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Let $f(\cdot)$ be a continuously differentiable function over $\mathbb{R}$, and $u\in L^2_{loc}(0,\infty)$, $a\in \mathbb{R}$, and $x(t)$ solves the integral of $$\dot{x}(t)=ax(t)+f(x(t))+u(t), \quad x(0)=x_0\in \mathbb{R}.$$

It is known that for every $x_0$ and $R$, where $\|u\|_2\le R$, there is a time $T$ such that the integral of above equation admits a solution $x\in C([0,T])$. Is this true that the set of all $u(t)$ for which an integral solution $x\in C([0,T])$ exists is in fact an open subset of $L^2(0,T)$ containing $$U=\{u(t): \|u\|_2\le R\}.$$


What about the same question over a general Banach space, and $a$ being an operator that generates a semi-group?


Note that in the example $\dot{x}=x^2-R+u$ with $x(0)=0$; we have a global solution for $|u|\le R$. We know there is no global solution for $u>R$. But for every $T$, pick $\epsilon<\frac{\pi^2}{4T^2}$ then there is this local solution $x(t)=\sqrt\epsilon\tan(\sqrt\epsilon t)$ defined over $[0,T]$ corresponding to $u=R+\epsilon$.

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  • $\begingroup$ Note that the blow-up time still continuously depends on $\|u\|_2$. Decrease $\|u\|_2$, the blow-up time gets increased. Let $T$ be a fixed number; we get $\|u\|_2 \le R$, for some $R$, there is a solution. What about now? Do we have an open set around each input $\|u\|_2=R$? $\endgroup$ – Sara Winslet Oct 15 '18 at 23:55
  • $\begingroup$ @ChristianRemling I clarified the question. $\endgroup$ – Sara Winslet Oct 16 '18 at 5:08
  • $\begingroup$ There is something unclear to me about the question. You ask whether the set of all $u(t)$ for which $x\in C([0,T])$ exists, which I'll denote by $B\subseteq L^2 _{\rm loc} (\mathbb{R})$ is open. But then you ask "in other words" whether $u_0 \in \partial U$ is an interior point. This is to claim that $U$ is open, which is requiring less then that $B$ is open. $\endgroup$ – Amir Sagiv Oct 16 '18 at 6:36
  • $\begingroup$ I removed that sentence to avoid any confusion. $\endgroup$ – Sara Winslet Oct 16 '18 at 6:39
  • $\begingroup$ Also for clarification: You fix a time $T > 0$ and the question is whether the set of all $u$ in $L^2(0,T)$ such that the solution $x_u$ of your equation exists on $[0,T]$, is open in $L^2(0,T)$. Is that what it is? $\endgroup$ – Hannes Oct 16 '18 at 8:14
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The implicit function theorem usually helps to arrive at such conclusions. You translate the (differential equation) to an operator on function spaces, in your example e.g. $$E \colon W^{1,2}(0,T) \times L^2(0,T) \to L^2(0,T)$$ with $$E(x,u) = ax + f(x) + u - \dot x.$$ (One can also set this up differently, for instance directly with the Duhamel formula, but the principle remains the same.) Then the differential equation is satisfied by $x$ for parameter $u$ on $(0,T)$ for your desired regularity exactly if $E(x,u) = 0$.

You then show that $E$ is continuously differentiable, and that, for given $(\bar x,\bar u)$, the linear (!) operator $\partial_x E(\bar x,\bar u)$ is continuously invertible. (This partial derivative in $(\bar x,\bar u)$ will then be an operator in $\mathcal L(W^{1,2}(0,T);L^2(0,T))$ which corresponds to the linearized equation of the original nonlinear equation.)

From this, the IFT tells you that if you have a pair $(\bar x,\bar u) \in W^{1,2}(0,T) \times L^2(0,T)$ satisfying the differential equation, so $E(\bar x,\bar u) = 0$, then there exist open neighborhoods $U_{\bar x}$ of $\bar x$ in $W^{1,2}(0,T)$ and $U_{\bar u}$ of $\bar u$ in $L^2(0,T)$ together with an implicit function $\varphi \colon U_{\bar u} \to U_{\bar x}$ such that $E(\varphi(u),u) = 0$ for all $u \in U_{\bar u}$. (And, indeed, that $(\varphi(u),u)$ are the unique pairs in the neighborhoods verifying this.)

The only thing left would be to establish that there indeed exists a solution $\bar x \in W^{1,2}(0,T)$ corresponding to $\bar u \in L^2(0,T)$ at all. (Usually the first candidate here would be $u\equiv0$.) Then the foregoing reasoning gives you an open set $U_{\bar u}$ around $\bar u$ such that the for every $u \in U_{\bar u}$ there exists a unique solution $x_u := \varphi(u)$ on $(0,T)$ for your equation.

Of course, this set $U_{\bar u}$ includes a ball $B_{R_\ast}(\bar u)$ around $\bar u$ in $L^2(0,T)$ for $R_\ast$ small enough. The other way around, if you happen to know that every $u \in M$ for some set $M$ gives rise to a solution $x_u$ on $(0,T)$, then you obtain open neighborhoods $U_u$ for every $u \in M$ such that the solutions $x_v$ associated to $v \in U_u$ again exist on $(0,T)$. Thus, $$U_{M} := \bigcup_{u \in M} U_u$$ is an open set containing $M$ whose elements all give rise to solutions on $(0,T)$.

Generally, the result to take away is that in the given situation, the set of parameters/controls $u$ producing a solution $x_u$ on a given time interval, is open. (In fact, the IFT also tells you that the solution mapping $\varphi$ is continuously differentiable which might also be of interest.)

The technique generalizes to the evolution equation setting.

My standard reference for the Banach space IFT would be the book of Serge Lang, there it is Ch. XIV, Thm. 2.1. (See the citation below.) The technique is usually used to show that the solution operator (here $\varphi$) in optimal control problems is continuously differentiable in order to apply first order optimality conditions to characterize solutions to optimal control problems, so I have also added two more references (the books by Tröltzsch and Hinze/Pinnau/Ulbrich/Ulbrich) for that. (The books are about optimal control of PDEs though, but that's where I'm from, so..)

Lang, Serge, Real and functional analysis., Graduate Texts in Mathematics. 142. New York: Springer-Verlag. xiv, 580 p. (1993). ZBL0831.46001.

Tröltzsch, Fredi, Optimal control of partial differential equations. Theory, procedures, and applications., Wiesbaden: Vieweg (ISBN 3-528-03224-3). x, 297 p. (2005). ZBL1142.49001.

Hinze, M.; Pinnau, R.; Ulbrich, M.; Ulbrich, S., Optimization with PDE constraints, Mathematical Modelling: Theory and Applications 23. Dordrecht: Springer (ISBN 978-1-4020-8838-4/hbk; 978-1-4020-8839-1/ebook). xi, 270 p. (2009). ZBL1167.49001.

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  • $\begingroup$ Shouldn't it be close interval $[0,T]$. $\endgroup$ – Sara Winslet Oct 16 '18 at 17:33
  • $\begingroup$ It makes no difference for the Sobolev or Lebesgue spaces, but $W^{1,2}(0,T)$ is continuously embedded into bounded $1/2$-Hölder functions on $(0,T)$ which admit a unique (Hölder-) extension to $[0,T]$. $\endgroup$ – Hannes Oct 16 '18 at 17:45
  • $\begingroup$ Yes; that's correct. Could also mention a reference in your answer. Thanks a lot. $\endgroup$ – Sara Winslet Oct 16 '18 at 17:47
  • $\begingroup$ Done, and you're welcome. $\endgroup$ – Hannes Oct 16 '18 at 18:29

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