http://dx.doi.org/10.4134/JKMS.2012.49.5.947
A MEMORY TYPE BOUNDARY STABILIZATION FOR AN EULER-BERNOULLI BEAM UNDER BOUNDARY OUTPUT
FEEDBACK CONTROL
Yong Han Kang, Jong Yeoul Park, and Jung Ae Kim
Abstract. In this paper, the memory type boundary stabilization for an Euler-Bernoulli beam with one end fixed and control at the other end is considered. We prove the existence of solutions using the Galerkin method and then investigate the exponential stability of solutions by using multiplier technique.
1. Introduction
In this paper, we consider the following Euler-Bernoulli beam with memory which has one end fixed and control input at the other end:
y tt (x, t) + y xxxx (x, t) − Z t
0
κ(t − τ )y xxxx (x, τ )dτ + g(y t (x, t)) = 0, (1.1)
x ∈ (0, L), t ≥ 0,
y(0, t) = y x (0, t) = y xx (L, t) = 0, t ≥ 0, y xxx (L, t) −
Z t 0
κ(t − τ )y xxx (L, τ )dτ = u(t) − e θ sin t, t ≥ 0, y(x, 0) = y 0 (x), y t (x, 0) = y 1 (x), x ∈ (0, L),
y out (t) = y t (L, t), t ≥ 0,
where κ represents the kernel of memory term, g : R → R is a given function, e θ is a positive constant, u : R + → R is the boundary control force applied at the free end of the beam and y out (t) stands for the measured signal of the system at time t. System (1.1) describes the transverse vibration of an extensible beam clamped at x = 0 and supported at x = L by a control force. The advantage of the adaptive stabilization is that stabilization and good control
Received December 2, 2010; Revised February 2, 2012.
2010 Mathematics Subject Classification. 49K20, 65N30, 93B52, 74G25, 74G30.
Key words and phrases. Euler-Bernoulli beam, output feedback control, memory, Galer- kin method.
This work was supported by research grants from the Catholic University of Daegu in 2012.
c
2012 The Korean Mathematical Society
947
performance can be automatically achieved even in the presence of various types of uncertainty. In this paper, we consider the stabilization of the system (1.1). To this end, we design the following adaptive output feedback controller:
u(t) = h(t)y t (L, t) + θ(t) sin t, t ≥ 0, (1.2)
h t (t) = ry t 2 (L, t), h(0) = h 0 > 0, t ≥ 0, r > 0, θ t (t) = y t (L, t) sin t, θ(0) = θ 0 , t ≥ 0,
where θ 0 is the initial condition of the estimator.
Under this adaptive controller, the closed-loop system (1.1) becomes
y tt (x, t) + y xxxx (x, t) − Z t
0
κ(t − τ )y xxxx (x, τ )dτ + g(y t (x, t)) = 0, (1.3)
x ∈ (0, L), t ≥ 0,
y(0, t) = y x (0, t) = y xx (L, t) = 0, t ≥ 0, y xxx (L, t) −
Z t 0
κ(t − τ )y xxx (L, τ )dτ = h(t)y t (L, t) + [θ(t) − e θ] sin t, t ≥ 0, y(x, 0) = y 0 (x), y t (x, 0) = y 1 (x), x ∈ (0, L),
h t (t) = ry t 2 (L, t), h(0) = h 0 > 0, t ≥ 0, r > 0, θ t (t) = y t (L, t) sin t, θ(0) = θ 0 .
The energy of the system (1.3) is given by E(t) = 1
2 Z t
0
(y t 2 (x, t) + y 2 xx (x, t))dx.
Ma [12, 13] studied the boundary stabilization for a nonlinear beam on elastic bearings without memory and boundary output feedback control. The exis- tence and uniform decay of solutions for an Euler-Bernoulli beam with memory was consider by Park and Kim [15]. Also, the Euler-Bernoulli beam equation with memory and boundary output feedback control was studied by Park, Kang and Kim [14]; they proved the existence and exponential stability of solutions for the following system:
y tt (x, t) + y xxxx (x, t) − Z t
0
κ(t − τ )y xxxx (x, τ )dτ + g(y t (x, t)) = 0, (1.4)
x ∈ (0, L), t ≥ 0,
y(0, t) = y x (0, t) = y xx (L, t) = 0, t ≥ 0, y xxx (L, t) −
Z t 0
κ(t − τ )y xxx (L, τ )dτ = h(t)y t (L, t), t ≥ 0, y(x, 0) = y 0 (x), y t (x, 0) = y 1 (x), x ∈ (0, L),
h t (t) = ry t 2 (L, t), h(0) = h 0 > 0, t ≥ 0, r > 0.
Conrad and Om¨er [3] proved the existence and uniform decay for a flexible beam with a tip mass. The boundary stabilization and boundary controllability for the beams were consider by authors [1, 2, 4-6, 9-11, 16, 17]. Guo and Luo [8] studied the stabilization and parameter estimation for a flexible-beam vibration with gain adaptive direct strain feedback control. Recently, Guo and Guo [7] consider the adaptive stabilization for a Kirchhoff-type nonlinear beam under boundary output feedback control. Motivated by [14] and some idea in [7], we can obtain our main results. The objective of this paper is to study the stabilization for a more general Euler-Bernoulli beam with memory under boundary output feedback control. Our choice of boundary feedback is motivated by the fact that boundary controls are easily implemented as the need to act only on the boundary of the spatial domain. This paper is organized as follows. In Section 2, using a constructive Galerkin approximation scheme, we show the existence and uniqueness of the solution for the system (1.3). The exponential decay that is dependent on the initial data is obtained in Section 3 by making use of the multiplier technique.
2. Existence result
In this section we prepare some notation and hypotheses which will be needed in the proof of our result. Let L 2 (0, L) be the usual Hilbert space with the inner product (·, ·) and the inner product induced norm k · k. Throughout this paper, we define V = {y ∈ H 2 (0, L) | y(0) = y x (0) = 0} equipped with the norm kyk V = ky xx k, W = {y ∈ V ∩ H 4 (0, L) | y xx (L) = 0} equipped with the norm kyk W = ky xx k + ky xxxx k and (u, v) = R L
0 u(x)v(x)dx. From the Poincar´e inequality, it follows that k · k V and k · k W are equivalent to the standard norms of H 2 (0, L) and H 4 (0, L), respectively. Now, we state the following hypotheses which will be assumed in this paper.
(H 1 ) For any y 0 ∈ W, y 1 ∈ V ,
y xxxx (x, 0) + g(y 1 ) = 0, x ∈ (0, L), (2.1)
y(0, 0) = y x (0, 0) = y xx (L, 0) = 0, y xxx (L, 0) = h 0 y t (L, 0).
(H 2 ) Let g : R → R be a continuously differentiable function and there exists a positive constant µ such that
(2.2) g(0) = 0, (g(r) − g(s))(r − s) ≥ µ|r − s| 2 , r, s ∈ R.
(H 3 ) Let the function κ : R + → R + be a positive and bounded C 2 -function such that
(2.3) ℓ = 1 −
Z ∞ 0
κ(r)dr > 0 and for some positive m i , i = 0, 1, 2
−m 0 κ(t) ≤ κ t (t) ≤ −m 1 κ(t), ∀t ≥ 0,
(2.4)
0 ≤ κ tt (t) ≤ m 2 κ(t), ∀t ≥ 0.
Considering the above hypotheses we have the following existence result by Galerkin method.
Theorem 2.1. Let y 0 ∈ W , y 1 ∈ V. Suppose that (H 1 ), (H 2 ) and (H 3 ) are satisfied. Then the problem (1.3) has a unique solution y in the sense that for any time T > 0,
y ∈ L ∞ (0, T ; W ), y t ∈ L ∞ (0, T ; V ), y tt ∈ L ∞ (0, T ; L 2 (0, L)), (2.5)
h ∈ C 1 [0, T ], θ ∈ C 1 [0, T ], y tt (x, t) + y xxxx (x, t) −
Z t 0
κ(t − τ )y xxxx (x, τ )dτ + g(y t (x, t)) = 0 in L 2 (0, T ; L 2 (0, L)),
y(0, t) = y x (0, t) = y xx (L, t) = 0, t ≥ 0, y xxx (L, t) −
Z t 0
κ(t − τ )y xxx (L, t)dx = h(t)y t (L, t) + [θ(t) − e θ] sin t, t ≥ 0, y(x, 0) = y 0 (x), y t (x, t) = y 1 (x), x ∈ (0, L),
h t (t) = ry t 2 (L, t), t ≥ 0, h(0) = h 0 > 0, r > 0, θ t (t) = y t (L, t) sin t, t ≥ 0, θ(0) = θ 0 .
By the Sobolev embedding theorem, it follows that y ∈ C((0, L) × [0, T ]).
Proof. Let us solve the variational problem with (2.5) which is given by: find y(t) ∈ V such that
(y tt (t), w) + (y xx (t), w xx ) − Z t
0
κ(t − τ )(y xx (τ ), w xx )dτ + (g(y t (t)), w) (2.6)
+{h(t)y t (L, t) + (θ(t) − e θ) sin t}w(L) = 0 for all w ∈ V.
Let {w j } be a complete orthogonal system of V for which {y 0 , y 1 } ∈ Span{w 1 , w 2 }. For each m ∈ N, we denote V m =Span{w 1 , w 2 , . . . , w m }. We search for a function
y m (t) = X m j=1
k j (t)w j
such that for any w ∈ V m , it satisfies the approximate equation (y tt m (t), w) + (y xx m (t), w xx ) −
Z t 0
κ(t − τ )(y xx m (τ ), w xx )dτ + (g(y m t (t)), w) (2.7)
+{h m (t)y t m (L, t) + (θ m (t) − e θ) sin t}w(L) = 0, h m t (t) = r h X m
j=1
k j t (t)w j (L) i 2
= r[y m t (L, t)] 2 ,
θ t m (t) = h X m
j=1
k t j (t)w j (L) i
sin t = y t m (L, t) sin t, h m (0) = h 0 > 0, θ m (0) = θ 0 ,
y m (0) = y 0m → y 0 in W, y t m (0) = y 1m → y 1 in V.
By standard methods in differential equations, we can prove the existence of a solution to (2.7) on some interval, (0, t m ), where t m = ∞ by using the first estimate below. In order to prove Theorem 2.1 it suffices to prove the following a priori estimates.
Estimate I. Replacing w by y t m in (2.7), we have 1
2 d dt
ky m t (t)k 2 + ky m xx (t)k 2
+ (g(y m t (t)), y m t (t)) (2.8)
= d dt
Z t
0
κ(t − τ )(y m xx (τ ), y m xx (t))dτ
− Z t
0
κ t (t − τ )(y xx m (τ ), y m xx (t))dτ
− κ(0)ky m xx (t)k 2 − {h m (t)[y t m (L, t)] 2 + [θ m (t) − e θ] sin t · y t m (L, t)}.
By hypotheses (2.4) and Cauchy-Schwartz inequality, we deduce that
Z t
0
κ t (t − τ )(y xx m (τ ), y m xx (t))dτ (2.9)
≤ ky m xx (t)k Z t
0
|κ t (t − τ )|ky xx m (τ )kdτ
≤ m 2 0
2 ky xx m (t)k 2 + 1 2
Z t
0
κ(t − τ )ky xx m (τ )kdτ 2
≤ m 2 0
2 ky xx m (t)k 2 + 1
2 kκk L1(0,∞)
Z t 0
κ(t − τ )ky xx m (τ )k 2 dτ.
Integrating (2.8) over (0, t) and using (2.2) and (2.9), we obtain ky t m (t)k 2 + ky m xx (t)k 2 + 1
r [h m (t)] 2 + [θ m (t) − e θ] 2 + 2µ Z t
0
ky t m (τ )k 2 dτ (2.10)
≤ ky t m (0)k 2 + ky m xx (0)k 2 + 1
r [h m (0)] 2 + [θ 0 − e θ] 2 + 2
Z t 0
κ(t − τ )(y m xx (τ ), y m xx (t))dτ
+
m 2 0 + kκk 2 L1(0,∞) − 2κ(0) Z t
0
ky xx (τ )k 2 dτ.
Using Schwartz inequality and Young’s inequality, we get
Z t
0
κ(t − τ )(y xx m (τ ), y xx m (t))dτ (2.11)
≤ ky m xx (t)kkκk L121(0,∞) Z t
0
κ(t − τ )ky m xx (τ )k 2 dτ
12≤ 1
8 ky m xx (t)k 2 + 2kκk L1(0,∞) kκk L
∞(0,∞)
Z t 0
ky xx m (τ )k 2 dτ.
Combining the inequalities (2.10), (2.11) and applying Gronwall’s Lemma, we see the first estimate:
ky m t (t)k 2 + ky m xx (t)k 2 + 1
r [h m (t)] 2 + [θ m (t) − e θ] 2 ≤ M 1 , (2.12)
where M 1 > 0 depends on the initial data y 0 , y 1 , h 0 , θ 0 .
From this, we obtain y m t (L, t) ∈ L ∞ (0, ∞). Therefore, the approximate so- lution can be extended to the whole interval [0, T ], where T = ∞.
Estimate II. First of all, we estimate the L 2 -norm of y tt m (0). Considering t = 0 and w = y m tt (0) in (2.7), then we get
ky tt m (0)k 2 + (y xx m (0), y xxtt m (0)) + (g(y t m (0)), y m tt (0)) + h 0 y m t (L, 0)y m tt (L, 0) = 0.
Since the compatibility condition (H 1 ), we obtain
(y m xx (0), y xxtt m (0)) = y m xx (L, 0)y xtt m (L, 0)− y m xx (0, 0)y m xtt (0, 0)− y xxx m (L, 0)y m tt (L, 0) + y m tt (0, 0)y xxx m (0, 0) + (y m xxxx (0), y m tt (0))
= − h 0 y t m (L, 0)y tt m (L, 0) − (g(y m t (0)), y m tt (0)).
Therefore, from the above equality, there exists the following result
(2.13) ky tt m (0)k = 0.
Finally, differentiating (2.7) and writing the equation with w = y tt m (t), we have
1 2
d
dt (ky tt m (t)k 2 + ky m xxt (t)k 2 ) + κ(0)ky xxt m (t)k 2 + (g ′ (y t m (t))y tt m (t), y tt m (t)) (2.14)
+
r[y m t (L, t)] 3 + h m (t)y m tt (L, t)
y m tt (L, t) +
y m t (L, t)sin 2 t + (θ m (t) − e θ) cos t
y tt m (L, t)
= d dt
Z t
0
κ t (t − τ )(y xx m (τ ), y xxt m (t))dτ
− Z t
0
κ tt (t − τ )(y m xx (τ ), y m xxt (t))dτ
− κ t (0)(y xx m (t), y m xxt (t)) + κ(0) d
dt (y xx m (t), y xxt m (t)).
From hypotheses (2.4), we deduce that
Z t
0
κ tt (t − τ )(y m xx (τ ), y m xxt (t))dτ (2.15)
≤ m 2 2
2 ky xxt (t)k 2 + 1
2 kκk L1(0,∞)
Z t 0
κ(t − τ )ky xx m (τ )k 2 dτ.
Since g ∈ C 1 (R) and y m t (t) is bounded by (2.12), there exists C 1 > 0 depends on the initial datas y 0 , y 1 , h 0 , θ 0 such that
|(g ′ (y m t (t))y m tt (t), y tt m (t))| ≤ C 1 ky m tt (t)k 2 .
(2.16)
Using (2.13), (2.15) and (2.16), we integrate (2.14) over (0, t) to obtain
1
2 ky m tt (t)k 2 + 1
2 ky m xxt (t)k 2 + κ(0) Z t
0
ky m xxt (τ )k 2 dτ + r
4 [y t m (L, t)] 4 (2.17)
≤ C 2 + m 2 2 2
Z t 0
ky m xxt (τ )k 2 dτ + 1
2 kκk 2 L1(0,∞)
Z t 0
ky xx m (τ )k 2 dτ
+ C 1
Z t 0
ky m tt (τ )k 2 dτ + κ(0)(y xx m (t), y m xxt (t)) − κ t (0) Z t
0
(y xx m (τ ), y xxt m (τ ))dτ +
Z t 0
κ t (t − τ )(y xx m (τ ), y xxt m (t))dτ + r
4 [y m t (L, 0)] 4 − Z t
0
h m (τ )[y m tt (L, τ )] 2 dτ
− 1
2 [y m t (L, t)] 2 sin 2 t + 1 2
Z t 0
[y t m (L, τ )] 2 sin 2τ dτ − y t m (L, t)[θ m (t) − e θ] cos t
+ y t m (L, 0)[θ 0 − e θ]+
Z t 0
y t m (L, τ ){y t m (L, τ ) sin τ cos τ −[θ m (τ )− e θ] sin τ }dτ, where C 2 > 0 depends on the initial data y 0 , y 1 , h 0 .
Now, since (2.4), for any η > 0 we see that
|κ(0)(y xx m (t), y xxt m (t))| ≤ (κ(0)) 2
4η ky m xx (t)k 2 + ηky xxt m (t)k 2 , (2.18)
κ t (0)
Z t 0
(y xx m (τ ), y xxt m (τ ))dτ ≤ m 2 0 (κ(0)) 2 4η
Z t 0
ky xx m (τ )k 2 dτ +η Z t
0
ky m xxt (τ )k 2 dτ (2.19)
and
Z t 0
κ t (t − τ )(y m xx (τ ), y m xxt (t))dτ (2.20)
≤ m 2 0
4η kκk L1(0,∞) kκk L
∞(0,∞)
Z t 0
ky m xx (τ )k 2 dτ + ηky xxt m (t)k 2 . From the inequality ab ≤ δ 1 a 2 + 4δ 1
1
b 2 , δ 1 > 0 and the Sobolev embedding theorem ky t (L, t)k 2 ≤ δ 2 ky xxt (t)k 2 , δ 2 > 0, we deduce
− 1
2 [y m t (L, t)] 2 sin 2 t − y t m (L, t)[θ m (t) − e θ] cos t (2.21)
≤ δ 1 [y m t (L, t)] 2 sin 2 t + δ 1 [y m t (L, t)] 2 cos 2 t + 1 4δ 1
[θ m (t) − e θ] 2
≤ e δ 1 ky xxt m (t)k 2 + 1 4δ 1
[θ m (t) − e θ] 2 , where e δ 1 = δ 1 δ 2
and using Young’s inequality we obtain
Z t
0
y m t (L, τ ){y m t (L, τ ) sin τ cos τ − [θ m (τ ) − e θ] sin τ }dτ
(2.22)
≤ Z t
0
[y t m (L, τ )] 2 dτ + 1 2
Z t 0
[θ m (τ ) − e θ] 2 dτ
≤ h m (t) r − h 0
r + 1 2
Z t 0
[θ m (τ ) − e θ] 2 dτ.
Thus from (2.17)-(2.22), we have
1
2 ky tt m (t)k 2 + 1
2 ky xxt m (t)k 2 + κ(0) Z t
0
ky m xxt (τ )k 2 dτ + r
4 [y t m (L, t)] 4 (2.23)
≤ C 2 + C 1
Z t 0
ky tt m (τ )k 2 dτ + (κ(0)) 2
4η ky xx m (t)k 2 + m 2 2
2 + η Z t
0
ky xxt m (τ )k 2 dτ + 2ηky m xxt (t)k 2
+ 1
2 kκk 2 L1(0,∞) + m 2 0 (κ(0)) 2 4η + m 2 0
4η kκk L1(0,∞) kκk L
∞(0,∞)
Z t
0
ky xx m (τ )k 2 dτ
+ r
4 [y t m (L, 0)] 4 − Z t
0
h m (τ )[y m tt (L, τ )] 2 dτ − 1
2 [y t m (L, t)] 2 sin 2 t + 1
2 Z t
0
[y m t (L, τ )] 2 sin 2τ dτ − y m t (L, t)[θ m (t) − e θ] cos t + y t m (L, 0)[θ 0 − e θ]
+ Z t
0
y m t (L, τ ){y t m (L, τ ) sin τ cos τ − [θ m (τ ) − e θ] sin τ }dτ
≤ C 2 + C 1
Z t 0
ky tt m (τ )k 2 dτ + (κ(0)) 2
4η ky xx m (t)k 2 + m 2 2
2 + η Z t
0
ky xxt m (τ )k 2 dτ + 2ηky m xxt (t)k 2 + 1
2 kκk 2 L1(0,∞) + m 2 0 (κ(0)) 2 4η + m 2 0
4η kκk L1(0,∞) kκk L
∞(0,∞)
Z t
0
ky xx m (τ )k 2 dτ + r
4 [y t m (L, 0)] 4 + 3h m (t)
2r + e δ 1 ky m xxt (t)k 2 + 1 4δ 1
[θ m (t) − e θ] 2 + 1
2 [y t m (L, 0)] 2 + 1
2 [θ 0 − e θ] 2 + 1 2
Z t 0
[θ m (τ ) − e θ] 2 dτ − 3h 0
2r .
From the inequalities (2.12) and (2.23) and choosing η > 0 sufficiently small, we get
1
2 ky m tt (t)k 2 + 1 2 − e δ 1
ky xxt m (t)k 2 + κ(0) Z t
0
ky xxt m (τ )k 2 dτ + r
4 [y m t (L, t)] 4
≤ C 3 + C 1
Z t 0
ky m tt (τ )k 2 dτ + m 2 2
2 + η Z t
0
ky xxt m (τ )k 2 dτ,
where C 3 > 0 depends on the initial data y 0 , y 1 , h 0 , θ 0 .
Taking e δ 1 with 1 2 − e δ 1 > 0 and using of Gronwall’s Lemma, we have the second estimate:
(2.24) ky m tt (t)k 2 + ky m xxt (t)k 2 + r[y m t (L, t)] 4 ≤ M 2 , where M 2 depends on the initial data y 0 , y 1 , h 0 , θ 0 .
Analysis of the nonlinear terms. By estimates (2.12) and (2.24), we deduce that
{y m } is bounded in L ∞ (0, T ; V ), (2.25)
{y t m } is bounded in L ∞ (0, T ; V ), {y tt m } is bounded in L ∞ (0, T ; L 2 (0, L)), {y t m (L, t)} is bounded in L 2 (0, T ), {h m } is bounded in L ∞ (0, T ),
{h m t } = {r(y m t (L, t)) 2 } is bounded in L ∞ (0, T ), {θ m } is bounded in L ∞ (0, T ),
{θ m t } = {y m t (L, t) sin t} is bounded in L ∞ (0, T ).
Therefore, there exists a subsequence of {y m }, still denoted by {y m } such that y m → y weakly star in L ∞ (0, T ; V ),
(2.26)
y m t → y t weakly star in L ∞ (0, T ; V ), y m tt → y tt weakly star in L ∞ (0, T ; L 2 (0, L)), y m t (L, t) → y t (L, t) weakly in L 2 (0, T ), h m → h weakly star in L ∞ (0, T ), h m t → h t weakly star in L ∞ (0, T ), θ m → θ weakly star in L ∞ (0, T ), θ t m → θ t weakly star in L ∞ (0, T ).
Due to the compact embedding V ֒→ L 2 (0, L), we can get a subsequence such that
(2.27) y m t → y t strongly in L 2 (0, T ; L 2 (0, L)).
From (H 2 ) and (2.27), we get
(2.28) g(y m t ) → g(y t ) a.e. in x ∈ (0, L), t > 0.
From the above convergence and due to the boundedness of the sequence {g(y m t )} in L 2 (0, T ; L 2 (0, L)), we conclude by Lion’s Lemma that
g(y t m ) → g(y t ) weakly in L 2 (0, T ; L 2 (0, L)).
Moreover, by the Sobolev embedding theorem and (2.26), we see that h ∈ C 1 [0, T ] and h m (t)y t m (L, t) → h(t)y t (L, t) weakly in L 2 (0, T ).
We also derive that
θ ∈ C 1 [0, T ] and θ m (t) sin t → θ(t) sin t weakly in L 2 (0, T ).
The above convergence results are sufficient to pass to the limit in the nonlin- ear terms of (2.7). Then it is a matter of routine to deduce the existence of global solutions in [0, T ]. The uniqueness can be proved by the straightforward
methods and Gronwall’s inequality.
3. Exponential stability
Having established global existence of solution to (1.3), we focus our atten- tion on exponential decay that can be obtained for the energy function.
We define the energy E(t) of problem (1.3) by E(t) = 1 2 ky t (t)k 2 + 1 2 ky xx (t)k 2 . Then the derivative of the energy is given by
E t (t) = − (g(y t (t)), y t (t)) + Z t
0
κ(t − τ )(y xx (τ ), y xxt (τ ))dτ − h(t)(y t (L, t)) 2
− [θ(t) − e θ] sin t · y t (L, t).
Defining
(3.1) (κy xx )(t) = Z t
0
κ(t − τ )ky xx (τ ) − y xx (t)k 2 dτ.
By simple computation, we obtain (κy xx ) t (t) = (κ t y xx )(t) − 2
Z t 0
κ(t − τ )(y xx (τ ), y xxt (t))dτ (3.2)
+ d dt
ky xx (t)k 2 Z t
0
κ(τ )dτ
− κ(t)ky xx (t)k 2 . From (3.2), we see that
Z t 0
κ(t − τ )(y xx (τ ), y xxt (t))dτ (3.3)
= − 1
2 (κy xx ) t (t) + 1
2 (κ t y xx )(t) + 1 2
d dt
ky xx (t)k 2 Z t
0
κ(τ )dτ
− 1
2 κ(t)ky xx (t)k 2 . Define the modified energy by
e(t) = 1
2 ky t (t)k 2 + 1
2 (κy xx )(t) + 1 2
1 − Z t
0
κ(τ )dτ
ky xx (t)k 2 (3.4)
+ 1
2 [θ(t) − e θ] 2 . Then, from (3.3) and (3.4) we obtain
e t (t) = − (g(y t (t)), y t (t)) + 1
2 (κ t y xx )(t) − 1
2 κ(t)ky xx (t)k 2 − h(t)(y t (L, t)) 2 (3.5)
≤ − (g(y t (t)), y t (t)) − m 1
2 (κy xx )(t) − 1
2 κ(t)ky xx (t)k 2
− h(t)(y t (L, t)) 2 ≤ 0, ∀t ≥ 0.
We observe that in view of hypotheses (2.3) and (2.4), we see that e(t) ≥ 0 and E(t) ≤ ℓ −1 e(t). Therefore it is enough to obtain the desired exponential stability for the modified energy e(t) which will be done below. In order to carry the proof of Theorem 3.2, we need the following Theorem 3.1.
Theorem 3.1. Let y be the solution given by Theorem 2.1 and e(t) be defined by (3.4). Then we have
t→∞ lim e(t) = 0, lim
t→∞ h(t) ≤ p
2re(0) + (h(0)) 2 , h(t) ≤ p
2re(0) + (h(0)) 2 , ∀t ≥ 0.
Now, we define the perturbed energy by
e ε (t) = e(t) + εψ(t), where ψ(t) = δ Z t
0
y(x, t)y t (x, t)dx with 0 < δ < 1 2 . (3.6)
Then, we obtain the following two propositions.
Proposition 3.1. There exists C 1′ > 0 such that
(3.7) |e ε (t) − e(t)| ≤ εC 1′e(t), ∀t ≥ 0 and ∀ε > 0.
Proof. The proof of this proposition is the same the proof of [15, Proposition
3.1], so we will omit this proof.
Proposition 3.2. There exist positive constants C 2 ′ , C 3 ′ and C 4 ′ such that d
dt e ε (t) ≤ − εC 2 ′ e(t) + εC 3 ′ (1 + h(t))(y(L, t)) 2 + εC 4 ′ [θ(t) − e θ] 2 ,
∀t ≥ 0 and ∀ε ∈ [0, ε 1 ].
Proof. Using (1.1), we deduce that d
dt ψ(t) = δ Z L
0
y 2 t (t)dx + δ Z L
0
y(t)y tt (t)dx (3.8)
= δ Z L
0
(y t (t)) 2 dx + δ Z t
0
y(t)(−y xxxx (t))dx
+ δ Z L
0
y(t) Z t
0
κ(t − τ )y xxxx (τ )dτ dx
− δ Z L
0
y(t)g(y t (t))dx.
By y(0, t) = y x (0, t) = y xx (L, t) = 0 and y ∈ L ∞ (0, T ; W ), we have the follow- ing two inequalities
Z L 0
y(t)(−y xxxx (t))dx
= − y(L, t)y xxx (L, t) + Z L
0
y x (t)y xxx (t)dx
= − y(L, t)y xxx (L, t) + y x (L, t)y xx (L, t) − Z L
0
(y xx (t)) 2 dx
= − y(L, t)y xxx (L, t) − Z L
0
(y xx (t)) 2 dx and Z L
0
y(t) Z t
0
κ(t − τ )y xxxx (τ )dτ dx
= y(L, t) Z t
0
κ(t − τ )y xxx (L, τ )dτ
− Z L
0
y x (t) Z t
0
κ(t − τ )y xxx (τ )dτ dx
= y(L, t) Z t
0
κ(t − τ )y xxx (L, τ )dτ
− y x (L, t) Z t
0
κ(t − τ )y xx (L, τ )dτ
+ Z L
0
y xx (t) Z t
0
κ(t − τ )y xx (τ )dτ dx
= y(L, t) Z t
0
κ(t − τ )y xxx (L, τ )dτ +
Z L 0
y xx (t) Z t
0
κ(t − τ )y xx (τ )dτ dx.
Thus, from (3.8) we get
d
dt ψ(t) = δ Z t
0
(y t (t)) 2 dx − δy(L, t) h
y xxx (L, t) − Z t
0
κ(t − τ )y xxx (L, τ )dτ i (3.9)
− δ Z L
0
(y xx (t)) 2 dx + δ Z L
0
y xx (t) Z t
0
κ(t − τ )y xx (τ )dτ dx
− δ Z L
0
y(t)g(y t (t))dx.
From the assumptions on g, Young’s inequality and Sobolev imbedding theo- rem, we see that
(3.10) δ Z t
0
y(t)g(y t (t))dx ≤ M 1 ′ ky t (t)k 2 + M 2 ′ ky xx (t)k 2 , where M 1 ′ and M 2 ′ are some positive constants.
Moreover, for any η > 0, from Fubini’s theorem, we have Z L
0
y xx (t) Z t
0
κ(t − τ )y xx (τ )dτ dx (3.11)
= Z t
0
κ(t − τ ) Z L
0
y xx (τ )y xx (t)dx dτ
= Z t
0
κ(t − τ ) Z L
0
(y xx (τ ) − y xx (t))y xx (t)dx
dτ
+ Z t
0
κ(t − τ ) Z L
0
(y xx (t)) 2 dx dτ
≤ ηky xx (t)k 2 + 1 4η
Z t 0
Z L 0
(κ(t − τ )) 2 (y xx (τ ) − y xx (t)) 2 dxdτ
+ Z t
0
κ(τ ) Z L
0
(y xx (t)) 2 dx dτ
≤ ηky xx (t)k 2 + 1
4η kκk L∞(0,∞) (κy xx )(t) + Z t
0
κ(τ )dτ ky xx (t)k 2 . Combining (3.9)-(3.11), we obtain
d
dt ψ(t) ≤ δ Z L
0
(y t (t)) 2 dx − δy(L, t) h
y xxx (L, t) − Z t
0
κ(t − τ )y xxx (L, τ )dτ i (3.12)
− δ Z t
0
(y xx (t)) 2 dx + M 1 ′ ky t (t)k 2 + M 2 ′ ky xx (t)k 2 + ηδky xx (t)k 2
+ δ
4η kκk L∞(0,∞) (κy xx )(t) + δ Z t
0
κ(τ )dτ ky xx (t)k 2
≤ (δ + M 1 ′ )ky t (t)k 2 − δy(L, t)[h(t)y t (L, t) + (θ(t) − e θ) sin t]
+
M 2 ′ + ηδ + δ Z t
0
κ(τ )dτ − δ
ky xx (t)k 2 + δ
4η kκk L∞(0,∞) (κy xx )(t)
= − δe(t) + M 1 ′ + 3
2 δ ky t (t)k 2 +
M 2 ′ + ηδ + δ 2
Z t 0
κ(τ )dτ − δ 2
ky xx (t)k 2
+ δ 2 + δ
4η kκk L∞(0,∞)
(κy xx )(t)
− δy(L, t)[h(t)y t (L, t) + (θ(t) − e θ) sin t] + δ
2 [θ(t) − e θ] 2 . On the other hand, from (2.2), (3.5), (3.6) and (3.12), we get
d
dt e ε (t) = d
dt e(t) + ε d dt ψ(t) (3.13)
= − (g(y t (t)), y t (t)) − m 1
2 (κy xx )(t) − 1
2 κ(t)ky xx (t)k 2
− h(t)(y t (L, t)) 2 − εδe(t) + ε M 1 ′ + 3
2 δ
ky t (t)k 2 + ε
M 2 ′ + ηδ + δ 2
Z t 0
κ(τ )dτ − δ 2
ky xx (t)k 2
+ ε δ 2 + δ
4η kκk L∞(0,∞)
(κy xx )(t)
− εδy(L, t)[h(t)y t (L, t) + (θ(t) − e θ) sin t] + εδ
2 [θ(t) − e θ] 2
≤ − εδe(t) − µ − ε
M 1 ′ + 3 2 δ
ky t (t)k 2
− m 1
2 − ε δ 2 + δ
4η kκk L∞(0,∞)
(κy xx )(t)
− 1
2 κ(t) − ε
M 2 ′ + ηδ + δ 2
Z t 0
κ(τ )dτ − δ 2
ky xx (t)k 2
− h(t) 1 − δε
2
(y t (L, t)) 2 + εδ
2 (1 + h(t))(y(L, t)) 2 + εδ[θ(t) − e θ] 2 . Now, we define
ε 1 = min n 2µ
2M 1 ′ + 3δ , 2m 1 η
δ(2η + kκk L∞(0,∞) ) , κ(t) 2M 2 ′ + 2ηδ + δ R t
0 κ(τ )dτ − δ , 2 δ
o . (3.14)
Considering ε ∈ [0, ε 1 ] and choosing δ > 0 such that (2η − ℓ)δ + 2M 2 ′ > δ, then from (3.13) and (3.14), we obtain
d
dt e ε (t) ≤ −εδe(t) + εδ
2 (1 + h(t))(y(L, t)) 2 + εδ[θ(t) − e θ] 2 . (3.15)
This completes the proof of Proposition 3.2.
Proof of Theorem 3.1. From Proposition 3.1, we see that (3.16) (1 − εC 1 ′ )e(t) ≤ e ε (t) ≤ (1 + εC 1 ′ )e(t), ∀t ≥ 0.
From (3.15) and (3.16), we have d
dt e ε (t) ≤ −εδ
1 + εC 1 ′ e ε (t) + εδ
2 (1 + h(t))(y(L, t)) 2 + εδ[θ(t) − e θ] 2 . (3.17)
Let C ε = εδ/(1 + εC 1 ′ ) and apply Gronwall’s inequality to (3.17), we get e ε (t) ≤ e −Cεt e ε (0) + εδ
2 Z t
0
e −Cε(t−τ ) (1 + h(τ ))(y(L, τ )) 2 dτ (3.18)
+ εδ Z t
0
e −Cε(t−τ ) [θ(t) − e θ] 2 dτ
≤ e −Cεt e ε (0) + εδ 2 sup
t≥0
[1 + |h(t)|]
Z t 0
e −Cε(t−τ ) (y(L, τ )) 2 dτ
+ εδ Z t
0
e −Cε(t−τ ) [θ(t) − e θ] 2 dτ.
Referring to the paper [7], we deduce that Z t
0
e −Cε(t−τ ) (y(L, τ )) 2 dτ
≤ Z
t20
e −Cε(t−τ ) (y(L, τ )) 2 dτ + Z t
t 2
e −Cε(t−τ ) (y(L, τ )) 2 dτ
≤ max
0≤τ ≤
t2
e −Cε(t−τ ) Z
2t
0
(y(L, τ )) 2 dτ + max
t2
≤τ ≤t e −C
ε(t−τ ) Z t
t 2
(y(L, τ )) 2 dτ
≤ e −Cε2 t Z
2t
0
(y(L, τ )) 2 dτ + Z ∞
t 2
(y(L, τ )) 2 dτ.
By Theorem 2.1 and W ֒→ L 2 (0, T ; L 2 (0, L)), y(L, t) ∈ L 2 (0, ∞), we obtain (3.19)
Z t 0
e −Cε(t−τ ) (y(L, τ )) 2 dτ → 0 as t → ∞.
Similarly, we see that (3.20)
Z t 0
e −Cε(t−τ ) [θ(τ ) − e θ] 2 dτ → 0 as t → ∞.
From (3.18), (3.19) and (3.20), we deduce
(3.21) lim
t→∞ e ε (t) = 0.
Let ε 0 = min{ε 1 , 1/(2C 1 ′ )}, where C 1 ′ is given in Proposition 3.1.
Since ε ≤ 1/(2C 1 ′ ) (ε ∈ (0, ε 0 ]) and from (3.16), we get
(3.22) 1
2 e(t) ≤ e ε (t) ≤ 3 2 e(t).
Therefore, from (3.21) and (3.22), we have
t→∞ lim e(t) = 0.
(3.23)
Now, we consider the Lyapunov functional U (t) for the system (1.3) as follows:
U (t) = e(t) + (h(t)) 2 2r . Then from (3.5), we see that
(3.24) U t (t) ≤ 0.
Since y t (L, ·) ∈ L 2 (0, ∞), we obtain sup
t≥0
n e(t) + (h(t)) 2 2r
o ≤ M 3 ′ ,
where M 3 ′ > 0 is a constant depending on the initial data.
From U (t) is decreasing, we get e(∞) + 1
2r (h(∞)) 2 ≤ e(0) + 1
2r (h(0)) 2 .
(3.25)
Thus, from (3.23) and (3.25), we have h(∞) ≤ p
2re(0) + (h(0)) 2 . Since h(t) is nondecreasing, we deduce that
(3.26) h(t) ≤ p
2re(0) + (h(0)) 2 .
Thus, the proof of Theorem 3.1 is completed.
We can now proceed to state our main exponential stability result.
Theorem 3.2. Let y be the solution of Theorem 2.1 and let e(t) be defined by (3.4). Then there exist constants K > 0 and ν > 0 depending on the initial data such that
E(t) ≤ Ke −νt , ∀t ≥ 0.
Proof. From (3.17), we have d
dt e ε (t) ≤ −C ε e ε (t) + εδ
2 (1 + h(t))(y(L, t)) 2 + εδ[θ(t) − e θ] 2 , (3.27)
where C ε = δε/(1 + εC 1 ′ ).
Using (3.26) and by integrating over (0, t) in (3.27) we obtain e ε (t) ≤ e ε (0) − C ε
Z t 0
e ε (τ )dτ + εδ 2
Z t 0
(1 + h(τ ))(y(L, τ )) 2 dτ
+ εδ Z t
0
[θ(t) − e θ] 2 dτ
≤ e ε (0) − C ε
Z t 0
e ε (τ )dτ + εδ
2 (1 + khk L∞(0,∞) ) Z t
0
(y(L, τ )) 2 dτ + εδ
Z t 0
[θ(t) − e θ] 2 dτ.
Since R ∞
0 (y(L, τ )) 2 dτ and R ∞
0 [θ(t) − e θ] 2 dτ are bounded, by Gronwall’s inequal- ity, we deduce that
e ε (t) ≤ (K 1 + e ε (0)) exp(−C ε t), where K 1 is some positive constant.
For sufficiently small ε
0 < ε < 2C 1′
1