1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Heat equation in the first quadrant.

  1. Dec 21, 2011 #1

    A_B

    User Avatar

    1. The problem statement, all variables and given/known data
    Solve the heat equation [itex]u_t=u_{xx}+u_{yy}[/itex] fot [itex]t>0[/itex] in the first quadrant of [itex]\mathbb{R}^2[/itex]. The boundary conditions are [itex]u(0,y,t)=u(x,0,t)=0[/itex] and the initial temperature distribution is
    [tex]
    f(x,y)=
    \begin{cases}
    1 \;\;\;\; \text{in the square } \; 0<x<1; \; 0<y<1 \\
    0 \;\;\;\; \text{elsewhere}
    \end{cases}
    [/tex]


    3. The attempt at a solution
    I can solve problems in finite domains, ie where enough boundary conditions are given to determine the countable infinite set of eigenvalues. This problem is on a semi-infinite domain, with insufficient boundary conditions to do so. I have come to understand that to solve such a problem, one replaces "the sum over eigenvalues" with an integral of the solutions obtained by separation of variables, over the (roots of the) eigenvalues.

    This is where I got (detailed derivations not typed out)

    Separation of variables (solutions of the form [itex]u=X(x)Y(y)T(t)[/itex] gives ODEs
    [tex]
    \begin{cases}
    -\ddot{X} = \lambda_1 \;\;\;\;\;\;\;\;\;\;\;\; X(0)=0\\
    -\ddot{Y} = \lambda_2 \;\;\;\;\;\;\;\;\;\;\;\; Y(0)=0\\
    \dot{T}=-(\lambda_1+\lambda_2)T
    \end{cases}
    [/tex]

    The "eigenvalues" can be shown to be all positive. So let [itex]\lambda_1 = \gamma_1^2[/itex] and [itex]\lambda_2 = \gamma_2^2[/itex]. With [itex]\gamma_1[/itex] and [itex]\gamma_2[/itex] both positive.

    The solutions are
    [tex]
    X(x)=\sin(\gamma_1 x)
    [/tex]
    [tex]
    Y(y)=\sin(\gamma_2 y)
    [/tex]
    [tex]
    T(t) = e^{-(\gamma_1^2 + \gamma_2^2)t}
    [/tex]

    So the solution for separated variables is
    [tex]
    u(x,y,t)=\sin(\gamma_1 x)\sin(\gamma_2 y) e^{-(\gamma_1^2 + \gamma_2^2)t}
    [/tex]

    Since there are solutions for every positive [itex]\gamma_1[/itex] and [itex]\gamma_2[/itex], we cant determine eigenvalues like in regular problems. We write the final solution as a weighed integral of the solution for separated variables, with respect to the gamma's.

    [tex]
    u(x,y,t) = \int_0^\infty\int_0^\infty d\gamma_1 d\gamma_2 \alpha(\gamma_1, \gamma_2) \sin(\gamma_1 x)\sin(\gamma_2 y) e^{-(\gamma_1^2 + \gamma_2^2)t}
    [/tex]


    The function [itex]\alpha(\gamma_1,\gamma_2)[/itex] is determined by the initial conditions:

    [tex]
    f(x,y) = u(x,y,0) = \int_0^\infty\int_0^\infty d\gamma_1 d\gamma_2 \alpha(\gamma_1, \gamma_2) \sin(\gamma_1 x)\sin(\gamma_2 y)
    [/tex]

    And here I'm stuck, how do I determine alpha?


    Any help is much appreciated,
    A_B
     
    Last edited: Dec 21, 2011
  2. jcsd
  3. Dec 24, 2011 #2
    Fourier transform ;)
     
  4. Dec 26, 2011 #3

    A_B

    User Avatar

    I solved it, the last formula expresses f(x,y) as a double Fourier sine transform of α(γ_1, γ_2). So taking the inverse Fourier sine transform of f(x,y) twice gives α.

    A_B
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Heat equation in the first quadrant.
  1. Heat Equation (Replies: 1)

  2. Heat equation (Replies: 2)

  3. The heat equation (Replies: 17)

Loading...