Numerical Exact Controllability of the 1D Heat Equation: Duality and Carleman Weights

Abstract

This article is devoted to the numerical computation of distributed null controls for the 1D heat equation. The goal is to compute a control that drives (a numerical approximation of) the solution from a prescribed initial state exactly to zero. We extend the earlier contribution of Carthel, Glowinski, and Lions, which is devoted to the computation of controls of minimal square-integrable norm. We start from some constrained extremal problems (involving unbounded weights in time), introduced by Fursikov and Imanuvilov, and we apply appropriate duality techniques. Then, we provide numerical approximations of the associated dual problems, and apply conjugate gradient algorithms. Finally, several experiments are presented, and we highlight the influence of the weights and analyze this approach in terms of robustness and efficiency. Also, the results are compared with others in a previous article of the authors, where primal methods were considered.

Appendix: Proofs of Propositions 3.1 and 3.2

Proof of Proposition 3.1

First, note that

$$\begin{aligned} J_{R}(\hat{y},\hat{v}) = {1 \over 2} \left( \displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}T_R(\rho )^2 |\hat{y}|^2 + \displaystyle \int \!\!\!\!\displaystyle \int _{q_{T}}\rho _{0}^2 |\hat{v}|^2 \right) \le J(\hat{y},\hat{v}) \end{aligned}$$

for all $R > 0$. Consequently, the solutions to the problems (16) satisfy

$$\begin{aligned} J_{R}(y_{R},v_{R}) = {1 \over 2} \left( \displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}T_R(\rho )^2 |y_{R}|^2 + \displaystyle \int \!\!\!\!\displaystyle \int _{q_{T}}\rho _{0}^2 |v_{R}|^2 \right) \le J(\hat{y},\hat{v}). \end{aligned}$$

This shows that $T_R(\rho )y_{R}$ is uniformly bounded in $L^2(Q_{T})$ and $\rho _{0}v_{R}$ is uniformly bounded in $L^2(q_{T})$. Therefore, at least for some subsequence one has

$$\begin{aligned} \rho _{0}v_{R} \rightarrow w \ \ \hbox {weakly in }L^2(q_{T}) \ \ \hbox {and} \ \ T_R(\rho )y_{R} \rightarrow z \ \ \hbox {weakly in }L^2(Q_{T}). \end{aligned}$$

(36)

Let us set $\tilde{y}= \rho ^{-1}z$ and $\tilde{v}= \rho _{0}^{-1}w$. Then it is clear from (36) that

$$\begin{aligned} \begin{array}{l} \displaystyle v_{R} = \rho _{0}^{-1}(\rho _{0}v_{R}) \rightarrow \tilde{v}\ \ \hbox {weakly in }L^2(q_{T})\\ \displaystyle y_{R} = T_R(\rho )^{-1}(T_R(\rho )y_{R}) \rightarrow \tilde{y}\ \ \hbox {weakly in }L^2(Q_{T}). \end{array} \end{aligned}$$

In fact, $\tilde{y}$ is the state associated to $\tilde{v}$ and $y_{R}$ converges strongly to $\tilde{y}$. For every $(y',v') \in \mathcal {C}(y_0,T)$, one has

$$\begin{aligned} \displaystyle J(\tilde{y},\tilde{v})&\le {1 \over 2} \liminf _{R\rightarrow +\infty } \left( \displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}T_R(\rho )^2 |y_{R}|^2 + \displaystyle \int \!\!\!\!\displaystyle \int _{q_{T}}\rho _{0}^2 |v_{R}|^2 \right) \nonumber \\&\le {1 \over 2} \lim _{R\rightarrow +\infty } \left( \displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}T_R(\rho )^2 |y^{\prime }|^2 + \displaystyle \int \!\!\!\!\displaystyle \int _{q_{T}}\rho _{0}^2 |v^{\prime }|^2 \right) = J(y^{\prime },v^{\prime }). \end{aligned}$$

(37)

Hence, $(\tilde{y},\tilde{v}) = (\hat{y},\hat{v})$. Finally, we also deduce from (36) that

$$\begin{aligned} \limsup _{R\rightarrow +\infty } \left( \displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}T_R(\rho )^2 |y_{R}|^2 + \displaystyle \int \!\!\!\!\displaystyle \int _{q_{T}}\rho _{0}^2 |v_{R}|^2 \right) \le J(\tilde{y},\tilde{v}), \end{aligned}$$

whence we see that (17) holds. $\square$

Proof of Proposition 3.2

In view of the structures of the previous extremal problems, the proof of this result can be obtained from standard results in optimal control theory; see for instance [25, 35, 36]. However, for completeness, we will provide a proof that uses Fenchel–Rockafellar theory.

From the decomposition (21), we can write, for any $(y,v) \in \mathcal {A}(y_{0},T)$ that $J_{R,\varepsilon }(y,v)=F_{R,\varepsilon }(Mv,Bv)+ G(v)$, where the functions $F_{R,\varepsilon }$ and $G$ are defined by

$$\begin{aligned} F_{R,\varepsilon }(z,z_{T}):= \frac{1}{2}\displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}T_R(\rho )^2 |z+\overline{y}|^2\,{\text {d}}x\,{\text {d}}t+ \frac{1}{2\varepsilon }\int _0^1 |z_{T}+\overline{y}(\cdot ,T)|^2 \,{\text {d}}x \end{aligned}$$

and

$$\begin{aligned} G(v):=\frac{1}{2}\displaystyle \int \!\!\!\!\displaystyle \int _{q_{T}}\rho _0^2 | v |^2\,{\text {d}}x\,{\text {d}}t. \end{aligned}$$

Let us introduce $V:=L^2(Q_T)\times L^2(q_T)$. The functions $F_{R,\varepsilon }:V \mapsto \mathbb {R}$ and $G:L^2(q_T) \mapsto \mathbb {R}$ are both convex and continuous, and we can apply the duality Theorem of W. Fenchel and T.R. Rockafellar; see Theorem 4.2 p. 60 in [18]. We deduce that

$$\begin{aligned} \inf _{(y,v)\in \mathcal {A}(y_{0},T)} J_{R,\varepsilon }(y,v)= -\inf _{(\mu ,\varphi _T)\in V} \biggl \{G^{\star }(M^{\star }\mu + B^{\star }\varphi _T) +F_{R,\varepsilon }^{\star }(-(\mu ,\varphi _T))\biggr \}, \end{aligned}$$

where $F_{R,\varepsilon }^{\star }$ and $G^{\star }$ are the convex conjugate functions of $F_{R,\varepsilon }$ and $G$, respectively.

Note that

$$\begin{aligned} F_{R,\varepsilon }^{\star }(\mu ,\varphi _T)&= \sup _{V} \biggl \{ \displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}\mu \,z \,{\text {d}}x\,{\text {d}}t + \int _0^1 \varphi _T \, z_{T} \,{\text {d}}x - F(z,z_{T})\biggr \} \\&= \frac{1}{2}\displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}T_R(\rho )^{-2} |\mu |^2 \,{\text {d}}x\,{\text {d}}t - \displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}\mu \,\overline{y} \,{\text {d}}x\,{\text {d}}t \\&\quad + \frac{\varepsilon }{2} \Vert \varphi _T\Vert ^2_{L^2} - \int _0^1 \varphi _T \, \overline{y}(\cdot ,T) \,{\text {d}}x \end{aligned}$$

for all $(\mu ,\varphi _T) \in V$. On the other hand, $G^{\star }(w)=G(\rho ^{-2}w)$ for all $w \in L^2(Q_T)$. Therefore,

$$\begin{aligned} G^{\star }(M^{\star }\mu + B^{\star }\varphi _T) + F_{R,\varepsilon }^{\star }(-(\mu ,\varphi _T))&= \frac{1}{2} \displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}T_R(\rho )^{-2} |\mu |^2 \,{\text {d}}x\,{\text {d}}t\\&+ \frac{1}{2 }\displaystyle \int \!\!\!\!\displaystyle \int _{q_{T}}\rho _0^{-2}|\varphi |^2\,{\text {d}}x\,{\text {d}}t + \frac{\varepsilon }{2} \Vert \varphi _T\Vert _{L^2}^2\\&+ \displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}\mu \, \overline{y} \,{\text {d}}x\,{\text {d}}t + \int _0^1 \varphi _T \, \overline{y}(\cdot ,T) \,{\text {d}}x, \end{aligned}$$

where we have used again the notation $\varphi =M^{\star }\mu +B^{\star }\varphi _T$.

Finally, multiplying the state equation of (25) by $\overline{y}$ and integrating by parts, we obtain that

$$\begin{aligned} \displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}\mu \, \overline{y} \,{\text {d}}x\,{\text {d}}t + \int _0^1 \varphi _T \, \overline{y}(\cdot ,T) \,{\text {d}}x = \int _0^1 \varphi (\cdot ,0) \, y_0 \,{\text {d}}x, \end{aligned}$$

whence

$$\begin{aligned}&G^{\star }(M^{\star }\mu + B^{\star }\varphi _T) + F_{R,\varepsilon }^{\star }(-(\mu ,\varphi _T))\\&\quad = {1 \over 2} \left( \displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}T_R(\rho )^{-2} |\mu |^2 \,{\text {d}}x\,{\text {d}}t + \displaystyle \int \!\!\!\!\displaystyle \int _{q_{T}}\rho _0^{-2} |\varphi |^2 \,{\text {d}}x\,{\text {d}}t\right) \\&\qquad + \displaystyle \int _0^1 \varphi (\cdot ,0)\,y_{0} \,{\text {d}}x + \frac{\varepsilon }{2}\Vert \varphi _T\Vert _{L^2}^2. \end{aligned}$$

This proves that (24) is the dual of (18).

It is also easy to verify that (18) and (24) are stable and possess unique solutions. Indeed, the hypotheses of Theorem 4.2 in [18] are satisfied for (18) (notice that this is not the case for (16), since the interior of the constraint set $\mathcal {C}(y_0,T)$ is empty).

Finally, let us deduce that the optimality conditions (26) hold. Let us set $(y,v) = (y_{R,\varepsilon },v_{R,\varepsilon })$ and $(\mu ,\varphi _{T}) = (\mu _{R,\varepsilon },\varphi _{T,R,\varepsilon })$. Then, since (24) and (18) are dual to each other, one has:

$$\begin{aligned} \displaystyle 0&= {1 \over 2} \displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}T_R(\rho )^2 |y|^2 \,{\text {d}}x\,{\text {d}}t + {1 \over 2} \displaystyle \int \!\!\!\!\displaystyle \int _{q_{T}}\rho _{0}^2 |v|^2 \,{\text {d}}x\,{\text {d}}t+ {1 \over 2\varepsilon } \Vert y(\cdot \,,T)\Vert _{L^2}^2 \\&\quad + {1 \over 2} \!\displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}\!\!\!T_R(\rho )^{-2} |\mu |^2 \,{\text {d}}x\,{\text {d}}t \!+\! {1 \over 2} \!\displaystyle \int \!\!\!\!\displaystyle \int _{q_{T}}\!\!\rho _{0}^{-2} |\varphi |^2 \,{\text {d}}x\,{\text {d}}t \\&\quad + (\varphi (\cdot \,,0),y_{0}) \!+\! {\varepsilon \over 2} \Vert \varphi _T \Vert _{L^2}^2 \\&= {1 \over 2} \displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}T_R(\rho )^2 |y +T_R(\rho )^{-2}\mu |^2 \,{\text {d}}x\,{\text {d}}t + {1 \over 2} \displaystyle \int \!\!\!\!\displaystyle \int _{q_{T}}\rho _{0}^2 |v - \rho _{0}^{-2}\varphi |^2 \,{\text {d}}x\,{\text {d}}t \\&\quad + {1 \over 2\varepsilon } \Vert y(\cdot \,,T) + \varepsilon \varphi _{T}\Vert _{L^2}^2 \!-\! \displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}T_R(\rho )^2 \mu \,y \,{\text {d}}x\,{\text {d}}t \!+\! \displaystyle \int \!\!\!\!\displaystyle \int _{q_{T}}\rho _{0}^2\,v\,\varphi \,{\text {d}}x\,{\text {d}}t \\&\quad - (y(\cdot \,,T),\varphi _{T})_{L^2} + (\varphi (\cdot \,,0),y_{0})_{L^2}. \end{aligned}$$

However, the terms in the last line cancel, since $\varphi = M^{\star }\mu + B^{\star }\varphi _{T}$. Consequently,

$$\begin{aligned}&\displaystyle \displaystyle \int \!\!\!\!\displaystyle \int _{Q_{T}}T_R(\rho )^2 |y + T_R(\rho )^{-2}\mu |^2 \,{\text {d}}x\,{\text {d}}t + \displaystyle \int \!\!\!\!\displaystyle \int _{q_{T}}\rho _{0}^2 |v - \rho _{0}^{-2}\varphi |^2 \,{\text {d}}x\,{\text {d}}t \displaystyle \\&\quad + {1 \over \varepsilon } \Vert y(\cdot \,,T) + \varepsilon \varphi _{T}\Vert _{L^2}^2 = 0. \end{aligned}$$

and we get (26). $\square$