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Cone-bounded feedback laws for m-dissipative operators on Hilbert spaces

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Abstract

This work studies the influence of some constraints on a stabilizing feedback law. An abstract nonlinear control system is considered for which we assume that there exists a linear feedback law that makes the origin of the closed-loop system globally asymptotically stable. This controller is then modified via a cone-bounded nonlinearity. Well-posedness and stability theorems are stated. The first theorem is proved thanks to the Schauder fixed-point theorem and the second one with an infinite-dimensional version of LaSalle’s invariance principle. These results are illustrated on a linear Korteweg-de Vries equation by some simulations and on a nonlinear heat equation.

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Notes

  1. A function \(x:[0,\infty )\rightarrow X\) is called a strong solution to (7) if \(x(t)\in D(A)\) for all \(t\ge 0\) and if it satisfies the initial value problem.

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Authors and Affiliations

  1. Univ. Grenoble Alpes, CNRS, Gipsa-lab, 38000, Grenoble, France

    Swann Marx & Christophe Prieur

  2. Université Lyon 1 CNRS UMR 5007 LAGEP, Lyon, France

    Vincent Andrieu

Authors
  1. Swann Marx
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  2. Vincent Andrieu
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  3. Christophe Prieur
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Correspondence to Swann Marx.

Appendices

A Precompacity of the KdV equation with a cone-bounded nonlinearity

This section is devoted to the proof of the precompactness of the canonical embedding from \(D(A_\sigma )=D(A)\), defined in (26), into \(X:=L^2(0,L)\). Let us state the lemma and prove it.

Lemma 2

The canonical embedding from \(D(A_\sigma )\), equipped with the graph norm, into \(X:=L^2(0,L)\) is compact.

Proof of Lemma 2

We follow the strategy of [21, 26] and [11]. Let us recall the definition of the graph norm

$$\begin{aligned} \begin{aligned} \Vert x\Vert ^2_{D(A_\sigma )}&:=\Vert x\Vert ^2_{L^2(0,L)}+\Vert A_\sigma x\Vert ^2_{L^2(0,L)}\\&=\,\int _0^L \left( |x(z)|^2+|-x^{\prime \prime \prime }(z)-x^{\prime }(z)-\sigma \left( {\mathfrak {1}}_{\varOmega }x\right) (z)|^2\right) \mathrm{d}z\\&=\,\int _0^L \left( |x(z)|^2+|x^{\prime \prime \prime }(z)+x^{\prime }(z)+\sigma \left( {\mathfrak {1}}_{\varOmega }x\right) (z)|^2\right) \mathrm{d}z. \end{aligned} \end{aligned}$$
(97)

Note that

$$\begin{aligned} \Vert \sigma \left( {\mathfrak {1}}_{\varOmega }x\right) \Vert _{L^2(0,L)}\le 2\Vert x\Vert _{L^2(0,L)}. \end{aligned}$$
(98)

From the definition of the graph norm, we get the following two inequalities

$$\begin{aligned} \Vert x\Vert ^2_{D(A_\sigma )}\ge \Vert x\Vert ^2_{L^2(0,L)} \end{aligned}$$
(99)

and, since, for all \((s,\tilde{s})\in {\mathbb {C}}^2\), it holds \(|s+\tilde{s}|^2\le 2|s|^2+2|\tilde{s}|^2\), we have

$$\begin{aligned} \begin{aligned} \Vert x\Vert ^2_{D(A_\sigma )}\ge&\,\frac{1}{2}\int _0^L |-\sigma \left( {\mathfrak {1}}_{\varOmega }x\right) (z)|^2\mathrm{d}z\\&+\,\frac{1}{2}\int _0^L |x^{\prime \prime \prime }(z)+x^\prime (z)+\sigma \left( {\mathfrak {1}}_{\varOmega }x\right) (z)|^2\mathrm{d}z\\ \ge&\,\frac{1}{4}\int _0^L |x^{\prime \prime \prime }(z)+x^\prime (z)|^2 \mathrm{d}z. \end{aligned} \end{aligned}$$
(100)

Noticing that \(\Vert x^{\prime \prime \prime }\Vert ^2_{L^2(0,L)}=\Vert x^{\prime \prime \prime }+x^\prime -x^\prime \Vert ^2_{L^2(0,L)}\), we have

$$\begin{aligned} \Vert x^{\prime \prime \prime }\Vert ^2_{L^2(0,L)} \le 2\Vert x^{\prime \prime \prime }+x^\prime \Vert ^2_{L^2(0,L)}+2\Vert x^\prime \Vert ^2_{L^2(0,L)}, \end{aligned}$$
(101)

and using that \(\Vert x^\prime \Vert ^2_{L^2(0,L)}=\Vert x^\prime +x^{\prime \prime \prime }-x^{\prime \prime \prime }+zx-zx\Vert ^2_{L^2(0,L)}\), we obtain

$$\begin{aligned} \begin{aligned} \Vert x^\prime \Vert ^2_{L^2(0,L)}\le&\, 2\Vert x^\prime +x^{\prime \prime \prime }\Vert ^2_{L^2(0,L)}+2\Vert x^{\prime \prime \prime }-zx+zx\Vert ^2_{L^2(0,L)}\\ \le&\, 2\Vert x^\prime +x^{\prime \prime \prime }\Vert ^2_{L^2(0,L)}+4\Vert x^{\prime \prime \prime }-zx\Vert ^2_{L^2(0,L)}+4\Vert zx\Vert ^2_{L^2(0,L)}\\ \le&\, 2\Vert z^\prime +x^{\prime \prime \prime }\Vert ^2_{L^2(0,L)}+4\Vert x^{\prime \prime \prime }\Vert ^2_{L^2(0,L)}\\&-8\int _0^L zx^{\prime \prime \prime }(z)x(z)\mathrm{d}z+8\Vert zx\Vert ^2_{L^2(0,L)}. \end{aligned} \end{aligned}$$

Deriving some integrations by parts, we get

$$\begin{aligned} \int _0^L zx^{\prime \prime \prime }(z)x(z)\mathrm{d}z=\frac{3}{2}\Vert x^\prime \Vert ^2_{L^2(0,L)}, \end{aligned}$$

and therefore

$$\begin{aligned} \begin{aligned} \Vert x^\prime \Vert ^2_{L^2(0,L)}\le 2\Vert x^\prime +x^{\prime \prime \prime }\Vert ^2_{L^2(0,L)}+4\Vert x^{\prime \prime \prime }\Vert ^2_{L^2(0,L)}-12\Vert x^{\prime }\Vert ^2_{L^2(0,L)}+8\Vert zx\Vert ^2_{L^2(0,L)}. \end{aligned}\nonumber \\ \end{aligned}$$
(102)

Hence,

$$\begin{aligned} \begin{aligned} 13\Vert x^\prime \Vert ^2_{L^2(0,L)}\le 2\Vert x^\prime +x^{\prime \prime \prime }\Vert ^2_{L^2(0,L)}+4\Vert x^{\prime \prime \prime }\Vert ^2_{L^2(0,L)}+8L^2\Vert x\Vert ^2_{L^2(0,L)}. \end{aligned} \end{aligned}$$
(103)

Plugging inequality (101) in (103), we have

$$\begin{aligned} \begin{aligned} 13\Vert x^\prime \Vert ^2_{L^2(0,L)}\le&\,2\Vert x^\prime +x^{\prime \prime \prime }\Vert ^2_{L^2(0,L)}+4\left( 2\Vert x^{\prime \prime \prime }+x^\prime \Vert ^2_{L^2(0,L)}+2\Vert x^\prime \Vert ^2_{L^2(0,L)}\right) \\&+\,8L^2\Vert x\Vert ^2_{L^2(0,L)}\\ \le&\,10\Vert x^\prime +x^{\prime \prime \prime }\Vert ^2_{L^2(0,L)}+8\Vert x^\prime \Vert ^2_{L^2(0,L)}+8L^2\Vert x\Vert ^2_{L^2(0,L)}. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert x^\prime \Vert ^2_{L^2(0,L)}\le 2\Vert x^\prime +x^{\prime \prime \prime }\Vert ^2_{L^2(0,L)}+\frac{8L^2}{5}\Vert x\Vert ^2_{L^2(0,L)}. \end{aligned}$$
(104)

Considering Equations (99) and (100), it leads us to the following inequality, for all \(x\in D(A)\),

$$\begin{aligned} \Vert x^\prime \Vert ^2_{L^2(0,L)}\le \varDelta \Vert x\Vert ^2_{D(A_\sigma )} \end{aligned}$$
(105)

where \(\varDelta \) is a term which depends only on L.

Thus, if we consider now a sequence \(\{x_n\}_{n\in \mathbb {N}}\) in \(D(A_\sigma )\) bounded for the graph norm of \(D(A_\sigma )\), we have from (105) that this sequence is bounded in \(H^1_0(0,L)\). Since the canonical embedding from \(H^1_0(0,L)\) to \(L^2(0,L)\) is compact, there exists a subsequence still denoted \(\{x_n\}_{n\in \mathbb {N}}\) such that \(x_n\rightarrow x\) in \(L^2(0,L)\). Thus x belongs to \(L^2(0,L)\) which allows us to state that \(D(A_\sigma )\) embeds compactly in X. It concludes the proof of Lemma 2. \(\square \)

B Nonlinear heat equation

1.1 B.1 Proof of the m-dissipativity of the nonlinear heat equation

This subsection is devoted to the proof of the following theorem.

Theorem 4

The operator defined by (39) is m-dissipative

Proof of Theorem 4

The proof of Theorem 4 is divided in two steps. First, the operator A is proved to be dissipative. Secondly, we prove that, for all \(f\in L^2(0,L)\), there exist \(x\in D(A)\) such that

$$\begin{aligned} x-Ax = f. \end{aligned}$$
(106)

Let us recall that the dissipativity and the existence of \(x\in D(A)\) such that (106) holds imply that A is a m-dissipative operator.

First step: Dissipativity of the operator A.

Note that we have

$$\begin{aligned} \langle Ax-A{\tilde{x}},x-{\tilde{x}}\rangle _{L^2(0,1)}= & {} \int _0^1 (x-{\tilde{x}})(x^{\prime \prime }-{\tilde{x}}^{\prime \prime }) \mathrm{d}z\nonumber \\&+\int _0^L (x-{\tilde{x}})(\sin (x)-\sin ({\tilde{x}}))\mathrm{d}z. \end{aligned}$$
(107)

Performing some integrations by parts leads to

$$\begin{aligned} \int _0^L (x-{\tilde{x}})(x^{\prime \prime }-{\tilde{x}}^{\prime \prime }) \mathrm{d}z=-\int _0^1 (x^\prime -{\tilde{x}}^\prime )^2 \mathrm{d}z. \end{aligned}$$
(108)

Moreover, using the fact that \(\sin \) is Lipschitz together with a Poincaré inequality, one has

$$\begin{aligned} \int _0^L (x-{\tilde{x}})(\sin (x)-\sin ({\tilde{x}}))\mathrm{d}z \le \int _0^L (x-{\tilde{x}})^2 \mathrm{d}z\le \frac{4}{\pi ^2} \int _0^1 (x^\prime - {\tilde{x}}^\prime )^2\mathrm{d}z.\qquad \end{aligned}$$
(109)

Hence, it is easy to see that

$$\begin{aligned} \langle Ax-A{\tilde{x}},x-{\tilde{x}}\rangle _{L^2(0,1)}\le 0. \end{aligned}$$
(110)

Second step: Existence of \(x\in D(A)\) such that (106) holds

To prove the existence of \(x\in D(A)\) such that (106) holds, one has to prove that there exists a solution to the following nonlinear ODE

$$\begin{aligned} \left\{ \begin{aligned}&x - x^{\prime \prime }+\sin (x)= f,\\&x(0)=x(1)=0. \end{aligned} \right. \end{aligned}$$
(111)

We aim at applying the Schauder fixed-point theorem to the following nonhomogeneous linear ODE

$$\begin{aligned} \left\{ \begin{aligned}&x - x^{\prime \prime }=-\sin (y)+ f,\\&x(0)=x(1)=0, \end{aligned} \right. \end{aligned}$$
(112)

where \(y\in L^2(0,1)\). It is easy to see that there exists a unique solution to (112).

Focus on the map

$$\begin{aligned} \begin{aligned} {\mathcal {T}}:L^2(0,1)&\rightarrow L^2(0,1)\\ y&\mapsto x ={\mathcal {T}}(y) \end{aligned} \end{aligned}$$
(113)

where \(x={\mathcal {T}}(y)\) is the unique solution to (112).

We define

$$\begin{aligned} C : = \lbrace x\in H^1_0(0,1)\mid \Vert x\Vert _{H^1_0(0,1)}\le M\rbrace . \end{aligned}$$
(114)

From the theorem of Rellich, the injection of \(H^1_0(0,1)\) in \(L^2(0,1)\) is compact, then C is bounded in \(H^1_0(0,1)\) and is relatively compact in \(L^2(0,1)\). Moreover, it is a closed subset of \(L^2(0,1)\). Thus C is a compact subset of \(L^2(0,1)\). In order to apply the Schauder theorem, we have to prove that \({\mathcal {T}}(L^2(0,1))\subset C\) for a suitable choice of M and \(\lambda \). Let us multiply the first line of (112) by z and then integrate between 0 and 1. After some integrations by parts, one has

$$\begin{aligned} \Vert x\Vert ^2_{L^2(0,1)}+\Vert x^\prime \Vert ^2_{L^2(0,1)}=\int _0^1 fx \mathrm{d}z -\int _0^1 \sin (y)x\mathrm{d}z. \end{aligned}$$
(115)

Hence, applying Cauchy Schwarz inequality leads to

$$\begin{aligned} \Vert x^\prime \Vert ^2_{L^2(0,1)}\le \frac{1}{2}\Vert f\Vert ^2_{L^2(0,1)} + \frac{1}{2} - \Vert x\Vert ^2_{L^2(0,1)}+\Vert x\Vert ^2_{L^2(0,1)}. \end{aligned}$$
(116)

Therefore, since \(\Vert x^\prime \Vert ^2_{L^2(0,1)}\) and \(\Vert x\Vert _{H^1_0(0,1)}\) are equivalent by the Poincaré inequality, one has

$$\begin{aligned} \Vert x\Vert _{H^1_0(0,1)}\le M, \end{aligned}$$
(117)

where

$$\begin{aligned} M:=\sqrt{\frac{1}{2}\left( \Vert f\Vert ^2_{L^2(0,1)} + 1\right) }. \end{aligned}$$

Hence, applying Theorem 3, it concludes the proof of Theorem 4. \(\square \)

1.2 B.2 Precompacity of the nonlinear heat equation with a cone-bounded nonlinearity

This subsection is devoted to the proof of the following lemma.

Lemma 3

The canonical embedding from \(D(A_\sigma )\), equipped with the graph norm, into \(X:=L^2(0,1)\) is compact.

Proof of Lemma 3

We follow the strategy of [21, 26] and [11]. Let us recall the definition of the graph norm

$$\begin{aligned} \begin{aligned} \Vert x\Vert ^2_{D(A_\sigma )}&:=\Vert x\Vert ^2_{L^2(0,1)}+\Vert A_\sigma x\Vert ^2_{L^2(0,1)}\\&=\,\int _0^1 \left( |x(z)|^2+|x^{\prime \prime }(z)+\sin (x)(z)-\sigma \left( x\right) (z)|^2\right) \mathrm{d}z \end{aligned} \end{aligned}$$
(118)

Note that we have

$$\begin{aligned} \Vert x\Vert ^2_{D(A_\sigma )}\ge \Vert x\Vert ^2_{L^2(0,1)} \end{aligned}$$
(119)

and

$$\begin{aligned}&\Vert x\Vert ^2_{D(A_\sigma )}\ge \, \frac{1}{4}\int _0^1 |\sigma (x)(z)|^2 \mathrm{d}z + \frac{1}{4}\int _0^1 |-\sin (x)(z)|^2\mathrm{d}z\nonumber \\&\qquad \qquad \qquad \,+\,\frac{1}{4}\int _0^1 |x^{\prime \prime }(z)+\sin (x)(z)-\sigma (x)(z)|^2 \mathrm{d}x\nonumber \\&\qquad \qquad \quad \,\ge \,\frac{1}{8}\int _0^1 |x^{\prime \prime }(z)|^2\mathrm{d}z. \end{aligned}$$
(120)

Hence,

$$\begin{aligned} \Vert x\Vert ^2_{D(A_\sigma )}\ge \frac{1}{8}\Vert x^{\prime \prime }\Vert ^2_{L^2(0,1)}. \end{aligned}$$
(121)

Noticing that \(\Vert x\Vert _{L^2(0,1)}^2 = \Vert x-x^{\prime \prime }+x^{\prime \prime }\Vert _{L^2(0,1)}^2\), we have

$$\begin{aligned} \begin{aligned} \Vert x\Vert _{L^2(0,1)}^2&= \Vert x+x^{\prime \prime }\Vert _{L^2(0,1)}^2 + \Vert x^{\prime \prime }\Vert ^2_{L^2(0,1)}\\&=\,\Vert x\Vert _{L^2(0,1)}^2 +\Vert x^{\prime \prime }\Vert ^2_{L^2(0,1)}+2\int _0^1 x(z)x^{\prime \prime }(z)~\mathrm{d}z+ \Vert x^{\prime \prime }\Vert ^2_{L^2(0,1)}. \end{aligned} \end{aligned}$$
(122)
$$\begin{aligned} -\int _0^1 x(z)x^{\prime \prime }(z)\mathrm{d}z = \Vert x^{\prime \prime }(z)\Vert ^2_{L^2(0,1)}. \end{aligned}$$
(123)

Performing an integration by parts, we obtain

$$\begin{aligned} \int _0^1 x(z)x^{\prime \prime }(z)\mathrm{d}z =-\Vert x^{\prime }(z)\Vert ^2_{L^2(0,1)}. \end{aligned}$$
(124)

Hence, using (121), the following inequality holds

$$\begin{aligned} \Vert x^{\prime }\Vert ^2_{L^2(0,1)}\le 8\Vert x\Vert ^2_{D(A_\sigma )}. \end{aligned}$$
(125)

Thus, if we consider now a sequence \(\{x_n\}_{n\in \mathbb {N}}\) in \(D(A_\sigma )\) bounded for the graph norm of \(D(A_\sigma )\), we have from (105) that this sequence is bounded in \(H^1_0(0,L)\). Since the canonical embedding from \(H^1_0(0,L)\) to \(L^2(0,L)\) is compact, there exists a subsequence still denoted \(\{x_n\}_{n\in \mathbb {N}}\) such that \(x_n\rightarrow x\) in \(L^2(0,L)\). Thus x belongs to \(L^2(0,L)\) which allows us to state that \(D(A_\sigma )\) embedds compactly in X. It concludes the proof of Lemma 3. \(\square \)

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Marx, S., Andrieu, V. & Prieur, C. Cone-bounded feedback laws for m-dissipative operators on Hilbert spaces. Math. Control Signals Syst. 29, 18 (2017). https://doi.org/10.1007/s00498-017-0205-x

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