I'm trying to obtain information about the following system of ODEs, where $c_1, c_2$ are constants and $h_1, h_2: I \subset \mathbb{R} \to \mathbb{R}$ are smooth positive functions defined on an open interval:
$$ \begin{cases} c_1h_2\left(1-(h_1)^2 \right) - h_1 (h_1)'' h_2 - (r_1 - 1) (h_1')^2 h_2 - r_2 h_1 h_1' h_2' = 0 \\ c_2h_1\left(1-(h_2)^2 \right) - h_2 (h_2)'' h_1 - (r_2 - 1) (h_2')^2 h_1 - r_1 h_2 h_2' h_1' = 0 \end{cases} $$
It is clear (dividing by $h_2$ or $h_1$ on the first or second, equation, respectively) that once one has initial value conditions and one determines $h_1$ (resp. $h_2$), then $h_2$ is determined as well. But I would like to know whether there is any kind of uniqueness of solutions in this situation: i.e., if $\psi_1, \psi_2$ also satisfy this system, is it true that $h_1 = \psi_1$ and $h_2 = \psi_2$? If not, are there any conditions we can impose on the system so that the solutions are unique?
I'd really appreciate any help. Thanks in advance!
Write the equations as $h_1'' = \ldots$ and $h_2'' = \ldots$, and the standard Existence and Uniqueness theorem for systems of differential equations applies. The only wrinkle is that to do this we must divide by $h_1 h_2$, so that things may go bad if $h_1$ or $h_2$ hits $0$. But on any interval where $h_1$ and $h_2$ are strictly positive, solutions are uniquely determined by initial conditions of the form $h_1(t_0) = a$, $h_2(t_0) = b$, $h_1'(t_0) = c$, $h_2'(t_0)=d$.