Let $x(t)\in\mathbb{R}$ be a positive function of $t\in\mathbb{R}$, $t\ge 0$ and suppose the following inequality holds for all $t\ge 0$ $$\tag{$\star$}\label{ineq} x(t)\le c e^{-at + b x(t) t^2}, $$ where $a,b,c$ are positive scalar.
My question: Do there exist conditions on the coefficients $a$, $b$ and $c$ guaranteeing that $$ \lim_{t\to \infty}x(t)=0\ ? $$
What makes the problem hard (at least to me) is the term $b x(t) t^2$ on the exponential in \eqref{ineq}. I tried to take the logarithm on both sides of \eqref{ineq}, but this does not seem to help much. So any suggestion/comment is very appreciated, thanks!
Take $x(t)=1$ and consider the set of $a,b,c$ that satisfy $x(t)\leq ce^{-at+bx(t)t^2}$. This reduces to $1\leq ce^{-at+bt^2}$, or $0\leq \ln c-at+bt^2$, which implies $bt^2-at-\ln c\geq 0$. The minimum of the left side occurs at $t=\frac{a}{2b}$, giving $\frac{a^2}{4b}-\frac{a^2}{2b}-\ln c = \frac{a^2(1-2b)}{2b}-\ln(c).$ There are plenty of solutions $(a,b,c)$ that satisfy this, for example if you fix a $c$, just ensure $a$ is large and $b=1/2$. But none of these solutions imply $x(t)\rightarrow 0$.