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Solution satisfying initial conditions for a pde of second order

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    I have found the general solution to a second order pde to be
    U(x,t) = f(3x + t) + g(-x + t) where f and g are arbitrary functions

    I have initial conditions
    U(x,0) = sin(x)
    Du/dt (x,0) = cos (2x)

    3. The attempt at a solution

    I have found that
    U(x,0) = f(3x) + g(-x) = sin(x)
    Du/dt(x,0) = f'(3x) + g'(-x) = cos(2x)

    From this point im not sure what to do. I have tried differentiating u(x,0) with respect to x which i think gives me
    Du/dx (x,0) = 3f'(3x) - g'(-x) = cos (x)
    I thought i would then equate the equations but im not actually sure ths helps.

    Could someone please point me in the right direction for this question

  2. jcsd
  3. Oct 24, 2011 #2
    Ok, didn't get it that other time huh? Then I'll work one through and you do the other one ok.


    I differentiate the top one with respect to x:


    Now add them:


    To get rid of that 3x in there, let u=3x or x=1/3u then:


    Then substitute that into that DE with 3x in there:

    [tex]12 \frac{df}{du}=2\cos(1/3u)[/tex]

    integrate it:


    so that:

    [tex]f(3x+t)=1/2 \sin((3x+t)/3)[/tex]

    I believe that's right anyway. Would need to back-substitute it into the PDE to check it.

    Ok, now you do the other one.
    Last edited: Oct 24, 2011
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