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PDE by Laplace Transform

  1. May 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Use the Laplace Transform to solve the PDE for u(x,t) with x>0 and t>0:
    x(du/dx) + du/dt = xt
    with IC: u(x,0) = 0 and BC: u(0,t) = 0

    2. Relevant equations

    3. The attempt at a solution

    After taking LT of the PDE wrt t, the PDE becomes
    x(dU/dx) + sU = x/(s2)

    Integrating factor :
    I = exp([tex]\int(s/x)dx[/tex]) = xs

    ODE becomes
    d/dx(Uxs) = xs/s2

    Integrating both sides:
    U = x/(s3+s) + A(s)/xs

    then I don't know how to find A(s), if I use BC, the factor 1/0 will come out...or is there some other way to calculate the PDE with LT?

    thanks
     
  2. jcsd
  3. May 22, 2009 #2

    gabbagabbahey

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    Unless A(s)=0 :wink:
     
  4. May 23, 2009 #3
    Thanks!
     
  5. May 24, 2009 #4
    so U(x,s) = x/(s3+s2)

    but then I don't know how to do the inverse LT to get u(x,t) such that it fits the PDE...

    thanks
     
  6. May 24, 2009 #5

    gabbagabbahey

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    Just use partial fractions:

    [tex]\frac{1}{s^3+s}=\frac{1}{s(s^2+1)}=\frac{A}{s}+\frac{Bs}{s^2+1}+\frac{C}{s^2+1}[/tex]

    Solve for A,B and C
     
  7. May 24, 2009 #6
    I finally solved it! Thanks very much!
     
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