Potential of Sphere Center: Solve Laplacian Equation

[rolandas9999]

Homework Statement

Given potential r=2m on the surface of the sphere which meets the Laplasian equation (triangle)u=0 and is u(2,theta,psi)=5sin(theta)sin(0.25psi). I need to find the potential of sphere center. Can anyone help me?

Homework Equations

The Attempt at a Solution

 

[Lavabug]

Write down Laplace's equation of your potential U in spherical coordinates. Your potential U is a function of 3 variables: r, theta and psi, so you can "probe" or attempt to find a solution of U by setting it equal to the product of 3 functions of 1 independent variable each: R(r).Phi(psi).Theta(theta). (I use uppercase to indicate the uppercase Greek letter).

So substitute U for your product solution R.P.T. in the laplacian, if you're careful your partials will become total derivatives.

Next step is to try to separate variables, try to get one side with only R and theta and another one with only psi (I think, its trial and error/experience). If you manage to do that, you can prove that differentiating both sides by the variable on the opposite side will give you zero (as you just separated the dependencies).

This allows you to set each of your sides equal to a constant (generally m^2 for spherical problems), giving you two separate ordinary differential equations that are easy to solve. (the one with 2 variables might require additional variable separation). You'll probably have to play around with one of them to identify a special function's solution like Bessel's, Legendre polynomials, etc. In these cases you have to ignore the parts of the solution that give you problems where you want them (ie: you don't want a solution that gives you a singularity at r = 0, which is where you want to solve the problem! Or one that diverges as r approaches infinity, since you very well know electric/gravitational potential has to go to zero as you get farther away!)

You solve each of those single-variable ode's for R(r), Theta(theta), and Phi(psi) and the general solution for your potential U is the product of all three solutions. Then you set your boundary and/or initial conditions to solve for your Fourier coefficients in your solution. Lastly, you find the solution at r = 0. Whew!

Hope this helps.

 

[Lavabug]

By the way, do you have the solution to the problem? I think I could use this problem to study for my exam too haha.

 

[rolandas9999]

Hi,

Thanks for the answer. Sorry, but I don't have the solution, so that's why I am asking if someone could help me. Thanks again.

 

[grey_earl]

It is even easier than that, lavabug. Since the poster already has u(2,\theta,\phi)=5\sin(\theta)\sin(\phi/4), separation of variables makes the ansatzu(r, \theta,\phi) = f(r) \sin(\theta)\sin(\phi/4) with f(2) = 5 and only an ODE for r remains.
So, rolandas9999, take the Laplacian in spherical coordinates (from your course, or Wikipedia), calculate \triangle u = 0 with the ansatz above, and solve for f(r). If it doesn't work out, show your tries and we will find any mistakes :)

 

[Lavabug]

Well I've done a very similar problem and that's the exact procedure I used, in fact the R(r) equation had a solution in the form of Legendre polynomials. You have to ignore the set of polynomials that diverges at r=0. The other two ODE's in the problem are an associated Legendre function and a plain homogenous 2nd order ODE with imaginary roots. The general solution is massive.

If you can wait about a week, I can give it a try. The key to these problems is patience and looking at solved examples to learn all the details of the procedure.

 

[rolandas9999]

Thanks for Your help very much. This thing is a bit difficult for me, so I will ask a bit more. So the Laplasian operator in spherical coordinates is
http://upload.wikimedia.org/math/8/2/7/8275dcad2a7a2f9587c996beb5cc8c37.png
What should I do with it?

 

[Lavabug]

Have you done easier PDE problems before? Start with solving for the laplacian in 1 or 2-D problems in cartesian coordinates (ie: steady state temperature or potential in a wire, or rectangular plate), then work your way up to problems with cylindrical and spherical coordinates.

 

[grey_earl]

rolandas9999: f is now u for you. Just plug the ansatz in I gave you and calculate. And the guy is named Laplace, he was a Frenchman.

Lavabug: If you need the general solution with some boundary conditions given, then yes. But the poster doesn't need all that. And in generally you don't have to ignore the other polynomials, only if your boundary condition says the solution should be regular at r = 0. Another typical condition then says that the solution should be one-to-one, so you have f(\phi+2\pi) = f(\phi), as well as f(-\theta,\phi) = f(\theta,\phi+\pi), which simplifies it quite a lot.

 

[Lavabug]

rolandas9999: f is now u for you. Just plug the ansatz in I gave you and calculate. And the guy is named Laplace, he was a Frenchman.

Lavabug: If you need the general solution with some boundary conditions given, then yes. But the poster doesn't need all that. And in generally you don't have to ignore the other polynomials, only if your boundary condition says the solution should be regular at r = 0. Another typical condition then says that the solution should be one-to-one, so you have f(\phi+2\pi) = f(\phi), as well as f(-\theta,\phi) = f(\theta,\phi+\pi), which simplifies it quite a lot.

Ah wow I see what you did there, you applied the boundary condition straight away therefore only had 2 DE's to solve, guess that's a smarter choice. I was taught to use a "brute force" approach and get the general solution, then start stripping it down to what the physical problem requires (regularity at r=0 for example) lol.

 

[grey_earl]

Well, I prefer to think first and calculate later, saves me much time in my PhD :)

 

[rolandas9999]

Thanks for the GREAT help grey earl! And Lavabug of course :)

 

[L-x]

I was taught to use a "brute force" approach and get the general solution, then start stripping it down to what the physical problem requires (regularity at r=0 for example) lol.

This approach always works, so it's good to be able to fall back on, as you've seen however is it usually not the quickest way.

For problems with spherical symmetry, try setting the constant to be (m+1)m instead of -m^2 when you separate the radial from the angular parts of the solution. When then separating theta from phi using -n^2 is going to be easiest.