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Thread: 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.


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

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

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

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

    But this doesn't seem to help me solve for $\displaystyle \omega$. Normally I would solve for one constant and then solve for \omega, plug this into $\displaystyle v(x,t)$ 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 $\displaystyle v$ was

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

    $\displaystyle v_{t}=v_{xx}-v-2\sin(x).$
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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.

    $\displaystyle v(t,x) = -\sin(x)$
    Is a solution to the PDE. So let $\displaystyle X(x) = \sin x \text{ and } T(t) = -1$.
    $\displaystyle \Rightarrow \left(-\sin x\right)_t = \left(-\sin x\right)_{xx} - 2\sin x$
    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
    $\displaystyle \dfrac{X''(x)}{X(x)} = \dfrac{T'(t)}{T(t)} = -\lambda$
    Now I would solve the DEQ
    $\displaystyle X''(x) = -\lambda X(x)$.
    From here I have cases
    Case I: $\displaystyle \lambda > 0$
    Let $\displaystyle \omega = \sqrt{\lambda}$
    $\displaystyle X(x) = A_1 \cos \omega x + A_2 \sin \omega x$
    $\displaystyle X'(x) = -A_1\omega \sin \omega x + A_2 \omega \cos \omega x$
    $\displaystyle X'(0) = A_2 \omega = 0 \Rightarrow A_2 = 0$
    $\displaystyle X'(\pi) = -A_1\omega \sin \omega \pi = 0 \Rightarrow \omega = n, \text{for } n = 0,1,\dots$
    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

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

    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.
    $\displaystyle \v_t - v_{xx} + v = f(x)$ $\displaystyle 0 < x < 1$ $\displaystyle t>0$
    $\displaystyle u_x(0,t) = u_x(\pi,t) = 0$
    $\displaystyle u(x,0) = g(x)$
    First we solve the DEQ
    $\displaystyle Lv = u_{xx} = -\lambda$
    Once we have a solution to this DEQ, we need to normalize the eigenfunctions. Now that we have the normalized eigenfunctions, $\displaystyle \phi_n(x)$ we take the innerproduct
    $\displaystyle \left<v_t,\phi_n(x)\right> = \left<v_{xx} - v + f(x),\phi_n(x)\right>$.
    Here we must remember that $\displaystyle v_n(t) = \left<v,\phi_n(x)\right>$. Thus we have
    $\displaystyle \int_0^1\! v_t \phi_n\, dt = \int_0^1\! u_{xx} \phi_n\, \dx - \int_0^1\! v \phi_n\, \dx + \int_0^1\! f \phi_n\, \dx$
    $\displaystyle \frac{d}{dt} \int_0^1\! v \phi_n \, dt = \left.v_x \phi_n\right|_0^1 - \int_0^1\! v_x \left(\phi_n\right)_x\, \dx - v_n + f_n$
    $\displaystyle \left(v_n\right)' = -\left.v \left(\phi_n\right)_x\right|_0^1 + \int_0^1\! v \left(\phi_n\right)_{xx}\, \dx - v_n + f_n$
    $\displaystyle \left(v_n\right)' = \int_0^1\! v \left(-\lambda \cdot \phi_n\right)\, \dx - v_n + f_n$
    $\displaystyle \left(v_n\right)' = \left(-\lambda - 1\right)v_n + f_n$
    $\displaystyle \left(v_n\right)' + \left(\lambda + 1\right)v_n= f_n$
    $\displaystyle \frac{d}{dt}\left[e^{\left(\lambda + 1\right)t}v_n\right] = e^{\left(\lambda + 1\right)t}f_n$
    $\displaystyle \int_0^t\! \frac{d}{ds}\left[e^{\left(\lambda + 1\right)t}v_n\right]\, ds = \int_0^t\! e^{\left(\lambda + 1\right)s}f_n\, ds$
    $\displaystyle e^{\left(\lambda + 1\right)t}v_n(t) - v_n(0) = \int_0^t\! e^{\left(\lambda + 1\right)s}f_n\, ds$
    $\displaystyle v_n(t) = e^{-\left(\lambda + 1\right)t}v_n(0) + \int_0^t\! e^{\left(\lambda + 1\right)\left(s - t\right)}f_n\, ds$.
    Then of course, we sum up all $\displaystyle v_n$.
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