Robust Stackelberg controllability for the Kuramoto–Sivashinsky equation

Abstract

In this article, the robust Stackelberg controllability (RSC) problem is studied for a nonlinear fourth-order parabolic equation, namely the Kuramoto–Sivashinsky equation. When three external sources are acting into the system, the RSC problem consists essentially in combining two subproblems: the first one is a saddle point problem among two sources. Such sources are called the “follower control” and its associated “disturbance signal.” This procedure corresponds to a robust control problem. The second one is a hierarchic control problem (Stackelberg strategy), which involves the third force, so-called leader control. The RSC problem establishes a simultaneous game for these forces in the sense that the leader control has as objective to verify a controllability property, while the follower control and perturbation solve a robust control problem. In this paper, the leader control obeys to the exact controllability to the trajectories. Additionally, iterative algorithms to approximate the robust control problem as well as the robust Stackelberg strategy for the nonlinear Kuramoto–Sivashinsky equation are developed and implemented.

A. Appendix

In this appendix, we mention the well-posedness results we used in this paper for both linearized and nonlinear equations. First, in order to consider external sources with lower regularity in space, we define solution by transposition for the linearized KS equation. Let us define

$$\begin{aligned}{\mathcal {Z}}:=C([0,T];H_0^2(0,1))\cap L^2(0,T;H^4(0,1))\cap L^\infty (0,T;W^{1,\infty }(0,1)).\end{aligned}$$

Hence, let $\overline{y}_0\in H_0^2(0,1)$ and let $\overline{y}\in {\mathcal {Z}}$ be a solution of the KS equation

$$\begin{aligned} {\left\{ \begin{array}{ll} \overline{y}_{t}+\overline{y}_{xxxx}+\overline{y}_{xx}+\overline{y}\overline{y}_x =0 &{} \text {in }Q,\\ \overline{y}(0,t)=\overline{y}(1,t)=\overline{y}_{x}(0,t)=\overline{y}_{x}(1,t)=0 &{} \text {on }(0,T),\\ \overline{y}(\cdot ,0)=\overline{y}_0&{} \text {in }(0,1). \end{array}\right. } \end{aligned}$$

(A.1)

First, we consider the following linearized system:

$$\begin{aligned} {\left\{ \begin{array}{ll} y_{t}+y_{xxxx}+y_{xx}+\overline{y}y_x+\overline{y}_{x}y =f &{} \text {in }Q,\\ y(0,t)=y(1,t)=y_{x}(0,t)=y_{x}(1,t)=0 &{} \text {on }(0,T),\\ y(\cdot ,0)=y_0&{} \text {in }(0,1). \end{array}\right. } \end{aligned}$$

(A.2)

Now, from [8, Section 2] we have the following definition.

Definition 1

Let $y_0\in H^{-2}_0(0,1)$ and $f\in L^1(0,T;W^{-1,1}(0,1))$. A solution of the system (A.2) is a solution $y\in L^2(Q)$ such that for any $g\in L^2(Q)$,

$$\begin{aligned} \iint \limits _{Q}y(x,t)g(x,t)\hbox {d}x\mathbf{d} t= & {} \langle y_0,w(0,\cdot )\rangle _{H^{-2}(0,1),H^2(0,1)} \nonumber \\&+\langle f,w\rangle _{L^1(0,T;W^{-1,1}(0,1)), L^\infty (0,T;W^{1,\infty }(0,1))}, \end{aligned}$$

(A.3)

where $w=w(x,t)\in {\mathcal {Z}}$ is the solution to

$$\begin{aligned} {\left\{ \begin{array}{ll} -w_{t}+w_{xxxx}+w_{xx}-\overline{y}w_x=g &{} \text {in }Q,\\ w(0,t)=w(1,t)=w_{x}(0,t)=w_{x}(1,t)=0 &{} \text {on }(0,T),\\ w(\cdot ,T)=0&{} \text {in }(0,1). \end{array}\right. } \end{aligned}$$

(A.4)

Lemma 6

Assume $\overline{y}\in {\mathcal {Z}}$. Then, for any $y_0\in H^{-2}_0(0,1)$ and $f\in L^1(0,T;W^{-1,1}(0,1))$, the linearized system (A.2) admits a unique solution $y\in C([0,T];H^{-2}(0,1))\cap L^2(0,T;L^2(0,1))$.

Remark 7

Note that both the regularity for the solution w of (A.4) and an exhaustive proof of Lemma 6 can be obtained in an easy way from [8, Proposition 2.1] and [22]. Due to that, we have omitted those details here.

The following lemma shows regularity results for (A.2) by considering data $(f,y_0)$ belong to more regular spaces like $L^2(Q)\times L^2(0,1)$ and $L^2(Q)\times H^2_0(0,1)$. We invite to the reader to review [7, Appendix A], [8, Proposition 2.1] and [22], for more details.

Lemma 7

Assume $\overline{y}\in {\mathcal {Z}}$.

• (a) For any $y_0\in L^{2}(0,1)$ and $f\in L^2(Q)$, the linearized system (A.2) admits a unique solution $y\in C([0,T];L^2(0,1))\cap L^2(0,T;H^2(0,1))$ with $y_t\in L^2(0,T;H^{-2}(0,1))$. Moreover, there exists a positive constant such that

  $$\begin{aligned} \Vert y\Vert _{C([0,T];L^2(0,1))\cap L^2(0,T;H^2(0,1))}\le C\Bigl (\Vert f\Vert _{L^2(Q)}+\Vert y_0\Vert _{L^{2}_0(0,1)}\Bigr ). \end{aligned}$$

  (A.5)

  Furthermore, if there is a constant $R>0$ such that $\Vert \overline{y}\Vert _{L^\infty (0,T;W^{1,\infty }(0,1))} \le R$, then the constant C only depends on R and T.

• (b) For $(y_0,f)\in H_0^{2}(0,1)\times L^2(Q)$, the linearized system (A.2) admits a unique solution y in $C([0,T];H_0^2(0,1))\cap L^2(0,T;H^4(0,1))$. Moreover, there exists a positive constant such that

  $$\begin{aligned} \Vert y\Vert _{C([0,T];H_0^2(0,1))\cap L^2(0,T;H^4(0,1))}\le C\Bigl (\Vert f\Vert _{L^2(Q)}+\Vert y_0\Vert _{H^{2}_0(0,1)}\Bigr ), \end{aligned}$$

  (A.6)

  Furthermore, if there is a constant $R>0$ such that $\Vert \overline{y}\Vert _{L^\infty (0,T;W^{1,\infty }(0,1))} \le R$, then the constant C only depends on R and T.

Now, we mention a result for coupled fourth-order system. Its proof is found in [7, Appendix A]. Let us consider the system:

$$\begin{aligned} {\left\{ \begin{array}{ll} y_{t}+y_{xxxx}+y_{xx}+\overline{y}y_x+\overline{y}_xy= g_1+ -\mu ^{-2}z &{} \text {in }Q,\\ -z_{t}+z_{xxxx}+z_{xx}-(y+\overline{y})z_{x}=g_2&{} \text {in }Q,\\ y(0,t)=y(1,t)=z(0,t)=z(1,t)=0 &{} \text {on }(0,T),\\ y_{x}(0,t)=y_{x}(1,t)=z_{x}(0,t)=z_{x}(1,t)=0 &{} \text {on }(0,T),\\ y(\cdot ,0)=y_{0}(\cdot ),\,\, z(\cdot ,T)=0&{} \text {in }(0,1). \end{array}\right. } \end{aligned}$$

(A.7)

Lemma 8

Assume that $\overline{y}\in L^\infty (Q)$. Then, there exists $\mu _0>0$ such that for every $\mu \ge \mu _0$, any $g_1,g_2\in L^2(Q)$ and any $y_0\in L^2(0,1)$, (y, z) is the unique solution of (A.7) in the space

$$\begin{aligned}(y,z)\in (L^\infty (0,T;L^{2}(0,1))\cap L^2(0,T;H^2(0,1)))^2.\end{aligned}$$

The next step in this appendix corresponds to the nonlinear problem

$$\begin{aligned} {\left\{ \begin{array}{ll} y_{t}+y_{xxxx}+y_{xx}+\overline{y}y_x+\overline{y}_{x}y+yy_x =f &{} \text {in }Q,\\ y(0,t)=y(1,t)=y_{x}(0,t)=y_{x}(1,t)=0 &{} \text {on }(0,T),\\ y(\cdot ,0)=y_0&{} \text {in }(0,1). \end{array}\right. } \end{aligned}$$

(A.8)

Lemma 9

• (a) Assume $\overline{y}\in L^\infty (0,T;W^{1,\infty }(0,1))$. There exists $\delta >0$ such that for any $(f,y_0)\in L^2(Q)\times L^2(0,1)$ satisfying

  $$\begin{aligned}\Vert y_0\Vert _{L^2(0,1)}+\Vert f\Vert _{L^2(Q)}\le \delta \end{aligned}$$

  problem (A.8) has a unique solution in $C([0,T];L^2(0,1))\cap L^2(0,T;H^2(0,1))$.

• (b) Let $\overline{y}=0$ in (A.8). There exists $\delta >0$ such that for any $(f,y_0)\in L^2(Q)\times H_0^2(0,1)$ satisfying

  $$\begin{aligned}\Vert y_0\Vert _{H_0^2(0,1)}+\Vert f\Vert _{L^2(Q)}\le \delta \end{aligned}$$

  problem (A.8) has a unique solution in $C([0,T];H_0^2(0,1))\cap L^2(0,T;H^4(0,1))$.

  Moreover, there exists a positive constant C depending only on T such that

  $$\begin{aligned} \Vert y\Vert _{C([0,T];H_0^2(0,1))\cap L^2(0,T;H^4(0,1))}\le C\Bigl (\Vert f\Vert _{L^2(Q)}+\Vert y_0\Vert _{H^{2}_0(0,1)}\Bigr ). \end{aligned}$$

  (A.9)

Remark 8

Although in [7, Theorem A.4] the authors have proved the first part of the above result by considering $f\in L^1(0,T;L^2(0,1))$ instead of $f\in L^2(0,T;L^2(0,1))$, their arguments can be easily adapted for proving this part of Lemma 9. The second part can be obtained from [22]. For this reason, we have omitted the proof of Lemma 9.

Remark 9

Observe that, from Lemma 9, part b), and the fact that $H_0^2(0,1)$ embeds continuously into $W^{1,\infty }(0,1)$, it follows that $y\in L^\infty (0,T;W^{1,\infty }(0,1))$.