When solving a 2nd order ODE, say \begin{equation}\tag{*}\begin{cases}\frac{dx}{dt}=f(x,y)\\\frac{dy}{dt}=g(x,y),\end{cases}\end{equation} it is common to eliminate time and solve the resulting 1st order ODE \begin{equation}\tag{**}\frac{dy}{dx}=\frac{g(x,y)}{f(x,y)}\end{equation} that gives a dependence $y=y(x)$.

I wonder what are the conditions for this approach to be valid at an equilibrium point $(x^*,y^*)$? At this point, the fraction $\frac{g(x^*,y^*)}{f(x^*,y^*)}=\frac{0}{0}$. Potentially, this indeterminacy can be resolved using a sort of multivalued L'Hopital rule, but that is quite tricky.

Intuitively, I understand that the answer depends on the structure of the eigenvalues of the linearization of (*) at $(x^*,y^*)$, but I cannot formulate this quite well.

Let, say, the linearized system have a saddle at $(x^*,y^*)$. The equation ($**$) has two solutions corresponding to the stable and the unstable manifolds (they seem to be both unstable as they go away from $(0,0)$). Does it imply that the DE ($**$) isn't well posed?


1 Answer 1


Indeed, it doesn't work at an equilibrium point, but you don't really need it there: $(x,y) = (x^*, y^*)$ is the solution with initial conditions $x(0)=x^*, y(0)=y^*$.
It's also not defined on the curve $f(x,y) = 0$ (although there, when $g(x,y) \ne 0$, you could look at $x$ as a function of $y$). However, it is OK everywhere else, and it can be useful to study limits of these solutions as $x \to x^*$.

  • $\begingroup$ Dear @Robert, I ask because I often saw this approach used for computing stable/unstable manifolds of a system of nonlinear DEs. In this case the solution goes exactly from the equilibrium. Sometimes this seems to be a trick rather than a formal approach. $\endgroup$
    – Dmitry
    Dec 25, 2018 at 16:24
  • $\begingroup$ I don't think so: the stable and unstable manifolds would have the same initial condition $y(x^*) = y^*$. The equilibrium should correspond to a limit. For example, the linear saddle-point system $\dot{x} = y$, $\dot{y} = x$ leads to $\dfrac{dy}{dx} = \dfrac{x}{y}$, not defined at $(0,0)$, but the solutions $y=x$ and $y = -x$ have limits $(0,0)$ as $x \to 0$. $\endgroup$ Dec 26, 2018 at 0:52

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