Global stabilization of a Korteweg-de Vries equation with saturating - - PowerPoint PPT Presentation

global stabilization of a korteweg de vries equation with
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Global stabilization of a Korteweg-de Vries equation with saturating - - PowerPoint PPT Presentation

Problem statement Main results Stability analysis Simulation Perspectives Global stabilization of a Korteweg-de Vries equation with saturating distributed control Swann MARX 1 A joint work with Eduardo CERPA 2 , Christophe PRIEUR 1 and Vincent


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Problem statement Main results Stability analysis Simulation Perspectives

Global stabilization of a Korteweg-de Vries equation with saturating distributed control

Swann MARX1 A joint work with Eduardo CERPA2, Christophe PRIEUR1 and Vincent ANDRIEU3.

1GIPSA-lab, Grenoble, France. 2Universidad Técnica Federico Santa María, Valparaíso, Chile. 3LAGEP

, Lyon, France and Bergische Universität Wuppertal, Wuppertal, Germany.

Coron fest

1/22

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Problem statement Main results Stability analysis Simulation Perspectives

Table of contents

1

Problem statement

2

Main results

3

Stability analysis

4

Simulation

5

Perspectives

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Problem statement Main results Stability analysis Simulation Perspectives

KdV equation with a distributed control The Korteweg-de Vries describes approximately long waves in water of relatively shallow depth. For all L > 0, it is described as follows      yt + yx + yxxx + yyx + u = 0 y(t, 0) = y(t, L) = yx(t, L) = 0 y(0, x) = y0 (KdV-u) Stabilization References : [Perla Menzala et al., 2002], [Rosier and Zhang, 2006], [Pazoto, 2005] In the following, we will focus on (KdV-u).

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Problem statement Main results Stability analysis Simulation Perspectives

KdV equation with a distributed control The Korteweg-de Vries describes approximately long waves in water of relatively shallow depth. For all L > 0, it is described as follows      yt + yx + yxxx + yyx + u = 0 y(t, 0) = y(t, L) = yx(t, L) = 0 y(0, x) = y0 (KdV-u) Stabilization References : [Perla Menzala et al., 2002], [Rosier and Zhang, 2006], [Pazoto, 2005] In the following, we will focus on (KdV-u). The paper [Cerpa, 2014] is a good introduction to the control of this equation.

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Problem statement Main results Stability analysis Simulation Perspectives

Case without control : critical length phenomenon      yt + yx + yxxx = 0 y(t, 0) = y(t, L) = yx(t, L) = 0 y(0, x) = y0(x) (LKDV) Critical length set for the linear KdV equation [Rosier, 1997] If L ∈

  • k2+kl+l2

3

  • k, l ∈ N∗
  • , there exist solutions of

(LKDV) for which the energy does not decay to zero.

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Problem statement Main results Stability analysis Simulation Perspectives

Case without control : critical length phenomenon Critical length set for the linear KdV equation [Rosier, 1997] If L ∈

  • k2+kl+l2

3

  • k, l ∈ N∗
  • , there exist solutions of

(LKDV) for which the energy does not decay to zero. With L = 2π, ye = 1 − cos(x) is an equilibrium solution. Indeed ye

t + ye x + ye xxx = 0

Thus, with y0(x) = 1 − cos(x), the solution does not decay to zero.

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Problem statement Main results Stability analysis Simulation Perspectives

Case without control : critical length phenomenon Critical length set for the linear KdV equation [Rosier, 1997] If L ∈

  • k2+kl+l2

3

  • k, l ∈ N∗
  • , there exist solutions of

(LKDV) for which the energy does not decay to zero. Local asymptotic stability of 0 with L = 2π [Chu, Coron and Shang, 2015] Let us assume that L = 2π and u = 0. Then 0 ∈ L2(0, L) is (locally) asymptotically stable for (KdV-u). Thus the nonlinearity yyx improves the stability. Note that the stability is local.

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Global stabilization of y = 0 with a distributed control without constraint : general case In [Pazoto, 2005] and [Rosier and Zhang, 2006], the authors use a control u(t, x) = a(x)y(t, x) with a defined as follows a =

  • 0 < a0 ≤ a(x) ≤ a1,

∀x ∈ ω, where ω is a nonempty open subset of (0, L). (loc-control) They prove that the origin of (KdV-u) is globally asymptotically stabilized with such a control.

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Problem statement Main results Stability analysis Simulation Perspectives

Saturation function : finite dimension Usual saturation For all s ∈ R, the function sat satisfies sat(s) =    −u0 if s ≤ −u0 s if − u0 ≤ s ≤ u0, u0 if s ≥ u0. where u0 denotes the saturation level. Saturating a controller can lead to catastrophic behavior for the stability of the system.

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Saturation in infinite dimension Saturation operator For any function s and all x ∈ [0, L], the operator sat satisfies sat(s)(x) = sat(s(x)) (SAT-loc)

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Illustration of the saturation

0.5 1 1.5 2 2.5 3 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 cos(x) and sat(cos(x)) x

FIGURE: x ∈ [0, π]. Red : sat(cos(x)) and u0 = 0.5, Blue : cos(x).

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Problem statement Main results Stability analysis Simulation Perspectives

Distributed control saturated System under consideration          yt + yx + yxxx + yyx + sat(ay) = 0, y(t, 0) = y(t, L) = 0, yx(t, L) = 0, y(0, x) = y0(x). (KdV-sat) Remark : A similar work has been done on the wave equation [Prieur, Tarbouriech and Gomes da Silva Jr, 2016] and the linear KdV equation [SM, Cerpa, Prieur and Andrieu, 2015].

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Well-posedness theorem Theorem (Well posedness (SM-Cerpa-Prieur-Andrieu)) For any initial conditions y0 ∈ L2(0, L), there exists a unique mild solution y ∈ B(T) := C(0, T; L2(0, L)) ∩ L2(0, T; H1(0, L)) to (KdV-sat). B(T) is endowed with the following norm yB(T) := max

t∈[0,T] y(t)L2(0,L) +

T y(t)2

H1(0,L)dt

1/2 . Recall : L2(0, L) =

  • f, fL2(0,L) :=

L

0 f(x)2dx < ∞

  • H1(0, L) =
  • f, fH1(0,L) := fL2(0,L) + f ′L2(0,L) < ∞
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Global asymptotic stability theorem Theorem (Global asymptotic stability(SM-Cerpa-Prieur-Andrieu)) There exist a positive value µ⋆, a class K∞ function α, such that, for any initial condition y0 ∈ L2(0, L), every solution y to (KdV-sat) satisfies, for all t ≥ 0 y(t, .)L2(0,L) ≤ α(y0L2(0,L))e−µ⋆t, Recall : α is said to be a class K∞ function if it is nonnegative, it is strictly increasing, limr→+∞ α(r) = +∞ α(0) = 0.

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Global asymptotic stability theorem Theorem (Global asymptotic stability(SM-Cerpa-Prieur-Andrieu)) There exist a positive value µ⋆, a class K∞ function α, such that, for any initial condition y0 ∈ L2(0, L), every solution y to (KdV-sat) satisfies, for all t ≥ 0 y(t, .)L2(0,L) ≤ α(y0L2(0,L))e−µ⋆t, Example : The function r → α(r) = r is a class K∞ function.

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Global asymptotic stability theorem Theorem (Global asymptotic stability(SM-Cerpa-Prieur-Andrieu)) There exist a positive value µ⋆, a class K∞ function α, such that, for any initial condition y0 ∈ L2(0, L), every solution y to (KdV-sat) satisfies, for all t ≥ 0 y(t, .)L2(0,L) ≤ α(y0L2(0,L))e−µ⋆t, We have in fact Globally asymptotically stable + Semi-globally exponentially stable

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Semi-global exponential stability Semi-global exponential stability The origin for the system (KdV-sat) is said to be semi-globally exponentially stable in L2(0, L) if for any r > 0 there exist two constants C = C(r) > 0 and µ = µ(r) > 0 such that for any y0 ∈ L2(0, L) such that y0L2(0,L) ≤ r the weak solution y = y(t, x) to (KdV-sat) satisfies y(t, .)L2(0,L) ≤ Cy0L2(0,L)e−µt ∀t ≥ 0.

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Problem statement Main results Stability analysis Simulation Perspectives

Sector condition Sector condition Let r be a positive value. Let a be defined by a =

  • 0 < a0 ≤ a(x) ≤ a1,

∀x ∈ ω, where ω is a nonempty open subset of (0, L). Given s ∈ L∞(0, L) satisfying, for all x ∈ [0, L], |s(x)| ≤ r, we have

  • sat(a(x)s(x)) − k(r)a(x)s(x)
  • s(x) ≥ 0,

∀x ∈ [0, L], (sec-cond-loc) with k(r) = min u0 a1r , 1

  • .

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Problem statement Main results Stability analysis Simulation Perspectives

Some estimates Bounded initial conditions Let r be a positive value such that y0L2(0,L) ≤ r. Some estimates y(T, .)2

L2(0,L) =y02 L2(0,L) −

T |yx(σ, 0)|2dt −2 T L sat(ay)ydxdt (Stability) y2

L2(0,T;H1(0,L)) ≤ 8T + 2L

3 y02

L2(0,L) + TK

27 y04

L2(0,L).

(Regularity)

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Claim Claim For any T > 0 and any r > 0 there exists a positive constant C = C(r, T) > 1 such that for any solution y to (KdV-sat) with an initial condition y0 ∈ L2(0, L) such that y0L2(0,L) ≤ r, it holds that y02

L2(0,L) ≤C

T |yx(t, 0)|2dt +2 T L sat(ay(t, x))y(t, x)dtdx

  • .

(Claim) Proving this Claim allows to prove the semi-global exponential stability of the origin of the equation when using u = sat(ay).

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Problem statement Main results Stability analysis Simulation Perspectives

Assuming the Claim Indeed, if this claim holds, we obtain with (Stability) y(kT, .)2

L2(0,L) ≤ γky02 L2(0,L)

∀k ≥ 0 where 1 − 1

C := γ ∈ (0, 1).

Once again with (Stability), we have y(t, .)L2(0,L) ≤ y(kT, .)L2(0,L) for kT ≤ t ≤ (k + 1)T. Therefore, if we assume the Claim, we have, for all t ≥ 0, y(t, .)2

L2(0,L) ≤ 1

γ y0L2(0,L)e

log γ T

t

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Problem statement Main results Stability analysis Simulation Perspectives

Proof of the Claim Another useful estimate Ty02

L2(0,L) ≤

T L |y(t, x)|2dxdt + T (T − t)|yx(t, 0)|2dt +2 T (T − t) L sat(ay)ydxdt. (Stability-2) We prove also the Claim with y2

L2(0,T;L2(0,L)) ≤C1

T |yx(t, 0)|2dt +2 T L sat(ay(t, x))y(t, x)dtdx

  • ,

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Problem statement Main results Stability analysis Simulation Perspectives

Proof of the Claim We proceed by contradiction. Suppose the claim fails to be

  • true. Then there exists a sequence of solution yn ∈ B(T) of

(KdV-sat) such that Bounded initial conditions yn(0, .)L2(0,L) ≤ r (IC-bounded) Contradiction argument lim

n→+∞

yn2

L2(0,T;L2(0,L))

T

0 |yn x (t, 0)|2dt + 2

T L

0 sat(ayn)yndxdt

= +∞. (Contra-argu)

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Problem statement Main results Stability analysis Simulation Perspectives

Using the sector condition Note that in a first hand we have, from the fact that the solution is bounded in L2(0, L) and from the hidden regularity (Regularity) yn2

L2(0,T;H1(0,L)) ≤ β := 8T + 2L

3 r 2 + TK 27 r 4. Therefore, L∞(0, L)-regularity ∀x ∈ [0, L], T |yn(t, x)|2dt ≤ Lyn2

L2(0,T;H1(0,L)) ≤ Lβ.

(L∞-reg)

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Problem statement Main results Stability analysis Simulation Perspectives

Using the sector condition Now let us consider Ωi ⊂ [0, T] defined as follows Ωi =

  • t ∈ [0, T], sup

x∈[0,L]

|y(t, x)| > i

  • .

In the following, we will denote by Ωc

i its complement. It is

defined by Ωc

i =

  • t ∈ [0, T], sup

x∈[0,L]

|y(t, x)| ≤ i

  • .

(Space-sec-cond)

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Problem statement Main results Stability analysis Simulation Perspectives

Using the sector condition Since the function t → supx∈[0,L] |yn(t, x)|2 is a nonnegative function, we have T sup

x∈[0,L]

|yn(t, x)|2dt ≥

  • Ωi

sup

x∈[0,L]

|yn(t, x)|2dt ≥ i2ν(Ωi), where ν(Ωi) denotes the Lebesgue measure of Ωi. Therefore, with T

0 |yn(t, x)|2dt ≤ Lβ, we obtain

ν(Ωi) ≤ Lβ i2 . We deduce from the previous equation that max

  • T − Lβ

i2 , 0

  • ≤ ν(Ωc

i ) ≤ T.

(Lebesgue-measure)

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Problem statement Main results Stability analysis Simulation Perspectives

Using the sector condition Moreover, since |y(t, x)| ≤ i in Ωc

i , then we can use the sector

condition

  • sat(a(x)s(x)) − k(r)a(x)s(x)
  • s(x) ≥ 0,

∀x ∈ [0, L], and we obtain, for all i ∈ N T L sat(ayn)yndtdx ≥

  • Ωc

i

L ak(i)(yn)2dtdx. (Stability-sec-cond)

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Using the sector condition Let λn := ynL2(0,T;L2(0,L)) and vn(t, x) = yn(t,x)

λn

. Notice that λn is bounded. Hence, up to extracting a subsequence, we may assume that λn → λ ≥ 0. Then vn fullfills          vn

t + vn x + vn xxx + λnvnvn x + sat(aλnvn)

λn = 0, vn(t, 0) = vn(t, L) = vn

x (t, L) = 0,

vnL2(0,T;L2(0,L)) = 1.

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Problem statement Main results Stability analysis Simulation Perspectives

Using the sector condition Due to the contradiction argument, that is lim

n→+∞

yn2

L2(0,T;L2(0,L))

T

0 |yn x (t, 0)|2dt + 2

T L

0 sat(ayn)yndxdt

= +∞, then we have T |vn

x (t, 0)|2dt + 2

T L sat(aλnvn) λn vndtdx → 0.

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Using the sector condition And since we have T L sat(aλnvn) λn vndtdx ≥

  • Ωc

i

L ak(i)(vn)2dtdx. then we obtain that T |vn

x (t, 0)|2dt + 2

  • Ωc

i

L k(i)a(vn)2dtdx → 0

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Conclusion Using a result of Aubin Lions, we prove that vn converges strongly to v ∈ L2(0, T; L2(0, L)). Moreover, we have, for all i ∈ N v(t, x) = 0, ∀x ∈ ω, ∀t ∈

  • i∈N

Ωc

i , and vx(t, 0) = 0, ∀t ∈ (0, T).

With max

  • T − Lβ

i2 , 0

  • ≤ ν(Ωc

i ) ≤ T,

we know that ν

  • i∈N

Ωc

i

  • = T.

Thus, for almost every t ∈ [0, T] v(t, x) = 0, ∀x ∈ ω, and vx(t, 0) = 0.

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Conclusion Moreover, v solves      vt + vx + vxxx + λvvx = 0, v(t, 0) = v(t, L) = vx(t, L) = 0, vL2(0,T;L2(0,L)) = 1. Thus v ∈ B(T) and in particular v is continuous. Thus v(t, x) = 0, ∀x ∈ ω, ∀t ∈ [0, T], and vx(t, 0) = 0, ∀t ∈ (0, T).

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Problem statement Main results Stability analysis Simulation Perspectives

Conclusion With a unique continuation argument given by [Saut and Scheurer, 1987], we prove that v(t, x) = 0, ∀x ∈ [0, L], ∀t ∈ [0, T] It is in contradiction with vL2(0,T;L2(0,L)) = 1. Thus, the claim is true and we have Semi-global exponential stability Let y be a solution to the KdV equation with a saturated control. Thus, for every initial condition y0 satisfying y0L2(0,L) ≤ r, there exist positive values µ := µ(r) and K := K(r) such that yL2(0,L) ≤ Ke−µty0L2(0,L), ∀t ≥ 0

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Problem statement Main results Stability analysis Simulation Perspectives

y0(x) = 1 − cos(x), L = 2π and ω = 1

3L, 2 3L

  • 2

4 6 2 4 6 8 −0.5 0.5 1 1.5 2 2.5

x yw(t, x) t

FIGURE: Solution y(t,x) with a localized feedback law without saturation

2 4 6 2 4 −0.5 0.5 1 1.5 2 2.5

x y(t, x) t

FIGURE: Solution y(t,x) with a localized feedback law saturated ; u0 = 0.5

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Problem statement Main results Stability analysis Simulation Perspectives

y0(x) = 1 − cos(x), L = 2π and ω = 1

3L, 2 3L

  • 2

4 6 2 4 6 8 −0.5 0.5 1

x satloc(ay)(t, x) t

FIGURE: Control f = sat(ay) where ω = 1

3L, 2 3L

  • , u0 = 0.5

1 2 3 4 5 6 1 2 3 4 5 6 7 8 9 10

t E y2

L2(0,L)

yw2

L2(0,L)

FIGURE: Energy functions of the solutions

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Perspectives Very few papers deal with stabilization of PDEs with bounded boundary controls (see [Lasiecka and Seidman, 2003]). For instance, is the origin for the following linear hyperbolic equation

  • zt + Λzx = 0,

z(t, 0) = Hz(t, 1) + Bsat(Kz(t, 1)), where λ(Λ) > 0, stable ?

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THANK YOU FOR YOUR ATTENTION ! HAPPY BIRTHDAY JEAN-MICHEL !

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Cerpa, E. (2014). Control of a Korteweg-de Vries equation : a tutorial. Mathematical Control and Related Fields, 4(1) :45–99. Lasiecka, I. and Seidman, T. I. (2003). Strong stability of elastic control systems with dissipative saturating feedback. Systems & Control Letters, 48 :243–252. Marx, S., Cerpa, E., Prieur, C., and Andrieu, V. (July 2015). Stabilization of a linear Korteweg-de Vries with a saturated internal control. In Proceedings of the European Control Conference, page To appear, Linz, AU. Pazoto, A. (2005). Unique continuation and decay for the Korteweg-de Vries equation with localized damping.

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ESAIM : Control, Optimisation and Calculus of Variations, 11 :3 :473–486. Perla Menzala, G., Vasconcellos, C. F ., and Zuazua, E. (2002). Stabilization of the Korteweg-de Vries equation with localized damping.

  • Quart. Appl. Math., 60(1) :111–129.

Prieur, C., Tarbouriech, S., and da Silva, J. G. (2014). Well-posedness and stability of a 1-D wave equation with saturating input. In Proceedings of the 53rd Conference on Decision and Control, pages 2846 – 2851, Los Angeles, CA. Rosier, L. (1997). Exact boundary controllability for the Korteweg-de Vries equation on a bounded domain.

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ESAIM : Control, Optimisation and Calculus of Variations, 2 :33–55. Rosier, L. and Zhang, B.-Y. (2006). Global stabilization of the generalized Korteweg–de Vries equation posed on a finite domain. SIAM Journal on Control and Optimization, 45(3) :927–956. Saut, J.-C. and Scheurer, B. (1987). Unique continuation for some evolution equations.

  • J. Differential Equations, 66 :118–139.

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