on a system of nonlinear wave equations associated with the helical flows of maxwell fluid

Nonlinear Analysis: Real World Applications 12 (2011) 3356–3372

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Nonlinear Analysis: Real World Applications

journal homepage: www.elsevier.com/locate/nonrwa

On a system of nonlinear wave equations associated with the helicalflows of Maxwell fluidLe Xuan Truong a, Le Thi Phuong Ngoc b, Cao Huu Hoa c, Nguyen Thanh Long d,∗

a Department of Mathematics, University of Economics of HoChiMinh City, 59C Nguyen Dinh Chieu Str., Dist. 3, HoChiMinh City, Vietnamb Nhatrang Educational College, 01 Nguyen Chanh Str., Nhatrang City, Vietnamc Department of Fundamental Sciences, Tra Vinh University, 126, Highway 53, ward 5, Tra Vinh City, Vietnamd Department of Mathematics and Computer Science, University of Natural Science, Vietnam National University Ho Chi Minh City, 227 Nguyen Van Cu Str.,Dist. 5, Ho Chi Minh City, Vietnam

a r t i c l e i n f o

Article history:Received 18 August 2010Accepted 31 May 2011

Keywords:System of nonlinear wave equationsMixed nonhomogeneous conditionsThe helical flows of Maxwell fluidFaedo–Galerkin methodExponential decay

a b s t r a c t

The paper is devoted to the study a system of nonlinear wave equations associated withthe mixed nonhomogeneous conditions. Existence of a weak solution is proved by usingthe Faedo–Galerkin method. Uniqueness, regularity and decay properties of solutions arealso discussed.

© 2011 Elsevier Ltd. All rights reserved.

1. Introduction

In this paper, we consider the initial-boundary value problem for the system of nonlinear wave equationsutt − a1

uxx +

1xux −

1x2

u

+ λ1ut + f1(u, v) = F1(x, t), 1 < x < R, 0 < t < T ,

vtt − a2

vxx +

1xvx

+ λ2vt + f2(u, v) = F2(x, t), 1 < x < R, 0 < t < T ,

(1.1)

ux(1, t) = b1u(1, t)+ h1(t), vx(1, t) = h2(t),u(R, t) = v(R, t) = 0, (1.2)u(x, 0) =u0(x), ut(x, 0) =u1(x),v(x, 0) = v0(x), vt(x, 0) = v1(x), (1.3)

where a1 > 0, a2 > 0, b1 > 0, λ1 > 0, λ2 > 0, R > 1 are given constants andu0,u1,v0,v1, F1, F2, f1, f2, h1, h2 are givenfunctions satisfying conditions specified later.

∗ Corresponding author.E-mail addresses: [email protected] (L.X. Truong), [email protected], [email protected] (L.T.P. Ngoc), [email protected],

[email protected] (C.H. Hoa), [email protected], [email protected] (N.T. Long).

1468-1218/$ – see front matter© 2011 Elsevier Ltd. All rights reserved.doi:10.1016/j.nonrwa.2011.05.033

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http://dx.doi.org/10.1016/j.nonrwa.2011.05.033

L.X. Truong et al. / Nonlinear Analysis: Real World Applications 12 (2011) 3356–3372 3357

In [1], Jamil and Fetecau studied the following problem

λutt + ut = ν

uxx +

1xux −

1x2

u, 1 < x < R, t > 0,

λVtt + Vt = ν

Vxx +

1xVx

, 1 < x < R, t > 0,

ux(1, t)− u(1, t) =fµt, Vx(1, t) =

gµt, t > 0,

u(R, t) = V (R, t) = 0, t > 0,u(x, 0) = ut(x, 0) = 0, 1 < x < R,V (x, 0) = Vt(x, 0) = 0, 1 < x < R,

(1.4)

where λ,µ, ν, f , g are the given constants. The problem (1.4) is a mathematical model describing the helical flows ofMaxwell fluid in the annular region between two infinite coaxial circular cylinders of radii 1 and R > 1. The authors haveobtained an exact solution for this problembymeans of finite Hankel transforms and presented under series form in terms ofBessel functions J0(x), Y0(x), J1(x), Y1(x), J2(x) and Y2(x), satisfy all imposed initial and boundary conditions. More recently,works on helical flows for ordinary and fractional derivative models appear in [2–6].

Themain goal of this note is to extend the results of Jamil and Fetecau [1] to findweak solutions of the problem (1.1)–(1.3).The proof is based on the Galerkin method associated with a priori estimates, the weak convergence and the compactnesstechniques.

We consider three main parts as follows.In Part 1, for under the conditions (u0,u1), (v0,v1) ∈ V × L2, with V = u ∈ H1(1, R) : u(R) = 0, f1, f2 ∈ C0(R2

; R)and some other conditions, we prove that the problem (1.1)–(1.3) has a unique weak solution (u, v) ∈ L∞(0, T ; V × V ),with (ut , vt) ∈ L∞(0, T ; L2 × L2), (utt , vtt) ∈ L1(0, T ; V ′

× V ′). In Part 2, we prove that the unique solution (u, v) belongsto L∞(0, T ; (V ∩ H2)2), with (ut , vt) ∈ L∞(0, T ; V × V ), (utt , vtt) ∈ L∞(0, T ; L2 × L2), if we assume (u0,u1), (v0,v1) ∈

(V ∩H2)×V , f1, f2 ∈ C1(R2; R) and some other conditions. Finally, in Part 3, with h1 = h2 = 0, by adding some conditions,

we obtain a unique solution (u(t), v(t)) exists onR+ inwhich ‖u′(t)‖+‖v′(t)‖+‖ux(t)‖+‖vx(t)‖+‖F(u(t), v(t))‖L1 decayexponentially to 0 as t → +∞, with f1 =

∂F∂u , f2 =

∂F∂v

. The results obtained here may be considered as the generalizationsof those in [1].

2. Preliminaries

PutΩ = (1, R),QT = Ω× (0, T ), T > 0.We omit the definitions of the usual function spaces: Cm(Ω), Lp(Ω),Wm,p(Ω).We define Wm,p

= Wm,p(Ω), Lp = W 0,p(Ω),Hm= Wm,2(Ω), 1 ≤ p ≤ ∞,m = 0, 1, . . . . The norm in L2 is denoted by

‖ · ‖. We also denote by (·, ·) the scalar product in L2 or a pair of dual scalar products of continuous linear functionals withan element of a function space. We denote by ‖ · ‖X the norm of a Banach space X and by X ′ the dual space of X . We denoteby Lp(0, T ; X), 1 ≤ p ≤ ∞ for the Banach space of the real functions u : (0, T ) → X measurable, such that

‖u‖Lp(0,T ;X) =

∫ T

0‖u(t)‖p

Xdt1/p

< ∞ for 1 ≤ p < ∞,

and‖u‖L∞(0,T ;X) = ess sup

0<t<T‖u(t)‖X for p = ∞.

Let u(t), u′(t) = ut(t), u′′(t) = utt(t), ux(t), uxx(t) denote u(x, t), ∂u∂t (x, t),

∂2u∂t2(x, t), ∂u

∂x (x, t),∂2u∂x2(x, t), respectively.

On H1, we shall use the following norm

‖v‖H1 = (‖v‖2+ ‖vx‖

2)1/2. (2.1)

We put

V = v ∈ H1(Ω) : v(R) = 0. (2.2)

V is a closed subspace of H1 and on V two norms ‖v‖H1 and ‖vx‖ are equivalent norms.Note that L2,H1 are also the Hilbert spaces with respect to the corresponding scalar products

⟨u, v⟩ =

∫ R

1xu(x)v(x)dx, ⟨u, v⟩ + ⟨ux, vx⟩, (2.3)

respectively. Thenorms in L2 andH1 inducedby the corresponding scalar products are denoted by‖·‖0 and‖·‖1, respectively.H1 is continuously and densely embedded in L2. Identifying L2 with (L2)′ (the dual of L2), we have H1 → L2 → (H1)′; onthe other hand, the notation ⟨·, ·⟩ is used for the pairing between H1 and (H1)′.

3358 L.X. Truong et al. / Nonlinear Analysis: Real World Applications 12 (2011) 3356–3372

We have the following lemmas.

Lemma 2.1. We have the following inequalities

(i) ‖v‖ ≤ ‖v‖0 ≤√R‖v‖, for all v ∈ L2,

(ii) ‖v‖H1 ≤ ‖v‖1 ≤√R‖v‖H1 , for all v ∈ H1.

(2.4)

Proof of Lemma 2.1. From the following inequalities

(i)∫ R

1v2(x)dx ≤

∫ R

1xv2(x)dx ≤ R

∫ R

1v2(x)dx, for all v ∈ L2,

(ii)∫ R

1(v2(x)+ v2x (x))dx ≤

∫ R

1x(v2(x)+ v2x (x))dx

≤ R∫ R

1(v2(x)+ v2x (x))dx, for all v ∈ H1,

(2.5)

(2.4) follows.

Lemma 2.2. The embedding H1 → C0(Ω) is compact and

‖v‖C0(Ω) ≤ α0‖v‖H1 for all v ∈ H1, (2.6)

where α0 =1

√2(R−1)

1 +

1 + 16(R − 1)2.

Proof of Lemma 2.2. It is well known that the embedding H1 → C0(Ω) is compact (see [7]). Since C1(Ω) is dense in H1,we only show that (2.6) holds for all v ∈ C1(Ω).

For all v ∈ C1(Ω), and x, y ∈ Ω , we have

v2(x) = v2(y)+ 2∫ x

yv(t)v′(t)dt. (2.7)

Integrate over t from 0 to s to get

(R − 1)v2(x) = ‖v‖2+ 2

∫ R

1dy

∫ x

yv(t)v′(t)dt

= ‖v‖2+ 2

∫ R

1dy

∫ x

1v(t)v′(t)dt − 2

∫ R

1dy

∫ y

1v(t)v′(t)dt

= ‖v‖2+ 2(R − 1)

∫ x

1v(t)v′(t)dt − 2

∫ R

1(R − t)v(t)v′(t)dt

≤ ‖v‖2+ 4(R − 1)

∫ R

1|v(t)v′(t)|dt.

(2.8)

By the inequality 2ab ≤ αa2 +1αb2, for all a, b ∈ R, α > 0, we deduce from (2.8), that

(R − 1)v2(x) ≤ ‖v‖2+ 2(R − 1)

[α‖v′

‖2+

1α

‖v‖2]

=

[1 + 2(R − 1)

1α

]‖v‖2

+ 2(R − 1)α‖v′‖2, for all α > 0.

(2.9)

Choose α > 0 such that 1 + 2(R − 1) 1α

= 2(R − 1)α, or α =1+

√1+16(R−1)2

4(R−1) . Hence (2.6) holds.

Lemma 2.3. The embedding V → C0(Ω) is compact and

(i) ‖v‖C0(Ω) ≤√R − 1‖vx‖ for all v ∈ V ,

(ii) ‖v‖ ≤R − 1√2

‖vx‖ for all v ∈ V ,

(iii) ‖v‖0 ≤

R2(R − 1)‖vx‖0 for all v ∈ V .

(2.10)


Proof of Lemma 2.3. The embedding V → H1 is continuous and the embedding H1 → C0(Ω) is compact, so theembedding V → C0(Ω) is compact.

(i) For all v ∈ V , and x ∈ [1, R],

|v(x)| =

− ∫ R

xv′(y)dy

≤

∫ R

1|v′(y)|dy ≤

√R − 1

∫ R

1

v′(y)2 dy1/2

=√R − 1‖vx‖. (2.11)

(ii) For all v ∈ V , and x ∈ [1, R],

v2(x) =

− ∫ R

xv′(y)dy

2 ≤ (R − x)∫ R

x|v′(y)|2dy = (R − x)‖vx‖2. (2.12)

Integrating over x from 1 to R, we obtain

‖v‖2=

∫ R

1v2(x)dx ≤

∫ R

1(R − x)‖vx‖2dx =

(R − 1)2

2‖vx‖

2. (2.13)

(iii) For all v ∈ V ,

‖v‖0 ≤√R‖v‖ ≤

R2(R − 1)‖vx‖ ≤

R2(R − 1)‖vx‖0. (2.14)

Remark 2.1. On L2, two norms v −→ ‖v‖ and v −→ ‖v‖0 are equivalent. So are two norms v −→ ‖v‖H1 andv −→ ‖v‖1 on H1, four norms v −→ ‖v‖H1 , v −→ ‖v‖1, v −→ ‖vx‖ and v −→ ‖vx‖0 on H1

0 , and four normsv −→ ‖v‖H1 , v −→ ‖v‖1, v −→ ‖vx‖ and v −→ ‖vx‖0 on V .

Remark 2.2. The weak formulation of the initial-boundary value problem (1.1)–(1.3) can be given in the followingmanner:Find (u, v) ∈ W = (u, v) ∈ L∞(0, T ; V × V ) : (ut , vt) ∈ L∞(0, T ; L2 × L2), (utt , vtt) ∈ L1(0, T ; V ′

× V ′), such that (u, v)satisfies the following variational equation⟨utt(t), w⟩ + a1a(u(t), w)+ a1h1(t)w(1)+ λ1⟨ut(t), w⟩ + a1

1x2

u(t), w+ ⟨f1(u, v), w⟩ = ⟨F1(t), w⟩,

⟨vtt(t), φ⟩ + a2b(v(t), φ)+ a2h2(t)φ(1)+ λ2⟨vt(t), φ⟩ + ⟨f2(u, v), φ⟩ = ⟨F2(t), φ⟩,

(2.15)

for all (w, φ) ∈ V × V , a.e., t ∈ (0, T ), together with the initial conditionsu(x, 0) =u0(x), ut(x, 0) =u1(x),v(x, 0) = v0(x), vt(x, 0) = v1(x), (2.16)

where a(·, ·), b(·, ·) are the symmetric bilinear forms on V × V defined by a(u, w) = b(u, w) + b1u(1)w(1), b(u, w) =

⟨ux, wx⟩, for all u, w ∈ V , with b1 > 0 is a given constant.

3. The existence and the uniqueness of a weak solution

Wemake the following assumptions:

(H0) a1, a2, b1, λ1, λ2 are positive constants;(H1) (u0,u1), (v0,v1) ∈ V × L2;(H2) F1, F2 ∈ L1(0, T ; L2);(H3) h1, h2 ∈ W 1,1(0, T );(H4) There exists F ∈ C1(R2

; R) such that(i) ∂F

∂u (u, v) = f1(u, v), ∂F∂v (u, v) = f2(u, v),(ii) There exists the constant C1, such that

−F(u, v) ≤ C1(1 + u2+ v2), ∀u, v ∈ R;

(H5) For everyM > 0, there exists a constant LM > 0, such that

|fi(u1, v1)− fi(u2, v2)| ≤ LM(|u1 − u2| + |v1 − v2|),

for all u1, v1, u2, v2 ∈ [−M,M], i = 1, 2.


Remark 3.1. We present an example in which the functions f1, f2 satisfy the assumptions (H4) and (H5). Consider thefollowing functions

f1(u, v) = 2

α2

(α2 + u2)2+

1β2 + v2

v2u,

f2(u, v) = 2

β2

(β2 + v2)2+

1α2 + u2

u2v,

where α, β are constants. It is obvious that (H5) holds, (H4) is also valid, since there exists a F ∈ C2(R2; R) defined by

F(u, v) =

1

α2 + u2+

1β2 + v2

u2v2

such that

∂F∂u(u, v) = 2

α2

(α2 + u2)2+

1β2 + v2

v2u = f1(u, v),

∂F∂v(u, v) = 2

β2

(β2 + v2)2+

1α2 + u2

u2v = f2(u, v),

F(u, v) =

1

α2 + u2+

1β2 + v2

u2v2 ≤ u2

+ v2, for all (u, v) ∈ R2.

On the other hand, by

2F(u, v) ≤ uf1(u, v)+ vf2(u, v) ≤ 4F(u, v), for all (u, v) ∈ R2,

the functions f1, f2 also satisfy the assumption (H ′′

4 ) in Section 5 as below.We have the following theorem.

Theorem 3.1. For every T > 0, let (H0)–(H4) hold. Then, there exists a weak solution (u, v) of the problem (1.1)–(1.3), such that

(u, v) ∈ L∞(0, T ; V × V ), (ut , vt) ∈ L∞(0, T ; L2 × L2), (utt , vtt) ∈ L1(0, T ; V ′× V ′). (3.1)

Furthermore, if (H5) is satisfied, then the solution is unique.

Proof of Theorem 3.1. The proof consists of Steps 1–4.Step 1. The Faedo–Galerkin approximation. Let wi be a denumerable base of V . We find the approximate solution of Prob.(1.1)–(1.3), in the form

um(t) =

m−j=1

cmj(t)wj, vm(t) =

m−j=1

dmj(t)wj, (3.2)

where the coefficient functions (cmj, dmj) satisfy the system of ordinary differential equations⟨u′′

m(t), wj⟩ + a1a(um(t), wj)+ a1h1(t)wj(1)+ λ1⟨u′

m(t), wj⟩ + a1

1x2

um(t), wj

+ ⟨f1(um(t), vm(t)), wj⟩

= ⟨F1(t), wj⟩,⟨v′′

m(t), wj⟩ + a2b(vm(t), wj)+ a2h2(t)wj(1)+ λ2⟨v′

m(t), wj⟩ + ⟨f2(um(t), vm(t)), wj⟩

= ⟨F2(t), wj⟩, 1 ≤ j ≤ m,

(3.3)

um(0) = u0m, u′

m(0) = u1m, vm(0) = v0m, v′

m(0) = v1m, (3.4)

where

u0m =

m−j=1

u(0)mj wj →u0 strongly in V ,

u1m =

m−j=1

u(1)mj wj →u1, strongly in L2,

v0m =

m−j=1

v(0)mj φj → v0 strongly in V ,

v1m =

m−j=1

v(1)mj φj → v1, strongly in L2.

(3.5)


From the assumptions of Theorem 3.1, the Cauchy problem for the system of ordinary differential equations (3.3) and(3.4) has a local solution (um(t), vm(t)) on [0, Tm]. This is a consequence of the Caratheodory Theorem, (see [8, p. 43]). Theextension of the local solution outside [0, Tm] to [0, T ] is a consequence of the estimates.Step 2. By multiplying (3.3)1 by c ′

mj(t), (3.3)2 by d′

mj(t), and summing up with respect to j and then integrating with respectto the time variable from 0 to t , we get after some rearrangements

Sm(t) = Sm(0)− 2∫ t

0[⟨f1(um(s), vm(s)), u′

m(s)⟩ + ⟨f2(um(s), vm(s)), v′

m(s)⟩]ds

+ 2∫ t

0⟨F1(s), u′

m(s)⟩ds + 2∫ t

0⟨F2(s), v′

m(s)⟩ds − 2a1

∫ t

0h1(s)u′

m(1, s)ds

− 2a2

∫ t

0h2(s)v′

m(1, s)ds − 2a1

∫ t

0

1x2

um(s), u′

m(s)ds

= Sm(0)+

6−j=1

Ij, (3.6)

where

Sm(t) = ‖u′

m(t)‖20 + ‖v′

m(t)‖20 + a1a(um(t), um(t))+ a2b(vm(t), vm(t))

+ 2λ1

∫ t

0‖u′

m(s)‖20ds + 2λ2

∫ t

0‖v′

m(s)‖20ds. (3.7)

On the other hand, we have, it follows from (3.7), that

Sm(t) ≥ Sm(t), (3.8)

where

Sm(t) = ‖u′

m(t)‖20 + ‖v′

m(t)‖20 + a∗(‖umx(t)‖2

0 + ‖vmx(t)‖20)+ 2λ∗

∫ t

0(‖u′

m(s)‖20 + ‖v′

m(s)‖20)ds, (3.9)

and

a∗ = mina1, a2, λ∗ = minλ1, λ2. (3.10)

By (3.5) and the embedding H1 → C0(Ω), there exists a positive constant C0 depending only on u0,u1,v0,v1 anda1, a2, b1, such that

Sm(0)+ |u0m(1)| + |v0m(1)| ≤ C0, for allm. (3.11)

On the other hand, we have

‖um(t)‖20 =

um(0)+

∫ t

0u′

m(s)ds2

0≤ 2‖u0m‖

20 + 2t

∫ t

0‖u′

m(s)‖20ds. (3.12)

Hence, by (3.9), (3.11) and (3.12), it follows that

‖um(t)‖20 + ‖vm(t)‖2

0 ≤ C0 + 2T∫ t

0Sm(s)ds. (3.13)

We shall estimate the following integrals on the right-hand side of (3.6).First integral. By assumptions (H4), and the embedding H1 → C0(Ω), we deduce from (3.9), (3.11) and (3.13) that

I1 = −2∫ t

0[⟨f1(um(s), vm(s)), u′

m(s)⟩ + ⟨f2(um(s), vm(s)), v′

m(s)⟩]ds

= −2∫ t

0

dds

∫ R

1xF(um(x, s), vm(x, s))dx

= 2∫ R

1xF(u0m(x), v0m(x))dx − 2

∫ R

1xF(um(x, t), vm(x, t))dx

≤ (R2− 1) sup

|y|,|z|≤√C0

|F(y, z)| + 2C1

∫ R

1x[1 + u2

m(x, t)+ v2m(x, t)]dx

≤ C0 + CT

∫ t

0Sm(s)ds, (3.14)

where CT is a bound depending on T . For short, in what follows, CT always is a constant with the same meaning.


Second integral. Using the assumption (H2), the Cauchy–Schwartz inequality yields

I2 = 2∫ t

0⟨F1(s), u′

m(s)⟩ds ≤ 2∫ t

0‖F1(s)‖0‖u′

m(s)‖0ds

≤

∫ t

0‖F1(s)‖0ds +

∫ t

0‖F1(s)‖0‖u′

m(s)‖20ds

≤ ‖F1‖L1(0,T ;L2) +

∫ t

0‖F1(s)‖0Sm(s)ds ≤ CT +

∫ t

0‖F1(s)‖0Sm(s)ds. (3.15)

Third integral. Similarly

I3 = 2∫ t

0⟨F2(s), v′

m(s)⟩ds ≤ ‖F2‖L1(0,T ;L2) +

∫ t

0‖F2(s)‖0Sm(s)ds

≤ CT +

∫ t

0‖F2(s)‖0Sm(s)ds. (3.16)

Fourth integral. By using integration by parts, we rewrite

I4 = −2a1

∫ t

0h1(s)u′

m(1, s)ds = 2a1h1(0)u0m(1)− 2a1h1(t)um(1, t)+ 2a1

∫ t

0h′

1(s)um(1, s)ds. (3.17)

Using Lemma 2.3 and the following inequalities

2ab ≤ εa2 +1εb2, for all a, b ≥ 0, ε > 0, (3.18)

and

|um(1, t)| ≤ ‖um(t)‖C0(Ω) ≤√R − 1‖umx(t)‖0 ≤

R − 1a∗

Sm(t) (3.19)

leads to

I4 ≤ 2a1|h1(0)| |u0m(1)| + 2a1|h1(t)|

R − 1a∗

Sm(t)+ 2a1

R − 1a∗

∫ t

0|h′

1(s)|Sm(s)ds

≤ 2a1|h1(0)|C0 +1εa21

R − 1a∗

h21(t)+ εSm(t)+ a21

R − 1a∗

∫ t

0|h′

1(s)|ds +

∫ t

0|h′

1(s)|Sm(s)ds

≤ 2a1|h1(0)|C0 +1εa21

R − 1a∗

‖h1‖2L∞(0,T ) + a21

R − 1a∗

‖h′

1‖L1(0,T ) +

∫ t

0|h′

1(s)|Sm(s)ds + εSm(t)

≤

1 +

1ε

[2a1C0|h1(0)| + a21

R − 1a∗

(‖h1‖2L∞(0,T ) + ‖h′

1‖L1(0,T ))

]+

∫ t

0|h′


≤

1 +

1ε

CT +

∫ t

0|h′

1(s)|Sm(s)ds + εSm(t), (3.20)

for all ε > 0, where CT ≥ 2a1C0|h1(0)| + a21R−1a∗(‖h1‖

2L∞(0,T ) + ‖h′

1‖L1(0,T )).Fifth integral. Similarly

I5 = −2a2

∫ t

0h2(s)v′

m(1, s)ds ≤ 2a2|h2(0)|C0 +1εa22

R − 1a∗

‖h2‖2L∞(0,T )

+ a22R − 1a∗

‖h′

2‖L1(0,T ) +

∫ t

0|h′


≤

1 +

1ε

[2C0a2|h2(0)| + a22

R − 1a∗

(‖h2‖2L∞(0,T ) + ‖h′

2‖L1(0,T ))

]+

∫ t

0|h′


≤

1 +

1ε

CT +

∫ t

0|h′

2(s)|Sm(s)ds + εSm(t), (3.21)

for all ε > 0, in which CT ≥ 2C0a2|h2(0)| + a22R−1a∗(‖h2‖

2L∞(0,T ) + ‖h′

2‖L1(0,T )).


Sixth integral. By Lemma 2.3, we deduce from (3.9) that

I6 = −2a1

∫ t

0

1x2

um(s), u′

m(s)ds ≤ 2a1

∫ t

0‖um(s)‖0‖u′

m(s)‖0ds

≤ 2a1

R2(R − 1)

∫ t

0‖umx(s)‖0‖u′

m(s)‖0ds ≤ 2a1

R

2a∗

(R − 1)∫ t

0Sm(s)ds. (3.22)

Combining (3.6), (3.8), (3.9), (3.11), (3.14)–(3.16) and (3.20)–(3.22), we obtain

(1 − 2ε)Sm(t) ≤12C (1)T (ε)+

12

∫ t

0C (2)T (s)Sm(s)ds, (3.23)

for all ε > 0, where12C (1)T (ε) = 2C0 + 2

2 +

1ε

CT ,

12C (2)T (s) =

2−i=1

(|h′

i(s)| + ‖Fi(s)‖0)+ 2a1

R

2a∗

(R − 1)+ CT , C (2)T ∈ L1(0, T ).(3.24)

Choosing ε =14 , by Gronwall’s lemma, we deduce from (3.23) that

Sm(t) ≤ C (1)T (ε) exp∫ t

0C (2)T (s)ds

≤ CT , ∀m ∈ N, ∀t ∈ [0, T ], ∀T > 0. (3.25)

Step 3. Limiting process. From (3.9), (3.13) and (3.25), there exists a subsequence of (um, vm) still also so denoted, such that(um, vm) → (u, v) in L∞(0, T ; V × V ) weakly∗,

(u′

m, v′

m) → (u′, v′) in L∞(0, T ; L2 × L2) weakly∗.(3.26)

Also by the compactness lemma of Lions [9, p. 57], we can deduce from (3.26) the existence of a subsequence, also stilldenoted by (um, vm), such that

(um, vm) → (u, v) strongly in L2(QT )× L2(QT ) and a.e., in QT . (3.27)

The continuity of f1 gives

f1(um, vm) → f1(u, v) a.e., (x, t) in QT . (3.28)

By ‖f1(um, vm)‖L2(QT )≤

√T (R − 1) sup|y|, |z|≤

√CT |f1(y, z)| < ∞, we deduce from (3.28) and Lions’s lemma [9, Lemma

1.3, p. 12] that

f1(um, vm) → f1(u, v) in L2(QT )weakly. (3.29)

Similarly

f2(um, vm) → f2(u, v) in L2(QT )weakly. (3.30)

Passing to the limit in (3.3), (3.4) by (3.26), (3.27), (3.29) and (3.30), we have (u, v) satisfying the problemddt

⟨ut(t), w⟩ + a1a(u(t), w)+ a1h1(t)w(1)+ λ1⟨ut(t), w⟩ + a1

1x2

u(t), w+ ⟨f1(u, v), w⟩ = ⟨F1(t), w⟩,

ddt

⟨vt(t), φ⟩ + a2b(v(t), φ)+ a2h2(t)φ(1)+ λ2⟨vt(t), φ⟩ + ⟨f2(u, v), φ⟩ = ⟨F2(t), φ⟩,

(3.31)

for all (w, φ) ∈ V × V , a.e., t ∈ (0, T ), together with the initial conditionsu(x, 0) =u0(x), ut(x, 0) =u1(x),v(x, 0) = v0(x), vt(x, 0) = v1(x). (3.32)

On the other hand, we have from (3.26), (3.29), (3.31)1 and (H2) that

utt = a1

uxx +

1xux −

1x2

u

− λ1ut − f1(u, v)+ F1 ∈ L1(0, T ; V ′). (3.33)

Similarly vtt ∈ L1(0, T ; V ′) and the existence of a weak solution is proved.


Step 4.Uniqueness of the solution. Assumenow that (H5) is satisfied. Let (ui, vi), i = 1, 2 be twoweak solutions of the problem(1.1)–(1.3), such that

(ui, vi) ∈ L∞(0, T ; V × V ), (u′

i, v′

i) ∈ L∞(0, T ; L2 × L2),(u′′

i , v′′

i ) ∈ L1(0, T ; V ′× V ′), i = 1, 2.

(3.34)

Then (u, v), with u = u1 − u2, v = v1 − v2, satisfy the variational problem⟨u′′(t), w⟩ + a1a(u(t), w)+ λ1⟨u′(t), w⟩ + a1

1x2

u(t), w+ ⟨f1(t), w⟩ = 0,

⟨v′′(t), φ⟩ + a2b(v(t), φ)+ λ2⟨v′(t), φ⟩ + ⟨f2(t), φ⟩ = 0, for all (w, φ) ∈ V × V ,

u(0) = v(0) = u′(0) = v′(0) = 0,

(3.35)

where f1(t) = f1(u1, v1)− f1(u2, v2), f2(t) = f2(u1, v1)− f2(u2, v2).For every s ∈ (0, T ), we define

Us(t) =

−

∫ s

tu(ξ)dξ, 0 ≤ t < s < T ,

0, 0 < s ≤ t ≤ T ,(3.36)

Vs(t) =

−

∫ s

tv(ξ)dξ, 0 ≤ t < s < T ,

0, 0 < s ≤ t ≤ T .(3.37)

It is clear that Us(T ) = Us(s) = Vs(T ) = Vs(s) = 0 and Us(t), Vs(t) ∈ V for each t ∈ [0, T ].Let (w, φ) = (Us(t), Vs(t)) in (3.35)1 and integrating over t from 0 to s, the result is∫ s

0⟨u′′(t),Us(t)⟩dt + a1

∫ s

0a(u(t),Us(t))dt + λ1

∫ s

0⟨u′(t),Us(t)⟩dt + a1

∫ s

0

1x2

u(t),Us(t)dt

+

∫ s

0⟨ f1(t),Us(t)⟩dt = 0,∫ s

0⟨v′′(t), Vs(t)⟩dt + a2

∫ s

0b(v(t), Vs(t))dt + λ2

∫ s

0⟨v′(t), Vs(t)⟩dt +

∫ s

0⟨ f2(t), Vs(t)⟩dt = 0.

(3.38)

Note that

(i)∫ s

0⟨u′′(t),Us(t)⟩dt = −

12‖u(s)‖2

0,

(ii) a1

∫ s

0a(u(t),Us(t))dt = −

a12a(Us(0),Us(0)),

(iii) λ1

∫ s

0⟨u′(t),Us(t)⟩dt = −λ1

∫ s

0‖u(t)‖2

0dt,

(iv) a1

∫ s

0

1x2

u(t),Us(t)dt = −

a12

1√xUs(0)

2

0,

(v)∫ s

0⟨f1(t),Us(t)⟩dt = −

∫ s

0

f1(t),

∫ s

tu(ξ)dξ

dt = −

∫ s

0

u(ξ),

∫ ξ

0f1(t)dt

dξ .

(3.39)

Then we get

‖u(s)‖20 + a1a(Us(0),Us(0))+ 2λ1

∫ s

0‖u(t)‖2

0dt + a1

1√xUs(0)

2

0= −2

∫ s

0⟨u(ξ),

∫ ξ

0f1(t)dt⟩dξ . (3.40)

On the other hand, by (H5), it follows that

− 2∫ s

0

u(ξ),

∫ ξ

0f1(t)dt

dξ ≤ 2

∫ s

0

u(ξ),

∫ ξ

0f1(t)dt

dξ ≤ 2

∫ s

0‖u(ξ)‖0

∫ ξ

0f1(t)dt

0dξ

≤ 2∫ s

0‖u(ξ)‖0dξ

∫ ξ

0‖f1(t)‖0dt

≤ β1

∫ s

0‖u(ξ)‖2

0dξ +1β1

∫ s

0

∫ ξ

0‖f1(t)‖0dt

2

dξ


≤ β1

∫ s

0‖u(ξ)‖2

0dξ +1β1

∫ s

0

ξ

∫ ξ

0‖f1(t)‖2

0dtdξ

≤ β1

∫ s

0‖u(ξ)‖2

0dξ +1

2β1T 2

∫ s

0‖f1(t)‖2

0dt

≤ β1

∫ s

0‖u(ξ)‖2

0dξ +1β1

T 2L2M

∫ s

0(‖u(t)‖2

0 + ‖v(t)‖20)dt, (3.41)

for all β1 > 0, where M =∑2

i=1(‖ui‖L∞(0,T ;V ) + ‖vi‖L∞(0,T ;V )).Choosing β1, with 0 < β1 < 2λ1, it follows from (3.40), (3.41) that

‖u(s)‖20 ≤

1β1

T 2L2M

∫ s

0(‖u(t)‖2

0 + ‖v(t)‖20)dt. (3.42)

Similarly, with 0 < β2 < 2λ2, we have also from (3.38)2 that

‖v(s)‖20 ≤

1β2

T 2L2M

∫ s

0(‖u(t)‖2

0 + ‖v(t)‖20)dt. (3.43)

Combine (3.42), (3.43) and Gronwall’s lemma to obtain ‖u(s)‖20 + ‖v(s)‖2

0 ≡ 0, i.e., u = u1 − u2 ≡ 0, v = v1 − v2 ≡ 0.The uniqueness of a weak solution follows.

Theorem 3.1 is proved completely.

4. The regularity of solutions

In this section, we study the regularity of solutions of the problem (1.1)–(1.3) corresponding to (u0,u1), (v0,v1) ∈

(V ∩ H2)× V . For this purpose, we assume that a1, a2, b1, λ1, λ2 are the positive constants.Henceforth, we strengthen the hypotheses as follows.

(H ′

1) (u0,u1), (v0,v1) ∈ (V ∩ H2)× V ;(H ′

2) (F1, F2), (F′

1, F′

2) ∈ L1(0, T ; L2 × L2);(H ′

3) h1, h2 ∈ W 2,1(0, T );(H ′

4) F ∈ C2(R2; R) satisfies (H4).

Then, we have the following theorem.

Theorem 4.1. Let T > 0. Suppose that (H0), (H ′

2)–(H′

4) hold and the initial data (u0,u1), (v0,v1) ∈ (V ∩ H2)× V satisfy thecompatibility conditionsu0x(1) = b1u0(1)+ h1(0), v0x(1) = h2(0). (4.1)

Then, there exists a unique weak solution (u, v) of the problem (1.1)–(1.3), such that(u, v) ∈ L∞(0, T ; (V ∩ H2)× (V ∩ H2)), (ut , vt) ∈ L∞(0, T ; V × V ),(utt , vtt) ∈ L∞(0, T ; L2 × L2).

(4.2)

Remark 4.1. (i) Noting that with the regularity obtained by (4.2), it follows that the problem (1.1)–(1.3) has a unique strongsolution (u, v) that satisfies

u, v ∈ L∞(0, T ; V ∩ H2) ∩ C0(0, T ; V ) ∩ C1(0, T ; L2),ut , vt ∈ L∞(0, T ; V ), utt , vtt ∈ L∞(0, T ; L2).

(4.3)

(ii) From (4.3) we can see that u, v, ux, vx, ut , vt , uxx, vxx, uxt , vxt , utt , vtt ∈ L∞(0, T ; L2) ⊂ L2(QT ). Also if (u0,u1),(v0,v1) ∈ (V ∩ H2)× V , then the weak solution (u, v) of the problem (1.1)–(1.3) belongs to [H2(QT ) ∩ L∞(0, T ; V ∩ H2)]2.So the solution is almost classical which is rather natural since the initial data (u0,v0) and (u1,v1) do not belong necessarilyto (C2(Ω))2 and (C1(Ω))2, respectively.

Proof of Theorem 4.1. The proof consists of Steps 1–4.Step 1. The Faedo–Galerkin approximation. Let wj be a denumerable base of V ∩ H2. We find the approximate solution(um(t), vm(t)) of the problem (1.1)–(1.3) in the form (3.2), where the coefficient functions (cmj, dmj) satisfy the system ofordinary differential equation (3.3) with initial conditions

um(0) =u0, u′

m(0) =u1, vm(0) = v0, v′

m(0) = v1. (4.4)


Step 2. A priori estimates I. Proceeding as in the proof of Theorem 3.1, using assumptions (H1)–(H4), we get

Sm(t) ≤ CT , (4.5)

for all t ∈ [0, T ], for allm, where

Sm(t) = ‖u′

m(t)‖20 + ‖v′

m(t)‖20 + a∗(‖umx(t)‖2

0 + ‖vmx(t)‖20)+ 2λ∗

∫ t

0(‖u′

m(s)‖20 + ‖v′

m(s)‖20)ds, (4.6)

and CT always indicates a bound depending on T .A priori estimates II.First of all, we are going to estimate ‖u′′

m(0)‖20 + ‖v′′

m(0)‖20.

Letting t → 0+ in Eq. (3.3)1, multiplying the result by c ′′

mj(0) and using the first compatibility (4.1), we get

‖u′′

m(0)‖20 − a1

u0xx +1xu0x −

1x2

u0, u′′

m(0)+ λ1⟨u1, u′′

m(0)⟩ + ⟨f1(u0,v0), u′′

m(0)⟩ = ⟨F1(0), u′′

m(0)⟩. (4.7)

This implies that

‖u′′

m(0)‖0 ≤ a1

u0xx +1xu0x −

1x2

u0

0+ λ1‖u1‖0 + ‖f1(u0,v0)‖0 + ‖F1(0)‖0 ≡ C01 for allm, (4.8)

where C01 is a constant depending only on a1, λ1,u0,v0,u1, ‖F1(0)‖0.Similarly, we have also

‖v′′

m(0)‖0 ≤ a2

v0xx +1xv0x

0+ λ2‖v1‖0 + ‖f2(u0,v0)‖0 + ‖F2(0)‖0 ≡ C02 for allm, (4.9)

where C02 is a constant depending only on a2, λ2,u0,v0,v1, ‖F2(0)‖0.Differentiating (3.3) with respect to t , the results are

⟨u′′′

m(t), wj⟩ + a1a(u′

m(t), wj)+ a1h′

1(t)wj(1)+ λ1⟨u′′

m(t), wj⟩ + a1

1x2

u′

m(t), wj

+

∂ f1∂u (um(t), vm(t))u′

m(t)+∂ f1∂v(um(t), vm(t))v′

m(t), wj

= ⟨F ′

1(t), wj⟩,

⟨v′′′

m (t), wj⟩ + a2b(v′

m(t), wj)+ a2h′

2(t)wj(1)+ λ2⟨v′′

m(t), wj⟩

+

∂ f2∂u (um(t), vm(t))u′

m(t)+∂ f2∂v(um(t), vm(t))v′

m(t), wj

= ⟨F ′

2(t), wj⟩, 1 ≤ j ≤ m.

(4.10)

By multiplying (4.10)1 by c ′′

mj(t), (4.10)2 by d′′

mj(t), and summing up with respect to j and then integrating with respectto the time variable from 0 to t , we have after some rearrangements

Xm(t) = Xm(0)− 2a1

∫ t

0h′

1(s)u′′

m(1, s)ds − 2a2

∫ t

0h′

2(s)v′′

m(1, s)ds − 2a1

∫ t

0

1x2

u′

m(s), u′′

m(s)ds

+ 2∫ t

0[⟨F ′

1(s), u′′

m(s)⟩ + ⟨F ′

2(s), v′′

m(s)⟩]ds

− 2∫ t

0

[∂ f1∂u(um(s), vm(s))u′

m(s)+∂ f1∂v(um(s), vm(s))v′

m(s), u′′

m(s)]

ds

− 2∫ t

0

[∂ f2∂u(um(s), vm(s))u′


m(s), v′′

m(s)]

ds

= Xm(0)+

6−j=1

Jj, (4.11)

where

Xm(t) = ‖u′′

m(t)‖20 + ‖v′′

m(t)‖20 + a1a(u′

m(t), u′

m(t))+ a2b(v′

m(t), v′

m(t))

+ 2λ1

∫ t

0‖u′′

m(s)‖20ds + 2λ2

∫ t

0‖v′′

m(s)‖20ds ≥ Xm(t), (4.12)

with

Xm(t) = ‖u′′

m(t)‖20 + ‖v′′

m(t)‖20 + a∗[‖u′

mx(t)‖20 + ‖v′

mx(t)‖20] + 2λ∗

∫ t

0(‖u′′

m(s)‖20 + ‖v′′

m(s)‖20)ds. (4.13)


By (4.4), (4.7), (4.9) and (4.12), there exists a positive constant C0 depending only on u0,u1,v0,v1 anda1, a2, λ1, λ2, ‖F1(0)‖0, ‖F2(0)‖0 such that

Xm(0)+ |u1(1)| + |v1(1)| = ‖u′′

m(0)‖20 + ‖v′′

m(0)‖20 + a1a(u1,u1)+ a2b(v1,v1)+ |u1(1)| + |v1(1)| ≤ C0,

for allm. (4.14)

We shall estimate, respectively the following integrals in the right-hand side of (4.11).First integral. By using integration by parts leads

J1 = −2a1

∫ t

0h′

1(s)u′′

m(1, s)ds = 2a1h′

1(0)u1(1)− 2a1h′

1(t)u′

m(1, t)+ 2a1

∫ t

0h′′

1(s)u′

m(1, s)ds. (4.15)

Using Lemma 2.3, (3.18) and

|u′

m(1, t)| ≤ ‖u′

m(t)‖C0(Ω) ≤√R − 1‖u′

mx(t)‖0 ≤

R − 1a∗

Xm(t), (4.16)

we get

J1 ≤ 2a1|h′

1(0)||u1(1)| + 2a1|h′

1(t)|

R − 1a∗

Xm(t)+ 2a1

R − 1a∗

∫ t

0|h′′

1(s)|Xm(s)ds

≤ 2a1|h′

1(0)|C0 +1εa21

R − 1a∗

|h′

1(t)|2+ εXm(t)+ a21

R − 1a∗

∫ t

0|h′′

1(s)|ds +

∫ t

0|h′′

1(s)|Xm(s)ds

≤ 2a1|h′

1(0)|C0 +1εa21

R − 1a∗

‖h′

1‖2L∞(0,T ) + a21

R − 1a∗

‖h′′

1‖L1(0,T ) +

∫ t

0|h′′

1(s)|Xm(s)ds + εXm(t)

≤

1 +

1ε

[2a1

h′

1(0)C0 + a21

R − 1a∗

(‖h′

1‖2L∞(0,T ) + ‖h′′

1‖L1(0,T ))

]+

∫ t

0|h′′


≤

1 +

1ε

CT +

∫ t

0|h′′

1(s)|Xm(s)ds + εXm(t), (4.17)

for all ε > 0, where CT ≥ 2a1|h′

1(0)|C0 + a21R−1a∗(‖h′

1‖2L∞(0,T ) + ‖h′′

1‖L1(0,T )).Second integral. Similarly

J2 ≤ 2a2|h′

2(0)|C0 +1εa22

R − 1a∗

‖h′

2‖2L∞(0,T ) + a22

R − 1a∗

‖h′′

2‖L1(0,T ) +

∫ t

0|h′′


≤

1 +

1ε

[2a2|h′

2(0)|C0 + a22R − 1a∗

(‖h′

2‖2L∞(0,T ) + ‖h′′

2‖L1(0,T ))

]+

∫ t

0|h′′


≤

1 +

1ε

CT +

∫ t

0|h′′

2(s)|Xm(s)ds + εXm(t), for all ε > 0. (4.18)

Third integral. By ((2.10)(iii)), we deduce from (4.16), that

J3 = −2a1

∫ t

0

1x2

u′

m(s), u′′

m(s)ds ≤ 2a1

∫ t

0‖u′

m(s)‖0‖u′′

m(s)‖0ds

≤ 2a1

R2(R − 1)

∫ t

0‖u′

mx(s)‖0‖u′′

m(s)‖0ds ≤ 2a1

R

2a∗

(R − 1)∫ t

0Xm(s)ds. (4.19)

Fourth integral. Using the assumption (H ′

2), the Cauchy–Schwartz inequality gives

J4 = 2∫ t

0[⟨F ′

1(s), u′′

m(s)⟩ + ⟨F ′

2(s), v′′

m(s)⟩]ds

≤ 2∫ t

0‖F ′

1(s)‖0‖u′′

m(s)‖0ds + 2∫ t

0

F ′

2(s)0 ‖v′′

m(s)‖0ds

≤ ‖F ′

1‖L1(0,T ;L2) + ‖F ′

2‖L1(0,T ;L2) +

∫ t

0(‖F ′

1(s)‖0 + ‖F ′

2(s)‖0)Xm(s)ds

≤ CT +

∫ t

0(‖F ′

1(s)‖0 + ‖F ′

2(s)‖0)Xm(s)ds. (4.20)


Fifth integral. Put

K1(T , f1) = sup|y|,|z|≤

(R−1)CT

a∗

[∂ f1∂u (y, z) +

∂ f1∂v (y, z)] . (4.21)

Hence, it follows from ((2.10)(iii)), (4.13) and (4.21), that

J5 = −2∫ t

0

[∂ f1∂u(um(s), vm(s))u′


m(s), u′′

m(s)]

ds

≤ 2K1(T , f1)∫ t

0(‖u′

m(s)‖0 + ‖v′

m(s)‖0)‖u′′

m(s)‖0ds

≤ 4K1(T , f1)

R

2a∗

(R − 1)∫ t

0Xm(s)ds. (4.22)

Sixth integral. Similarly

J6 = −2∫ t

0

[∂ f2∂u(um(s), vm(s))u′


m(s), v′′

m(s)]

ds

≤ 4K1(T , f2)

R

2a∗

(R − 1)∫ t

0Xm(s)ds, (4.23)

where K1(T , f2) = sup|y|,|z|≤

(R−1)CT

a∗

∂ f2∂u (y, z) +

∂ f2∂v (y, z).Combining (4.11)–(4.14), (4.17)–(4.20), (4.22) and (4.23) and choosing ε =

14 , we obtain

Xm(t) ≤ D(1)T +

∫ t

0D(2)T (s)Xm(s)ds, (4.24)

whereD(1)T = 2C0 + 22CT ,

D(2)T (s) = 4[a1 + 2K1(T , f1)+ 2K1(T , f2)]

R

2a∗

(R − 1)+ 2|h′′

1(s)| + 2|h′′

2(s)| + 2‖F ′

1(s)‖0 + 2‖F ′

2(s)‖0,

D(2)T ∈ L1(0, T ).

(4.25)

By Gronwall’s lemma, we deduce from (4.24), that

Xm(t) ≤ D(1)T exp∫ t

0D(2)T (s)ds

≤ CT , ∀m ∈ N, ∀t ∈ [0, T ]. (4.26)

Step 3. Limiting process. From (4.5), (4.6), (4.13) and (4.26), there exists a subsequence of (um, vm) still also so denoted, suchthat

(um, vm) → (u, v) in L∞(0, T ; V × V )weakly∗,

(u′

m, v′

m) → (u′, v′) in L∞(0, T ; V × V )weakly∗,

(u′′

m, v′′

m) → (u′′, v′′) in L∞(0, T ; L2 × L2)weakly∗.

(4.27)

By the compactness lemma of Lions [9, p. 57], it follows from (4.27) the existence of a subsequence also still denoted by(um, vm) such that

(um, vm) → (u, v) strongly in L2(QT )× L2(QT ) and a.e. in QT ,

(u′

m, v′

m) → (u′, v′) strongly in L2(QT )× L2(QT ) and a.e. in QT .(4.28)

Note that

‖f1(um, vm)− f1(u, v)‖L2(QT )≤ K1(T , f1)‖um − u‖L2(QT )

+ ‖vm − v‖L2(QT ). (4.29)

By (4.28)1, (4.29), we obtain

f1(um, vm) → f1(u, v) strongly in L2(QT ). (4.30)


Similarly

f2(um, vm) → f2(u, v) strongly in L2(QT ). (4.31)

Passing to the limit in (3.3) by (4.27), (4.28), (4.30) and (4.31), we have (u, v) satisfying the problem⟨u′′(t), w⟩ + a1a(u(t), w)+ a1h1(t)w(1)+ λ1⟨u′(t), w⟩ + a1

1x2

u(t), w+ ⟨f1(u, v), w⟩ = ⟨F1(t), w⟩,

⟨v′′(t), φ⟩ + a2b(v(t), φ)+ a2h2(t)φ(1)+ λ2⟨v′(t), φ⟩ + ⟨f2(u, v), φ⟩ = ⟨F2(t), φ⟩,

(4.32)

for all (w, φ) ∈ V × V , a.e., t ∈ (0, T ), together with the initial conditions (1.3).On the other hand, we have from (4.27) and (4.32)1 and (H ′

2) that

uxx =1a1

[utt + λ1ut + f1(u, v)− F1] −1xux +

1x2

u ∈ L∞(0, T ; L2). (4.33)

Thus u ∈ L∞(0, T ; V ∩ H2). Similarly v ∈ L∞(0, T ; V ∩ H2) and the existence of a weak solution is proved completely.Using the same arguments as in the proof of Theorem 3.1, we also prove that theweak solution of the problem (1.1)–(1.3)

is unique. The proof of Theorem 4.1 is completed.

5. Exponential decay of solutions

In this section, we study the decay of the solution of the problem (1.1)–(1.3) corresponding to h1 = h2 = 0. Forthis purpose, we also assume that the constants (a1, a2, b1, λ1, λ2) and the functions (u0,u1,v0,v1), (F1, F2) satisfy theassumptions (H0), (H ′

1), (H′

2), respectively.Henceforth, we strengthen the hypotheses as below.

(H ′′

4 ) There exist F ∈ C2(R2; R) and the constants d1, d2 > 0 such that

(i) ∂F∂u (u, v) = f1(u, v), ∂F∂v (u, v) = f2(u, v),

(ii) F(u, v) ≥ 0 for all (u, v) ∈ R2,(iii) d1F(u, v) ≤ uf1(u, v)+ vf2(u, v) ≤ d2F(u, v), for all (u, v) ∈ R2.

Let (u, v) be a weak solution of the problem (1.1)–(1.3) that satisfies (4.2). In order to obtain the decay result, we use thefunctional

L(t) = E(t)+ δψ(t), (5.1)

where δ is a positive constant and

E(t) = ‖u′(t)‖20 + ‖v′(t)‖2

0 + a1a(u(t), u(t))+ a2b(v(t), v(t))+ a1

1xu(t)

2

0+ 2

∫ R

1xF(u(x, t), v(x, t))dx, (5.2)

ψ(t) = ⟨u(t), u′(t)⟩ + ⟨v(t), v′(t)⟩ +λ1

2‖u(t)‖2

0 +λ2

2‖v(t)‖2

0, (5.3)

E1(t) = ‖u′(t)‖20 + ‖v′(t)‖2

0 + ‖ux(t)‖20 + ‖vx(t)‖2

0 +

∫ R

1xF(u(x, t), v(x, t))dx. (5.4)

Lemma 5.1. There exist the positive constants β1, β2 such that

β1E1(t) ≤ L(t) ≤ β2E1(t), (5.5)

when 0 < δ < min2, 4a∗

R(R−1)2

.

Proof. Note that

a∗(‖ux‖20 + ‖vx‖

20) ≤ a1a(u, u)+ a2b(v, v) ≤ a∗(‖ux‖

20 + ‖vx‖

20), (5.6)

for all u, v ∈ V , where a∗= maxa1, [1 + b1(R − 1)]a2.


By using the inequalities (2.10)1 and (5.6), it follows from (5.1)–(5.4), that

L(t) = ‖u′(t)‖20 + ‖v′(t)‖2

0 + a1a(u(t), u(t))+ a2b(v(t), v(t))+ a1

1xu(t)

2

0

+ 2∫ R

1xF(u(x, t), v(x, t))dx + δ⟨u(t), u′(t)⟩ + δ⟨v(t), v′(t)⟩ +

δλ1

2‖u(t)‖2

0 +δλ2

2‖v(t)‖2

0

≥ ‖u′(t)‖20 + ‖v′(t)‖2

0 + a∗(‖ux(t)‖20 + ‖vx(t)‖2

0)+ 2∫ R

1xF(u(x, t), v(x, t))dx

−δ

2

[R2(R − 1)2‖ux(t)‖2

0 + ‖u′(t)‖20

]−δ

2

[R2(R − 1)2‖vx(t)‖2

0 + ‖v′(t)‖20

]=

1 −

δ

2

‖u′(t)‖2

0 + ‖v′(t)‖20

+

[a∗ −

δR4(R − 1)2

](‖ux(t)‖2

0 + ‖vx(t)‖20)

+ 2∫ R

1xF(u(x, t), v(x, t))dx ≥ β1E1(t), (5.7)

which implies that

L(t) ≥ β1E1(t), (5.8)

where

β1 = min2, 1 −

δ

2, a∗ −

δR4(R − 1)2

, (5.9)

with 0 < δ < min2, 4a∗

R(R−1)2

.

Similarly, we have

L(t) = ‖u′(t)‖20 + ‖v′(t)‖2

0 + a1a(u(t), u(t))+ a2b(v(t), v(t))+ a1

1xu(t)

2

0

+ 2∫ R

1xF(u(x, t), v(x, t))dx + δ⟨u(t), u′(t)⟩ + δ⟨v(t), v′(t)⟩ +

δλ1

2‖u(t)‖2

0 +δλ2

2‖v(t)‖2

0

≤ ‖u′(t)‖20 + ‖v′(t)‖2

0 + a∗(‖ux(t)‖20 + ‖vx(t)‖2

0)+ a1R2(R − 1)2‖ux(t)‖2

0

+ 2∫ R

1xF(u(x, t), v(x, t))dx +

δ

2

[R2(R − 1)2‖ux(t)‖2

0 + ‖u′(t)‖20

]+δ

2

[R2(R − 1)2‖vx(t)‖2

0 + ‖v′(t)‖20

]+δλ1

2R2(R − 1)2‖ux(t)‖2

0 +δλ2

2R2(R − 1)2‖vx(t)‖2

0

=

1 +

δ

2

‖u′(t)‖2

0 +

1 +

δ

2

‖v′(t)‖2

0 +

[a∗

+ (2a1 + δ + δλ1)R4(R − 1)2

]‖ux(t)‖2

0

+

[a∗

+ (1 + λ2)δR4(R − 1)2

]‖vx(t)‖2

0 + 2∫ R

1xF(u(x, t), v(x, t))dx ≤ β2E1(t), (5.10)

where

β2 = max2, 1 +

δ

2, a∗

+ (2a1 + δ + δλ1)R4(R − 1)2, a∗

+ (1 + λ2)δR4(R − 1)2

. (5.11)

Lemma 5.1 is proved.

Lemma 5.2. The functional E(t) defined by (5.2) satisfies

E ′(t) ≤ −(2λ1 − ε1)‖u′(t)‖20 − (2λ2 − ε1)‖v

′(t)‖20 +

1ε1(‖F1(t)‖2

0 + ‖F2(t)‖20), (5.12)

for all ε1 > 0.

Proof. Multiplying (1.1) by (u′(x, t), v′(x, t)) and integrating over (1, R) leads to

E ′(t) = −2λ1‖u′(t)‖20 − 2λ2‖v′(t)‖2

0 + 2⟨F1(t), u′(t)⟩ + 2⟨F2(t), v′(t)⟩. (5.13)


On the other hand, for ε1 > 0, we have2⟨F1(t), u′(t)⟩ ≤ ε1‖u′(t)‖2

0 +1ε1

‖F1(t)‖20,

2⟨F2(t), v′(t)⟩ ≤ ε1‖v′(t)‖2

0 +1ε1

‖F2(t)‖20.

(5.14)

Combining (5.13), (5.14) and (5.12) follows. Lemma 5.2 is proved.

Lemma 5.3. The functional ψ(t) defined by (5.3) satisfies

ψ ′(t) ≤ ‖u′(t)‖20 + ‖v′(t)‖2

0 − (a∗ − ε2)(‖ux(t)‖20 + ‖vx(t)‖2

0)

+18ε2

R(R − 1)2(‖F1(t)‖20 + ‖F2(t)‖2

0)− d1

∫ R

1xF(u(x, t), v(x, t))dx, (5.15)

for all ε2 > 0.

Proof. Multiplying the Eq. (1.1) by (u(x, t), v(x, t)) and integrating over (1, R), the result is

ψ ′(t) = ‖u′(t)‖20 + ‖v′(t)‖2

0 − a1a(u(t), u(t))− a2b(v(t), v(t))− a1

1xu(t)

2

0+ ⟨F1(t), u(t)⟩ + ⟨F2(t), v(t)⟩

− ⟨f1(u(t), v(t)), u(t)⟩ − ⟨f2(u(t), v(t)), v(t)⟩. (5.16)

By (5.6), we have

− a1a(u(t), u(t))− a2b(v(t), v(t)) ≤ −a∗(‖ux(t)‖20 + ‖vx(t)‖2

0). (5.17)

On the other hand, for ε2 > 0, by some estimations as in the proof of Lemma 5.2, we deduce the following inequalities⟨F1(t), u(t)⟩ ≤ ‖F1(t)‖0‖u(t)‖0 ≤ ‖F1(t)‖0

R2(R − 1)‖ux(t)‖0

≤ ε2‖ux(t)‖20 +

18ε2

R(R − 1)2‖F1(t)‖20,

⟨F2(t), v(t)⟩ ≤ ε2‖vx(t)‖20 +

18ε2

R(R − 1)2‖F2(t)‖20.

(5.18)

In order to estimate the last term of (5.16), we use the hypotheses (H ′′

4 , (iii)) and obtain

− ⟨f1(u(t), v(t)), u(t)⟩ − ⟨f2(u(t), v(t)), v(t)⟩ ≤ −d1

∫ R

1xF(u(x, t), v(x, t))dx. (5.19)

Combining (5.16)–(5.19), it follows that (5.15) holds. The Lemma 5.3 is proved.

Theorem 5.4. Assume that

‖F1(t)‖20 + ‖F2(t)‖2

0 ≤ η1 exp(−η2t) for all t ≥ 0, (5.20)

where η1, η2 are two positive constants. Then, there exist positive constants γ , C1 such that

E1(t) ≤ C1 exp(−γ t) for all t ≥ 0. (5.21)

Proof. It follows from (5.1), (5.5), (5.12) and (5.15), that

L′(t) ≤ −(2λ1 − ε1 − δ)‖u′(t)‖20 − (2λ2 − ε1 − δ)‖v′(t)‖2

0 − δ(a∗ − ε2)(‖ux(t)‖20 + ‖vx(t)‖2

0)

+

[1ε1

+δ

8ε2R(R − 1)2

](‖F1(t)‖2

0 + ‖F2(t)‖20)− δd1

∫ R

1xF(u(x, t), v(x, t))dx, (5.22)

for all ε1, ε2, δ > 0.Choose δ, ε2 small enough such that

0 < ε2 < a∗,

0 < δ < min2λ1, 2λ2, δ1, with δ1 ≡ min2,

4a∗

R(R − 1)2

.

(5.23)


When δ is fixed, we choose the positive constant ε1 satisfying

0 < ε1 < min2λ1 − δ, 2λ2 − δ. (5.24)

we deduce from (5.4), (5.5), (5.20) and (5.22)–(5.24) that there exist the positive constants γ1, γ , with γ < η2, such that

L′(t) ≤ −γ L(t)+ γ1 exp(−η2t) for all t ≥ 0. (5.25)

Combining (5.4), (5.5) and (5.25), we get (5.21). The Theorem 5.4 is completely proved.

Acknowledgments

The authors wish to express their sincere thanks to the referees for the suggestions and valuable comments. The authorsare also extremely grateful for the support given byVietnam’s National Foundation for Science and TechnologyDevelopment(NAFOSTED) under Project 101.01-2010.15.

References

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on a system of nonlinear wave equations associated with the helical flows of maxwell fluid

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