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Homework Help: Heat equation - separation of variables

  1. Jul 24, 2010 #1
    1. The problem statement, all variables and given/known data

    du/dt=d2u/dx2, u(0,t)=0, u(pi,t)=0

    u(x,0) = sin^2(x) 0<x<pi

    Find the solution

    Also find the solution to the initial condition:

    du/dt u(x,0) = sin^2(x) 0<x<pi

    3. The attempt at a solution

    From separation of variables I obtain

    u(x,t) = B.e^(-L^2t).sin(Lx)

    For the boundary condition u(pi,t)=0, u(x,t) = Bm . e^(-m^2t) . sin(mx)

    Finally for u(x,0) = sin^2x

    u(x,0) = 4(cos(m.pi)-1).e^(-m^2t).sinmx / pi(m^3-4m)
    found through superposition and fourier sum of sines

    What I dont see is how to solve the initial condition because when I diff. above wrt t and set t-0 I obtain

    du/dt u(x,0) = -m^2 4(cos(m.pi)-1).e^(-m^2t).sinmx / pi(m^3-4m) = sin^2x

    How can this be set to sin^2x by choosing a value of m?
  2. jcsd
  3. Jul 24, 2010 #2


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    I'm not sure what you are talking about here. You understand that these are two different problems, don't you? You solution "for u(x,0) = sin^2x

    u(x,0) = 4(cos(m.pi)-1).e^(-m^2t).sinmx / pi(m^3-4m)" is a sum isn't it?

    That is
    [tex]u(x, t)= \sum_{m=1}^\infty e^{-m^2t}(4(cos(m.pi)-1).e^(-m^2t).sinmx / pi(m^3-4m))[/tex].

    Now, with initial condition [tex]u_t(x, 0)= sin^2(x)[/tex] you must have
    [tex]u_t(x, 0)= \sum_{n=1}^\infty -m^2 B_m sin(mx)= sin^2(x)[/tex]

    Find Fourier sine coefficients for [itex]sin^2(x)[/itex] and solve for [itex]B_m[/itex].
  4. Jul 24, 2010 #3


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    Just to be clear, separation of variables gives you

    [tex]u(x,t) = e^{-L^2t}(A\cos Lx + B\sin Lx)[/tex]

    Then applying the boundary condition u(0,t)=0 gives you A=0, and applying the condition u(π,t)=0 requires L be an integer.
  5. Jul 24, 2010 #4
    Understood now. Thanks a lot for clearing that up
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