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1st order PDE, quadratic in derivatives, two variables analytic solution?

  1. Sep 30, 2011 #1
    I have the PDE:

    (v_r)^2+(v_z)^2=p^2 where v=v(r,z), p=p(r,z).

    I have some boundary conditions, of sorts:
    p=c*r*exp(r/a)exp(z/b) for some constants a,b,c, at r=infinity and z=infinity
    p=0 at f=r, where
    (f_r)^2=p*r/v-v*v_r
    (f_z)^2=p*r/v+v*v_r

    Is it possible that one could obtain an analytic solution (with some unknown constants of course)? Or if one has to use numerical integration, what would be the best method?

    Thanks in advance!
     
  2. jcsd
  3. Oct 6, 2011 #2

    Hootenanny

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    Yes, as far as I'm aware an analytical solution does exist for suitable p(z,r) and boundary conditions. However, I am concerned about the condition at infinity:
    Are you sure about this?
     
  4. Oct 6, 2011 #3
    Hah, oops, I meant
    p=c*r*exp(-r/a)exp(-|z|/b) for some constants a,b,c, at r=infinity and z=infinity.

    The z dependence of p at infinity could also be exp(-z^2/b), sech(z/b) or sech^2(z/b). At this point I'm more interested in getting a solution than exactly how fast it falls off.

    Would one use method of characteristics for an analytic solution? I think our whole class was a bit lost in PDEs, so I'm having a bit of trouble getting anything out.
     
  5. Oct 6, 2011 #4

    Hootenanny

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    Those conditions at infinity are much better! :approve:

    Yes, the method of characteristics should prove sufficient here. If you are unsure on how to apply the method to fully non-linear problems, take a look at this PDF: http://www.stanford.edu/class/math220a/handouts/firstorder.pdf from page 16 onwards. The first example is an equation of the same form as you have here.

    Anyway, take a stab at it and let us know if you get stuck.
     
    Last edited: Oct 6, 2011
  6. Oct 6, 2011 #5
    Thankyou! I'll see how I go :)
     
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