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Strong stability of elastic control systems with dissipative saturating feedback
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Systems & Control Letters 48 (2003) 243 – 252 www.elsevier.com/locate/sysconle
Strong stability of elastic control systems with dissipative saturating feedback Irena Lasieckaa;∗;1 , Thomas I. Seidmanb;2 b Department
a Department of Mathematics, University of Virginia, Charlottesville, VA 22904, USA of Mathematics and Statistics, University of Maryland, Baltimore County, Baltimore, MD 21250, USA
Received 29 August 2001; received in revised form 8 February 2002; accepted 1 April 2002 This paper is dedicated to the memory of the late J.L. Lions, whose seminal work in applied mathematics led, in particular, to the embedding of control-theoretic problems in the context of the modern theory of partial di2erential equations and inspired us all.
Abstract We will consider, with a focus on saturating feedback control laws, two problems associated with damping in a bounded acoustic cavity ⊂ R3 . Our objective is to verify (compare (Discrete Continuous Dynamical Systems 7 (2001) 319, Math. Control Signals Systems 2 (1989) 265) that these are strongly stable: for every
1. Introduction
equation
We will be considering, as indicating some relevant technical diCculties in the analysis, the question of stabilization for two paradigmatic examples of elastic systems when the feedback is dissipative but ‘saturating’. Both of these examples involve the wave
ztt − Kz = 0
∗ Corresponding author. Tel.: +1-804-924-8896; fax: +1-804982-3084. E-mail addresses: [email protected] (I. Lasiecka), seidman@ math.umbc.edu (T.I. Seidman). URLs: http://www.math.virginia.edu/∼il2v, http://www.math. umbc.edu/∼seidman 1 Partially supported by the NSF Grant DMS 9804056. 2 Supported by the TRW Foundation.
in ⊂ R3
(1.1)
e.g., modelling acoustic vibration in a bounded enclosure ) with interaction available only at part of the boundary. In each of our examples the uncontrolled system is conservative and linear and it will be possible to show that a corresponding linear feedback does stabilize the system. Our principal concern here is to see that the saturating control law, coinciding with the linear feedback for small enough data, still provides global stabilization: the relevant physical energy goes to 0 as t → ∞. As the saturation is inherently nonlinear, our context will be the theory of contraction semigroups of nonlinear operators on Hilbert space.
c 2002 Elsevier Science B.V. All rights reserved. 0167-6911/03/$ - see front matter PII: S 0 1 6 7 - 6 9 1 1 ( 0 2 ) 0 0 2 6 9 - 4
244
I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
For our
on 9;
(1.2)
the control speci
where the control ’ is to have support restricted to some speci
F(y) := min{1; M=y} y if y 6 M; = My=y if y ¿ M
E(t) = Ekinetic + Epotential
to ensure that ’ does not exceed the saturation threshold M . Our major thrust in this paper is to show—for this and correspondingly for our second example—that the linear control law (1.5) is asymptotically stable and that we can overcome the technical diCculties to compare the e2ects of (1.5) and its saturating version to show that the latter also provides stabilization. For showing the asymptotic stability of (1.5), we follow [2] in applying [1]. For consideration of the saturating version it had been our intention to follow [9], but it seems more convenient to present, in the next section, a new abstract comparison theorem which makes the comparison directly, without the intervening reference to the uncontrolled situation used in [9]. (It is worth noting at this point the emphasis in our Theorem 1 on output stabilization for ‘nice’ initial data.) Our second example corresponds to damping of the acoustic vibration in through interaction with a plate occupying a (2-D, Pat) portion of the boundary 9— the well-known structural acoustic model [8,4]. Here the feedback is through coupling of the acoustic vibration in with the vibration in the plate . We again use ‘velocity feedback’ to dampen those vibrations and so the entire system, but for the structural acoustic model this will appear as distributed control on (part of) (i.e., in the system equations) rather than as boundary control on ⊂ 9 as in our
:= 12 [zt 2 + ∇z2 R] ¡ ∞
(1.3)
and wish to ensure that—always—we have E(t) → 0 for the feedback-controlled solutions. Formally, we may di2erentiate E(·), use (1.1), and then apply the Divergence theorem to the term zt ; Kz to get, in view of (1.2), dE = zt z = zt ; ’ : (1.4) dt 9 For our
restricted to :
(1.5) 2
This gives dE=dt = −(trace of zt on ) so at least the control action would be dissipative. Apart from showing that we then would eventually have stabilization, there are three considerable diCculties with this: • justi
ztt = Kz
in ⊂ R3 ;
z = wt
on ⊂ 9;
z + cz = 0
(1.6)
on = 9 \ ;
wtt + K2 w + zt = −’ w = 0; Kw = 0
in (⊂ R2 );
at 9 ⊂ R2 :
(1.7)
I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
Remark 1. Note that the Laplacians K appearing in (1.7) are di2erent in the equations for z and for w, with the
245
oped in the next section. Indeed, the strong stability results obtained for these examples do not follow from standard approaches. To wit, a major diCculty in Example 1 is the presence of nontrivial steady states in a classical formulation of the abstract wave equation. This of course could be eliminated (as is often done in the literature) by changing the problem: modifying the boundary conditions on 9 or by creating a “hole” in the domain with zero Dirichlet boundary conditions prescribed on a portion of the boundary. However, in our tratment we do not do this, aiming for the ultimate goal of stabilizing the acoustic energy only. In the second example the diCculty is more serious as it is related to an intrinsic lack of compactness of the resolvent operator. 2. General theory We
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I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
˜ with 6 ; A 6 : (2) for ∈ D(A) ; A → 0 implies () → 0, ˜ with 6 ; A ˜ 6 : (3) for ∈ D(A) (·) is uniformly continuous: X → R+ ., ˜ if and only if (4) when () 6 : ∈ D(A) ∈ D(A) ˜ = A. — and then A ˜ is strongly stable, i.e., S(t)x ˜ (H3) S(·) 0 → 0 for each x0 ∈ X. Then S(·) is also strongly stable. Proof. Our principal task will be to prove that: (C) If x0 ∈ D(A), then (S(t)x0 ) → 0 as t → ∞. To see from (C) that S(t)x0 → 0 for x0 ∈ D(A), note that it gives (S(t)x0 ) ¡ from some time s on so, by (H2-4), A˜ then coincides with A and the conver˜ − s)x(s) follows from (H3). gence to 0 of S(t)x0 = S(t Strong stability now follows by a density argument: if we are given arbitrary x0 ∈ X, then for any ¿ 0 we can choose x0 ∈ D(A) with x0 − x0 ¡ =2 and note,
(2.8)
and by (H2-3) we have ∀ ¿ 0; ∃ = ( ) = ( ; ; ) such that if ˜ ˜ 6 ; A ; | 6 ; A; then − 6 ⇒ | () − ( )| 6 :
(2.9)
3 The equality depends, of course, on the single-valuedness of A. We also note that we are here taking Dt+ x to be the forward derivative (lim[x(t + h) − x(t)]=h for h 0+).
Supposing we have ˜ (s) ˆ ¡ for some sˆ ¿ 0 so, by ˜ with Ax( ˜ s) (H2-4), we have x(s) ˆ ∈ D(A) ˆ = Ax(s) ˆ ˜ s) ˜ − s)x( and Ax( ˆ 6 . Then we set x(t) ˆ := S(t ˆ s), ˆ ˆ (t) := (x(t)) ˆ for t ¿ s; ˆ note that ˆ (s) ˆ = ˜ (s) ˆ ¡ and, again using the Komura–Kato Theorem, we have ˜ (so ˆ is well deDt+ xˆ = A˜ xˆ pointwise—x(t) ˆ ∈ D(A) ˜
(2.10)
so, as x; Ax ¿ 0, there is a sequence sk → ∞ at which x; Ax → 0 so, by (H2-2), we have ˜ (sk ) → 0. Without loss of generality, we assume each ˜ (sk ) ¡ ¡ and introduce xˆ = xˆk by taking sˆ = sk ˜ − sk )x(sk ) in the notation above, i.e., xˆk (t) := S(t ˆ ˆ and (t) = k (t) := (x(t)) ˆ for t ¿ sk . Note that each ˆ k is continuous in t for t ¿ sk , since xˆk (·) is continuous—indeed, t t · 6 x(t) ˆ − x(s) ˆ 6 x ˆ Ax ˆ s
6 |t − s|
s
(2.11)
and we have (H2-3). There is then a maximal interval Ik = [sk ; tk ] on which ˆ 6 . On this interval xˆ · = A˜ xˆ = Axˆ by (H2-4) so, by uniqueness of the solution, x(t)=S(t ˆ −sk )x(s ˆ k )=x(t) on Ik . By continuity, there is necessarily a subinterval IV k = [sVk ; tVk ] ⊂ Ik such that, noting that ˆ k = ˜ on Ik , ˜ (sVk ) = 6 ˜ (t) 6 2 = ˜ (tVk ) on IV k : Note that, by (2.8), having ˜ ¿ requires that x; Ax ¿ = ( ) ¿ 0 on IV k . Also, by (2.9), ˆ (tVk )− ˆ (sVk ) = requires that x( ˆ tVk ) − x( ˆ sVk ) ¿ = ( ) whence, by (2.11), we have tVk − sVk ¿ =. Combining these shows that the integral over IV k of x; Ax is not less than = ¿ 0, independent of k. Since, without loss of generality, we can assume that the intervals V I k are disjoint, this would contradict the bound on x; Ax dt of (2.10) —i.e., we must have ˜ → 0 as in (C).
I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
For each of our systems the state space X will have the form X = U × H where U is a Hilbert space of ‘con
(2.12)
so A (hence also A) is monotone. To show that A is maximal monotone, it is suCcient to show that (A+1) is surjective. Since (2.12) does not give coercivity, we proceed indirectly and consider a ‘reduced’ probcont: lem involving a related operator M : V → V∗ with ∗ V ,→ H ,→ V . This M will be monotone and coercive, which will enable us to show that—A generates a contraction semigroup S(·) on X. A similar construction applies to the linear problem, with F(·) replaced by the identity on Y. To verify the hypothesis (H3) above we will, in each case, apply a result by Arendt and Batty [1]. ˜ be a C0 of linear operators on Theorem 2. Let S(·) ˜ a Hilbert space X with in8nitesimal generator −A. ˜ If %(A) ∩ iR is at most countable and contains no ˜ is strongly asymptotically point spectrum, then S(·) ˜ stable, i.e., S(t)x0 → 0 as t → ∞ for every x0 ∈ X. Our argument is then completed by verifying the ˜ hypothesis of Theorem 2 on %(A)—indeed, we will ˜ show for our examples that %(A) ∩ iR = ∅—obviously to be considered in the complexi
247
Our key steps in the analysis of each of these examples will thus be the following: (1) Reformulate the problem as a
in
with z = −aF(azt ) on 9; (3.13)
where a(·) is a bounded measurable function on 9 with nontrivial support and F, de
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I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
Remark 2. We note that, on the strength of Holmgren’s uniqueness theorem, the detectability condition holds true whenever is an open set in 9. Theorem 3. Assume the detectability condition (D). Then we have E(t) → 0 as t → ∞ for every solution of (3.13) with initial data of 8nite energy, where E(t) is de8ned by (1.3). Proof. As Step 1 we will reformulate (3.13) as a
(3.14)
where we note that L2grad ()={∇h: h ∈ V} is a closed subspace of L2 ( → R3 ) and so is a Hilbert space in its own right; cf., e.g. [12]. Indeed, this follows from the fact that L2grad () coincides with the range of ∇ : H 1 () → [L2 ()]3 and, since this map can be taken to be an isometry after factoring out constants, the range is closed. We write the state as x = (; u)T ∈ X, with and u corresponding to ∇z and zt , respectively, so 12 x2X is just the energy E of (1.3). Wave equation (3.13) then takes the form t = ∇u ut = ∇
on with · = −aF(au) on 9: (3.15)
Note that if we are given the usual (
0
We wish to consider (3.15) as x˙ + Ax = 0. It is preferable to give a weak formulation for A, since just writing A=
0
−∇
−∇
0
would hide the boundary con-
ditions, including the feedback of particular interest
to us, in the domain speci
0 −∇ ˜ Ax X = x˜ · x x; −∇ 0 = [∇u˜ · − xi ˜ · ∇u] + auF(au) ˜
˜ x and, as this last makes continuous sense whenever x; are in Z, we de
(3.16)
and, now simply specifying D(A) as the pre-image of X under A, use this to de
=
[˜ · f + ug] ˜
˜ u) for all (; ˜ T ∈ Z. With u˜ = 0 this gives = ∇u + f so, setting Mu := u˜ → [∇u˜ · ∇u + uu] ˜ +
auF(au) ˜ ∈ V∗
for u ∈ V;
[∇u˜ · f − ug] ˜ ∈ V∗ ; f˜ := u˜ →
˜ Since we have the reduced problem of solving Mu=f. ∗ we easily see that M : V → V is coercive on V as well as continuous and monotone, this can be solved for u ∈ V ⊂ H and we then set = ∇u + f to get the desired x = (; u)T ∈ D(A) ⊂ X. This proves that (A + 1) is surjective to X so A is maximal monotone and −A generates a contraction semigroup S(·) on X. The same argument shows that −A˜ (where, we recall, −A˜ corresponds to the linear problem) generates ˜ on X so we have veri
I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
the hypothesis (H1) of Theorem 1. The observation operator for this example is C : Z → Y := L2 () ⊂ L2 (9) : x = (; u)T → a[trace of u]:
(3.17)
so : x → y2Y =Cx2Y is uniformly continuous on bounded subsets of Z—although it is not well-de
=
[˜V · f + ug] ˜V
˜ u) for all (; ˜ T ∈ Z—where, of course, the overlines indicate conjugation since we must now work in the complexi
− ir a2 uu ˜V ∈ V∗ for u ∈ V; [∇u˜ · f − ug] ˜V ∈ V∗ : f˜ := u˜ →
We easily verify that (Mr +s1) : V → V∗ is coercive as well as continuous and monotone when s ¿ r 2 , so there exists (Mr + s1)−1 : V∗ → V ,→ H. Since the embedding V := H 1 () ,→ H := L2 () is compact, this shows Mr has compact resolvent and so is continuously invertible on H, whence (A˜ − ir1) is continuously invertible on X, if Mr is injective. Thus, we wish to show u = 0 when Mr u = 0. The imaginary part of u; Mr u [V : V∗ ] is a2 |u|2 so Mr u = 0 gives y = au = 0 ∈ Y (i.e., u vanishes on supp(a)) and the homogeneous equation becomes simply the weak form
249
of the eigenfunction equation: −Ku = r 2 u with u = 0. Since we are here considering r = 0 so u = const:, the assumption (D) then gives u = 0. This shows that ˜ for 0 = r ∈ R we have ir ∈ %(A). When r = 0 above, we must proceed di2erently: we have ∇u + f = 0 and will now use this to eliminate u. Note that f ∈ U so there is some h ∈ V with f = ∇h; thus ∇(u+h)=0 gives u=−h+c where the constant c is to be determined. We also note that we are seeking ∈ U in the form =∇v by the de
with v = a2 (h − c):
It that this has a solution if and only if is well-known g + v = 0, so we havethe unique determination that c = [ g + a2 h]=[ a2 ]: The resulting v is not unique, but this indeterminacy is nugatory: = ∇v is unique in U. This establishes a map A˜ −1 : X → ˜ ⊂ X and, since A˜ is clearly a closed linear D(A) operator, the map A˜ −1 is continuous, showing that ˜ and so completing Step 4. 0 ∈ %(A) This veri
(4.18)
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I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
where a is a nontrivial smooth function on ⊂ 9 with support in the interior of and the function c is assumed positive. We consider
As before, if we (formally) apply the Divergence The orem we obtain dE=dt = − yF(y) 6 0 (now with y := K[awt ]), so our feedback is again dissipative: the energy is nonincreasing. Our main aim is to ensure that we have E(t) → 0 as t → 0 for all
2
:= {w ∈ H (): w|9 = 0};
H := L () × L (); Z := U × U:
+
czz ˜ +
[(Kw)(Kw) ˜ + vv]; ˜
(4.21)
noting that with nontrivial c ¿ 0 on this induces a norm on X. We will write
z u = ∈ U and ! = ∈H w v so, as is usual for a second-order system, we are taking the state to be x = (; !)T = (z; w; u; v)T ∈ X with u; v to be identi
−u −v −Kz
2
K w + u| + aKF(K[av]) (4.22)
with the domain D(A) given by D(A) := Kz ∈ L2 (); x ∈ Z: −cz on z = v on ;
2 (K[av]) ∈ L (); K2 w + aKF
Kw = 0 at 9
:
Remark 5. Note that the domain D(A) is not compactly imbedded in X so A does not have compact resolvent. To see this it suCces to consider the set D˜ ≡ {(z; w; u; v)∈BX (0; M ); supp w ∈ 0 ; v + w = 0;
2
2
(4.19)
where, for this example, 1 |zt |2 + |wt |2 ; Ekinetic ≡ 2 1 2 2 2 Epotential ≡ |∇z| + cz + |Kw| : 2
H∗2 ()
The inner product of the Hilbert space X will be [∇z˜ · ∇z + uu] ˜ x; ˜ x :=
Kz = 0; z + cz = 0; on ; z = v; on }
X := U × H; (4.20)
(4.23)
Clearly D˜ ⊂ D(A) with D˜ bounded in X yet not compact in X [6].
I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
The weak formulation of this is x; ˜ Ax X = [∇u˜ · ∇z − ∇z˜ · ∇u]
+
+
the reduced system for = (z; w)T , which we can now write as: M = f˜ ∈ V∗ where [∇ u ˜ · ∇z + uz] ˜ + cuz ˜
u˜ + [(Kv)(Kw) ˜ + vw] ˜ M : → v˜ + (K[av]) ˜ F(K[a(w − w∗ )]);
c[uz ˜ − zu] ˜ +
[vu ˜ − uv] ˜
[(Kv)(Kw) ˜ − (Kw)(Kv)] ˜
+
K[av] ˜ F(K[av])
f˜ :
and, as this last makes continuous sense whenever x; ˜x are in Z := U × U, we de
[∇u˜ · ∇z − ∇z˜ · ∇u]
u˜ v˜
→
u(u ˜ ∗ + z∗ )
+
cuw ˜ ∗+
v(v ˜ ∗ + z∗ | + w∗ ):
As in the previous section, M : V → V∗ is coercive, continuous, and monotone on V—hence surjective [3]—and we use this to show that A is monotone, hence—A generates a contraction semigroup S(·) on X; again the same argument also shows that—A˜ ˜ generates a contraction semigroup S(·) on X. We have veri
+ c[uz ˜ − zu] ˜ + [vu ˜ − uv] ˜ + [(Kv)(Kw) ˜ − (Kw)(Kv)] ˜ + K[av˜ ] F(K[av])]
251
C : Z → Y := L2 () : x = (z; w; u; v)T (4.24)
and note that we again have (2.12) with y = K[av] ∈ Y = L2 () well-de
−v + w = w∗ ∈ H∗2 (); −cz on ; 2 −Kz + u = u∗ ∈ L (); z = v on ;
→ y := K[av]
(4.25)
and : x → y2Y = Cx2Y is uniformly continuous on bounded subsets of Z since that bounds v in H∗2 () and we have assumed the coeCcient function a(·) is smooth enough. The remainder of the veri
K2 w + u| + aK F(K[av]) + v = v∗ ∈ L2 ();
K2 w − r 2 w + ir[aK2 aw + z| ] = aK2 aw∗ + irw∗ ;
w = 0 = Kw at 9;
w = 0 = Kw at 9:
The
Again, this may be expressed in terms of a continuous linear operator Mr : V → V∗ . Without
(4.26)
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I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
explicitly writing the full form of Mr , we note that, for = (z; w)T ∈ V, we have ; (Mr + s1) H = [ − z(Kz) V + (s − r 2 )|z|2 ]
+
[w(K V 2 w) + (s − r 2 )|w|2 ]
+ ir
=
2
2
2
[|∇z| + (s − r )|z| ] +
+r +
2 w[aK V w + z]
c|z|2
References
c Im{wz| V } 2
2
2
[|Kw| + (s − r )|w| ] + ir
We also have z ≡ 0 on 0 from the boundary condition of the
|K(aw)|2 : (4.27)
Since trace : H 1 () → L2 (9) is compact, we have an estimate z 6 zH 1 () + C zL2 () and one easily sees that (Mr +s1) is monotone and coercive for large enough s ∈ R. As for the previous example, this shows Mr has compact resolvent (for arbitrary r ∈ R) so invertibility will follow from injectivity. Setting r = 0 in (4.27), we see that this is immediate in that case—M0 = 0 gives 0 = ; M0 = 2V so = 0, completing Step 4. For Step 3 we wish to show that the hypothesis of Theorem 4 regarding the support of a(·) is suCcient to ensure the requisite detectability. This will now involve two separate applications of the classical Holmgren Uniqueness Theorem, once for each component of the coupled system. Taking the imaginary part of (4.27) when s = 0 and Mr = 0, we see that this gives K(aw) ≡ 0 on and this, with the boundary condition w =0=Kw at 9, gives aw on so w ≡ 0 on 0 , whence also K2 w ≡ 0 on 0 . Using this in the second equation of (4.26), now taken as homogeneous and interpreted pointwise, then gives z ≡ 0 on 0 .
[1] W. Arendt, W. Batty, Tauberian theorems and stability of one-parameter semigroups, Trans. Amer. Math. Soc. 8 (1988) 837–852. [2] G. Avalos, I. Lasiecka, The strong stability of a semigroup arising from a coupled hyperbolic/parabolic system, Semigroup Forum 57 (1998) 278–292. [3] V. Barbu, Nonlinear Semigroups and Di2erential Equations in Banach Spaces, Noordho2, Leiden, 1976. [4] J. Beale, Spectral properties of an acoustic boundary conditions, Indiana Univ. Math. J. 9 (1976) 895–917. [5] H. Brezis, Monotonicity methods in Hilbert spaces and some applications to nonlinear partial di2erential equations, in: E.H. Zarontonello (Ed.), Contributions to Nonlinear Functional Analysis, Academic Press, New York, 1971. [6] P. Grisvard, Characterisation de quelques espaces d’interpolation, Arch. Rational Mech. Anal. 26 (1967) 40–63. [7] Y. Komura, Nonlinear semigroups in Hilbert space, J. Math. Soc. Japan 19 (1967) 493–507. [8] P.M. Morse, K.U. Ingard, Theoretical Acoustics, McGraw-Hill, New York, 1968. [9] T.I. Seidman, H. Li, A note on stabilization with saturating feedback, Discrete Continuous Dynamical Systems 7 (2001) 319–328. [10] R. Showalter, Monotone Operators in Banach Spaces and Nonlinear Partial Di2erential Equations, American Mathematical Society, Providence, RI, 1997. [11] M. Slemrod, Feedback stabilization of a linear control system in Hilbert space with an a priori bounded control, Math. Control Signals Systems 2 (1989) 265–285. [12] R. Temam, Navier–Stokes Equations, North-Holland, Amsterdam, 1979.
Systems & Control Letters 48 (2003) 243 – 252 www.elsevier.com/locate/sysconle
Strong stability of elastic control systems with dissipative saturating feedback Irena Lasieckaa;∗;1 , Thomas I. Seidmanb;2 b Department
a Department of Mathematics, University of Virginia, Charlottesville, VA 22904, USA of Mathematics and Statistics, University of Maryland, Baltimore County, Baltimore, MD 21250, USA
Received 29 August 2001; received in revised form 8 February 2002; accepted 1 April 2002 This paper is dedicated to the memory of the late J.L. Lions, whose seminal work in applied mathematics led, in particular, to the embedding of control-theoretic problems in the context of the modern theory of partial di2erential equations and inspired us all.
Abstract We will consider, with a focus on saturating feedback control laws, two problems associated with damping in a bounded acoustic cavity ⊂ R3 . Our objective is to verify (compare (Discrete Continuous Dynamical Systems 7 (2001) 319, Math. Control Signals Systems 2 (1989) 265) that these are strongly stable: for every
1. Introduction
equation
We will be considering, as indicating some relevant technical diCculties in the analysis, the question of stabilization for two paradigmatic examples of elastic systems when the feedback is dissipative but ‘saturating’. Both of these examples involve the wave
ztt − Kz = 0
∗ Corresponding author. Tel.: +1-804-924-8896; fax: +1-804982-3084. E-mail addresses: [email protected] (I. Lasiecka), seidman@ math.umbc.edu (T.I. Seidman). URLs: http://www.math.virginia.edu/∼il2v, http://www.math. umbc.edu/∼seidman 1 Partially supported by the NSF Grant DMS 9804056. 2 Supported by the TRW Foundation.
in ⊂ R3
(1.1)
e.g., modelling acoustic vibration in a bounded enclosure ) with interaction available only at part of the boundary. In each of our examples the uncontrolled system is conservative and linear and it will be possible to show that a corresponding linear feedback does stabilize the system. Our principal concern here is to see that the saturating control law, coinciding with the linear feedback for small enough data, still provides global stabilization: the relevant physical energy goes to 0 as t → ∞. As the saturation is inherently nonlinear, our context will be the theory of contraction semigroups of nonlinear operators on Hilbert space.
c 2002 Elsevier Science B.V. All rights reserved. 0167-6911/03/$ - see front matter PII: S 0 1 6 7 - 6 9 1 1 ( 0 2 ) 0 0 2 6 9 - 4
244
I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
For our
on 9;
(1.2)
the control speci
where the control ’ is to have support restricted to some speci
F(y) := min{1; M=y} y if y 6 M; = My=y if y ¿ M
E(t) = Ekinetic + Epotential
to ensure that ’ does not exceed the saturation threshold M . Our major thrust in this paper is to show—for this and correspondingly for our second example—that the linear control law (1.5) is asymptotically stable and that we can overcome the technical diCculties to compare the e2ects of (1.5) and its saturating version to show that the latter also provides stabilization. For showing the asymptotic stability of (1.5), we follow [2] in applying [1]. For consideration of the saturating version it had been our intention to follow [9], but it seems more convenient to present, in the next section, a new abstract comparison theorem which makes the comparison directly, without the intervening reference to the uncontrolled situation used in [9]. (It is worth noting at this point the emphasis in our Theorem 1 on output stabilization for ‘nice’ initial data.) Our second example corresponds to damping of the acoustic vibration in through interaction with a plate occupying a (2-D, Pat) portion of the boundary 9— the well-known structural acoustic model [8,4]. Here the feedback is through coupling of the acoustic vibration in with the vibration in the plate . We again use ‘velocity feedback’ to dampen those vibrations and so the entire system, but for the structural acoustic model this will appear as distributed control on (part of) (i.e., in the system equations) rather than as boundary control on ⊂ 9 as in our
:= 12 [zt 2 + ∇z2 R] ¡ ∞
(1.3)
and wish to ensure that—always—we have E(t) → 0 for the feedback-controlled solutions. Formally, we may di2erentiate E(·), use (1.1), and then apply the Divergence theorem to the term zt ; Kz to get, in view of (1.2), dE = zt z = zt ; ’ : (1.4) dt 9 For our
restricted to :
(1.5) 2
This gives dE=dt = −(trace of zt on ) so at least the control action would be dissipative. Apart from showing that we then would eventually have stabilization, there are three considerable diCculties with this: • justi
ztt = Kz
in ⊂ R3 ;
z = wt
on ⊂ 9;
z + cz = 0
(1.6)
on = 9 \ ;
wtt + K2 w + zt = −’ w = 0; Kw = 0
in (⊂ R2 );
at 9 ⊂ R2 :
(1.7)
I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
Remark 1. Note that the Laplacians K appearing in (1.7) are di2erent in the equations for z and for w, with the
245
oped in the next section. Indeed, the strong stability results obtained for these examples do not follow from standard approaches. To wit, a major diCculty in Example 1 is the presence of nontrivial steady states in a classical formulation of the abstract wave equation. This of course could be eliminated (as is often done in the literature) by changing the problem: modifying the boundary conditions on 9 or by creating a “hole” in the domain with zero Dirichlet boundary conditions prescribed on a portion of the boundary. However, in our tratment we do not do this, aiming for the ultimate goal of stabilizing the acoustic energy only. In the second example the diCculty is more serious as it is related to an intrinsic lack of compactness of the resolvent operator. 2. General theory We
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I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
˜ with 6 ; A 6 : (2) for ∈ D(A) ; A → 0 implies () → 0, ˜ with 6 ; A ˜ 6 : (3) for ∈ D(A) (·) is uniformly continuous: X → R+ ., ˜ if and only if (4) when () 6 : ∈ D(A) ∈ D(A) ˜ = A. — and then A ˜ is strongly stable, i.e., S(t)x ˜ (H3) S(·) 0 → 0 for each x0 ∈ X. Then S(·) is also strongly stable. Proof. Our principal task will be to prove that: (C) If x0 ∈ D(A), then (S(t)x0 ) → 0 as t → ∞. To see from (C) that S(t)x0 → 0 for x0 ∈ D(A), note that it gives (S(t)x0 ) ¡ from some time s on so, by (H2-4), A˜ then coincides with A and the conver˜ − s)x(s) follows from (H3). gence to 0 of S(t)x0 = S(t Strong stability now follows by a density argument: if we are given arbitrary x0 ∈ X, then for any ¿ 0 we can choose x0 ∈ D(A) with x0 − x0 ¡ =2 and note,
(2.8)
and by (H2-3) we have ∀ ¿ 0; ∃ = ( ) = ( ; ; ) such that if ˜ ˜ 6 ; A ; | 6 ; A; then − 6 ⇒ | () − ( )| 6 :
(2.9)
3 The equality depends, of course, on the single-valuedness of A. We also note that we are here taking Dt+ x to be the forward derivative (lim[x(t + h) − x(t)]=h for h 0+).
Supposing we have ˜ (s) ˆ ¡ for some sˆ ¿ 0 so, by ˜ with Ax( ˜ s) (H2-4), we have x(s) ˆ ∈ D(A) ˆ = Ax(s) ˆ ˜ s) ˜ − s)x( and Ax( ˆ 6 . Then we set x(t) ˆ := S(t ˆ s), ˆ ˆ (t) := (x(t)) ˆ for t ¿ s; ˆ note that ˆ (s) ˆ = ˜ (s) ˆ ¡ and, again using the Komura–Kato Theorem, we have ˜ (so ˆ is well deDt+ xˆ = A˜ xˆ pointwise—x(t) ˆ ∈ D(A) ˜
(2.10)
so, as x; Ax ¿ 0, there is a sequence sk → ∞ at which x; Ax → 0 so, by (H2-2), we have ˜ (sk ) → 0. Without loss of generality, we assume each ˜ (sk ) ¡ ¡ and introduce xˆ = xˆk by taking sˆ = sk ˜ − sk )x(sk ) in the notation above, i.e., xˆk (t) := S(t ˆ ˆ and (t) = k (t) := (x(t)) ˆ for t ¿ sk . Note that each ˆ k is continuous in t for t ¿ sk , since xˆk (·) is continuous—indeed, t t · 6 x(t) ˆ − x(s) ˆ 6 x ˆ Ax ˆ s
6 |t − s|
s
(2.11)
and we have (H2-3). There is then a maximal interval Ik = [sk ; tk ] on which ˆ 6 . On this interval xˆ · = A˜ xˆ = Axˆ by (H2-4) so, by uniqueness of the solution, x(t)=S(t ˆ −sk )x(s ˆ k )=x(t) on Ik . By continuity, there is necessarily a subinterval IV k = [sVk ; tVk ] ⊂ Ik such that, noting that ˆ k = ˜ on Ik , ˜ (sVk ) = 6 ˜ (t) 6 2 = ˜ (tVk ) on IV k : Note that, by (2.8), having ˜ ¿ requires that x; Ax ¿ = ( ) ¿ 0 on IV k . Also, by (2.9), ˆ (tVk )− ˆ (sVk ) = requires that x( ˆ tVk ) − x( ˆ sVk ) ¿ = ( ) whence, by (2.11), we have tVk − sVk ¿ =. Combining these shows that the integral over IV k of x; Ax is not less than = ¿ 0, independent of k. Since, without loss of generality, we can assume that the intervals V I k are disjoint, this would contradict the bound on x; Ax dt of (2.10) —i.e., we must have ˜ → 0 as in (C).
I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
For each of our systems the state space X will have the form X = U × H where U is a Hilbert space of ‘con
(2.12)
so A (hence also A) is monotone. To show that A is maximal monotone, it is suCcient to show that (A+1) is surjective. Since (2.12) does not give coercivity, we proceed indirectly and consider a ‘reduced’ probcont: lem involving a related operator M : V → V∗ with ∗ V ,→ H ,→ V . This M will be monotone and coercive, which will enable us to show that—A generates a contraction semigroup S(·) on X. A similar construction applies to the linear problem, with F(·) replaced by the identity on Y. To verify the hypothesis (H3) above we will, in each case, apply a result by Arendt and Batty [1]. ˜ be a C0 of linear operators on Theorem 2. Let S(·) ˜ a Hilbert space X with in8nitesimal generator −A. ˜ If %(A) ∩ iR is at most countable and contains no ˜ is strongly asymptotically point spectrum, then S(·) ˜ stable, i.e., S(t)x0 → 0 as t → ∞ for every x0 ∈ X. Our argument is then completed by verifying the ˜ hypothesis of Theorem 2 on %(A)—indeed, we will ˜ show for our examples that %(A) ∩ iR = ∅—obviously to be considered in the complexi
247
Our key steps in the analysis of each of these examples will thus be the following: (1) Reformulate the problem as a
in
with z = −aF(azt ) on 9; (3.13)
where a(·) is a bounded measurable function on 9 with nontrivial support and F, de
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I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
Remark 2. We note that, on the strength of Holmgren’s uniqueness theorem, the detectability condition holds true whenever is an open set in 9. Theorem 3. Assume the detectability condition (D). Then we have E(t) → 0 as t → ∞ for every solution of (3.13) with initial data of 8nite energy, where E(t) is de8ned by (1.3). Proof. As Step 1 we will reformulate (3.13) as a
(3.14)
where we note that L2grad ()={∇h: h ∈ V} is a closed subspace of L2 ( → R3 ) and so is a Hilbert space in its own right; cf., e.g. [12]. Indeed, this follows from the fact that L2grad () coincides with the range of ∇ : H 1 () → [L2 ()]3 and, since this map can be taken to be an isometry after factoring out constants, the range is closed. We write the state as x = (; u)T ∈ X, with and u corresponding to ∇z and zt , respectively, so 12 x2X is just the energy E of (1.3). Wave equation (3.13) then takes the form t = ∇u ut = ∇
on with · = −aF(au) on 9: (3.15)
Note that if we are given the usual (
0
We wish to consider (3.15) as x˙ + Ax = 0. It is preferable to give a weak formulation for A, since just writing A=
0
−∇
−∇
0
would hide the boundary con-
ditions, including the feedback of particular interest
to us, in the domain speci
0 −∇ ˜ Ax X = x˜ · x x; −∇ 0 = [∇u˜ · − xi ˜ · ∇u] + auF(au) ˜
˜ x and, as this last makes continuous sense whenever x; are in Z, we de
(3.16)
and, now simply specifying D(A) as the pre-image of X under A, use this to de
=
[˜ · f + ug] ˜
˜ u) for all (; ˜ T ∈ Z. With u˜ = 0 this gives = ∇u + f so, setting Mu := u˜ → [∇u˜ · ∇u + uu] ˜ +
auF(au) ˜ ∈ V∗
for u ∈ V;
[∇u˜ · f − ug] ˜ ∈ V∗ ; f˜ := u˜ →
˜ Since we have the reduced problem of solving Mu=f. ∗ we easily see that M : V → V is coercive on V as well as continuous and monotone, this can be solved for u ∈ V ⊂ H and we then set = ∇u + f to get the desired x = (; u)T ∈ D(A) ⊂ X. This proves that (A + 1) is surjective to X so A is maximal monotone and −A generates a contraction semigroup S(·) on X. The same argument shows that −A˜ (where, we recall, −A˜ corresponds to the linear problem) generates ˜ on X so we have veri
I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
the hypothesis (H1) of Theorem 1. The observation operator for this example is C : Z → Y := L2 () ⊂ L2 (9) : x = (; u)T → a[trace of u]:
(3.17)
so : x → y2Y =Cx2Y is uniformly continuous on bounded subsets of Z—although it is not well-de
=
[˜V · f + ug] ˜V
˜ u) for all (; ˜ T ∈ Z—where, of course, the overlines indicate conjugation since we must now work in the complexi
− ir a2 uu ˜V ∈ V∗ for u ∈ V; [∇u˜ · f − ug] ˜V ∈ V∗ : f˜ := u˜ →
We easily verify that (Mr +s1) : V → V∗ is coercive as well as continuous and monotone when s ¿ r 2 , so there exists (Mr + s1)−1 : V∗ → V ,→ H. Since the embedding V := H 1 () ,→ H := L2 () is compact, this shows Mr has compact resolvent and so is continuously invertible on H, whence (A˜ − ir1) is continuously invertible on X, if Mr is injective. Thus, we wish to show u = 0 when Mr u = 0. The imaginary part of u; Mr u [V : V∗ ] is a2 |u|2 so Mr u = 0 gives y = au = 0 ∈ Y (i.e., u vanishes on supp(a)) and the homogeneous equation becomes simply the weak form
249
of the eigenfunction equation: −Ku = r 2 u with u = 0. Since we are here considering r = 0 so u = const:, the assumption (D) then gives u = 0. This shows that ˜ for 0 = r ∈ R we have ir ∈ %(A). When r = 0 above, we must proceed di2erently: we have ∇u + f = 0 and will now use this to eliminate u. Note that f ∈ U so there is some h ∈ V with f = ∇h; thus ∇(u+h)=0 gives u=−h+c where the constant c is to be determined. We also note that we are seeking ∈ U in the form =∇v by the de
with v = a2 (h − c):
It that this has a solution if and only if is well-known g + v = 0, so we havethe unique determination that c = [ g + a2 h]=[ a2 ]: The resulting v is not unique, but this indeterminacy is nugatory: = ∇v is unique in U. This establishes a map A˜ −1 : X → ˜ ⊂ X and, since A˜ is clearly a closed linear D(A) operator, the map A˜ −1 is continuous, showing that ˜ and so completing Step 4. 0 ∈ %(A) This veri
(4.18)
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I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
where a is a nontrivial smooth function on ⊂ 9 with support in the interior of and the function c is assumed positive. We consider
As before, if we (formally) apply the Divergence The orem we obtain dE=dt = − yF(y) 6 0 (now with y := K[awt ]), so our feedback is again dissipative: the energy is nonincreasing. Our main aim is to ensure that we have E(t) → 0 as t → 0 for all
2
:= {w ∈ H (): w|9 = 0};
H := L () × L (); Z := U × U:
+
czz ˜ +
[(Kw)(Kw) ˜ + vv]; ˜
(4.21)
noting that with nontrivial c ¿ 0 on this induces a norm on X. We will write
z u = ∈ U and ! = ∈H w v so, as is usual for a second-order system, we are taking the state to be x = (; !)T = (z; w; u; v)T ∈ X with u; v to be identi
−u −v −Kz
2
K w + u| + aKF(K[av]) (4.22)
with the domain D(A) given by D(A) := Kz ∈ L2 (); x ∈ Z: −cz on z = v on ;
2 (K[av]) ∈ L (); K2 w + aKF
Kw = 0 at 9
:
Remark 5. Note that the domain D(A) is not compactly imbedded in X so A does not have compact resolvent. To see this it suCces to consider the set D˜ ≡ {(z; w; u; v)∈BX (0; M ); supp w ∈ 0 ; v + w = 0;
2
2
(4.19)
where, for this example, 1 |zt |2 + |wt |2 ; Ekinetic ≡ 2 1 2 2 2 Epotential ≡ |∇z| + cz + |Kw| : 2
H∗2 ()
The inner product of the Hilbert space X will be [∇z˜ · ∇z + uu] ˜ x; ˜ x :=
Kz = 0; z + cz = 0; on ; z = v; on }
X := U × H; (4.20)
(4.23)
Clearly D˜ ⊂ D(A) with D˜ bounded in X yet not compact in X [6].
I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
The weak formulation of this is x; ˜ Ax X = [∇u˜ · ∇z − ∇z˜ · ∇u]
+
+
the reduced system for = (z; w)T , which we can now write as: M = f˜ ∈ V∗ where [∇ u ˜ · ∇z + uz] ˜ + cuz ˜
u˜ + [(Kv)(Kw) ˜ + vw] ˜ M : → v˜ + (K[av]) ˜ F(K[a(w − w∗ )]);
c[uz ˜ − zu] ˜ +
[vu ˜ − uv] ˜
[(Kv)(Kw) ˜ − (Kw)(Kv)] ˜
+
K[av] ˜ F(K[av])
f˜ :
and, as this last makes continuous sense whenever x; ˜x are in Z := U × U, we de
[∇u˜ · ∇z − ∇z˜ · ∇u]
u˜ v˜
→
u(u ˜ ∗ + z∗ )
+
cuw ˜ ∗+
v(v ˜ ∗ + z∗ | + w∗ ):
As in the previous section, M : V → V∗ is coercive, continuous, and monotone on V—hence surjective [3]—and we use this to show that A is monotone, hence—A generates a contraction semigroup S(·) on X; again the same argument also shows that—A˜ ˜ generates a contraction semigroup S(·) on X. We have veri
+ c[uz ˜ − zu] ˜ + [vu ˜ − uv] ˜ + [(Kv)(Kw) ˜ − (Kw)(Kv)] ˜ + K[av˜ ] F(K[av])]
251
C : Z → Y := L2 () : x = (z; w; u; v)T (4.24)
and note that we again have (2.12) with y = K[av] ∈ Y = L2 () well-de
−v + w = w∗ ∈ H∗2 (); −cz on ; 2 −Kz + u = u∗ ∈ L (); z = v on ;
→ y := K[av]
(4.25)
and : x → y2Y = Cx2Y is uniformly continuous on bounded subsets of Z since that bounds v in H∗2 () and we have assumed the coeCcient function a(·) is smooth enough. The remainder of the veri
K2 w + u| + aK F(K[av]) + v = v∗ ∈ L2 ();
K2 w − r 2 w + ir[aK2 aw + z| ] = aK2 aw∗ + irw∗ ;
w = 0 = Kw at 9;
w = 0 = Kw at 9:
The
Again, this may be expressed in terms of a continuous linear operator Mr : V → V∗ . Without
(4.26)
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I. Lasiecka, T.I. Seidman / Systems & Control Letters 48 (2003) 243 – 252
explicitly writing the full form of Mr , we note that, for = (z; w)T ∈ V, we have ; (Mr + s1) H = [ − z(Kz) V + (s − r 2 )|z|2 ]
+
[w(K V 2 w) + (s − r 2 )|w|2 ]
+ ir
=
2
2
2
[|∇z| + (s − r )|z| ] +
+r +
2 w[aK V w + z]
c|z|2
References
c Im{wz| V } 2
2
2
[|Kw| + (s − r )|w| ] + ir
We also have z ≡ 0 on 0 from the boundary condition of the
|K(aw)|2 : (4.27)
Since trace : H 1 () → L2 (9) is compact, we have an estimate z 6 zH 1 () + C zL2 () and one easily sees that (Mr +s1) is monotone and coercive for large enough s ∈ R. As for the previous example, this shows Mr has compact resolvent (for arbitrary r ∈ R) so invertibility will follow from injectivity. Setting r = 0 in (4.27), we see that this is immediate in that case—M0 = 0 gives 0 = ; M0 = 2V so = 0, completing Step 4. For Step 3 we wish to show that the hypothesis of Theorem 4 regarding the support of a(·) is suCcient to ensure the requisite detectability. This will now involve two separate applications of the classical Holmgren Uniqueness Theorem, once for each component of the coupled system. Taking the imaginary part of (4.27) when s = 0 and Mr = 0, we see that this gives K(aw) ≡ 0 on and this, with the boundary condition w =0=Kw at 9, gives aw on so w ≡ 0 on 0 , whence also K2 w ≡ 0 on 0 . Using this in the second equation of (4.26), now taken as homogeneous and interpreted pointwise, then gives z ≡ 0 on 0 .
[1] W. Arendt, W. Batty, Tauberian theorems and stability of one-parameter semigroups, Trans. Amer. Math. Soc. 8 (1988) 837–852. [2] G. Avalos, I. Lasiecka, The strong stability of a semigroup arising from a coupled hyperbolic/parabolic system, Semigroup Forum 57 (1998) 278–292. [3] V. Barbu, Nonlinear Semigroups and Di2erential Equations in Banach Spaces, Noordho2, Leiden, 1976. [4] J. Beale, Spectral properties of an acoustic boundary conditions, Indiana Univ. Math. J. 9 (1976) 895–917. [5] H. Brezis, Monotonicity methods in Hilbert spaces and some applications to nonlinear partial di2erential equations, in: E.H. Zarontonello (Ed.), Contributions to Nonlinear Functional Analysis, Academic Press, New York, 1971. [6] P. Grisvard, Characterisation de quelques espaces d’interpolation, Arch. Rational Mech. Anal. 26 (1967) 40–63. [7] Y. Komura, Nonlinear semigroups in Hilbert space, J. Math. Soc. Japan 19 (1967) 493–507. [8] P.M. Morse, K.U. Ingard, Theoretical Acoustics, McGraw-Hill, New York, 1968. [9] T.I. Seidman, H. Li, A note on stabilization with saturating feedback, Discrete Continuous Dynamical Systems 7 (2001) 319–328. [10] R. Showalter, Monotone Operators in Banach Spaces and Nonlinear Partial Di2erential Equations, American Mathematical Society, Providence, RI, 1997. [11] M. Slemrod, Feedback stabilization of a linear control system in Hilbert space with an a priori bounded control, Math. Control Signals Systems 2 (1989) 265–285. [12] R. Temam, Navier–Stokes Equations, North-Holland, Amsterdam, 1979.