Non linear differential equation

  1. I can't type in latex so in this post d^2a is the secpnd derivative of a, while (da)^2 is the square of the derivative.

    This equation arose from the G_thetatheta compinent of the einstein tensor. Iwas solving tfor the shcwarzchildmetric where where the cosmos constant is nonzero.

    the equation is:

    e^2a(d^2a +2(da)^2 +2/r(da))= L where L is the cosmological constantand r is the radial coordinate. Umm how do we solve this?
     
  2. jcsd
  3. [tex] e^{2a}(da'' +2da^2 + \frac{2da}{r}) = L [/tex]

    is that right?
     
    Last edited: Jun 8, 2007
  4. Yes it is. can you helpme?

    essentially I determined from G_tt and G_rr that b=-a just like with the ordinary metric. When I plugged it into the G_thetathata equation, it was still fairly ugly, and icouldnt find a solution by inspection. Non linears are annoying.
     
    Last edited: Jun 8, 2007
  5. Is it not

    [tex]a^{\prime \prime} + 2 a^{\prime 2} + \frac{2a}{r} = L e^{-2a}[/tex]

    and, presumably the derivative is with respect to r, yes?
     
  6. yes it is. It may be that. It is possible that I've made a small mistake in my calculations of miswrote someon paper. How do we solve that? Can it be done by inspection? non linearity scares me.

    could someone help me solve it?
     
    Last edited: Jun 10, 2007
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