Need help solving this set of differential equations of motion

Question

I'm a masters student and I have a set of Differential equations I need to solve for my research. For context they are equations of motion for scalar fields coupled to gravity, though I suppose that's not really relevant to the maths involved.

Specifically, they take the form $$\chi''(\rho)+2A'(\rho)\chi'(\rho)-(\chi'(\rho))^2-12=0$$ $$A''(\rho)-A'(\rho)\chi'(\rho)+2(A'(\rho))^2-24=0$$ $$2(A'(\rho))^2-2(\chi'(\rho))^2-24=0$$

Now, I already know that there are solutions where $\chi'$ and A' are constant ($\chi'=2$, $A'=4$). However, I know from existing literature using a different number of dimensions that there should also be another solution taking the form of a sum of (natural) logarithms of hyperbolic functions, such that when $\rho\rightarrow\infty$, $\chi$ and A tend to the constant solutions. The problem is that for the life of me I cannot figure out how to find this additional solution. I have tried writing out trial solutions of the form $$\chi=\chi_0+\chi_1\log(\cosh(\chi_2\rho))+\chi_3\log(\sinh(\chi_4\rho))$$ $$A=A_0+A_1\log(\cosh(A_2\rho))+A_3\log(\sinh(A_4\rho))$$ But despite my best efforts I haven't been able to decipher any kind of solution or determine the constants so far. I feel a little embarrassed because I should probably be able to solve this kind of thing by now but I desperately need some help here. Im not even really sure where to start, apart from redefining $\chi_4$ and $A_4$ in terms of the other constants. I'm using mathematica if that helps at all. Apologies if this is a stupid question.

Answer 1

Let $(\psi,B)=(\chi',A')$; then \begin{gather*} \psi'+2B\psi-\psi^2=12 \\ B'-B\psi+2B^2=24 \\ B^2-\psi^2=12 \tag{1} \end{gather*} We can try to remove some nonlinearities by taking linear combinations. Adding, $$\psi'+B'+B\psi+2B^2-\psi^2=36$$ Now substituting the hyperbola condition (1): $$\psi'+B'+B(\psi+B)=24$$

But at this point, our equation is secretly 1-D. Note that $$2B=(B-\psi)+(B+\psi)=\frac{12}{B+\psi}+B+\psi\tag{2}$$ Thus if $u=B+\psi$, then we have the following autonomous ODE: $$24=u'+\frac{1}{2}\left(\frac{12}{u}+u\right)u=u'+6+\frac{u^2}{2}$$ Canceling, we discover a disguised Riccati equation $$36=2u'+u^2$$

Riccati equations have a trick: let $u=\frac{2v'}{v}$; then $$36=\frac{4(v''v-v'^2)+4v'^2}{v^2}=\frac{4v''}{v}$$ Thus $$v=\alpha\sinh{\sqrt{9}t}+\beta\cosh{\sqrt{9}t}$$

Correspondingly, $$\psi+B=u=\frac{2v'}{v}=6\cdot\frac{\alpha\cosh{3t}+\beta\sinh{3t}}{\alpha\sinh{3t}+\beta\cosh{3t}}$$ Solving for $B$ and $\psi$ via (2) and matching your boundary conditions is something I leave to you.

Answer 2

Substituting eqn. (3) into the first yields that \begin{align} \chi''\pm2\chi'\sqrt{(\chi')^2+24}=(\chi')^2+12. \end{align} Taking $\chi'=p(\chi)$ \begin{align} pp'\pm2p\sqrt{p^2+24}=p^2+12,\\ \chi=\int\frac{p\ \mathrm dp}{p^2\mp2p\sqrt{p^2+24}+12}+c_1. \end{align} Which, to my surprise, has the elementary solution \begin{align} \chi=\frac{1}{12}\Big[\log\left(\left|\frac{2p^3+48p\pm(p^2+12)\sqrt{p^2+24}}{-2p^3-48p\pm(p^2+12)\sqrt{p^2+24}}\right|\right)+4\log\left(\left|\frac{\pm\sqrt{p^2+24}-p}{\pm\sqrt{p^2+24}+p}\right|\right)-\log\left(\left|p^4+24p^2-48\right|\right)\Big]+c_1. \end{align}

(As computed by <integral-calculator.com>), my rudimentary check against a numerically computed integral in Desmos seemed to agree.

Recall from a few lines ago that $p=\chi'$. This resulting equation may make it seem that all hope is lost, but that isn't so. Denote our integral as $F(p)$, i.e. \begin{align} \chi=c_1+F(\chi'). \end{align} The solution to such an equation can be found through the great technique known as 'integration by differentiation' which I first found in Handbook of exact solutions for ordinary differential equations / Andrei D. Polyanin, Valentin F. Zaitsev.--2nd e.

\begin{align} \rho=\int\frac{F(s)'}{s}\mathrm ds+c_2,\quad \chi=F(s)+c_1, \end{align} perhaps you can derive this yourself? It's a satisfying one. A hint: take a derivative.

This integral has an elementary solution as well! This never happens I swear. Interestingly WolframAlpha does not give a solution. I will spare you the couple lines of equation it yields. Let's denote the solution $G(s)$, so that \begin{align} \rho=G(s)+c_2,\quad \chi=F(s)+c_1. \end{align}

Following the same recipe for $A$ yields a more complicated yet still elementary integral \begin{align} A=\int\frac{q\ \mathrm dq}{-2q^2\mp iq\sqrt{q^2+24}+24}+k_1\equiv\tilde F(q)+k_1,\quad [q=A'], \end{align} with the implicit solution \begin{align} \rho=\tilde G(t)+k_2, \quad A=\tilde F(t)+k_1, \end{align} where $\tilde G(t)=\smallint (\tilde F(t)'/t)\mathrm dt$.

The solution for $A(\rho)$ is similar to that for $\chi(\rho)$, just a bit longer. I believe Mathematica would be useful for the integrals. Let me know if there are any errors.