. Aug 6, 2014

### skrat

Here is how I started:

$\nabla ^2 \phi =0$

Now let's say that $\phi =X(x)Y(y)Z(z)$, so $\nabla ^2 \phi = \frac{\partial^2 }{\partial x^2}\phi +\frac{\partial^2 }{\partial y^2}\phi +\frac{\partial^2 }{\partial z^2}\phi=$
$=X^{''}(x)Y(y)Z(z)+X(x)Y^{''}(y)Z(z)+X(x)Y(y)Z^{''}(z)=0$ dividing whole expression with $X(x)Y(y)Z(z)$ leaves me with

$\frac{1}{X(x)}X^{''}(x)+\frac{1}{Y(y)}Y^{''}(y)+\frac{1}{Z(z)}Z^{''}(z)=0$ which can only be true if each term is in fact a constant, so

$\frac{1}{X(x)}X^{''}(x)=a^2$
$\frac{1}{Y(y)}Y^{''}(y)=b^2$ and
$\frac{1}{Z(z)}Z^{''}(z)=-c^2$

One of them obviously has to be negative, because $a^2+b^2-c^2=0$.

Now solving each DE above, leaves me with
$X(x)=C_1sin(ax)+C_2cos(ax)$
$Y(y)=C_3sin(bx)+C_4cos(bx)$ and
$Z(z)=C_5sinh(cx)+C_6cosh(cx)$

So now I know that $\phi =X(x)Y(y)Z(z)=[C_1sin(ax)+C_2cos(ax)][C_3sin(bx)+C_4cos(bx)][C_5sinh(cx)+C_6cosh(cx)]$

I guess the next step (if so far everything looks fine) would be to determine the value of all those constants. Right?

8. Aug 7, 2014

Nobody? :/

9. Aug 8, 2014

### skrat

Ok, the way I started, the problem has no solution, beacuse $\nabla ^2 \phi =0$ everywhere except on the electrodes, unless also $U=0$ which is nonsense of course.

So I guess I have to finally define the shape of the electrodes.

10. Aug 8, 2014

### HallsofIvy

In what sense are these "boundary conditions"? These are at single points.

11. Aug 8, 2014

### skrat

Yes, these boundary conditions were written when I thought that it is ok to imagine that the electrodes are infinitesimally small - points.
But if that would be true, than the problem has no solution; unless also $U=0$.

So I guess I have to consider some dimension to the electrodes, which will of course also change my boundary conditions.

12. Aug 9, 2014

### skrat

For example, let's say that electrodes are circles with radius $R$, where $R<<r$. $r$ is the distance between the electrodes.
I will also use notation $a=\frac r 2 - R$

This gives me a new set of boundary conditions:

$\phi (x=\pm a,y=0)=-\frac U 2$
$\phi (x=\pm \infty , y=\pm \infty )=0$

Where I also ignored the $z$ component. I think I can do that since the plate is infinite.

Than $\nabla ^2 \phi =0$ or $\frac{1}{X(x)}X^{''}(x)+\frac{1}{Y(y)}Y^{''}(y)=0$

So $\frac{1}{X(x)}X^{''}(x)=-\frac{1}{Y(y)}Y^{''}(y)=-\lambda ^2$

Therefore $X(x)=C_1sin(\lambda x)+C_2cos(\lambda x)$

Now how does one continue?

13. Aug 9, 2014

### Orodruin

Staff Emeritus
The problem here is that this separation of variables is not going to lead you to the eigenfunctions of the problem as these are dependent on a geometry that is not described as simple coordinate surfaces. You could perhaps try some Green's function methods with mirror charges.

14. Aug 9, 2014

### skrat

Can you be more specific? Maybe some links, so I can educate myself about that method?

15. Aug 9, 2014

### Orodruin

Staff Emeritus
I would start with the Wikipedia page on Green's functions. It should also be described in any textbook on solving PDE problems in physics.

16. Aug 13, 2014

### skrat

Ok, I have never worked with the Green's function before so I might need even more help than I normally would. But already normally I need quite a lot of help :D

So, If I understood wikipedia right, the goal is to solve
$\phi(x)=\int _V G(x,{x}')\rho({x}')d^3{x}'+\int _S\left [ \phi({x}'){\nabla }'G(x,{x}')-G(x,{x}'){\nabla }'\phi({x}') \right ]d{\hat{\sigma }}'$

Where $\phi$ is electric potential, $\rho$ is the charge density, $G$ is Greens function.

But I have to do that in two dimensions, somehow.

So, following the example on wiki: http://en.wikipedia.org/wiki/Green's_function

$Lu(x,y)=\nabla ^2u(x,y)=f(x)=0$

${g}''(x,y)=\delta (x-s,y-t)$

etc.

Are things so far ok?

17. Aug 13, 2014

### Orodruin

Staff Emeritus
Well, the Green's function for the two-dimensional Poisson equation all of R^2 is well known and you should be able to find it in any reference on the subject. However, you should aim to find one that is also zero on the boundary of your region as this will allow you to eliminate one of the surface terms. This can be done mirroring techniques.

18. Aug 13, 2014

### skrat

Ok, why are we talking about Poisson equation now? You meant that in general or does my Laplace equation somehow become Poisson?

According to http://www-users.math.umn.edu/~olver/am_/leq.pdf on page 24

$G(x,y,\xi ,\eta)=-\frac{1}{4\pi }log[(x-\xi )^2+(y-\eta )^2]+z(x,y)$ where $z(x,y)=\frac{1}{4\pi }log[(x-\xi )^2+(y-\eta )^2]$ is harmonic with the same boundary conditions as the logarithmic potential function.

Noooow, my conditions sat that $\phi (x=\pm \infty, y=\pm \infty)=0$. Buuuut this Greens function is not 0 on the boundary. :/

19. Aug 13, 2014

### Orodruin

Staff Emeritus
Because the Poisson equation is the Laplace equation with an inhomogeneity. The inhomogeneity is the $\rho$ (charge density) that will be put into the volume integral so this also disappears and you will end up with only one of the surface integrals.

It is impossible to impose zero potential at infinity in two dimensions (simply because the Green's function is logarithmic). However, as in your case if you pick the potentials of the two electrodes to U/2 and -U/2, you will end up having 0 at infinity anyway because of the symmetry. In fact, it will show already in your Green's function once you have performed the correct mirroring arguments.

20. Aug 18, 2014

### skrat

Hmm... I read some online literature, but... Can you be more specific about "correct mirroring arguments"?

Or a good link would hopefully do. I really have to solve this Laplace. -.- Honestly I am embarrassed how people knew this in 19th century and I am having so much trouble with it in 21st century.

21. Aug 24, 2014

### skrat

Ok, there is something I don't understand..

In general I have a Laplace equation $\nabla ^2 u(x,y)=0$. And a solution can be found using $u(x,y)=\int \int G(x,y)f(x,y)dxdy$.

Now If I am not mistaken according to wikipedia, than $G(x,y)=\frac{1}{2\pi }ln(\sqrt{x^2+y^2})$. But I must somehow include the electrodes in my Green function, therfore I think it should hopefully be something like $G(x,y)=\frac{1}{2\pi }(ln\sqrt{(x-x_1)^2+y^2}+ln\sqrt{(x+x_2)^2+y^2})$. Where in $x_1$ and $x_2$ are my electrodes.

If this is ok, then let me ask what $f(x,y)$ is in $u(x,y)=\int \int G(x,y)f(x,y)dxdy$?

I used a bit odd notation this time, but if I understood everything ok, than $u(x,y)$ is infact the potential $\phi (x,y)$ I am looking for. But in order to continue from this point, I would really have to understand what $f(x,y)$ is...

Last edited: Aug 24, 2014
22. Aug 27, 2014

### Orodruin

Staff Emeritus
I suggest by looking at the problem of solving for the Green's function in the 2 dimensional plane with a disk removed and with boundary condition $G(\vec r, \vec r') = 0$ when $\vec r$ is on the disk boundary, call this region $\Omega$. Luckily, there is an analytical solution for this.

We know that the Green's function in the plane is given by
$$G_0(\vec r, \vec r') = \frac{1}{2\pi} \ln(|\vec r - \vec r'|).$$
Thus, this will also be a solution to the Poisson equation in $\Omega$ but we may also add any function $f(\vec r, \vec r')$ to this Green's function as long as it fulfills $\nabla^2 f = 0$ in $\Omega$. Let us assume that
$$f(\vec r, \vec r') = C - G_0(\vec r, \vec r''),$$
where $\vec r''(\vec r')$ is inside the disk and depends on $\vec r'$. In this way, the Poisson equation outside of the disk is not affected. We thus have
$$G(\vec r, \vec r') = \frac{1}{2\pi}\ln\left(\frac{|\vec r - \vec r'|}{A|\vec r - \vec r''|}\right).$$
For symmetry reasons, we must have $\vec r'' = k \vec r'$, where $k$ may depend on $|\vec r'|$. In order for this Green's function to be zero when $|\vec r| = R$ (the disk radius), we must have
$$(\vec r - \vec r')^2 = R^2 + \rho^2 - 2R \rho\cos\theta = A^2(\vec r - \vec r'')^2 = A^2(R^2 + k^2 \rho^2 - 2R\rho k \cos\theta),$$
where $\theta$ is the angle between $\vec r$ and $\vec r'$ and $|\vec r'| = \rho$. It follows that
$$A^2 k = 1 \quad \Rightarrow \quad R^2 + \rho^2 = A^2 (R^2 + A^{-4}\rho^2)$$
which has the solution
$$A = \frac{\rho}{R}, \quad k = \frac{R^2}{\rho^2}.$$
The Green's function is thus
$$G(\vec r, \vec r') = \frac 1{2\pi} \ln\left(\frac{|\vec r - \vec r'| R}{|\vec r'|\cdot\left|\vec r - \vec r' (R^2/{\vec r'}^2) \right|}\right).$$
In order to write down the solution, this Green's function should be negatively mirrored in a plane, resulting in a Green's function that is zero in the plane and on the disk. Once that is done, the solution can be written down using the formula in post #16 and the knowledge that the boundaries are the reflection plane (with potential zero) and the boundary of the disk (with potential $U/2$). This is a bit sketchy and your expression will not be pretty, but it is an analytic solution (albeit on integral form).

23. Aug 27, 2014

### skrat

There are two things I don't understand in your last post:
I am really having troubles to understand (or to imagine) what $\Omega$ and my disk are now. At first I thought $\Omega$ is just a disk with radius $R$, where $R-> \infty$ but after carefully reading the whole post I seriously started to doubt that my interpretation of $\Omega$ is ok.

How on earth is $A^2k=1$?
It is obvious that $A^2k=1$ if you look at the third term in the expression, but if you look at the second one than $A^2k^2=1$ ... ?

Also ${\vec{r}}'$ is vector to one of the electrodes?

Thank you very much for your idiotproof explanation. I will continue with this problem (hopefully not completely wrong) tomorrow trying to mirror the function correctly.

24. Aug 27, 2014

### Orodruin

Staff Emeritus
$\Omega$ is going to be one of your electrodes in the end. I therefore made it circular with an undetermined radius.

You just answered your own question. The expression has to hold regardless of $\cos\theta$. Therefore, the terms proportional to $\cos\theta$ must be equal and those constant with $\theta$ must be equal. This gives you two equations and two unknowns. Note that $k$ may depend on $\rho$ (and does so).

$\vec r'$ is the position vector to the source term introduced when searching for the Green's function. That is, $\nabla^2 G = \delta(\vec r - \vec r')$.

When you insert the Green's function into your expression in #16 (or rather, the gradient of the Green's function), then $\vec r'$ is going to be the position vector at the surface that you are integrating over.

However, I realize now that I read this that this mirroring will not actually work. The second mirroring is going to break the equipotential of the first on the circle. I am now unsure how to proceed properly ... Sorry to have led you down this road. Hopefully you will find similar methods to be of use at some point.

25. Aug 27, 2014

### skrat

Aaaaaaah this can't be! This problem has no solutions. It can not be solved! :O

BTW one more reply to your post in #22: I have never said I need analytic solution. I wanted analytic solution because I believe I have never tried to get a numerical solution of any problem, therefore I don't have a clue on how to even start solving this numerical.

Anyway, Orodruin I HIGHLY appreciate your help here! Thanks for your patience and idiot proof explanations. If nothing more I have learned a lot about Green's functions. :D