# A differential equation

Think of $t$ and $r$ as two independent variables.

• Suppose $E$ be a function of $r$ and $V~$ be a function of $(t,r)$ such that both go to $0$ at $r=0$.

• There exists a positive function $M(r)$ such that $M(0)=0$ and $V(t,r) = -\dfrac{M(r)}{R(t,r)}$ where $R$ is another positive function such that $R(0,r)=r$.

• Let $p(r) = \dfrac{E(r)}{V(0,r)}$ be a function regular at $r=0$ such that $p(0) \in (-\infty,1)$.

• Also define a function $a$ of $r$ such that, $a(r) = \dfrac{M(r)}{\dfrac{4}{3}\pi r^3}$. Then $a$ is also a positive definite function with a well-defined value at $r=0$.

• Define $\alpha = a(0)$

Now look at this differential equation,

$$\frac{\dot{R}^2}{2} + V(t,r) = E(r)$$

Apparently this differential equation has a solution of the form,

$$\frac{t}{t_0} = \sqrt{\frac{\alpha}{a(r)}}\frac{F(p(r))}{F(p(0))} \left [1 - \left ( \dfrac{R(t,r)}{r} \right)^{\dfrac{3}{2}}~\cdot~\dfrac{F\left(~~ \dfrac{p(r)R(t,r)}{r} \right) }{F(p(r))} \right ]$$

where $t_0 = \sqrt {\dfrac{3}{8\pi \alpha}} F(p(0))$

and the function $F$ is defined over the interval $(-\infty,1)$ as,

$$F(x) = \left\{ \begin{array}{c c} -\frac{\sqrt{1-x}}{x} - \frac{1}{(-x)^{\frac{3}{2}}} \tanh^{-1} \left [ \sqrt{\frac{x}{x-1}} \right ] & x&lt;0 \\ \frac{2}{3} & x =0 \\ \frac{1}{x^{\frac{3}{2}}}tan^{-1} \left [ \sqrt{\frac{x}{1-x}} \right ] - \frac{\sqrt{1-x}}{x} & 0&lt;x&lt;1 \end {array} \right.$$

How does one get the above solution?

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What do you mean by a "positive definite function"? And how does that reconcile with the notion that $M(0) = 0$? – Willie Wong♦ Oct 11 '10 at 22:45
@Willie Thanks for pointing out the typo. I have corrected it. Any help with solving this differential equation? – Anirbit Oct 13 '10 at 6:26
I've fixed your latex: the problem was that you needed to \-escape a \{ and all \\, and that the < sign has to be written in some cases using an HTML escape (for silly reasons! Of course, in comments the rules are different...) – Mariano Suárez-Alvarez♦ Oct 13 '10 at 20:26

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What is $\dot{R}$, is it $\partial R/\partial t$? Also, there seems to be a lot of interdependence among the definitions; can you tell us what is given for a particular instance of the problem, and what is to be determined? Better yet, if you could provide the original source or motivation for this problem, it might make things clearer. – Rahul Narain Oct 13 '10 at 20:45

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@Mariano Thanks for correcting the LaTeXing. @Rahul Yes $\dot{R}$ is partial derivative of $R$ with respect to it. I didn't understand the second part of your query. I think I have completely defined all the quantities in question. You see any ambiguities? This is taken from a paper and I am giving you the reference if that helps, "Strength of naked singularities in Tolman-Bondi spacetimes" by R.P.A.C Newman in Class. Quantum Grav.3 (1986) 527-539 The third page of the paper has this. Will be happy to get back any help. – Anirbit Oct 14 '10 at 4:33

## 1 Answer

I'm sort of just guessing here (partly based on the solution already found). By explicitly plugging in $V(t,r) = - \frac{M(r)}{R(t,r)}$, you arrive at the ordinary differential equation (for each fixed $r$) for $R$ as

$$\dot{R}^2 - \frac{2M}{R} = 2 E$$

where $M$ and $E$ are constants in time. Now, re-scale the original equation by $t = \lambda s$ and $R = \mu \rho$. Then $\partial_t R = \frac{\mu}{\lambda} \partial_s \rho$. Then you can solve $(\frac{\mu}{\lambda})^2 = \frac{2M}{\mu} = 2E$ to reduce the equation to

$$\dot{\rho}^2 - \frac{1}{\rho} = 1$$

(the weights $\mu$ and $\lambda$ will, roughly speaking, give you the weights $p(r)$ and $a(r)$ in your question). Now note that this scaling degenerates if $E = 0$. In the case that $E = 0$, the equation can be solved by quadrature:

$$\dot{x}^2 = x^{-1} \Rightarrow \sqrt{x} dx = dt \Rightarrow x^{3/2} \sim t$$

This gives the solution to the homogeneous case. In the inhomogeneous case, you take that as a sort of integrating factor: assume that $\rho^{3/2} f(\rho) \sim t$, this implies that

$$\dot{\rho} \left( \rho^{3/2} f(\rho) \right)' = 1$$

we plug this into the equation, which we first re-arrange as

$$\dot{\rho} = \sqrt{\frac{1}{\frac{\rho}{1+\rho}}}$$

and we conclude that

$$\sqrt{\frac{\rho}{1+\rho}} = \frac{d}{d\rho}( \rho^{3/2} f(\rho))$$

and you solve this by directly integrating it. I think this $f$ you find should be exactly the $F$ you wrote down above (I didn't check it myself). Note that due to the singular weight $\rho$ which degenerates as $\rho \to 0$, you will have to separately integrate in the regime where $\rho > 0$ and $\rho < 0$. The existence and uniqueness of solution is guaranteed by the theory of Fuchsian ODEs.

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Thanks a lot for this help. Are you defining $f$ as, $f(\rho) = \frac{kt}{\rho ^ \frac{3}{2}}$ ? (for some constant $k$) Since there are no other replies I am giving away the bounty to you. (Though I am yet to understand your solution completely) – Anirbit Oct 19 '10 at 6:18
Actually I think I defined it with $k = 1$. The constant $k$ can be absorbed into the (implicit) definition of $\rho$, by rescaling $\rho$; so it is not too important. The reason I wrote $\sim t$ instead of $=t$ is just because there can be a constant term: $f(\rho) = (kt + C) / \rho^{3/2}$. who doesn't change the derivative. (In fact, I think in the final answer, to match the boundary conditions there may need to be a $C$.) – Willie Wong♦ Oct 19 '10 at 10:06
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