To address the problems of facial feature point recognition clarity and recognition efficiency in different human motion conditions, a facial feature point recognition method using Genetic Neural Network (GNN) algorithm was proposed. As the technical platform, weoll be using the Hikey960 development board. The optimized BP neural network algorithm is used to collect and classify human motion facial images, and the genetic algorithm is introduced into neural network algorithm to train human motion facial images. Combined with the improved GNN algorithm, the facial feature points are detected by the dynamic transplantation of facial feature points, and the detected facial feature points are transferred to the face alignment algorithm to realize facial feature point recognition. The results show that the efficiency and accuracy of facial feature point recognition in different human motion images are higher than 85% and the performance of anti-noise is good, the average recall rate is about 90% and the time-consuming is short. It shows that the proposed method has a certain reference value in the field of human motion image recognition.
Citation: Qingwei Wang, Xiaolong Zhang, Xiaofeng Li. Facial feature point recognition method for human motion image using GNN[J]. Mathematical Biosciences and Engineering, 2022, 19(4): 3803-3819. doi: 10.3934/mbe.2022175
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To address the problems of facial feature point recognition clarity and recognition efficiency in different human motion conditions, a facial feature point recognition method using Genetic Neural Network (GNN) algorithm was proposed. As the technical platform, weoll be using the Hikey960 development board. The optimized BP neural network algorithm is used to collect and classify human motion facial images, and the genetic algorithm is introduced into neural network algorithm to train human motion facial images. Combined with the improved GNN algorithm, the facial feature points are detected by the dynamic transplantation of facial feature points, and the detected facial feature points are transferred to the face alignment algorithm to realize facial feature point recognition. The results show that the efficiency and accuracy of facial feature point recognition in different human motion images are higher than 85% and the performance of anti-noise is good, the average recall rate is about 90% and the time-consuming is short. It shows that the proposed method has a certain reference value in the field of human motion image recognition.
This paper is concerned with the initial boundary value problem
{utt(x,t)−uxx(x,t)+μ1(t)ut(x,t)+μ2(t)ut(x,t−τ(t))=0in Ω×]0,+∞[,u(0,t)=u(L,t)=0on ]0,+∞[,u(x,0)=u0(x),ut(x,0)=u1(x)on Ω,ut(x,t−τ(0))=f0(x,t−τ(0))in Ω×]0,τ(0)[, | (1) |
where
This problem has been first proposed and studied in Nicaise and Pignotti [22] in case of constant coefficients
With a weight depending on time,
W. Liu in [19] studied the weak viscoelastic equation with an internal time varying delay term. By introducing suitable energy and Lyapunov functionals, he establishes a general decay rate estimate for the energy under suitable assumptions.
F. Tahamtani and A. Peyravi [29] investigated the nonlinear viscoelastic wave equation with source term. Using the Potential well theory they showed that the solutions blow up in finite time under some restrictions on initial data and for arbitrary initial energy.
Global existence and asymptotic behavior of solutions to the viscoelastic wave equation with a constant delay term was considered by M. Remil and A. Hakem in [28].
Global existence and asymptotic stability for a coupled viscoelastic wave equation with time-varying delay was studied in [3] by combining the energy method with the Faedo-Galerkin's procedure.
The stabilization problem by interior damping of the wave equation with boundary or internal time-varying delay was studied in [23] by introducing suitable Lyapunov functionals.
Energy decay of solutions for the wave equation with a time varying delay term in the weakly nonlinear internal feedbacks was considered in [11].
For problems with delay in different contexts we cite [9,10,30,32] with references therein. In absence of delay (
Time delay is the property of a physical system by which the response to an applied force is delayed in its effect, and the central question is that delays source can destabilize a system that is asymptotically stable in the absence of delays, see [7]. In fact, an arbitrarily small delay may destabilize a system that is uniformly asymptotically stable in the absence of delay unless additional control terms have been used, see for example [8,12,31]
By energy method in [24] was studied the stabilization of the wave equation with boundary or internal distributed delay. By semigroup approach in [27] was proved the well-posedness and exponential stability for a wave equation with frictional damping and nonlocal time-delayed condition. Transmission problem with distributed delay was studied in [18] where was established the exponential stability of the solution by introducing a suitable Lyapunov functional.
Here we consider a wave equation with non-constant delay and nonlinear weights, thus, the present paper is a generalization of the previous ones. The remaining part of this paper is organized as follows. In the section 2 we introduce some notations and prove the dissipative property of the full energy of the system. In the section 3, for an approach combining semigroup theory (see [21] and [4]) with the energy estimate method we prove the existence and uniqueness of solution. In section 4 we present the result of exponential stability.
We will need the following hypotheses:
(H1)
|μ′1(t)μ1(t)|≤M1,0<α0≤μ1(t),∀t≥0, | (2) |
where
(H2)
|μ2(t)|≤βμ1(t), | (3) |
|μ′2(t)|≤M2μ1(t), | (4) |
for some
We now state a lemma needed later.
Lemma 2.1 (Sobolev-Poincare's inequality). Let
‖Ψ‖q≤c∗‖Ψx‖2,forΨ∈H10(]0,L[). |
Lemma 2.2 ([13][16]). Let
∫+∞SE1+σ(t)dt≤1ωEσ(0)E(S), 0≤S<+∞. |
Then
E(t)=0 ∀t≥Eσ(0)ω|σ|, if−1<σ<0,E(t)≤E(0)(1+σ1+ωσt)1σ ∀t≥0, ifσ>0,E(t)≤E(0)e1−ωt ∀t≥0, ifσ=0. |
As in [23], we assume that
τ(t)∈W2,+∞([0,T]), for T>0 | (5) |
and there exist positive constants
0<τ0≤τ(t)≤τ1, ∀t>0 | (6) |
and
τ′(t)≤d<1, ∀t>0. | (7) |
We introduce the new variable
z(x,ρ,t)=ut(x,t−τ(t)ρ), x∈Ω,ρ∈]0,1[,t>0. | (8) |
Then
τ(t)zt(x,ρ,t)+(1−τ′(t)ρ)zρ(x,ρ,t)=0, x∈Ω, ρ∈]0,1[, t>0 |
and problem (1) takes the form
{utt(x,t)−uxx(x,t)+μ1(t)ut(x,t)+μ2(t)z(x,1,t)=0inΩ×]0,+∞[,τ(t)zt(x,ρ,t)+(1−τ′(t)ρ)zρ(x,ρ,t)=0inΩ×]0,1[×]0,+∞[,u(0,t)=u(L,t)=0on]0,+∞[,u(x,0)=u0(x),ut(x,0)=u1(x)onΩ,z(x,ρ,0)=ut(x,−τ(0)ρ)=f0(x,−τ(0)ρ)inΩ×]0,1[. | (9) |
We define the energy of the solution of problem (9) by
E(t)=12‖ut‖2L2(Ω)+12‖ux‖2L2(Ω)+ξ(t)τ(t)2∫Ω∫10z2(x,ρ,t)dρdx, | (10) |
where
ξ(t)=ˉξμ1(t) | (11) |
is a non-increasing function of class
β√1−d<ˉξ<2−β√1−d. | (12) |
Our first result states that the energy is a non-increasing function.
Lemma 2.3. Let
E′(t)≤−μ1(t)(1−ˉξ2−β2√1−d)‖ut‖2L2(Ω)−μ1(t)(ˉξ(1−τ′(t))2−β√1−d2)‖z(x,1,t)‖2L2(Ω)≤0. | (13) |
Proof. Multiplying the first equation (9) by
12ddt(‖ut‖2L2(Ω)+‖ux‖2L2(Ω))+μ1(t)‖ut‖2L2(Ω)+μ2(t)∫Ωz(x,1,t)utdx. | (14) |
Now multiplying the second equation (9) by
τ(t)ξ(t)∫Ω∫10zt(x,ρ,t)z(x,ρ,t)dρdx=−ξ(t)2∫Ω∫10(1−τ′(t)ρ)∂∂ρ(z(x,ρ,t))2dρdx. |
Consequently,
ddt(ξ(t)τ(t)2∫Ω∫10z2(x,ρ,t)dρdx)=−ξ(t)2∫Ω∫10(1−τ′(t)ρ)∂∂ρ(z(x,ρ,t))2dρdx+ξ′(t)τ(t)2∫Ω∫10z2(x,ρ,t)dρdx=ξ(t)2∫Ω(z2(x,0,t)−z2(x,1,t))dx+ξ(t)τ′(t)2∫Ω∫10z2(x,1,t)dρdx+ξ′(t)τ(t)2∫Ω∫10z2(x,ρ,t)dρdx. | (15) |
From (10), (14) and (15) we obtain
E′(t)=ξ(t)2‖ut‖2L2(Ω)−ξ(t)2‖z(x,1,t)‖2L2(Ω)+ξ(t)τ′(t)2‖z(x,1,t)‖2L2(Ω)+ξ′(t)τ(t)2∫Ω∫10z2(x,ρ,t)dρdx−μ1(t)‖ut‖2L2(Ω)−μ2(t)∫Ωz(x,1,t)utdx. | (16) |
Due to Young's inequality, we have
μ2(t)∫Ωz(x,1,t)utdx≤|μ2(t)|2√1−d‖ut‖2L2(Ω)+|μ2(t)|√1−d2‖z(x,1,t)‖2L2(Ω). | (17) |
Inserting (17) into (16), we obtain
E′(t)≤−(μ1(t)−ξ(t)2−|μ2(t)|2√1−d)‖ut‖2L2(Ω)−(ξ(t)2−ξ(t)τ′(t)2−|μ2(t)|√1−d2)‖z(x,1,t)‖2L2(Ω)+ξ′(t)τ(t)2∫Ω∫10z2(x,ρ,t)dρdx≤−μ1(t)(1−ˉξ2−β2√1−d)‖ut‖2L2(Ω)−μ1(t)(ˉξ(1−τ′(t))2−β√1−d2)‖z(x,1,t)‖2L2(Ω)≤0. |
Lemma 2.4. Let
‖ut(x,t)‖2L2(Ω)<−1σE′(t), |
where
Proof. From Lemma 2.3, we have that
−E′(t)≥μ1(t)(1−ˉξ2+β2√1−d)‖ut‖2L2(Ω)+μ1(t)(ˉξ(1−τ′(t))2+β√1−d2)‖z(x,1,t)‖2L2(Ω)≥0 |
and from (H1), we obtain
0≤a0(1−ˉξ2+β2√1−d)‖ut‖2L2(Ω)≤μ1(t)(1−ˉξ2+β2√1−d)‖ut‖2L2(Ω)≤−E′(t) |
and the lemma is proved.
For the semigroup setup we
{Ut=A(t)U,U(0)=U0=(u0,u1,f0(⋅,−,τ(0)))T, | (18) |
where the operator
AU=(v,uxx−μ1(t)v−μ2(t)z(x,1,t),−1−τ′(t)ρτ(t)zρ(x,ρ,t))T. | (19) |
We introduce the phase space
H=H10(Ω)×L2(Ω)×L2(Ω×]0,1[) |
and the domain of
D(A(t))={(u,v,z)T∈H/v=z(⋅,0) in Ω}, | (20) |
where
H=H2(Ω)∩H10(Ω)×H10(Ω)×L2(Ω;H10(]0,1[)). |
Notice that the domain of the operator
D(A(t))=D(A(0)),∀t>0. | (21) |
⟨U,ˉU⟩H=∫Ωuxˉuxdx+∫Ωvˉvdx+ξ(t)τ(t)∫Ω∫10zˉzdρdx, | (22) |
for
Using this time-dependent inner product and the next theorem we will get a result of existence and uniqueness.
Theorem 3.1. Assume that
(i)
(ii) (21) holds,
(iii) for all
(iv)
Then, problem (18) has a solution
Our goal is then to check the above assumptions for problem (18).
First, we prove
The proof is the same as the one Lemma
Let
0=⟨(u,v,z)T,(f,g,h)T⟩H=∫Ωuxfxdx+∫Ωvgdx+ξ(t)τ(t)∫Ω∫10zhdρdx, |
for all
We first take
∫Ω∫10zhdρdx=0. |
Since
The above orthogonality condition is then reduced to
0=∫Ωuxfxdx,∀(u,v,z)T∈D(A(0)). |
By restricting ourselves to
0=∫Ωuxfxdx,∀(u,0,0)T∈D(A(0)). |
Since
We consequently
D(A(0) is dense in H. | (23) |
Secondly, we notice that
‖Φ‖t‖Φ‖s≤ec2τ0|t−s|,∀t,s∈[0,T], | (24) |
where
‖Φ‖2t−‖Φ‖2secτ0|t−s|=(1−ec2τ0|t−s|)(‖ux‖2L2(Ω)+‖v‖2L2(Ω))+(ξ(t)τ(t)−ξ(s)τ(s)ecτ0|t−s|)∫Ω∫10z2(x,ρ,t)dρdx. |
It is clear that
τ(t)=τ(s)+τ′(r)(t−s), |
where
Hence
ξ(t)τ(t)≤ξ(s)τ(s)+ξ(s)τ′(r)(t−s), |
which implies
ξ(t)τ(t)ξ(s)τ(s)≤1+|τ′(r)|τ(s)|t−s|. |
Using (5) and
ξ(t)τ(t)ξ(s)τ(s)≤1+cτ0|t−s|≤ecτ0|t−s|, |
which proves (24) and therefore
Now we calculate
⟨A(t)U,U⟩t=∫Ωvxuxdx+∫Ω(uxx−μ1(t)v−μ2(t)z(⋅,1))vdx−ξ(t)∫Ω∫10(1−τ′(t)ρ)zρ(x,ρ)z(x,ρ)dρdx. |
Integrating by parts, we obtain
⟨A(t)U,U⟩t=−μ1(t)‖v‖2L2(Ω)−μ2(t)∫Ωz(⋅,1)vdx−∫Ω∫10(1−τ′(t)ρ)∂∂ρz2(x,ρ)dρdx. |
Since
(1−τ′(t)ρ)∂∂ρz2(x,ρ)=∂∂ρ((1−τ′(t)ρ)z2(x,ρ))+τ′(t)z2(x,ρ), |
we have
∫10(1−τ′(t)ρ)∂∂ρz2(x,ρ)dρ=(1−τ′(t))z2(x,1)−z2(x,0)+τ′(t)∫10z2(x,ρ)dρ. |
So we get
⟨A(t)U,U⟩t=−μ1(t)‖v‖2L2(Ω)−μ2(t)∫Ωz(x,1)vdx+ξ(t)2‖z(x,0)‖2L2(Ω)−ξ(t)(1−τ′(t))2‖z(x,1)‖2L2(Ω)−ξ(t)τ′(t)2∫Ω∫10z2(x,ρ)dρdx. |
Therefore, from (16) and (17), we deduce
⟨A(t)U,U⟩t≤−μ1(t)(1−ˉξ2−β2√1−d)‖v‖2L2(Ω)−μ1(t)(ˉξ(1−τ′(t))2−β√1−d2)‖z(x,1,t)‖2L2(Ω)+ξ(t)|τ′(t)|2τ(t)τ(t)∫Ω∫10z2(x,ρ)dρdx. |
Then, we have
⟨A(t)U,U⟩t≤−μ1(t)(1−ˉξ2−β2√1−d)‖v‖2L2(Ω)−μ1(t)(ˉξ(1−τ′(t))2−β√1−d2)‖z(x,1,t)‖2L2(Ω)+κ(t)⟨U,U⟩t, |
where
κ(t)=√1+τ′(t)22τ(t). |
From the (13), we obtain
⟨A(t)U,U⟩t−κ(t)⟨U,U⟩t≤0, | (25) |
which means that the operator
Moreover,
ddtA(t)U=(0,0,τ″(t)τ(t)ρ−τ′(t)(τ′(t)ρ−1)τ(t)2zρ)T, |
with
ddt˜A(t)∈L∞∗([0,T],B(D(A(0)),H)), | (26) |
the space of equivalence classes of essentially bounded, strongly measurable functions from
Now, we will show that
(λI−A(t))U=F, |
that is verifying following system of equations
{λu−v=f1,λv−uxx+μ1(t)v−μ2(t)z(⋅,1)=f2,λz+1−τ′(t)ρτ(t)zρ=f3. | (27) |
Suppose that we have found
v=λu−f1. | (28) |
It is clear that
z(x,0)=v(x), for x∈Ω. | (29) |
Following the same approach as in [22], we obtain, by using equation for
z(x,ρ)=v(x)e−ϑ(ρ,t)+τ(t)e−ϑ(ρ,t)∫ρ0f3(x,s)eϑ(s,t)ds, |
if
z(x,ρ)=v(x)eζ(ρ,t)+eζ(ρ,t)∫ρ0τ(t)f3(x,s)1−sτ′(s)e−ζ(s,t)ds, |
otherwise, where
From (28), we obtain
z(x,ρ)=λu(x)e−ϑ(ρ,t)−f1(x,ρ)e−ϑ(ρ,t)+τ(t)e−ϑ(ρ,t)∫ρ0f3(x,s)eϑ(s,t)ds, | (30) |
if
z(x,ρ)=λu(x)eζ(ρ,t)−f1(x,ρ)eζ(ρ,t)+eζ(ρ,t)∫ρ0τ(t)f3(x,s)1−sτ′(s)e−ζ(s,t)ds, | (31) |
otherwise.
In particular, if
z(x,1)=λu(x)e−ϑ(1,t)−f1(x,1)e−ϑ(1,t)+τ(t)e−ϑ(1,t)∫10f3(x,s)eϑ(s,t)ds, | (32) |
and if
z(x,1)=λu(x)eζ(1,t)−f1(x,1)eζ(1,t)+eζ(1,t)∫10τ(t)f3(x,s)1−sτ′(s)e−ζ(s,t)ds. | (33) |
By using (27) and (28), the function
λ2u−uxx+μ1(t)v+μ2(t)z(⋅,1)=f2+λf1. | (34) |
Solving the equation (34) is equivalent to finding
∫Ω(λ2uη+uxηx+μ1(t)vη+μ2(t)z(⋅,1)η)dx=∫Ω(f2+λf1)ηdx, | (35) |
for all
Consequently, the equation (35) is equivalent to the problem
Υ(u,η)=L(η), | (36) |
where the bilinear form
Υ:H10(Ω)×H10(Ω)→R |
and the linear form
L:H10(Ω)→R |
are defined by
Υ(u,η)=∫Ω(λ2uη+uxηx)dx+∫Ωλu(μ1(t)+μ2(t)N1)ηdx |
and
L(η)=∫Ω(μ1(t)f1η+μ2(t)N2)ηdx+∫Ω(f2+λf1)ηdx, |
where
N1={e−ϑ(1,t),ifτ′(t)=0,eζ(1,t),ifτ′(t)≠0 |
and
N2={−f1(x,1)e−ϑ(1,t)+τ(t)e−ϑ(1,t)∫10f3(x,s)eϑ(s,t)ds,ifτ′(t)=0,−f1(x,1)ezeta(1,t)+ezeta(1,t)∫10τ(t)f3(x,s)1−sτ′(t)e−ζ(s,t)ds,ifτ′(t)≠0. |
It is easy to verify that
u∈H10(Ω). |
Applying the classical elliptic regularity, it follows from (35) that
u∈H2(Ω). |
Therefore, the operator
λI−˜A(t)=(λ+κ(t))I−A(t) is surjective, | (37) |
for any
Then, (24), (25) and (37) imply that the family
{˜Ut=˜A(t)˜U,˜U(0)=U0=(u0,u1,f0(⋅,−,τ(0)))T | (38) |
has a unique solution
U(t)=e∫t0κ(s)ds˜U(t) |
because
Ut(t)=κ(t)e∫t0κ(s)ds˜U(t)+e∫t0κ(s)ds˜Ut(t)=e∫t0κ(s)ds(κ(t)+˜A(t))˜U(t)=A(t)e∫t0κ(s)ds˜U(t)=A(t)U(t), |
which concludes the proof.
The existence and uniqueness are obtained by the following result.
Theorem 3.2 (Global solution). For any initial datum
U∈C([0,+∞[,H) |
for problem (18).
Moreover, if
U∈C([0,+∞[,D(A(0)))∩C1([0,+∞[,H). |
Proof. A general theory for equations of type (18) has been developed using semigroup theory [14], [15] and [26]. The simplest way to prove existence and uniqueness results in to show that the triplet
In this section we shall investigate the asymptotic behavior of problem (1). The stability result will be obtained using Lemma 2.2.
Theorem 4.1 (Stability Result). Let
u∈C([0,+∞[,H10(Ω))∩C1([0,+∞[,L2(Ω)), |
z∈C([0,+∞[,L2(Ω)×]0,1[). |
Proof. From now on, we denote by
Given
∫TSEq∫Ωu(utt−uxx+μ1(t)ut+μ2(t)z(x,1,t))dxdt=0. |
Notice that
uttu=(utu)t−u2t, |
using integration by parts and the boundary conditions we know that
0=[Eq(t)∫Ωuutdx]TS−∫TSqEq−1(t)E′(t)∫Ωuutdxdt−∫TSEq(t)‖ut‖2L2(Ω)dt+∫TSEq(t)‖ux‖2L2(Ω)dt+∫TSEq(t)∫Ωμ1(t)uutdxdt+∫TSEq(t)∫Ωμ2(t)uz(x,1,t)dxdt. | (39) |
Similarly, we multiply the second equation of (9) by
0=∫TS∫Ω∫10Eq(t)ξ(t)e−2ρτ(t)z(τ(t)zt+(1−ρτ′(t))zρ)dρdxdt |
=12∫Ω∫10∫TSEq(t)ξ(t)e−2ρτ(t)∂∂tz2dtdρdx+12∫TSEq(t)ξ(t)∫Ω∫10e−2ρτ(t)(1−ρτ′(t))∂∂ρz2dρdxdt. |
Using integration by parts and the boundary conditions we know that
0=[ξ(t)τ(t)2Eq(t)∫Ω∫10e−2ρτ(t)z2dρdx]TS−12∫TSqEq−1(t)E′(t)ξ(t)τ(t)∫Ω∫10e−2ρτ(t)z2dρdxdt−12∫TSqEq(t)ξ′(t)τ(t)∫Ω∫10e−2ρτ(t)z2dρdxdt+12∫TSEq(t)ξ(t)∫Ω[e−2ρτ(t)(1−τ′(t))z2(x,1,t)−z2(x,0,t)]dxdt+∫TSEq(t)ξ(t)τ(t)∫Ω∫10e−2ρτ(t)z2dρdxdt. | (40) |
Since
∫TSqEq(t)ξ′(t)τ(t)∫Ω∫10e−2ρτ(t)z2dρdxdt≤0. | (41) |
Moreover, as
−12∫TSEq(t)ξ(t)∫Ωe−2ρτ(t)(1−τ′(t))z2(x,1,t)dxdt≤0, | (42) |
then, from (40), (41) and (42), we have that
∫TSEq(t)ξ(t)τ(t)∫Ω∫10e−2ρτ(t)z2dρdxdt≤−[ξ(t)τ(t)2Eq(t)∫Ω∫10e−2ρτ(t)z2dρdx]TS+12∫TSqEq−1(t)E′(t)ξ(t)τ(t)∫Ω∫10e−2ρτ(t)z2dρdxdt−12∫TSEq(t)ξ(t)∫Ωz2(x,0,t)dxdt. | (43) |
Using the definition of
γ0∫TSEq+1dt≤−[Eq(t)∫Ωuutdx]TS−[ξ(t)τ(t)2Eq(t)∫Ω∫10e−2ρτ(t)z2dρdx]TS+q∫TSEq−1(t)E′(t)∫Ωuutdxdt+q∫TSξ(t)τ(t)2Eq−1(t)E′(t)∫Ω∫10e−2ρτ(t)z2dρdxdt |
+2∫TSEq(t)‖ut‖2L2(Ω)dt−∫TSEq(t)∫Ωμ1(t)uutdxdt−∫TSEq(t)∫Ωμ2(t)uz(x,1,t)dxdt+12∫TSξ(t)Eq(t)e−2ρτ(t)∫Ωz2(x,0,t)dxdt, | (44) |
where
Using the Young and Sobolev-Poincaré inequalities and Lemma 2.3, we find that
−[Eq(t)∫Ωuutdx]TS≤Eq(S)∫Ωu(x,S)ut(x,S)dx−Eq(T)∫Ωu(x,T)ut(x,T)dx≤cEq+1(S). |
Now, we known that
−[ξ(t)τ(t)2Eq(t)∫Ω∫10e−2ρτ(t)z2dρdx]TS≤ξ(S)τ(S)2Eq(S)∫Ω∫10e−2ρτ(S)z2(x,ρ,S)dρdx≤cEq(S)ξ(S)τ(S)∫Ω∫10z2(x,ρ,S)dρdx≤cEq+1(S). |
By (13), we have
∫TSEq−1(t)E′(t)∫Ωuutdxdt≤c∫TS(−E′(t))Eq(t)dt≤cEq+1(S). |
Similarly,
∫TSEq−1(t)E′(t)ξ(t)τ(t)2∫Ω∫10e−2ρτ(t)z2dρdxdt≤cEq+1(S). |
From Lemma 2.4, we deduce that
∫TSEq(t)‖ut‖2L2(Ω)dt≤−c∫TSEq(t)E′(t)dt≤cEq+1(S). |
Now, we get that
|∫TSEq(t)∫Ωμ1(t)uutdxdt|≤μ1(0)|∫TSEq(t)∫Ωuutdxdt|≤c(ε1)∫TSEq(t)∫Ωu2tdxdt+ε1∫TSEq(t)∫Ωu2xdxdt≤c(ε1)∫TSEq(t)(−E′(t))dt+ε1∫TSEq(t)E(t)dt≤c(ε1)Eq+1(S)+ε1∫TSEq+1(t)dt | (45) |
and from (H2) we obtain that
|∫TSEq(t)∫Ωμ2(t)uz(x,1,t)dxdt|≤βμ1(0)|∫TSEq(t)∫Ωφz(x,1,t)dxdt|≤c(ε2)Eq+1(S)+ε2∫TSEq+1(t)dt. | (46) |
Finally,
12∫TSEq(t)ξ(t)∫Ωz2(x,0,t)dxdt≤ˉξμ1(0)2∫TSEq(t)‖ut‖2L2(Ω)dt≤c∫TSEq(t)(−E′(t))dt≤cEq+1(S). |
Choosing
∫TSEq+1dt≤1γEq+1(S). |
Since
∫TSEq+1dt≤1γE(0)Eq(S). |
We choose
E(t)≤E(0)e1−γt. |
This ends the proof of Theorem 4.1.
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