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2nd order linear hyperbolic PDE?

  1. Oct 5, 2009 #1
    uxx - x2 uyy = 0 (assume x>0)

    Is there any systematic method (e.g. change of variables) to solve this hyperbolic equation?

    dy/dx = [B + sqrt(B2 - AC)]/A
    => dy/dx = x
    => 2y -x2 = c

    dy/dx = [B - sqrt(B2 - AC)]/A
    => dy/dx = -x
    => 2y + x2 = k

    So the characteristic curves are 2y -x2 = c and 2y + x2 = k

    Now does this imply that the general solution is u = f(2y-x2)+g(2y+x2) ?

  2. jcsd
  3. Oct 5, 2009 #2
    u = f(2y-x2)+g(2y+x2)

    Is uxx - x2 uyy identically zero? Then it must be a general solution :wink:
  4. Oct 6, 2009 #3
    Hi, I tried to verify it by computing a bunch of tedious second derivatives, but somehow I didn't get 0... I may have made a mistake in my calculations...
  5. Oct 6, 2009 #4
    The most direct way to solve this PDE is the Fourier method, that is, supposing that solution has the form

    u(x,y)=\int_{-\infty}^{\infty}F(x,k)exp(iyk) dk

    we obtain ODE for F

    F_{xx}+x^2k^2F=0 ,

    which solution is as follows

    F(x,k)=x^{1/2}[F1(k)J_{1/4}(x^2k/2)+F2(k)K_{1/4}(x^2k/2)] ,

    where J_{1/4} and K_{1/4} are modified Bessel functions; F1 and F2 are arbitrary functions.
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