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Clever coordinate substitution for linear PDE

  1. Oct 17, 2012 #1
    Hi! I am currently working with a linear PDE on the form

    [itex]\frac{\partial f}{\partial t} + A(v^2 - v_r^2)\frac{\partial f}{\partial \phi} + B\cos(\phi)\frac{\partial f}{\partial v} = 0[/itex].

    [itex]A[/itex] and [itex]B[/itex] are constants. I wish to find a clever coordinate substitution that simplifies, or maybe even solves the problem. So far I have tried with an action-angle approach that reduces the dimensionality of the problem. I first assume a general substitution [itex]\lbrace\phi, v\rbrace \rightarrow \lbrace r(\phi, v), s(\phi, v)\rbrace[/itex], and choose

    [itex]r(\phi, v) = A\left(\frac{1}{3}v^3 - v_r^2 v\right) - B\sin(\phi)[/itex].

    This choice then cancels all the [itex]\partial/\partial r[/itex] terms, and one is left with

    [itex]\frac{\partial f}{\partial t} + \left[A(v^2 - v_r^2)\frac{\partial s}{\partial\phi} + B\cos(\phi)\frac{\partial s}{\partial v}\right]\frac{\partial f}{\partial s} = 0[/itex].

    Now I am left with the choice of [itex]s(\phi, v)[/itex]. Is there maybe a way to choose [itex]s[/itex] such that the expression in front of [itex]\partial f/\partial s[/itex] becomes constant? Am I even on the right track, or are there much more clever ways to solve the problem?

    Thanks in advance,
    Simon
     
  2. jcsd
  3. Oct 17, 2012 #2
    Sorry, I forgort to mention also that [itex]v_r[/itex] is constant.
     
  4. Oct 17, 2012 #3

    Mute

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    You basically need to solve the PDE

    $$A(v^2 - v_r^2)\frac{\partial s}{\partial\phi} + B\cos(\phi)\frac{\partial s}{\partial v} = \mbox{const}.$$

    Note that, however, solving this equation is the same as solving the time-independent version of your original equation. e.g., let ##f(t,v,\phi) = g(v,\phi)\exp(\lambda t)##; your original equation will reduce to the same equation you would need to solve to get ##s(v,\phi)##.
     
  5. Oct 19, 2012 #4
    Ah, yes, I see that.. =P But I think I actually have found a way to solve the problem. First to separate out the time dependence as you suggested. Then use the coordinate substitution ##r(\phi, v)## as I wrote in the first post, and simply put ##s(\phi, v) = v##. Then of course ##B\cos(\phi)## has to be rewritten in terms of ##r## and ##v##. I end up with an ODE that is solveable. The final result I got was

    $$f(t, \phi, v) = f_0(r(\phi, v))\exp\left(\lambda\left[t \pm \int_0^v{\frac{d u}{\sqrt{B^2 - \left[A\left(\frac{u^3 - v^3}{3} - v_r(u - v)\right) + B\sin(\phi)\right]^2}}}\right]\right),$$

    where ##\lambda## is a constant, ##f_0## is an arbitrary function of ##r(\phi,v)##, and the ##\pm## depends on ##\phi## (minus if ##|\phi| < \pi/2##). Well, the integral doesn't have an analytical solution as far as I know, and Mathematica didn't think so either, so I guess this is as far as one can get. And since the equation is linear one can add an arbitrary sum of these solutions as well. Anyway, thanks for the help Mute. :)
     
  6. Oct 20, 2012 #5
    Maybe I'm missing something, but it seems to me this equation can be solved using the method of characteristics.
     
  7. Oct 21, 2012 #6
    Yes, it was actually the method that I used to solve it (before I knew there was a name for the method =P). The substitution

    $$r(\phi, v) = A\left(\frac{1}{3}v^3 - v_r^2 v\right) - B\sin(\phi)$$
    is the characteristic of the ##\phi, v##-differential equation.
     
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