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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 $\mathbb{R}_+$.

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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.

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  • $\begingroup$ Thanks for your response. I have some remarks:1) the condition $f\in D(A)$ is not requiered for the existence of the mild solution (see e.g. A. Pazy theorem 1.4 pp. 185) 2) I don't see in your response where you have used the assumption '$F$ locally continuous lipschitzian'. 3)Indeed, as I mentionned in the edit, I am rather intersted in the existence of a solution defined on $\mathbb{R}_+$. \\My question is to know if such solution always exists? $\endgroup$ Aug 31, 2015 at 20:09
  • $\begingroup$ 1) The condition $f\in D(A)$ is needed if you want $d(e^{tA}f)/dt=Ae^{tA}f \in C([0,T];{\mathcal X})$. 2) See the estimate $\|F(y,u)-F(\bar y,\bar u)\|_{\mathcal X} \leq \ldots$ introducing $L$. 3) In general, the solution will blow up in finite time. If you want more, then you have to make additional assumptions. $\endgroup$
    – ifw
    Aug 31, 2015 at 21:39
  • $\begingroup$ Have you any idea about an additional condition which ensures that $T=+\infty$? $\endgroup$ Aug 31, 2015 at 21:46
  • $\begingroup$ For 1) I think that the the function $t\mapsto e^{tA}f$ is differentiable for every $f\in\mathcal{X}$. $\endgroup$ Aug 31, 2015 at 21:58
  • $\begingroup$ Concerning your second comment: There is no general answer. You have to be more specific on your problem first. Concerning your third comment: No, this is not true. The derivative is $Ae^{tA}f=e^{tA}(Af)$ if it exists. $e^{tA}$ acts on $\mathcal X$, so you need $Af\in\mathcal X$, i.e., $f\in D(A)$. Consult Pazy. $\endgroup$
    – ifw
    Aug 31, 2015 at 22:13

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