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Characteristic method + Jacobian

  1. Sep 16, 2007 #1


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    If I am using the method of characteristics to solve a PDE [tex]\Psi(x,t)[/tex] (first order, semi-linear), and after using the method of characteristics I find that the Jacobian
    [tex]|\frac{\partial{(x,t)}}{\partial{(\sigma,\eta)}}| = 0[/tex]
    (where [tex]\sigma[/tex] and [tex]\eta[/tex] are parameters for the curve) does this imply that no solution exists, or just that the solution is not unique? I could not find this in my textbook.
  2. jcsd
  3. Sep 17, 2007 #2
    Could you post the problem please?
  4. Sep 17, 2007 #3


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    [tex]x\frac{\partial\psi}{\partial{x}} + t\frac{\partial\psi}{\partial{t}} + (x + 2t)\psi = 3xt \textnormal{ With intitial Condition:} \psi (x,4x)=x[/tex]

    This leads me to:

    x(\sigma,\eta) = \eta e^{\sigma}[/tex]
    t(\sigma,\eta) = 4\eta e^{\sigma}[/tex]

    (I can fill in those steps if necessary too)
  5. Sep 18, 2007 #4
    No, that wont be necesary. The problem is that your initial contidion is over a characteristic, and you can't construct a solution. Remember that the method of characteristics uses the initial condition as curve of departure (parametrized by [itex]\eta[/itex]) and then tries to move further from it as far as the ode system allows it (parametrized by [itex]\sigma[/itex]). In your case, this can't be done, as the characteristic is colinear with the initial condition, wich means the normal vector for the parametrized surface is zero (so the surface is not well defined).

    Let me give it a little more tought, and i'll get back to you.
    Last edited: Sep 18, 2007
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