Prove that an autonomous ODE f(x)=x' has no nonconstant periodic solutions Prove that an autonomous ODE $f(x)=x'$ has no nonconstant periodic solutions.
I guess I could prove it by contradiction by saying $x(t+T) = x(t)$ implies $x(t) =$ constant.
 A: I think you are trying to prove that a first-order autonomous ODE $dx/dt = f(x)$ cannot have a periodic solution.  Of course it's pretty easy to think of a second-order autonomous ODE  that does have a periodic solution, so first-order autonomous systems of differential equations can likewise have periodic solutions.
For the single equation case, think about a periodic function (other than the trivial case of a constant function).  If the function is to be differentiable, it must be continuous.
Find two points on a continuous nonconstant periodic function where the function value is the same, but the function is decreasing at one argument and increasing at the other.
This would contradict the autonomous differential equation, right?
Once more, with rigor:
We show that if $x(t)$ is a real periodic solution of the autonomous first-order
differential equation:
$$x' = f(x)$$
then $x(t)$ is a constant function.
Suppose not for the sake of contradiction: $x$ has an interval of periodicity
$[0,T]$ and attains a bounded range $[X_{min},X_{max}]$.  WLOG we assume, by
translation in t if necessary, that $x(0) = X_{min} = x(T)$, and for the
sake of specificity that $x(t_*) = X_{max}$ for some $t_* \in (0,T)$.  Of course those
extremal values might also be attained elsewhere in $[0,T]$.
Since $X_{min}, X_{max}$ are extrema of $x(t)$, function $f$ must vanish there:
$$f(X_{min}) = 0 = f(X_{max})$$
Unless $f$ takes a nonzero value somewhere on $[X_{min},X_{max}]$, all solutions
$x(t)$ of the differential equation in this range would be constant.  So choose
$x_* \in (X_{min},X_{max})$ such that $f(x_*)$ is nonzero.
By the Intermediate Value Thm. continuity of $x$ implies $x(t) = x_*$ for some
points in $(0,t_*)$ and $(t_*,T)$.  In fact there is an open interval around
$t_*$ where $x(t) \gt x_*$, so we can pick a greatest $t_{left} \in (0,t_*)$
and a least $t_{right} \in (t_*,T)$ where $x$ takes the value $x_*$.
It follows that $x$ is increasing on the right side of $t_{left}$ and decreasing
on the left side of $t_{right}$.  But this forces the derivative $x'$ at both
points to be zero, contradicting our choice of $x_*$ s.t. $f(x_*)$ is nonzero.
QED
In a sense this is interesting because it assumes nothing about regularity of
$f$.  Under standard assumptions
solutions to initial value problems for the DE are (locally) unique.  But as
$x(t) = X_{min}$ would be a constant solution of the IVP for $x(0) = X_{min}$,
such uniqueness precludes nonconstant solutions.
A: I will give a slightly less rigorous, but perhaps more intuitively satisfying proof.
We have the differential equation $\frac{dx}{dt} = f(x)$. We want to find a periodic solution such that $f(x_t) = f(x_{t+T})$ for some $T>0$.
Since $f(x_t)$ is autonomous, $f$ does not explicitly depend on $t$.  Therefore, by chain rule, $\frac{df(x)}{dt} = 0\cdot \frac{dx}{dt} = 0$.
And since $f(x_t) = \frac{dx}{dt}$, we have $\frac{df(x_t)}{dt} = \frac{d^2x}{dt^2} = 0$.
Then, $x(t)$ is a linear function.  Now, the only linear function for which there exists a non-zero $T$ such that $x(t) = x(t+T)$ is a constant linear function.  Then, $x(t)$ must be a constant function, and we are done.
