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Parabolic pde with additional term

  1. Sep 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Solve
    [itex]\frac{\partial v}{\partial t} = k\frac{\partial^2 v}{\partial x^2} - v
    [/itex]

    [itex] 0\leq x \leq L [/itex]

    [itex] t > 0 [/itex]


    2. Relevant equations
    [itex] v(x,0) = f(x), v(0,t)=0, \frac{\partial v}{\partial x}(L,t) = -v(L,t) [/itex]



    3. The attempt at a solution

    I've already attempted to solve this using separation of variables, but I'm not sure if that is a valid approach as the equation is not homogeneous (?) due to the v(x,0) = f(x) term. I've also attempted using green's theorem but I keep getting stuck, though this may be due to a math error. Can anyone give me any insight as to where I may be going wrong?
     
  2. jcsd
  3. Sep 13, 2014 #2

    haruspex

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    If you can separate the variables, it will produce a valid solution. The question is whether all solutions can be achieved that way. The critical issue is whether a linear sum of two solutions produces a solution. If so, arbitrary solutions can be represented as an infinite sum of solutions obtained by separation of variables.
    Does this equation have that property?
     
  4. Sep 13, 2014 #3

    pasmith

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    The equation is homogeneous and linear, so separation of variables should work. (Not being homogeneous is in any case not fatal to separation of variables, you just have to use a "particular integral plus complementary functions" approach; not being linear is fatal.)

    The eigenfunctions are [tex]
    v(x,t) = T(t)X(x) = T(t)\sin(\lambda x)
    [/tex] where the possible values of [itex]\lambda[/itex] are determined by the condition that [tex]
    X(L) + X'(L) = 0,
    [/tex] and since you're given arbitrary initial data the fact that you can't find [itex]\lambda[/itex] analytically is not important. Sturm-Liouville theory tells you that the resulting eigenfunctions will be orthogonal with respect to the inner product [tex]
    \int_0^L f(x)g(x)\,dx[/tex] as usual. You can find [itex]\|X\|^2 = \int_0^L X(x)^2\,dx[/itex] in terms of [itex]L[/itex] and [itex]\lambda[/itex], but obviously if you can't find [itex]\lambda[/itex] analytically you can't find [itex]\|X\|^2[/itex] analytically.
     
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