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So the problem comes from cosmology and I want to solve for the unknown function $a(t)$, which is the scale factor for the universe. So I have an integral involving $a$: $$ (1)\quad\int\left[\frac{\Omega}{a(t)}+\frac{1-\Omega}{\sqrt{a(t)}}\right]^{-1/2}da=H_0t,\ $$ where $\Omega$ and $H_0$ are constants, which, when you do the intergral subject to the restraint $a(0)=0$, yeilds: $$ (2)\quad\frac{-4\sqrt{(1-\Omega)\sqrt{a(t)}+\Omega}\;\Big(3(\Omega-1)^2 a(t)+4\Omega(\Omega-1)\sqrt{a(t)}+8\Omega^2\Big)+32 \Omega^{5/2}}{15(1-\Omega)^3}=H_0t. $$ This is the expression I want to solve for $a(t)$. It's not so pretty, but basically it surmounts to solving an expression of the form $$ (3)\quad\sqrt{k_1\sqrt{x}+k_2}\;\Big(k_3x+k_4\sqrt{x}+k_5\Big)+k_6=0 $$ in $x$ with $\{k_1,\dots,k_6\}$ being constants.

This, however seams quite difficult, though it seams there ought to a solution. Solving (3) with just either of the factors on the left hand side is straight forward, so solving $$ (4)\quad\sqrt{(1-a)\sqrt{x}+a}=kt\quad\text{or}\quad(5)\quad\big(ax+b\sqrt{x}+c\big)=kt $$ is easy. For (4) you just square both sides, rearrange and square again. Likewise (5) is easy to solve since it can be turned into an ordinary second order equation by writing $ax+b\sqrt{x}+c'=0$ with $c'=c-kt$ then you can isolate the $b\sqrt{x}$ term and square both sides at which point you're just dealing with a second degree polynomial.

Looking at (3) it sort of takes the form of the product of two "polynomials" both containing rational powers of $x$, however it's not really the case since $$ \sqrt{a\sqrt{x}+b}\neq\sqrt{a}\sqrt[4]{x}+\sqrt{b} $$ but even if it had been the case it wouldn't have been of much help since I still don't see a way to eliminate the rational powers of $x$. Since (4) and (5) can be solved (and also for $k=0$), determining the roots of (3) (at least for $k_6=0$) is again easy, but since I cant just let the right hand side of (2) vanish, I feel like I'm at a loss.

For the value of my constants I have $\Omega=.3$ and the Hubble constant is $H_0=67$, which, when plugged into (2) gives $$ (6)\quad.78\sqrt{.7\sqrt{a(t)}+.3}\;\Big(1.47a(t)-.84\sqrt{a(t)}+.72\Big)-.31=67t. $$ I would like to find a direct algebraic solution to (2), expressing $a$ i terms of $\Omega$, $H_0$ and $t$, but most of all I would like to have a numerical expression for $a(t)$ from (6), since I want to plot it's graph to do data analysis.

Any help would be much appreciated, thanks in advance :)

Erik Olesen
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  • Over what range of values of $t$ would you like a numerical solution? – Plutoro Jul 13 '15 at 02:53
  • sorry for not mentioning that, $t>0$. – Erik Olesen Jul 13 '15 at 02:55
  • But what is your time scale like? Doing cosmology, a numerical solution from $t=0$ to $t=1$ is not very helpful if $t$ is in seconds. Since any numerical solution is going to be a good approximation only on a bounded interval, what is a good upper limit for $t$, at least for a first try? – Plutoro Jul 13 '15 at 03:00
  • Or more precisely on the range from $0<t<t_0$, ($t_0$ being now) which in this model is $0.743H_0^{-1}$, which is $10.855\times 10^{9}yr$ or $3.423\times 10^{17}s$, as is the time unit in this case – Erik Olesen Jul 13 '15 at 03:01
  • Well the natural thing to do is to use the Hubble time $H_0^{-1}=14.61 Gyr=14.6\times 10^{14}s$ as a time unit. So $t\in(0,0.8H_0^{-1})$ – Erik Olesen Jul 13 '15 at 03:09
  • but that however changes things a bit I suppose – Erik Olesen Jul 13 '15 at 03:10
  • Everything on the left hand side is dimensionless and $t$ as expressed here is cosmic time. In (6) when I say $H_0=67,km/s/Mpc$, that's expressed in seconds. – Erik Olesen Jul 13 '15 at 03:13
  • So analysis that I'm doing is comparing this particular model to data from type 1A supernovas with redshifts ranging from .0075 to 1.775, giving an approximate cosmic time from 3.78Gyr to 14Gyr. – Erik Olesen Jul 13 '15 at 04:02

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Letting $x=z^2$ and squaring (3) we get the equation $$ (k_1\,z+k_2)(k_3\,z^2+k_4\,z+k_5)^2=k_6^2. $$ This a fifth degree polynomial equation. The general quintic equation is not solvable in radicals, but can be solved in terms of elliptic functions. I expect the solution to be so complicated to become useless.

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Julián Aguirre
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