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Partial differential equation with conditions

  1. Oct 26, 2012 #1
    I'm not sure how to solve this:

    du/dt = 3 [itex]\frac{d^{2}u}{dx^{2}}[/itex]

    These are the conditions:
    u(0,t)= -1
    u(pi,t)= 1
    u(x,0) = -cos 7x

    Suggestion:
    I should use steady state solution to get a homogeneous initial condition.

    Starting with separtion of variables

    u(x,t) = G(x)H(t)

    And du/dt = GH' and d^2u/dx^2 = G"H

    So GH'= c^2 G"H

    and H'/(c^2)H = G"/G

    Then H'/(c^2)H = k = G"/G

    thus H' - (c^2)Hk = 0 and G" - Gk = 0.

    Once here how do I make sure u(x,t) follows the conditions?
     
  2. jcsd
  3. Oct 26, 2012 #2

    LCKurtz

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    Science Advisor
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    Gold Member

    I wouldn't start it that way. You need to make the boundary conditions homogeneous before you separate the variables. I would start with the substitution$$
    u(x,t) = v(x,t) + \Psi(x)$$ Substituting that into your system gives$$
    v_t = 3v_{xx}+3\Psi''(x)$$ $$
    v(0,t) + \Psi(0)= -1$$ $$
    v(\pi, t) +\Psi(\pi) = 1$$ $$
    v(x,0) + \Psi(x) = -\cos x$$
    Now if you let ##3\Psi''(x) = 0,\, \Psi(0) = -1,\, \Psi(\pi)= 1##, you can solve for ##\Psi(x)## and you are left with the homogeneous system$$
    v_t = 3v_{xx}$$ $$
    v(0,t) = 0$$ $$
    v(\pi, t) = 0$$ $$
    v(x,0) = -\cos x-\Psi(x)$$Solve this system in the usual way with separation of variables. Once you know ##v(x,t)##, your solution will be ##u(x,t)=v(x,t)+\Psi(x)##.
     
    Last edited: Oct 26, 2012
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