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Math Help - Series Solution of PDE Non-Homogeneous BCS

  1. #1
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    Series Solution of PDE Non-Homogeneous BCS

    I thought I had these problems down, but I found one for which I am having trouble.


    u_t = u_{xx} - u,\quad 0<x<\pi,\; t>0
    u_x(0,t) = 1,\; u_x(\pi,t) = -1,\; t>0
    u(x,0) = 0,\; 0<x<\pi

    My attempt:
    Let  u(x,t) = v(x,t) + w(x,t) where w(x,t) satisfies the BCs
    w_x(0,t) = 1,\; w_x(\pi,t) = -1,\; t>0.
    So, I picked w(x,t) = \sin x.
    Now the new BCs and IC for v(x,t) are
    u_t = u_{xx} - u,\quad 0<x<\pi,\; t>0
    u_x(0,t) = 0,\; u_x(\pi,t) = 0,\; t>0
    u(x,0) = -\sin x,\; 0<x<\pi.

    Now assume v(x,t) = X(x)T(t) then
    \dfrac{T'(t)}{T(t)} = \dfrac{X''(x)}{X(x)} - 1 = -\lambda
    X''(x) = \left(-\lambda + 1\right) X(x)

    Case I: \lambda > 1
    Let \omega = \sqrt{\lambda - 1}
    X(x) = A_1 \cos \omega x + A_2 \sin \omega x
    X'(x) = -A_1 \omega \sin \omega x + A_2 \omega \cos \omega x

    But this doesn't seem to help me solve for \omega. Normally I would solve for one constant and then solve for \omega, plug this into v(x,t) then use the IC to solve for the last constant.
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  2. #2
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    Can you show me how you got that the DE for v was

    v_{t}=v_{xx}-v? Because I'm getting

    v_{t}=v_{xx}-v-2\sin(x).
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  3. #3
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    That is right. So, how do I proceed from there?
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  4. #4
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    So, I would still have the same issue since I am suppose to solve the homogeneous problem and then solve the particular solution.
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  5. #5
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    If this is your new equation, I'm not so sure it's separable, is it? Perhaps you could try an eigenvector expansion or something.
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  6. #6
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    Sure it is separable.

    v(t,x) = -\sin(x)
    Is a solution to the PDE. So let X(x) = \sin x \text{ and } T(t) = -1.
    \Rightarrow \left(-\sin x\right)_t = \left(-\sin x\right)_{xx} - 2\sin x
    and it satisfies the initial condition and the boundary conditions.
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  7. #7
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    So, what do you get when you plug in your "separable" ansatz?
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  8. #8
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    Well, as far as I understand I would still want to first solve the homogeneous part, so I have
    \dfrac{X''(x)}{X(x)} = \dfrac{T'(t)}{T(t)} = -\lambda
    Now I would solve the DEQ
    X''(x) = -\lambda X(x).
    From here I have cases
    Case I: \lambda > 0
    Let \omega = \sqrt{\lambda}
    X(x) = A_1 \cos \omega x + A_2 \sin \omega x
    X'(x) = -A_1\omega \sin \omega x + A_2 \omega \cos \omega x
     X'(0) = A_2 \omega = 0 \Rightarrow A_2 = 0
    X'(\pi) = -A_1\omega \sin \omega \pi = 0 \Rightarrow \omega = n, \text{for } n = 0,1,\dots
    I guess I made a mistake some where earlier. Thank you for your help.
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  9. #9
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    What I might suggest is to use

    u = e^{-t} v(x,t)

    to transform to the usual heat equation and then deal with the BC's
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  10. #10
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    I think I had to do it the way I was doing it. Not sure if they would care if I tried it your way though.
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  11. #11
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    You're welcome for whatever help I could give you. Did you finish the problem?
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  12. #12
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    Okay, so I figured this problem out a while back but I just didn't post back. However, for completeness I think I should post the procedure to solve this non-homogeneous PDE.
    \v_t - v_{xx} + v = f(x) 0 < x < 1 t>0
    u_x(0,t) = u_x(\pi,t) = 0
    u(x,0) = g(x)
    First we solve the DEQ
    Lv = u_{xx} = -\lambda
    Once we have a solution to this DEQ, we need to normalize the eigenfunctions. Now that we have the normalized eigenfunctions, \phi_n(x) we take the innerproduct
    \left<v_t,\phi_n(x)\right> = \left<v_{xx} - v + f(x),\phi_n(x)\right>.
    Here we must remember that v_n(t) = \left<v,\phi_n(x)\right>. Thus we have
    \int_0^1\! v_t \phi_n\, dt = \int_0^1\! u_{xx} \phi_n\, \dx - \int_0^1\! v \phi_n\, \dx + \int_0^1\! f \phi_n\, \dx
    \frac{d}{dt} \int_0^1\! v \phi_n \, dt = \left.v_x \phi_n\right|_0^1 - \int_0^1\! v_x \left(\phi_n\right)_x\, \dx - v_n + f_n
    \left(v_n\right)' = -\left.v \left(\phi_n\right)_x\right|_0^1 + \int_0^1\! v \left(\phi_n\right)_{xx}\, \dx - v_n + f_n
    \left(v_n\right)' = \int_0^1\! v \left(-\lambda \cdot \phi_n\right)\, \dx - v_n + f_n
    \left(v_n\right)' = \left(-\lambda - 1\right)v_n + f_n
    \left(v_n\right)' + \left(\lambda + 1\right)v_n= f_n
    \frac{d}{dt}\left[e^{\left(\lambda + 1\right)t}v_n\right] = e^{\left(\lambda + 1\right)t}f_n
    \int_0^t\! \frac{d}{ds}\left[e^{\left(\lambda + 1\right)t}v_n\right]\, ds = \int_0^t\! e^{\left(\lambda + 1\right)s}f_n\, ds
    e^{\left(\lambda + 1\right)t}v_n(t) - v_n(0) = \int_0^t\! e^{\left(\lambda + 1\right)s}f_n\, ds
    v_n(t) = e^{-\left(\lambda + 1\right)t}v_n(0) + \int_0^t\! e^{\left(\lambda + 1\right)\left(s - t\right)}f_n\, ds.
    Then of course, we sum up all v_n.
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