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Two systems are topologically equivalent if there exists a homeomorphism of the phase space which takes phse curves of one system to phase curves of another.

Theorem Let $f\colon \mathbb{R}^n\to\mathbb{R}^n, \lambda\colon\mathbb{R}^n\to\mathbb{R}$ smooth, $\lambda>0$. Then the two systems $\dot{x}=f(x)$ and $\dot{x}=\lambda(x)f(x)$ are topologically equivalent and the homeomorphism is the identity map.

Unfortunately, I do not understand the following proof:

Proof:

(The proof consists of proving that a solution $x=x(t)$ which is a solution of the first system, is also a solution of the second system.)

We show that there exists a monotonic function $s(t)$ of $t$, such that $x(s(t))$ is a solution of the second system.

Let $x=x(t)$ be a solution of the first system. We want $x(s(t))$ to be a solution of the second system, hence

$$ \frac{dx(s(t))}{ds}\cdot\frac{ds(t)}{dt}=\lambda(x(s(t))\cdot f(x(s(t)).~~~(*) $$

We cannot - as one might think at first sight - cancel $\frac{dx(s(t))}{ds}$ from the left side and $f(x(s(t))$ from the right side. But we can write the following:

$$ \left(\frac{ds(t)}{dt}-\lambda(x(s(t))\right)\cdot f(x(s(t)) =0.~~~~(**) $$

Then the first or the second factor is zero.

The second factor is zero if the solution $x(t)$ is constant. That is, if $x$ is an equilibrium, we are done.

Let the first factor be zero. Then we can separate variables and get $$ \int_{0}^s \frac{ds}{\lambda(x(s(t))}=\int_{t_0}^t \, dt\Leftrightarrow t=t_0+\int_{0}^s\frac{dw}{\lambda(x(w))}=F(s) $$ and we get $s(t)=F^{-1}(t)$.


There are three main points I do not understand.

1) What does the expression $\frac{dx(s(t))}{ds}$ mean, to what are we derivating? $s$ is a function...

(2) Why can't we delete/ cancel $\frac{dx(s(t))}{ds}$ from the left side and $f(x(s(t))$ from the right side?

(3) How do we get from $(*)$ to $(**)$?

(4) And how do we know that $s(t)=F^{-1}(t)$ is within the existence interval of the solution?

Very confusing....

Maybe you know a source where I can read that more clearly as we did it in the reading...

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1) This is just the chain rule. See https://en.wikipedia.org/wiki/Chain_rule. $s$ itself becomes a free variable in that case.

2) Because $f(x(s(t)))$ might be 0. If we try to cancel, it would be undefined.

3) $dx(t)/dt = f(x(t))$ is the same thing as $dx(s)/ds = f(x(s))$. Just think $s$ is a free variable like $t$, and indeed it is as you can select $s$ arbitrarily and the equation is satisfied.

4) $t = F(s)$ means that $s = F^{-1}(t)$. Event though it is not stated explicitly in this equation, $s$ itself depends on $t$, so we can write $s(t) = F^{-1}(t)$.

I understand that your main confusion is about using a function as a free variable. Well, a free variable is, by definition, that you can select the values of it freely, and some other function may depend on it. Selecting values freely includes selecting values using a rule (or a function), so you can also use a function as a free variable, as is done in (*).

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