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$$ \dot x(t)=Ax(t)+Bu(t),\quad t\in[0,T]\\ x(0)=x_0\\ u\in L^\infty([0,T],\mathbb R^m)\text{ and }x\in L^\infty([0,T],\mathbb R^n) $$

It is known that $\lim_{a\to 0}||u_a-u_0||_{L^2}=0$, where $u_0$ is a fixed trajectory, $u$ uniquely determined by $a$. I also know that for each $u$, there exists unique $x^u$ for the above problem.

My question:

what additional condition shall I impose on the above problem to prove $$\lim_{a\to 0}||x^{u_a}-x^{u_0}||_{L^\infty}=0$$

Thanks in advance.

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    $\begingroup$ Sorry, I cannot understand your notation. what do you denote by $u_a$? And do I understand you correctly: $u$ is a given function and $x_0$ is a given initial data while your unknown is $x$? $\endgroup$
    – DeM
    Nov 9, 2015 at 16:58
  • $\begingroup$ @DelioM. Thanks for your attention! Yes, your understanding is correct. Additionally, $u$ can be uniquely determined by the real parameter $a$, hence I denote such as $u_a$. $\endgroup$
    – user143763
    Nov 9, 2015 at 17:14
  • $\begingroup$ Ok, so let me rephrase your question: You have a family $(u_a)_{a\in \mathbb R}$ of $L^\infty$-functions and hence a family $(v_a)_{a\in \mathbb R}=(Bu_a)_{a\in \mathbb R}$ of $L^\infty$-functions, if you are assuming that $B$ is a bounded linear operator on $L^\infty$ (are you?). So the question is whether the solutions to the problem with inhomogeneous terms $v_a$ converges to the solution of the problem with inhomogeneous term $v_0$, right? $\endgroup$
    – DeM
    Nov 9, 2015 at 17:25
  • $\begingroup$ @DelioM. Yes, you are right. Sorry for my misleading statement. $\endgroup$
    – user143763
    Nov 9, 2015 at 17:35
  • $\begingroup$ Do you know the general solution for this equation. I think writing down $x^{u_a}$ and $x^{\hat{u}}$ is a good starting point. $\endgroup$
    – obareey
    Nov 9, 2015 at 18:13

1 Answer 1

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Ok, then the solution is given by the variation-of-constants-formula $$x(t)=e^{tA}x_0 +\int_0^t e^{(t-s)A}Bu_a(s)\ ds$$ so your question amounts to asking whether $$\sup_{0\le t\le T}\left\|\int_0^t e^{(t-s)A}B(u_a-u_0)(s)\ ds \right\|$$ (for any norm $\|\cdot\|$ on $\mathbb R^n$) goes to $0$ as $a$ goes to $0$. Now this is certainly the case, because this term can be estimated by $$ \begin{split} \left\|\int_0^t e^{(t-s)A}B(u_a-u_0)(s)\ ds \right\|&\le \int_0^t \left\|e^{(t-s)A}B(u_a-u_0)(s)\right\|\ ds \\ &\le M\|B\|\int_0^t \|(u_a-u_0)(s)\|ds \\ &\le M\|B\| T \|u_a-u_0\|_\infty \to 0 \end{split}$$ where $M$ is the upper bound of the semigroup generated by $A$ (any semigroup has an upper bound on time intervals of finite length).

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  • $\begingroup$ Your answer is great! But I am a bit confused about the last inequality, how to relate the given condition $||u_a−u_0||_{L^2}\to 0$ with your $||u_a−u_0||_\infty\to 0$? Thanks! $\endgroup$
    – user143763
    Nov 10, 2015 at 6:33
  • $\begingroup$ The $L^\infty(0,T)$-norm is stronger than the $L^2(0,T)$-norm, so you cannot deduce $\|u_a-u_0\|_\infty\to 0$ from $\|u_a-u_0\|_2\to 0$. Imposing $\|u_a-u_0\|_\infty\to 0$ would be such an additional condition you are looking for. (If you are familiar with vector-valued convolutions, then you do indeed get your sought-after estimate without any additional condition using Young's inequality $\|f*g\|_1\le \|f\|_1\|g\|_1$ and the imbedding of $L^2(0,T)$ in $L^1(0,T)$. $\endgroup$
    – DeM
    Nov 10, 2015 at 9:12
  • $\begingroup$ Yes, then such a condition is required. Additionally, if no condition is imposed, can I have $||x^{u_a}-x^{u_0}||_{L^2}\to 0$? $\endgroup$
    – user143763
    Nov 10, 2015 at 11:44
  • $\begingroup$ @user143763 Well, again: The $L^\infty(0,T)$-norm is stronger than the $L^2(0,T)$-norm, so your convergence in $L^\infty$-norm also implies convergence in $L^2$-norm. $\endgroup$
    – DeM
    Nov 10, 2015 at 14:19
  • $\begingroup$ Thanks so much. Do you mean $||e^{\cdot A}*B(u_a-u_0)||_1\leq||e^{\cdot A}||_1||B(u_a-u_0)||_1$? Can you kindly show the detailed proof with vector-valued convolutions? Thanks. $\endgroup$
    – user143763
    Nov 10, 2015 at 16:27

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