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Laplace's equation with unusual boundary conditions

  1. Mar 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve Laplace's equation

    [tex]u_{xx} + u_{yy} = 0[/tex]

    on the semi-infinite domain -∞ < x < ∞, y > 0, subject to the boundary condition that [itex]u_y = (1/2)x u[/itex] on y=0, with u(0,0) = 1. Note that separation of variables will not work, but a suitable transform can be applied.


    2. Relevant equations



    3. The attempt at a solution

    I'm very scattered about how to attempt this problem. My feeling is that you want to apply some transform in y to the equation. On the boundary, you would hope that taking the derivative with respect to y gives you some equation involving u(0,0), or at least u(0,x), which you could use along with the right hand side of the boundary equation - similar to what you get in a Laplace transform. The other obvious thing seems to be to solve the differential equation on the boundary to get u = exp(xy/2) + c(x), where c(0) = 0, but since you can't figure out anything else about c(x) from there, and u would not decay for large y, this seems just as futile.

    The only other idea that occurs to me is to take a Fourier transform of u in x but with respect to y (so where you normally have exp(ikx) dx, you'd have exp(iyx) dx). But then the inverse transform, which involves an integral from -∞ to ∞, wouldn't work, since we only use y > 0.

    Does anyone have some insight into this problem? Thanks!
     
  2. jcsd
  3. Mar 10, 2012 #2
    Start with u_y=1/2x*u, differentiate by y, you get u_yy=1/2x*u_y=1/4x^2*u on y=0. Then use the Laplace equation you get -u_xx=1/4x^2*u on y=0, you solve this ODE for your boundary condition, then use this boundary condition to solve your PDE. But a quick solution to the ODE yields something really unfamiliar, maybe there's another way ...
     
  4. Mar 10, 2012 #3
    That's a very clever idea! I like it. Like you said, though, the solution to that is weird - plus since you now have a second order ODE, wouldn't you need two boundary conditions? We have one: u(0,0) = 1. It seems like the only other boundary condition that would make sense is that the solution has to vanish at infinity, but then you couldn't impose the condition that it vanishes at -infinity, right?

    The solution to that ODE is in terms of parabolic cylinder functions of order -1/2 and in fact imaginary.
     
  5. Mar 11, 2012 #4
    As I thought it over what I did was invalid, the BC is only valid on the x axis, so I cannot differentiate with respect to y. I tried to treat u as the real part of an analytic function, but no luck. This is a curious problem though. Let me know if you are finally able to solve it. Good luck.
     
  6. Mar 11, 2012 #5
    It seems like you'd want to apply a transformation of some sort to the equation that also hits the boundary condition.
     
  7. Mar 11, 2012 #6
    Using a Fourier representation of u(x,0) on real axis, I was able to extend u to y>0 according to Laplace equation:
    u(x,0)=∫ U(ω) exp(iωx) dω, and u(x,y)=∫ U(ω) exp(iωx-ωy) dω, sub into BC, I got u_x=i/2*x*u, and solving for U and u, I finally got u=exp(i/4*(x-i*y)^2). Only problem is this solution blows up as y→inf, maybe there's another solution that decays to zero (and blows up as y→-inf)
     
  8. Mar 12, 2012 #7
    Wow, thank you for all the thought you've put into this! I'm confused though about how you went from the real axis into y > 0
     
  9. Mar 12, 2012 #8
    Basically separation of variables, u=f(x)g(y), so that f''/f=-g''/g=-ω^2, then f(x)=exp(±iωx) and g(y)=exp(±ωy), and I made some choice of the signs. If you consider the other choice of signs, you might get another solution so that you can construct a solution that decay to 0 at inf.
     
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