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PDE Separation of Variables with Nonzero Boundary Conditions

  1. Nov 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve the diffusion equation:

    [itex]u_{xx}-\alpha^2 u_{t}=0[/itex]

    With the boundary and initial conditions:

    [itex]u(0,t)=u_{0}[/itex]
    [itex]u(L,t)=u_{L}[/itex]
    [itex]u(x,0=\phi(x)[/itex]

    3. The attempt at a solution

    I want to solve using separation of variables...
    I start by assuming a solution of the form:

    [itex]u(x,t)=X(x)T(t)[/itex]

    Differentiating and dropping the notation so f(F) = F :

    [itex]u_{x}=X'T[/itex]
    [itex]u_{xx}=X''T[/itex]
    [itex]u_{t}=XT'[/itex]

    Substituting back into the PDE,

    [itex]X''T-\alpha^2 XT'=0[/itex]
    [itex]X''T=\alpha^2 XT[/itex]

    [itex]\frac{X''}{X}=\alpha^2 \frac{T'}{T}[/itex]

    Let each side be equal to a constant, [itex]-\lambda^2[/itex]
    So now we have two ODE's:

    [itex]X''+\lambda^2 X=0[/itex]
    [itex]T'+(\frac{\lambda}{\alpha})^2 T=0[/itex]

    These ODE's are easy to solve:

    [itex]T(t)=Ae^{-(\frac{\lambda}{\alpha})^2 t}[/itex]
    [itex]X(x)=Bsin(\lambda x)+Ccos(\lambda x)[/itex]

    So now, let AB=A and AC=B so the PDE has the general solution:

    [itex]u_{\lambda}(x,t)=e^{-(\frac{\lambda}{\alpha})^2 t}[Asin(\lambda x)+Bcos(\lambda x)][/itex]

    So now, I want to match my boundary conditions. All the other PDE's I've solved have zero-valued boundary conditions, so I'm not exactly sure what to do here.

    [itex]u(0,t)=u_{0}=e^{-(\frac{\lambda}{\alpha})^2 t}[Asin(0)+Bcos(0)][/itex]
    [itex]u(0,t)=u_{0}=Be^{-(\frac{\lambda}{\alpha})^2 t}[/itex]

    This is where I'm confused.
    To satisfy the B.C., my intuition says [itex]B=u_{0}[/itex] and [itex]e^{-(\frac{\lambda}{\alpha})^2 t}[/itex] must be a multiple of [itex]\pi[/itex].
    Even if this is correct, I'm not sure how to express it as a constraint on [itex]\lambda[/itex].
     
  2. jcsd
  3. Nov 24, 2012 #2

    LCKurtz

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    Before you separate variables, make the substitution ##u(x,t) = v(x,t) +\psi(x)##. Substitute that into your problem:$$
    v_{xx} + \psi''(x) + v_t = 0$$ $$
    v(0,t) + \psi(0) = u_0$$ $$
    v(L,t) + \psi(L) = u_L$$ $$
    v(x,0) + \psi(x) = \phi(x)$$
    Now if you let ##\psi''(x)=0,\ \psi(0)=u_0,\ \psi(L)=u_L##, you can solve for ##\psi(x)## and what is left is a system in ##v(x)## you can separate variables on. Your boundary conditions will be homogeneous and the last equation will be ##v(x,0)=\phi(x) - \psi(x)##. Your final answer will be ##u(x,t) = v(x,t)+\psi(x)##.
     
  4. Nov 24, 2012 #3
    LCKurtz,

    I can't find anything like that in my textbook. Did you do that so the solution would satisfy both the nonhomogenous and homogenous parts? I suppose you added the extra function of x because the nonhomogenous boundary conditions only depend on x?
     
  5. Nov 24, 2012 #4

    LCKurtz

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    I don't know about your text, but it is a standard method. You let the ##\psi(x)## take care of the nonhomogeneous part. So did you try it on your problem?
     
  6. Nov 24, 2012 #5
    I will be trying it soon.
     
  7. Nov 25, 2012 #6
    Got it! It was trivial after I did the substitution you mentioned. Thanks again!
     
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