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Homework Help: Heat equation in 2dim

  1. May 25, 2006 #1
    Im trying to solve the heat equation in 2dim on a plate.
    0=<x=<L, 0=<y=<L. With homogenous dirichlet conditions on the boundary and the initial condition:
    T(x,y)=T0sin(pi*x/L)sin(pi*y/L)

    With separation of variables i get the solution
    [tex]
    T(x,y,t)=\sum_{m=0}^\infty\sum_{n=0}^\infty B_{mn}*exp[\frac{-(m^2+n^2)\pi^2kt}{L^2}]*sin[\frac{m\pi x}{L}]*sin[\frac{m\pi y}{L}]
    [/tex]

    m,n integers and k the constant from the heat equation.

    Now the initial condition determine the constants B_mn

    [tex]
    B_{mn} = \frac{4T_0}{L^2}*\int_{0}^{L}\int_{0}^{L} sin[\frac{\pi x}{L}]*sin[\frac{\pi y}{L}] sin[\frac{m\pi x}{L}]*sin[\frac{n\pi y}{L}] dx dy
    [/tex]

    But an integral like
    [tex]
    \int_{0}^{L}sin\frac{\pi x}{L}sin\frac{m\pi x}{L}dx
    [/tex]

    is zero for m not equal to 1.
    So m=1 and n=1...???
    and with this i dont get a fourier series as the solution...
    What am i doing wrong ?
     
    Last edited: May 25, 2006
  2. jcsd
  3. May 25, 2006 #2

    HallsofIvy

    User Avatar
    Science Advisor

    "Bmn= 0" is not an initial condition! Your initial condition is that
    [tex]T(x,y,0)=\sum_{n=0}^\infty B_{mn}sin[\frac{m\pi x}{L}]*sin[\frac{m\pi y}{L}]= T_0sin(\frac{\pi x}{L})sin(\frac{pi y}{L})[/tex]

    Now, apply the Fourier integral to both sides of that.
     
  4. May 25, 2006 #3
    I never said that Bmn= 0 is an initial condition, i said that the integral is zero for m NOT equal to one, because of the orthogonality of sin-functions.

    Im not sure what you mean by applying the Fourier integral, i have only used that with Fourier transforms.

    This is how i get the expression for Bmn:
    I have the general solution

    [tex]T(x,y,t)=\sum_{m=0}^\infty\sum_{n=0}^\infty B_{mn}*exp[\frac{-(m^2+n^2)\pi^2kt}{L^2}]*sin[\frac{m\pi x}{L}]*sin[\frac{n\pi y}{L}][/tex]

    Then i expand the initial condition in the eigenfunctions

    [tex](sin[\frac{m\pi x}{L}]*sin[\frac{n\pi y}{L}])_{m,n=1}^{\infty}=\Psi
    [/tex]

    The initial condition is

    [tex]f(x,y)=T_0sin(\frac{\pi x}{L})sin(\frac{\pi y}{L})[/tex]

    [tex]T(x,y,0)=\sum_{n=0}^\infty\sum_{m=0}^\infty B_{mn}sin[\frac{m\pi x}{L}]*sin[\frac{m\pi y}{L}]= f(x,y)[/tex]

    I multiply with the eigenfunctions and integrate over the domin to get an expression for Bmn:

    [tex]B_{mn}=\frac{(f,\Psi)}{(\Psi,\Psi)}[/tex]

    where the normalisation integral is

    [tex](\Psi,\Psi) = \int_{0}^{L}\int_{0}^{L}sin^2[\frac{m\pi x}{L}]*sin^2[\frac{n\pi y}{L}] dxdy=L^2/4[/tex]

    and

    [tex](f,\Psi) = T_0*\int_{0}^{L}\int_{0}^{L} sin[\frac{\pi x}{L}]*sin[\frac{\pi y}{L}]*sin[\frac{m\pi x}{L}]*sin[\frac{n\pi y}{L}]dxdy[/tex]

    this integral is

    [tex](f,\Psi) = T_0\int_{0}^{L}sin[\frac{\pi x}{L}]*sin[\frac{m\pi x}{L}]dx\int_{0}^{L} sin[\frac{\pi y}{L}]*sin[\frac{n\pi y}{L}]dy[/tex]

    but for m not equal to 1

    [tex]\int_{0}^{L}sin[\frac{\pi x}{L}]*sin[\frac{m\pi x}{L}]dx=\frac{1}{2}\int_{0}^{L}cos[\frac{(1-m) \pi x}{L}]dx+\int_{0}^{L}cos[\frac{(1+m)\pi x}{L}]dx=\frac{Lm*sin(m\pi)}{\pi-m^2\pi}=0[/tex]

    and the same for the integral over y.
    Therefor m and n has to be 1 and i dont get a series solution!

    Im probably doing some silly misstakes, can you find them?
     
    Last edited: May 26, 2006
  5. May 31, 2006 #4
    hmm, i was probably right from the beginning.
    I got confused because i had never solved a problem before where this happens.

    Is it like this? :
    When you have a certain "frequency" as the initial condition you cant get a solution for later times as a superposition over other frequencies.
    Thats why in this case only the terms with m=1,n=1 survived in the sum.
     
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