Boundary Controllability of a Simplified Stabilized Kuramoto-Sivashinsky System

Abstract

In this paper, we study the controllability of a nonlinear system of coupled second- and fourth-order parabolic equations. This system can be regarded as a simplification of the well-known stabilized Kuramoto-Sivashinsky system. Using only one control applied on the boundary of the second-order equation, we prove that the local-null controllability of the system holds if the square root of the diffusion coefficient of the second-order equation is an irrational number with finite Liouville-Roth constant.

Appendices

Appendix A: A Well-Posedness Result

Here, we give a proof of Proposition 2.1.

Proof

Since $(-\partial _{xx})^{2}:H^{2}(0,\pi )\cap H_{0}^{1}(0,\pi )\subset (H^{2}(0, \pi )\cap H_{0}^{1}(0,\pi ))^{\prime }\to (H^{2}(0,\pi )\cap H_{0}^{1}(0, \pi ))^{\prime}$ and $-d\partial _{xx}:L^{2}(0,\pi )\subset (H^{2}(0,\pi )\cap H_{0}^{1}(0, \pi ))^{\prime}\to (H^{2}(0,\pi )\cap H_{0}^{1}(0,\pi ))^{\prime}$ are strictly positive self-adjoint operators, we have the following two results due to [4, Thm. 3.1, pp. 143].

If $f\in L^{2}(t_{1},t_{2};(H^{2}(0,\pi )\cap H_{0}^{1}(0,\pi ))^{\prime})$ and $u_{1}\in L^{2}(0,\pi )$, then

$$ \left \lbrace \textstyle\begin{array}{l@{\quad}l} u_{t}(t,x) + u_{xxxx}(t,x) = f ,&t\in (t_{1},t_{2}),\,x\in (0,\pi ), \\ u(t,0)=u_{xx}(t,0) = 0,&t\in (t_{1},t_{2}), \\ u(t,\pi )=u_{xx}(t,\pi ) = 0,&t\in (t_{1},t_{2}), \end{array}\displaystyle \right . $$

(A.1)

admits a unique solution $u\in C([t_{1},t_{2}],L^{2}(0,\pi ))\cap L^{2}(t_{1},t_{2};H^{2}(0, \pi )\cap H_{0}^{1}(0,\pi ))$ satisfying $u(t_{1},\cdot )=u_{1}$, and we have

$$ \left \lVert u\right \rVert _{L^{2}(t_{1},t_{2};H^{2}\cap H_{0}^{1})}^{2}+ \left \lVert u\right \rVert _{C([t_{1},t_{2}],L^{2})}^{2}\le K_{1} \left (\left \lVert u_{1}\right \rVert _{L^{2}}^{2}+\left \lVert f \right \rVert _{L^{2}(t_{1},t_{2};(H^{2}\cap H_{0}^{1})^{\prime})}^{2} \right ), $$

(A.2)

for some $K_{1}>0$ independent of $t_{1}$, $t_{2}$.

The second result tells that if $h \in L^{2}(t_{1},t_{2})$ and $v_{1}\in H^{-1}(0,\pi )$, then

$$ \left \lbrace \textstyle\begin{array}{l@{\quad}l} v_{t}(t,x) - d v_{xx}(t,x) = 0, &t\in (t_{1},t_{2}),\,x\in (0,\pi ), \\ v(t,0) = h(t),\, v(t,\pi )=0,&t\in (t_{1},t_{2}), \end{array}\displaystyle \right . $$

(A.3)

admits a unique solution $v\in C([t_{1},t_{2}],H^{-1}(0,\pi ))\cap L^{2}(t_{1},t_{2};L^{2}(0, \pi ))$ satisfying $v(t_{1},\cdot )=v_{1}$. Moreover, we have the following energy estimate

$$ \left \lVert v\right \rVert _{L^{2}(t_{1},t_{2};L^{2})}^{2}+\left \lVert v\right \rVert _{C([t_{1},t_{2}],H^{-1})}^{2}\le K_{2}\left ( \left \lVert v_{1}\right \rVert _{H^{-1}}^{2} + \left \lVert h\right \rVert _{L^{2}(t_{1},t_{2})}^{2} \right ), $$

with $K_{2}>0$ independent of $t_{1}$, $t_{2}$.

We will now exploit the cascade structure of the system. Notice that if $v\in L^{2}(t_{1},t_{2}; L^{2}(0,\pi ))$, then in particular $v\in L^{2}(t_{1},t_{2};(H^{2}(0,\pi )\cap H_{0}^{1}(0,\pi ))^{\prime})$ and

$$ \left \lVert v\right \rVert _{L^{2}(t_{1},t_{2};(H^{2}\cap H_{0}^{1})^{ \prime})}^{2}\le \left \lVert v\right \rVert _{L^{2}(t_{1},t_{2};L^{2})}^{2}. $$

We take $v$ to be the solution of (A.3), and $u$ to be the solution of (A.1) with right hand side $f+v$. Since

$$ \left \lVert v+f\right \rVert _{L^{2}(t_{1},t_{2};(H^{2}\cap H_{0}^{1})^{ \prime})}^{2}\le 2\left (\left \lVert f\right \rVert _{L^{2}(t_{1},t_{2};(H^{2} \cap H_{0}^{1})^{\prime})}^{2}+\left \lVert v\right \rVert _{L^{2}(t_{1},t_{2};(H^{2} \cap H_{0}^{1})^{\prime})}^{2}\right ), $$

we get

$$ \left \lVert v\right \rVert _{L^{2}(t_{1},t_{2};L^{2})}^{2}+\left \lVert v\right \rVert _{C([t_{1},t_{2}],H^{-1})}^{2}\le K_{2}\left ( \left \lVert h\right \rVert _{L^{2}(t_{1},t_{2})}^{2} +\left \lVert v_{1} \right \rVert _{H^{-1}}^{2} \right ), $$

(A.4)

and

$$\begin{aligned} \left \lVert u\right \rVert _{L^{2}(t_{1},t_{2};H^{2}\cap H_{0}^{1})}^{2}+ \left \lVert u\right \rVert _{C([t_{1},t_{2}],L^{2})}^{2} &\le K_{1} \left (\left \lVert u_{1}\right \rVert _{L^{2}}^{2}+\left \lVert f+v \right \rVert _{L^{2}(t_{1},t_{2};(H^{2}\cap H_{0}^{1})^{\prime})}^{2} \right ) \\ &\le 2K_{1}\left (\left \lVert u_{1}\right \rVert _{L^{2}}^{2}+\left \lVert f\right \rVert _{L^{2}(t_{1},t_{2};(H^{2}\cap H_{0}^{1})^{ \prime})}^{2}\right ) \\ &\quad +2K_{2}K_{1}\left (\left \lVert h\right \rVert _{L^{2}(t_{1},t_{2})}^{2} +\left \lVert v_{1}\right \rVert _{H^{-1}}^{2} \right ), \end{aligned}$$

(A.5)

where we have used (A.4) in the last line. Writing $y=(u,v)$ and $y_{1}=(u_{1},v_{1})$, we can combine (A.4) and (A.5) and take $C_{1}=\max \{K_{2},(2K_{1}+1)K_{2}\}$ to deduce

$$ \begin{aligned} \left \lVert y\right \rVert _{L^{2}(t_{1},t_{2};(H^{2} \cap H_{0}^{1})\times L^{2})}^{2}&+\left \lVert y\right \rVert _{C([t_{1},t_{2}],L^{2} \times H^{-1})}^{2} \\ &\le C\left (\left \lVert h\right \rVert _{L^{2}(t_{1},t_{2})}^{2} + \left \lVert y_{1}\right \rVert _{L^{2}\times H^{-1}}^{2} +\left \lVert f\right \rVert _{L^{2}(t_{1},t_{2};(H^{2}\cap H_{0}^{1})^{ \prime})}^{2}\right ). \end{aligned} $$

This concludes the proof. □

Appendix B: Sketch of the Proof of Proposition 2.2

We define $T_{k}=T\left (1-\dfrac{1}{q^{k}}\right )$. Notice that with this

$$ \rho _{0}(T_{k+2})=\rho _{1}(T_{k})K(T_{k+2}-T_{k+1}). $$

Consider the notation $y=(u,v)$. We recursively define $y_{k+1}$ as $z(T_{k+1})$, where $z=(z_{1},z_{2})$ solves (2.1) on $(T_{k},T_{k+1})$ with initial condition $z(T_{k})=0$ and control $h=0$. By Proposition 2.1 we have

$$ \left \lVert z\right \rVert _{L^{2}(T_{k},T_{k+1};(H^{2}\cap H_{0}^{1}) \times L^{2})}^{2}+\left \lVert z\right \rVert _{C([T_{k},T_{k+1}],L^{2} \times H^{-1})}^{2}\le C_{1}\left \lVert f\right \rVert _{L^{2}(T_{k},T_{k+1};(H^{2} \cap H_{0}^{1})^{\prime})}^{2}, $$

with this

$$ \left \lVert y_{k+1}\right \rVert _{L^{2}\times H^{-1}}^{2}=\left \lVert z(T_{k+1})\right \rVert _{L^{2}\times H^{-1}}^{2}\le \left \lVert z\right \rVert _{C([T_{k},T_{k+1}],L^{2}\times H^{-1})}^{2} \le C_{1}\left \lVert f\right \rVert _{L^{2}(T_{k},T_{k+1};(H^{2} \cap H_{0}^{1})^{\prime})}^{2}. $$

Define $h$ on $(T_{k},T_{k+1})$ as the minimal norm control that drives (2.1) to zero at time $T_{k+1}$ with initial condition $y_{k}$ at time $T_{k}$ and source term $f\equiv 0$. We denote the corresponding solution by $w$. Since we have assumed that system (2.1) with $f\equiv 0$ is null-controllable with control cost $K(\cdot )$, we have, because of Remark 3.3,

$$ \left \lVert h\right \rVert _{L^{2}(T_{k},T_{k+1})}^{2}\le K^{2}(T_{k+1}-T_{k}) \left \lVert y_{k}\right \rVert _{L^{2}\times H^{-1}}^{2} $$

and therefore

$$\begin{aligned} \left \lVert h\right \rVert _{L^{2}(T_{k+1},T_{k+2})}^{2} \le& K^{2}(T_{k+2}-T_{k+1}) \left \lVert y_{k+1}\right \rVert _{L^{2}\times H^{-1}}^{2} \\ \le& C_{1}K^{2}(T_{k+2}-T_{k+1}) \left \lVert f\right \rVert _{L^{2}(T_{k},T_{k+1};(H^{2}\cap H_{0}^{1})^{ \prime})}^{2}. \end{aligned}$$

We define $y=z+w$. Since $z(T_{k}^{-})+w(T_{k}^{-})=z(T_{k}^{+})+w(T_{k}^{+})$, $y$ is continuous at $T_{k}$ for every $k\geq 0$. With this definition, it is also clear that $y$ solves (2.1) for $t\in [0,T)$.

Notice that, due to the fact that $\rho _{0}$ and $\rho _{1}$ are nonincreasing

$$ \begin{aligned} \left \lVert h\right \rVert _{L_{\rho _{0}}^{2}(T_{k+1},T_{k+2})}^{2}& \le \dfrac{1}{\rho _{0}^{2}(T_{k+2})}\left \lVert h\right \rVert _{L^{2}(T_{k+1},T_{k+2})}^{2} \\ &\le C_{1}\dfrac{K^{2}(T_{k+2}-T_{k+1})}{\rho _{0}^{2}(T_{k+2})}\left \lVert f\right \rVert _{L^{2}(T_{k},T_{k+1};(H^{2}\cap H_{0}^{1})^{ \prime})}^{2} \\ &\le \dfrac{C_{1}}{\rho _{1}^{2}(T_{k})}\left \lVert f\right \rVert _{L^{2}(T_{k},T_{k+1};(H^{2} \cap H_{0}^{1})^{\prime})}^{2}\le C_{1}\left \lVert f\right \rVert _{L_{\rho _{1}}^{2}(T_{k},T_{k+1};(H^{2}\cap H_{0}^{1})^{ \prime})}^{2}, \end{aligned} $$

(B.1)

and

$$ \left \lVert h\right \rVert _{L_{\rho _{0}}^{2}(0,T_{1})}^{2}\le \dfrac{1}{\rho _{0}^{2}(T_{1})}\left \lVert h\right \rVert _{L^{2}(0,T_{1})}^{2} \le \dfrac{K^{2}(T_{1})}{\rho _{0}^{2}(T_{1})}\left \lVert y_{0} \right \rVert _{L^{2}\times H^{-1}}^{2}. $$

(B.2)

Proposition 2.1 yields

$$\begin{aligned} &\left \lVert y\right \rVert _{L_{\rho _{0}}^{2}(T_{k+1},T_{k+2};(H^{2} \cap H_{0}^{1})\times L^{2})}^{2}+\left \lVert \dfrac{y}{\rho _{0}} \right \rVert _{C([T_{k+1},T_{k+2}],L^{2}\times H^{-1})}^{2} \\ &\quad \le C_{1}^{2}\left (1+\dfrac{1}{K^{2}(T_{k+2}-T_{k+1})}\right ) \left \lVert f\right \rVert _{L_{\rho _{1}}^{2}(T_{k},T_{k+1};(H^{2} \cap H_{0}^{1})^{\prime})}^{2} \\ &\qquad +\dfrac{C_{1}}{K^{2}(T_{k+2}-T_{k+1})}\left \lVert f\right \rVert _{L_{\rho _{1}}^{2}(T_{k+1},T_{k+2};(H^{2}\cap H_{0}^{1})^{ \prime})}^{2} \end{aligned}$$

(B.3)

and also

$$ \begin{aligned} &\left \lVert y\right \rVert _{L_{\rho _{0}}^{2}(0,T_{1};(H^{2} \cap H_{0}^{1})\times L^{2})}^{2}+\left \lVert \dfrac{y}{\rho _{0}} \right \rVert _{C([0,T_{1}],L^{2}\times H^{-1})}^{2} \\ &\quad \le \dfrac{C_{1}(K^{2}(T_{1})+1)}{\rho _{0}^{2}(T_{1})}\left \lVert y_{0}\right \rVert _{L^{2}\times H^{-1}}^{2}+ \dfrac{C_{1}\rho _{1}^{2}(0)}{\rho _{0}^{2}(T_{1})}\left \lVert f \right \rVert _{L_{\rho _{1}}^{2}(0,T_{1};(H^{2}\cap H_{0}^{1})^{ \prime})}^{2}. \end{aligned} $$

(B.4)

It follows from Proposition 2.1 that $y$ is continuous on $[0,T]$, with values in $L^{2}\times H^{-1}$. Because $\rho _{0}$ is positive on $[0,T)$, $\frac{y}{\rho _{0}}$ is also continuous on $[0,T)$. Since the right hand side of (B.3) converges to 0 as $k\to \infty$, it follows that $\frac{y(t)}{\rho _{0}}\to 0$ in $L^{2}\times H^{-1}$ as $t\to T^{-}$. With this, we may extend $\frac{y}{\rho _{0}}$ to $[0,T]$ and it remains continuous, in particular, this implies that $(u(T),v(T))=0$. This tells us that the control $h$ constructed drives the state to zero, and also gives a bound on how quickly the convergence takes place, namely, faster than $\rho _{0}$ approaches 0.

Combining estimates (B.1)–(B.2) with (B.3)–(B.4) we can deduce the existence of a constant $C_{2}>0$ such that (2.2) holds. This ends the proof.  □

Appendix C: Liouville-Roth Constant of the Reciprocal of a Number

Lemma C.1

Let $x>0$. Then $x$ and $1/x$ have the same Liouville-Roth constant.

Proof

Assume without loss of generality that the Liouville-Roth constant of $1/x$ is strictly greater than that of $x$ (this implies $x$ has finite Liouville-Roth constant). Fix $r$ strictly in between both Liouville-Roth constants and $\varepsilon >0$ such that $r+\varepsilon$ is also in this interval. By definition, there exist infinitely many pairs $(p,q)\in \mathbb{Z}^{2}$ with $q>0$ such that

$$ 0< \left |\frac{1}{x}-\frac{p}{q}\right |< \frac{1}{q^{r+\varepsilon}}, $$

so we may construct a sequence $(p_{n},q_{n})$ of distinct such pairs. Since for a given $q_{n}$, only finitely many values of $p_{n}$ can satisfy the equation, we know by the pigeonhole principle that $q_{n}\to \infty$ as $n\to \infty$. With this, $\frac{p_{n}}{q_{n}}\to \frac{1}{x}$, which implies $p_{n}\to \infty$ and $\frac{q_{n}}{p_{n}}\to x$. Notice that

$$ 0< \left |x-\frac{q_{n}}{p_{n}}\right |p_{n}^{r}=\frac{xq_{n}}{p_{n}} \left |\frac{p_{n}}{q_{n}}-\frac{1}{x}\right |q_{n}^{r+\varepsilon} \left (\frac{p_{n}}{q_{n}}\right )^{r+\varepsilon} \frac{1}{p_{n}^{\varepsilon}}\to 0, $$

since $\frac{xq_{n}}{p_{n}}\left (\frac{p_{n}}{q_{n}}\right )^{r+ \varepsilon}\to x^{2-r-\varepsilon}$, $\left |\frac{p_{n}}{q_{n}}-\frac{1}{x}\right |q_{n}^{r+\varepsilon}<1$ and $\frac{1}{p_{n}^{\varepsilon}}\to 0$. It follows that there are infinitely many pairs $(p,q)$ such that

$$ 0< \left |x-\frac{q}{p}\right |< \frac{1}{p^{r}}, $$

which contradicts the fact that $r$ is greater that the Liouville-Roth constant of $x$. □