We consider the following continuous-time nonlinear evolution problem
\begin{equation}
 \begin{cases} \dot{y}(t)=Ay(t)+F(y(t),u(t)),\quad t\geq0\\y(0)=f\in\mathcal{X}\end{cases}
\end{equation}
where $A$ designs a linear operator of domain $D(A)$ defined on a Banach space $\mathcal{X}$ generating a $C\_0$-semigroup $T(t)$, $u(t)$ is a **piecewise continuous** control defined on $\mathbb{R}\_+$ and taking its values on a Banach space $\mathcal{X}'$, and the function $F:\mathcal{X}\times\mathcal{X}'\to\mathcal{X}$ is **locally Lipschitz continuous with respect to both variables**.


I would like to know if we can affirm that the above problem admits a (mild or classical) solution defined on $\mathbb{R}\_+$. 



Solutions of a nonlinear evolution problem
The problem is locally well-posed, i.e., the problem admits a unique
solution $y\in C([0,T];{\mathcal X})$ for some (in general small) $T>0$. In addition, it holds that $\dot{y}\in L^\infty((0,T);{\mathcal X})$ if $f\in D(A)$. 


This can be proven by standard arguments. Indeed, denote the nonlinearity by $F$
(as $f$ is the initial value). There are constants $K\geq1$, $L>0$,
and $T>0$ such that $\|e^{tA}\|\_{\mathcal L(\mathcal X)} \leq K e^{K t}$
for all $t\geq0$,
\begin{align\*}
 & \|f\|\_{\mathcal X} + LK\left(K+2\right) T \leq K, \quad \|F(0,0)\|\_{\mathcal X}\leq K, \quad
 \|u\|\_{L^\infty((0,T);{\mathcal X}')}\leq K, \\
 & \|F(y,u)-F(\bar y,\bar u)\|\_{\mathcal X} \leq L\left(\|y-\bar y\|\_{\mathcal X}+\|u-\bar u\|\_{\mathcal X'}\right)
\end{align\*}
if $\max\{\|y\|\_{\mathcal X},\|\bar y\|\_{\mathcal X}\}\leq K^2$, $\max\{\|u\|\_{{\mathcal X}'},\|\bar u\|\_{{\mathcal X}'}\}\leq K$, and $LKT<1$. Define
$$
\mathbb X=\{y\in C([0,T];{\mathcal X})\mid y(0)=f,\,\|e^{-K
 t}y\|\_{L^\infty((0,T);{\mathcal X})}\leq K^2\}
$$
which is non-empty and a complete metric space furnished with the metric
$(y,\bar y)\mapsto \|e^{-Kt} y-e^{-Kt}\bar
y\|\_{L^\infty((0,T);{\mathcal X})}$. Define further
$$
 (\Phi y)(t) = e^{tA}f + \int\_0^t e^{(t-s)A}F(y(s),u(s))\,ds, \quad 0\leq t\leq T.
$$ 
Then $\Phi\colon \mathbb X\to\mathbb X$ and $\|e^{-Kt}\Phi y-e^{-Kt}\Phi \bar y\|\_{L^\infty((0,T);{\mathcal X})}\leq LKT \, \|e^{-Kt}y-e^{-Kt}\bar y\|\_{L^\infty((0,T);{\mathcal X})}$ for $y,\bar y\in\mathbb X$. So the contration mapping principle applies and demonstrates that $\Phi$ has a unique fixed point $y\in\mathbb X$ which is the desired solution.


