Stabilization to trajectories for parabolic equations

Abstract

Both internal and boundary feedback exponential stabilization to trajectories for semilinear parabolic equations in a given bounded domain are addressed. The values of the controls are linear combinations of a finite number of actuators which are supported in a small region. A condition on the family of actuators is given which guarantees the local stabilizability of the control system. It is shown that a linearization-based Riccati feedback stabilizing controller can be constructed. The results of numerical simulations are presented and discussed.

Appendix

1.1 Proof of Lemma 2.2

Observe that given z solving (22), with $\gamma =0$, by rescaling time $t=\frac{\tau }{\nu }$ and defining $\breve{z}(\tau )=z(\frac{\tau }{\nu })$, $\breve{a}(\tau )={\hat{a}}(\frac{\tau }{\nu })$, $\breve{b}(\tau )={\hat{b}}(\frac{\tau }{\nu })$, and $\breve{h}(\tau )=h(\frac{\tau }{\nu })$, we find

$$\begin{aligned} \partial _\tau \breve{z}-\varDelta \breve{z}+ \textstyle \frac{\breve{a}}{\nu } \breve{z} +\nabla \cdot \left( \textstyle \frac{\breve{b}}{\nu }\breve{z}\right) +\textstyle \frac{\breve{h}}{\nu }=0, \qquad \breve{z}|_\varGamma =0, \qquad \breve{z}(\nu s_0)=z_0. \end{aligned}$$

Then, by standard arguments, we can find

$$\begin{aligned} \left| \breve{z}(\nu s)\right| _{H}^2&\le \mathrm {e}^{\frac{C^2}{\nu ^2}C_{\mathcal {W}}(\nu s-\nu s_0)} \left( \left| \breve{z}(\nu s_0)\right| _{H}^2+\textstyle \frac{1}{\nu ^2}\left| \breve{h}\right| _{L^2((\nu s_0,\nu s),V')}^2\right) ,\\ 2\nu \left| \breve{z}\right| _{L^2(\nu I,V)}^2&= \left| \breve{z}(\nu s_0)\right| _{H}^2-\left| \breve{z}(\nu s_1)\right| _{H}^2 +\int _{\nu s_0}^{\nu s_1}\left\langle \textstyle \frac{\breve{a}}{\nu } \breve{z} +\nabla \cdot \left( \textstyle \frac{\breve{b}}{\nu }\breve{z}\right) +\textstyle \frac{\breve{h}}{\nu },\breve{z}\right\rangle _{V',V}\,\mathrm {d}\tau \\&\le C_1\left| \breve{z}\right| _{L^\infty (\nu I,H)}^2 +3\left| \textstyle \frac{\breve{h}}{\nu }\right| _{L^2(\nu I,V')}^2+ \left| \breve{z}\right| _{L^2(\nu I,V)}^2, \end{aligned}$$

with $C_1:=\left( 1+3C^2\left| \textstyle \frac{\breve{a}}{\nu }\right| _{L^\infty (\nu I,L^d)}^2 +3C^2\left| \textstyle \frac{\breve{b}}{\nu }\right| _{L_w^\infty (\nu I,L^\infty )}^2\right)$. Therefore,

$$\begin{aligned} \left| {z}(s)\right| _{H}^2&\le \mathrm {e}^{\frac{C^2}{\nu }C_{\mathcal {W}}(s-s_0)} \left( \left| {z}(s_0)\right| _{H}^2+\textstyle \frac{1}{\nu }\left| h\right| _{L^2((s_0,s),V')}^2\right) ,\\ \left| z\right| _{L^2(I,V)}^2&\le \left( \textstyle \frac{1}{2\nu }+\textstyle \frac{3C^2}{2\nu ^3}\left| (a,b)\right| _{\mathcal {W}}^2 \right) \left| {z}\right| _{L^\infty (I,H)}^2 +\textstyle \frac{3}{2\nu ^2}\left| {h}\right| _{L^2(I,V')}^2,\\ \left| {\textstyle \frac{\partial }{\partial t}} z\right| _{L^2(I,V')}&\le \left( \nu +C\left| a\right| _{L^\infty (I,L^d)}+C\left| b\right| _{L_w^\infty (I,L^\infty )}\right) \left| {z}\right| _{L^2(I,V)} +\left| {h}\right| _{L^2(I,V')}, \end{aligned}$$

which imply the statement of Lemma 2.2. $\square$

1.2 Proof of Proposition 2.1

We construct an extension for ${\hat{b}}={\hat{b}}(t,x)$ independent of t. We consider a different local system of space coordinates w, in order to flatten the boundary. We use some basic concepts from Riemannian manifolds, see [44, chapter 1, Section 13 and chapter 2, Section 2].

Change of coordinates. Up to a translation and rotation, we may suppose that locally the boundary $\varGamma =\partial \varOmega$ is the graph of a smooth function $\varLambda$, with $\overline{w}:=(w_1,w_2,\ldots ,w_{d-1})$,

$$\begin{aligned} \overline{w}&\mapsto G_\varLambda (\overline{w}):=(\overline{w},\varLambda (\overline{w}))\in \varGamma ,\\ \overline{w}\in {\mathbb {D}}_r^{d-1}&:=\{\overline{w}\in {\mathbb {R}}^{d-1}\mid w_1^2+w_2^2+\cdots +w_{d-1}^2\le r\}, \end{aligned}$$

with (small) $r>0$. Locally, a tubular neighborhood is given by

$$\begin{aligned} \mathcal {T}=\mathcal {T}_{r,l}:=\left\{ x\in {\mathbb {R}}^d\mid x=G_\varLambda (\overline{w})+w_d\mathbf {n}_{G_\varLambda (\overline{w})},\quad w:=(\overline{w},w_{d})\in {\mathbb {D}}_r^{d-1}\times (-l,l)\right\} , \end{aligned}$$

(A.1)

with (small) $l>0$. Where $\mathbf {n}_{G_\varLambda (\overline{w})}$ stands for the unit outward normal vector at $G_\varLambda (\overline{w})\in \varGamma$.

Let us denote $\mathcal {O}^-={\mathbb {D}}_r^{d-1}\times (-l,0)$ and $\mathcal {O}^+={\mathbb {D}}_r^{d-1}\times (0,l)$. Notice that the points outside $\varOmega$ correspond to those in $\mathcal {O}^+$. We may suppose that $x_0=(0,G_\varLambda (0))\in \varGamma _{1}\subseteq \varGamma \textstyle \bigcap \mathcal {T}$, that is, in Proposition 2.1 we may take ${\widetilde{\omega }}$ corresponding to a subset of $\mathcal {O}^+$.

The new coordinates $(w_1,w_2,\ldots ,w_d)$ induce the vector fields

$$\begin{aligned} \textstyle \frac{\partial }{\partial _{w_i}}=\sum \limits _{i=1}^{d}\frac{\partial x_j}{\partial _{w_i}}\frac{\partial }{\partial _{x_j}},\quad i=1,2,\ldots ,d, \end{aligned}$$

(A.2)

defined in $\mathcal {O}:={\mathbb {D}}_r^{d-1}\times (-l,l)$. We can see the neighborhood $\mathcal {T}$ (endowed with the usual Euclidean scalar product) as the Riemannian manifold $(\mathcal {O},g)$ by taking the metric tensor

$$\begin{aligned} g=\sum \limits _{i=1}^{d}\sum \limits _{j=1}^{d}g_{ij}\mathrm {d}w_j\otimes \mathrm {d}w_j, \qquad \text{ with }\quad g_{ij}=\left( {\textstyle \frac{\partial }{\partial _{w_i}}},{\textstyle \frac{\partial }{\partial _{w_j}}}\right) _{{\mathbb {R}}^d} \end{aligned}$$

where $\left( \cdot ,\cdot \right) _{{\mathbb {R}}^d}$ stands for the usual Euclidean scalar product in ${\mathbb {R}}^d$.

The divergence of a vector field $V=\sum \limits _{i=1}^{d}V_i\textstyle \frac{\partial }{\partial _{w_i}}$ reads, in the new coordinates,

$$\begin{aligned} \nabla _w\cdot V =(-1)^{d}\frac{1}{\sqrt{{\bar{g}}}}\displaystyle \sum _{j=1}^{d}\textstyle \frac{\partial (V_j\sqrt{{\bar{g}}})}{\partial _{w_j}}, \end{aligned}$$

where ${\bar{g}}:=\det [g_{ij}]$ stands for the determinant of the matrix whose entries are the coefficients of the metric tensor. Recall that in our setting $\sqrt{{\bar{g}}}$ coincides with the Jacobian $\left| \det \left[ {\frac{\partial x}{\partial w}}\right] \right|$ of the smooth diffeomorphism $w\mapsto x$, because

$$\begin{aligned}{}[g_{ij}]=\left[ \sum _{k=1}^d\textstyle \frac{\partial x_k}{\partial w_i}\textstyle \frac{\partial x_k}{\partial w_j}\right] =\left[ {\frac{\partial x}{\partial w}}\right] \left[ {\frac{\partial x}{\partial w}}\right] ^\top , \end{aligned}$$

where $\left[ {\frac{\partial x}{\partial w}}\right] ^\top$ is the transpose matrix of $\left[ {\frac{\partial x}{\partial w}}\right]$.

The extension. For a given vector field $V^-=\sum \limits _{i=1}^{d}V_i^-\textstyle \frac{\partial }{\partial _{w_i}}$, defined in $\mathcal {O}^-$, we consider the following vector field defined in $\mathcal {O}^+$:

$$\begin{aligned} V^+:=\sum \limits _{i=1}^{d}V_i^+\textstyle \frac{\partial }{\partial _{w_i}},\quad \text{ with }\quad {\left\{ \begin{array}{ll} V_i^+(\overline{w},s)=-\mathcal {Q}(w)V_i^-(\overline{w},-s),&{} \text{ if } i\ne d,\\ V_d^+(\overline{w},s)=\mathcal {Q}(w) V_d^-(\overline{w},-s),&{} \end{array}\right. } \end{aligned}$$

(A.3)

where $s\in (0,l)$ and

$$\begin{aligned} \mathcal {Q}(w):=\frac{\sqrt{{\bar{g}}}|_{(\overline{w},-s)}}{\sqrt{{\bar{g}}}|_{(\overline{w},s)}}. \end{aligned}$$

Then, denoting the mapping $\sigma :\mathcal {O}^+\rightarrow \mathcal {O}^-$, $(\overline{w},s)\mapsto (\overline{w},-s)$, we find

$$\begin{aligned}&\sqrt{{\bar{g}}}|_{(\overline{w},s)} (-1)^{d}(\nabla _w\cdot V^+)|_{(\overline{w},s)} =\displaystyle \sum _{j=1}^{d}\left. \textstyle \frac{\partial (V^+_j\sqrt{{\bar{g}}})}{\partial _{w_j}}\right| _{(\overline{w},s)}\\&\quad =\left. \textstyle \frac{\partial ((V^-_d\sqrt{{\bar{g}}})\circ \sigma )}{\partial _{w_d}}\right| _{(\overline{w},s)} -\displaystyle \sum _{j=1}^{d-1}\left. \textstyle \frac{\partial ((V^-_d\sqrt{{\bar{g}}})\circ \sigma )}{\partial _{w_j}}\right| _{(\overline{w},s)} =-\displaystyle \sum _{j=1}^{d}\left. \textstyle \frac{\partial (V^-_d\sqrt{{\bar{g}}})}{\partial _{w_j}}\right| _{(\overline{w},-s)}, \end{aligned}$$

which gives us

$$\begin{aligned} (\nabla _w\cdot V^+)|_{(\overline{w},s)}=-\mathcal {Q}(w)(\nabla _w\cdot V^-)|_{(\overline{w},-s)}. \end{aligned}$$

(A.4)

For $S\subseteq \mathcal {O}$, let us denote $\mathcal {W}_{2}(S):=\{v\in L^\infty (S,{\mathbb {R}}^d)\mid (\nabla _w\cdot v)\in L^r_g(S,{\mathbb {R}})\}$, with $r\in \{2,\infty \}$ (cf. (29)). In particular, since $\mathcal {Q}(w)$ is a smooth function, we observe that $V^+\in \mathcal {W}_{2}(\mathcal {O}^+)$ if $V^-\in \mathcal {W}_{2}(\mathcal {O}^-)$. Here, $(f,h)_{L^2_g(S,{\mathbb {R}})}:=\int _S fh\, \mathrm {d}\mathcal {O}$, $\mathrm {d}\mathcal {O}:=\sqrt{{\bar{g}}}\mathrm {d}w_1\wedge \mathrm {d}w_2\wedge \cdots \wedge \mathrm {d}w_d$. It is also clear that the linear mapping $V^-\mapsto V^+$ is continuous. Finally, we prove that the function defined by

$$\begin{aligned} V^-\mapsto \overline{V},\quad \text{ with }\quad \overline{V}(w):={\left\{ \begin{array}{ll} V^-(w),&{}\text{ if } w\in \mathcal {O}^-,\\ V^+(w),&{}\text{ if } w\in \mathcal {O}^+, \end{array}\right. } \end{aligned}$$

maps $\mathcal {W}_{2}(\mathcal {O}^-)$ into $\mathcal {W}_{2}(\mathcal {O})$. We need to prove that $\nabla _w\cdot \overline{V}\in L^2_g(\mathcal {O},{\mathbb {R}})$. For a smooth function $\phi \in \mathcal {D}(\mathcal {O}):=C_\mathrm{c}^\infty (\mathcal {O},{\mathbb {R}})$ with support contained in $\mathcal {O}$ we find, in the distribution sense, with $\varGamma _0:=\{w\in \mathcal {O}\mid w_d=0\}$,

$$\begin{aligned}&\langle \nabla _w\cdot \overline{V},\phi \rangle _{\mathcal {D}(\mathcal {O})',\mathcal {D}(\mathcal {O})} :=-\langle \overline{V},\nabla _w\phi \rangle _{\mathcal {D}(\mathcal {O})',\mathcal {D}(\mathcal {O})}=-(\overline{V},\nabla _w\phi )_{L^2_g(\mathcal {O},{\mathbb {R}})}\\&\quad =(\nabla _w\cdot \overline{V},\phi )_{L^2_g(\mathcal {O}^-,{\mathbb {R}})}+(\nabla _w\cdot \overline{V},\phi )_{L^2_g(\mathcal {O}^+,{\mathbb {R}})} \\&\qquad -\int _{\varGamma _0}\!\! \phi g\left( V^-,\textstyle \frac{\partial }{\partial _{w_d}}\right) -\phi g\left( V^+,\textstyle \frac{\partial }{\partial _{w_d}}\right) \,\mathrm {d}_g\varGamma _0. \end{aligned}$$

Notice that $\textstyle \frac{\partial }{\partial _{w_d}}$ is the unit outward normal at $\varGamma _0\subset \partial \mathcal {O}^-$.

Now, since $g\left( V^-,\textstyle \frac{\partial }{\partial _{w_d}}\right) -g\left( V^+,\textstyle \frac{\partial }{\partial _{w_d}}\right) =V^-_d-V^+_d$, from $V^+_d(\overline{w},s)=\mathcal {Q}(w)V^-_d(\overline{w},-s)$ for all $s>0$ and $\mathcal {Q}(\overline{w},0)=1$, we can conclude that $V^-_d-V^+_d$ necessarily vanishes at $\varGamma _0$. Hence, the boundary term vanishes, and we can conclude that $\nabla _w\cdot V\in L^2_g(\mathcal {O},{\mathbb {R}})$. We may write

$$\begin{aligned} \langle \nabla _w\cdot V,\phi \rangle _{\mathcal {D}(\mathcal {O})',\mathcal {D}(\mathcal {O})}&=(\nabla _w\cdot V,\phi )_{L^2_g(\mathcal {O}^-,{\mathbb {R}})}+(\nabla _w\cdot V,\phi )_{L^2_g(\mathcal {O}^+,{\mathbb {R}})}\\&=(\nabla _w\cdot V,\phi )_{L^2_g(\mathcal {O},{\mathbb {R}})}. \end{aligned}$$

Therefore, if in addition we have $\nabla _w\cdot V^-\in L^\infty _g(\mathcal {O}^-,{\mathbb {R}})$ then from (A.4) it follows that $\nabla _w\cdot \overline{V}\in L^\infty _g(\mathcal {O},{\mathbb {R}})$.

It is also clear that the mapping $V^-\mapsto \overline{V}$ maps $\mathcal {W}_{2}(\mathcal {O}^-)$ into $\mathcal {W}_{2}(\mathcal {O})$ continuously.

In the original coordinates, the extension above reads: given ${\hat{b}}={\hat{b}}_i\frac{\partial }{\partial _{x_i}}$, we firstly rewrite ${\hat{b}}$ in the new coordinates ${\hat{b}}={\hat{b}}_i^w\frac{\partial }{\partial _{w_i}}=:V^-$, next we extend $V^-$ to $\overline{V}=\overline{V}_i\frac{\partial }{\partial _{w_i}}$ (through $V^+$ as above), finally we rewrite $\overline{V}$ in the original coordinates: $\overline{V}=\overline{V}_i^o\frac{\partial }{\partial _{x_i}}=:\bar{b}$.

The continuity of $({\tilde{a}},{\tilde{b}})\mapsto ({\tilde{a}},\overline{b})$ from $\mathcal {W}_\mathrm{st}$ into ${\widetilde{\mathcal {W}}}_\mathrm{st}$ follows straightforwardly. $\square$

Remark A.2

In [40, Proposition 4.2] we find, for $d=3$, the result we present here in Proposition 2.1. Our proof borrows the idea from [40, Appendix]. We still present the proof in here because in [40], when computing the vector fields $\frac{\partial }{\partial _{w_i}}$ for $i=1,2,\ldots ,d-1$, as in (A.2) above, the terms $w_d\frac{\partial (\mathbf {n}_{G_{\varLambda (\overline{w})}})_j}{\partial _{w_i}}\frac{\partial }{\partial _{x_j}}$ have been missed, see [40, Eq. (A.2)].

1.3 Proof of Proposition 2.2

From [33, chapter 4, Section 2.5, Theorem 2.3] we know that $u\mapsto (u_0,u|_\varGamma )$, maps the space $W({\mathbb {R}}_0,H^2(\varOmega ,{\mathbb {R}}),L^2(\varOmega ,{\mathbb {R}}))$ continuously onto the product space

$$\begin{aligned} \left\{ (z,g)\in H^1(\varOmega ,{\mathbb {R}})\times \left( L^2({\mathbb {R}}_0,H^\frac{3}{2}(\varGamma ,{\mathbb {R}}))\textstyle \bigcap H^\frac{3}{4}({\mathbb {R}}_0,L^2(\varGamma ,{\mathbb {R}}))\right) \,\Bigr |\, g(0)=z|_\gamma \right\} . \end{aligned}$$

In particular, this implies that $G^2(J,\varGamma ) = L^2(J,H^\frac{3}{2}(\varGamma ,{\mathbb {R}}))\bigcap H^\frac{3}{4}(J,L^2(\varGamma ,{\mathbb {R}}))$.

From the discussion in Remark 2.3, it follows that $(\vartheta \cdot )\in \mathcal {L}(G^2(J,\varGamma ))$. It is also not difficult to check that

$$\begin{aligned} {\vartheta {P}}_M\vartheta \in \mathcal {L}(L^2(J,H^\frac{3}{2}(\varGamma ,{\mathbb {R}})))\textstyle \bigcap \mathcal {L}(H^1(J,L^2(\varGamma ,{\mathbb {R}}))). \end{aligned}$$

Indeed, we may suppose, without loss of generality, that the family of functions $\varPsi _i$ is orthonormal in $L^2(\varGamma ,{\mathbb {R}})$, and in that case, for a Hilbert space ,

$$\begin{aligned} \left| {\vartheta {P}}_M{\vartheta {\zeta }}\right| _{L^2(J,X)}^2&=\!\int _{J}\left| \vartheta \sum _{i=1}^M({\vartheta {\zeta }}(s),\varPsi _i)_{L^2(\varGamma ,{\mathbb {R}})}\varPsi _i\right| _{X}^2\,\!\!\!\!\!\mathrm {d}s\\&\le C_{\vartheta {M}}\sum _{i=1}^M\!\int _{J}\left| \zeta (s)\right| _{L^2(\varGamma ,{\mathbb {R}})}^2\left| \varPsi _i\right| _{X}^2\,\!\mathrm {d}s\\&\le C_{\vartheta {M}}\max _{1\le i\le M}\left| \varPsi _i\right| _{X}^2\left| \zeta \right| _{L^2(J,L^2(\varGamma ,{\mathbb {R}}))}^2\le C\left| \zeta \right| _{L^2(J,X)}^2,\\ \left| \textstyle \frac{\partial }{\partial t}{\vartheta {P}}_M\vartheta \zeta \right| _{L^2(J,X)}^2&=\int _{J}\left| \vartheta \textstyle \frac{\partial }{\partial s}\displaystyle \sum _{i=1}^M(\vartheta \zeta (s),\varPsi _i)_{L^2(\varGamma ,{\mathbb {R}})}\varPsi _i\right| _{X}^2\!\!\!\,\mathrm {d}s \le C\left| \textstyle \frac{\partial }{\partial t}\zeta \right| _{L^2(J,X)}^2. \end{aligned}$$

Now, from ${\vartheta {P}}_M\vartheta \in \mathcal {L}(L^2(J,L^2(\varGamma ,{\mathbb {R}})))\textstyle \bigcap \mathcal {L}(H^1(J,L^2(\varGamma ,{\mathbb {R}})))$, by an interpolation argument, it follows that ${\vartheta {P}}_M\vartheta \in \mathcal {L}(H^\frac{3}{4}(J,L^2(\varGamma ,{\mathbb {R}})))$. See [32, chapter 1, Section 5.1] and [33, chapter 4, Section 2.1]. Finally, ${\vartheta {P}}_M\vartheta \in \mathcal {L}(L^2(J,H^\frac{3}{2}(\varGamma ,{\mathbb {R}})))\textstyle \bigcap \mathcal {L}(H^\frac{3}{4}(J,L^2(\varGamma ,{\mathbb {R}})))$ implies, by Proposition 2.3, that ${\vartheta {P}}_M\vartheta \in \mathcal {L}(G^2(J,\varGamma ))$.

Finally, we prove that $\vartheta Q_{{\widetilde{M}}}P_M\vartheta \in \mathcal {L}(G^2(J,\varGamma ))$. Since ${\vartheta {P}}_M\vartheta \in \mathcal {L}(G^2(J,\varGamma ))$, it is enough to prove that $Q_{{\widetilde{M}}}\in \mathcal {L}({\vartheta {P}}_M\vartheta G^2(J,\varGamma ), G^2(J,\varGamma ))$. Notice that

$$\begin{aligned} {\vartheta {P}}_M\vartheta G^2(J ,\varGamma )\subseteq L^2(J ,{\vartheta {P}}_MH^\frac{3}{2}(\varGamma ))\textstyle \bigcap H^{\frac{3}{4}}(J ,{\vartheta {P}}_ML^2(\varGamma )), \end{aligned}$$

and $P_MH^\frac{3}{2}(\varGamma )=\mathcal {S}_\varPsi =P_ML^2(\varGamma )$. Since the space $\mathcal {S}_\varPsi =\mathrm{span}\{\varPsi _i\mid i\in \{1,2,\ldots ,M\}\}$ is finite-dimensional, it remains to observe that $Q_{{\widetilde{M}}}\in \mathcal {L}\left( L^2(J ,{\mathbb {R}}^M)\textstyle \bigcap H^{\frac{3}{4}}(J ,{\mathbb {R}}^M)\right)$, which follows from $Q_{{\widetilde{M}}}\in \mathcal {L}\bigl (L^2(J ,{\mathbb {R}})\bigr )\textstyle \bigcap \mathcal {L}\bigl (H^{\frac{3}{4}}(J ,{\mathbb {R}})\bigr )$. Finally, observe that looking at $H^{\frac{3}{4}}(J ,{\mathbb {R}})=:H^{\frac{3}{4}}_\mathrm{f}(J ,{\mathbb {R}})$ as the domain of $(-\varDelta _{J }+1)^\frac{3}{8}$ (cf. proof of Lemma 2.8) we find the identities $\left| Q_{{\widetilde{M}}}\right| _{\mathcal {L}\bigl (L^2(J ,{\mathbb {R}})\bigr )}^2=1 =\left| Q_{{\widetilde{M}}}\right| _{\mathcal {L}\bigl (H^{\frac{3}{4}}_\mathrm{f}(J ,{\mathbb {R}})\bigr )}^2$. $\square$