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Diff. eq. for the Einstein metric inside a body

  1. Aug 8, 2008 #1
    Solving the Einstein eq. i found the following differential eq. that would descrive the metric inside a body:
    where D1(y) and D2(y) are known function of y and 0<=x<=1.
    I try to solve numerically but looks like there is a cusp; any suggestion how to approach that eq. or some links where that eq. is treated?
  2. jcsd
  3. Aug 9, 2008 #2
    If we change the x with x=exp(-t) we have the following diff. eq.:
    where 0<=t<=[tex]\infty[/tex].

    Any hint? Do you know where this kind of eq. has been analyzed?

  4. Aug 11, 2008 #3

    Yes this is very difficult task you want to solve, but for my opinion is unsolvable in analytical way!
    You don't know what kind of functions are [tex]D_{1}(y), D_{2}(y)[/tex] this is main problem and even not knowing whether you can solve analytical or numerical, but for solving numerical this is very difficult task. If you know what function are [tex]D_{1}(y), D_{2}(y)[/tex] than is trivial task as you already know :smile:
  5. Aug 11, 2008 #4
    Thanks Snoopy. Actualli I know the D1(y)=A1*(B1-ye)/ye+1/2 and D2(y)=A2(1-ye)(B2-ye)/y2e-1... (Ai, Bi, e are costants) but the problems is that when i try to solve that equation numerically y->0 for a finite value of t and that has no physical meaning (the metric degenerates at a finite (>0) distance from the center of the star). Why u said it's trivial if i would know D1 and D2?
  6. Aug 11, 2008 #5
    No problem gijeqkeij.
    It is only expression we use to simplify our life :smile:

    Trivial task is only when"thinking" in numerical solving problem without analytical solution.
    On the other hand you can see, with your note, that [tex]D(y), y \neq y(x) [/tex] but you are want to solve DE [tex] y''(t)+C_{1}\; y'(t)= C_{2}[/tex]
    where [tex]C_{1}, C_{2}[/tex] are only constants.
    If I understand you correctly [tex]D_{1,2}(y),\; D_{1,2} \neq D_{1,2}(y(t)) [/tex], if so you just have to solve DE second degree analytical :wink:

    Glad to help
  7. Aug 12, 2008 #6
    Sorry Snoopy I guess I was unclear. The eq. to solve is: [tex] y''(t)+C_{1}(y(t))\; y'(t)= C_{2}(y(t))[/tex]. I thought that kind of eq. received some attention in literature and I wonder if somebody can help me in finding papers that analyze that type of DE. Thanks
  8. Aug 12, 2008 #7
    Hy :smile:

    You wrote something like this:
    [tex]D_{1}(y) = A_{1}\frac{B_{1}-y^{e}}{y^{e}}+\frac{1}{2}[/tex] and
    [tex]D_{2}(y) = A_{2} (1-y^{e})\frac{B_{2}-y^{e}}{y^{2e-1}}\;\;\; ... [/tex]

    where are [tex]A_{1}, A_{2}, B_{1},B_{2}, e=exp(1) \in const[/tex]. OK this DE cannot be solve analytical :smile:

    Numerical approach :wink:

    [tex]x^{2}y''(x)+\left[2xA_{1}(B_{1}y(x)^{-e}-1)+x \right]y'(x) +A_{2}(B_{2}-y^{e})(y^{1-e}-y^{1-2e}) = 0[/tex]

    For knowing [tex]y(x=?)= ..., y'(x= ?)= ...[/tex] than make DE something like this:
    [tex]y'_{1} = y_{2} [/tex]
    [tex]y'_{2} = -\frac{1}{x^{2}}\left[\left(2A_{1}x(B_{1}y^{-e}_{1}-1)+x\right)y_{2}+A_{2}(B_{2}-y^{e}_{1})(y^{1-e}-y^{1-2e})\right][/tex]

    This is best to solve with Runge-Kutta 5 degree or with ode45 in MatLab or Octave.
    I attach file with Runge-Kutta method in file.m (MatLab, Octave).
    Write like this in MatLab:
    [tex]f=inline([\; 'y(1)' \; ;\; '\frac{1}{x^2}\left[..... \right]'\;], 'x' , 'y' )[/tex]

    In Runge-Kutta file will output vector dimension 2*n where n is steps make in calculation.
    Where you want do solve this equation on interval [tex]I[/tex] second column in function RungeKutta5, [tex]y0(x=?)=[y(x=?) , y'(x=?)][/tex] and for n put 100 or more.

    I hope I helped you :smile:


    Attached Files:

  9. Aug 13, 2008 #8
    Thanks Snoopy... that help but the problem is that a numerical solution bring to a singularity somewhere for t=t* (finite value). I wonder if somebody has analyzed that kind of eq. to identify what are (if any) the conditions for it's integrability.
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