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Thread: Non-Homogeneous PDE

  1. #1
    Junior Member
    Sep 2008

    Non-Homogeneous PDE

    Use the method of eigenfunction expansion to solve u_t = u_{xx} + e^{-t} \sin (3x) with boundary conditions  u(0,t) = u(\pi,t) = 0 and initial condition u(x,0) = f(x)

    I think I might have stuffed my solution here, but I don't know.
    Would someone please be able to help me out?

    Base a trial solution on the solution to the homogeneous problem

    \displaystyle u_t = u_{xx} \ \Rightarrow \ u(x,t) = \sum_1^{\infty} b_n \sin(\frac{n \pi x}{L})

    \displaystyle \frac{\partial u}{\partial t} = \sum_1^{\infty} \dot{b_n(t)} \sin (\frac{n \pi x}{L}) , \frac{\partial u}{\partial x} = \sum_1^{\infty} b_n(t) \frac{n \pi}{L} \cos (\frac{n \pi x}{L}) , \frac{\partial^2 u}{\partial x^2} = - \sum_1^{\infty} b_n(t) \frac{n^2 \pi^2}{L^2} \sin (\frac{n \pi x}{L})

    Now represent q(x,t) as a series of the particular eigenfunctions


    \displaystyle q(x,t) = \sum_1^{\infty} c_n (t) \sin (\frac{n \pi x}{L})

    i.e. Let \displaystyle c_n(t) = \frac{2}{L} \int_0^L \sin (\frac{n \pi x}{L}) q(x,t) \ dx)

    Sub into PDE \displaystyle u_t = u_{xx} + q(x,t)

    \displaystyle  \sum_1^{\infty} \dot{b_n(t)} \sin (\frac{n \pi x}{L})  =  - \sum_1^{\infty} b_n(t) \frac{n^2 \pi^2}{L^2} \sin (\frac{n \pi x}{L}) + \sum_1^{\infty} c_n (t) \sin (\frac{n \pi x}{L})

    \displaystyle \sum_1^{\infty} \big( \dot{b_n(t)} + b_n(t) \frac{n^2 \pi^2}{L^2} - c_n(t) \big) \sin (\frac{n \pi x}{L}) = 0

    The only way for this to be true for all x is if

    \displaystyle \dot{b_n}(t) +b_n(t) \frac{n^2 \pi^2}{L^2} = c_n(t)

    This is a first order use an integrating factor

    \displaystyle e^{\int \frac{n^2 \pi^2}{L^2} dt} = e ^{\frac{n^2 \pi^2}{L^2}t}

    \displaystyle e^{k_n t} where \displaystyle k_n = \frac{n^2 \pi^2}{L^2}

    \displaystyle \dot{b_n} + k_n b_n = c_n

    Multiply by e^{k_n t} \ \Rightarrow

    \displaystyle \dot{b_n}e^{k_n t} + k_n e^{k_n t} b_n = c_n e^{k_n t}

    \displaystyle \frac{d}{dt} \Big( b_n e^{k_n t} \Big) = c_n e^{k_n t}

    \displaystyle \int_0^t \frac{d}{d \tau} \Big( b_n(\tau) e^{k_n \tau} \Big) \ d \tau = \int_0^t c_n (\tau) e^{k_n \tau} \ d \tau

    \displaystyle b_n(t) e^{k_n t} - b_n(0) = \int_0^t c_n (\tau) e^{k_n \tau} \ d \tau

    In order for u(x,0) = f(x) choose

    \displaystyle b_n(0) = \frac{2}{L} \int_0^L \sin \Big( \frac{n \pi x}{L} \Big) f(x) dx

    Skipping a few lines of calculation...

    \displaystyle u(x,t) =  \sum_1^{\infty} b_n(0) \sin \Big( \frac{n \pi x}{L}\Big) e^{-k_n t} + \sum_1^{\infty} \int_0^t c_n(\tau) e^{- k_n (t - \tau)} \ d \tau  \sin \Big( \frac{n \pi x}{L} \Big)
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  2. #2
    MHF Contributor
    Jester's Avatar
    Dec 2008
    Conway AR
    I think what you have here is OK but an overkill. First off, you interval is [0,\pi] with the BCs u(0,t) = u(\pi,t) = 0 so your eigenfunction expansion should be

     u(x,t) = \displaystyle \sum_1^{\infty} b_n \sin( n x). Next when you sub. into your PDE you obtain

    \displaystyle \sum_1^{\infty} \dot{b_n} \sin( n x) = \sum_1^{\infty} - n^2 b_n \sin( n x) + e^{-t} \sin 3 x

    which gives

    \dot{b_n} = -n^2 b_n \;\text{for}\; n \ne 3<br />

    <br />
\dot{b_n} = -n^2 b_n + e^{-t} \; \text{for}\; n = 3<br />

    Each of which you can solve separately. This should get you started.
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