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Using initial conditions in a second order PDE

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data

    I have a PDE for which i have found the general solution to be u(x,y) = f1(3x + t) + f2(-x + t)
    where f1 and f2 are arbitrary functions. I have initial conditions u(x,0) = sin (x) and partial derivative du/dt (x,0) = cos (2x)


    2. Relevant equations

    u(x,y) = f1(3x + t) + f2(-x + t)
    u(x,0) = sin (x)
    du/dt (x,0) = cos (2x)

    3. The attempt at a solution

    I have substituted u(x,0) and du/dt (x,0) into the general solution which gives me;

    u(x,0) = f1(3x) + f2(-x) = sin (x)
    du/dt(x,0) = f1'(3x) + f2'(-x) = cos (2x)

    but i am unsure as to where to go from here

    Thanks for any help
     
  2. jcsd
  3. Oct 20, 2011 #2
    I'll use h and v:

    [tex]h'(3x)-v'(-x)=\cos(x)[/tex]
    [tex]h'(3x)+v'(-x)=\cos(2x)[/tex]

    It's not hard to eliminate v'(-x) right?

    then you'd have a regular DE:

    [tex]2h'(3x)+\cos(x)+\cos(2x)[/tex]

    but then you got that 3x in there. What about making a change of independent variable by letting u=3x then convert the DE from one in terms of x (the one above) into one involvind u by remembering:

    [tex]\frac{dh}{du}=\frac{dh}{dx}\frac{dx}{du}[/tex]

    now make this substitution into the one involving x, get the one involving u, solve it, then wherever there is a u in the answer, replace it by 3x.
     
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