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Good day!

We have an equation $y'+Ay=Bu$ where $y=\{\theta,\varphi\}$, $A, B$ are nonlinear operators.

$u \in L^\infty(\Gamma)$, $\theta, \varphi \in W = \{y \in L^2(0,T;V) : y'\in L^2(0,T;V')\}$, $V = H^1(\Omega)$.

If $v = \{v_1, v_2\}$ then $ (Bu, v) = \int\limits_\Gamma \beta u v_1 d\Gamma + \int\limits_\Gamma \gamma u^4 v_2 d\Gamma $.

We consider an optimal control problem with the control $u$:

$$ J(u) = \| \theta - \theta_d \|_{L^2(0,T;L^2(\Omega))}^2 \to \inf $$ or maybe $$ J(u) = \| \theta - \theta_d \|_{L^2(0,T;L^2(\Omega))}^2 + \mu\|u\|_{L^2(\Gamma)}^2 \to \inf $$

(Here $\theta$ is a solution of the equation $y'+Ay=Bu$.)

The set of admissible controls is $$ U_{ad} = \{ u: 0 \leq u \leq M \}. $$

Our goal is to prove the existence of the optimal control. Let $J(u_k) \to j = \inf J$. Since $U_{ad}$ isn't compact we should use weak convergence. $\{u_k\}$ is bounded therefore $u_k \to u$ weakly in $L^2(\Omega)$ (or its subsequence, we also may use any $L^p(\Omega), p\geq 1$). Also $y_k \to y$ weakly in $W$ (because $\{y_k\}$ is bounded in $W$) therefore $y_k \to y$ strongly in $L^2(0,T;H)$.

Next we should prove that the pair $\{y, u\}$ satisfies our equation $y'+Ay=Bu$. Thus we need to prove that $u_k^4 \to u^4$ weakly. But in the general case it isn't so.

How to prove the existence of the optimal control?

Thanks for your help!

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  • $\begingroup$ @AthanagorWurlitzer Thanks a lot for your answer! Yes, $\Omega$ is bounded. Could you explain why the fact that $u_k \to u$ weakly in $L^8$ implies that $u_k^4 \to u^4$ weakly in $L^2$? $\endgroup$
    – jokersobak
    Mar 26, 2014 at 12:34
  • $\begingroup$ What I meant was that you can, by extracting a first subsequence which converges weakly in $L^2$, and from that subsequence, which is still bounded in $L^8$, extract another subsequence so that $u^4$ converges weakly as well in $L^2$. $\endgroup$
    – username
    Mar 26, 2014 at 15:00
  • $\begingroup$ @AthanagorWurlitzer It's clear that if $\{u_k^4\}$ is bounded in $L^2$ then (subsequence of) $u_k^4 \to \chi$ in $L^2$ weakly. But we need that $\chi = u^4$ where $u_k \to u$ weakly in $L^2$ (or in $L^8$). $\endgroup$
    – jokersobak
    Mar 26, 2014 at 18:41
  • $\begingroup$ hum, yes, silly me. And no $H^1$ bound comes from $A$? Then I don't see. $\endgroup$
    – username
    Mar 26, 2014 at 19:39
  • $\begingroup$ @AthanagorWurlitzer We have a priori estimates for the solution $y$: $y$ is bounded in $L^\infty(Q) \cap W$ where $Q = \Omega \times (0,T)$, $W = \{y \in L^2(0,T;V) : y' \in L^2(0,T;V') \}$, $V = H^1(\Omega)$. Thus $y$ is bounded in $L^2(0,T;H^1(\Omega))$. I think that without the term $\mu\|u\|^2$ in $J$ proving of the existence may be impossible. I will think further. Thank you! $\endgroup$
    – jokersobak
    Mar 26, 2014 at 23:49

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