Appl Math Optim (2012) 66:81–122 DOI 10.1007/s00245-012-9165-1

Existence and Asymptotic Behavior of the Wave Equation with Dynamic Boundary Conditions Philip Jameson Graber · Belkacem Said-Houari

Published online: 7 March 2012 © Springer Science+Business Media, LLC 2012

Abstract The goal of this work is to study a model of the strongly damped wave equation with dynamic boundary conditions and nonlinear boundary/interior sources and nonlinear boundary/interior damping. First, applying the nonlinear semigroup theory, we show the existence and uniqueness of local in time solutions. In addition, we show that in the strongly damped case solutions gain additional regularity for positive times t > 0. Second, we show that under some restrictions on the initial data and if the interior source dominates the interior damping term and if the boundary source dominates the boundary damping, then the solution grows as an exponential function. Moreover, in the absence of the strong damping term, we prove that the solution ceases to exists and blows up in finite time. Keywords Wave equation · Dynamic boundary condition · Source · Damping · Blow up · Finite time · Exponential growth

1 Introduction We consider the following semilinear damped wave equation with dynamic boundary conditions:

Communicating Editor: Irena Lasiecka. P.J. Graber Department of Mathematics, University of Virginia, Charlottesville, VA 22904, USA e-mail: [email protected] B. Said-Houari () Division of Mathematical and Computer Sciences and Engineering, King Abdullah University of Science and Technology (KAUST), Thuwal, Kingdom of Saudi Arabia e-mail: [email protected]

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⎧ utt − u − αut + φ(ut ) = f1 (u), ⎪ ⎪ ⎪ ⎨ u(x, t) = 0,   α∂ut ⎪ utt (x, t) = − ∂u ⎪ ∂ν (x, t) + ∂ν (x, t) + ρ(ut ) + f2 (u), ⎪ ⎩ u(x, 0) = u0 (x), ut (x, 0) = u1 (x),

x ∈ , t > 0, x ∈ 0 , t > 0, x ∈ 1 , t > 0,

(1.1)

x ∈ ,

where u = u(x, t), t ≥ 0, x ∈ ,  denotes the Laplacian operator with respect to the x variable,  is a regular and bounded domain of RN , (N ≥ 1), ∂ = 0 ∪ 1 , mes(0 ) > 0, 0 ∩ 1 = ∅ and ∂/∂ν denotes the unit outer normal derivative. Hybrid systems with dynamical boundary control have attracted interest, with motivation being practical applications related to stabilization and active control of large elastic structures. Nowadays the wave equation with dynamic boundary conditions is used in a wide field of applications. See [30] for some applications. Problems similar to (1.1) arise (for example) in modeling of longitudinal vibrations in a homogeneous bar in which there are viscous effects. The term ut , indicates that the stress is proportional not only to the strain, but also to the strain rate. See [10] for more details. From the mathematical point of view, these problems do not neglect acceleration terms on the boundary. Such type of boundary conditions are usually called dynamic boundary conditions. They are not only important from the theoretical point of view but also arise in several physical applications. For instance in one space dimension, problem (1.1) can be seen as a model of the dynamic evolution of a viscoelastic rod that is fixed at one end and has a tip mass attached to its free end. The dynamic boundary conditions represents the Newton’s law for the attached mass (see [3, 9, 14] for more details). In the two dimension space, as showed in [23] and in the references therein, these boundary conditions arise when we consider the transverse motion of a flexible membrane  whose boundary may be affected by the vibrations only in a region. Also some dynamic boundary conditions as in problem (1.1) appear when we assume that  is an exterior domain of R3 in which homogeneous fluid is at rest except for sound waves. Each point of the boundary is subjected to small normal displacements into the obstacle (see [5] for more details). This type of dynamic boundary conditions are known as acoustic boundary conditions. For a comparison between the dynamic boundary conditions studied here and acoustic boundary conditions, see [16]. In 1988 a pioneering contribution was made by Littman & Markus [28] where they considered a system which describe an elastic beam, linked at its free end to a rigid body. The whole system is governed by the Euler-Bernoulli PDE with dynamic boundary conditions. They used the classical semigroup methods to establish existence and uniqueness results while the asymptotic stabilization of the structure is achieved by the use of feedback boundary damping. This work led to some other contributions on linear and nonlinear hybrid systems for the Euler-Bernoulli beam equation and the wave. See [2, 18, 19, 31, 38]. In [37], the author considered a hybrid system consisting of a cable linked at its end to a rigid body. The whole dynamic is described by the system ⎧ 0 < x < 1, t > 0, ⎪ ⎨ utt − (a(x)ux )x = 0, (aux )(0, t) = y(t), t > 0, (1.2) ⎪ ⎩ Mutt (1, t) + (aux )(1, t) = 0, t > 0,

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where a(x) is the tension force of the cable and M represents the load. By choosing a boundary feedback of the form  (1.3) −y(t) + αu(0, t) = −f ut (0, t) , where f (s)s > 0, the author showed that system (1.2)–(1.3) is asymptotically stable. Uniform decay estimates of energy are also established. In [18] the author introduced the model utt − uxx − utxx = 0,

x ∈ (0, L), t > 0,

(1.4)

which describes the damped longitudinal vibrations of a homogeneous flexible horizontal rod of length L when the end x = 0 is rigidly fixed while the other end x = L is free to move with an attached load. Thus she considered Dirichlet boundary condition at x = 0 and dynamic boundary conditions at x = L, namely utt (L, t) = −[ux + utx ](L, t),

t > 0.

(1.5)

By rewriting the whole system within the framework of the abstract theories of the so-called B-evolution theory, an existence of a unique solution in the strong sense has been shown. An exponential decay result was also proved in [19] for a problem related to (1.4)–(1.5), which describe the weakly damped vibrations of an extensible beam. See [19] for more details. Subsequently, Zang & Hu [43], considered the problem utt − p(ux )xt − q(ux )x = 0,

x ∈ (0, 1), t > 0,

with u(0, t) = 0,

p(ux )t + q(ux )(1, t) + kutt (1, t) = 0,

t ≥ 0.

By using the Nakao inequality, and under appropriate conditions on p and q, they established both exponential and polynomial decay rates for the energy depending on the form of the terms p and q. The interested reader is also refereed to the works of Pellicer & Solà-Morales [34–36] where an alternative model for the classical springmass damper system has been studied and some dynamic boundary conditions were used. Higher dimensional hybrid structures have also become an interesting subject of research. In [41] You & Lee studied the stabilization of two-dimensional membrane vibration on a rectangular bounded region with mass/force dynamics along one edge. By reducing the mathematical model to an abstract evolutionary equation and using spectral eigenspace analysis they showed that the evolutionary system is approximately controllable and strongly stabilizable by linear boundary damping feedback control. Recently, Gerbi & Said-Houari studied in [21] and [22] problem (1.1) with f2 = φ = 0, f1 (u) = |u|p−2 u and a nonlinear boundary damping term of the form ρ(ut ) = |ut |m−2 ut . A local existence result was obtained by combining the Faedo–Galerkin method with the contraction mapping theorem. Concerning the asymptotic behavior, the authors showed that the solution of such problem is unbounded and grows up

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exponentially when time goes to infinity if the initial data are large enough and the damping term is nonlinear. The blow up result was shown when the damping is linear (i.e. m = 2). Also, they proved in [22] that under some restrictions on the exponents m and p, we can always find initial data for which the solution is global in time and decay exponentially to zero even if m > 2. Our main goal in this paper is to extend the results in [21] to our system (1.1). The main difficulty in the analysis of the model under consideration is the presence of the double interaction between source and damping terms, both in the interior of the domain  and on the boundary 1 . First, in Sect. 3 and using the semigroup theory, we are able to prove the existence and uniqueness of local in time solutions with values defined in the finite energy space H10 () × L2 () × L2 (1 ). To analyze the dynamics in this model, both in the interior and on the boundary, the system is treated as a coupled system of partial differential equations, with the variables u, ut satisfying the wave equation coupled with the variable w = ut |1 on the boundary. See [24] and [25] for similar semigroup formulations. The behavior of solutions greatly depends on the strong damping term αut . In case α = 0 the system (1.1) is “hyperbolic,” while if α > 0 the system is “parabolic.” This means that in the latter case, for times t > 0 we can expect solutions to obtain greater regularity than if the strong damping were absent. Well-posedness of solutions is thus divided into two distinct parts. For the hyperbolic case, we are left only with the standard tools of nonlinear semigroup theory, where the sources f1 (u), f2 (u) are viewed as a locally Lipschitz perturbation of a maximal monotone operator (see Sect. 3.1). For the parabolic case, we instead use the robust perturbation theory for analytic (holomorphic) semigroups, as one would expect for a parabolic system. See Sect. 3.2 for details. Second, in Sect. 4 we prove a crucial energy identity for finite energy solutions. In general, energy inequality holds for finite energy solutions by standard weak lowersemicontinuity arguments. However, for the purposes of studying long-time behavior and in particular blow-up of generalized finite energy solutions, one needs identity as opposed to inequality. Third, in Sects. 5–7 we study long-time behavior of solutions, with a particular focus on growth of the energy. In all of these sections we restrict our attention to those solutions satisfying the energy identity as discussed in Sect. 4. In Sect. 5, we show that under some restrictions on the initial data and if: (i) the interior source dominates the interior damping term (i.e. r < p) and (ii) the boundary source dominates the boundary damping (i.e. q ≤ k), then the norm





u(t) + u(t)

(1.6) p k, 1

of any solution satisfying the energy identity grows as an exponential function. The rate of growth obtained is shown to decrease as the strong damping is increased, i.e. as α gets larger. Our result holds without assuming any polynomial structure on the damping terms. To prove our result, we adapt (very carefully) the method in [21] with the necessary modification imposed by the nature of our problem. Then in Sect. 6, in the absence of the strong damping term, that is for α = 0, and for the same class of initial data as in Sect. 5, we show that the solution ceases to exists and the norm (1.6) tends to infinity when t approaches a finite value T ∗ .

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Finally, in Sect. 7, we show that the absence of the interior source term (i.e. f1 = 0), leads to an exponential growth of the norm u(t)k,1 even in the case α > 0. In Sect. 8, and using the method in [20], we show that if the interior damping term dominates the interior source and if the boundary damping dominates the boundary source, then the solution is global “in time”. Last of all, in Sect. 9 we list some open problems for future consideration.

2 Preliminaries In this section, we present some material that we shall use in order to present our results. We denote  H10 () = u ∈ H 1 ()/ u|0 = 0 .

By (., .) we denote the scalar product in L2 () i.e. (u, v)(t) =  u(x, t)v(x, t)dx. Also we mean by  · q the Lq () norm for 1 ≤ q ≤ ∞, and by  · q,1 the Lq (1 ) norm. We assume that 0 has non-empty interior so that the Poincaré inequality holds. Thus (∇·, ∇·) will be taken as the scalar product on H10 (). By ., . we denote the scalar product on L2 (1 ). We will also use the embedding (see [1, Theorem 5.8]):  2(N −1) 1 k N −2 , if N ≥ 3, 2 ≤ k ≤ k where k = (2.1) H0 () → L (1 ), +∞, if N = 1, 2, and also  H10 () → Lp (),

2≤p≤p

where p =

2N N −2 ,

+∞,

if N ≥ 3, if N = 1, 2.

In general we will assume that p ≤ p and k ≤ k so that H10 () → Lk (1 ) and H10 () → Lp (). We define the state space for finite energy solutions: H := H10 () × L2 () × L2 (1 ) with scalar product defined component-wise. In what follows we shall also need the notation defined here: • • • •

1

γ0 : H 1 () → H 2 () is the trace map of order zero, i.e. γ0 (u) = u| ; 1 ∂ u| ; γ1 : H 1 () → H − 2 () is the trace map of order one, i.e. γ1 (u) = ∂ν  · ∞ denotes the supremum norm. Cr ([0, T ); X) denotes the space of right-continuous functions with values in a Banach space X.

We will also assume that the functions φ and ρ are continuous, monotone, increasing functions on R equal to zero at zero.

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3 Local Well-Posedness In this section we prove a local well-posedness result for system (1.1). The existence of the strong damping term αut has a profound impact on the nature of the dynamics. If α = 0, then the system is hyperbolic, while if α > 0, the system is parabolic. It is well-known that solutions to parabolic systems experience significant gains in regularity. We will therefore need to distinguish between the cases α = 0 and α > 0 when developing the local well-posedness theory. 3.1 Hyperbolic Case α = 0 In the case α = 0 there is no strong damping and the system (1.1) is hyperbolic. We thus apply the techniques for proving existence and uniqueness of local solutions for the wave equation. Our goal is to prove the following: Theorem 3.1 Assume that f1 : H10 () → L2 () and f2 : H10 () → L2 (1 ) are locally Lipschitz. Let U (0) = (u(0), ut (0), ut |1 (0)) ∈ H . Then there exists a maximum time Tm > 0 depending on U (0)H such that system (1.1) has a unique solution U (·) ∈ C([0, Tm ); H ), where U (t) = (u(t), ut (t), ut |1 (t)). In addition, suppose U (0) satisfies the following initial regularity:  u(0) − φ ut (0) ∈ L2 (), ut (0) ∈ H10 (), (3.1)  ∂u(0)/∂ν + ρ ut |1 (0) ∈ L2 (1 ). Then the solution U (·) satisfies    u(·) − φ ut (·) ∈ Cr [0, Tm ); L2 () , ut (·) ∈ Cr [0, Tm ); H 1 () ,   ∂u(·)/∂ν + ρ ut |1 (·) ∈ Cr [0, Tm ); L2 (1 ) . Remark 3.2 Suppose that the functions f1 and f2 have polynomial structure given by f1 (s) = |s|p−2 s and f2 (s) = |s|k−2 s. Then f1 and f2 satisfy the assumption in ¯ ¯ Theorem 3.1 if 2 ≤ p ≤ 1 + p2¯ and 2 ≤ k ≤ 1 + k2 . The exponents p2¯ and k2 are the critical exponents, and sources possessing this level of polynomial growth are known as subcritical (if p, k are below the critical level) or critical (if p, k are at the critical level). For example, if the dimension N = 3, then the allowable range is limited to 2 ≤ p ≤ 4 and 2 ≤ k ≤ 3. If p, k are above the critical level, then the sources are super-critical. In this case, local well-posedness is called into question in the hyperbolic case. A suitable remedy would be to impose growth conditions on the damping functions φ and ρ (see [7]), but we will not address this here. Remark 3.3 For general finite energy solutions, ut does not have a well-defined trace since at each time t it has only L2 () regularity. Thus we must say that the third component of the solution ut |1 is not technically a well-defined trace unless U (t) is a strong solution satisfying (3.1). The pair (ut , ut |1 ) may be considered as a strong limit in L2 () × L2 (1 ) of function pairs in H 1 () × L2 (1 ) where the second component is the trace of the first. See [15] for a related description of the state space for a heat equation with Wentzell boundary conditions.

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The proof of Theorem 3.1 will proceed in two steps. First, assume f1 and f2 are globally Lipschitz sources with Lipschitz constants L1 and L2 , respectively, and show that we have existence of a semigroup, hence global in time solutions. Second, assume f1 and f2 are locally Lipshitz, take a sequence of truncations to globally Lipschitz sources to get a sequence of global in time solutions, and show that on some finite time interval these solutions converge to a local in time solution of (1.1). Let us now proceed with the proof of Theorem 3.1. Step 1. We assume f1 and f2 are globally Lipschitz sources with Lipschitz constants L1 and L2 , respectively. The first step is to show the existence of a semigroup. Define the operator A : D(A ) ⊂ H → H by ⎛ ⎞ ⎛ ⎞ u v ⎜ ⎟ ⎜ ⎟ A ⎝ v ⎠ = ⎝ u − φ(v) + f1 (u) ⎠ w −γ1 u − ρ(w) + f2 (γ0 u) with domain  u  D(A ) =

v w

∈ H : v ∈ H10 (), u − φ(v) ∈ L2 (), γ0 v = w, 

γ1 u + ρ(w) ∈ L (1 ) . 2

We want to prove the following: Lemma 3.4 The operator A generates a strongly continuous semigroup S(t). It follows that (i) for each (u0 , u1 , u1 |1 ) ∈ D(A ) the function (u(t), ut (t), ut (t)|1 ) = S(t)(u0 , u1 , u1 |1 ) is a strong solution to the system (1.1); and (ii) for each (u0 , u1 , w1 ) ∈ H , the function S(t)(u0 , u1 , w1 ) is a solution to (1.1) in the generalized sense, i.e. as a strong limit of strong solutions: take (un0 , un1 , un1 |1 ) = (un0 , un1 , w1 ) ∈ D(A ) converging to (u0 , u1 , w1 ) in H , then S(t)(un0 , un1 , un1 |1 ) → S(t)(u0 , u1 , w1 ) uniformly on bounded intervals. Proof The operator A is densely defined. By nonlinear semigroup theory, in order to prove Lemma 3.4 it suffices to show that A is maximally ω-dissipative. We use the following calculations to see that A is ω-dissipative. Taking y1 = (u1 , v1 , w1 ) ∈ D(A ), y2 = (u2 , v2 , w2 ) ∈ D(A ), and y = (u, v, w) = y1 − y2 , we see that (A y1 − A y2 , y1 − y2 )H      = (∇v, ∇u) + u − φ(v1 ) − φ(v2 ) + f1 (u1 ) − f1 (u2 ) , v       + −γ1 u − ρ(w1 ) − ρ(w2 ) + f2 (γ0 u1 ) − f2 (γ0 u2 ) , w = (∇v, ∇u) + (u, v) − γ1 u, w    − φ(v1 ) − φ(v2 ), v − (ρ(w1 ) − ρ(w2 ), w    + f1 (u1 ) − f1 (u2 ), v + f2 (γ0 u1 ) − f2 (γ0 u2 ), w .

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Now using the compatibility condition v|1 = w given in the definition of D(A ) and by Green’s formula, we have    (A y1 − A y2 , y1 − y2 )H = − φ(v1 ) − φ(v2 ), v − (ρ(w1 ) − ρ(w2 ), w    + f1 (u1 ) − f1 (u2 ), v + f2 (γ0 u1 ) − f2 (γ0 u2 ), w . Using the monotonicity of φ and ρ and using the Lipschitz bounds on f1 and f2 , we obtain (A y1 − A y2 , y1 − y2 )H ≤

C 2 L21 C 2 L22 1 1 ∇u22 + v22 + ∇u22 + w22,1 , 2 2 2 2

(3.2)

where C is a constant determined by the Poincaré inequality and the trace theorem. By taking   1 C 2 L21 C 2 L22 + ω = max , 2 2 2 we see that A − ωI is dissipative. In order to see that A − ωI is maximally dissipative, we must show that for some λ > ω we have that λI − A is surjective. So for (x1 , x2 , x3 ) ∈ H , we want to solve ⎛ ⎞ ⎛ ⎞ λu − v x1 ⎜ ⎟ ⎜ ⎟ (3.3) ⎝ λv − u + φ(v) − f1 (u) ⎠ = ⎝ x2 ⎠ λw + γ1 u + ρ(w) − f2 (γ0 u) x3 for (u, v, w) ∈ D(A ). Define A : D(A) ⊂ L2 () → L2 () by  Aφ = −φ, D(A) = φ ∈ L2 () : φ ∈ L2 (), γ1 φ = 0 . Also define the Neumann map N : L2 () → H 1 () by letting N φ be the unique solution to the elliptic boundary value problem ψ = 0,

in ,

∂ψ = φ, ∂ν

on 1 ,

ψ = 0,

on 0 .

Then the system (3.3) can be written as a single equation for v ∈ H10 ():    1 1 1 λv + Av + AN λγ0 v + ρ(γ0 v) − f2 γ0 v + γ0 x 1 λ λ λ   1 1 v + x1 + φ(v) − f1 λ λ

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1 = x2 − Ax1 + AN x3 λ where we consider the duality pairing H10 (), (H10 ()) . Let    1 1 B(v) = AN λγ0 v + ρ(γ0 v) − f2 γ0 v + γ0 x 1 , λ λ   1 1 λ v + x1 , C(v) = v − f1 2 λ λ

89

(3.4)

(3.5) (3.6)

B, C : H10 () → (H10 ()) . To show that (3.4) has a solution, it suffices to show that for λ sufficiently large,√B(·) + C(·) + φ(·) is maximal monotone. For λ large enough, in particular if λ > L2 , then the function   1 1 · + γ0 x 1 λ · +ρ(·) − f2 λ λ is increasing (without any further restriction on ρ, which is assumed increasing), and so B is maximal monotone (it may be written √ as the subgradient of a convex integrand on ) (see [8, p. 33]). Similarly, for λ > 2L1 we have that   1 1 λ · −f1 · + x1 2 λ λ is increasing, hence C(·) is Lipschitz, monotone, and coercive. Since φ(·) is monotone (and therefore generates a maximal monotone operator) it follows from standard perturbation results that B(·) + C(·) + φ(·) is maximal monotone for sufficiently large λ. Therefore (3.4) has a unique solution v. It remains to obtain u, w. Set w = v|1 and u = λ1 v + λ1 x1 . Then (3.4) becomes   λv + Au + AN λw + ρ(w) − f2 (u) − x3 + φ(v) − f1 (u) = x2

(3.7)

where the equality holds in (H10 ()) . This implies that λv − u + φ(v) − f1 (u) = x2

(3.8)

in L2 () (using the fact that H01 () is dense in L2 ()). This is the second line of (3.3). Observe that v, f1 (u), x2 ∈ L2 () by assumption (recall that f1 is bounded from H 1 () → L2 ()); hence we obtain the regularity u − φ(v) ∈ L2 (). Finally, by the trace theorem and the fact that H 1/2 (1 ) ⊂ L2 (1 ), (3.7) now implies that γ1 u + λw + ρ(w) − f2 (u) − x3 = 0

(3.9)

in L2 (1 ), which is the third line of (3.3). Since w, f2 (u), x3 ∈ L2 (1 ) this also implies the regularity γ1 u + ρ(w) ∈ L2 (1 ). Thus (u, v, w) ∈ D(A ) solves (3.3). This completes the proof of Lemma 3.4. 

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Step 2. We now prove Theorem 3.1. The following argument uses the truncation method employed in [11, 26]. We truncate f1 and f2 to obtain globally Lipschitz functions: in particular, for i = 1, 2 let  fi (u) if ∇u2 ≤ K, fi,K (u) = Ku fi ( ∇u2 ) if ∇u2 > K. As shown in [13], f1,K and f2,K are globally Lipschitz with constants L1 (K), L2 (K) respectively. Consider the problem (1.1) with f1 , f2 replaced by their respective truncations; for each T > 0, this has a unique solution on [0, T ]. Using the Lipschitz bound on f1,K , f2,K , the calculations already carried out in (3.2) give the following energy inequality for strong solutions:









∇u(t) 2 + ut (t) 2 + ut (t) 2 2

2

2,1

2



2

2 ≤ ∇u(0) + ut (0) + ut (0)

2

2

2,1

2

2

2

L2 (K) + L2 (K) ∇u(s) + f1 (0) + f2 (0)

t

+ 0

1

2



2 + ut (s)

2,1

2

2

2,1



2 + + ut (s) 2

ds.

Hence there exists a constant ω(K) such that









∇u(t) 2 + ut (t) 2 + ut (t) 2 2

2

2,1

2



2

2 ≤ ∇u(0) + ut (0) + ut (0)

2

2

2,1







∇u(s) 2 + + ut (s) 2 + ut (s) 2

t

+ ω(K) 0

2

2

2,1

+ 1ds.

Assume without loss of generality that ω(K) ≥ 1. Add 1 + et = 2 + sides and apply Gronwall’s inequality to get

t 0

es ds to both









∇u(t) 2 + ut (t) 2 + ut (t) 2 2 2 2,1

2

2

2 

≤ ∇u(0) 2 + ut (0) 2 + ut (0) 2, + 2 eω(K)t . 1

Set

(3.10)

  K 1 log TK = , ω(K) ∇u(0)22 + ut (0)22 + ut (0)22,1 + 2

assuming of course that K is greater than the denominator. Then (3.10) implies that ∇u(t)22 ≤ K for t ≤ TK . It then follows that on [0, TK ] solutions to the original problem (with sources f1 and f2 ) coincide with solutions to the truncated problem (with sources f1,K and f2,K ). Thus we obtain a unique solution

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U (·) ∈ C([0, TK ]; H ) to the original problem. We may repeat this procedure incrementally with larger and larger K to obtain a maximum Tm for which there exists a unique solution in C([0, Tm ); H ). 3.2 Parabolic Case α > 0 It is well-known that the wave equation with strong damping (provided by the term αut ) generates an analytic semigroup using an appropriate functional analysis setup [12]. On the other hand, parabolic problems with dynamic Wentzell-type boundary conditions have been studied, for instance in [15]. More recently, Mugnolo has studied abstract damped wave equations with dynamic boundary conditions, proving that they generate analytic semigroups [32]. Thus we expect to obtain similar results for the system (1.1) viewed as a semilinear parabolic equation on an appropriately defined state space. In particular, we would expect to see the following phenomena: • For t > 0, solutions become more regular than the given initial data from the state space (the so-called “smoothing” effect). • Growth conditions on the nonlinear functions f1 , f2 governing the sources may be relaxed, allowing for “super-critical” sources. • The underlying linear system, disregarding both sources and nonlinear damping terms, generates an analytic (holomorphic) semigroup. In this subsection we examine each of these phenomena under various conditions on (a) the nonlinear sources, (b) the nonlinear damping, and (c) the regularity of initial data. In particular, we will see that it is possible to obtain finite energy solutions for super-critical sources. In addition, assuming certain conditions on the nonlinear damping and initial data (specified below), one obtains more regular solutions than would be expected in the hyperbolic case. These results are summarized in the following theorem. Theorem 3.5 (i) Assume that f1 and f2 are locally Lipschitz from H10 () to (H10 ()) and (H 1/2 (1 )) , respectively. Then (1.1) has a unique local solution   U (·) = u(·), ut (·), ut |1 (·) ∈ C [0, Tm ); H for some maximal time Tm ∈ (0, ∞]. In addition, we have ut (·) ∈ L2 (0, Tm ; H10 ()). (ii) Let θ ∈ [1, 3/2). Assume f, φ : H θ () → L2 () and f2 , ρ : H θ () → 2 L (1 ) are locally Lipschitz. Let the initial condition U (0) satisfy the added regularity   u(0) ∈ H θ () ∩ H10 (), (u + αut )(0), (u + αut )|1 (0) ∈ D (−A0 )θ−1/2 where A0 is the operator given below in (3.19). (In particular, D((−A0 )θ−1/2 ) is continuously embedded in H θ () × H θ−1/2 (1 ).) Then there exists a T > 0 such that the system (1.1) has a unique solution (u, ut , ut |1 ) which satisfies  u ∈ C [0, T ); H θ () ∩ H10 () ,

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  u + αut ∈ C [0, T ); H θ () ∩ H10 () ∩ C (0, T ); H 3/2 () ∩ H10 () ,  (u + αut ) ∈ C (0, T ); L2 () . Remark 3.6 Suppose that the sources f1 , f2 have polynomial structure f1 (s) = |s|p−2 s, f2 (s) = |s|k−2 s. Then f1 , f2 satisfy the assumptions in case (i) whenever ¯ and they satisfy the assumptions in case (ii) whenever 2 ≤ p ≤ p¯ and 2 ≤ k ≤ k, −2θ 2 ≤ p ≤ 2N and 2 ≤ k ≤ 2NN−2θ−1 for N ≥ 3, or 2 ≤ p, k < ∞ for N = 2. In N −2θ −2θ both cases sources may be “super-critical,” i.e. the exponents p and k may exceed the ¯ critical exponents 1 + p2¯ and 1 + k2 , respectively. In particular, note that in dimension N = 3, case (i) allows for p up to 6 and k up to 4, while case (ii) allows for sources to have arbitrary polynomial bounds. Contrast with Remark 3.2. Remark 3.7 We assume in Theorem 3.5 certain growth conditions on the nonlinear damping functions φ and ρ, rather than simply the sources f1 and f2 . This assumption arises from our method of proof, in which we view the problem as a semilinear parabolic problem. In this way we derive the smoothing property entirely from the linear part, rather than from monotonicity. (By “smoothing property” we refer to the statement that solutions are more regular for times t > 0 than at initial time t = 0.) It is well known that a subgradient of a proper convex lower semicontinuous functional generates a nonlinear semigroup which also has the smoothing property, much like an analytic semigroup ([39, Proposition 3.2]). In our case the nonlinear semigroup obtained in Theorem 3.5(i) is not generated by a subgradient, but rather by a linear generator of an analytic semigroup (which we show in part (ii)) added to a subgradient (the nonlinear monotone damping). To the best of our knowledge, there is no precise way of showing that an operator of this form generates a nonlinear semigroup with the smoothing property. For this reason it remains an open question whether our assumptions on the functions φ and ρ may be relaxed or eliminated. Note well that for the purposes of the present paper, the issue is moot, due to the assumptions made in the study of longtime behavior. (See Sects. 5–7.) 3.2.1 Proof of Theorem 3.5(i) Part (i) of Theorem 3.5 allows for super-critical sources by relying on the smoothing effect provided by the strong damping. The proof of this result is as follows: we first show existence of a semigroup for the homogeneous monotone problem, then use a fixed point argument to show existence and uniqueness of the nonlinear perturbed problem. Define the operator A : D(A ) ⊂ H → H (not to be confused with A from the previous subsection) by ⎛

⎞ ⎛ ⎞ u v ⎜ ⎟ ⎜ ⎟ A ⎝ v ⎠ = ⎝ u + αv − φ(v) ⎠ w −γ1 u − αγ1 v − ρ(w)

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93

with domain

⎧⎛ ⎞ ⎫ ⎨ u v ∈ H10 (), (u + αv) − φ(v) ∈ L2 (), ⎬ D(A ) = ⎝ v ⎠ ∈ H : . ⎩ γ0 v = w, γ1 (u + αv) + ρ(w) ∈ L2 (1 ) ⎭ w

With U = (u, ut , ut |1 ) we treat the system (1.1) as a “semi-monotone” equation 

U  (t) = A U (t) + F (U (t)), U (0) = U0 ∈ H ,

where

t > 0,

(3.11)

⎛ ⎞ 0  F U (t) = ⎝ f1 (u(t)) ⎠ . f2 (u(t))

(3.12)

Equation (3.11) can be rewritten by the “variation of parameters” formula t

U (t) = S(t)U0 +

 S(t − s)F U (s) ds =: F (U )(t).

(3.13)

0

Let us show that F defines a contraction on the space C([0, T ]; BR (H )), where BR (H ) := {U ∈ H : U H ≤ R}, and R and T are to be specified later. This can be done in three steps. Step 1: Show that A generates a semigroup. The following can be proved using the same argumentation as in Sect. 3.1: Lemma 3.8 The operator A generates a strongly continuous semigroup of contractions S(t). It follows that (i) for each (u0 , u1 , u1 |1 ) ∈ D(A ) the function (u(t), ut (t), ut (t)|1 ) = S(t)(u0 , u1 , u1 |1 ) is a strong solution to the system (1.1); and (ii) for each (u0 , u1 , w1 ) ∈ H , the function S(t)(u0 , u1 , w1 ) is a solution to (1.1) in the generalized sense, i.e. as a strong limit of strong solutions: take (un0 , un1 , un1 |1 ) = (un0 , un1 , w1 ) ∈ D(A ) converging to (u0 , u1 , w1 ) in H , then S(t)(un0 , un1 , un1 |1 ) → S(t)(u0 , u1 , w1 ) uniformly on bounded intervals. Moreover, the solution satisfies ut ∈ L2 (0, T ; H 1 ()). Proof Dissipativity and maximality follow from the same arguments as those given in Sect. 3.1. Note well that the semigroup is a contraction semigroup, since the absence of sources means the operator is dissipitative (without translation by a constant). Equation (3.10) becomes









∇u(t) 2 + ut (t) 2 + ut (t) 2 2

2

2,1

2



2

2 ≤ ∇u(0) + ut (0) + ut (0)

2

2



∇ut (t) 2

t

+α 2,1

0

,

2

(3.14)

which implies the additional regularity ut ∈ L2 (0, T ; H 1 ()). This completes the proof of Lemma 3.8. 

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Step 2: Estimate inhomogeneous solutions. For this intermediate step, we treat f1 , f2 as sources independent of the solution U (t). Set t

U (t) = S(t)U0 +

 S(t − s) 0, f1 (s), f2 (s) ds

0

for some f1 ∈ L2 (0, T ; L2 ()), f2 ∈ L2 (0, T ; L2 (1 )). Then U is a solution to the system ⎧ utt − u − αut + φ(ut ) = f1 , x ∈ , t > 0, ⎪ ⎪ ⎪ ⎨ u(x, t) = 0, x ∈ 0 , t > 0,  ∂u α∂ut  (3.15) ⎪ ⎪ ⎪ utt = − ∂ν + ∂ν + ρ(ut ) + f2 , x ∈ 1 , t > 0, ⎩ u(x, 0) = u0 (x), ut (x, 0) = u1 (x), x ∈ . Using energy methods, which is to say multiply both sides of the first equation by ut and integrate by parts, we obtain the following energy inequality, which holds for all generalized (semigroup) solutions by lower semicontinuity of the norm (cf. the proof of Theorem 3.1):



U (t) 2 + 2α H

T 0

2

= U (0) H + 2

∇ut 22 dt + 2 T

T

  φ(ut ), ut + ρ(ut ), ut dt

0

(f1 , ut ) + f2 , ut dt

0

2

≤ U (0) H + 2f1 L2 (0,T ;(H 1

0 ())

)

ut L2 (0,T ;H 1

0 ())

+ 2f2 L2 (0,T ;(H 1/2 (1 )) ) ut L2 (0,T ;H 1/2 (1 ))

2

 ≤ U (0) H + Cα f1 2L2 (0,T ;(H 1 ()) ) + f2 2L2 (0,T ;(H 1/2 (

1 ))

0

T

+α 0

)

∇ut 22 dt.

If we set U0 = (0, 0, 0) we can see that





t 0

2

 2 S(t − s)F (s)ds

≤ Cα f1 L2 (0,T ;(H1

0

()) )

+ f2 2L2 (0,T ;(H 1/2 (

1 ))

)

.

(3.16) This proves that the map (f1 , f2 ) → 0 S(· − s)(0, f1 (s), f2 (s))ds is a locally bounded operator from L2 (0, T ; (H10 ()) ) × L2 (0, T ; (H 1/2 (1 )) ) → C([0, T ]; H ). Step 3: Complete the fixed point argument. Returning to (3.13), using the result of Step 2, we have the estimate





F (U )(t) ≤ U0 H + C 1/2 f1 (u) 2 α H L (0,T ;(H 1 ()) )

·



+ f2 (u)



L2 (0,T ;(H 1/2 (1 )) )

0

Appl Math Optim (2012) 66:81–122

95

≤ U0 H + Cα1/2 (L1,R + L2,R )uL2 (0,T ;H 1

0 ())

+ CT

≤ U0 H + Cα1/2 (L1,R + L2,R )T 1/2 uC([0,T ];H 1

0 ())

+ CT ,

where L1,R , L2,R are the local Lipschitz constants for f1 , f2 respectively and





CT = Cα1/2 f1 (0) L2 (0,T ;(H 1 ()) ) + f2 (0) L2 (0,T ;(H 1/2 ( )) ) . 1

0

(Note that CT → 0 as T → 0.) Moreover,





F (U )(t) − F (U˜ )(t) ≤ C 1/2 f1 (u) − f1 (u) ˜ L2 (0,T ;(H 1 α H



+ f2 (u) − f2 (u) ˜

0 ())

)



L2 (0,T ;(H 1/2 (1 )) )

≤ Cα1/2 (L1,R + L2,R )u − u ˜ L2 (0,T ;H 1

0 ())

≤ Cα1/2 (L1,R + L2,R )T 1/2 u − u ˜ C([0,T ];H 1

0 ())

.

Take R = 2U0 H , then let T be small enough so that Cα1/2 (L1,R + L2,R )T 1/2 R + CT ≤ R/2,

and Cα1/2 (L1,R + L2,R )T 1/2 ≤ 1/2.

Then the above estimates prove that F gives a well-defined contraction. By the Contraction Mapping Principle, F has a unique fixed point, which is a solution of (3.13). This completes the proof of Theorem 3.5(i). 3.2.2 Generation of an Analytic Semigroup (Proof of Theorem 3.5(ii)) In this subsection we consider the linear part of the model in question: ⎧ utt − u − αut = 0, x ∈ , t > 0, ⎪ ⎪ ⎪ ⎨ u(x, t) = 0, x ∈ 0 , t > 0, ∂u ∂u t ⎪ utt + ∂ν + α ∂ν = 0, x ∈ 1 , t > 0, ⎪ ⎪ ⎩ u(x, 0) = u0 (x), ut (x, 0) = u1 (x), x ∈ .

(3.17)

Make the substitution z = u + αut . For convenience in notation, let β = 1/α. Then we have the system ⎧ zt = αz + βz − βu, x ∈ , t > 0, ⎪ ⎪ ⎪ ⎪ ut = βz − βu, x ∈  ∪ 1 , t > 0, ⎪ ⎪ ⎪ ⎨ u(x, t) = z(x, t) = 0, x ∈ 0 , t > 0, (3.18) ∂z ⎪ zt = −α ∂ν + βz − βu, x ∈ 1 , t > 0, ⎪ ⎪ ⎪ ⎪ u(x, 0) = u0 (x), z(x, 0) = z0 (x), x ∈ , ⎪ ⎪ ⎩ z(x, 0) = ζ0 (x), x ∈ 1 . Our goal in this subsection is to show that this system generates an analytic semigroup. Once we have established this fact, we can use the robust theorems available for analytic semigroups in order to obtain well-posedness for the semilinear problem.

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Appl Math Optim (2012) 66:81–122

Let us start by focusing on the underlying z-dynamics in the model (3.18). Let X0 = L2 () × L2 (1 ). Define A0 : D(A0 ) ⊂ X0 → X0 by A0 (z, ζ ) = α(z, −γ1 z),   (z, ζ ) ∈ H 3/2 () ∩ H10 () × H 1 (1 ) : ζ = z|1 , z ∈ L2 () .

(3.19)

Note that A0 corresponds to the dynamics ⎧ zt = αz, ⎪ ⎪ ⎪ ⎪ ⎪ z ⎨ = 0, ∂z zt = −α ∂ν , ⎪ ⎪ ⎪ z(x, 0) = z0 (x), ⎪ ⎪ ⎩ z(x, 0) = ζ0 (x),

x ∈ , t > 0, x ∈ 0 , t > 0, x ∈ 1 , t > 0, x ∈ , x ∈ 1 .

(3.20)

It can be seen from [15] that A0 is the generator of an analytic semigroup on L2 () × L2 (1 ). For convenience, we will show this directly in three steps. It is enough to show that (i) A0 is symmetric, (ii) A0 is dissipative, and (iii) A0 is maximally dissipative. From these three facts it follows that A0 is self-adjoint and thus also the generator of a C0 analytic semigroup of contractions. (i) To see that A0 is symmetric, just apply Green’s formula twice to (z, ζ ), (z , ζ  ) ∈ D(A0 ). Let (., .) denote the inner product in L2 () and ., . denote the inner product in L2 (1 ). Then      β A0 (z, ζ ), z , ζ  L2 (),L2 ( ) = z, z − γ1 z, ζ  1      = − ∇z, ∇z + γ1 z, γ0 z − γ1 z, ζ   = − ∇z, ∇z    = z, z − γ0 z, γ1 z      = z, z − ζ, γ1 z = β (z, ζ ), A0 z , ζ  . (ii) The above calculation also shows that A0 is dissipative, since (A0 (z, ζ ), (z, ζ ))X0 = −α|∇z|2L2 () for (z, ζ ) ∈ D(A0 ). (iii) We now show that A0 is maximally dissipative by solving the following eigenvalue problem for λ > 0:  λz − αz = x1 ∈ L2 (), (3.21) λζ + αγ1 z = x2 ∈ L2 (1 ). This can be solved using standard elliptic theory. Let V = {(u, v) ∈ H10 () × L2 (1 ) : u|1 = v} with the norm



(u, v) = V



1/2 |∇u|2 d + 

|v|2 d 1

.

Appl Math Optim (2012) 66:81–122

97

Then V is a closed subspace of H10 () × L2 (1 ). We get the following weak formulation of the eigenvalue problem in the space V : λ(z, φ) + α(∇z, ∇φ) + λζ, ψ = (x1 , φ) + x2 , ψ ∀(φ, ψ) ∈ V .

(3.22)

Since the left-hand side is a bounded and coercive bilinear form in V × V and the right-hand side is a bounded linear functional in V , the weak form problem has a unique solution (z, ζ ) ∈ V . We now show that (z, ζ ) is a solution to (3.21) by the usual “bootstrapping” method. Since z ∈ H 1 (), the a priori definition of z ∈ H −1 () = (H01 ()) is (z, φ) = −(∇z, ∇φ) for φ ∈ H01 (). Taking φ ∈ H01 () and ψ = φ|1 = 0 in (3.22) yields (λz − αz − x1 , φ) = 0 for all φ ∈ H01 (). Since H01 () is densely embedded in L2 (), it follows that λz − αz − x1 = 0 in L2 (); hence z ∈ L2 () and the first equation in (3.21) is satisfied. Then γ1 z is welldefined in L2 (1 ) and (3.22) becomes −αγ1 z + λζ, ψ = x2 , ψ for all (φ, ψ) ∈ V , hence for all ψ ∈ L2 (1 ) (by the trace theorem and the fact that H 1/2 (1 ) is dense in L2 (1 )). Thus the second equation in (3.21) holds. By proving (i), (ii) and (iii) we have established that A0 generates an analytic semigroup. In particular, −A0 is positive self-adjoint. For future reference, we make note of the fractional powers of −A0 . The above calculations show that |(−A0 )1/2 (z, ζ )|2X0 = αz2 1 . By the trace theorem, if z ∈ H s () for s > 1/2 H () 0

then ζ = γ0 z ∈ H s−1/2 () and ζ H s−1/2 () ≤ CzH s () . Hence D((−A0 )1/2 ) is continuously embedded in the space H10 () × H 1/2 (1 ), and by interpolation theory (see [27, Theorem 5.1, Theorem 9.6]) we have that D((−A0 )s ) is continuously embedded in H 1/2+s () × H s (1 ) for s ∈ [1/2, 1]. Now return to the model in (3.18). Let ζ = z|1 . We have the following coupled dynamics, given in terms of a matrix of operators: ⎛ ⎞ ⎛ ⎞⎛ ⎞ −βI βI 0 u u ∂ ⎝ ⎠ ⎝ z = −βI α + βI 0 ⎠ ⎝ z ⎠ . (3.23) ∂t ζ −βγ −αγ βI ζ 0

1

We may decompose the above matrix of operators into a self-adjoint operator and a lower-order operator. Symbolically, we decompose the matrix ⎛ ⎞ −βI βI 0 G = ⎝ −βI α + βI 0 ⎠ (3.24) −βγ0 −αγ1 βI into G = A + B, where ⎛ ⎞ ⎛ −βI 0 0 −βI α 0 ⎠ = ⎝ 0 A=⎝ 0 0 −αγ1 0 0

⎞

⎛

⎞ 0 ⎠, 0 ⎠. A0 βI (3.25) Let us specify what we mean more rigorously. Fix θ ∈ [1, 3/2), let Y = H θ () ∩ H10 (), and let X = Y × X0 . Define A : D(A) ⊂ X → X by  D(A) = Y × D(A0 ) (3.26) A(u, z, ζ ) = −βu, A0 (z, ζ ) , 0

0

0 and B = ⎝ −βI −βγ0

βI βI 0

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Appl Math Optim (2012) 66:81–122

and let B : D(B) ⊂ X → X by B(u, z, ζ ) = β(z, z − u, ζ − u|1 ),

 D(B) = Y × Y × L2 (1 ) .

(3.27)

Observe that A is dissipative and self-adjoint, hence the generator of an analytic semigroup of contractions. In particular, |(−A)s (u, z, ζ )|2X = β 2s u2H θ () + |(z, ζ )|2D((−A0 )s ) . It follows that B is an (−A)s -bounded operator whenever s = θ − 1/2: for (u, z, ζ ) ∈ D(B) we have





B(u, z, ζ ) ≤ β zH θ () + z2 + u2 + ζ 2, + u2, (3.28) 1 1 X





s ≤ C (−A) (u, z, ζ ) X + (u, z, ζ ) X , (3.29) for C chosen large enough. By perturbation theory for analytic semigroups (see [33, Corollary 3.2.4]), G = A + B with D(G) = D(A) generates an analytic semigroup. Finally, returning to the original linear dynamics in (3.17), we make the change of variable (u, z, w) → (u, ut , ut |1 ) using the transformation ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ u I 0 0 u u ⎝ ut ⎠ = ⎝ −βI βI 0 ⎠ ⎝ z ⎠ =: L ⎝ z ⎠ . (3.30) −βγ0 0 βI ut |1 w w Observe that L is an isomorphism on X, with inverse given by ⎛ ⎞ I 0 0 L−1 = ⎝ I αI 0 ⎠ . γ0 0 αI

(3.31)

Thus A = LGL−1 is the generator of an analytic semigroup on X, with domain  u  D(A) =

v w

∈ H : u + αv ∈ H10 (), (u + αv)|1 = u|1 + αw, 

(u + αv) ∈ L () . 2

(3.32)

Theorem 3.5(ii) now follows from Theorem 6.3.1 and Theorem 6.3.3 in [33].

4 Energy Identity The results of the remaining sections depend not only on the existence of solutions but also on an energy identity. Note well that whereas a standard weak lower semicontinuity argument allows us to deduce a corresponding energy inequality, one does not necessarily have an energy identity for all finite energy solutions. We will show that under certain assumptions on the sources and damping, we can pass the energy identity through to finite energy solutions in addition to strong (or regular) solutions.

Appl Math Optim (2012) 66:81–122

99

An additional consequence of this section will be that generalized solutions are also weak or variational solutions, in the sense that they satisfy the variational form (4.5) below; however, we will not address the issue of uniqueness of weak solutions here (see [6, 7] for more insight on this question). Assumption 4.1 We assume the following restrictions: Sources: we shall assume that the functions f1 and f2 are of a polynomial structure. That is, f1 (s) = |s|p−2 s,

f2 (s) = |s|k−2 s,

where k, p ≥ 2 are such that H10 () → Lp (),

H10 () → Lk (1 );

Damping: we assume that the functions ρ and φ are monotone, continuous and there exist four positive constants mq , Mq , cr and Cr such that mq |s|q ≤ ρ(s)s ≤ Mq |s|q ,

∀s ∈ R,

(4.1)

and cr |s|r ≤ φ(s)s ≤ Cr |s|r ,

∀s ∈ R.

(4.2)

Exponents: we assume either that (i) α > 0, or else (ii) we have the restriction p ≤ ¯ p(r−1) ¯ + 1 and k ≤ k(q−1) , where p¯ and k¯ are given in (2.1). r q Then the energy functional E associated to the system (1.1) takes the form

2 1

2

2 1

1

E(t, u, ut ) = E(t) = ut (t) 2 + ut (t) 2, + ∇u(t) 2 1 2 2 2

p 1

k 1

− u(t) p − u(t) k, . 1 p k

(4.3)

Multiplying the first equation in (1.1) but ut and integrating by parts using Green’s formula, we obtain the following energy identity:

∇ut (t) 2 dt +

T

E(T ) + α S

2

T S

T

φ(ut )ut dxdt + 

S

ρ(ut )ut ddt = E(S) 1

(4.4) for 0 ≤ S ≤ T . However, this derivation is merely formal in the case of general finite energy solutions, since we do not have sufficient regularity a priori to perform the necessary integration by parts. The remainder of this section will be devoted to showing that by passing to the limit on strong solutions, we obtain the above energy identity for generalized finite energy solutions in addition to strong solutions. Remark 4.2 Assumption 4.1 can be extended to somewhat more general assumptions, particularly as regards the sources f1 (u), f2 (u) [6, 7]. However, for the purposes of

100

Appl Math Optim (2012) 66:81–122

this paper, it does not seem necessary to overly complicated the matter, since all of our present results on long time behavior take the above assumption as given. Notice the third assumption: if α > 0, no additional growth restrictions are needed, but if α = 0 then we compensate for the loss of smoothing by restricting the growth of the source terms. To the best of our knowledge, some compensation of this form is necessary for the hyperbolic case. This highlights another added benefit of the smoothing provided by the strong damping term αut . Assume first that α > 0. Suppose we are given a strong solution U = (u, ut , ut |1 ) with the additional regularity specified in Theorem 3.1, and consider test functions ψ ∈ C([0, T ]; H10 ()) ∩ C 1 ([0, T ]; L2 ()). (Here T > 0 is some given time such that T < Tm , where Tm is the maximum life-time of solutions.) Integration by parts yields  T (ut , ψ) + ut , ψ 0 + T

=

T

   ∇[u + αut ], ∇ψ + φ(ut ), ψ + ρ(ut ), ψ dt

0

   (ut , ψt ) + ut , ψt + f1 (u), ψ + f2 (u), ψ dt.

(4.5)

0

Choosing ψ = ut yields (4.4) for strong solutions. Note that (4.2) and (4.1) also imply that T 0

T

|ut |r dxdt + mq

cr

0



|ut |q d1 dt 1

≤ E(0) − E(T )

2 1

2 1

≤ ut (0) 2 + ut (0) 2, + 1 2 2

p 1

k 1

+ u(T ) p + u(T ) k, 1 p k

2 1

2 1

≤ ut (0) 2 + ut (0) 2, + 1 2 2 + Cp u

p C([0,T ];H1 ()) 0

1

∇u(0) 2 2 2

1

∇u(0) 2 2 2

+ Ck ukC([0,T ];H 1

0 ())

.

(4.6)

Now suppose we have a generalized solution U = (u, ut , ut |1 ). By definition, U is a strong limit in the energy space C([0, T ]; H) of strong solutions Un = (un , unt , unt |1 ). In addition to the finite energy regularity U ∈ C([0, T ]; H), we also have unt  ut weakly in Lr (0, T ; ) ∩ Lq (0, T ; 1 ) by (4.6) applied to unt . Then r (4.2) and (4.1) imply that (on a subsequence) ψ(unt )  ψ0 weakly in L r−1 (0, T ; ) q

and ρ(unt |1 )  ρ0 weakly in L q−1 (0, T ; ). Moreover, applying (4.5) to Un − Um with ψ = unt − um t we get





 1 

un − um 2 + un − um 2 + ∇ un − um 2 T + α t t 2 t t 2,1 2 0 2

T

0



∇ un − um 2 dt t t 2

Appl Math Optim (2012) 66:81–122

101

T

+

  n   n  m n  m m φ unt − φ um t , ut − ut + ρ ut − ρ ut , ut − ut dt

0 T

=

    n  m n  m f1 un − f1 um , unt − um t + f2 u − f2 u , ut − ut dt.

0

(4.7)

p

Taking into account that (a) u → f1 (u) : Lp () → L p−1 (), v → f2 (v) : Lk (1 ) → k L k−1 (1 ) are locally Lipschitz, that (b) the embeddings H10 () → Lp () and H10 () → Lk (1 ) hold, and that (c) un → u in H 1 (), it follows that there is some uniform constant L (depending on  and u) such that T

    n  m n  m f1 un − f1 um , unt − um t + f2 u − f2 u , ut − ut dt

0

T

≤L 0

 

∇ un − um ∇ un − um , t t 2 2

(4.8)

where L is a constant depending on supn supt∈[0,T ] un (t)H 1

0 ()

finite. Using Young’s inequality, we obtain α 2

, which must be

T

0



∇ un − um 2 dt t t 2 T

+ 0

  n   n  m n  m m φ unt − φ um t , ut − ut + ρ ut − ρ ut , ut − ut dt

2 n

2

 n 2  1 

m m







≤ unt (0) − um t (0) 2 + ut (0) − ut (0) 2,1 + ∇ u (0) − u (0) 2 2 T 

∇ un − um 2 dt → 0 (4.9) + Cα 2 0

as m, n → ∞. In this way we have ∇unt → ∇ut in L2 (0, T ; ) and, by Lemma 1.3 on p. 42 in [4] that φ0 = φ(ut ), ρ0 = ρ(ut ) and (on a subsequence), T 0

 φ unt , unt dt →

T

φ(ut ), ut dt,

T

0

0

  ρ unt , unt dt →

T

 ρ(ut ), ut dt.

0

Finally, let En (t) denote the energy corresponding to the solution Un . Then due to the growth restrictions on the nonlinear sources imposed by Assumption 4.1, we see that En (t) → E(t) (in particular, un → u in H 1 () implies convergence in Lp () and Lk (1 ), as well). Since the energy identity holds for strong solutions, we have

∇un (t) 2 dt +

T

En (T ) + α 0

t

2

T 0



 φ unt unt dxdt +

= En (0). Letting n → ∞ we obtain the desired result.

T 0

1

 ρ unt unt ddt

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Appl Math Optim (2012) 66:81–122

¯ For the case α = 0, we have the assumption p ≤ p(r−1) + 1 and k ≤ r (4.8) instead becomes (after some elementary computations)

¯ k(q−1) . q

Then

T

    n  m n  m f1 un − f1 um , unt − um t + f2 u − f2 u , ut − ut dt

0





 



∇ un − um un − um + un − um

t t r t t q, . 2

T

≤L 0

1

(4.10)

r q Considering that unt , um t are bounded in L (0, T ; ) and L (0, T ; 1 ), this quantity goes to zero. Thus we obtain T

  n   n  m n  m m φ unt − φ um t , ut − ut + ρ ut − ρ ut , ut − ut dt

0

≤





  1 

un (0) − um (0) 2 + un (0) − um (0) 2 + ∇ un (0) − um (0) 2 t t t t 2 2, 2 1 2 T 







∇ un − um un − um + un − um

→ 0. (4.11) +L t t r t t q, 2 1

0

The rest of the proof proceeds as before.

5 Exponential Growth of Solutions In this section, we prove that under appropriate assumptions on the initial data and on the source terms, the solution to problem (1.1) is unbounded and grows as an exponential function. To this end, we take Assumption 4.1 as given. Next, let B1 be the best constant in the embedding H10 () → Lp () and B2 be the best constants in the embedding H10 () → Lk (1 ). That is  B1−1 = inf ∇u2 : u ∈ H10 () : up = 1 ,  B2−1 = inf ∇u2 : u ∈ H10 () : uk,1 = 1 . Define the function F as p

B Bk 1 F (x) = x 2 − 1 x p − 2 x k 2 p k and let α1 be the first positive zero of the function F  (x). It can be easily checked that α1 is a point of local maximum. (See [11] for more details). Accordingly, let us define α1 as p

p p−2

1 = B1 α1

+ B2k α1k−2 ,

B p Bk 1 and E1 = F (α1 ) = α12 − 1 α1 − 2 α1k . 2 p k

(5.1)

Appl Math Optim (2012) 66:81–122

103

Define E2 as follows E2 =

⎧ ⎨ (k−2)α12 ,

if p ≥ k, 2p ⎩ (p−2)α12 , if k ≥ p. 2k

(5.2)

Then, we have E2 < E1 . Our main result in this section reads as follows: Theorem 5.1 Assume that Assumption 4.1 hold. Let 2 ≤ r < p ≤ p and 2 ≤ q < k ≤ k. Then, any solution of (1.1) satisfying E(0) < E2 ,

∇u0 2 ≥ α1 ,

(5.3)

grows up exponentially when t tends to infinity. That is, there exist constants C, η > 0 such that p

up + ukk,1 ≥ Ceηt . The constant η depends on α in the sense that η → 0 as α → ∞, i.e. the rate of exponential growth obtained gets slower as the amount of strong damping increases. The proof of Theorem 5.1 will be done through some Lemmas. The following lemma will play an essential role in the proof of our main result, and it is inspired by the work in [11] where the authors proved a similar lemma for the wave equation. Then, we have: Lemma 5.2 Let u be a weak solution of (1.1). Assume that ∇u0 2 > α1 .

(5.4)

Then there exists a constant α2 > α1 such that



∇u(t) ≥ α2 , ∀t ∈ [0, Tmax ) 2

(5.5)

E(0) < E1

and

and

1

u(t) p + p p

p k

1

u(t) k ≥ B1 α p + B2 α k , 2 k, 1 k p k 2

∀t ∈ [0, Tmax ).

(5.6)

The proof of Lemma 5.2 can be done along the same line as in [40] and [11]. We omit the details. Proof of Theorem 5.1 We define H (t) = E2 − E(t).

(5.7)

Of course by (4.4) and (5.4), we deduce that H (t) is a non-decreasing function. So, by using (4.3) and, (5.7) we get 0 < H (0) ≤ H (t)

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Appl Math Optim (2012) 66:81–122

= E2 − E(t)

p 1

k 1 1

≤ E1 − ∇u22 + u(t) p + u(t) k, . 1 2 p k

(5.8)

From (4.3) and (5.5), we obtain 1 E1 − ∇u22 2 1 = F (α1 ) − ∇u22 2 p

B p Bk 1 < F (α1 ) − α12 = − 1 α1 − 2 α1k < 0, 2 p k

∀t ≥ 0.

Hence we finally obtain the following inequality:



1

u(t) p + 1 u(t) k , p k,1 p k

0 < H (0) ≤ H (t) ≤

∀t ≥ 0.

(5.9)

Now, for ε small to be chosen later, we then define the auxillary function: L(t) = H (t) + ε

ut udx + ε

(5.10)

ut ud.



1

First, we have the following lemma. Lemma 5.3 Let u be a solution of (1.1). Then, under the assumptions in Theorem 5.1, there exists a constant η1 > 0 independent of t, such that  dL(t) p ≥ η1 ut 22 + ut 22,1 + H (t) + up + ukk,1 + E2 , dt

∀t ≥ 0. (5.11)

The constant η1 depends on α in the sense that η1 → 0 as α → ∞. On the other hand, there exists  > 0, such that for all 0 < ε < , we have L(0) = H (0) + ε

u0 u1 (x)dx + ε

u0 u1 (x)d > 0.



1

Proof By taking the time derivative of (5.10), we obtain: dL(t) = α∇ut 22 + dt +ε

φ(ut )ut dx + 

1

utt udx + ε 

1

ρ(ut )ut d + εut 22

utt udσ + εut 22,1 .

Using problem (1.1), equation (5.12) takes the form: dL(t) = α∇ut 22 + dt

φ(ut )ut dx + 

1

ρ(ut )ut d + εut 22

(5.12)

Appl Math Optim (2012) 66:81–122

105

p

k



− ε∇u22 + ε u(t) p + ε u(t) k, + εut 22,1 1

−ε

φ(ut )udx − ε 

ρ(ut )ud − εα 1

∇ut ∇udx.

(5.13)



Exploiting (4.3) and (5.7), we get −∇u22 = 2H (t) − 2E2 + ut 22 + ut 22,1 −

2

u(t) p − p p

2

u(t) k . (5.14) k,1 k

Plugging (5.14) into (5.13), we get dL(t) = α∇ut 22 + dt

φ(ut )ut dx + 

1

ρ(ut )ut d + 2εut 22 + 2εut 22,1

  

p

2

2



u(t) k u(t) p + ε 1 − + 2εH (t) − 2εE2 + ε 1 − k,1 p k −ε



φ(ut )udx − ε 

ρ(ut )ud − εα 1

∇ut ∇udx.

(5.15)



On the other hand, we have 

 p  p B1 p B2k k B1 p B2k k −1 α + α . α + α −2εE2 = −2εE2 p 2 k 2 p 2 k 2   p  B1 p B2k k −1 1 1 p k up + uk,1 . α + α ≥ −2εE2 . p k p 2 k 2

(5.16)

Inequality (5.16) together with (5.15) yield dL(t) ≥ α∇ut 22 + dt

φ(ut )ut dx + 

1

ρ(ut )ut d + 2εut 22 + 2εut 22,1

p

k



+ 2εH (t) + εc1 u(t) + εc2 u(t)

p

−ε

φ(ut )udx − ε 

k,1

ρ(ut )ud − εα 1

∇ut ∇udx,

(5.17)



where c1 and c2 are as follows:  p  B1 p B2k k −1 2 2 − E2 α2 + α2 , p p p k  p  B1 p B2k k −1 2 2 α2 + α2 c2 = 1 − − E 2 . k k p k

c1 = 1 −

Our next goal is to show that c1 > 0 and c2 > 0. To this end, we distinguish two cases: Case 1 p ≥ k.

106

Appl Math Optim (2012) 66:81–122 (k−2)α 2

In this case E2 takes the value E2 = 2p 1 and it is obvious that c1 > c2 . Therefore, since α2 > α1 , then to prove that c2 > 0, it is suffices to prove that  1−

2 k



p

B1 p B2k k α + α p 1 k 1



2 ≥ E2 k =

2 (k − 2)α12 . k 2p

(5.18)

We recall that our parameter α1 is defined through the relation (5.1), then we have p

B1 p B2k k 1 2 α + α ≥ α . p 1 k 1 p 1

(5.19)

Consequently, (5.19) yields (5.18). Moreover, if p ≥ k, we have c1 ≥ c2 > 0. Case 2 k ≥ p. (p−2)α 2

In this case, E2 = 2k 1 and we can show with the same method that c2 ≥ c1 > 0. To estimate the last three terms in (5.17), we make use of (4.1), (4.2) and the following Young’s inequality XY ≤

λα X α λ−β Y β + , α β

(5.20)

X, Y ≥ 0, λ > 0, α, β ∈ R+ such that 1/α + 1/β = 1, then we get ρ(ut )ud ≤ Mq

|ut |q−2 ut ud

1

1

≤ Mq

λq q − 1 −q/(q−1) q q uq,1 + Mq λ ut q,1 . q q

(5.21)

Similarly, φ(ut )udx ≤ Cr 

|ut |r−2 ut udx 

μr r − 1 −r/(r−1) ≤ Cr urr + Cr μ ut rr , r r

μ > 0.

(5.22)

Also, ∇ut ∇udx ≤

α 

α ∇ut 22 + αδ∇u22 , 4δ

δ > 0.

(5.23)

Inserting the estimates (5.21)–(5.23) into (5.17), using (4.1) and (4.2), we find   ε dL(t) q ≥α 1− ∇ut 22 + mq ut q,1 + cr ut rr + 2εut 22 + 2εut 22,1 dt 4δ

Appl Math Optim (2012) 66:81–122

107 p

+ 2εH (t) + εc1 up + εc2 ukk,1 − εαδ∇u22 μr r − 1 −r/(r−1) urr − εCr μ ut rr r r λq q − 1 −q/(q−1) q q − εMq uq,1 − εMq ut q,1 . λ q q − εCr

(5.24)

Since p > r and k > q, we then use the embedding Lk (1 ) → Lq (1 ) and Lp () → Lr (), we get q/k  q uq,1 ≤ C ukk,1 and

 p r/p , urr ≤ C up

where C is a positive constant which may vary from line to line. Exploiting the algebraic inequality κ ν ≤ (κ + 1) ≤ (1 + 1/ω)(κ + ω),

∀κ ≥ 0, 0 < ν ≤ 1, ω ≥ 0

(5.25)

with κ = u(t)kk,1 , d = 1 + 1/H (0), ω = H (0) and ν = q/k, then we get  q/k  ukk,1 ≤ d ukk,1 + H (0)  ≤ d ukk,1 + H (t) ,

∀t ≥ 0.

(5.26)

Similarly, we have 

p r/p

up

 p ≤ d up + H (t) ,

∀t ≥ 0.

(5.27)

Plugging (5.26) and (5.27) into (5.24), we obtain     dL(t) ε q − 1 −q/(q−1) q 2 ut q,1 ≥α 1− ∇ut 2 + mq − εMq λ dt 4δ q   r − 1 −r/(r−1) ut rr + 2εut 22 + 2εut 22,1 μ + cr − εCr r     μr λq μr p − CdMq H (t) + ε c1 − CdCr up + ε 2 − CdCr r q r   λq ukk,1 − εαδ∇u22 . + ε c2 − CdMq (5.28) q Using once again inequality (5.14), then (5.28) takes the form     dL(t) ε q − 1 −q/(q−1) q 2 ut q,1 ≥α 1− ∇ut 2 + mq − εMq λ dt 4δ q   r − 1 −r/(r−1) μ ut rr + ε(2 + αδ)ut 22 + cr − εCr r

108

Appl Math Optim (2012) 66:81–122

  μr λq − CdMq H (t) + ε(2 + αδ)ut 22,1 + ε 2 + 2αδ − CdCr r q   2αδ μr p − up + ε c1 − CdCr r p   2αδ λq − ukk,1 − 2εαδE2 + ε c2 − CdMq q k which in turn by using (5.16) can be rewritten as     dL(t) q − 1 −q/(q−1) ε q ut q,1 ≥α 1− ∇ut 22 + mq − εMq λ dt 4δ q   r − 1 −r/(r−1) + cr − εCr ut rr + ε(2 + αδ)ut 22 μ r + ε(2 + αδ)ut 22,1   μr λq − CdMq H (t) + 2αδE2 + ε 2 + 2αδ − CdCr r q    p  4αδ 4αδE2 B1 p B2k k −1 μr p up − − α2 + α2 + ε c1 − CdCr r p p p k    p  4αδ 4αδE2 B1 p B2k k −1 λq ukk,1 . − − α2 + α2 + ε c2 − CdMq q k k p k (5.29) Let γ = αδ. At this point, we choose λ, μ and γ small enough such that ⎧ μr λq ⎪ ⎪ 2 + 2γ − CdCr r − CdMq q ≥ c3 , ⎪ ⎨ p r B2k k −1 4γ E2  B1 p − ≥ c3 , c1 − CdCr μr − 4γ p p p α2 + k α2 ⎪ ⎪ p ⎪ k  q ⎩ B2 k −1 4γ E2 B1 p c2 − CdMq λq − 4γ ≥ c3 , k − k p α2 + k α2 where c3 is some constant, small but fixed with respect to the parameters. Once the constants λ, μ and δ are fixed, we may pick ε small enough such that ⎧ ε 1 − 4δ ≥ 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ mq − εMq q−1 λ−q/(q−1) ≥ 0, q r−1 −r/(r−1) ⎪ ⎪ c − εC ≥ 0, r r r μ ⎪ ⎪ ⎪ ⎩ L(0) > 0. Setting η1 = min{ε(2 + γ ), εc3 , 2γ }, we see that (5.11) holds. Finally, we note that η1 = η1 (α) → 0 as α → ∞. In particular, in the above calculations ε ≤ 4δ = 4γ /α, and thus η1 ≤ c4 /α for some constant c4 . The proof of Lemma 5.3 is finished. 

Appl Math Optim (2012) 66:81–122

109

Second, we have the next result. Lemma 5.4 Let u be a solution of (1.1). Then, under the assumptions in Theorem 5.1, there exists a constant η2 > 0 independent of t, such that  p L(t) ≤ η2 ut 22 + ut 22,1 + H (t) + up + ukk,1 + E2 , ∀t ≥ 0. (5.30) Proof It is clear that, by Young’s inequality and Poincaré’s inequality, we get, for some γ > 0   L(t) ≤ γ H (t) + ut 22 + ut 22,1 + ∇u22 . (5.31) Since H (t) > 0, we have for all t ≥ 0, 1 1 1 p ∇u22 ≤ up + ukk,1 + E2 . 2 p k Thus, the inequality (5.31) becomes:   p L(t) ≤ η2 H (t) + ut 22 + ut 22,1 + up + ukk,1 + E2 ,

(5.32)

for some η2 > 0. 

Thus, the proof of Lemma 5.4 is finished.

To complete the proof of Theorem 5.1, we have from the two inequalities (5.11) and (5.30), we finally obtain the differential inequality: dL(t) ≥ ηL(t) dt

for some η > 0,

(5.33)

in particular by setting η = η1 /η2 . Integrating the previous differential inequality (5.33) between 0 and t gives the following estimate for the function L: L(t) ≥ L(0)eηt .

(5.34)

Due to the particular choice of η, it follows from Lemma 5.3 that η → 0 as α → ∞. Now, from the definition of the function L (and for small values of the parameter ε), it follows that: 1 1 p L(t) ≤ up + ukk,1 . (5.35) p k From the two inequalities (5.34) and (5.35) we conclude the exponential growth of  the norm up + uk,1 , which concludes the proof of Theorem 5.1. 6 Blow up in Finite Time for α = 0 In this section, we prove that in the absence of the strong damping −ut , (i.e. α = 0), then the solution of problem (1.1) blows up in finite time. By this we mean that up + uk,1 → ∞ as t → T ∗ for some 0 < T ∗ < ∞. We state our main result in this section.

110

Appl Math Optim (2012) 66:81–122

Theorem 6.1 Let α = 0. Assume that Assumption 4.1 hold. Let 2 ≤ r < p ≤ p and 2 ≤ q < k ≤ k. Then, any solution of (1.1) satisfying E(0) < E2 ,

∇u0 2 ≥ α1 ,

(6.1)

blows up in finite time. That is up + uk,1 → ∞ as t → T ∗ for some 0 < T ∗ < ∞. Remark 6.2 Under the assumptions in Theorem 6.1, local-in-time solutions may fail to exist if p and k are super-critical, that is, unless satisfy the assumptions given in Theorem 3.1. On the other hand, certain conditions on the exponents r and q could perhaps remedy this situation (see [7]), although we have not proved it for this model. See Remark 3.2. Proof of Theorem 6.1 To prove Theorem 6.1, we suppose that the solution exists for all time and we reach to a contradiction. Following the idea introduced in [20] and developed in [29] and [40], we define a function Lˆ which is a perturbation of the total energy of the system and satisfies the differential inequality ˆ d L(t) ≥ ξ Lˆ 1+ν (t) dt

(6.2)

in [0, ∞), where ν > 0. Inequality (6.2) leads to a blow up of the solutions in finite ˆ −ν ξ −1 ν −1 , provided that L(0) ˆ time T ∗ ≥ L(0) > 0. Thus, our steps in the prove is to find the functional Lˆ and prove inequality (6.2). We define the functional Lˆ as follows ˆ = H 1−σ (t) +  L(t)

ut udx +  

ut ud,

(6.3)

1

where the functional H is defined in (5.7), σ is satisfying   k−q p−r p−2 k−2 0 < σ ≤ min , , , , k(q − 1) p(r − 1) 2p 2k

(6.4)

and  is a small positive constant to be chosen later. Taking the time derivative of ˆ L(t) and following the same steps as in the proof of Lemma 5.3, we get ˆ d L(t) ≥ (1 − σ )H −σ (t)H  (t) + 2ut 22 + 2ut 22,1 + 2H (t) dt μr r − 1 −r/(r−1) p μ + c1 up + c2 ukk,1 − Cr urr − Cr ut rr r r λq q − 1 −q/(q−1) q q λ − Mq uq,1 − Mq ut q,1 . (6.5) q q It is clear that from (5.7), (4.1) and (4.2), we get H  (t) ≥ mq ut q,1 + cr ut rr . q

(6.6)

Appl Math Optim (2012) 66:81–122

111

Next, for large positive constants M1 and M2 to be chosen later, we select λ−q/(q−1) = M1 H −σ (t) and μ−r/(r−1) = M2 H −σ (t), then using (5.7) together with (6.6), we get from (5.17) that   ˆ

q d L(t) q −1 ≥ (1 − σ )mq − M1 Mq H −σ (t) ut (t) q, 1 dt q  

r r −1 + (1 − σ )cr − M2 Cr H −σ (t) ut (t) r r

p

k

2



+ 2 ut (t) 2 + 2ut 22,1 + 2H (t) + c1 u(t) p + c2 u(t) k,

1

− Mq

−(q−1) M1

q

1

−(r−1)

− Cr

M2

r

q

H σ (q−1) (t) u(t) q,

r

H σ (r−1) (t) u(t) r .

(6.7)

Our objective now is to analyze the last two terms on the right-hand side in (5.24). Indeed, since k > q, we then use the embedding Lk (1 ) → Lq (1 ) and inequality (5.9) to get H

σ (q−1)







1

u(t) q

u(t) p + 1 u(t) k p k, q,1 1 p k σ (q−1) 







q/k 1

u(t) k

u(t) p + 1 u(t) k ≤C p k, k, 1 1 p k σ (q−1) 



1

u(t) p + 1 u(t) k ≤C p k,1 p k

p

k q/k 

× u(t) + u(t)

q u(t) q, ≤



σ (q−1)

1



p

p

k 

≤ C u(t) + u(t)

p

k,1

σ (q−1)+q/k

k,1

(6.8)

.

p

Using the algebraic inequality (5.25) with κ = u(t)p + u(t)kk,1 , d = 1 + 1/H (0), ω = H (0) and ν = ν ≤ 1 and therefore

kσ (q−1)+q , k

then the condition (6.4) implies that 0 <



p

k



σ (q−1)+q/k 



u(t) p + u(t) k ≤ d u(t) p + u(t) k, + H (0) p k,1 1

p

k 

≤ d u(t) p + u(t) k, + H (t) . 1

(6.9)

112

Appl Math Optim (2012) 66:81–122

Similarly, by using the embedding Lp () → Lr (), we obtain σ (r−1)+r/p 

r





1

u(t) p + 1 u(t) k H σ (r−1) (t) u(t) r ≤ C p k,1 p k

p

k 

≤ Cd u(t) p + u(t) k, + H (t) . 1

(6.10)

Plugging (6.9) and (6.10) into (6.7), we get   ˆ

q

2 q −1 d L(t) ≥ (1 − σ )mq − M1 Mq H −σ (t) ut (t) q, + 2 ut (t) 2 1 dt q  

r r −1 H −σ (t) ut (t) r + 2ut 22,1 + (1 − σ )cr − M2 Cr r   −(q−1) −(r−1) M1 M2 Cd − Cr Cd H (t) +  2 − Mq q r   −(q−1) −(r−1)

p

M M +  c 1 − Mq 1 Cd − Cr 2 Cd u(t) p q r   −(q−1)

k

M M −(r−1) Cd − Cr 2 Cd u(t) k, . +  c 2 − Mq 1 (6.11) 1 q r At this point, we take M1 and M2 large enough such that ⎧ −(q−1) −(r−1) M1 M2 ⎪ ⎪ 2 − CM d − CC d > 0, ⎪ q r q r ⎪ ⎪ ⎨ −(q−1) −(r−1) M M c1 − CMq 1 q d − CCr 2 r d > 0, ⎪ ⎪ ⎪ ⎪ −(q−1) −(r−1) ⎪ M2 ⎩ c − M C M1 d − CC d > 0. 2 q r q r Once M1 and M2 are fixed, we may choose  small enough such that  (1 − σ )mq − M1 Mq q−1 q > 0, (1 − σ )cr − M2 Cr r−1 r > 0, and ˆ L(0) > 0. Consequently, there exists a positive constant ηˆ > 0, such that ˆ

p

k

2



d L(t) ≥ ηˆ ut (t) 2 + ut 22,1 + H (t) + u(t) p + u(t) k, , 1 dt

∀t ≥ 0. (6.12)

On the other hand, it is clear from the definition (6.3), we have   1 Lˆ 1−σ (t) ≤ C(, σ ) H (t) +

 ut udx 

1 1−σ





+

ut ud 1

1 1−σ

 .

(6.13)

Appl Math Optim (2012) 66:81–122

113

By the Cauchy-Schwarz inequality and Hölder’s inequality, we have: 1 

 ut udx ≤ 



1

2

u2t dx

u2 dx

2



1 

 ≤C 

u2t dx

1

2

p

|u|p dx

,



where C is the positive constant which comes from the embedding Lp () → L2 (). This inequality implies that there exists a positive constant C1 > 0 such that: 







1 1−σ

≤ C1

ut udx

|u| dx p



1 (1−σ )p









u2t dx

1 2(1−σ )

 .

Applying Young’s inequality to the right hand-side of the preceding inequality, there exists a positive constant also denoted C > 0 such that: 



1 1−σ

ut udx





≤C

|u| dx p



τ (1−σ )p





+





u2t dx

θ 2(1−σ )

 (6.14)

,

for 1/τ + 1/θ = 1. We take θ = 2(1 − σ ), hence τ = 2(1 − σ )/(1 − 2σ ), to get 



1 1−σ

ut udx





≤C



|u|p dx

2 (1−2σ )p



+ 

 u2t dx . p

By using (6.4) and the algebraic inequality (5.25) with κ = up , d = 1 + 1/H (0), 2 ω = H (0) and ν = p(1−2σ ) , the condition (6.4) on σ ensures that 0 < ν ≤ 1 and we get   κ ν ≤ d κ + H (0) ≤ d κ + H (t) . Therefore, there exists a positive constant denoted C2 such that for all t ≥ 0, 



1 1−σ

ut udx 

p



2   ≤ C2 H (t) + u(t) p + ut (t) 2 .

(6.15)

Following the same method as above, we can show that there exists C3 > 0 such that 

 ut ud 1

1 1−σ

k



2   ≤ C3 H (t) + u(t) k, + ut (t) 2, . 1

1

(6.16)

Collecting (6.13), (6.15) and (6.16), we obtain

p

k

2



1 Lˆ 1−σ (t) ≤ ηˆ 1 ut (t) 2 + ut 22,1 + H (t) + u(t) p + u(t) k, , 1

for some ηˆ 1 > 0.

∀t ≥ 0, (6.17)

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Appl Math Optim (2012) 66:81–122

Combining (6.12) and (5.30), then, there exists a positive constant ξ > 0, as small as , such that for all t ≥ 0, 1

Lˆ  (t) ≥ ξ Lˆ 1−σ (t).

(6.18)

ˆ Thus, inequality (6.2) holds. Therefore, L(t) blows up in a finite time T ∗ . On the other hand, from the definition of the function L (t) (and for small values of the parameter ), it follows that





ˆ ≤ κ u(t) p + u(t) k L(t) (6.19) p k, 1

where κ is a positive constant. Consequently, from the inequality (6.19) we conclude p that the norm u(t)p + u(t)kk,1 of the solution u, blows up in the finite time T ∗ , which implies the desired result.  7 Exponential Growth for f1 = 0 The goal of this section is to prove that even in the absence of the interior source (i.e. f1 = 0), then the boundary source still lead to an exponential growth of the norm u(t)k,1 . We assume that the interior damping is linear, that is φ(ut ) = ut . In this case our energy functional takes the form



˜ = 1 ut (t) 2 + 1 ut 22, + 1 ∇u22 − ˜ u, ut ) = E(t) E(t, 2 1 2 2 2

1

u(t) k . k,1 k

(7.1)

Let us define now the following constants −k/(k−2) α˜ 1 = B2 ,

and E˜ 1 =



 1 1 2 − α˜ . 2 k 1

(7.2)

Therefor, similar to Lemma 5.2, we have Lemma 7.1 Let u be a weak solution of (1.1) with f1 = 0. Assume that ˜ E(0) < E˜ 1

∇u0 2 > α˜ 1 .

(7.3)

Then there exists a constant α˜ 2 > α˜ 1 such that



∇u(t) ≥ α˜ 2 , ∀t ∈ [0, Tmax ) 2

(7.4)

and



u(t)

k,1

and

≥ B2 α˜ 2 ,

∀t ∈ [0, Tmax ).

(7.5)

Theorem 7.2 Let 2 < q ≤ k ≤ k. Assume that the Assumption 4.1 holds with φ(ut ) = ut . Then, any solution of (1.1) with f1 = 0 and satisfying ˜ E(0) < E˜ 1 ,

∇u0 2 ≥ α˜ 1 ,

grows up exponentially when t tends to infinity.

(7.6)

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115

Remark 7.3 A refinement of the method introduced in [21] can be used to prove an exponential growth of the norm u(t)p in the absence of the boundary source. Proof of Theorem 7.2 Let us define ˜ H˜ (t) := E˜ 1 − E(t).

(7.7)

Consequently, we have as in (5.9)

k 1

0 < H˜ (0) ≤ H˜ (t) ≤ u(t) k, , 1 k

∀t ≥ 0.

(7.8)

˜ Next, we define the functional L(t) as ˜ = H˜ (t) + ε L(t)

ut udx + ε 

(7.9)

ut ud. 1

Following the same method (up to inequality (5.15)) as in the proof of Lemma 5.3, we arrive at ˜ d L(t) = α∇ut 22 + ut 22 + dt

1

ρ(ut )ut d + 2εut 22 + 2εut 22,1

 

2

u(t) k + 2ε H˜ (t) − 2ε E˜ 1 + ε 1 − k,1 k −ε

ut udx − ε

∇ut ∇udx,

(7.10)

k 2

−∇u22 = 2H˜ (t) − 2E˜ 1 + ut 22 + ut 22,1 − u(t) k, , 1 k

(7.11)



ρ(ut )ud − εα 1



where we have used

instead of (5.14). Using the estimate (7.5), we have    

k

k

2

2 −k



˜ ˜ u(t) k, u(t) k, ≥ 1 − − 2E1 (B2 α˜ 2 ) −2E1 + 1 − 1 1 k k

k

= c˜0 u(t) k, . (7.12) 1

We have c˜0 > 0, since α˜ 2 > arrive at

−k/(k−2) B2 .

˜ d L(t) ≥ α∇ut 22 + ut 22 + dt

1

Thus, inserting (7.12) into (7.10) then we

ρ(ut )ut d + 2εut 22 + 2εut 22,1

k

+ 2ε H˜ (t) + c˜0 ε u(t) k, − ε 1

− εα

∇ut ∇udx. 

ut udx − ε 

ρ(ut )ud 1

(7.13)

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Now, for any λ˜ > 0, Young’s inequality gives



2 ˜ ut udx ≤ λu 2+

1 ut 22 . 4λ˜

(7.14)

Taking into account (5.21), (7.14), (4.1), (5.23) and (5.26) (with d˜ = 1 + 1/H˜ (0) instead of d) then (7.13), takes the form     ˜ ε d L(t) ε ut 22 ≥α 1− ∇ut 22 + 1 + 2ε − ˜ dt 4δ 4λ   q − 1 −q/(q−1) q ut q,1 λ + mq − εMq q   λq ˜ ˜ 2 + 2εut 2,1 + ε 2 − Mq C d H (t) q   q

k

λ 2 2 ˜ + ε c˜0 − Mq C d˜ u(t) k, − ε λu 2 − εαδ∇u2 . (7.15) 1 q Since

 u22 ≤ C∇u22 ≤ C ∇u22 + 2H˜ (t) ,

which gives by exploiting (7.7)  

k 2

u22 + ∇u22 ≤ C1 2E˜ 1 − ut 22 − ut 22,1 + u(t) k, , 1 k

(7.16)

where C1 = C + 1. Inserting (7.16) into (7.15) gives     ˜ ε d L(t) ε ≥α 1− ∇ut 22 + 1 + 2ε − − ε(λ˜ + αδ)C1 ut 22 dt 4δ 4λ˜   q − 1 −q/(q−1) q λ ut q,1 + mq − εMq q    λq ˜ ˜ 2 ˜ + ε 2 − (λ + αδ)C1 ut 2,1 + ε 2 − Mq C d H (t) q   q



λ 2 k + ε c˜0 − Mq C d˜ − (λ˜ + αδ)C1 u(t) k, 1 q k − ε(λ˜ + αδ)C1 2E˜ 1 . Using once again the inequality (7.5), then we arrive at     ˜ d L(t) ε ε 2 ˜ ≥α 1− ∇ut 2 + 1 + 2ε − − ε(λ + αδ)C1 ut 22 dt 4δ 4λ˜   q − 1 −q/(q−1) q λ ut q,1 + ε(λ˜ + αδ)C1 2E˜ 1 + mq − εMq q

(7.17)

Appl Math Optim (2012) 66:81–122

117

   λq + ε 2 − (λ˜ + αδ)C1 ut 22,1 + ε 2 − Mq C d˜ H˜ (t) q    q

k

2 λ + 4E˜ 1 (B2 α˜ 2 )−k u(t) k, . + ε c˜0 − Mq C d˜ − (λ˜ + αδ)C1 1 q k (7.18) At this point, we have to choose our constants carefully in (7.18). Indeed, let us pick λ, λ˜ and δ small enough such that ⎧ 2 − (λ˜ + αδ)C1 > 0, ⎪ ⎪ ⎨ q 2 − Mq λq C d˜ > 0, ⎪ ⎪  q ⎩ c˜0 − Mq λq C d˜ − (λ˜ + αδ)C1 2k + 4E˜ 1 (B2 α˜ 2 )−k > 0. Once the above constants are fixed, we may take ε small enough such that ⎧ ε 1 − 4δ > 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 1 + 2ε − ε − ε(λ˜ + αδ)C1 > 0, ˜ 4λ

−q/(q−1) > 0, ⎪ ⎪ mq − εMq q−1 ⎪ q λ ⎪ ⎪ ⎩ ˜ L(0) > 0.

Consequently, there exists a positive constant η˜ 1 such that the estimate (7.18) becomes ˜

k

 d L(t) ≥ η˜ 1 ut 22 + ut 22,1 + H˜ (t) + u(t) k, + E˜ 1 , 1 dt

∀t > 0.

(7.19)

On the other hand, by Young’s inequality and Poincaré’s inequality, we get, for some γ˜ > 0   ˜ ≤ γ˜ H˜ (t) + ut 22 + ut 22, + ∇u22 . L(t) (7.20) 1 Since H˜ (t) > 0, we have for all t ≥ 0, 1 1 ∇u22 ≤ ukk,1 + E˜ 1 . 2 k Thus, the inequality (7.20) becomes:



  ˜ ≤ η˜ 2 ut 22 + ut 22, + H˜ (t) + u(t) k + E˜ 1 , L(t) k, 1 1

(7.21)

for some η˜ 2 > 0. (7.22)

Finally, combining (7.19) and (7.22), then our desired result holds. This completes the proof of Theorem 7.2.  8 Global Existence In this section, we show that if the interior damping term dominates the interior source (i.e. r ≥ p) and if the boundary damping dominates the boundary source (i.e. q ≥ k),

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then the local solution can be extended to be global in time. Thus, our global existence result reads as follows: Theorem 8.1 Assume that Assumption 4.1 hold. Let 2 ≤ p ≤ p and 2 ≤ k ≤ k. Suppose that p≤r

and

k ≤ q.

(8.1)

Then, the life span T of the solution is T = ∞. Proof Inspired by [20], we introduce the auxiliary function

2 1 1

1 (t) = ut (t) 2 + ut 22,1 + ∇u22 2 2 2

p 1

k 1

+ u(t) p + u(t) k, . 1 p k

(8.2)

Using the fact that $

$ d

u(t) p = $u(t)$p−2 u(t)ut (t)dx, p dt  $

$

d

k $u(t)$k−2 u(t)ut (t)d, u(t) k, = 1 dt 1 and (4.4), we get

2  (t) = −α ∇ut (t) 2 −

φ(ut )ut dx − 

ρ(ut )ut d 1

$ $ $u(t)$p−2 u(t)ut (t)dx + 2

+2 

$ $ $u(t)$k−2 u(t)ut (t)d. 1

Exploiting (4.1) and (4.2), we infer that

2

q

r



 (t) ≤ −α ∇ut (t) 2 − mq ut (t) q, − cr ut (t) r 1

$ $ $u(t)$p−2 u(t)ut (t)dx + 2

+2 

$ $ $u(t)$k−2 u(t)ut (t)d. 1

Hölder’s inequality leads to $ $ $ $ $ $ $ $u(t)$p−2 u(t)ut (t)dx $ ≤ up−1 p ut p , $ $  $ $ $ $ $ $ $ $u(t)$k−2 u(t)ut (t)d $ ≤ uk−1 ut k, . 1 k,1 $ $ 1

Applying Young’s inequality, we get $ $ $ $ $ $ $ $u(t)$p−2 u(t)ut (t)dx $ ≤ C()upp + ut pp $ $ 

(8.3)

Appl Math Optim (2012) 66:81–122

and

$ $ $ $

1

119

$ $ $ $ $u(t)$k−2 u(t)ut (t)d $ ≤ C()uk + ut k . k,1 k,1 $

Inserting the above estimates into (8.3), we get

2

q

r



 (t) ≤ −α ∇ut (t) 2 − mq ut (t) q, − cr ut (t) r 1

p

k



p



k + C() u(t) p +  ut (t) p + C() u(t) k, +  ut (t) k, . (8.4) 1

1

The assumption (8.1) together with first inequality in (5.25) gives









ut (t) p ≤ C ut (t) p = C ut (t) r p/r p r r

r 

≤ C 1 + ut (t) r .

(8.5)

Similarly, we have



ut (t) k

k,1



q  ≤ 1 + ut (t) q, . 1

Plugging (8.5) and (8.6) into (8.4) and choosing  sufficiently small, we get   (t) ≤ C 1 + (t) .

(8.6)

(8.7)

Applying Gronwall’s lemma to (8.7), we deduce that  ∈ L∞ (0, Tmax ). This together with the continuation principle gives our desired result. This completes the proof of Theorem 8.1. 

9 Open Problems For the sake of convenience to the reader, we list here some of the open questions raised in the present work. • In Theorem 3.1 we imposed the condition that sources be “sub-critical,” meaning we imposed a limit on the polynomial growth of sources, which is below the maximum growth allowed for the associated potential well energy to be defined. More ¯ concretely, we imposed the conditions 2 ≤ p ≤ 1 + p2¯ , 2 ≤ k ≤ 1 + k2 , where p and k were the exponents governing the polynomial growth of f1 and f2 , respectively. Potential well theory, on the other hand, should allow us to go up to p = p¯ and ¯ k = k. In [7], Bociu & Lasiecka have proved full Hadamard well-posedness for solutions to the wave equation on a regular bounded domain with nonlinear sources and damping on the boundary and in the interior, where the sources are “supercritical.” Their proof relies on using the damping in order to prevent the sources from causing global non-existence of solutions. It seems plausible that similar techniques could be extended to other models, including ours, but this remains an open

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question. The difficulty of extending this technique may lie in incorporating the dynamics on the boundary into an already delicate interaction between the boundary and the interior sources and damping. • In the proof of Theorem 3.5 we used the perturbation theory for analytic (holomorphic) semigroups, and for this it was necessary to treat the problem as semilinear rather than as a nonlinear monotone problem. As a consequence, the nonlinear damping played no role in the proof other than acting as relatively bounded nonlinear perturbations. For this reason the damping terms had to satisfy certain growth restrictions which were in fact not required for generation of a nonlinear semigroup. Thus, while the smoothing property was gained via perturbation theory for analytic semigroups, the monotone structure of the problem was lost. It remains open to see whether the monotonicity can be exploited to obtain more robust results. In general it is known that if B = ∂ is the subgradient of a proper convex lower semicontinuous functional  on a Hilbert space X, then −B generates a nonlinear semigroup S(t) with the smoothing property, i.e. S(t)x ∈ D(B) for t > 0 and all x ∈ X ([39], Proposition IV.3.2). Suppose also that −A generates an analytic semigroup on H ; this semigroup also has the smoothing property [33]. However, to the best of our knowledge there is no general method for showing that the sum −(A + B) generates a semigroup with the smoothing property, even if A + B is monotone. • It remains open to see whether nonlinear sources can cause finite time blow-up for solutions to the strongly damped wave equation with dynamic boundary conditions, such as we have studied in this paper. This question has been addressed for the wave equation in recent works by Gazzola & Squassina [17] and by Yu [42]. The difficulty in extending these results lies in the interaction between damping and sources in the interior and on the boundary, while also incorporating the dynamic boundary conditions. However, in the particular case where the damping terms φ and ρ are linear functions and for f2 = 0, the question of the blow up in finite time has been solved recently by Gerbi & Said-Houari in [22] by using the concavity method. The same method in [22] can be adapted to extend the result in [22] to the case f2 = 0. • It is an interesting open problem to prove Theorem 7.2 for a nonlinear function φ satisfying (4.2). Acknowledgements The first author wishes to thank the Virginia Space Grant Consortium and the Jefferson Scholars Foundation for their support. The second author wants to thank KAUST for its support. Both authors are very grateful to Prof. Irena Lasiecka for many fruitful discussions.

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