J. Evol. Equ. 12 (2012), 141–164 © 2011 Springer Basel AG 1424-3199/12/010141-24, published online November 2, 2011 DOI 10.1007/s00028-011-0127-x
Journal of Evolution Equations
Uniform boundary stabilization of a wave equation with nonlinear acoustic boundary conditions and nonlinear boundary damping P. Jameson Graber
Abstract. We consider a wave equation with nonlinear acoustic boundary conditions. This is a nonlinearly coupled system of hyperbolic equations modeling an acoustic/structure interaction, with an additional boundary damping term to induce both existence of solutions as well as stability. Using the methods of Lasiecka and Tataru for a wave equation with nonlinear boundary damping, we demonstrate well-posedness and uniform decay rates for solutions in the finite energy space, with the results depending on the relationship between (i) the mass of the structure, (ii) the nonlinear coupling term, and (iii) the size of the nonlinear damping. We also show that solutions (in the linear case) depend continuously on the mass of the structure as it tends to zero, which provides rigorous justification for studying the case where the mass is equal to zero.
1. Introduction The purpose of this paper is to study a wave equation with nonlinear acoustic boundary conditions. The model, which arises in the study of acoustic/structure interaction, consists of a wave equation restricted to a bounded domain coupled with a pointwise damped harmonic oscillator equation governing the structure of a frame, with boundary conditions on the interface. In contrast to other studies of acoustic/structure interaction models, the present paper takes into account a nonlinear coupling between two sets of dynamics. Under such conditions, our goal is to provide the following results: 1. existence and uniqueness of finite energy solutions; 2. uniform decay rates for the energy as time goes to infinity; and 3. continuous dependence of solutions on the mass of the structure as it tends to zero. The paper is organized into five sections. In the first section, we give a description of the problem and the main results. In the second section, we give an appropriate abstract functional analysis setup of the problem. The third section is devoted to showing existence of a semigroup and well-posedness of suitably defined weak solutions. Keywords: Acoustic boundary conditions, wave equation, nonlinear boundary damping, boundary stabilization, coupled systems, nonlinear semigroup theory. Research partially supported by the Jefferson Scholars Foundation and the Virginia Space Grant Consortium.
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The fourth section proves uniform decay rates given suitable nonlinear damping. In the fifth section we analyze the effect of making the mass of the structure in question smaller and even negligible (equal to zero), thereby making the coupled system more stable. 1.1. Description of the problem Let be a bounded domain in Rn with a boundary of class C 2 . Assume = 0 ∪ 1 (disjoint). Let ν be the unit normal vector pointing to the exterior of . Consider the following boundary value problem: u tt (x, t) − u(x, t) + α(x)u(x, t) = 0,
x ∈ , t > 0;
(1)
u(x, t) = 0,
x ∈ 0 , t > 0;
(2)
u t (x, t) + m(x)z tt (x, t) + f (x)z t (x, t) + g(x)z(x, t) = 0,
x ∈ 1 , t > 0;
(3)
∂u (x, t) + θ (u t (x, t)) − h(x)η(z t (x, t)) = 0, ∂ν u(x, 0) = u 0 (x), u t (x, 0) = u 1 (x),
x ∈ 1 , t > 0;
(4)
x ∈ ;
(5)
z(x, 0) = z 0 (x), z t (x, 0) = z 1 (x),
x ∈ 1 ;
(6)
where is the Laplacian operator, and α : → R, f, g, h, m : 1 → R, η, θ : R → R are given functions. The PDE system above is an established model of a wave equation with acoustic boundary conditions, introduced by Morse and Ingard [18] and developed by Beale and Rosencrans [5–7]. In this model, the portion of the boundary denoted 1 is a locally reacting surface, with each point on the surface acting like a damped harmonic oscillator in response to excess pressure from the fluid in the interior (Eq. 3) [7]. The coupling between the acoustic pressure and the displacement of the boundary is given by Eq. (4). In this model, the boundary is taken to be porous, and the coupling between the interior and boundary dynamics is assumed to be nonlinear rather than linear; this is modeled by the term η(z t (x, t)), where η is a monotone increasing function. The function h(x) indicates that the porosity of the boundary is variable in space. The term θ (u t (x, t)) represents frictional damping added to the boundary as a feedback control, where θ is a monotone increasing function. In this paper, we study the existence, uniqueness, and uniform decay of finite energy solutions to the above PDE system. Existence and uniqueness is established using nonlinear semigroup theory; the framework for studying decay rates is provided by a paper by I. Lasiecka and D. Tataru on the semi-linear wave equation with boundary damping, which provides a way of eliminating the need for growth bounds on θ near the origin [16]. For the present model, the most critical interaction is between (i) the mass term m(x) in Eq. (3), (ii) the nonlinear coupling term η(z t (x, t)) in Eq. (4), and (iii) the nonlinear frictional damping term θ (u t (x, t)) in Eq. (4). The interplay between these terms determines crucial information about the problem’s well-posedness as well as long-term behavior.
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For the moment assume θ = 0 (no damping and that the coupling is linear, i.e., η(s) = s. This reduces (1)–(6) to a model which has already been considered in many references [5–11,13,19]. From a linear semigroup point of view, the interesting feature of this problem is that the resolvent is not compact [5], and therefore approaches to studying stability based on Nagy–Foias theory or the LaSalle Invariance Principle are inapplicable. This characteristic has been noted in other acoustic/structure interaction models [2,3,15]. Nevertheless, if we assume the mass of the structure to be negligible (i.e., m ≡ 0), then one can show strong stability of solutions and even exponential decay rates under appropriate geometric assumptions on the boundary [13] (see also [10,19]). In such a case, there is no reason to include the presence of a nonlinear damping force. On the other hand, if the structure has non-negligible mass (i.e., m > 0), then in general solutions are unstable [5,11]. See also [1], which studies a similar boundary condition and shows that the presence of the z tt term in the dynamics causes the solutions to be unstable. For this reason, one of the main goals of the present work is to stabilize solutions in the presence of non-negligible mass m by adding sufficiently large boundary damping. The problem is quite different when the coupling function η is nonlinear. Whereas in the linear case the damping in the harmonic oscillator on the boundary translates into boundary damping on the total energy of the system, in the nonlinear case this translation is at best imperfect. From the point of view of semigroup theory, even when η is monotone, the problem is no longer monotone. If we assume a massless structure (m ≡ 0), then one has well-posedness of finite energy solutions [12]; however, unlike the linear case, the nonlinear model has the possibility of an exponential blow-up of the energy. If we assume non-negligible mass (m > 0), even well-posedness is called into question. The most drastically unstable situation occurs when the mass is positive yet has no positive lower bound. Physically this may occur when a structure is “lopsided,” so that most of its mass is concentrated in a particular region. To cope with the instabilities caused by a nonlinear coupling as well as the mass of the structure, we require a nonlinear damping (θ (u t )) which is sufficiently “steep” (i.e., strongly monotone) in order to compensate. The degree of the required “steepness” increases with respect to the severity of the particular case (see Theorems 1 and 3 below). In addition to existence, uniqueness, and uniform stability of finite energy solutions, we also consider the precise mathematical connection between the two cases m ≡ 0 and m > 0. It is clear from the preceding remarks that assuming zero mass is conducive to obtaining both well-posedness and stability of finite energy solutions. From the mathematical point of view, this is because the assumption m ≡ 0 means that the dynamics on z are first-order in time, and thus the coupling term η(z t ) no longer behaves as a potential term. From a physical point of view, however, it is more realistic to assume positive mass, as all real-world structures possess some mass. In the linear model, there is a disparity between theoretical results showing stability [10,13,19] for m ≡ 0 and those proving lack of stability [5,11] for m > 0. To resolve this disparity, in Sect. 5 we prove a theorem showing continuous dependence of solutions on the
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parameter m as m → 0. This result provides rigorous justification for studying the case m ≡ 0, since solutions to the unstable positive mass problem can be shown to behave in a stable way, up to a given finite time, by choosing the mass term sufficiently small. 1.2. Preliminaries In this subsection, first we give the notation for defining the phase space; then we define weak solutions; and finally we state the well-posedness theorem. Let H10 () = {u ∈ H 1 () : u|0 = 0}. Let (·, ·) denote the usual inner product in L 2 () with corresponding norm | · | and let ·, · denote the usual inner product in L 2 () with corresponding norm | · |2 . For u, v ∈ H10 () we let ((u, v))α = (∇u, ∇v) + (αu, v) with corresponding norm u 2α = ((u, u))α . If k : 1 → R and φ, ψ ∈ L 2 (1 ) we let φ, ψk = kφ, ψ with corresponding norm | · |k (if defined). ASSUMPTION 1. The inner product ((·, ·))α on H10 () is equivalent to the usual inner product. The functions f, g, h, and m are all positive and bounded (a.e.), and there exists a positive constant f 0 such that f ≥ f 0 . H1
We will make the following assumption on the nonlinear functions η and θ . ASSUMPTION 2. Both η and θ are continuous, monotone increasing functions on R with η(0) = θ (0) = 0. We have η0 (s − t)2 ≤ (η(s) − η(t))(s − t) ≤ η1 (s − t)2 for some 0 < η0 ≤ η1 . The damping function θ is coercive, i.e., θ (s)s > 0 for s = 0, and has linear growth away from the origin, i.e., θ0 s 2 ≤ θ (s)s ≤ θ1 s 2 for |s| ≥ 1, for some 0 < θ0 ≤ θ1 . Observe that η(s) ˜ = η(s) − s is also a Lipschitz function. We denote by η˜ 1 the sharpest Lipschitz constant such that |η(t) ˜ − η(s)| ˜ ≤ η˜ 1 |t − s| for all s, t ∈ R. We now define functions p, q which describe the decay rates we will show. These functions are defined in [16], and we reproduce the definitions here for convenience. Let H (s) be a real-valued function defined for s ≥ 0, which is concave, strictly increasing, and satisfies H (0) = 0 and H (sθ (s)) ≥ s 2 + θ 2 (s) for |s| ≤ N , for some N > 0.
(7)
Let H D (x) := H (Dx), x ≥ 0, where D is a positive constant yet to be specified. Let p(x) = p(x, c, K ) := (cI + H D )−1 (K x),
(8)
where c ≥ 0, K > 0 are given constants. Then, p is a positive, continuous, strictly increasing function with p(0) = 0. Let q(x) := x − (I + p)−1 (x), x > 0. Then, q is a positive increasing function.
(9)
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1
In what follows, γ0 : H 1 () → H 2 () is the trace map of order zero, i.e., γ0 (u) = 1 ∂ u| ; γ1 : H 1 () → H − 2 () is the trace map of order one, i.e., γ1 (u) = ∂ν u| ; and | · |∞ denotes the supremum norm. For X a topological vector space, Cw ([0, T ], X ) denotes the space of vector-valued functions u : [0, T ] → X that are weakly continuous with respect to the topology given on X ; similarly, Cr ([0, T ], X ) denotes the space of vector-valued functions u : [0, T ] → X that are right continuous with respect to the topology given on X . Using the notation given above, we define the “finite energy space” that will be associated with “weak solutions”: H := H10 () × L 2 () × L 2 (1 ) × L 2 (1 ) where the norm on H is given by (u, v, z, w) 2H = u 2α + |v|2 + |z|2gh + |w|2hm . We now give the following definition of weak solutions of the system (1)–(6). DEFINITION 1. A pair of functions u : × [0, ∞) → R, z : 1 × [0, ∞) → R is called a weak variational solution to the system (1)–(6) provided that, given T > 0 arbitrary, • u ∈ Cw ([0, T ]; H10 ()), u t ∈ Cw ([0, T ]; L 2 ()), u tt ∈ L ∞ ([0, T ]; H −1 ()); • z ∈ Cw ([0, T ]; L 2 (1 )), z t ∈ Cw ([0, T ]; L 2 (1 )), u t , θ (u t ) ∈ L 2 ([0, T ]×1 ), and • for every φ ∈ H10 (), ψ ∈ L 2 (1 ), we have (u tt , φ) + ((u, φ))α = hη(z t ) − θ (γ0 u t ), γ0 φ,
(10)
z tt , ψhm + z t , ψ f h + z, ψgh = −γ0 u t , ψh ,
(11)
u(0) = u 0 , u t (0) = u 1 , z(0) = z 0 , z t (0) = z 1
(12)
REMARK 1. Relation (12) describing the initial conditions is understood with respect to weak topology, i.e., limt→0 ((u(t), φ)) = ((u 0 , φ)), ∀φ ∈ H10 (), etc. 1.3. Main results THEOREM 1. (Weak solutions) Suppose that Assumptions 1 and 2 hold, and that one of the following holds: (a) η(s) = s for all s ∈ R; or (b) there exists m 0 > 0 such that m ≥ m 0 a.e. and θ satisfies the strong coercivity condition (θ (s) − θ (t))(s − t) ≥ (s − t)2 for some fixed > 0 (small); or (c) m has no positive lower bound, but instead θ satisfies the following coercivity condition: (θ (s) − θ (t))(s − t) ≥ C f,h,η (s − t)2 for all s, t ∈ R, where C f,h,η >
|h|∞ η˜ 12 4 f0
is a fixed constant.
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Then, for each set of initial conditions (u 0 , u 1 , z 0 , z 1 ) ∈ H, there exists a unique pair of weak solutions u, z to the system (1)–(6) such that (u(t), u t (t), z(t), z t (t)) is continuous in time and depends continuously on (u 0 , u 1 , z 0 , z 1 ). Theorem 1 allows us to construct the map St : H → H given by St (u 0 , u 1 , z 0 , z 1 ) := (u(t), u t (t), z(t), z t (t)) which defines a continuous semiflow on the phase space H. The pair (St , H) is a well posed dynamical system [20,22]. COROLLARY 1. (Semigroup solutions) St is a strongly continuous nonlinear semigroup on H. Semigroup solutions (or generalized solutions) coincide with weakvariational solutions, defined in Definition 1. In addition, the following energy inequality holds: |St (x)|H ≤ Ceωt |x|H for some constants C ≥ 1 and ω ≥ 0. THEOREM 2. (Regularity) Assume the hypotheses of Theorem 1. Let (u 0 , u 1 , z 0 , z 1 ) ∈ H be a set of initial conditions such that u 1 ∈ H10 (), u 0 ∈ L 2 (), and γ1 u 0 = hη(z 1 ) − θ (γ0 u 1 ) on 1 . Then, the unique solution u, z obtained satisfies u t ∈ L ∞ ([0, ∞); H10 ()), u tt ∈ L ∞ ([0, ∞); L 2 ()), z tt ∈ L ∞ ([0, ∞); L 2 (1 )), u ∈ Cr ([0, ∞); L 2 ()) and ∂u ∂ν ∈ Cr ([0, ∞); L 2 (1 )), which implies u ∈ Cr ([0, ∞); H 3/2 ()). REMARK 2. We note that for u 0 ∈ H10 () such that u 0 ∈ L 2 () one obtains from Green’s formula, the trace theorem and the Riesz representation theorem that ∂ −1/2 ( ). Thus, ∂ u in Theorem 2 is well defined a priori. 1 ∂ν u 0 ∈ H ∂ν 0 THEOREM 3. (Decay rates) Assume the hypotheses of Theorem 1. Suppose that 1. either η(s) = s for all s ∈ R, or else θ satisfies the coercivity condition θ (s)s ≥ C f,h,η s 2 for all s ∈ R; and 2. (geometric condition) there exists x0 ∈ Rn such that μ(x) · ν(x) ≤ 0 for x ∈ 0 , where μ(x) = x − x0 and ν(x) is the unit normal vector. Then for any solution u, z with initial data (u 0 , u 1 , z 0 , z 1 ) ∈ H, we have 1 E(t) := (u(t), u t (t), z(t), z t (t)) 2H ≤ R 2
t −1 for all t > T0 , T0
(13)
where T0 > 0 is some constant, R(t) is a solution to the ordinary differential equation d R(t) + q(R(t)) = 0, R(0) = E(0), dt
(14)
and q(s) is given by (9), (8) with the constant K in (8) depending in general on E(0) 1 and c = |1 ×[0,T (θ0 + θ1−1 ). 0 ]|
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2. Generation of a semigroup In this section, we show that the system (1)–(6) generates a strongly continuous semiflow under suitable conditions. To achieve this, we use the K¯omura-Kato theorem from nonlinear semigroup theory. First, we must define the operator which is to be the generator of the semiflow. Write (1)–(6) abstractly in the form ⎛
⎞ ⎛ 0 I u ⎟ ⎜ ∂ ⎜ − α 0 u ⎜ t⎟=⎜ 0 ∂t ⎝ z ⎠ ⎝ 0 1 zt 0 − m γ0 (·)
⎞⎛ ⎞ 0 0 u ⎜ ut ⎟ 0 0 ⎟ ⎟⎜ ⎟. 0 I ⎠⎝ z ⎠ zt − mg − mf
(15)
The action of the operator A is given by the matrix of operators that appears in (15). The remaining boundary conditions are encoded in the domain of A, which is given by
D(A) = (u, v, z, w) ∈ H : v ∈ H10 (), u ∈ L 2 (), γ1 u = hη(w)−θ (γ0 v) . (16) We now prove that A is dissipative up to translation by a constant. Let yi = (u i , vi , z i , wi ) ∈ D(A), i = 1, 2; y = (u, v, z, w) = y1 − y2 . Note that y is not assumed to be in D(A), which is nonlinear. We need to show that (Ay1 − Ay2 , y1 − y2 )H ≤ 0. By taking inner products in each component space of H and applying Green’s formula, we have (Ay1 − Ay2 , y)H = ((v, u))α + (u − αu, v) + w, zgh − γ0 v + f w + gz, wh = γ1 u, γ0 v − γ0 v + f w, wh = η(w1 ) − η(w2 ) − w, γ0 vh − θ (γ0 v1 ) − θ (γ0 v2 ), γ0 v − |w|2f h , where we have used the trace condition given in (16). Now there are three cases, corresponding to conditions (a), (b), and (c) in Theorem 1. For the first case, if η(s) = s for all s, then (Ay1 − Ay2 , y)H = −θ (γ0 v1 ) − θ (γ0 v2 ), γ0 v − |w|2f h ≤ 0. For the second case, we assume m ≥ m 0 and (θ (s) − θ (t))(s − t) ≥ (s − t)2 . Then by Cauchy–Schwartz and Young’s inequality, we have (Ay1 − Ay2 , y)H ≤ η˜ 1 |w|h |γ0 v|h − |γ0 v|2 ≤
|h|∞ η˜ 12 |w|2mh . 4 m 0
For the third case, we assume (θ (s) − θ (t))(s − t) ≥ C f,h,η (s − t)2 where C f,h,η > |h|∞ η˜ 12 4 f0 .
Then by Cauchy–Schwartz and Young’s inequality, we have (Ay1 − Ay2 , y)H ≤ η˜ 1 |w|h |γ0 v|h − C f,h,η |γ0 v|2 − f 0 |w|2h ≤ 0.
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In the first and third cases, A is dissipative, and in the second case A is dissipative up to translation by a constant. Now, we must show maximality. We must solve, for some λ > 0, λy − Ay = x, where x ∈ H is arbitrary. Writing y = (u, v, z, w) ∈ D(A) and x = (x1 , x2 , x3 , x4 ) we find that u satisfies the elliptic boundary value problem λ2 u − u + αu = λx1 + x2 in u = 0 on 0
(17)
∂ u = hη(w) − θ (γ0 v) on 1 ∂ν where v = λu − x1 ; w =
−λγ0 v + λx4 − gx3 . λ2 m + λ f + g
We define A : L 2 () → L 2 () by
∂ A = −u + αu with domain D(A) = u ∈ H 2 () ∩ H10 () : u|1 = 0 ∂ν and the corresponding “Neumann map” N : L 2 () → L 2 () by ( − α)N φ = 0 in , N φ|0 = 0,
∂ N φ|1 = φ. ∂ν
Then, we see that v solves the following equation, where each side is understood as a map from H10 into the dual space (H10 ) :
−λγ0 v + λx4 − gx3 1 1 λv + Av + AN θ (γ0 v) − hη = − Ax1 + x2 . (18) λ λ2 m + λ f + g λ
It is well known that N ∗ A∗ v = γ0 v for v ∈ H 1 (). Therefore, because η is Lipschitz ∗ ∗ ·+λx −gx 4 3 and monotone, we have that B := AN [θ (N ∗ A∗ ·) − η( −N λ2Am+λ )] is maximal f +g
monotone for λ large. Since A is bounded as a mapping H10 → (H10 ) , it follows from standard perturbation results that λ1 A + B is maximal monotone. Hence, there exists a solution v ∈ H10 () to (18), and u, z, w can be defined based on v. It is clear that y = (u, v, z, w) ∈ D(A) solves λy − Ay = x, as desired. We now invoke the K¯omura-Kato Theorem [21], and A generates a strongly continuous semigroup of contractions. 3. Equivalence of weak and semigroup solutions The result of the previous section is that the initial value problem yt = Ay; y(0) = y0 ∈ H has a unique continuous in time solution in the generalized sense; that is, we may write y(t) = limn→∞ yn (t), where yn is the solution to the initial value problem
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with yn (0) ∈ D(A) and the convergence is in the H topology. Such solutions yn are “strong” or “regular” solutions defined in Theorem 2. Our task now is to show that these generalized solutions correspond with weak solutions. To do this, it suffices to prove that every generalized solution is a weak solution, and then to show that weak solutions are unique. 3.1. A priori bounds Let y = (u, u t , z, z t ) be a strong solution to the initial value problem. Multiply (1) by u t and integrate by parts, then multiply (3) by z t and integrate over 1 , add the two equations, and integrate over [0, T ]; letting E(t) := 21 (u(t), u t (t), z(t), z t (t)) 2H be the energy of the system, we obtain h[z t − η(z t )]u t + f hz t2 + θ (u t )u t = E(0), (19) E(t) + 1 (t)
where 1 (t) = [0, t] × 1 . Note that T is arbitrary. We may also define a nonlinear energy functional 1 2 2 2 hm(z t (t)), E(t) := ( u(t) α + |u t (t)| + |z(t)|gh ) + 2 1 where (t) =
t
η(s) ds. Then, we have the energy relation E(t) + ghz[η(z t ) − z t ] + f hη(z t )z t + θ (u t )u t = E(0). 0
1 (t)
(20)
Setting η¯0 = min{1, η0 }, η¯1 = max{1, η1 }, we have η¯0 E(t) ≤ E(t) ≤ η¯1 E(t). In order to obtain a priori bounds on the energy, we consider three cases, as in the proof of generation of a semigroup. In the first case, where η(s) = s (linear acoustic boundary conditions) Eq. (19) gives E(T ) ≤ E(0). In the second case, where m ≥ m 0 , apply Cauchy–Schwartz and Young’s Inequality in (20) to obtain ω E(t) ≤ E(0) + 2 by picking ω = This implies E(t)
|g|∞ η˜ 12 ghz + ωη0 m 0 1 (t)
2
1
hm(z t ) ≤ E(0) + ω
t
E(s) ds
0
|g|∞ η˜ 12 ωt η0 m 0 . Applying Grownwall’s Lemma, we have E(t) ≤ E(0)e . ≤ ηη¯¯01 E(0)eωt . In the third case, where m is possibly degenerate |h| η˜ 2 that that θ (s)s ≥ C f,h,η s 2 for all s ∈ R, where C f,h,η ≥ 4∞f0 1 .
on 1 , we assume Applying Cauchy–Schwartz and Young’s Inequality in (19) we obtain E(T ) ≤ E(0). 3.2. Existence of weak solutions
First, we show that if y = (u, u t , z, z t ) is a strong solution to the initial value problem above, then it is a weak solution. The pair u, z is a pointwise solution to (1)–(6).
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Multiply (1) by φ ∈ H10 () and (3) by ψ ∈ L 2 (1 ) and integrate by parts (Green’s formula) to obtain (10) and (11). Most of the desired regularity properties are clear; it remains merely to show that u t , θ (u t ) ∈ L 2 (1 ) (where 1 = 1 (T ) = 1 × [0, T ]. In each of three cases (corresponding to the three cases above), we obtain estimates on either 1 θ (u t )u t or 1 u 2t ; we then use the concave function H given by (7) to obtain u t , θ (u t ) ∈ L 2 (1 ). In the first case when η(s) = s Eq. (19) implies 1 θ (u t )u t ≤ E(0). In the second case, assuming m ≥ m 0 , Eq. (20) implies
1
T
θ (u t )u t ≤ E(0) + ω
E(t)dt = E(0)eωT .
0
In the third case, assuming θ (s)s ≥ C f,h,η s 2 , (19) implies cases, then, 1 θ (u t )u t ≤ C T E(0). Now let
1
θ (u t )u t ≤ E(0). In all
A = {(x, t) ∈ 1 = 1 × (0, T ) : |u t (x, t)| ≥ N } B = 1 \ A . From Assumption 2, we gain the estimate
A
θ 2 (u t ) + u 2t ≤
A
(θ0 + θ1−1 )θ (u t )u t .
(21)
From (7), Assumption 2, and Jensen’s inequality we obtain
B
θ
2
(u t ) + u 2t
≤
B
H (θ (u t )u t ) ≤ |1 |H
1 |1 |
1
θ (u t )u t .
(22)
Combining inequalities (21) and (22) with the fact that 1 θ (u t )u t ≤ C T E(0) and H is strictly increasing, we obtain a constant C(E(0)) such that 1
θ 2 (u t ) + u 2t ≤ C(E(0)),
hence u t , θ (u t ) ∈ L 2 (1 ) as desired. Let y = (u, u t , z, z t ) be a “generalized” solution to the initial value problem above and let y n = (u n , u nt , z n , z tn ) be a sequence of strong solutions converging to y in H. To show that y is a weak solution, it suffices to show the following delicate point: that u t , θ (u t ) ∈ L 2 (1 ) and that we can pass to the limit on the term θ (γ0 u nt ), γ0 φ in Eq. (10). Passing to the limit on other terms is straightforward (note that η is Lipschitz) and we get the desired regularity. To obtain the desired result on u t , θ (u t ), observe that the above argument for strong solutions proves that u nt , θ (u nt ) are uniformly bounded in L 2 (1 ); hence (on a subsequence) we have weak convergence of these sequences in
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L 2 (1 ). Now, let E n (t) represent the energy of the solution y n . We have the following energy identity, which can be estimated as before: n m [θ (u nt ) − θ (u m E n (T ) − E m (T ) − (E n (0) − E m (0)) + t )][u t − u t ] 1 n m 2 h[η(z tn ) − η(z tm ) − (z tn − z tm )][u nt − u m = t ] − f h[z t − z t ] 1
|h|∞ η˜ 12 ≤ 4 f0
1
2 (u nt − u m t ) ,
Note that |E n (t) − E m (t)| → 0 as m, n → ∞. If we assume that η(s) = s is linear, then the estimate on the right-hand side is unnecessary and we obtain 1 [θ (u nt ) − n m θ (u m t )][u t − u t ] → 0 as m, n → ∞. Otherwise, we assume the coercivity condition on θ holds, namely (θ (s) − θ (t))(s − t) ≥ C f,h,η (s − t)2 for all s, t ∈ R, 2 then we have shown that 1 (u nt − u m t ) → 0 as m, n → ∞, which in turn implies n m n m 1 [θ (u t ) − θ (u t )][u t − u t ] → 0. Since θ is monotone, Lemma 1.3 on p. 42 in [4] implies that (on a subsequence) u n → u and θ (u n ) → θ (u) in L 2 (1 ), thus giving the desired result. 3.3. Uniqueness of weak solutions Let y 1 = (u 1 , u 1t , z 1 , z t1 ), y2 = (u 2 , u 2t , z 2 , z t2 ) be weak solutions with y 1 (0) = = y0 = (u 0 , u 1 , z 0 , z 1 ) ∈ H. Let (u, u t , z, z t ) = y = y 1 − y 2 . Then, we have the following variational formulation for y: for every φ ∈ H10 (), ψ ∈ L 2 (1 ) we have y 2 (0)
(u tt , φ) + ((u, φ))α = hη(z t1 ) − hη(z t2 ) − θ (γ0 u 1t ) + θ (γ0 u 2t ), γ0 φ,
(23)
z tt , ψhm + z t , ψ f h + z, ψgh = −γ0 u t , ψh ,
(24)
u(0) = 0, u t (0) = 0, z(0) = 0, z t (0) = 0.
(25)
The “correct” test function for (23) is u t , but as this is not a valid test function (its values are not in the space H10 ()) we take finite differences. Define, for > 0, D u(t) =
u(t + ) − u(t − ) . 2
where functions u(t) are extended to negative real axis by constancy. This is to say u(t) = u(0), t ≤ 0. Now, we substitute φ = D u, ψ = z t in Eqs. (23) and (24). After taking differences and rearranging we get 1 d (|z(t)|2gh + |z t (t)|2hm ) = η(z t1 (t)) 2 dt − η(z t2 (t)) − z t (t) − (θ (γ0 u 1t (t)) − θ (γ0 u 2t (t))), D (γ0 u)(t)h − |z t (t)|2f h . (26)
(u tt (t), D u(t)) + ((u(t), D u(t)))α +
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We pass to the limit → 0 in the following way. Let T > 0 be arbitrary. By [14], p. 208, Proposition 4.3, we have the following limits:
T
1 [|u t (T )|2 − |u t (0)|2 ];
→0 0 2 T 1 ((u(t), D u(t)))dt = [||u(T )||2 − ||u(0)||2 ]; lim
→0 0 2 lim
(u tt (t), D u(t))dt =
lim D u = u t in L 2 (0, T ; L 2 ());
→0
lim D γ0 u = γ0 u t in L 2 (0, T ; L 2 (1 ));
→0
since, by definition of weak solutions, u ∈ Cw ([0, T ]; H10 ()), u t ∈ Cw ([0, T ]; L 2 ()) ∩ L 2 ([0, T ] × 1 ), and u tt ∈ L ∞ ([0, T ]; H −1 ()). The latter two equalities imply, in particular,
T
lim
→0 0
= 0
η(z t1 (t)) − η(z t2 (t)) − z t − (θ (γ0 u 1t (t)) − θ (γ0 u 2t (t))), D γ0 uh
T
η(z t1 (t)) − η(z t2 (t)) − z t − (θ (γ0 u 1t (t)) − θ (γ0 u 2t (t))), γ0 u t h .
Therefore, we integrate (26) with respect to time t from 0 to T , and let → 0 to obtain the following energy identity: ˜ ) := 1 [|u t (T )|2 + u(T ) 2α + |z(T )|2gh + |z t (T )|2hm ] E(T 2 T ˜ = E(0) + η(z t1 (t)) − η(z t2 (t)) − z t (t), (γ0 u t )(t)h dt, − 0
(27)
0
T
(θ (γ0 u 1t (t)) − θ (γ0 u 2t (t))), (γ0 u t )(t)h + |z t (t)|2f h dt
(28)
˜ where in fact E(0) = 0. Now in the case where η(s) = s, we have η(z ˜ t ) = z t , in which case by the monotonicity of θ we may bound the right-hand side of (27) by ˜ E(t) = 0. If η is nonlinear, we use the coercivity of θ to obtain an estimate of the form ˜ ) ≤ E(0) ˜ E(T +C
T
˜ E(t)dt
0 C T = 0. Hence y = y 1 − y 2 = 0, ˜ ) ≤ E(0)e ˜ and by Gronwall’s inequality we have E(T and we have uniqueness.
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4. Proof of Theorem 3 The general method of proving uniform decay rates will be as follows. We define the following energy functional for solutions of the PDE system (1)–(6): 1
u(t) 2α + |u t (t)|2 + |z(t)|2gh + |z t (t)|2hm E(t) = E u,z (t) = 2 1 |∇u(x, t)|2 + α(x)(u(x, t))2 + u t (x, t)2 dx = 2 1 + g(x)h(x)(z(x, t))2 + h(x)m(x)(z t (x, t))2 d. 2 1 Recall that we have the following energy identity: E(T ) + θ (u t )u t + h[z t − η(z t )]u t + f hz t2 = E(0), 1
(29)
where 1 = 1 × [0, T ]. This was already calculated for strong solutions and holds for weak solutions by a density argument (cf. the proof of existence of weak solutions). We are assuming in this proof that either η(s) = s or else θ satisfies the coercivity |h| η˜ 2
condition θ (s)s ≥ C f,h,η s 2 for all s ∈ R, where C f,h,η > 4∞f0 1 . This implies the a priori bound E(T ) ≤ E(0) for all T > 0. Our goal is to find a decay rate of this energy functional. It suffices to obtain an estimate of the form E(T ) + p(E(T )) ≤ E(0), for some T > 0, which implies E(t) ≤ R
(30)
t − 1 E(0) for t > T, T
where R(t) is the solution to the differential equation d R(t) + q(R(t)) = 0, R(0) = E(0). dt See the proof of Theorem 2 in [16]. Here, p and q are defined in (8) and (9). The proof proceeds in three steps. The first step is to bound the energy using the damping plus lower order terms. The second step is to deal with the lower order terms and obtain a bound on the energy using only the dissipation terms θ 2 (u t ), u 2t , z t2 . Finally, we use the concave function H from Eq. (7) and provide appropriate constants in the definition of p to obtain an energy estimate of the form (30). Step 1: Proposition 3.1 from [16] (see also Lemma 7.2 in [17]) gives us T −β [ u(t) 2α + |u t (t)|2 ]dt ≤ C sup ( u(t) 2α + |u t (t)|2 ) β
T
+ 0
1
∂u ∂ν
2
t∈[0,T ]
+ u 2t ddt + C T |u|2L 2 [0,T ;H 1/2+ρ ()] ,
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where C is a constant that does not depend on time, β > 0 and 1/2 > ρ > 0 are small enough, arbitrary but fixed. (Here, we have used the fact that (αu, u) ≤ |α|∞ |u|2 ≤ |α|∞ |u|2H 1/2+ρ .) Using the fact that f, h are bounded below and above and taking into account (29), we write T −β [ u(t) 2α + |u t (t)|2 ]dt ≤ C1 E(T ) β
+
1
On the other hand,
θ
2
(u t ) + u 2t
β
+
T T −β
0
+
+ C T |u|2L 2 [0,T ;H 1/2+ρ ()] .
f hz t2
[ u(t) 2α + |u t (t)|2 ]dt ≤ 2β E(0),
so taking into account (29) we get a constant C2 such that T [ u(t) 2α + |u t (t)|2 ]dt ≤ C2 E(T ) 0
+
1
θ
2
(u t ) + u 2t
+
f hz t2
+ C T |u|2L 2 [0,T ;H 1/2+ρ ()] .
(31)
Now, multiply Eq. (3) by z, integrate over 1 × [0, T ] and integrate by parts in time. Using Young’s inequality as well as the bounds on f, g, h, m we get, for > 0, T ghz 2 = − hu t z + hmz tt z + h f z t z 1
0 T
=− ≤ ≤
0 T
1
1
1
0
1
hu t z
+ hmz t2
+ h f zt z −
T
ghz 2 + C( ) 0
ghz 2 + C( )
1
1
1
T hmz t z 0
u 2t + f hz t2 + C3 sup ghz 2 + hmz t2 t∈[0,T ]
u 2t + f hz t2 + C3 E(0)
Pick small enough and pick C3 large enough so that 2 ghz ≤ C3 [E(0) + u 2t + f hz t2 ].
0
T
1
1
ghz 2 + hmz t2 to Eq. (31) to obtain, for some constant C4 , 2 2 2 E(t) dt ≤ C4 E(T ) + θ (u t ) + u t + f hz t + C T |u|2L 2 [0,T ;H 1/2+ρ ()] .
We can thus add
T 0
1
1
Since for every > 0 we can pick C( ) such that T 2
u 2α + C( )|u|2 |u| L 2 [0,T ;H 1/2+ρ ()] ≤ 0
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we can thus pick a constant C5 such that T E(t) dt ≤ C5 E(T ) + θ 2 (u t ) + u 2t + f hz t2 + C T 1
0
Now, since T E(T ) ≤ enough so that
T 0
0
u2.
E(t) dt and C5 does not depend on T , we can fix T large
E(T ) ≤ C5
T
1
θ
2
(u t ) + u 2t
+
f hz t2
+ CT 0
T
u2.
(32)
T Step 2: To deal with the lower order term 0 u 2 , we will need a lemma, proved using a nonlinear version of a “compactness–uniqueness” type argument. LEMMA 1. Let T > 0 be large enough such that (32) holds for any pair of soluT (E u,z (0)) (depending on the initial tions to (1)–(6). Then there exists a constant C energy) such that if u, z are solutions to (1)–(6), then 0
T
T (E u,z (0)) u2 ≤ C
T 0
1
θ 2 (u t ) + u 2t + f hz t2 ddt.
Proof. Suppose to the contrary. Then there exists a sequence of solutions u l , zl such that T 2 0 ul =∞ (33) lim T 2 + f h(z )2 2 l→∞ θ ((u ) ) + (u ) l t l l t t 0 1 while the energy remains bounded, i.e., E ul ,zl (0) ≤ M uniformly in l for some constant M. We will denote El (t) = E ul ,zl (t), the energy for the solution u l , zl at time t; by the energy inequality (29) it follows that El (t) ≤ M for all l = 1, 2, 3, . . . and t ≥ 0. This implies that, on a subsequence, u l → u weakly in H 1 (Q), weakly∗ in L ∞ (0, T ; H 1 ()) (34) u l → u strongly in L 2 (1 ) ∩ L 2 (Q) T where 1 = 1 × [0, T ] and Q = × [0, T ]. Since 0 u l2 ≤ T El (0) ≤ T M, it T follows that 0 1 θ 2 ((u l )t ) + (u l )2t + f h(zl )2t → 0. Case 1: Assume u = 0. By (33), we have that θ ((u l )t ), (u l )t , (zl )t → 0 in L 2 (). Note that since η is Lipschitz, we also obtain η((zl )t ) → 0 in L 2 (). So inserting u l , zl into (1)–(6) and taking l → ∞, we obtain for u ⎧ ⎨ u tt − u + αu = 0 in ∂ (35) u = 0, u t = 0 on 1 ⎩ ∂ν u = 0 on 0 . By standard uniqueness results for the wave equation, we have u = 0, a contradiction.
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Case 2: Assume u = 0. Let Ml = |u l | L 2 (Q) , and let u˜l = M1l u l , z˜l = M1l zl . Then |u˜ l | L 2 () = 1, but on the other hand Ml → 0 because u = 0. Also, (33) becomes 1
lim T
l→∞
0
1 2 2 1 M 2 θ ((u l )t ) + (u˜ l )t l
+ f h(˜zl )2t
=∞
(36)
which proves that M1l θ ((u l )t ), (u˜ l )t , (˜zl )t → 0 in L 2 (), hence also M1l η((zl )t ) → 0 in L 2 (), because η is Lipschitz. Now combining (32) and (29), we find that for any solution u, z we have T T 1 u 2α ≤ E(t) ≤ E(0) ≤ (C5 + 1) θ 2 (u t ) + u 2t + f hz t2 + C T u2. 2 0 0 1 Apply this to u l , zl and divide by Ml2 to obtain 1 u˜ l 2α ≤ (C5 + 1) 2
T 0
1
1 2 θ ((u l )t ) + (u˜ l )2t + f h(˜zl )2t + C T . Ml2
Considering (36), this implies that u˜ l α is bounded, hence u˜ l → u˜ weakly in H 1 (Q), weakly∗ in L ∞ (0, T ; H 1 ()) u˜ l → u˜ strongly in L 2 (1 ) ∩ L 2 (Q)
(37)
Putting u l , zl in (1)–(6), dividing by Ml , and taking l → ∞, we see that u˜ is a solution to (35), and therefore we again obtain u˜ = 0. But |u˜l | L 2 (Q) = 1 for all l implies |u| ˜ L 2 (Q) = 1, a contradiction. Applying Lemma 1, Eq. (32) becomes T E(T ) ≤ C T θ 2 (u t ) + u 2t + f hz t2 . 0
1
(38)
for some constant C T . We are now in a position to obtain an estimate of the form (30). Before we proceed, we first note that there exists a constant D f,h,η > 0 such that 2 θ (u t )u t + f hz t ≤ D f,h,η θ (u t )u t + h[z t − η(z t )]u t + f hz t2 . (39) 1
1
If η(s) = s, then this is obvious (take D f,h,η = 1). Otherwise use the fact that θ (u t )u t ≥ C f,h,η u 2t , where C f,h,η > |h|∞ η˜ 12 4(1−δ) f 0
|h|∞ η˜ 12 4 f0 .
Fix δ > 0 small enough that C f,h,η >
=: C f,h,η,δ , then use Young’s inequality to see that
θ (u t )u t + h[z t − η(z t )]u t + f hz t2 ≥ θ (u t )u t − C f,h,η,δ u 2t + δ f hz t2 ≥ thus proving the desired inequality.
C f,h,η − C f,h,η,δ θ (u t )u t + δ f hz t2 , C f,h,η
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Step 3: We are now ready to finish the proof of Theorem 3. Let A = {(x, t) ∈ 1 = 1 × (0, T ) : |u t (x, t)| ≥ N } B = 1 \ A . From Assumption 2, we gain the estimate 2 2 θ (u t ) + u t ≤ (θ0 + θ1−1 )θ (u t )u t .
(40)
From (7) and (H-5)(b), we obtain θ 2 (u t ) + u 2t ≤
(41)
A
A
B
B
H (θ (u t )u t ).
By Jensen’s inequality, the fact that H is strictly increasing, and Eq. (39), we have 1 H (θ (u t )u t ) ≤ |1 |H θ (u t )u t |1 | 1 2 θ (u t )u t + h[z t − η(z t )]u t + f hz t , ≤ |1 |H D
B
(42)
1
D
where D is now taken to be D = |f,h,η , where |1 | denotes the measure of 1 and 1| D f,h,η is defined above. Combining inequalities (40), (41), and (42) together with (38) and (39), and again using that H is increasing, we obtain some constant M depending only on θ0 , θ1 such that E(T ) ≤ C T M
T
1
0
+ C T |1 |H D
θ (u t )u t + h[z t − η(z t )]u t + f hz t2
T 0
1
θ (u t )u t + h[z t − η(z t )]u t +
f hz t2
.
Let the constants K and c in p(x) = p(x, c, K ) (Eq. 8) be given by K =
1 M and c = . C T |1 | |1 |
Then, we have p(E(T )) ≤ 0
T
1
θ (u t )u t + f hz t2 = E(0) − E(T ),
as desired. By the proof of Theorem 2 in [16] (see Lemma 3.3 and what follows), Theorem 3 is established.
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5. The case m = 0 5.1. Well-posedness and exponential decay for m = 0 Consider the modified system u tt (x, t) − u(x, t) + α(x)u(x, t) = 0,
x ∈ , t > 0;
(43)
u(x, t) = 0,
x ∈ 0 , t > 0;
(44)
u t (x, t) + f (x)z t (x, t) + g(x)z(x, t) = 0,
x ∈ 1 , t > 0;
(45)
x ∈ 1 , t > 0;
(46)
x ∈ ;
(47)
x ∈ 1 ;
(48)
∂u (x, t) + θ (u t (x, t)) − h(x)η(z t (x, t)) = 0, ∂ν u(x, 0) = u 0 (x), u t (x, 0) = u 1 (x), z(x, 0) = z 0 (x),
The model given here was studied in [12] in the case without boundary damping, i.e., θ = 0, where it was discovered that the nonlinear semigroup need not be stable or even contractive, due to the nonlinearity of the η(z t ) term. Our goal is to show that with sufficient boundary damping, we obtain uniform decay rates for solutions to this model, under less restrictive assumptions than in Theorem 1. For completeness, we prove that the system (43)–(48) is well posed, then show uniform decay of the solutions. Let us first define a modified state space: H0 = H10 () × L 2 () × L 2 (1 ) with energy (norm) given by (u, v, z) 2H0 := u 2α + |v|2 + |z|2gh . We define weak solutions to the problem (43)–(48) in the usual way: DEFINITION 2. A pair of functions u : × [0, ∞) → R, z : 1 × [0, ∞) → R is called a weak variational solution to the system (43)–(48) provided that, given T > 0 arbitrary, • u ∈ Cw ([0, T ]; H10 ()), u t ∈ Cw ([0, T ]; L 2 ()), u tt ∈ L ∞ ([0, T ]; H −1 ()); • z ∈ Cw ([0, T ]; L 2 (1 )), h 1/2 γ (u t ) ∈ L 2 ([0, T ] × 1 ), and • for every φ ∈ H10 (), ψ ∈ L 2 (1 ) we have 1 [−u t − gz] − θ (u t ), φ , (u tt , φ) + ((u, φ)) + (αu, φ) = hη (49) f f z t , ψ + gz, ψ = −u t , ψ, (50) u(0) = u 0 , u t (0) = u 1 , z(0) = z 0
(51)
THEOREM 4. Suppose that Assumptions 1 and 2 hold. Then for each set of initial conditions (u 0 , u 1 , z 0 ) ∈ H, there exists a unique pair of weak solutions u, z to the system (43)–(48) such that (u(t), u t (t), z(t)) is continuous in time and depends continuously on (u 0 , u 1 , z 0 ).
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REMARK 3. Note that we need no strong monotonicity assumption on θ in Theorem 4 (cf. Theorem 1). THEOREM 5. (Decay rates) Assume the hypotheses of Theorem 4. Suppose that 1. θ satisfies the coercivity condition θ (s)s ≥ C f,h,η s 2 for all s ∈ R; and 2. (geometric condition) there exists x0 ∈ Rn such that μ(x) · ν(x) ≤ 0 for x ∈ 0 , where μ(x) = x − x0 and ν(x) is the unit normal vector. Then for any solution u, z with initial data (u 0 , u 1 , z 0 , z 1 ) ∈ H, we have 1 E(t) := (u(t), u t (t), z(t)) 2H ≤ R 2
t −1 for all t > T0 , T0
(52)
where T0 > 0 is some constant, R(t) is a solution to the ordinary differential equation d R(t) + q(R(t)) = 0, R(0) = E(0), dt
(53)
and q(s) is given by (9), (8) with the constant K in (8) depending in general on E(0) 1 and c = |1 ×[0,T (θ0 + θ1−1 ). 0 ]| The proof of Theorem 4 again uses nonlinear semigroup theory. Although the system treated here is very similar to that treated in [12], we give a new proof of well-posedness here both for the sake of completeness and in order to present a somewhat better proof which eliminates some cumbersome calculations. We define an operator A0 (the generator of a nonlinear semigroup) as follows: ⎞ 0 I 0 ⎟ ⎜ 0 ⎠, A0 = ⎝ − α 0 0 − 1f γ0 − gf ⎛
D(A0 ) = (u, v, z, w) ∈ H : v ∈ V (), u ∈ L 2 (),
h γ1 u = η(−γ0 v − gz) − θ (γ0 v) . f
(54)
(55)
We now prove that A0 is dissipative up to translation by a constant. Let yi = (u i , vi , z i ) ∈ D(A0 ), i = 1, 2; y = (u, v, z) = y1 − y2 . Note that y is not assumed to be in D(A), which is nonlinear. We need to show that (A0 y1 − A0 y2 , y1 − y2 )H ≤ 0. We also let wi = − 1f γ0 vi − gf z i , i = 1, 2 and let w = w1 − w2 . By taking inner products in each component space of H0 and applying Green’s formula, the trace condition given in (55), and the bounds on η, we have
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(A0 y1 − A0 y2 , y)H0 = ((v, u))α + (u − αu, v) + w, zgh = γ1 u, γ0 v + w, zgh = η(w1 ) − η(w2 ), γ0 vh − θ (γ0 v1 ) − θ (γ0 v2 ), γ0 v + w, zgh ≤ −η(w1 ) − η(w2 ), f wh − η(w1 ) − η(w2 ), gzh + w, zgh ≤ −η0 f 0 |w|2h + η˜ 1 |w|gh |z| f h ≤
|g|∞ η˜ 12 2 |z| , 4η0 f 0 gh
which is what we needed to show. Now, we need to prove maximality, i.e., we need to show that for x ∈ H0 the equation (λ − A0 )y = x has a solution y ∈ D(A0 ) for some λ > 0. Setting y = (u, v, z), x = (x1 , x2 , x3 ) we see that v solves the equation 1 1 f N ∗ A∗ v + gx3 /λ ∗ ∗ = − Ax1 + x2 , λv + Av + AN θ (N A v) − hη − 2 λ f + g f /λ λ (56) where A and N are defined as before: A : L 2 () → L 2 () is given by
∂ 2 1 u|1 = 0 A = −u + αu with domain D(A) = u ∈ H () ∩ H0 () : ∂ν and the corresponding “Neumann map” N : L 2 () → L 2 () by ( − α)N φ = 0 in , N φ|0 = 0,
∂ N φ|1 = φ. ∂ν
Again, N ∗ A∗ v = γ0 v for v ∈ H 1 (). Now because η is Lipschitz and monotone, ∗ ∗ 3 /λ we have that B := AN [θ (N ∗ A∗ ·) − hη(− f N f 2A+g·+gx )] is maximal monotone for f /λ
λ large. Since A is bounded as a mapping H10 → (H10 ) , it follows from standard perturbation results that λ1 A + B is maximal monotone. Hence, there exists a solution v ∈ H10 () to (56), and u, z can be defined based on v. It is clear that y = (u, v, z) ∈ D(A0 ) solves λy − A0 y = x, as desired. Let E(t) = E u,z (t) = 21 u(t) 2α + |u(t)|2 + |z(t)|2gh be the energy functional associated with solutions to (43)–(48). To see that every strong solution is also a weak solution, multiply (43) and (45) by appropriate test functions and integrate by parts to get the appropriate variational form; the following energy relation implies that the solution is also a weak solution: t gh[η(z t ) − z t ]z + f hη(z t )z t + θ (u t )u t (57) E(0) − E(t) = 0
=
t 0
1
1
h[z t − η(z t )]u t + f hz t2 + θ (u t )u t .
(58)
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Note that (57) implies the following a priori bound for all weak solutions: |g|∞ η˜ 12 t |g|∞ η˜ 12 t |z(s)|2gh ds ≤ E(s) ds, E(t) ≤ E(0) + 4 f 0 η0 0 2 f 0 η0 0
(59)
|g| η˜ 2
hence E(t) ≤ E(0)eωt , ω = 2 f∞0 η01 by Gronwall. It is clear that a “generalized” semigroup solution obtained as a limit of strong solutions is also a weak solution. To see that weak solutions are unique, note that (59) also applies to the difference of solutions; hence solutions with the same initial condition must be equal for all time. The proof of Theorem 4 is complete. The proof of Theorem 5 follows closely the proof of Theorem 3. Proposition 3.1 from [16] and the energy relation (58) combined with Lemma 1 from [13] give the estimate T 2 2 2 θ (u t ) + u t + f hz t + C T u2. (60) E(T ) ≤ C1 1
0
Then, Lemma 1 can be proved for the case m = 0 in the exact same way as before, thus giving the estimate T θ 2 (u t ) + u 2t + f hz t2 . (61) E(T ) ≤ C T 0
1
Then using the coercivity estimate θ (u t )u t ≥ C f,h,η u 2t , one can finish as in Step 3 of the proof of Theorem 3. Theorem 5 follows. 5.2. Continuous dependence on m In this subsection, we prove that in the linear case, the case m = 0 behaves much the same as the case where m is sufficiently small. Continuous dependence of solutions on the parameters has been considered before by Beale [5]. However, he did not consider the case m → 0, which is what we are mainly concerned with here. In the following, assume that θ = 0 and η(s) = s, so that we are no longer dealing with nonlinear but linear systems in (1)–(6) and (43)–(48). Let Sm (t) be the semigroup given by Theorem 1, that is the semiflow generated by the system (1)–(6), with corresponding state space Hm and generator Am . Note that for each m, Hm = H10 () × L 2 () × L 2 (1 ) × L 2 (1 ), but with a different norm imposed on the fourth coordinate depending on m. We recall that the norm on Hm is given by 2 2 2 2 (u, v, z, w) Hm = |∇u| + αu + v + ghz 2 + hmw 2 .
1
Let S0 (t) be the semigroup given by Theorem 4, that is the semiflow generated by (43)–(48), with state space given by H0 and generator A0 . Define a subspace V0 of H0 to be the set of all (u, v, z) ∈ H0 such that w = − 1f γ0 v − gf z ∈ L 2 (1 ), and let Q(u, v, z) = (u, v, z, w) for such vectors in V0 . Let P be the projection operator from Hm down to H0 given by P(u, v, z, w) = (u, v, z). We show the following:
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THEOREM 6. Assume the hypotheses of Theorem 1 which guarantee generation of a semigroup Sm (t); additionally, assume that there exists g0 > 0 such that g ≥ g0 . Let m n be a sequence of functions in L ∞ (), with corresponding semigroups Sm n (t), such that |m n |∞ → 0. Then for each y0 ∈ V0 we have P Sm n (t)Qy0 − S0 (t)y0 H0 → 0 uniformly on bounded intervals as n → ∞. REMARK 4. Given y ∈ V0 , it follows that Qy ∈ Hm for any m. Theorem 6 should be interpreted as follows. Initial data from H0 satisfying an appropriate compatibility condition can be fed into the initial value problem (1)–(6), giving a solution which approximates the solution to (43)–(48) as m becomes small. Proof. Let y0 = (u 0 , v0 , z 0 ) ∈ D(A20 ), and let (u(t), v(t), z(t)) = S0 (t)(u 0 , v0 , z 0 ). Let y(t) = (u(t), v(t), z(t), z (t)), where z (t) = − 1f γ0 v(t) − gf z(t) is the time derivative of z(t). Observe that ⎞ ⎞⎛ 0 I 0 0 u(t) ⎟ ⎜ ⎜ − α 0 0 0 ⎟ ⎟ ⎜ v(t) ⎟ Am y(t) = ⎜ ⎠ ⎝ ⎝ 0 0 0 I z(t) ⎠ g f 1 z (t) 0 − m γ0 (·) − m − m ⎛ ⎞ v(t) ⎜ ⎟ ( − α)u(t) ⎟ =⎜ ⎝ ⎠ z (t) g f 1 − m γ0 v(t) − m z(t) − m z (t) ⎞ ⎛ v(t) ⎜ ( − α)u(t) ⎟ ⎟ =⎜ ⎠ ⎝ z (t) ⎛
0
It follows that y (t) = Am y(t) + (0, 0, 0, z (t)). By the “variation of parameters” formula, t Sm (t − s)(0, 0, 0, z (s))ds. y(t) = Sm (t)y(0) + 0
Hence,
y(t) −
Sm (t)y(0) 2Hm
t
|z (s)|2hm ds |m|∞ t ≤ z (s)|2gh ds g0 0
≤
0
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|m|∞ t ≤ A20 S0 (s)y0 2H0 ds g0 0 |m|∞ ≤ t A20 y0 2H0 . g0 Since S0 (t)y0 − P Sm (t)y(0) H0 ≤ y(t) − Sm (t)y(0) Hm , the result follows for all y0 ∈ D(A20 ). Since S0 (t), Sm (t) are contraction semigroups and D(A20 ) is dense in H0 , the result holds for all y0 ∈ V0 by a standard argument. REFERENCES [1] [2] [3] [4] [5] [6] [7] [8]
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R. E. Showalter, Monotone operators in Banach space and nonlinear partial differential equations, vol. 49 of Mathematical Surveys and Monographs, AMS, 1997. R. Temam, Infinite Dimensional Dynamical Systems in Mechanics and Physics, Springer, BerlinHeidelberg-New York, 1988. P. Jameson Graber Department of Mathematics University of Virginia Charlottesville VA 22904 USA E-mail:
[email protected]