Existence for the optimal control of $a(t)$: prove a inequality I have to prove that an optimal control ($a(t)$) exists using a theorem of existence (when the control set is compact).  One of the hypothesis I need to control is that:
$$|\dot{x}|\le c(1+|x|), \quad \forall x.$$
In my specific problem I have
$$|\dot{x}|=\left|\left(\begin{matrix} \dot{x_1}\\ \dot{x_2}\end{matrix}\right)\right|=\left|\left(\begin{matrix} \dot{v(t)}\\ \dot{s(t)}\end{matrix}\right)\right|=\left|\left(\begin{matrix} a(t)-c_0v(t)-c_1v^2(t)\\ v(t)\end{matrix}\right)\right|$$
How can I increase this function in order to prove the first inequality?
I wrote $$\left|\left(\begin{matrix} a(t)-c_0v(t)-c_1v^2(t)\\ v(t)\end{matrix}\right)\right|\le \sqrt{[a(t)-c_0v(t)-c_1v^2(t)]^2+v^2(t)}$$ but I cannot go on. Any ideas? Thank you.
EDIT: It's a minimum time problem. $c_0>0$, $c_1>0$, $v(0)=v_0$, $v(T)=v_f$, $s(0)=0$, $s(T)=L$ ($L>0$), control set$=[-a_{brake},a_{push}]$.
 A: In Cesari-type existence theorems for optimal control problems it is typically assumed that there exists some constant $c$ such that $|\dot x(t)|\leq c(1+|x(t)|)$ holds for all $t$ and all feasible state trajectories $x(\cdot)$. In your case it holds that 
$$x(t)=\left(\begin{array}{c}v(t)\\s(t)\end{array}\right)\qquad\dot x(t)=\left(\begin{array}{c}a(t)-c_0v(t)-c_1v(t)^2\\v(t)\end{array}\right).$$
The inequality that you have to verify therefore reads as
$$\sqrt{[a(t)-c_0v(t)-c_1v(t)^2]^2+v(t)^2}\leq c\left[1+\sqrt{v(t)^2+s(t)^2}\right].$$
I claim that there does not exist a constant $c$ such that this holds, at least if no further parameter restrictions are imposed. Since $v(0)$ is unrestricted, let me choose $v(0)=0$. Furthermore, I assume that $a_{brake}>c_0^2/(2c_1)$. Suppose that $a(t)=-a_{brake}$ for all $t$ (which is feasible). Then the velocity evolves according to
$$\dot v(t)=-a_{brake}-c_0v(t)-c_1v(t)^2.$$
It is easy to see that the right-hand side of this differential equation is negative for all $t$. It is also known that a differential equation of this form (due to the quadratic term $c_1v(t)^2$) does not have a global solution. Indeed, there will exist a finite time $\bar t$ such that $$\lim_{t\to\bar t_-}v(t)=-\infty.$$ Now look at the inequality you want to verify. Since the term under the root on the left-hand side is of order 4, whereas the term under the root on the right-hand side is of order 2, the left-hand side converges to $+\infty$ faster than the right-hand side, as $t$ approaches $\bar t$.  This implies that there cannot be any $c$ such that the inequality holds. 
