Integration and hyperbolic function problem

  • Thread starter JD_PM
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  • #1
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Homework Statement:
While studying Cosmology I came across a particular differential equation I do not see how to solve. Please read below for details
Relevant Equations:
N/A
This question arose while studying Cosmology (section 38.2 in Lecture Notes in GR) but it is purely mathematical, that is why I ask it here.

I do not see why the equation

$$H^2 = H_0^2 \left[\left( \frac{a_0}{a}\right)^3 (\Omega_M)_0 + (\Omega_{\Lambda})_0 \right] \tag{1}$$

Has the following solution

$$a(t) = a_0 \left( \frac{(\Omega_M)_0}{(\Omega_{\Lambda})_0}\right)^{1/3} \left(\sinh \left[(3/2)\sqrt{(\Omega_{\Lambda})_0}H_0t\right]\right)^{2/3}$$

Where (and ##\dot a## represents the derivative of ##a## wrt time)

$$H:=\left( \frac{\dot a}{a}\right)^2, \ \ \ \ H_0:=\left( \frac{\dot a_0}{a_0}\right)^2$$


I found a similar problem here. I suspect that, to start off, we should find a smart change of variables but which one?

Might you please guide me towards the solution?

Any help is appreciated.

Thank you :biggrin:
 

Answers and Replies

  • #2
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Perhaps I can offer a suggestion you may find useful as you study DEs: They're difficult in and of themselves so sometimes it is helpful to strip them of unnecessary trappings and just study the underlying pure, clean and simpler-looking equation. With that in mind, is your equation basically:

$$\left(\frac{y'}{y}\right)^4=d\left[b\left(\frac{a}{y}\right)^3+c\right]$$
for ##a,b,c,d## some constants?
 
  • #3
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With that in mind, is your equation basically:

$$\left(\frac{y'}{y}\right)^4=d\left[b\left(\frac{a}{y}\right)^3+c\right]$$
for ##a,b,c,d## some constants?

Alright, I realized I made a typo here

$$H:=\left( \frac{\dot a}{a}\right)^2, \ \ \ \ H_0:=\left( \frac{\dot a_0}{a_0}\right)^2$$

It should be

$$H:=\left( \frac{\dot a}{a}\right), \ \ \ \ H_0:=\left( \frac{\dot a_0}{a_0}\right)$$


So the DE we want to solve is

$$\left(\frac{y'}{y}\right)^2=d\left[b\left(\frac{a}{y}\right)^3+c\right]$$

OK at this point.

Should we start off by introducing a change of variables? Maybe it is a good idea (just guessing) to rewrite it a bit, by multiplying by ##y^2## both sides

$$y\left(y'\right)^2=d\left[ba^3+cy^3\right]$$
 
  • #4
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So the DE we want to solve is

$$\left(\frac{y'}{y}\right)^2=d\left[b\left(\frac{a}{y}\right)^3+c\right]$$

Isn't that then a separable equation?
 
  • #5
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I have the following suggestions. First, for economy of notation, define the constants,
$$
\Gamma_0=H_0^2a_0^3(\Omega_{M})_0
$$
$$
\beta_0=H_0^2(\Omega_{\Lambda})_0
$$
We then have the DE,
$$
(\frac{\dot a}{a})^2=\frac{1}{a^3}[\Gamma_0 + a^3\beta_0]
$$
$$
\dot a=\frac{1}{\sqrt a}[\Gamma_0 + a^3\beta_0]^{\frac{1}{2}}
$$
which leads to,
$$
\frac{1}{\sqrt{\Gamma_0}}\int \frac{\sqrt a da}{\sqrt {1 + a^3(\frac{\beta_0}{\Gamma_0}})}=t
$$
We make the substitution,
$$
u^2=a^3(\frac{\beta_0}{\Gamma_0})
$$
$$
da=\frac{2}{3}(\frac{\Gamma_0}{\beta_0})^{\frac{1}{3}}\frac{du}{u^{\frac{1}{3}}}
$$
thus,
$$
\frac{2}{3}\frac{1}{\sqrt {\beta_0}}\int \frac{du}{\sqrt {1+u^2}}=t
$$
$$
\sinh^{-1}(u)=\frac{3}{2}\sqrt{\beta_0}t
$$
$$
u=\sinh (\frac{3}{2}\sqrt{\beta_0}t)
$$
$$
a^{\frac{3}{2}}(\frac{\beta_0}{\Gamma_0})^{\frac{1}{2}}=\sinh (\frac{3}{2}\sqrt{\beta_0}t)
$$
$$
a=(\frac{\Gamma_0}{\beta_0})^{\frac{1}{3}}[\sinh (\frac{3}{2}\sqrt{\beta_0}t)]^{\frac{2}{3}}

$$
 
  • #6
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164
@aheight @Fred Wright thank you very much for your help.

Isn't that then a separable equation?

Yes (let me use ##e## as constant instead ;))

$$\frac{y'}{y}=\left(e\left[b\left(\frac{a}{y}\right)^3+c\right]\right)^{1/2} \Rightarrow \frac{dy}{y\left(e\left[b\left(\frac{a}{y}\right)^3+c\right]\right)^{1/2}} = dt$$
 

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