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.
Best Answer
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.