If $x$ is a solution to the equation $\dot y = f(y) \quad f \in C^\infty (\mathbb{R}^d)$, and $x(0) = x(T)$ for some T, prove that $x(t) = x(t+T)$.

Question

In the book of The Elements of Mechanics by Gallovotti, at page 16, it is asked that

If $x$ is a solution to the equation $\dot y = f(y) \quad f \in C^\infty (\mathbb{R}^d)$, and $x(0) = x(T)$ for some T, prove that $x(t) = x(t+T)$. Would this also be true if $f \in C^1(\mathbb{R}) ?$ [Hint: Use uniqueness.]

However, we do know that

$$x(t) = x(0) + \int_{\tau = 0}^{\tau=t} f(x(\tau)),$$

and if we looked at the difference $$x(t+T) - x(t) = \int_{\tau=t}^{\tau=t+T} f(x(\tau)),$$

I cannot see any reason why this integral should be zero.

Answer 1

Set

$z(t) = x(t + T); \tag 1$

then

$z(0) = x(0 + T) = x(T) = x(0); \tag 2$

also

$\dot z(t) = \dot x(t + T) = f(x(t + T)) = f(z(t)); \tag{2.5}$

now since

$f(y) \in C^\infty(\Bbb R^n) \tag 3$

is differentiable, it follows that $f(y)$ is locally Lipschitz continuous at each point $y_0 \in \Bbb R^n$; and from this it further follows, via the Picard-Lindeloef theorem, that through any $y_0 \in \Bbb R^n$ there is precisely one solution to

$\dot y = f(y), \; y(0) = y_0; \tag 4$

therefore, in light of (2), which says that both $x(t)$ and $z(t)$ satisfy

$x(0) = z(0) = y_0 \tag 5$

for some $y_0$, we conclude that

$x(t) = z(t) = x(t + T). \tag 6$

Since $f(y) \in C^1(\Bbb R^n)$ is sufficient for $f(y)$ Lipschitz, the result holds in this case as well.

Answer 2

Let $y_1(t) = x(t)$ and $y_2(t) = x(t+T)$. Then both $y_1$ and $y_2$ satisfy \begin{align} y' = f(y) \end{align} with the same initial condition. Hence by uniqueness we are done.