Southeast Asian Bulletin of Mathematics (2000) 24: 183–199
Southeast Asian Bulletin of Mathematics c
Springer-Verlag 2000
On the Existence and the Uniform Decay of a Hyperbolic Equation with Non-Linear Boundary Conditions M.M. Cavalcanti, V.N. Domingos Cavalcanti, and J.A. Soriano Department of Mathematics, University Estadual de Maringá, 87020-900 Maringá-PR, Brazil E-mail:
[email protected]
L.A. Medeiros Instituto de Matemática, UFRJ, CP. 68530, 21945-970 Rio de Janeiro, Brazil
AMS Subject Classification (2000): 35L05, 35L70, 35B40 Abstract. In this paper, we study a hyperbolic model based on the equation ytt − 1y +
n X j =1
bj (x, t)
∂yt =0 ∂xj
with nonlinear boundary conditions given by ∂y + f (y) + g(yt ) = 0. ∂ν We prove the existence and the uniqueness of global solutions. Also, we obtain the uniform decay of the energy without control of its derivative sign. Keywords: Galerkin approximation, uniform decay, non-linear boundary conditions, potentials
1. Introduction This paper studies the existence, uniqueness, and the uniform decay of strong solutions of the problem n X ∂yt − 1y + bj (x, t) = 0 in × (0, ∞) y tt ∂xj j =1 y = 0 on 01 × (0, ∞) (∗) ∂y + f (y) + g(yt ) = 0 on 00 × (0, ∞) ∂ν y(x, 0) = y 0 (x); yt (x, 0) = y 1 (x) in ,
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where is a bounded domain of R n , n ≥ 1, with smooth boundary 0 = 00 ∪ 01 . Here, 00 and 01 are closed and disjoint and ν is the unit normal vector pointing towards the exterior of , and bj , f, g are, respectively, the coefficient and nonlinear functions satisfying some general properties (see assumptions (A.1)–(A.3) below). In recent years, significant progress has been made concerning stabilization, dissipation, and controllability of evolution distributed systems. New techniques were developed which allow us to stabilize a system through its boundary as well as to control it from an initial to a final state. This new tendency of stabilizing or controling a system through its boundary received a strong impulse after the introduction of the Hilbert Uniqueness Method (HUM) developed by Lions [17]. There is not much literature on the existence and asymptotic behavior of evolution equations with nonlinear boundary conditions. We refer to the following works: Chen and Wong [5], Lagnese and Leugering [11], Zuazua [22], You [21], Cipollati, Machtyngier, and San Pedro Siqueira [6], Lasiecka and Tataru [15], and Favini, Horn, Lasieska, and Tataru [7], among others. The objective of this paper is to prove the existence and uniqueness of strong solutions to problem (∗) and, moreover, to show that those solutions decay to zero uniformly when t goes to infinity, under the assumption that a nonlinear feedback and an implicit damping coming from the potentials bj occur on the boundary. The proof of the existence is based on the Galerkin method. However, this method introduces technical difficulties which can be bypassed by considering a change of variables which transforms (∗) into an equivalent problem with trivial initial data. To obtain the asymptotic behavior, we use the perturbed energy method developed by Komornik and Zuazua [8] and construct a suitable Lyapunov functional which allows us to show that the energy Z Z 1 1 2 |y(x, t)| dx + |∇y(x, t)|2 dx (1.1) E(t) = 2 2 decays to zero exponentially when t goes to infinity. Our paper is organized as follows. In Sec. 2, we shall present some notations which will be used throughout this paper and will state the main result. In Sec. 3, we will prove the existence and uniqueness of strong solutions, and in Sec. 4, we will obtain the asymptotic behavior mentioned above.
2. Assumptions and Main Results For the rest of this article, x 0 will be a fixed point in R n . Then consider m(x) = x − x 0 ,
(2.1)
and assume that the partition {00 , 01 } of the boundary 0 with 0 0 ∩ 0 1 = ∅ has the form 00 = {x ∈ 0 : m(x) · ν(x) ≥ m0 > 0} and 01 = {x ∈ 0 : m(x) · ν(x) < 0} , (2.2) where m0 is a constant. Define the following: Z Z 2 |u(x)|2 dx, u(x)v(x) dx, |u| = (u, v) =
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185
Z (u, v)00 =
00
Z u(x)v(x) dx, |u|200 =
00
|u(x)|2 dx,
||u||∞ = ess sup ||u(t)||L∞ () , t≥0
V = u ∈ H () : u = 0 on 01 , 1
and R(x 0 ) = max kx − x 0 k. x∈
(2.3)
Next, we state the general hypotheses. A.1. Assumptions on the coefficients. The following assumptions are made on the coefficients: ∂bj ∈ L1 0, ∞; L∞ () ; j = 1, · · · , n. (H.1) bj ∈ W 1,∞ 0, ∞; C 1 () , ∂xj Also, assume that there exists a positive constant δ such that b · ν ≥ δ > 0 in 00 × (0, ∞),
(H.2)
where b = (b1 , . . . , bn ). A.2. Assumptions on f . f : R → R is a C 1 function such that f (s)s ≥ 0 for s ∈ R.
(H.3)
|f (s)| ≤ C0 (1 + |s|γ +1 ); s ∈ R,
(H.4)
There exists C0 > 0 such that where 0 < γ ≤ 1/(n − 2); if n ≥ 3 or γ > 0 if n = 1, 2, and 0 f (s) ≤ C0 (1 + |s|γ ); s ∈ R. Defining
Z F (s) =
s
(H.5)
f (λ) dλ,
0
there exist α, C > 0 such that
and
C |s|γ +2 ≤ F (s) ≤ α sf (s); s ∈ R
(H.6)
γ γ f (ξ ) − f (ξˆ ) (η − η) ˆ ≥ −D1 ( ξ + ξˆ ) ξ − ξˆ η − ηˆ
(H.7)
for some D1 > 0 and for all ξ, ξˆ , η, ηˆ ∈ R. A.3. Assumptions on g. g : R → R is a non-decreasing C 1 function such that g(s)s ≥ 0.
(H.8)
C1 |s|ρ+2 ≤ g(s)s ≤ C2 |s|ρ+2 ; s ∈ R
(H.9)
There exist C1 , C2 > 0 such that where 0 < ρ ≤ 1/(n − 2) if n ≥ 3 or ρ > 0 if n = 1, 2.
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Remark. The functions f (s) = |s|γ s and g(s) = |s|ρ s satisfy all the above conditions. A.4. Assumptions on the initial data. We consider the initial data verifying {y 0 , y 1 } ∈ V × H 2 () and
∂y 0 + f (y 0 ) + g(y 1 ) = 0 on 00 . ∂ν
(H.10)
Remark. We observe that according to the choice of γ and ρ, one has V ,→ L2(γ +1) (00 ) and V ,→ L2(ρ+1) (00 ). Consequently, the assumptions (H.4) and (H.9) imply that f (y 0 ) and g(y 1 ) belong to L2 (00 ). To obtain the exponential decay of the energy we make the following assumptions: A.5. Assumptions related with the uniform decay. Assume that bi mj = bj mi in × (0, ∞) for all i, j ∈ {1, . . . , n},
(H.11)
|f (s)| ≤ C0 |s|γ +1 , s ∈ R
(H.12)
k1 |s| ≤ g(s)s ≤ k2 |s| , s ∈ R
(H.13)
2
2
for some k1 , k2 > 0. Remark. In order to give an example of a function which verifies the assumptions (H.2) and (H.11) simultaneously, we can consider b(x, t) = m(x) θ (t); (x, t) ∈ × (0, ∞), where θ(t) is a regular function with θ (t) ≥ θ0 > 0 and kθ k∞ , kθ 0 k∞ are sufficiently small. However, in the unidimensional case, assumption (H.11) is unnecessary. Assumptions (H.11)–(H.13) are needed since we use the multiplier technique combined with a suitable Lyapunov functional defined in (4.5) below. Now, we in are a position to state our main result. Theorem 2.1 Under Assumptions (A.1)–(A.4), problem (∗) possesses a unique strong solution, y : × (0, ∞) → R, such that 0 ∞ 00 ∞ 2 (2.4) y ∈ L∞ loc (0, ∞; V ), y ∈ Lloc (0, ∞; V ) and y ∈ Lloc 0, ∞; L () . Moreover, assume that (A.5) holds and kbk∞ , kb0 k∞ , and kdiv(b)k∞ are sufficiently small. Then the energy Z Z Z 0 1 y (x, t) 2 dx + 1 |∇y(x, t)|2 dx + E(t) = F y(x, t) dx (2.5) 2 2 00 associated with the strong solution y decays exponentially, that is, for some positive constants C and γ , one has E(t) ≤ C exp(−γ t); for all t ≥ 0.
(2.6)
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3. Existence of Solutions In this section we prove the existence and uniqueness of the strong solutions of (∗). A variational formulation of problem (∗) leads to the equation y 00 (t), w + ∇y(t), ∇w + f (y), w 00 + g(y 0 ), w
00
+
n X
bj (t)
j =1
∂y 0 (t), w = 0 ∂xj
for all w ∈ V . Strong solutions to problem (∗) with nonlinear boundary conditions cannot be obtained by making use of the “special basis” of eigen-functions of the operator −1. Differentiating the above expression with respect to t does not help because of the technical difficulties when estimating y 00 (0). In order to avoid these difficulties, we transform (∗) into an equivalent problem with initial value equal to zero. In fact, by changing variables v(x, t) = y(x, t) − φ(x, t), (3.1) where φ(x, t) = y 0 (x) + ty 1 (x), we obtain
n X ∂y 0 00 − 1v + b = Fˆ in × (0, ∞) v j ∂xj j =1 v = 0 on 01 × (0, ∞) ∂v + f (v + φ) + g v 0 + φ 0 = G on 00 × (0, ∞) ∂ν v(0) = v 0 (0) = 0 in ,
where Fˆ = 1φ −
n X j =1
bj
∂φ ∂φ 0 and G = − . ∂xj ∂ν
(3.2)
(3.3)
Note that if v is a solution of (3.2) on [0,T], then y = v + φ is a solution of (∗) in the same interval. From the estimates obtained below, we are now able to prove that, for any T > O, 2 |1v(t)|2 + ∇v 0 (t) ≤ C(T ), for all t ∈ [0, T ]. From (3.1) we obtain the same inequality given above for the solution y. Then by using standard methods, we can extend y to the interval (0, ∞). Hence, it is sufficient to prove that (3.2) has a solution in [0,T], which can be done by employing the Galerkin method. Let (ων )ν∈N be a Hilbertian basis in V ∩ H 2 () (c.f. [2]), Vm the subspace generated by ω1 , . . . , ωm , and let m X gj m (t)ωi ∈ Vm vm (t) = i=1
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be the solution of the Cauchy problem 00 (t), w + ∇vm (t), ∇w + f (vm (t) + φ(t)) , w 00 vm 0 g(vm (t) + φ 0 (t)), w
00
+
n X
bj (t)
j =1
= Fˆ (t), w + G(t), w
00
0 ∂vm (t), w ∂xj
(3.4)
; ∀w ∈ Vm ,
0 (0) = 0, vm (0) = vm
which has a local solution in some interval [0, tm ). The extension of these solutions to the interval [0, T ] is a consequence of the first estimate we obtain below. 3.1. A Priori Estimates 3.1.1. The First Estimate 0 (t) in (3.4), we obtain Considering w = vm
d dt +
2 1 1 0 v (t) + |∇vm (t)|2 + 2 m 2
0 g(vm
+φ
0
0 ), vm
+φ
0
00
+
Z 00
F (vm + φ) d0
n X j =1
∂v 0 0 (t) bj (t) m (t), vm ∂xj
0 0 (t) + G(t), vm (t) 00 = Fˆ (t), vm + f (vm + φ) , φ 0
00
(3.5)
0 + g(vm + φ 0 ), φ 0 (t)
00
.
Next, we estimate some terms in (3.5). 0 /∂x )(t), v 0 (t) . Applying the Gauss formula, we Estimate for I1 := 6jn=1 bj (t)(∂vm j m deduce Z Z 0 2 0 2 1 1 d0. div(b) vm dx + (3.6) I1 = − (b · ν) vm 2 2 00 Estimate for I2 = f (vm + φ) , φ 0 0 . From assumption (H.4), noting that 1/ (γ + 2)/ 0 (γ + 1) + 1/(γ + 2) = 1 and employing Young’s inequality, we infer Z |I2 | ≤ C0
00
(1 + |vm + φ|γ +1 ) φ 0 d0
Z
≤ k1 +
00
|vm + φ|γ +1 φ 0 d0
γ +2 γ +2 ≤ k1 + k2 ||vm (t) + φ(t)||γ +2,00 + k2 φ 0 (t) γ +2,00 , where k1 and k2 are positive constants.
(3.7)
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2(γ +1) (0 ) ,→ Lγ +2 (0 ) and therefore, the norms Remark. We observe that 0 0 V0 ,→ L ||vm (t) + φ(t)||γ +2,00 , φ (t) γ +2,0 make sense. 0
0 + φ 0 ), φ 0 (t) . In a similar way, considering assumption (H.9) Estimate for I3 := g(vm 00 and taking Young’s inequality into account, we have Z 0 v + φ 0 ρ+1 φ 0 d0 |I3 | ≤ C2 (3.8) m 00
ρ+2 ρ+2 0 (t) + φ 0 (t) ρ+2,00 + k3 (η) φ 0 (t) ρ+2,00 , ≤ η vm where η is an arbitrary positive constant. 0 (0) = 0, and taking (H.2), (H.6), Integrating (3.5) over (0, t), noting that vm (0) = vm (H.9), (3.6), (3.7), and (3.8) into account, the result is 2 1 1 0 γ +2 vm (t) + |∇vm (t)|2 + C ||vm (t) + φ(t)||γ +2,00 (3.9) 2 2 Z t Z t 0 v (s) + φ 0 (s) ρ+2 ds + δ − η v 0 (s) 2 ds +(C1 − η) m m ρ+2,00 00 2 0 0 Z Z Z t 2 0 2 1 0 1 t γ +2 dx ds, ≤ k0 + vm (s) + k2 ||vm (s) + φ(s)||γ +2,00 ds + div(b) vm 2 2 0 0 where k0 is a positive constant. Choosing η > 0 sufficiently small and employing Gronwall’s lemma, from (3.9) we obtain the first estimate 0 v (t) 2 + |∇vm (t)|2 + ||vm (t) + φ(t)||γ +2 (3.10) m γ +2,00 Z
t
+ 0
0 v (s) + φ 0 (s) ρ+2 ds + m ρ+2,0 0
Z
t 0
0 v (s) 2 ds ≤ L1 , m 0 0
where L1 is a positive constant independent of m ∈ N and t ∈ [0, T ]. 3.1.2. The Second Estimate 00 (0) in (3.4) and First of all, we will estimate y 00 (0) in L2 -norm. Considering w = vm 0 (0) = 0, from (3.3), it follows that observing that vm (0) = vm
00 2 v (0) + f (y 0 ), v 00 (0) + g(y 1 ), v 00 (0) m m m 0 0 0
0
n X ∂y 1 00 ∂y 0 00 00 = 1y 0 , vm (0) − , vm (0) + − . , vm (0) bj (0) ∂xj ∂ν 00 j =1
Making use of assumption (H.10) and from (3.11), we conclude that ! n 1 0 X 00 2 ∂y v (0) ≤ 1y + bj (0) v 00 (0) . m m ∂xj j =1
(3.11)
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Consequently, there exists L2 > 0 such that 00 v (0) ≤ L2 ; ∀m ∈ N. m
(3.12)
Now, taking the derivative of (3.4) with respect to t, we obtain 000 0 vm (t), w + ∇vm (t), ∇w + Z +
00
g
0
0 (vm
+φ
0
00 )vm w d0
+
Z 00
0 f 0 (vm + φ)(vm + φ 0 )w d0
(3.13)
X n 0 00 ∂vm ∂vm 0 (t), w + (t), w bj (t) bj (t) ∂xj ∂xj j =1
n X j =1
= Fˆ 0 (t), w + G0 (t), w 00 . 00 (t) in (3.13), we obtain Substituting w = vm
d dt Z +
00
g
0
Z 2 1 00 2 1 0 0 00 f 0 (vm + φ)(vm + φ 0 )vm d0 v (t) + ∇vm (t) + 2 m 2 00
0 (vm
(3.14)
X n n 0 00 X 00 2 ∂vm ∂vm 0 00 00 + φ ) vm d0 + (t), vm (t) + (t), vm (t) bj (t) bj (t) ∂xj ∂xj 0
j =1
00 00 (t) + G0 (t), vm (t) = Fˆ 0 (t), vm
j =1
00
.
Next, we estimate the terms in (3.14). 00 /∂x )(t), v 00 (t) . Applying the Gauss formula, we Estimate for I4 := 6jn=1 bj (t)(∂vm j m deduce Z Z 00 2 00 2 1 dx + 1 d0. div(b) vm (3.15) I4 = − (b · ν) vm 2 2 00 Estimate for I5 := we obtain
R
0 +φ 0 )v 00 d0. Taking assumption (H.5) into account, f 0 (vm +φ)(vm m Z 00 0 d0. |I5 | ≤ + φ 0 vm 1 + |vm + φ|γ vm
00
00
Now, noting that 1/ 2(γ + 1)/γ + 1/ 2(γ + 1) + 1/2 = 1 and also that V ,→ L2(γ +1) (00 ), then making use of the inequality ab ≤ (1/4η)a 2 + ηb2 (η > 0 arbitary) and from the generalized Hölder inequality, we obtain |I5 | ≤
2 00 2 1 0 (t) 00 vm (t) + φ 0 (t) 00 + η vm 4η
γ
0 00 + vm (t) + φ(t) 2(γ +1),00 vm (t) + φ 0 (t) 2(γ +1),00 vm (t) 00 , that is,
0 2 00 2 |I5 | ≤ k4 (η) ∇vm (t) + ∇φ 0 (t) + η vm (t) 00 00 0 +k5 |∇vm (t) + ∇φ(t)|γ ∇vm (t) + ∇φ 0 (t) vm (t) 0 . 0
(3.16)
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The last inequality and the first estimate yield 0 2 00 2 |I5 | ≤ k6 (η) ∇vm (t) + ∇φ 0 (t) + 2η vm (t) 00 ,
(3.17)
where k6 (η) is a positive constant which depends on η. Combining (H.2), (3.14), (3.15), (3.17) and noting that g 0 (s) ≥ 0, we deduce 2 00 2 δ d 1 00 2 1 0 (t) 0 vm (t) + ∇vm (t) + − 3η vm 0 dt 2 2 2 Z 00 2 1 dx + k6 (η) ∇v 0 (t) + ∇φ 0 (t) 2 div(b) vm ≤ m 2 2 00 2 1 0 2 1 0 2 0 + k7 ∇vm (t) + vm (t) + Fˆ (t) + G (t) 00 , 2 2
(3.18)
where k7 is a positive constant which comes from the estimate 2 00 2 0 0 00 (t), vm (t) ≤ k7 ∇vm (t) + vm (t) t . b0 (t) · ∇vm Integrating (3.18) over (0, t), choosing η sufficiently small, taking (3.12) into account, and employing Gronwall’s lemma, we arrive at the second estimate 00 2 0 v (t) + ∇v (t) 2 + m m
Z
00 2 v (s) ds ≤ L3 , m 00
t
0
(3.19)
where L3 is a positive constant independent of m ∈ N and t ∈ [0, T ]. 3.2
Analysis of the Nonlinear Terms
In order to pass to the limit in the approximate problem (3.4), we need to obtain estimates for the nonlinear boundary terms on the left side of (3.4) containing f and g. Note that they are not included in the estimates (3.10) and (3.19). Thus, we will obtain it directly here. In fact, from the first estimate and considering the assumptions (H.54 and (H.9), we obtain Z
T
00
0
Z
T 0
Z
Z 00
|f (vm + φ)|
(γ +2)/(γ +1)
Z
T
d0 dt ≤ C1
0 g(v + φ 0 ) (ρ+2)/(ρ+1) d0 dt ≤ C1 m
00
0
Z 0
Z
T
Z 00
1 + |vm + φ|γ +2 d0 dt ≤ L, 0 ρ+2 (1 + vm + φ 0 )d0 dt ≤ L,
where L > 0. Then taking 60,T = 00 × (0, T ), we have f (vm + φ) → χ weakly in L(γ +2)/(γ +1) 60,T ,
(3.20)
0 + φ 0 ) → η weakly in L(ρ+2)/(ρ+1) 60,T . g(vm
(3.21)
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From the first and the second estimates and noting that ||v||H 1/2 (00 ) ≤ C |∇v| ; for all v ∈ V , we deduce the following: {vm } is bounded in L2 0, T ; H 1/2 (00 ) , 0 vm is bounded in L2 0, T ; H 1/2 (00 ) , 00 vm is bounded in L2 0, T ; L2 (00 ) .
(3.22) (3.23) (3.24)
Noting that the imbedding H 1/2 (00 ) ,→ L2 (00 ) is continuous and compact, from the Aubin–Lions theorem (c.f. [17, Theorem 5.1]), we obtain (3.25) vm → v strongly in L2 0, T ; L2 (00 ) , 0 vm → v 0 strongly in L2 0, T ; L2 (00 ) .
(3.26)
vm → v a.e. in 60,T ,
(3.27)
0 vm
(3.28)
Therefore, → v a.e. in 60,T .
Thus, from (3.20), (3.21), (3.27), (3.28) and thanks to Lions’s lemma (see [17, Lemma 1.3]), we conclude that (3.29) f (vm + φ) → f (v + φ) weakly in L(γ +2)/(γ +1) 60,T , 0 + φ 0 ) → g(v 0 + φ 0 ) weakly in L(ρ+2)/(ρ+1) 60,T . g(vm
(3.30)
We observe that in view of (H.5) and (H.10) and since V ,→ L2(γ +1) (00 ) and V ,→ L2(ρ+1) (00 ) we have from the first estimate Z
T
T 0
|f (vm + φ)| d0dt ≤ C 2
00
0
Z
Z
Z Z 00
0 g(v + φ 0 ) 2 d0dt ≤ m
T
Z 00
0
Z
T
0
Z 00
1 + |vm + φ|2(γ +1) d0dt ≤ L,
(3.31)
2(γ +1) 0 + φ0 1 + vm d0dt ≤ L.
(3.32)
Then, from (3.29)–(3.32), we have f (vm + φ) → f (v + φ) weakly in L2 0, T ; L2 (00 ) ,
(3.33)
0 g(vm + φ 0 ) → g(v 0 + φ 0 ) weakly in L2 0, T ; L2 (00 ) ,
(3.34)
which is enough to pass to the limit in the approximate problem given by (3.4). Using standard arguments, we obtain y 00 − 1y +
n X j =1
bj (x, t)
∂y 0 = 0 in L2loc 0, ∞; L2 () , ∂xj
(3.35)
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193
y(0) = y 0 ; y 0 (0) = y 1 . Also, from the generalized Green formula, we deduce ∂y (3.36) + f (y) + g(y 0 ) = 0 in L2loc 0, ∞; L2 (00 ) . ∂ν Since ∂y/∂ν = −f (y)−g(y 0 ) ∈ L2loc 0, ∞; L2 (00 ) and 0 0 ∩0 1 = ∅, the regularity of elliptic problems implies that y ∈ L2loc (0, ∞; H 3/2 () ∩ V ).
(3.37)
3.3. Uniqueness Let y1 and y2 be solutions to problem (∗). Then, defining z = y1 − y2 , from (3.35) and (3.36) we obtain n X ∂z0 (t), w bj (t) z00 (t), w + ∇z(t), ∇w + ∂xj j =1
= f (y2 ) − f (y1 ), w
00
+ g(y20 ) − g(y10 ), w
00 .
Substituting w = z0 (t) in the above equality and observing that g is non-decreasing, it results that d dt
X n 1 0 2 1 ∂z0 2 0 z (t) + |∇z(t)| + (t), z (t) bj (t) 2 2 ∂xj
(3.38)
j =1
≤ f (y2 ) − f (y1 ), z0 (t)
00
.
From assumptions (H.2), (H.7), and (3.38) it follows that
2 1 0 2 1 δ 2 z (t) + |∇z(t)| + z0 (t) 0 0 2 2 2 Z Z 0 2 |y2 |γ + |y1 |γ |z(t)| z0 (t) d0. ≤ div(b) z dx + D1 d dt
(3.39)
00
Using analogous arguments like those used in the second estimate, we obtain d dt
2 1 0 2 1 δ z (t) + |∇z(t)|2 + − η z0 (t) 0 0 2 2 2 2 ≤ C z0 (t) + |∇z(t)|2 ,
where η is arbitrary positive constant. Integrating the last inequality over (0, t) and making use of Gronwall’s lemma, we t u conclude that |z0 (t)| = |∇z(t)| = 0. This concludes the first part of the proof.
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4. Uniform Decay From (3.35) and (3.36), a simple computation gives us Z Z Z 2 2 1 1 g(y 0 )y 0 d0 + div(b) y 0 dx − E 0 (t) = − (b · ν) y 0 d0, 2 2 00 00 where
1 E(t) = 2
Z
1 |y | dx + 2 0 2
Z
Z
|∇y|2 dx +
00
F (y) dx
is the energy related to problem (∗). From assumptions (H.2) and (H.13), we deduce Z Z 0 2 2 δ y d0 + 1 div(b) y 0 dx. E 0 (t) ≤ − k1 + 2 00 2 Let λ > 0 be a constant such that Z Z 2 |v| dx ≤ λ |∇v|2 dx; for all v ∈ V .
(4.1)
(4.2)
(4.3)
For an arbitrary ε > 0, define the perturbed energy by Eε (t) = E(t) + εψ(t), where
ψ(t) = 2 y 0 (t), m · ∇y(t) + (n − 1) y 0 (t), y(t) + b(t) · ∇y(t), m · ∇y(t) + (n − 1) b(t) · ∇y(t), y(t) .
(4.4) (4.5)
It is easy to see that there exists a positive constant θ1 such that |Eε (t) − E(t)| ≤ εθ1 E(t).
(4.6)
Next, we are proving that there exists θ2 > 0 such that Eε0 (t) ≤ −θ2 E(t).
(4.7)
Indeed, differentiating each term in (4.5) with respect to t substituting y 00 = 1y − in the expression obtained and taking the assumption (H.11) into account yields the result (4.8) ψ 0 (t) = 2 1y(t), m · ∇y(t) + 2 y 0 (t), m · ∇y 0 (t) + (n − 1) 1y(t), y(t)
6jn=1 bj (∂y 0 /∂xj )
2 +(n − 1) y 0 (t) + b0 (t) · ∇y(t), m · ∇y(t) + (n − 1) b(t) · ∇y(t), y 0 (t) . Next, we will estimate some terms on the right-hand side of (4.8). Estimate for I1 := 2 1y(t), m ·∇y(t) . Using Green’s and Gauss’ theorems, we deduce Z Z ∂y 2 2 I1 = (n − 2) |∇y(t)| − (m · ν) |∇y| d0 + 2 (4.9) (m · ∇y) d0. ∂ν 0 0
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Estimate for I2 := 2 y 0 (t), m · ∇y 0 (t) . By Gauss’ theorem, we obtain Z m · ∇ y 02 dx I2 =
2 = −n y 0 (t) +
Z 00
(4.10)
2 (m · ν) y 0 d0.
Estimate for I3 := (n − 1) 1y(t), y(t) . Making use of Green’s formula, noting that ∂y = −f (y) − g(y 0 ) on 00 ∂ν and considering assumption (H.6), it follows that Z n−1 F (y) d0 − (n − 1) g(y 0 ), y 0 . I3 ≤ −(n − 1) |∇y|2 − 0 α 00
(4.11)
Then, combining (4.8)–(4.11) and putting L = min{1/α, 1}, we infer ψ 0 (t) ≤ −LE(t) + b0 (t) · ∇y(t), m · ∇y(t) + (n − 1) b(t) · ∇y(t), y 0 (t) (4.12) Z Z ∂y 2 − (m · ν) |∇y| d0 + 2 (m · ∇y) d0 ∂ν 0 0 Z 2 + (m · ν) y 0 d0 − (n − 1) g(y 0 ), y 00 . 00
But from the Cauchy–Schwarz inequality, we have 1/2 b0 (t) · ∇y(t), m · ∇y(t) ≤ 2 b0 ∞ R 1/2 (x 0 )E(t),
(4.13)
1/2 (n − 1) b(t) · ∇y(t), y 0 (t) ≤ (n − 1) b ∞ E(t),
(4.14)
where ||b||∞ =
n X j =1
2 ess sup bj (t) L∞ () . t≥0
Then the inequalities (4.12)–(4.14) yield 1/2 1/2 ψ 0 (t) ≤ −LE(t) + 2 b0 ∞ R 1/2 (x 0 )E(t) + (n − 1) b ∞ E(t) Z
Z
∂y (m · ∇y) d0 ∂ν 0 0 Z 2 (m · ν) y 0 d0 − (n − 1) g(y 0 ), y 00 . +
−
(m · ν) |∇y|2 d0 + 2
00
On the other hand, ∂y/∂xk = (∂y/∂ν)νk on 01 implies ∂y m · ∇y = (m · ν) ∂ν
and |∇y| = 2
∂y ∂ν
2 on 01 .
(4.15)
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Consequently, we obtain Z −
Z
0
Z 2
0
(m · ν) |∇y|2 d0 = − ∂y (m · ∇y) d0 = − 2 ∂ν
Z 00
(m · ν) |∇y|2 d0 −
01
(m · ν) Z
Z 00
f (y) (m · ∇y) d0 − 2
00
∂y ∂ν
2 d0, (4.16)
g(y 0 ) (m · ∇y) d0 (4.17)
∂y 2 (m · ν) d0. +2 ∂ν 01 R By combining (4.15)–(4.17) and noting that 01 (m · ν) (∂y/∂ν)2 d0 ≤ 0, it follows that 1/2 1/2 ψ 0 (t) ≤ −LE(t) + 2 b0 ∞ R 1/2 (x 0 ) + (n − 1) b ∞ E(t) (4.18) Z Z Z (m · ν) |∇y|2 d0 − 2 f (y) (m · ∇y) d0 − 2 g(y 0 ) (m · ∇y) d0 −
Z
00
Z +
00
00
00
2 (m · ν) y 0 d0 − (n − 1) g(y 0 ), y 00 .
Next, we estimate some terms on the right-hand side of (4.18). Estimate for I4 := (n − 1) g(y 0 ), y inequality ab (η > 0 arbitrary), we obtain |I4 | ≤
00
. From assumption (H.13) and making use of the ≤ (1/4η)a 2 + ηb2
(n − 1)2 C02 k2 0 2 y (t) 0 + 2ηE(t), 0 4η
(4.19)
where C0 is a positive constant such that |v|00 ≤ C0 |∇v| . R Estimate for I5 := −2 00 f (y) (m · ∇y) d0. From assumption (H.12) and making use of analogous considerations as in (4.19), we deduce 2(γ +1) C0 R(x 0 ) |I5 | ≤ y(t) 2(γ +1),00 + η η C0 k0 R(x 0 ) |∇y(t)|2(γ +1) + η ≤ η ≤
Z 00
|∇y|2 d0
(4.20)
Z
2γ +1 C0 k0 R(x 0 ) [E(t)]γ +1 + η η
00
|∇y|2 d0
Z 00
|∇y|2 d0,
where k0 > 0 comes from the imbedding V ,→ L2(γ +1) (00 ). Estimate for I6 := 2 we can write
R
g(y 0 ) (m · ∇y) d0. In the same way, from assumption (H.13), Z 2 |∇y|2 d0. |I6 | ≤ k22 R(x 0 ) y 0 (t) 0 + η (4.21) 00
0
00
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Thus, from (4.18)–(4.21) and noting that m · ν ≥ m0 > 0 on 00 , we arrive at 1/2 1/2 ψ 0 (t) ≤ −LE(t) + 2 b0 ∞ R 1/2 (x 0 ) + (n − 1) b ∞ E(t)
(4.22)
Z
2γ +1 C0 k0 R(x 0 ) [E(t)]γ E(t) η 00 !Z 2C2k 0 2 (n − 1) 2 0 y d0. + R(x 0 ) + + k22 R(x 0 ) 4η 00 |∇y|2 d0 +
+2ηE(t) − (m0 − 2η)
Suppose, for a moment, that there exists M > 0 such that |E(t)| ≤ M; for all t ≥ 0.
(4.23)
Then, from (4.22) and (4.23), we can write ψ 0 (t) ≤ − L − (N + 2η + J ) E(t) Z −(m0 − 2η) where
00
Z |∇y|2 d0 + K(η)
00
(4.24)
0 2 y d0,
1/2 1/2 N = 2 b0 ∞ R 1/2 (x 0 ) + (n − 1) b ∞ , (n − 1)2 C02 k2 + k22 R(x 0 ), 4η
K(η) = R(x 0 ) + J (η) =
2γ +1 C0 k0 R(x 0 )M . η
Choosing N, η and C0 sufficiently small, we obtain ψ 0 (t) ≤ −BE(t) + K
Z 00
0 2 y d0,
(4.25)
where B is a positive constant. Combining (4.2), (4.4), and (4.25), we obtain Eε0 (t) ≤ −
k1 +
δ 2
Z −ε
00
0 2 y d0 − (εB − ||div(b)||∞ ) E(t).
Choosing ε < k1 + (δ/2) and considering ||div(b)||∞ sufficiently small, we conclude the desired in (4.7). To conclude the proof, we still need to prove (4.23). Indeed, from (4.2), we obtain E 0 (t) ≤ 4(t)E(t), where 4(t) =
n X j =1
||bj (t)||L∞ () .
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Then
d dt
Z t E(t) exp − 4(s) ds ≤ 0. 0
Integrating the last inequality over (0, t), we deduce Z E(t) ≤ exp
∞
4(s) ds
E(0) = M.
0
Combining (4.6) and (4.7), we obtain the uniform decay. The proof is now complete. t u Further remarks. We have just proved the existence and uniform decay of strong solutions to problem (∗) when y 0 and y 1 are smooth. Now, if f = 0 or if f is a linear function, by a density argument and a procedure analogous to the one in the uniqueness, we can extend our results for weak solutions. Indeed, given y 0 , y 1 in V × L2 (), it is sufficient to consider 0 yµ ⊂ H01 () ∩ H 2 () such that yµ1 → y 1 in L2 ()
and yµ1
∂u ⊂ D(−1) = u ∈ V ∩ H (); = 0 on 00 , such that ∂ν 2
yµ1 → y 1 in V . Then repeating the same arguments used in the first estimate and the uniqueness of strong solutions, we obtain a sequence yµ : Q → R of strong solutions of the problem ytt − 1y + b · ∇yt = 0 in × (0, ∞) y = 0 on 01 × (0, ∞) ∂y + g(yt ) = 0 on 00 × (0, ∞) ∂ν y(0) = y 0 ; yt (0) = y 1
(P)
and a function y : Q → R such that yµ → y in C 0 ([0, T ]; V ) yµ0 → y 0 in C 0 [0, T ]; L2 () , yµ0 → y 0 in L2 0, T ; L2 (00 ) , g(yµ0 ) → g(y 0 ) weakly in L(ρ+2)/(ρ+1) (60,T ). The above convergences are sufficient to pass to the limit in order to obtain a weak solution to problem (P) as well as to obtain the energy decay given by (2.6). Acknowledgement. The authors would like to thank the referee for his helpful hints which resulted in this revised version.
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