# Solutions to the laplace equation

1. Jan 6, 2008

### ehrenfest

[SOLVED] solutions to the laplace equation

1. The problem statement, all variables and given/known data
http://mathworld.wolfram.com/LaplacesEquation.html
I don't understand why the solutions to the Laplace equation are different in different coordinate system. Obviously, the solutions will look different when you write them out as equations, but when you plot them and look at them in Mathematica, shouldn't they be coordinate system independent?

EDIT: maybe the answer is this: in each coordinate system, there is some really convenient basis that is expressible in a small number of terms with the variables in that coordinate system. however, the "space" of the solutions is the same in all coordinate systems. is that right?
2. Relevant equations

3. The attempt at a solution

Last edited: Jan 6, 2008
2. Jan 6, 2008

### CompuChip

It is because the gradient operator is different in each coordinate system.
I think you'd expect, for example, that transforming the polar coordinates solution to Cartesian coordinates should give the solution in Cartesian coordinates found directly from the equation, right?
The problem is, that doing the coordinate transformation the Laplacian also changes form (there are extra coordinate-dependent terms in front of the derivatives, such as 1/r^2 and 1/(r sin theta)) and rather than have the solution transform consistently, we want the combination with the Laplacian to transform consistently. This means the solution will have to look different, to compensate for the differential operator looking different.

3. Jan 6, 2008

### ehrenfest

Is what I wrote in the EDIT right?

4. Feb 17, 2008

### ehrenfest

anyone?

5. Feb 17, 2008

### malawi_glenn

I dont understand what you wrote in EDIT, but I totaly agree with CompuChip's answer.

6. Feb 17, 2008

### CompuChip

I also don't understand the part about
I sort of agree with this quote though (if you meant it the way I think you did):
You could compare this to bases in vector spaces: one vector can have completely different components ("look different") depending on your choice of basis, though it is always the same vector. So if you want to look at this way conceptually, that's fine. Note however, that in this case, the solution actually is different. That is, if you have a solution in spherical coordinates, you can transform it to Cartesian coordinates, but it will not be the same as a solution in Cartesian coordinates (and if you apply the Laplacian in Cartesian coordinates, it will not satisfy your differential equation). If you also transform the Laplacian from spherical to Cartesian coordinates however, and apply the transformed Laplacian to the transformed solution, then you will get the same as when you applied the original (spherical) Laplacian to the original (spherical) solution.

7. Feb 17, 2008

### ehrenfest

When you say "looks different", do you mean how it looks when you write it down? I am not talking about that. A solution to Laplace's equation is a twice-differentiable function from $\mathbb{R}^3$ to $\mathbb{R}$ regardless of the coordinate system . When I talk about how a solution "looks", I mean how it looks to someone who can visualize four dimensions and can thus literally "see" the solution just like you and I can see solutions plotted on the Cartesian plane. My claim is this: if you take any solution to Laplace's equation in spherical coordinates and "look" at it in 4D, then you find a solution to Laplace's equation in rectangular coordinates that "looks" exactly the same in 4D. The vector space I am talking about is the actual plots of the solutions in 4D. And my claim is that the vector space is identical in every coordinate system; only the convenient basis is different. PLEASE CONFIRM THIS.

8. Feb 17, 2008

### ehrenfest

anyone?

9. Feb 18, 2008

### CompuChip

I'm not going to expand very much on this thread anymore, as I think everything there is to say has already been said in post #2, but why don't you calculate solutions f(x, y) in Cartesian coordinates and g(r, θ) in polar coordinates (in two dimensions) and then compare g(r, θ) to f(r cos(θ), r sin(θ)) ?
You can even plot it numerically.

10. Feb 18, 2008

### ehrenfest

Can someone else please take a look at my argument in post #7? I think it is right but this is really bothering me. This is a very elementary issue.

11. Feb 18, 2008

### Marco_84

hey guys.. i think that u are getting in trouble 4 nothing...
it is normal that when you solve laplace equation.... in any kind of "good" coordinates the behavior.... (the graphic) of your solution is the same.
The analitical formula is different obviously.... but, since there should be dippheomorphisms between the coordinates, every one can use what he wants.

in formulas:
this is our chauchy problem:

$$\triangle f=0; + B.C.$$

where triangle mean the laplacian in any kind of coordinates...
whene you get $$G$$(cartesian) or $$\widetilde{G}$$ (spherical) as a solution with the proper chauchy problem,.

At each value of x,y,z or $$\theta,r,\phi$$ we have:

$$G(x,y,z)=G(x(\theta,r,\phi),y(\theta,r,\phi),z(\theta,r,\phi))=\widetilde{G}(\theta,r,\phi)$$

to visualize it, its enough to live in flat-land and plot it with any kind of tool.

regards
marco

12. Feb 18, 2008

### Marco_84

in other words:

$$\frac{1}{r}=\frac{1}{\sqrt{x^2+y^2+z^2}}$$

ciao

13. Feb 18, 2008

### ehrenfest

I think that confirms my proposition.