Sol,
the solutions I have seen start like so:
We have, of course, the gravitational potential:
[latex] \bigtriangledown ^ 2\phi = 4\pi G\rho [/latex]
and the equation for hydrostatic equilibrium:
[latex]\bigtriangledown P + \rho \bigtriangledown \phi = 0 [/latex]
These are tied together by the equation of state, which I believe relates the speed of sound in the mass distribution to the pressure:
[latex] P = \rho {{c}_{s}}^{2} [/latex]
If we begin assuming that:
[latex] \rho = \rho (r) [/latex]
Then I can barely understand how to get to:
[latex]\left (\frac{{{c}_{s}}^{2}}{\rho (r)} \right)\left(\frac{d \rho (r)}{d r} \right) = - \frac {4 \pi G}{{r}^{2}}\int_{r}^{0}{r}^{2} \rho (r) dr [/latex]
Now, I can integrate the expression on the far right, and descriptions of this integral that I have seen in on-line versions of these calculation steps often just call this M(r).
This equation overall is too complex for me, and on-line versions of this calculation use numerical methods to solve this.
A typical numerical solution looks like the attached graph, solution1 (from reference #1).
A typical numerical solution also takes the form of:
[latex] \rho (r) = \frac {{{c}_{s}}^{2}}{2 \pi G {r}^{2}} [/latex] (from reference #1)
Now, I cannot do the numerical solutions, or solve the equation further myself.
I did however, try to get "further along" by assuming the following forms for [latex] \rho (r) [/latex], and seeing how far I got:
[latex] \rho (r) = \frac {\alpha }{{r}^{2}} [/latex]
and:
[latex] \rho (r) = \frac {\alpha }{r} [/latex]
where [latex] \alpha [/latex] is just an arbitrary constant.
I hope that this math does not make you shudder, but here goes.
First, plugging in [latex] \rho (r) = \frac {\alpha }{{r}^{2}} [/latex], we have:
[latex] \left (\frac{{{c}_{s}}^{2} {r}^{2}}{\alpha} \right)\left(\frac{-2 \alpha}{{r}^{3}} \right) = - \frac {4 \pi G}{{r}^{2}}\int_{0}^{r}\alpha dr [/latex]
Combinations, and evaluation of the integral leads to:
[latex] \left (\frac{- 2{{c}_{s}}^{2}}{r} \right) = - \frac {4 \pi G \alpha} {r} [/latex]
This reduces to a solution for my arbitrary constant of:
[latex] \alpha = \frac {{{c}_{s}}^{2}}{2 \pi G} [/latex]
This seems to me to confirm the validity of the numerical solution.
If I try plugging in [latex] \rho (r) = \frac {\alpha }{r} [/latex], I get:
[latex] \left (\frac{{{c}_{s}}^{2}{r}}{\alpha} \right)\left(\frac{- \alpha}{{r}^{2}} \right) = - \frac {4 \pi G}{{r}^{2}}\int_{0}^{r}\alpha r dr} [/latex]
Combinations, and evaluation of the integral, provides:
[latex] \left (\frac{- {{c}_{s}}^{2}}{r} \right) = - 2 \pi G \alpha [/latex]
Which gives:
[latex] \alpha = \frac {{{c}_{s}}^{2}}{2 \pi G r} [/latex]
So, in this case, my arbitrary constant is forcing the density to be dependent upon [latex] {r}^{-2} [/latex] again!
Now, my funky checks of the density function do not prove the uniqueness of the solution that is dependent upon [latex] {r}^{-2} [/latex], but I think that it shows that for these equations, it is a viable solution.
I am sorry if my calculus is so poor, this is the best way I could present the solution, and try to work through at least some checks of the numerical solution on my own.
Hopefully I haven't done anything too outrageous, although I know I played fast and loose with my integrands and differential variables, I think.
References:
#1
http://www.mpia-hd.mpg.de/homes/dullemon/lectures/starplanet/chap_05_cloudcores.pdf
#2 Primordial Stellar Evolution, E.K.L. Upton, as presented in Stars and Stellar Systems Vol. VII: Nebulae and Interstellar Matter, U of Chicago Press, 1968
