While self-studying a set of notes about Cosmology, I have encountered the following claim:
We have the linear growth equation:
$$\delta''(\vec{k},\tau)+\mathcal{H}(\tau)\delta'(\vec{k},\tau)-\dfrac{3}{2}\Omega_m(\tau)\mathcal{H}^2(\tau)\delta(\vec{k},\tau)=0 \label{1}\tag{1}$$
We can easily confirm the above differential equation has a growing and a decaying mode solution:
$$\delta(\vec{k},\tau)=D_+(\tau)\delta_{+,0}(\vec{k})+D_-(\tau)\delta_{-,0}(\vec{k}) \label{2}\tag{2}$$
where $D_-(\tau)=D_{-,0}H=D_{-,0}\mathcal{H}/a$. The growing mode solution can then be obtained as:
$$D_+(\tau)=D_{+,0}H(\tau)\int^{a(\tau)}_0\dfrac{da'}{\mathcal{H}^3(a')} \label{3}\tag{3}$$
First, I will explain the notation and the relationships between the different variables, that have nothing to do with mathematics and come from the physics background of the question:
- $t$ and $\tau$ are two different measurements of time, respectively named coordinate time and conformal time.
- $a=a(t)=a(\tau)$ is a scalar that depends on time.
- $H=\dfrac{1}{a}\dfrac{da}{dt}$ and $\mathcal{H}=\dfrac{1}{a}\dfrac{da}{d\tau}$, where $d\tau=\dfrac{dt}{a}$, and therefore $\mathcal{H}=aH$.
- The following equality holds: $$\dfrac{\partial\mathcal{H}}{\partial\tau}=\mathcal{H}^2\bigg(1-\dfrac{3}{2}\Omega_m\bigg)$$
I am completely lost here. First, I don't know how to solve the differential equation itself, and then, it is not clear to me either how to prove that the expression given for $D_+(\tau)$ is a solution. I assume that the notes use a very confusing notation, since I understand that the $a'$ inside the integral is not $da/d\tau$, but simply a way to rename $a$ to that it is not represented by the same letter than the upper limit of the integral itself. So we would actually have, if I am correct:
$$D_+(\tau)=D_{+,0}H(\tau)\int^{a(\tau)}_0\dfrac{da}{\mathcal{H}^3(a)}$$
In order to prove that this is a solution of \eqref{1}, we would need to find its first and second derivatives with respect to $\tau$. However, when trying to do this by applying Leibniz rule for the derivation of an integral, given in the case of multiple parameters by:
$$\dfrac{\partial F}{\partial\lambda_k}=\int^{b(\lambda)}_{a(\lambda)}\dfrac{\partial f}{\partial\lambda_k}dx+\dfrac{\partial b}{\partial\lambda_k}\cdot f(\lambda,b(\lambda))-\dfrac{\partial a}{\partial\lambda_k}\cdot f(\lambda,a(\lambda))\ \ \ \ \text{where}\ \ \ \ F(\lambda)=\int^{b(\lambda)}_{a(\lambda)}f(\lambda,x)dx$$
I encounter a problem: we don't have here an independent integration variable and parameter, but rather an integration variable $a$ that depends on the parameter $\tau$. So I have no idea how to proceed.
I assume that the reasoning behind \eqref{2} is related to the fact that there are no derivatives with respect to $\vec{k}$ in \eqref{1}, so the solution can be written as a constant that only depends on $\vec{k}$ multiplied by the solution that depends on $\tau$, but I don't know why we can separate that solution into a growing and a decaying function of $\tau$.
Any help would be greatly appreciated. I've dwelled on this for weeks and can't find an explanation or a way to solve the differential equation.
Thank you very much in advance!
As far as I can see, you may write
\begin{align} I &= \int_{0}^{a(\tau)}\frac{da'}{\mathcal{H}^{3}(a')} = \int_{0}^{\tau}\mathcal{H}^{-3}(a'(\tau'))\frac{d\tau'}{d\tau'/da'} \\ &= \int_{0}^{\tau}\mathcal{H}^{-2}(a'(\tau'))a'(\tau')\;d\tau'= \int_{0}^{\tau} f(\tau')\;d\tau' \end{align}
so
\begin{align} \frac{dI}{d\tau} &= f(\tau) = \mathcal{H}^{-2}(\tau)a(\tau) \end{align}
Alternative expressions for \eqref{1} in David Weinberg's course notes on Astronomy 873 (first eq. on third page) and Rien van de Weijgaert's Large Scale Structure (2009) course notes (eq. 42) may be worth checking.