Proof by contradiction that there are no periodic solutions to $\dot x =f(x)$

Question

I need to finish the proof (by contradiction) that there do not exist any periodic (oscillating) solutions to the system $\dot x =f(x)$.

The proof starts out as follows:

Suppose on the contrary that $x(t)$ is a nontrivial periodic solution - I.e., that $x(t)=x(t+T)$ for some $T>0$ and $x(t)\neq x(t+s)$ for all $0<s<t$.

I'm supposed to derive a contradiction by considering $$ \int_{t}^{t+T}f(x)\frac{dx}{dt}dt$$

I saw a solution where the person said that $$\int_{t}^{t+T}f(x)\frac{dx}{dt}dt = \int_{x(t)}^{x(t+T)}f(x)dx=0$$ and i do not understand why that's true. Next, they said that $$\int_{t}^{t+T}f(x)\frac{dx}{dt}dt=\int_{t}^{t+T}f(x)^2 dt \geq 0, \quad \text{for}\, t^{*}\in (t, T+t)$$ which I also don't understand.

And finally, they said that this tells us that $f(x(t^{*}))=0$, which implies that the only solution is the trivial solution.

Now, I am not sure how those two things imply that $f(x(t^{*}))=0$, and then how that in turn implies that the only solution is the trivial solution.

Im assuming that all of these steps come perhaps from some form of the fundamental theorem of calculus, but I'm not exactly sure how it allows us to do these things. So, if someone could please explain these steps to me, I would appreciate it very much.

Note: I am not at all interested in alternative proofs for this result. They exist already aplenty on MSE for someone who is interested to find them. I am interested only in finishing/understanding the proof the way it is presented here.

Answer 1

Your notation is terrible: the variable of integration should never appear in an endpoint of the integral. What is meant is

$$ \int_t^{T+t} f(x(s)) \dfrac{d}{ds} x(s) \; ds $$ which, by the usual substitution formula for integrals, is

$$ \int_{x(t)}^{x(T+t)} f(u) \; du $$

But $x(T+t) = x(t)$ because of periodicity, so this integral is $0$.

Now using the differential equation $\dfrac{d}{ds} x(s) = f(x(s))$, the first integral becomes $$ \int_t^{T+t} f(x(s))^2 \; ds $$

Since we are dealing with real values, $f(x(s))^2 \ge 0$, and thus if $T > 0$ the integral also $\ge 0$. If $f(x(s)) \ne 0$ at some point, by continuity $f(x(s))^2 > 0$ in some interval, and then the integral over that interval $>0$. Since $f(x(s))^2 \ge 0$, the integral over any larger interval $>0$ as well. So the only way your integral can be $0$ is if $f(x(s)) = 0$ for all $s$ in the interval $[t, T+t]$. But then $\dot{x} = 0$, so $x$ is constant.