2004-Fez conference on Differential Equations and Mechanics Electronic Journal of Differential Equations, Conference 11, 2004, pp. 175–185. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp)

DECAY ESTIMATES FOR SOLUTIONS OF SOME SYSTEMS FOR ELASTICITY WITH NONLINEAR BOUNDARY FEEDBACK

NAJI YEBARI

Abstract. We study the energy decay rate for the Euler-Bernoulli and Kirchhoff plate equations. Under suitable growth assumptions on the nonlinear dissipative boundary feedback functions, we obtain some new results.

1. Introduction and main results

We consider the following Euler-Bernoulli beam equation with nonlinear boundary feedback controls:

y_{tt} + y_{xxxx} = 0 in (0, 1), t > 0,
y(0, t) = y_{x}(0, t) = 0 (clamped at x = 0), t > 0,
−y_{xx}(1, t) = h(y_{xt}(1, t)) (moment), t > 0,
y_{xxx}(1, t) = g(y_{t}(1, t)) (force), t > 0,
y(., 0) = y_{0} ∈ H_{E}^{2}, y_{t}(., 0) = y_{1} ∈ L^{2}(0, 1),

(1.1)

where H_{E}^{2} = {u ∈ H^{2}(0, 1); u(0) = u′(0) = 0}, x stands for the position and t the time. The flexural rigidity of the beam and the mass density are assumed to be equal to one. One end is controlled by a point force and point bending moment which are assumed to be nonlinear functions of the observation. By observation we mean velocity and angular velocity of the transversal deflexion at the end.

Throughout this paper, g and h are assumed to satisfy the following hypothesis

g ∈ C^{0}(ℝ), nondecreasing, g(0) = 0, g(s)s > 0 ∀s ≠ 0 ,  (1.2)
h ∈ C^{0}(ℝ), nondecreasing, h(0) = 0, h(s)s > 0 ∀s ≠ 0 .  (1.3)

The total energy of system (1.1) is

Ẽ(t) = \frac{1}{2} ∫_{0}^{1}(y_{t}^{2}(x, t) + y_{xx}^{2}(x, t))dx.  (1.4)

Formally,

\frac{d}{dt}Ẽ(t) = −y_{t}(1, t) g(y_{t}(1, t)) − y_{xt}(1, t) h(y_{xt}(1, t))  (1.5)

2000 Mathematics Subject Classification. 74KB05, 74K10, 74K20, 93D15.

Key words and phrases. Beams; linear elasticity; plates; stabilization of systems by feedback.

\begin{matrix}c \\ ◯ \end{matrix}2004 Texas State University - San Marcos.

Published October 15, 2004.

Assumptions (1.2) and (1.3) imply that the energy Ẽ(t) is non-increasing and is a Lyapounov function. It is well-known that for any initial data (y_{0}, y_{1}) ∈ H = H_{E}^{2} × L^{2}(0, 1), Problem (1.1) admits a unique weak solution y such that (y(t), y_{t}(t)) ∈ H and y ∈ C^{0}(ℝ^{+}; H^{4}(0, 1))∩C^{1}(ℝ^{+}; L^{2}(0, 1)) (see Conrad and Pierre [2]). The study of the strong asymptotic stability of the solution of (1.1) in H, using the invariance principle of Lasalle has been proved by Conrad and Pierre [2] in the case where g and h are multivalued maximal monotone graphs. Our aim in this study is to estimate the rate of decay of the energy Ẽ(t) when the nonlinear feedback functions g and h satisfy suitable growth conditions. However, the extension of this method to the case where g and h are maximal monotone graphs seems difficult. Our two main results are as follows.

Theorem 1.1. Assume (1.2) and (1.3) hold. Then for every solution y of system (1.1), we have

(i) If there exist positive constants C_{1}, C_{2}, C_{3}, C_{4} such that for all x ∈ ℝ,

\begin{matrix}C_{1}|x| ≤ |g(x)| ≤ C_{2}|x|, \\ C_{3}|x| ≤ |h(x)| ≤ C_{4}|x|, \end{matrix}  (1.6)

then given any M > 1, there exists a constant ω > 0 such that

Ẽ(t) ≤ MẼ(0)e^{−ωt} ∀t > 0.

(ii) If there exist positive constants C_{1}, C_{2}, C_{3}, C_{4}, and p, q in [1, +∞[ such that max(p, q) = p ∨ q > 1 such that for all x ∈ ℝ,

\begin{matrix}C_{1} min(|x|, |x|^{p}) ≤ |g(x)| ≤ C_{2}|x|, \\ C_{3} min(|x|, |x|^{q}) ≤ |h(x)| ≤ C_{4}|x|, \end{matrix}  (1.7)

then, given any M > 1, there exists a constant ω > 0 depending continuously on Ẽ(0) such that

Ẽ(t) ≤ MẼ(0)(1 + ωt)^{− \frac{2}{(p∨q)−1}} ∀t ≥ 0.

Theorem 1.2. Assume (1.2) and (1.3) hold. Then for every solution y of system (1.1) we have (i)If there exist positive constants C_{1}, C_{2}, C_{3}, C_{4} and p, q ∈]0, 1] with min(p, q) = p ∧ q < 1 such that for all x ∈ ℝ,

\begin{matrix}C_{1}|x| ≤ |g(x)| ≤ C_{2} max(|x|, |x|^{p}), \\ C_{3}|x| ≤ |h(x)| ≤ C_{4} max(|x|, |x|^{q}), \end{matrix}  (1.8)

then, given any M > 1, there exists a constant ω > 0 depending continuously on Ẽ(0) such that

Ẽ(t) ≤ MẼ(0)(1 + ωt)^{− \frac{2(p∧q)}{1−(p∧q)}} , ∀t ≥ 0.

(ii) If there exists positive constants C_{1}, C_{2}, C_{3}, C_{4} and (p, q) ∈]0, 1] × [1, +∞[ with \frac{1}{p} ∨ q > 1 such that for all x ∈ ℝ,

\begin{matrix}C_{1}|x| ≤ |g(x)| ≤ C_{2} max(|x|, |x|^{p}), \\ C_{3} min(|x|, |x|^{q}) ≤ |h(x)| ≤ C_{4}|x|, \end{matrix}  (1.9)

then given any M > 1, there exists a constant ω > 0 depending continuously on Ẽ(0) such that

Ẽ(t) ≤ MẼ(0)(1 + ωt)^{−_{\frac{2}{( \frac{1}{p} ∨q)−1}}} ∀t ≥ 0.

Theorems 1.1 and 1.2 will also be valid for the following model of a Kirchhoff plate equation (see Ciarlet [1] and Lagnese [5]), in star-shaped domain by nonlinear boundary feedback:

y_{tt} + Δ^{2}y = 0 in Ω×]0, +∞[,

Δy + (1 − μ)B_{1}y = v_{1} on Γ×]0, +∞[,

\frac{∂}{∂ν} Δy + (1 − μ)\frac{∂}{∂τ} B_{2}y = v_{2} on Γ×]0, +∞[,

y(0) = y_{0} ∈ H^{2}(Ω), y_{t}(0) = y_{1} ∈ L^{2}(Ω).

(1.10)

We assume that Ω is a bounded strongly star-shaped domain of ℝ^{2} with respect to x_{0} ∈ Ω and having smooth boundary Γ = ∂Ω of class C^{2}, which means there exists a positive constant δ such that

m(x).ν(x) ≥ δ^{−1} ∀x ∈ Γ,  (1.11)

where ν(x) = (ν_{1}(x), ν_{2}(x)) is the unit outer normal vector to Γ, m(x) = x − x_{0}, and the dot “.” denotes the scalar product in ℝ^{2}. τ(x) = (−ν_{2}(x), ν_{1}(x)) is a unit tangent vector. We denote by \frac{∂}{∂ν} (resp. \frac{∂}{∂τ}, ) the normal derivative (resp. tangent derivative). The constant 0 < μ < 1/2 is the Poisson coefficient and the boundary operators B_{1}, B_{2} are defined by

B_{1}y = 2ν_{1}ν_{2} \frac{∂^{2}y}{∂x_{1}∂x_{2}} −ν_{1}^{2} \frac{∂^{2}y}{∂x_{2}^{2}} −ν_{2}^{2} \frac{∂^{2}y}{∂x_{1}^{2}} , B_{2}y = (ν_{1}^{2}−ν_{2}^{2})\frac{∂^{2}y}{∂x_{1}∂x_{2}} +ν_{1}ν_{2}(\frac{∂^{2}y}{∂x_{1}^{2}} − \frac{∂^{2}y}{∂x_{2}^{2}} ),

where y = y(x_{1}, x_{2}, t) is the vertical displacement of the point x = (x_{1,}x_{2}) ∈ Ω at the time t of the plate.

The aim of this paper is to study the uniform energy decay rate of the system (1.10) by the nonlinear feedback laws v_{1} and v_{2} given as follows

v_{1}(t) = −β \frac{∂y}{∂ν} − h(\frac{∂y_{t}}{∂ν}), v_{2}(t) = αy + g(y_{t}) on Γ×]0, +∞[,  (1.12)
(α, β) ∈ (L^{∞}(Γ))^{2}, 0 < α_{0} ≤ α(x) ≤ α_{1}, 0 < β_{0} ≤ β(x) ≤ β_{1} ∀x ∈ Γ. (1.13)

It is well-known that for any initial data (y_{0}, y_{1}) ∈ V = H^{2}(Ω) × L^{2}(Ω), the system (1.10) has a unique weak solution y such that (y(t), y_{t}(t)) ∈ V and y ∈ C^{0}(ℝ^{+}, H^{2}(Ω)) ∩ C^{1}(ℝ^{+}, L^{2}(Ω)) (see Rao [7]).

We introduce the energy associated with the system (1.10) as follows

E(t) = \frac{1}{2}(∫_{Ω} |y_{t}|^{2}dx + a(y, y) + ∫_{Γ}(α|y|^{2} + β|\frac{∂y}{∂ν} |^{2})dΓ),  (1.14)

where

a(y, y) = ∫_{Ω}((\frac{∂^{2}y}{∂x_{1}^{2}} )^{2} + (\frac{∂^{2}y}{∂x_{2}^{2}} )^{2} + 2μ\frac{∂^{2}y}{∂x_{1}^{2}} \frac{∂^{2}y}{∂x_{2}^{2}} + 2(1 − μ)(\frac{∂^{2}y}{∂x_{1}∂x_{2}} )^{2})dx.

Using Green’s formula, the derivative of the energy E(t) is

\frac{dE}{dt} = − ∫_{Γ}(g(y_{t})y_{t} + h(\frac{∂y_{t}}{∂ν})\frac{∂y_{t}}{∂ν})dΓ.  (1.15)

Assumptions (1.2) and (1.3) imply that the energy is non-increasing and is a Lyapounov function. It is easy to prove the strong stabilization by applying Holmgren’s theorem (see Lions [6]) and Lasalle’s invariance principle (see [8]). The problem of estimating the rate of decay of the energy E(t) has been studied extensively by many authors, among which we can mention Lagnese [5], Komornik - Zuazua [4] and Zuazua [9]. In the nonlinear cases, under suitable growth conditions on the functions g and h, Rao [7] has established exponential or rational rates of decay of the energy for any positive functions α, β given by (1.13). We obtain here an improvement of these results in the sense that we prove the estimates when the nonlinear feedback functions g and h satisfy more general growth conditions. Now, we state our main results.

Theorem 1.3. Assume (1.2), (1.3), (1.11), (1.13). Then for every solution y of system (1.10) with feedback laws (1.12), we have

(i) In addition, if we assume that g and h such that (1.7) holds. Then, given any M > 1, there exists a constant ω > 0 depending continuously on E(0) such that

E(t) ≤ ME(0)(1 + ωt)^{− \frac{2}{(p∨q)−1}} , ∀t ≥ 0.

(ii) In addition, if we assume that g and h such that (1.8) holds. Then, given any M > 1, there exists a constant ω > 0 depending continuously on E(0) such that

E(t) ≤ ME(0)(1 + ωt)^{− \frac{2(p∧q)}{1−(p∧q)}} , ∀t ≥ 0.

(iii) In addition, if we assume that g and h such that (1.9) holds. Then, given any M > 1, there exists a constant ω > 0 depending continuously on E(0) such that

E(t) ≤ ME(0)(1 + ωt)^{−_{\frac{2}{( \frac{1}{p} ∨q)−1}}} ∀t ≥ 0.

2. Proof of Theorem 1.1

Let y be a smooth solution of (1.1). We define the functional

ρ(t) = ρ_{1}(t) + ρ_{2}(t) + ρ_{3}(t),

where

ρ_{1} = 4 ∫_{0}^{1} xy_{t}y_{x}dx, ρ_{2} = \frac{1}{4}y(1, t) ∫_{0}^{1} x^{2}(3 − 2x)y_{t}dx,

ρ_{3} = \frac{1}{4}y_{x}(1, t) ∫_{0}^{1} x^{2}(x − 1)y_{t}dx.

We can show that there exist positive constants K_{0}, K_{1} and K_{2} such that for any t ≥ 0, the following estimates hold

|ρ(t)| ≤ K_{0}Ẽ(t),  (2.1)
\frac{dρ}{dt}(t) ≤ −Ẽ(t) + K_{1}(y_{xt}^{2}(1, t) + y_{t}^{2}(1, t)) + K_{2}(g^{2}(y_{t}(1, t)) + h^{2}(y_{xt}(1, t))). (2.2)

Given ε > 0, we introduce the perturbed energy by

Ẽ_{ε}(t) = Ẽ(t) + ερ(t)[Ẽ(t)]^{\frac{(p∨q)−1}{2}} .  (2.3)

This together with the non-increasing of the energy Ẽ(t) implies that for any M > 1

M^{− \frac{1}{2}} [Ẽ_{ε}(t)]^{\frac{(p∨q)+1}{2}} ≤ [Ẽ(t)]^{\frac{(p∨q)+1}{2}} ≤ M^{1/2}[Ẽ_{ε}(t)]^{\frac{(p∨q)+1}{2}}  (2.4)

with

ε ≤ K_{0}^{−1}[Ẽ(0)]^{\frac{1−(p∨q)}{2}} (1 − M^{\frac{−1}{(p∨q)+1}} ).  (2.5)

Now, we calculate the derivative of the perturbed energy Ẽ_{ε}(t)

Ẽ′_{ε}(t) = Ẽ′(t) + ε\frac{(p ∨ q) − 1}{2}ρ(t)Ẽ′(t)[Ẽ(t)]^{\frac{(p∨q)−3}{2}} + ερ′(t)[Ẽ(t)]^{\frac{(p∨q)−1}{2}} , (2.6)

on the other hand, from (1.6), (1.7) and (2.2) one obtains

ρ′(t) ≤ −Ẽ(t) + K_{3}y_{t}^{2}(1, t) + K_{4}y_{xt}^{2}(1, t)  (2.7)

where K_{3} = K_{1} + K_{2}C_{2}^{2} and K_{4} = K_{1} + K_{2}C_{4}^{2}. Plugging (2.1) and (2.7) into equation (2.6), one obtains

Ẽ′_{ε}(t) ≤ (−1+ε\frac{(p ∨ q) − 1}{2}K_{0}[Ẽ(0)]^{\frac{(p∨q)−1}{2}} )(−Ẽ′(t))+F_{1}+F_{2}−ε[Ẽ(t)]^{\frac{(p∨q)+1}{2}} (2.8)

where F_{1} = εK_{3}[Ẽ(t)]^{\frac{(p∨q)−1}{2}} y_{t}^{2}(1, t) and F_{2} = εK_{4}[Ẽ(t)]^{\frac{(p∨q)−1}{2}} y_{xt}^{2}(1, t)). Now we distinguish the case p ∨ q = 1 and p ∨ q > 1.

(i) Case p ∨ q = 1. In this case (2.8) yields

Ẽ′_{ϵ}(t) ≤ (−1+εK_{3}/C_{1})y_{t}(1, t)g(y_{t}(1, t)+(−1+εK_{4}/C_{3})y_{xt}(1, t)h(y_{xt}(1, t))−εẼ(t).

If we choose ε ≤ min(C_{1}/K_{3}, C_{3}/K_{4}, K_{0}^{−1}[Ẽ(0)]^{\frac{1−(p∨q)}{2}} (1 − M^{\frac{−1}{(p∨q)+1}} )), this implies

Ẽ′_{ε}(t) ≤ −εẼ(t) ≤ −εM^{\frac{−1}{2}} Ẽ_{ε}(t),

so we obtain

Ẽ(t) ≤ MẼ(0)e^{−εM ^{\frac{−1}{2}} t}, ∀t > 0.

(ii) Case p ∨ q > 1. If y_{xt}^{2}(1, t) > 1, it follows from hypothesis (1.2), (1.3) and (1.7) that

F_{2} ≤ ε(K_{4}/C_{3})[Ẽ(0)]^{\frac{(p∨q)−1}{2}} y_{xt}(1, t)h(y_{xt}(1, t)).  (2.9)

However, while y_{xt}^{2}(1, t) ≤ 1, we have min (|y_{xt}(1, t)|, |y_{xt}(1, t)|^{q}) = |y_{xt}(1, t)|^{q} and

|y_{xt}(1, t)|^{(p∨q)+1} ≤ |y_{xt}(1, t)|^{q+1} ≤ (1/C_{3})y_{xt}(1, t)h(y_{xt}(1, t)),

by Young’s inequality, we have for any δ > 0,

\begin{aligned}F_{2} & ≤ ε\frac{(p ∨ q) − 1}{(p ∨ q) + 1}δ^{− \frac{(p∨q)+1}{(p∨q)−1}} [Ẽ(t)]^{\frac{(p∨q)+1}{2}} \\ & + ε\frac{2}{C_{3}(p ∨ q + 1)}(K_{4}δ)^{\frac{(p∨q)+1}{2}} y_{xt}(1, t)h(y_{xt}(1, t)). \end{aligned}  (2.10)

Combining (2.9) and (2.10), one has

F_{2} ≤ ε\frac{(p ∨ q) − 1}{(p ∨ q) + 1}δ^{− \frac{(p∨q)+1}{(p∨q)−1}} [Ẽ(t)]^{\frac{(p∨q)+1}{2}} + εK_{5}y_{xt}(1, t)h(y_{xt}(1, t)),  (2.11)

where

K_{5} = \frac{2}{C_{3}(p ∨ q + 1)}(K_{4}δ)^{\frac{(p∨q)+1}{2}} + (K_{4}/C_{3})[Ẽ(0)]^{\frac{(p∨q)−1}{2}} .

Similarly, we can show that

F_{1} ≤ ε\frac{(p ∨ q) − 1}{(p ∨ q) + 1}δ^{− \frac{(p∨q)+1}{(p∨q)−1}} [Ẽ(t)]^{\frac{(p∨q)+1}{2}} + εK_{6}y_{t}(1, t)g(y_{t}(1, t)),  (2.12)

with

K_{6} = \frac{2}{C_{1}(p ∨ q + 1)}(K_{4}δ)^{\frac{(p∨q)+1}{2}} + (K_{3}/C_{1})[Ẽ(0)]^{\frac{(p∨q)−1}{2}} .

Inserting (1.5),(2.11) and (2.12) into (2.8), we obtain

Ẽ′_{ε}(t) ≤ ε(−1 + 2\frac{(p ∨ q) − 1}{(p ∨ q) + 1}δ^{− \frac{(p∨q)+1}{(p∨q)−1}} )[Ẽ(t)]^{\frac{(p∨q)+1}{2}} + (−1 + ελ_{0})y_{t}(1, t)g(y_{t}(1, t))
+ (−1 + ελ_{1})y_{xt}(1, t)h(y_{xt}(1, t)),

where λ_{i} = K_{6−i} + K_{0} \frac{(p∨q)−1}{2}[Ẽ(0)]^{\frac{(p∨q)−1}{2}} for i = 0, 1. This implies that

Ẽ′_{ε}(t) ≤ −με[Ẽ(t)]^{\frac{(p∨q)+1}{2}} ,  (2.13)

provided δ is chosen such that for some μ > 0, 2\frac{(p∨q)−1}{(p∨q)+1}δ^{− \frac{(p∨q)+1}{(p∨q)−1}} ≤ 1 − μ and ε is

chosen as follows −1+ε(K_{6−i}+K_{0} \frac{(p∨q)−1}{2}[Ẽ(0)]^{\frac{(p∨q)−1}{2}} ) ≤ 0 for i = 0, 1. Combining (2.4) and (2.13), we get

Ẽ′_{ε}(t) ≤ −μεM^{\frac{−1}{2}} [Ẽ_{ε}(t)]^{\frac{(p∨q)+1}{2}} .  (2.14)

Finally, solving the differential inequality (2.14) and using (2.4) we obtain

Ẽ(t) ≤ MẼ(0)(1 + ωt)^{− \frac{2}{(p∨q)−1}} ,

with ω = \frac{(p∨q)−1}{2}μεM^{\frac{−(p∨q)}{(p∨q)+1}} [Ẽ(0)]^{\frac{(p∨q)−1}{2}} . This completes the proof of theorem 1.1.

3. Proof of Theorem 1.2

(i) First, by the conditions (1.8) and (2.2), we can deduce the following estimate

ρ′(t) ≤ −Ẽ(t) + K(g^{2}(y_{t}(1, t)) + h^{2}(y_{xt}(1, t))),  (3.1)

with K = K_{2} + K_{1}(1/C_{1}^{2} + 1/C_{2}^{2}). Next, given ε > 0, we introduce the perturbed energy by

Ẽ_{ε}(t) = Ẽ(t) + ερ(t)[Ẽ(t)]^{\frac{1−(p∧q)}{2(p∧q)}} .  (3.2)

Then, for any M > 1, we have (2.4) and (2.5) with 1/p instead of p. We have also (2.6) with 1/p instead of p. Plugging (1.5), (2.1) and (3.1) into equation (2.6) one obtains

Ẽ′_{ε}(t) ≤ (−1 + εK_{0} \frac{1 − (p ∧ q)}{2(p ∧ q)}[Ẽ(0)]^{\frac{1−(p∧q)}{2(p∧q)}} )(−Ẽ′(t)) + D_{1} + D_{2} − ε[Ẽ(t)]^{\frac{1+(p∧q)}{2(p∧q)}} .

(3.3)

where D_{1} = εK(Ẽ(t))^{\frac{1+(p∧q)}{2(p∧q)}} g^{2}(y_{t}(1, t) and D_{2} = εK(Ẽ(t))^{\frac{1+(p∧q)}{2(p∧q)}} h^{2}(y_{xt}(1, t). If y_{xt}^{2}(1, t) ≥ 1, it follows from hypothesis (1.2), (1.3) and (1.8), we have

D_{2} ≤ εC_{4}K[Ẽ(0)]^{\frac{1−(p∧q)}{2(p∧q)}} h(y_{xt}(1, t))y_{xt}(1, t).  (3.4)

However, while y_{xt}^{2}(1, t) < 1, we have max(|y_{xt}(1, t)|, |y_{xt}(1, t)|^{q}) = |y_{xt}(1, t)|^{q}, so

|h(y_{xt}(1, t))|^{\frac{1+(p∧q)}{(p∧q)}} ≤ C_{4}y_{xt}(1, t)h(y_{xt}(1, t),

by Young’s inequality we obtain

D_{2} ≤ (ε/4)[Ẽ(t)]^{\frac{1+(p∧q)}{2(p∧q)}} + εC_{4}(4K)^{\frac{1+(p∧q)}{2(p∧q)}} y_{xt}(1, t)h(y_{xt}(1, t)).  (3.5)

Setting θ_{i} = C_{4−i}((4K)^{\frac{1+(p∧q)}{2(p∧q)}} +K (Ẽ(0))^{\frac{1−(p∧q)}{2(p∧q)}} ) for i = 0 and i = 2 and combining (3.4) and (3.5), one obtains

D_{2} ≤ (ε/4)[Ẽ(t)]^{\frac{1+(p∧q)}{2(p∧q)}} + εθ_{0}y_{xt}(1, t)h(y_{xt}(1, t)).  (3.6)

Similarly we can show that

D_{1} ≤ (ε/4)[Ẽ(t)]^{\frac{1+(p∧q)}{2(p∧q)}} + εθ_{2}y_{t}(1, t)g(y_{t}(1, t))  (3.7)

Inserting (1.5), (3.6) and (3.7) into (3.3), it follows

Ẽ′_{ε}(t) ≤ −\frac{ε}{2}[Ẽ(t)]^{\frac{1+(p∧q)}{2(p∧q)}} + C(ε)(y_{t}(1, t)g(y_{t}(1, t)) + y_{xt}(1, t)h(y_{xt}(1, t))),

where C(ε) = ε(θ_{0} + θ_{2}) − 1 + εK_{0} \frac{1−(p∧q)}{2(p∧q)}(Ẽ(0))^{\frac{1−(p∧q)}{2(p∧q)}} . Using (2.4)–(2.5) with 1/p instead of p, we can show that

Ẽ(t) ≤ MẼ(0)(1 + ωt)^{− \frac{2(p∧q)}{1−(p∧q)}} ,  (3.8)

where ω = ε\frac{1−(p∧q)}{4(p∧q)}M^{− \frac{1}{1+(p∧q)}} (Ẽ(0))^{\frac{1−(p∧q)}{2(p∧q)}} and ε is chosen such that

ε ≤ min (((θ_{0} + θ_{2}) + K_{0} \frac{1 − (p ∧ q)}{2(p ∧ q)}(Ẽ(0))^{\frac{1−(p∧q)}{2(p∧q)}} )^{−1},
K_{0}^{−1}(Ẽ(0))^{\frac{(p∧q)−1}{2(p∧q)}} (1 − M^{− \frac{(p∧q)}{(p∧q+1)}} )).

(ii) First, by the conditions (1.9) and (2.2), we have

ρ′(t) ≤ −Ẽ(t) + Ay_{xt}^{2}(1, t) + Bg^{2}(y_{t}(1, t)),  (3.9)

with A = K_{1} + K_{2}C_{4}^{2} and B = (K_{1}/C_{1}^{2}) + K_{2}. Next, given ε > 0, we introduce the perturbed energy by

Ẽ_{ε}(t) = Ẽ(t) + ερ(t)[Ẽ(t)]^{\frac{( \frac{1}{p} ∨q)−1}{2}} .  (3.10)

Then, for any M > 1, we have (2.4) and (2.5) with 1/p instead of p. We have also (2.6) with 1/p instead of p. Plugging (2.1) and (3.9) into (2.4) one obtains

Ẽ′_{ε}(t) ≤ (−1 + εK_{0} \frac{(\frac{1}{p} ∨ q) − 1}{2}[Ẽ(0)]^{\frac{( \frac{1}{p} ∨q)−1}{2}} )(−Ẽ′(t)) + E_{1} + E_{2} − ε[Ẽ(t)]^{\frac{( \frac{1}{p} ∨q)+1}{2}}

(3.11)

where

E_{1} = εA(Ẽ(t))^{\frac{( \frac{1}{p} ∨q)−1}{2}} y_{xt}^{2}(1, t), E_{2} = εB(Ẽ(t))^{\frac{( \frac{1}{p} ∨q)−1}{2}} g^{2}(y_{t}(1, t))).

If y_{xt}^{2}(1, t) > 1, it follows from hypothesis (1.2), (1.3) and (1.9), that

E_{1} ≤ (A/C_{3})ε[Ẽ(0)]^{\frac{( \frac{1}{p} ∨q)−1}{2}} y_{xt}(1, t)h(y_{xt}(1, t))  (3.12)

However, while y_{xt}^{2}(1, t) ≤ 1, we have min(|y_{xt}(1, t)|, |y_{xt}(1, t)|^{q}) = |y_{xt}(1, t)|^{q},

|y_{xt}(1, t)|^{1+( \frac{1}{p} ∨q)} ≤ |y_{xt}(1, t)|^{q+1} ≤ (1/C_{3})y_{xt}(1, t)h(y_{xt}(1, t)),

and by Young’s inequality, for any δ > 0, we have

E_{1} ≤ ε/β′C_{3})(Aδ)^{β′}y_{xt}(1, t)h(y_{xt}(1, t)) + (ε/β)δ^{−β}[Ẽ(t)]^{\frac{( \frac{1}{p} ∨q)+1}{2}} .  (3.13)

Combining (3.11) and (3.12), one has

E_{1} ≤ (ε/β)δ^{−β}[Ẽ(t)]^{\frac{( \frac{1}{p} ∨q)+1}{2}} + εA(δ) y_{xt}(1, t)h(y_{xt}(1, t)),  (3.14)

with A(δ) = (A/C_{3})[Ẽ(0)]^{\frac{( \frac{1}{p} ∨q)−1}{2}} + (1/β′C_{3})(Aδ)^{β′}. If y_{t}^{2}(1, t) ≥ 1, it follows from hypothesis (1.2), (1.3) and (1.9), that

E_{2} ≤ εBC_{2}[Ẽ(0)]^{\frac{( \frac{1}{p} ∨q)−1}{2}} y_{t}(1, t)g(y_{t}(1, t)).  (3.15)

However, while y_{t}^{2}(1, t) < 1, we have max(|y_{t}(1, t)|, |y_{t}(1, t)|^{p}) = |y_{t}(1, t)|^{p},

|g(y_{t}(1, t))|^{1+( \frac{1}{p} ∨q)} ≤ (C_{2})^{( \frac{1}{p} ∨q)}y_{t}(1, t)g(y_{t}(1, t))

and by Young’s inequality, we can deduce that

E_{2} ≤ ε(4B)^{\frac{( \frac{1}{p} ∨q)−1}{2}} (C_{2})^{\frac{1}{p} ∨q}y_{t}(1, t)g(y_{t}(1, t)) + ε4^{−β}[Ẽ(t)]^{\frac{( \frac{1}{p} ∨q)+1}{2}} .  (3.16)

Combining (3.14) and (3.15), we obtain

E_{2} ≤ ε(C_{2}B[Ẽ(0)]^{\frac{( \frac{1}{p} ∨q)−1}{2}} +(4B)^{\frac{( \frac{1}{p} ∨q)−1}{2}} (C_{2})^{\frac{1}{p} ∨q})y_{t}(1, t)g(y_{t}(1, t)))+\frac{ε}{4}[Ẽ(t)]^{\frac{( \frac{1}{p} ∨q)+1}{2}} .

(3.17)

Inserting (3.13) and (3.16) into (2.6), one obtains

\begin{aligned}Ẽ′_{ε}(t) & ≤ −ε(1 − \frac{1}{4} − (δ^{−β}/β))[Ẽ(t)]^{\frac{( \frac{1}{p} ∨q)+1}{2}} + (−1 + εσ_{1})y_{t}(1, t)g(y_{t}(1, t)) \\ & + (−1 + εσ_{2})y_{xt}(1, t)h(y_{xt}(1, t)) \end{aligned} (3.18)

where

σ_{1} = (C_{2}B + K_{0} \frac{(\frac{1}{p} ∨ q) − 1}{2})[Ẽ(0)]^{\frac{( \frac{1}{p} ∨q)−1}{2}} + (4B)^{β′}(C_{2})^{\frac{1}{p} ∨q})

and

σ_{2} = A(δ) + K_{0} \frac{(\frac{1}{p} ∨ q) − 1}{2}[Ẽ(0)]^{\frac{( \frac{1}{p} ∨q)−1}{2}} .

By choosing δ so that \frac{1}{4} + (δ^{−β}/β) = \frac{1}{2}, and

ε ≤ min(σ_{1}^{−1}, σ_{2}^{−1}, K_{0}^{−1}[Ẽ(0)]^{\frac{1−( \frac{1}{p} ∨q)}{2}} (1 − M^{\frac{−1}{( \frac{1}{p} ∨q)+1}} )).

This with (3.17) and using (2.4) with 1/p instead of p, we obtain

Ẽ(t) ≤ MẼ(0)(1 + ωt)^{− \frac{2}{( \frac{1}{p} ∨q)−1}} ,

with ω = \frac{ε}{4}((\frac{1}{p} ∨ q) − 1)M^{− \frac{\frac{1}{p} ∨q}{( \frac{1}{p} ∨q)+1}} [Ẽ(0)]^{\frac{( \frac{1}{p} ∨q)−1}{2}} . The proof of theorem 1.2 is thus complete.

4. Proof of Theorem 1.3

(i) It follows from (1.7) that (1.7) is also true when p is replaced by p ∨ q and q is replaced by p ∨ q. So we get the desired estimate, by applying Rao’s theorem 1.1 (ii) (see [7]). (ii) (1.8) implies that (1.8) is also true when p is replaced by p ∧ q and q is replaced by p ∧ q. This is exactly the condition (1.28)-(1.29) with exponent p ∧ q < 1 of Rao’s theorem 1.2 (see [7]), consequently, the proof of this part is thus complete.

(iii) We adopt the method used by Rao [7] but in our case we have two nonlinear feedback functions g and h satisfying more general growth conditions. We introduce the same functional defined in [7],

ρ(t) = ∫_{Ω} y_{t}(m.∇y)dx + C_{0} ∫_{Ω} y_{t}φdx,  (4.1)

where C_{0} is a positive constant and φ is the solution of the problem

Δ^{2}φ = 0 in Ω,

φ = y, \frac{∂φ}{∂ν} = \frac{∂y}{∂ν} on Γ,

we verify that the following estimates hold

∫_{Ω} φ^{2}dx ≤ γ^{2} ∫_{Γ} ((y)^{2} + (\frac{∂y}{∂ν})^{2})dΓ, a(φ, y) = a(φ, φ) ≥ 0,

where γ is a constant depending only on the domain Ω. We can show that there exist positive constants K_{0}, K_{1}, K_{2} are such that for any t ≥ 0,

|ρ(t)| ≤ K_{0}E(t), ∀t ≥ 0,  (4.2)
ρ′(t) ≤ −E(t) + K_{1} ∫_{Γ}((y_{t})^{2} + (\frac{∂y_{t}}{∂ν})^{2})dΓ + K_{2} ∫_{Γ}(g^{2}(y_{t}) + h^{2}(\frac{∂y_{t}}{∂ν}))dΓ. (4.3)

This with (1.9) implies

ρ′(t) ≤ −E(t) + A ∫_{Γ} |\frac{∂y_{t}}{∂ν} |^{2}dΓ + B ∫_{Γ} g^{2}(y_{t})dΓ,  (4.4)

where A = K_{1} + KC_{4}^{2} and B = (K_{1}/C_{1}^{2}) + K_{2}.

Then, for any M > 1, Plugging (4.2) and (4.4) into (2.6) with p replaced by 1/p, one obtains

E′_{ε}(t) ≤ (−1 + εK_{0} \frac{(\frac{1}{p} ∨ q) − 1}{2}[E(0)]^{\frac{( \frac{1}{p} ∨q)−1}{2}} )(−E′(t)) − ε[E(t)]^{\frac{( \frac{1}{p} ∨q)+1}{2}} + L, (4.5)

where

L = ε[E(t)]^{\frac{( \frac{1}{p} ∨q)−1}{2}} (A ∫_{Γ} |\frac{∂y_{t}}{∂ν} |^{2}dΓ + B ∫_{Γ} g^{2}(y_{t})dΓ).

¿From (1.2), (1.3) and (1.9) we have

(g(s))^{2} ≤ C_{2}g(s)s ∀|s| ≥ 1; |g(s)|^{( \frac{1}{p} ∨q)+1} ≤ (C_{2})^{\frac{1}{p} ∨q}g(s)s ∀|s| ≤ 1; (4.6)
|s|^{2} ≤ (1/C_{3})h(s)s ∀|s| ≥ 1; |s|^{( \frac{1}{p} ∨q)+1} ≤ (1/C_{3})h(s)s ∀|s| ≤ 1.  (4.7)

Using (4.6) and (4.7), we have

L ≤ Q + εC[E(0)]^{\frac{( \frac{1}{p} ∨q)−1}{2}} ( ∫_{\{| \frac{∂y_{t}}{∂ν} |≥1\}} h(\frac{∂y_{t}}{∂ν})\frac{∂y_{t}}{∂ν} dΓ + ∫_{\{|y_{t}|≥1\}} g(y_{t})y_{t}dΓ), (4.8)

where

Q = ε[E(t)]^{\frac{( \frac{1}{p} ∨q)−1}{2}} (A ∫_{\{| \frac{∂y_{t}}{∂ν} |≤1\}}(\frac{∂y_{t}}{∂ν})^{2}dΓ + B ∫_{\{|y_{t}|≤1\}} g^{2}(y_{t})dΓ)

and C = (A + B)(C_{2} + 1/C_{3}), With the exponents β = \frac{( \frac{1}{p} ∨q)+1}{( \frac{1}{p} ∨q)−1}, β′ = \frac{( \frac{1}{p} ∨q)+1}{2},

applying Young’s inequality to Q, it follows from Hôlder’s inequality and (4.6), (4.7) that for any parameter δ > 0,

\begin{aligned}Q & ≤ 2εδ^{−β}/β[E(t)]^{\frac{( \frac{1}{p} ∨q)+1}{2}} + (ε/β′)|Γ|^{β′/β} \\ & × [(Aδ)^{β′}/C_{3} ∫_{\{| \frac{∂y_{t}}{∂ν} |≤1\}} h(\frac{∂y_{t}}{∂ν})\frac{∂y_{t}}{∂ν} dΓ + (Bδ)^{β′}C_{2}^{( \frac{1}{p} ∨q)} ∫_{\{|y_{t}|≤1\}} g(y_{t})y_{t}dΓ]. \end{aligned}

(4.9)

Inserting (4.6) into (4.8) gives

L ≤ εη(−E′(t)) + 2εδ^{−β}/β[E(t)]^{\frac{( \frac{1}{p} ∨q)+1}{2}}  (4.10)

where η = C(E(0))^{\frac{( \frac{1}{p} ∨q)−1}{2}} +(1/β′)|Γ|^{β′/β}((Aδ)^{β′}/C_{3} +(Bδ)^{β′}(C_{2})^{( \frac{1}{p} ∨q)}). Inserting (4.10) into (4.5), it follows

\begin{aligned}E′_{ε}(t) & ≤ ( − 1 + ε(η + K_{0} \frac{(\frac{1}{p} ∨ q) − 1}{2}(E(0))^{\frac{( \frac{1}{p} ∨q)−1}{2}} ))(−E′(t)) \\ & − ε(1 − 2δ^{−β}/β)(E(t))^{\frac{( \frac{1}{p} ∨q)+1}{2}} . \end{aligned}  (4.11)

By choosing δ so that δ = (4/β)^{(1/β)}, and ε as

ε ≤ min((η +K_{0} \frac{(\frac{1}{p} ∨ q) − 1}{2}(E(0))^{\frac{( \frac{1}{p} ∨q)−1}{2}} )^{−1}, K_{0}^{−1}(E(0))^{\frac{1−( \frac{1}{p} ∨q)}{2}} (1−M^{\frac{−1}{(( \frac{1}{p} ∨q)+1)}} ) ).

This with (4.11) and using (2.4) with p replaced by 1/p, we obtain

E(t) ≤ ME(0)(1 + ωt)^{−_{\frac{2}{( \frac{1}{p} ∨q)−1}}} ,

with

ω = \frac{ε}{4}((\frac{1}{p} ∨ q) − 1)M^{− \frac{\frac{1}{p} ∨q}{( \frac{1}{p} ∨q)+1}} (E(0))^{\frac{( \frac{1}{p} ∨q)−1}{2}} .

The proof of theorem 1.3 is thus complete.

References

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[3] F. Conrad and B. Rao, Decay of solutions of the wave equation in a star-shaped domain with nonlinear boundary feedback - Asymptotic Analysis 7, (1993) pp. 159-177.

[4] V. Komornik and E. Zuazua, A direct method for the boundary stabilization of the wave equation, J. Math. pures app. 69, (1990) pp. 33-54.

[5] J. E. Lagnese, Boundary stabilization of thin Plates, Siam Publications Philadelphia, Pennsylvania, (1989).

[6] J. L. Lions, Contrôlabilité exacte, perturbations et stabilisation de systèmes distribués, Vol. 1, Masson, Paris, (1988).

[7] B. Rao, Stabilisation of Kirchhoff Plate equation in a star-shaped domain by nonlinear boundary feedback, Nonlinear Analysis, Theory Methods and Applications, Vol. 20, 6, (1993) pp. 605-626.

[8] M. Slemrod, Feedback stabilization of a linear system in Hilbert space with an a priori bounded control, Math. Control Signals Systems, 2, (1989) pp. 265-285.

[9] E. Zuazua, Uniform stabilization of the wave equation by nonlinear boundary feedback, Siam J. Control and Optimization, 28, (1989) pp. 265-268.

Naji Yebari Département de Mathématiques, Université Abdelmalek Essaadi, B. P. 2121, Tetouan, Maroc

E-mail address: nyebari@hotmail.com