Could someone help me interpret the following proposition? I've been struggling to comprehend it. Thanks in advance.

Proposition: If $J \subset \mathbb{R}$ is an open interval containing the origin and $\gamma:J \rightarrow \mathbb{R^n}$ is a solution of the differential equation $\dot{x}=f(x)$ with $\gamma(0) = x_0 \in U$, then the function $B:J \rightarrow \mathbb{R}$ given by $$B(t) = \int_0^t\frac{1}{g(\gamma(s))}ds$$ is invertible on its range $K \subseteq \mathbb{R}$. If $\rho:K \rightarrow J$ is the inverse of $B$, then the identity $$\rho'(t) = g(\gamma(\rho(t)))$$ holds for all $t\in K$, and the function $\sigma:K \rightarrow \mathbb{R^n}$ given by $\sigma(t) = \gamma(\rho(t))$ is the solution of the differential equation $\dot{x}=g(x)f(x)$ with initial condition $\sigma(0) = x_0$.

$g:U \rightarrow \mathbb{R}$ is a positive, smooth function where $U$ is an open set.

  • $\begingroup$ What is $g$, and what are the assumptions on this function? By coincidence, $g$ could be $\equiv=0$ on the solution curve. $\endgroup$ Feb 9, 2014 at 19:02
  • $\begingroup$ @Christian Blatter See edit $\endgroup$
    – user85362
    Feb 9, 2014 at 19:06

1 Answer 1


This is about changing the time scale of the ODE. The trajectories as geometric objects stay the same, only the parametrization changes. That is, if $γ(t)$ is a solution of the ODE, then $γ(ρ(s))$ reaches the same points provided that $ρ$ is bijective.

Assuming that the parametrization is differentiable, the derivative of the composition is $\frac d{ds}γ(ρ(s))=γ'(ρ(s))ρ'(s)$, and one can prescribe any rule to $ρ'(s)$, for instance $ρ'(s)=g(γ(ρ(s)))$.

The statement involving $B$ is then an application of the inverse function theorem, if $B(ρ(s))=s$, then $1=B'(ρ(s))ρ'(s)=B'(ρ(s))g(γ(ρ(s)))$, and with $t=ρ(s)$, $1=B'(t)g(γ(t))$.


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