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Khaled Zennir (Ar-Rass)

Lakhdar Kassah Laouar (Algeria)

Energy decay result for a nonlinear wave p−Laplace equation with a delay term

Abstract We consider the nonlinear (in space and time) wave equation with delay term in the internal feedback. Under conditions on the delay term and the term without delay, we study the asymptotic behavior of solutions using the multiplier method and general weighted integral inequalities.

2010 Mathematics Subject Classification: 35L70 (93D15).

Key words and phrases: Nonlinear wave equation Time delay term Decay rate

Multiplier method p−Laplacian .

1. Introduction It is well known that the p−Laplace equations are a degenerate equations in divergence form. It has been much studied during the last years and their results are by now rather developed, especially with delay. In the classical theory of the evolution equations several main parts of mathematics are joined in a fruitful way, it is very remarkable that the wave p−Laplace equation occupies a similar position, when it comes to nonlinear problems.

In this paper we investigate the decay properties of solutions for the initial boundary value problem of a nonlinear wave equation of the form

d

dtl(u0)) − div(φp(∇u)) + µ1g(u0(x, t))

2g(u0(x, t − τ )) = 0 in Ω×]0, +∞[,

u(x, t) = 0 on Γ×]0, +∞[,

u(x, 0) = u0(x), u0(x, 0) = u1(x) in Ω,, u0(x, t − τ ) = f0(x, t − τ ) in Ω×]0, τ [,

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where Ω is a bounded domain in Rn, n ∈ N, with a smooth boundary

∂Ω = Γ, τ > 0 is a time delay, µ1 and µ2 are positive real numbers, and the initial data (u0, u1, f0) belong to a suitable space. ∇ is the gradient operator such that |∇u|2 =Pni=1∂x∂u

i

2

, the function φ is defined by

φx(y) = |y|x−2y. (2)

usually for x ­ 2.

For p = 2, when g is linear, it is well known that if µ2 = 0, that is, in the absence of a delay, the energy of problem (1) exponentially decays to

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zero (see for instance [7,8,14,19]). On the contrary, if µ1 = 0, that is, there exists only the delay part in the interior, the system (1) becomes unstable (see for instance [10]). In [10], the authors showed that a small delay in a boundary control can turn such a well-behaved hyperbolic system into a wild one and therefore, the delay becomes a source of instability. To stabilize a hyperbolic system involving input delay terms, additional control terms will be necessary (see [20–22]). In [20] the authors examined the problem (1) with p = 2 and determined suitable relations between µ1and µ2, for which stability or, alternatively, instability takes place. More precisely, they showed that the energy is exponentially stable if µ2< µ1 and they found a sequence of delays for which the corresponding solution will be unstable if µ2 ­ µ1. The main approach used in [20], is an observability inequality obtained by means of a Carleman estimate. The same results were shown if both the damping and the delay act in the boundary domain. We also recall the result by Xu, Yung and Li in [22], where the authors proved the same result as in [20] for the one-dimension space by adopting the spectral analysis approach.

When g is nonlinear and in the case µ2 = 0, p = 2, the problem of existence and energy decay have been previously studied by several authors (see [2,4,13–15]) and many energy estimates have been derived for arbitrary growing feedbacks (polynomial, exponential or logarithmic decay). The decay rate of a global solution depends on the growth near zero of g(s) as it was proved in [5,13–15,23,24].

Our purpose in this paper is to give energy decay estimates of soluti- ons to the problem (1) for a nonlinear damping and a delay term, in the p−Laplace type to extend results obtained by A. Benaissa et al. [4]1. We use the multiplier method and some properties of convex functions.

2. Preliminaries and Notations We omit the space variable x of u(x, t), u0(x, t) and for the simplicity reason denote u(x, t) = u and u0(x, t) = u0, when no confusion arises. The constants c used throughout this paper are positive generic constants which may be different in various occurrences also the functions considered are all real valued, here u0 = du(t)/dt and u00 = d2u(t)/dt2. We use familiar function spaces W0m,p.

First, let us assume the following hypotheses:

(H1) g : R → R is an odd non-decreasing function of the class C0(R) such that there exist 1(sufficiently small), c1, c2, c3, α1, α2> 0 and a convex and increasing function H : R+→ R+ of the class C1(R+) ∩ C2(]0, ∞[) satisfying H(0) = 0, and H linear on [0, 1] or (H0(0) = 0, H00 > 0 on

1Laboratory of ACEDPs, Djilali Liabes University, Sidi Bel Abbes, Algeria

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]0, 1]), such that

c1|s|l−1 ¬ |g(s)| ¬ c2|s|p, p ­ l − 1 if |s| ­ 1, (3)

|s|l+ |g|1+1p(s) ¬ H−1(sg(s)) if |s| ¬ 1, (4)

|g0(s)| ¬ c3, (5)

α1 sg(s) ¬ G(s) ¬ α2 sg(s), (6)

where

G(s) = Z s

0

g(r) dr (H2)

α2µ2 < α1µ1. (7) We first state some Lemmas which will be needed later.

Lemma 2.1 (Sobolev–Poincar´e’s inequality) Let q be a number with 2 ¬ q < +∞ (n = 1, 2, ..., p) or 2 ¬ q ¬ pn/(n − p) (n ­ p + 1). Then there is a constant c = c(Ω, q) such that

kukq¬ ck∇ukp for u ∈ W01,p(Ω).

Lemma 2.2 ( [11,12]) Let E : R+ → R+ be a non-increasing differentiable function and Ψ : R+ → R+ a convex and increasing function such that Ψ(0) = 0. Assume that

Z T s

Ψ(E(t)) dt ¬ E(s) ∀0 ¬ s ¬ T.

Then E satisfies the following estimate:

E(t) ¬ ψ−1(h(t) + ψ(E(0))) ∀t ­ 0, (8) where ψ(t) =Rt1Ψ(s)1 ds for t > 0, h(t) = 0 for 0 ¬ t ¬ Ψ(E(0))E(0) , and

h−1(t) = t + ψ−1(t + ψ(E(0)))

Ψ (ψ−1(t + ψ(E(0)))) ∀t ­ E(0) Ψ(E(0)).

Following the paper [20] , we introduce the new variable ρ and function z(x, ρ, t) = ut(x, t − τ ρ), x ∈ Ω, ρ ∈ (0, 1), t > 0, (9) which satisfies

τ z0(x, ρ, t) + zρ(x, ρ, t) = 0 in Ω × (0, 1) × (0, +∞). (10)

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The original problem is rewritten with the help of the new function z. Thus, it becomes a system of two equations for two functions u and z, with an additional variable ρ:

l(u0))0− div(φp(∇u)) + µ1g(u0(x, t))

2g(z(x, 1, t)) = 0 in Ω×]0, +∞[ , τ z0(x, ρ, t) + zρ(x, ρ, t) = 0 in Ω×]0, 1[×]0, +∞[ ,

u(x, t) = 0 on ∂Ω × [0, +∞[ ,

z(x, 0, t) = u0(x, t) on Ω × [0, +∞[ , u(x, 0) = u0(x) u0(x, 0) = u1(x) in Ω ,

z(x, ρ, 0) = f0(x, −ρτ ) in Ω×]0, 1[ .

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Let ξ be a positive constant such that τµ2(1 − α1)

α1 < ξ < τµ1− α2µ2

α2 . (12)

The energy of u at time t of the problem (11) is defined by E(t) = l − 1

l ku0(t)kll+1

pk∇xu(t)kpp+ ξ Z

Z 1 0

G(z(x, ρ, t)) dρ dx. (13) The first lemma proved here is the following energy estimate.

Lemma 2.3 Let (u, z) be a solution of the problem (11). Then, the energy functional defined by (13) satisfies

E0(t) ¬ −



µ1−ξα2

τ − µ2α2

 Z

u0g(u0) dx

ξ

τα1− µ2(1 − α1)

 Z

z(x, 1, t)g(z(x, 1, t)) dx

¬ 0. (14)

Proof Multiplying the first equation in (11) by u0, integrating over Ω and using the integration by parts, we get

d dt(l − 1

l ku0kll+1

pk∇xukpp) + µ1 Z

u0g(u0) dx + µ2 Z

u0g(z(x, 1, t)) dx = 0.(15) We multiply the second equation in (11) by ξg(z) and integrate the result over Ω × (0, 1) to obtain

ξ Z

Z 1 0

z0g(z(x, ρ, t)) dρ dx = −ξ τ

Z

Z 1 0

∂ρG(z(x, ρ, t)) dρ dx

= −ξ τ

Z

(G(z(x, 1, t)) − G(z(x, 0, t))) dx.(16)

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Then ξ d

dt Z

Z 1 0

G(z(x, ρ, t)) dρ dx = −ξ τ

Z

G(z(x, 1, t)) dx + ξ τ

Z

G(u0) dx. (17) From (15), (17) and using the Young inequality we get

E0(t) =



µ1−ξα2

τ

 Z

u0g(u0) dx

ξ τ

Z

G(z(x, 1, t)) dx − µ2

Z

u0(t)g(z(x, 1, t)) dx. (18) Let us denote G to be the conjugate function of the convex function G, i.e., G(s) = supt∈R+(st − G(t)). Then G is the Legendre transform of G which is given by (see Arnold [3, pp. 61–62] and Lasiecka [6,9,16,17])

G(s) = s(G0)−1(s) − G[(G0)−1(s)] ∀s ­ 0, (19) and satisfies the following inequality

st ¬ G(s) + G(t) ∀s, t ­ 0. (20) Then by the definition of G we get

G(s) = sg−1(s) − G(g−1(s)).

Hence

G(g(z(x, 1, t))) = z(x, 1, t)g(z(x, 1, t)) − G(z(x, 1, t))

¬ (1 − α1)z(x, 1, t)g(z(x, 1, t)). (21) Making use of (18), (20) and (21), we have

E0(t) ¬ −



µ1 ξα2

τ

 Z

u0g(u0) dx − ξ τ

Z

G(z(x, 1, t)) dx + µ2

Z

G(u0) + G(g(z(x, 1, t)))dx

¬ −



µ1 ξα2

τ − µ2α2

 Z

u0g(u0) dx − ξ τ

Z

G(z(x, 1, t)) dx + µ2

Z

G(g(z(x, 1, t))) dx. (22)



Using (6) and (12), we obtain E0(t) ¬ −



µ1−ξα2

τ − µ2α2

 Z

u0g(u0) dx

ξ

τα1− µ2(1 − α1)

 Z

z(x, 1, t)g(z(x, 1, t)) dx ¬ 0.

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3. Global existence We are now ready to employ the Galerkin method to prove the global existence Theorem.

Theorem 3.1 Let (u0, u1, f0) ∈ W2,p∩ W01,p× W01,l(Ω) × W01,l(Ω; W1,2(0, 1)) and assume that the hypotheses (H1)–(H2) hold. Then the problem (1) ad- mits a unique solution

u ∈ L([0, ∞); W2,p∩ W01,p), u0 ∈ L([0, ∞); W01,l), (23) Proof (of Theorem 3.1.) Let T > 0 be fixed and denote by Vk the space generated by {w1, w2, . . . , wk}. Now, we define for 1 ¬ j ¬ k the sequence φj(x, ρ) as follows:

φj(x, 0) = wj.

Then, we can extend φj(x, 0) by φj(x, ρ) over L2(Ω × [0, 1]) and denote Zk the space generated by {φ1, φ2, . . . , φk}.

We construct approximate solutions (uk, zk)(k = 1, 2, 3, . . . , ) in the form uk(t) =Pkj=1gjkwj,

zk(t) =Pkj=1hjkφj,

where gik and hik (j = 1, 2, . . . , m) are determined by the following ordinary differential equations:

((|u0k(t)|l−2u0k(t))0, wj) + (|∇xuk(t)|p−2xuk(t), ∇xwj) + µ1(g(u0k), wj) 1(g(zk(·, 1)), wj) = 0, 1 ¬ j ¬ k,

zn(x, 0, t) = u0k(x, t),

uk(0) = u0k=Pkj=1(u0, wj)wj → u0 in W2,p∩ W01,p as m → +∞, u0k(0) = u1k=Pkj=1(u1, wj)wj → u1 in W01,l as m → +∞,

(24)

and

(τ zkt+ z, φj) = 0, 1 ¬ j ¬ k, (25) zk(ρ, 0) = z0k=

k

X

j=1

(f0, φjj → f0 in W01,l(Ω; W01,2(0, 1)) as k → +∞.(26)



By virtue of the theory of ordinary differential equations, and following the techniques due to Lions [18] in order to deal with the convergence of the nonlinear terms, the system (24)–(26) has a unique local solution which is extended to a maximal interval [0, Tk[.

Next, using a standard compactness argument for the limiting procedure, we obtain a priori estimates for the solution, so that it can be extended to the global solution.

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Since the sequences u0k, u1kand z0k converge, then from (14) we can find a positive constant C independent of k such that

Ek(t) + c1 Zt

0

u0k(x, s)g(u0k(x, s))ds + c2 Zt

0

zk(x, 1, s)g(zk(x, 1, s))ds ¬ C. (27)

where

Ek(t) = l − 1

l ku0k(t)kll+1

pk∇xuk(t)kpp Z

Z 1 0

G(zk(x, ρ, t)) dρ dx ¬ C. (28)

c1 = µ1−ξα2

τ − µ2α2 and c2 = ξα1

τ − µ2(1 − α1)

These estimates imply that the solution (uk, zk) exists globally in [0, +∞[.

For the regularity, differentiating (24), (25) with respect to t, multiplying by g00jk(t), h0jk(t) and summing over j from 1 to k, taking the sum and using Cauchy-Schwarz, Young’s inequalities and Gronwall’s Lemma to conclude that

u0k is bounded in L(0, +∞; W01,l(Ω)), (29) z0k is bounded in L(0, +∞; W1,2(Ω × (0, 1))). (30) In the next estimate, replacing wj by −∆pwj in (24), multiplying the result by gjm0 (t), summing over j from 1 to k and replacing φj by −∆pφj in (25), multiplying the resulting equation by hjk(t), summing over j from 1 to k.

Using suitable calculus to conclude that

uk is bounded in L(0, +∞; W2,p(Ω) ∩ W01,p(Ω)), (31) zk is bounded in L(0, +∞; W01,l(Ω; W1,2(0, 1))). (32) Applying Dunford-Petti’s theorem and Aubin-Lions’ theorem [18], we can extract a subsequence (uν) of (uk) such that

u0ν → u0 strongly in Ll(Ω).

Therefore,

u0ν → u0 a.e in Ω. (33)

Similarly we obtain

zν → z a.e in Ω. (34)

For the nonlinear terms, we need the next lemmas.

Lemma 3.2 For each T > 0, g(u0), g(z(x, 1, t)) ∈ L1(Ω) and kg(u0)kL1(Ω), kg(z(x, 1, t))kL1(Ω)¬ K1, where K1 is a constant independent of t.

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Proof By (H1) we have

g(u0k(x, t)) → g(u0(x, t)) a.e. in Ω,

0 ¬ g(u0k(x, t))u0k(x, t) → g(u0(x, t))u0(x, t) a.e. in Ω.

Hence, by Fatou’s lemma we have Z T

0

Z

u0(x, t)g(u0(x, t)) dx dt ¬ K for T > 0. (35) By the Cauchy–Schwartz inequality and using (35), we have

Z T 0

Z

|g(u0(x, t))| dx dt ¬ c|Ω|12R0T Ru0g(u0) dx dt

1

2 (36)

¬ c|Ω|12K12 ≡ K1. (37)



Lemma 3.3 g(u0k) → g(u0) in L1(Ω × (0, T )) and g(zk) → g(z) in L1(Ω × (0, T )).

Proof Let E ⊂ Ω × [0, T ] and set E1=



(x, t) ∈ E; g(u0k(x, t)) ¬ 1 p|E|



, E2= E \ E1,

where |E| is the measure of E. If M (r) := inf{|s|; s ∈ R and |g(s)| ­ r}, Z

E

|g(u0k)| dxdt ¬ q

|E| + M

 1 p|E|

!−1 Z

E2

|u0kg(u0k)| dxdt.

We have supkRE|g(u0k)| dxdt → 0 as |E| → 0. From Vitali’s convergence theorem we deduce that g(u0k) → g(u0) in L1(Ω × (0, T )), hence

g(u0k) → g(u0) weak in L2(Ω).

Similarly, we have

g(zk0) → g(z0) weak in L2(Ω), and this implies that

Z T 0

Z

g(u0k)v dx dt → Z T

0

Z

g(u0)v dx dt for all v ∈ L2(0, T ; W01,l), (38) Z T

0

Z

g(zk)v dx dt → Z T

0

Z

g(z)v dx dt for all v ∈ L2(0, T ; W01,2) (39)

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as m → +∞. It follows at once (38), (39) that for each fixed v ∈ L2(0, T ; W01,l) and w ∈ L2(0, T ; W01,2(Ω × (0, 1)))

RT 0

R

((|u0k|l−2u0k)0− div(|∇xu0k|p−2xu0k) + +µ1g(u0k) + µ2g(zk))v dx dt

R0TR((|u0k|l−2u0k)0− div(|∇xu0k|p−2xu0k) + µ1g(u0) + µ2g(z))v dx dt RT

0

R1 0

R



τ zk0 + ∂ρzk



w dx dρ dt →R0TR01R



τ z0+∂ρ z



w dx dρ dt

as m → +∞. Hence

Z T 0

Z

((|u0k|l−2u0k)0− div(|∇xu0k|p−2xu0k)

+ µ1g(u0) + µ2g(z))v dx dt = 0, v ∈ L2(0, T ; W01,l), Z T

0

Z 1 0

Z

(τ u0+

∂ρz)w dx dρ dt = 0, w ∈ L2(0, T ; W01,2(Ω × (0, 1))).

Thus the problem (1) admits a global weak solution u. 

4. Energy decay The next main result reads as.

Theorem 4.1 Let (u0, u1, f0) ∈ W2,p∩ W01,p× W01,l(Ω) × W01,l(Ω; W1,2(0, 1)) and assume that the hypotheses (H1)–(H2) hold. Then, for some constants ω, 0 we have

E(t) ¬ ψ−1(h(t) + ψ(E(0))) ∀t > 0, (40) where ψ(t) =Rt1ωϕ(τ )1 dτ for t > 0, h(t) = 0 for 0 ¬ t ¬ ωϕ(E(0))E(0) ,

h−1(t) = t + ψ−1(t + ψ(E(0)))

ωϕ (ψ−1(t + ψ(E(0)))) ∀t > 0,

ϕ(s) =

( s if H is linear on [0, 1],

sH0(0s) if H0(0) = 0, and H00> 0 on ]0, 1].

Proof (of Theorem4.1.) Multiplying the first equation of (11) by ϕ(E)E u, we

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obtain

0 = Z T

S

ϕ(E) E

Z

ul(u0))0− div(φp(∇u))

+ µ1g(u0(x, t)) + µ2g(z(x, 1, t))dx dt

=

ϕ(E) E

Z

u|u0|l−2u0dx

T S

Z T

S

ϕ(E) E

0Z

u|u0|l−2u0dxdt

−3l − 2 l

Z T S

ϕ(E) E

Z

u0ldx dt + 2

Z T S

ϕ(E) E

Z

l − 1 l u0l+1

p|∇u|p

 dxdt + µ1

Z T S

ϕ(E) E

Z

ug(u0) dx dt + µ2

Z T S

ϕ(E) E

Z

ug(z(x, 1, t)) dxdt + (1 −2

p) Z T

S

ϕ(E) E

Z

|∇u|pdxdt.

Similarly, we multiply the second equation of (11) by ϕ(E)E e−2τ ρg(z(x, ρ, t)), we have

0 = Z T

S

ϕ(E) E

Z

Z 1 0

e−2τ ρg(z)(τ z0+ zρ) dxdρdt

=

ϕ(E) E

Z

Z 1

0

τ e−2τ ρG(z) dxdρ

T

S

− τ Z T

S

ϕ(E) E

0Z

Z 1 0

e−2τ ρG(z) dxdρdt +

Z T S

ϕ(E) E

Z

Z 1 0



∂ρ(e−2τ ρG(z)) + 2τ e−2τ ρG(z)



dxdρdt

=

ϕ(E) E

Z

Z 1 0

τ e−2τ ρG(z) dxdρ

T S

− τ Z T

S

ϕ(E) E

0Z

Z 1 0

e−2τ ρG(z) dxdρdt +

Z T S

ϕ(E) E

Z

(e−2τG(z(x, 1, t)) − G(z(x, 0, t))) dxdt + 2τ

Z T S

ϕ(E) E

Z 1 0

Z

e−2τ ρG(z) dxdρdt.

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Since, p ­ 2, summing to obtain, for A = min{2, 2τ e−2τ/2ξ}

A Z T

S

ϕ(E) dt ¬ −

ϕ(E) E

Z

u|u0|l−2u0dx

T S

+ Z T

S

(ϕ(E) E )0

Z

u|u0|l−2u0dxdt + 3l − 2

l Z T

S

ϕ(E) E

Z

u0ldxdt − µ1 Z T

S

ϕ(E) E

Z

ug(u0) dx dt

− µ2 Z T

S

ϕ(E) E

Z

ug(z(x, 1, t)) dxdt

ϕ(E) E

Z

Z 1 0

τ e−2τ ρG(z)dxdρ

T S

+ τ Z T

S

ϕ(E) E

0Z

Z 1 0

e−2τ ρG(z)dxdρdt

Z T

S

ϕ(E) E

Z

(e−2τG(z(x, 1, t)) − G(z(x, 0, t))) dxdt. (41) Since E is non-increasing, using the Holder,Cauchy–Schwartz, Poincare and Young’s inequalities with exponents l−1l , l, to get

ϕ(E) E

Z

u|u0|l−2u0dx

T S

= ϕ(E(S)) E(S)

Z

u(S)|u0(S)|l−2u0(S)dx

ϕ(E(T )) E(T )

Z

u(T )|u0(T )|l−2u0(T )dx

¬ Cϕ(E(S)),

Z T S

ϕ(E) E

0Z

u|u0|l−2u0dxdt

¬ c Z T

S

ϕ(E) E

0 E dt

¬ cϕ(E(S)),

ϕ(E) E

Z

Z 1 0

e−2τ ρG(z) dxdρ

T S

= ϕ(E(S)) E(S)

Z

Z 1 0

e−2τ ρG(z(x, ρ, S)) dxdρ,

ϕ(E(T )) E(T )

Z

Z 1 0

e−2τ ρG(z(x, ρ, T )) dxdρ

¬ Cϕ(E(S)),

Z T S

ϕ(E) E

0 Z

Z 1 0

e−2τ ρG(z) dxdρdt ¬ c Z T

S



ϕ(E) E

0 Edt

¬ cϕ(E(S)),

Z T S

ϕ(E) E

Z

e−2τG((x, 1, t)) dxdt ¬ c Z T

S

ϕ(E)

E (−E0) dt

¬ cϕ(E(S)),

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Z T S

ϕ(E) E

Z

G(z(x, 0, t))dxdt = Z T

S

ϕ(E) E

Z

G(u0(x, t)) dxdt

¬ c Z T

S

ϕ(E)

E (−E0) dt

¬ cϕ(E(S)), We conclude

A Z T

S

ϕ(E)dt ¬ cϕ(E(S)) + µ1 Z T

S

ϕ(E) E

Z

|u||g(u0)| dx dt + 3l − 2

l Z T

S

ϕ(E) E

Z

u0ldxdt + µ2

Z T S

ϕ(E) E

Z

|u||g(z(x, 1, t))| dxdt. (42) In order to apply the results of Lemma 2.2, we estimate the terms of the right-hand side of (42) .

We distinguish two cases.

4.1. H is linear on [0, 1]. We have c1|s|l−1 ¬ |g(s)| ¬ c2|s|p for all s ∈ R, and then, using (6) and noting that s 7→ ϕ(E(s))E(s) is non-increasing,

Z T S

ϕ(E) E

Z

|u0|ldxdt ¬ c Z T

S

ϕ(E) E

Z

u0g(u0)dxdt ¬ cϕ(E(S)), Using the Poincar´e, Young inequalities and the energy inequality from Lemma 2.3, we obtain, for all  > 0,

Z T S

ϕ(E) E

Z

|ug(u0)|dxdt ¬  Z T

S

ϕ(E) E

Z

up+1dxdt +c

Z T S

ϕ(E) E

Z

g1+1p(u0)dxdt

¬ c Z T

S

ϕ(E)dt + c Z T

S

ϕ(E) E

Z

u0g(u0)dxdt

¬ c Z T

S

ϕ(E)dt + cϕ(E(S)),

Z T S

ϕ(E) E

Z

|ug(z(x, 1, t))|dxdt ¬  Z T

S

ϕ(E) E

Z

up+1dxdt +c

Z T S

ϕ(E) E

Z

g1+1p(z(x, 1, t))dxdt

¬ c Z T

S

ϕ(E)dt + c

Z T S

ϕ(E) E

Z

z(x, 1, t)g(z(x, 1, t))dxdt

¬ c Z T

S

ϕ(E)dt + cϕ(E(S)).

(13)

Inserting these two inequalities into (42), choosing  > 0 small enough, we deduce that

Z T S

ϕ(E(t))dt ¬ cϕ(E(S)).

Using Lemma 2.2 for E in the particular case where ϕ(s) = s, we deduce from (8) that

E(t) ¬ ce−ωt.

4.2. H0(0) = 0 and H00 > 0 on ]0, 1]. For all t ­ 0, we consider the following partition of Ω

1t = {x ∈ Ω : |u0| ­ 1},2t = {x ∈ Ω : |u0| ¬ 1}, Ω˜1t = {x ∈ Ω : |z(x, 1, t)| ­ 1}, Ω˜2t = {x ∈ Ω : |z(x, 1, t)| ¬ 1}.

Using (3), (6) and the fact that s 7→ ϕ(s)s is non-decreasing, we obtain c

Z T S

ϕ(E) E

Z

1t

(|u0|l+ g1+1p(u0))dxdt ¬ c Z T

S

ϕ(E) E

Z

u0g(u0)dxdt ¬ cϕ(E(S)) and

Z T S

ϕ(E) E

Z

˜1t

g1+1p(z(x, 1, t))dxdt ¬ c Z T

S

ϕ(E) E

Z

z(x, 1, t)g(z(x, 1, t))dxdt

¬ cϕ(E(S)).

On the other hand, since H is convex and increasing, H−1 is concave and increasing. Therefore (4) and the reversed Jensen’s inequality for a concave function imply that

Z T S

ϕ(E) E

Z

2t

(|u0|l + gp/(p−1)(u0)) dxdt ¬ Z T

S

ϕ(E) E

Z

2t

H−1(u0g(u0)) dxdt

¬ Z T

S

ϕ(E)

E |Ω|H−1 1

|Ω|

Z

u0g(u0)dxdt. (43) Let us assume H to be the conjugate function of the convex function H, i.e., H(s) = supt∈R+(st − H(t)). Then H is the Legendre transform of H, which is given by (see [3], [6,9,16,17])

H(s) = s(H0)−1(s) − H[(H0)−1(s)] ∀s ­ 0 (44) and satisfies the following inequality

st ¬ H(s) + H(t) ∀s, t ­ 0. (45) Due to our choice ϕ(s) = sH0(0s), we have

H

ϕ(s) s



= 0sH0(0s) − H(0s) ¬ 0ϕ(s). (46)

(14)

Making use of (43), (45) and (46), we have Z T

S

ϕ(E) E

Z

2t

(|u0|l + gp/(p−1)(u0)) dxdt ¬ c Z T

S

H

ϕ(E) E

 dt +c

Z T S

Z

u0g(u0)dt ¬ 0 Z T

S

ϕ(E)dt + cE(S),

Z T S

ϕ(E) E

Z

˜2t

(|u0|l + gp/(p−1)(z(x, 1, t))) dxdt

¬ Z T

S

ϕ(E) E

Z

˜2t

H−1(z(x, 1, t)g(z(x, 1, t))) dxdt

¬ Z T

S

ϕ(E)

E |Ω|H−1 1

|Ω|

Z

z(x, 1, t)g(z(x, 1, t))dxdt

¬ c Z T

S

H

ϕ(E) E

 dt + c

Z T S

Z

z(x, 1, t)g(z(x, 1, t))dt

¬ 0 Z T

S

ϕ(E)dt + cE(S). (47)

Then, choosing 0 > 0 small enough and using (42), we obtain in both cases for all S ­ 0

Z +∞

S

ϕ(E(t))dt ¬ cE(S) + ϕ(E(S))

¬ c



1 +ϕ(E(S) E(S)



E(S) ¬ cE(S). (48) Using Lemma2.2in the particular case where Ψ(s) = ωϕ(s), we deduce from (8) our estimate (40). The proof of Theorem4.1 is now complete.  Remark 4.1 The nonlinearity in space and time makes our problem more general and very useful in the practical point of view.

Acknowledgments.

The author wishes to thank deeply the anonymous referee for his/her useful remarks and his/her careful reading of the proofs presented in this paper.

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Control Optim. 19:1 (1981), 114–122.

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