We have the following:
Proposition Let $x: [0,T) \to \mathbb{R}$ be continuous and differentiable and satisfies $x' = f(t,x)$, then for any increasing sequence of times $T_n \nearrow T$ we have that $\limsup_{n} |x(T_n) | < \infty$.
The above proposition implies that a solution cannot "blow up in finite future time", which I think is the proper interpretation of the question asked.
The proof of the proposition is simple: intuitively using that $(x^2)' = 2xx' = 2xf$ we have that by the condition given in the hypothesis, once $|x|$ reaches a value that is larger than $R$, it can only decrease (and not increase). This forces the value of $x$ to remain bounded and hence cannot blow up in finite time. To formalise the proof we write as follows.
Proof (rewritten in 2024 to be more clear):
By way of contradiction, suppose $\limsup_n |x(T_n)| = +\infty$, then there is a subsequence $T_{n_j}$ such that $|x(T_{n_j})|$ is increasing and diverges to infinity. WLOG we pass to the subsequence and still call it $T_n$ for convenience.
Since $|x(T_{n})|$ now increases and diverges to infinity, there is some $N$ such that for all $n \geq N$, $|x(T_{n})|> 2R$. Restrict our attention to the interval $[T_N, T_{N+1}]$. Let $\tau$ be the supremum within this interval
$$ \tau = \sup \{ t\in [T_N,T_{N+1}] | |x(t)| \leq 2R \} $$
so that if $|x(t)| > 2R$ throughout the interval, $\tau$ is defined to be $T_N$.
The reason for this definition is that we ensure two things:
- At no times beyond $\tau$ is $|x(t)|< 2R$.
- We guarantee that $|x(\tau)| < |x(T_{N+1})|$; this is because by construction $|x(T_{N+1})| > |x(T_N)| > 2R$.
By continuity of $x$, there is an interval of times near $T_{N+1}$ where $x$ takes values larger than $2R$ also, and so we also find that
- $\tau < T_{N+1}$ strictly.
Apply now the mean value theorem to $|x|^2$ on the interval $[\tau,T_{N+1}]$, we have that there exists $\tau'\in [\tau,T_{N+1}]$ such that $$2x x' |_{t = \tau'} = (|x|^2)' = \frac{|x(T_{N+1})|^2 - |x(\tau)|^2}{T_{N+1} - \tau} > 0 $$
On the other hand we have $|x(\tau')| \geq 2R$ as $\tau' \geq \tau$. Hence we also have
$$ xx'|_{t = \tau'} = xf < 0 $$
by the hypothesis of the problem. And we arrived at the contradiction.
A comment about existence: solutions to your differential equation do not need to exist if $f$ is sufficiently bad. It is well known that a derivative of a everywhere differentiable, continuous function cannot be too discontinuous. In particular, let $\chi(x)$ denote the function that equals $1$ when $x < 0$, and $-1$ when $x \geq 0$. Now let $\phi(t)$ be a nowhere continuous function such as the indicator of the rational numbers. Then for $f(x,t) = \phi(t) \chi(x)$ it is impossible for any continuous and differentiable function $x = x(t)$ to satisfy $x' = f(t,x)$.
So strictly speaking under the conditions given it is false that every solution exists for all positive time.
