# Preserving orbits by multiplication with a non-vanishing function

I'm reading through some notes from a past course of mine, where a system of ODE's of the form$$\begin{array}{c} x'=h(x,y)f(x,y)\\ y'=h(x,y)g(x,y) \end{array}$$ appears, such that $x,y\in\mathbb{R}$ and $h(x,y)>0$ for all values of $x,y$. Now it says that one can divide the RHS by $h(x,y)$ (since it's positive) and the orbit of the system remain unchanged, so that the above system is equivalent to

$$\begin{array}{c} x'=f(x,y)\\ y'=g(x,y) \end{array}$$Is this true ?

EDIT (due to Hans Lundmarks comment): Can someone please provide me with a reference or proof for this ?

• The orbits are preserved, the time parametrization is not. Nov 9, 2014 at 15:48
• @HansLundmark Could you please expand this as an answer, as to how one should prove this ? Or even a reference to a proof would be perfect.
– resu
Nov 9, 2014 at 16:27

Suppose that we have a solution $x(t),y(t)$ to $$x'=f(x,y)$$ $$y'=g(x,y).$$ Define $\tilde f(t)$ to satisfy the differential equation $$\tilde f'(t)=h(x(\tilde f(t)),y(\tilde f(t))).$$ Notice that, since we can regard $h$, $x$ and $y$ as known already, we can let, for some constant $c_1$: $$I(z)=\int_{c_1}^z\frac{d\lambda}{h(x(\lambda),y(\lambda))}$$ and notice that, from the differential equation $I(\tilde f(t))=t+c_2$ - meaning $\tilde f$ is the inverse function of $I$. Notice that $I$ is strictly increasing, since $h$ is positive, implying that $I$ is at least injective. To show surjectivity, suppose that $\lim_{z\rightarrow\infty}I(z)=k$ for some finite $k$. Then, it would follow, that the limit of the integrand would be $\lim_{z\rightarrow\infty}\frac{1}{h(x(z),y(z))}=0$ implying that $\lim_{(x,y)\rightarrow (x(k),y(k))}h(x,y)=\infty$, which would mean that $h$ would tend to infinity somewhere on the orbit of $(x,y)$ - impossible for a continuous function. A similar argument handles the case of $z\rightarrow-\infty$.

This implies that $I$ is bijective, and hence has a bijective inverse $\tilde f$ which solves the desired equation. Then, let $$x_f(t)=x(\tilde f(t))$$ $$y_f(t)=y(\tilde f(t))$$ we have that, from the chain rule $$x_f'(t)=\tilde f'(t)x'(\tilde f(t))$$ $$y_f'(t)=\tilde f'(t)y'(\tilde f(t))$$ but if we replace $x'$, $y'$, and $\tilde f'$ by the differential equations they solve and replace $x(\tilde f(t))$ with $x_f$, we get $$x_f'=h(x_f,y_f)f(x_f,y_f)$$ $$y_f'=h(x_f,y_f)g(x_f,y_f)$$ which means that $x_f$ and $y_f$ solve the modified differential equation, but since they are composition of $x$ and $y$ with a bijective $\tilde f$, they must have the same orbits.

• Thanks for the answer! I have number of unclarities: Is it an assumption, that $\tilde{f}'(t)=h(x(\tilde{f}(t)),y(\tilde{f}(t)))$ (I used $\tilde{f}$ to avoid confusion, since we already have an $f$ flying around) even possess a solution, or is this always (i.e. for every choice of $f$ and $g$ such that the initial system is solvable) the case that $\tilde{f}'(t)=h(x(\tilde{f}(t)),y(\tilde{f}(t)))$ possess a solution ? And is the bijectivity of $\tilde{f}$ also an assumption ? (Solution of some ODE's usually aren't bijective.)
– resu
Nov 9, 2014 at 18:32
• Lastly: You showed one direction, i.e. that the orbits of the system without $h$ satisfy the system with $h$. I assume that the other direction ("orbits of the system with $h$ are the same of the system without $h$", can be proved analogously by using the solution $\tilde{\tilde{f}}$ of the $h^{-1}$ equation: $\tilde{\tilde{f}}'(t)=h^{-1}(x(\tilde{\tilde{f}}(t)),y(\tilde{\tilde{f}}(t)))$.
– resu
Nov 9, 2014 at 18:36
• (What I'm basically asking in my first two question are the precise conditions under which our argument holds.)
– resu
Nov 9, 2014 at 18:50
• I upgraded the post, working from the assumption that $h$ is continuous (and you should be able to see how it would fail if $h$ isn't); The fact that $\tilde f$ is bijective is sufficient for both directions. Nov 9, 2014 at 18:57

You already got a detailed answer, but here are a few references to book sources in case you're interested:

• Oh thanks, its always best to have a solid reference, were all the details are to be found.
– resu
Nov 18, 2014 at 18:46