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Heat equation Fourier Series

  1. Jul 5, 2008 #1
    I'm trying to solve a basic heat equation [tex]\frac{\partial T}{\partial t}=\alpha \frac{\partial^2 T}{\partial x^2}[/tex] I manage to get [tex]T=X(x)\tau(t)[/tex]
    Then [tex]\tau(t)=A*e^{-\alpha*\lambda^2*t} and X(x)=C*sin(\lambda x) where \lambda=\pi/Ln n=1,2,3,...[/tex]

    From here I don't know how or why I get to a Fourier Serie. Like this [tex] u(t,x) = \sum_{n = 1}^{+\infty} D_n \left(\sin \frac{n\pi x}{L}\right) e^{-\frac{n^2 \pi^2 kt}{L^2}} \\ and \\ D_n = \frac{2}{L} \int_0^L f(x) \sin \frac{n\pi x}{L} \, dx. [/tex]

    Can someone explain why I have this Fourier Serie?

    Thanks Link
     
  2. jcsd
  3. Jul 5, 2008 #2

    Hootenanny

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    From here, recall that when you seperated the variables you stated that:
    Which leads to:

    [tex]T = u_n(x,t) = A_n*e^{-\alpha\left(\frac{n\pi}{L}\right)^2 t}\left(C_n*sin(\frac{n\pi}{L} x)\right)[/tex]

    This is the general solution for each value of [itex]n\in\mathbb{Z}^+[/itex]. However, the general solution to the heat equation is a superposition of all possible modes (or values of n). In otherwords, to find the general solution to the original PDE you need to sum over all possible n:

    [tex]u(x,t) = \sum^{\infty}_{n=1}\left\{ A_n*e^{-\alpha\left(\frac{n\pi}{L}\right)^2 t}\left[C_n*sin\left(\frac{n\pi}{L} x\right)\right]\right\}
    [/tex]

    Combining the coefficients such that [itex]A_n\cdot C_n = D_n[/itex]:

    [tex]u(x,t) = \sum^{\infty}_{n=1}D_ne^{-\alpha\left(\frac{n\pi}{L}\right)^2 t}\sin\left(\frac{n\pi}{L} x\right)
    [/tex]

    In general the Fourier series of a function my be written as the sum of sines and cosines, but since your function is odd (i.e. only has sines and exponents) the coefficients of the cosine component are identically equal to zero. Therefore, you are simply left with the coefficients of the sine component which are given by Euler's formula:

    [tex]b_n = \frac{2}{L}\int_0^Lf(x)\sin\left(\frac{2n\pi}{T}x\right)[/tex]

    Where T is the period, which in this case is 2L. Hence:

    [tex]D_n = \frac{2}{L}\int_0^Lf(x)\sin\left(\frac{n\pi}{L}x\right)[/tex]

    I hope this helps.
     
    Last edited: Jul 5, 2008
  4. Jul 6, 2008 #3
    Thanks a lot Hootenanny.
     
  5. Jul 7, 2008 #4
    I'm assuming f(x)=u(0,x) which is the initial condition of the PDE. So the solution u(t,x) must satsify IC (well I'm more used to u(x,t) but what the heck...)

    [tex]
    u(t,x) = \sum_{n = 1}^{+\infty} D_n \left(\sin \frac{n\pi x}{L}\right) e^{-\frac{n^2 \pi^2 kt}{L^2}}
    [/tex]
    so by setting t=0, you got
    [tex]
    u(0,x) = f(x)=\sum_{n = 1}^{+\infty} D_n \left(\sin \frac{n\pi x}{L}\right) }
    [/tex]
    Multiply both sides by sin(mPI/L)
    [tex]
    f(x) \sin \frac{m\pi x}{L}=\sum_{n = 1}^{+\infty} D_n \left(\sin \frac{m\pi x}{L} \sin \frac{n\pi x}{L}\right) }
    [/tex]
    perform a term by term integration on RHS over 0=<x=<L
    [tex]
    \int _{0}^{L}f(x) \sin \frac{m\pi x}{L}dx= \sum_{n = 1}^{+\infty} \int _{0}^{L}D_n \left(\sin \frac{m\pi x}{L} \sin \frac{n\pi x}{L}\right) }dx
    [/tex]
    because
    [tex]
    \int _{0}^{L}\left(\sin \frac{m\pi x}{L} \sin \frac{n\pi x}{L}\right) }dx = 0 \quad when \quad m \neq n
    [/tex]
    and
    [tex]
    \int _{0}^{L}\sin ^{2} \left(\frac{m\pi x}{L}\right) }dx = \frac{L}{2} \quad when \quad m=n \neq0
    [/tex]
    It reduced to
    [tex]
    \int _{0}^{L}f(x) \sin \frac{m\pi x}{L}dx=D_{n}\frac{L}{2}
    [/tex]
    now replace m with n (because m=n)
    [tex]
    D_{n}=\frac{2}{L}\int _{0}^{L}f(x) \sin \frac{n\pi x}{L}dx
    [/tex]

    Hope that helps, at least partially.
     
    Last edited: Jul 7, 2008
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