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Non linear differential equation

by Terilien
Tags: differential, equation, linear
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Terilien
#1
Jun8-07, 03:12 PM
P: 140
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?
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ice109
#2
Jun8-07, 08:07 PM
P: 1,705
Quote Quote by Terilien View Post
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?
[tex] e^{2a}(da'' +2da^2 + \frac{2da}{r}) = L [/tex]

is that right?
Terilien
#3
Jun8-07, 08:59 PM
P: 140
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.

Matthew Rodman
#4
Jun10-07, 08:18 PM
P: 66
Non linear differential equation

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?
Terilien
#5
Jun10-07, 08:43 PM
P: 140
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?


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