Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Poisson, non-homogeneous

  1. Jul 13, 2012 #1
    Solve:
    Δu=-1, u(1, theta)=sin(theta). 0<r<1, -pie<theta<pie, u finite at r=0


    What I've done: u=u1+u2. Δu2=0, with u(1, theta)=sin(theta). So eventually u2=rsin(theta).
    The u1 problem however I am not sure how to solve. Eigenfunction expansion doesnt seem like it would work (though not exactly sure why, if so).
    Help?
     
  2. jcsd
  3. Jul 15, 2012 #2
    Let u = u1 + u2, where u1 solves Laplaces equation with the boundary condition you've given (hint: use eigenfunctions). Then let u2 be any function such that Δu2 = -1 and u2 = 0 on the boundary (i.e. at r=1)
     
  4. Jul 15, 2012 #3
    As I said, not sure how to solve the non-homogeneous part u2=-1
     
  5. Jul 15, 2012 #4
    Also as I said this is a circular problem; you can tell me generalities but I don't see how they'll help. I can find no single relevant example of this in my text book or ANYWHERE online, involving a nonhomogeneous poisson problem in this circular situation. Yet my upcoming masters exam practice tests have an abundant number of these with terribly written solutions.
    In certain situations, though I have no idea how to tell when, it seems "circularly symmetic" comes into play, thus leading to the disregarding of the 3rd term in the polar laplacian.
     
  6. Jul 16, 2012 #5
    Fair enough, I'm happy to go through this:

    Firstly, I'm assuming you're fine with u1? Just look at the general solution to Laplace's Equation in plane polars and pick the eigenfunctions which are proportional to sin(theta) whilst eliminating the one which is singular at the origin. Then to work out your constant just plug in r=1 and you're done.

    As for u2, well, you have: Δu2 = -1, u2 = 0 at r = 1 and u2 finite at the origin.
    From the boundary conditions, you can tell that this is going to be circularly symmetric (otherwise one of your boundary conditions would depend on theta).
    Now, write out Δu2 in plane polars but exclude the last term as u2 does not depend on theta. Then try a solution of the form u2 = Ar^n and you should find that for Δu2 = -1, n = 2 and A = -1/4. Now your last problem is that you just need to satisfy u2 = 0 on the boundary so you can just add in an appropriate constant (as Δ acting on a constant will be zero) and as I'm sure is quite obvious, plug in r=1 and the constant you want is gonna be 1/4. Thus you have u2 = 1/4 (1 - r^2) and you can add this to u1 to get your solution :)

    [Note: I may have made a mistake whilst typing that, I was just doing it off the top of my head as opposed to writing it down and then copying it out, but I hope you get the idea!]
     
  7. Jul 16, 2012 #6
    Ahhhhhhhhhhhhh, sorry I may have completely forgotten everything I learnt in second year undergrad, but I'm almost certain that the solution is gonna simply be:
    u = rsin(theta) + 1/4 (1 - r^2)
     
  8. Jul 16, 2012 #7
    Obviously you are rigth. Your solution is the right one and the simplest one.
    Moreover, there was a mistake in my complicated developments. So, I delete my previous post.
     
  9. Jul 16, 2012 #8
    Haha, thanks, I was worried I'd forgotten how to do these for sec!
     
  10. Jul 16, 2012 #9
    Ok, if I may ask, what was your reasoning for choosing a soln of the form Ar^n?

    Do we always drop the last part of the polar laplacian if the boundary conditions don't depend on theta? By boundary conditions, do we just mean the condition at r=1?
     
  11. Jul 16, 2012 #10

    Right, u2 is just a particular solution so frankly any function that satisfies Δu2 = -1 with u2 = 0 at r=1 will do. This is usually just a case of trial and error but because maths is (almost) always really nice, you'll find that often some reasonable function will always work (e.g. a polynomial, trig function or exponential). In this case, we knew we wanted Δu2 = -1 so the RHS is a constant and we know that it's possible for me to get a constant on the LHS if we use a function of the form Ar^n (and it just so happens n=2)

    Boundary Conditions are exactly what they say on the tin, it's the information you've been given on the boundary of your problem. In this case your solution lies in the circle of radius 1 and your boundary is at r=1.
    For u1, the boundary condition was u = sin(theta) at r=1 so you have an eigenfunction forcing on the boundary and hence u is proportional to that particular eigenfunction i.e. [ar + b/r]sin(theta). You then impose u finite at the origin so b=0 and you work out what a is by plugging in r=1 (so obviously a=1 by your boundary condition)
    For u2, the boundary condition was u = 0 at r=1 and as this doesn't depend on theta, you know that u2 is gonna be independent of theta (thus the last term in the plane polar laplacian drops out).

    Basically, Boundary conditions determine what your solution is gonna look like. You could write out the full eigenfunction expansion if you wanted but you'll find that all the coefficients of the ones that aren't in your boundary condition will vanish.

    Hope that helps :)
     
  12. Jul 16, 2012 #11
    Thanks Mario.
    Any chance of you knowing where I can find similar examples like this to do?
     
  13. Jul 16, 2012 #12
    Not sure where you could find practice material. I found these kinds of problems cropped up a lot in my second year of undergrad in a mathematical methods course as well as a course on electromagnetism and a course on fluid dynamics.

    Feel free to have a look at the exam papers I did, however the questions aren't particularly straight forward and may end up not being of much help at all:
    http://maths.cam.ac.uk/undergrad/pastpapers/
    (you want to look at the IB papers and look at questions from Methods, Electromagnetism and Fluid Dynamics)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Poisson, non-homogeneous
Loading...