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Compatibilty of the Dirichlet boundary condition

  1. Mar 1, 2013 #1
    Hi,

    I am trying to solve a Poisson equation [itex]\nabla^2 \phi = f[/itex] in [itex]\Omega[/itex], with Dirichlet boundary condition [itex]\phi = 0[/itex] on [itex]\partial \Omega[/itex]. My problem is that I am trying to understand the condition under which a solution exists. All the text I consulted says that the problem is solvable.

    However, I am working on contrived example for which I dont see how a solution is possible, yet I am unable to explain it. Consider a function and its first two derivatives,
    [tex]
    F(x) = -\frac1 4 e^{-2x} (2x + 1) \\
    \frac{dF}{dx} = x e^{-2x} \\
    \frac{d^2F}{dx^2} = e^{-2x} (1-2x)
    [/tex]
    Clearly, [itex]F(x) \neq 0[/itex] for [itex]x = 0,1[/itex]. I am attaching the plots of these functions [itex]F(x)[/itex] in black, [itex]\frac{dF}{dx}[/itex] in red, and [itex]\frac{d^2F}{dx^2}[/itex] in green.

    Now, suppose, I solve the Poisson equation said above, with [itex]\nabla^2 \phi = e^{-2x} (1-2x)[/itex] for [itex]0 < x < 1[/itex], I hope to recover [itex]\phi = F[/itex] uniquely upto a harmonic. However, the given that [itex]\phi = 0[/itex] for [itex]x = 0, 1[/itex], I dont see how this can produce a continuous [itex]\phi[/itex], which matches the black curve.

    I think this is because the information I pass to the system is corrupt, however, no textbook tells me any requirement on the compatibility between the source function and the boundary condition. Any insights are appreciated.
     

    Attached Files:

    Last edited: Mar 1, 2013
  2. jcsd
  3. Mar 2, 2013 #2

    kai_sikorski

    User Avatar
    Gold Member

    You've found a particular solution to the problem, but I think you're forgetting about the two homogenous solutions that you can use to match your boundary conditions.
     
  4. Mar 2, 2013 #3
    Hi !
    Solution in attachment :
     

    Attached Files:

  5. Mar 3, 2013 #4
    Thanks both of you.

    JJacquelin, I think you used [itex](1-2x)[/itex] instead of [itex](1+2x) [/itex], and therefore calculation of [itex]c_1,c_2[/itex] are wrong. But I got the general idea. Thanks a lot.
     
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