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1-dimensional heat equation

  1. May 7, 2013 #1

    u_t=u_xx, x is in [0,1], t>0

    u(0,t)=u(1,t)=0, t>0
    u(x,0)=sin(pi*x)-sin(3*pi*x), x is in (0,1)

    i think its solution is of the form

    u(x,t)=sigma(n=1 to infinity){a_n*sin(n*pi*x)*exp(-n^2*pi^2*t)

    where a_n=2*integral(0 to 1){ (sin(pi*x)-sin(3*pi*x)) * sin(n*pi*x) }

    but i have a_n = 0, for all n..

    i don't know where is my mistake..
  2. jcsd
  3. May 7, 2013 #2
    Can you write your initial condition as a sin series?

    [itex]u(x,0)=\sum b_n sin(n \pi x) [/itex]

    How does this compare to your sin series for


    Can you relate [itex]a_n[/itex] to [itex]b_n[/itex]?
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