6
$\begingroup$

Consider the system of ODEs: \begin{equation} \varphi''\varphi'^{q-1}\psi'^{p-2}=\varphi^{p-1}\psi^{q-1}, \end{equation} \begin{equation} \varphi'^2+\psi'^2=1, \end{equation} where $\varphi>0$, $\psi'>0$ and $p\geq2,\ q\geq1$ are integers.

This problem has a geometric background: consider the linear $SO(p)\times SO(q)$-action on $\mathbb{R}^{p+q}$, and hypersurfaces with constant Gauss-Kronecker curvature $\pm1$ in $\mathbb{R}^{p+q}$ invariant under this action. By choosing appropriate parametrization $(\varphi(t),\psi(t))$ of the generating curve, we obtain the above equations. The sign of the curvature is not important here, as both choices of the sign lead to similar equations. We only discuss one case here and the other case with reverse sigh should be similar.

A special case where $p=3,\ q=1$ has been discussed in the previous question:Analysis of solutions to a nonlinear ODE, and Robert Bryant managed to solve the equation explicitly by separation of variables. In the general situation, we do not expect to solve the equations completely, but we still hope to have certain ways of analyzing the properties of the solution. If the general case is too hard, any attempts on other special cases are also helpful. (e.g. we are trying the case of $p=q=2$ now.)

$\endgroup$
2
  • $\begingroup$ Are you still interested in this question? I've got a solution for the general case of $q=1$. $\endgroup$ Commented Jan 13, 2022 at 6:19
  • $\begingroup$ Thank you. We also solved the general case of $q=1$. We are interested in the cases of $p\geq 2,\ q\geq 2$. $\endgroup$
    – Yuhang Liu
    Commented Sep 20, 2023 at 2:19

0

You must log in to answer this question.