# Interpreting a proposition

• MHB
• kalish1
In summary, the given proposition states that if a solution of a differential equation is transformed by a certain function, the resulting function is invertible and satisfies an identity involving the original solution and a modified form of the differential equation with the same initial condition.

#### kalish1

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 $\sigma(0) = x_0$.

I have crossposted this on: differential equations - Interpreting a proposition - Mathematics Stack Exchange

And you received an answer on SE:
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))##.