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PDE, heat equation with mixed boundary conditions

  1. Nov 4, 2013 #1
    1. The problem statement, all variables and given/known data
    solve the heat equation over the interval [0,1] with the following initial data and mixed boundary conditions.


    2. Relevant equations
    [itex]\partial _{t}u=2\partial _{x}^{2}u[/itex]
    [itex]u(0,t)=0, \frac{\partial u}{\partial x}(1,t)=0[/itex]

    with B.C

    [itex]u(x,0)=f(x)[/itex]
    where f is piecewise with values:
    [itex]0, 0<x\leq \frac{1}{2}[/itex]
    [itex]3, \frac{1}{2}<x<1[/itex]




    3. The attempt at a solution


    after separation of variables where [itex]u(x,t)=h(x)\phi (t)[/itex]:

    [itex]h''(x)=-\frac{\lambda }{2}h(x)[/itex]
    [itex]\phi'(t)=-\lambda \phi(t)[/itex]

    gen. solution to h is
    [itex]h(x)=a\sin \sqrt{\frac{\lambda }{2}}[/itex] the constant with the cos term is 0 from initial value

    I'm stuck trying to find the eigenvalue
    [itex]h'(1)=\frac{\lambda }{2}a\cos\sqrt{\frac{\lambda }{2}}=0[/itex]
    [itex]\sqrt\frac{\lambda }{2}=\arccos 0[/itex]

    I'm not sure what expression with n to use for arccos of 0. npi/2 won't work, or (n+1)pi/2, is this the right procedure though?

    Any help is greatly appreciated!

    EDIT:

    I'm trying [itex]\frac{\pi }{2}+n\pi[/itex] now to solve for the eigenvalue
     
  2. jcsd
  3. Nov 4, 2013 #2
    So for the general solution of u I have [itex]u(x,t)=\sum_{n=1}^{\infty}A_{n}\sin [\frac{\pi}{2}(1+2n)x]\exp -2t[\frac{\pi}{2}(1+2n)]^2[/itex]

    and the coefficient A_n given by

    [itex]A_{n}=\frac{12}{\pi(1+2n)}\cos \pi(1+2n)[/itex]

    There was another cosine term with pi/2 in the argument that was always zero for any n, but is this an acceptable way to leave the coefficient expression?
     
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