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Sturm-Liouville Like Equation with Boundary Conditions on Second Derivative

  1. Sep 13, 2010 #1
    Hello,

    I am facing a diffusion equation..
    [tex]\frac{du(x,t)}{dt} = D \frac{d^2u}{dx^2}[/tex]
    .. with slightly exotic boundary conditions:
    [tex]u(0,t) = 0[/tex]
    [tex]\frac{d^2u(a,t)}{dx^2}+ \alpha \frac{du(a,t)}{dx} = 0[/tex]

    I expected the solution to be relatively easy to find, since separation of variables quickly gives a Storm-Liouville equation. However, the different second boundary condition has so far trashed all of my efforts to find a solution.

    Is someone able to give some advice on this? Can I rely on the existence of eigenfunctions at all or do I need to get at this from an entirely different direction?
     
  2. jcsd
  3. Sep 18, 2010 #2
    What I would do is write that second boundary condition in a slightly different form.
    We see that the u(x,t) satisfies the diffusion equation for all x (in its domain of course). Hence for x=a too. With this it seems to be fruitful if we write the boundary condition as:

    [tex]
    \left(\frac{d^2u(x,t)}{dx^2}+ \alpha \frac{du(x,t)}{dx}\right) \bigg |_{x=a} =
    \left(\frac{1}{D}\frac{du(x,t)}{dt}+\alpha \frac{du(x,t)}{dx}\right) \bigg |_{x=a}=0
    [/tex]

    And so the boundary condition is only with first derivatives.
     
  4. Sep 18, 2010 #3
    Actually that is the original form of the boundary condition. I rewrote it in terms of the spacial derivative so that I could use separation of variables. Did I miss something obvious? To me this form seems even less tractable...
     
  5. Sep 18, 2010 #4
    You can use separation of variables anyway.
    Solve the diffusion equation in the usual manner with separation of variables, the time part as usual will have an exponential dependence then using the first boundary condition you can rule out one of the solutions to the spatial equation (the sine stays or the sinh), then apply the second boundary condition, and you shall get a transcendental equation something like k*tan(k) = ... this quantizes the constant you got when using the separation of variables, but you cannot solve this equation. From here you can superimpose the solution with the different quantas and you are done.
     
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