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In this article, we aim to prove the well posedness of a one point initial boundary value problem for a nonlinear hyperbolic singular integro- differential partial differential equation

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ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

NONLINEAR SINGULAR HYPERBOLIC INITIAL BOUNDARY VALUE PROBLEMS

SAID MESLOUB, IMED BACHAR Communicated by Mokhtar Kirane

Abstract. In this article, we aim to prove the well posedness of a one point initial boundary value problem for a nonlinear hyperbolic singular integro- differential partial differential equation. Our proofs are mainly based on a fixed point theorem and some a priori bounds. The solvability of the problem is proved when the given data are related to a weighted Sobolev space. An additional result may allow us to study the regularity of the solution.

1. Introduction

A big number of physical phenomena can be modeled and formulated by some nice partial differential equations. It happens sometimes that such phenomena cannot be modeled by classical partial differential equations while describing the system as a function at a given time; it fails to take into account the effect of past history, such as in thermo-elasticity and heat diffusion. In such situations, a memory term on the form of an integral should be included the equation. This integral term can be considered as a damping term in the equation. A range of elliptic integro- differential equations were studied in the works of numerous researchers, including for example Bokalo and Dmytriv [7], Barles and Imbert [4], Caffarelli and Silvestre [8], Chipot and Guesmia [10], Balachandran and Park [3], Bakalo and Dmytriv [7].

Integro-differential equations of hyperbolic and parabolic type are studied by many authors see for example Adolfsson [1], Baker [2], Sloan and Thomee [18], Berrimi and Messaoudi [5], Cannarsa and Sforza [9] and Weiking [19]. In the present work, we consider a hyperbolic integro-differential equation with an unknown function that enters both into the differential part and into the integral part of the equation.

It can occur in visco-elasticity see Renardi and Hrusa [15, 16] and other areas as fluid dynamics, biological models, and chemical kinetics. Well-posedness of problem (1.1)–(1.3) is studied, in fact, we prove the existence and uniqueness of solutions for a posed nonlinear hyperbolic integro-differential equation with a Neumann and Dirichlet conditions. The used tool is a variant of fixed point theorems.

2010Mathematics Subject Classification. 35L10, 35L20, 35L70.

Key words and phrases. Fixed point theorem; a priori bound; existence and uniqueness;

integro-differential equation.

c

2017 Texas State University.

Submitted May 10 2017. Published November 8, 2017.

1

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In the rectangleQT = (0,1)×(0, T), where 0< T <∞, we consider the nonlinear singular second order hyperbolic integro-differential equation

Lv=∂2v

∂t2 −1 x

∂x(x∂v

∂x)− d

dtmaxZ x 0

ξv(ξ, t)dξ,0

=f(x, t),

(1.1) With (1.1), we associate the following initial conditions

v(x,0) =Z1(x), vt(x,0) =Z2(x), x∈(0,1), (1.2) and the boundary conditions

vx(1, t) = 0, v(1, t) = 0, t∈(0, T), (1.3) whereZ1(x),Z2(x) andf(x, t) are given functions which will be specified later on.

In Section 2, we shall give some function spaces and tools which will be used repeatedly below. Section 3 is devoted to the proof of uniqueness of solution of the given problem in the classical Sobolev spaceWγ,21,1(QT). In section 4, we established the existence of solution and the proof was mainly based on the Schauder fixed point theorem. In the last section a priori bound is obtained and can help to establish some regularity results for the solution of problem (1.1)–(1.3)

2. Preliminaries

For the investigation of problem (1.1)–(1.3), we need the following functions spaces: Let L2γ(QT), L2σ(QT), L2ρ(QT) be the weighted Hilbert spaces of square integrable functions onQT withγ=x+x2,σ=x2, andρ=x. The inner products in L2γ(QT), L2σ(QT), L2ρ(QT) are respectively denoted by (·,·)L2

γ(QT), (·,·)L2 σ(QT), (·,·)L2

ρ(QT) such that (u, v)L2

A(QT)= Z 1

0

A(x)uvdx, A(x) =γ, σ, ρ. (2.1) LetL2(0, T; Wγ,21,1(0,1)) be the space of (classes) of functionumeasurable on [0, T] for the Lebesgue measure and having their values inWγ,21,1(0,1),and such that

kukL2(0,T;Wγ,21,1(0,1))=Z T 0

ku(·, t)k2W1,1

γ,2(0,1)dt1/2

<∞

The spaceL2(0, T; Wγ,21,1(0,1)) is a Hilbert space having the inner product (u, v)L2(0,T;Wγ,21,1(0,1))=

Z T

0

(u, v)W1,1

γ,2(0,1)dt, (2.2)

where

(u, v)W1,1 γ,2(0,1)=

Z 1

0

(x+x2)(uv+uxvx+utvt)dx. (2.3) The spaceWγ,21,1(QT), is the set of functionsu∈L2γ(QT) such thatux, ut∈L2γ(QT).

In general the elements ofWγ,2m,n(QT), withm, nnonnegative integers are functions having x−derivatives up to mth order in L2γ(QT), and t−derivative up to nth or- der in L2γ(QT). We also use weighted spaces in (0,1) such as: L2γ(0,1), L2σ(0,1), L2ρ(0,1),Wγ,21,1(0,1) andWγ,21 (0,1).

The following inequalities are needed:

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(1) Cauchyε-inequality which holds for allε >0 and for arbitraryAand B.

AB≤ ε 2A2+ 1

2εB2, (2.4)

(2) A Poincar´e type inequality (see [14]).

x(ξu)k2L2(0,1)≤1 2kuk2L2

ρ(QT), kΛ2x(ξu)k2L2(0,1)≤1

2kΛx(ξu)k2L2(QT),

(2.5) where

Λx(ξu) = Z x

0

ξu(ξ, t)dξ, Λ2x(ξu) = Z x

0

Z ξ

0

ηu(η, t)dηdξ.

(3) Gronwall’s inequality (see [14, Lemma 4.1]).

3. Uniqueness of solution

Theorem 3.1. LetZ1(x)∈Wγ,21 ((0,1)),Z2(x)∈L2γ((0,1)and f(x, t)∈L2γ(QT).

Then the posed problem (1.1)–(1.3) has at most one solution in Wγ,21,1(QT), if it exists.

Proof. Let v1 and v2 be two solutions of problem (1.1)–(1.3) and let η(x, t) = V1(x, t)−V2(x, t), where

Vj(x, t) = Z t

0

vj(x, τ)dτ, j= 1,2, (3.1) then the functionη(x, t) satisfies

Lη=∂2η

∂t2 −1 x

∂x x∂η

∂x

1(x, t)−γ2(x, t), (3.2) ηx(1, t) = 0, η(1, t) = 0, t∈(0, T), (3.3) η(x,0) = 0, ηt(x,0) = 0, x∈(0,1), (3.4) where

γj(x, t) = max(Λx(ξvj),0), j = 1,2. (3.5) Consider the identity

∂η

∂t,∂2η

∂t2 −1 x

∂x(x∂η

∂x)

L2ρ(QT)

+ 2∂η

∂t,∂2η

∂t2 −1 x

∂x(x∂η

∂x)

L2σ(QT)

+∂η

∂x,∂2η

∂t2 −1 x

∂x(x∂η

∂x)

L2σ(QT)

=∂η

∂t, γ1−γ2

L2ρ(QT)

+ 2∂η

∂t, γ1−γ2

L2σ(QT)

+∂η

∂x, γ1−γ2

L2σ(QT)

. (3.6)

Integrating by parts, and taking into account conditions (3.3) and (3.4), we deduce the following expressions for the terms on the left-hand side of (3.6)

∂η

∂t,∂2η

∂t2

L2ρ(QT)

= 1

2kηt(x, T)k2L2

ρ((0,1)), (3.7)

−∂η

∂t,1 x

∂x x∂η

∂x

L2ρ(QT)=1

2kηx(x, T)k2L2

ρ((0,1)), (3.8)

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2∂η

∂t,∂2η

∂t2

L2σ(QT)

=kηt(x, T)k2L2

σ((0,1)), (3.9)

−2∂η

∂t,1 x

∂x x∂η

∂x

L2σ(QT)=kηx(x, T)k2L2

σ((0,1))+ 2(∂η

∂t,∂η

∂x)L2

ρ(QT), (3.10) ∂η

∂x,∂2η

∂t2

L2σ(QT)=kηtk2L2 ρ(QT)+

Z 1

0

x2ηx(x, T)ηt(x, T)dx, (3.11)

−∂η

∂x,1 x

∂x x∂η

∂x

L2σ(QT)= 0. (3.12) Substituting formulas (3.7)–(3.12) into (3.6), we obtain

1

2kηt(x, T)k2L2

ρ((0,1))+1

2kηx(x, T)k2L2

ρ((0,1))+kηt(x, T)k2L2 σ((0,1))

+kηx(x, T)k2L2

σ((0,1))+kηtk2L2 ρ(QT)

=∂η

∂t, γ1−γ2

L2ρ(QT)

+ 2∂η

∂t, γ1−γ2

L2σ(QT)

+∂η

∂x, γ1−γ2

L2σ(QT)

− Z 1

0

x2ηx(x, T)ηt(x, T)dx−2(∂η

∂t,∂η

∂x)L2 ρ(QT).

(3.13)

By using Cauchyε−inequality and the Poincar´e type inequality (2.5), we estimate the terms on the right-hand side of (3.13) as follows

− Z 1

0

x2ηx(x, T)ηt(x, T)dx

≤ε1

2 kηt(x, T)k2L2

σ((0,1))+ 1 2ε1

x(x, T)k2L2 σ((0,1)),

(3.14)

−2∂η

∂t,∂η

∂x

L2ρ(QT)≤ε2xk2L2

ρ(QT)+ 1 ε2

tk2L2

ρ(QT), (3.15) ∂η

∂t, γ1−γ2

L2ρ(QT)

≤ ε3

2kηtk2L2

ρ(QT)+ 1 4ε3tk2L2

ρ(QT), (3.16) 2∂η

∂t, γ1−γ2

L2σ(QT)≤ε4tk2L2

σ(QT)+ 1 ε4

tk2L2

σ(QT), (3.17) ∂η

∂x, γ1−γ2

L2σ(QT)

≤ ε5 2kηtk2L2

σ(QT)+ 1 2ε5

xk2L2

σ(QT). (3.18) Letε1 = 1,ε2 = 8, ε3= 1, ε4= 1, ε5 = 1, and combine (3.14)–(3.18) and (3.13), we obtain

t(x, T)k2L2

ρ((0,1))+kηx(x, T)k2L2

ρ((0,1))+kηt(x, T)k2L2 σ((0,1))

+kηx(x, T)k2L2

σ((0,1))+kηtk2L2 ρ(QT)

≤64 kηtk2L2

ρ(QT)+kηxk2L2

ρ(QT)+kηtk2L2

σ(QT)+kηxk2L2 σ(QT)

.

(3.19)

Application of Gronwall’s lemma [14, Lemma 4.1] to (3.19), yields kηtk2L2

ρ(QT)= 0⇒ηt=v1−v2= 0 for all (x, t)∈QT,

hence, we conclude the uniqueness of the solution.

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4. Existence of the solution

Theorem 4.1. Let Z1(x)∈Wγ,21 ((0,1)), Z2(x)∈L2γ((0,1) andf(x, t)∈L2γ(QT) be given and satisfy

kZ1k2W1

γ,2((0,1))+kZ2k2L2

γ((0,1))+kfk2L2

γ(QT)≤C22, (4.1) forC2>0 small enough, and that

∂Z1(1)

∂x = 0, Z1(1) = 0, ∂Z2(1)

∂x = 0, Z2(1) = 0, (4.2) Then problem (1.1)–(1.3)has a unique solution v∈Wγ,21,1(QT).

Proof. Consider the class of functions K(A) =

v∈L2γ(QT) :kvkL2(0,T;W2,γ1,1(0,1))≤A, kvtkL2γ(QT)≤A}, (4.3) satisfying conditions (1.2) and (1.3), where A is a positive constant which will be specified later on. For anyu∈K(A), problem

2v

∂t2 −1 x

∂x x∂v

∂x

=Iu+f(x, t), (4.4)

v(x,0) =Z1(x), vt(x,0) =Z2(x), x∈(0,1) (4.5) vx(1, t) = 0, v(1, t) = 0, t∈(0, T), (4.6) where

Iu= d

dtmax(Λx(ξu),0), (4.7)

has a unique solution inWγ,21,1(QT).

Define a mapping M such that v = Mu. If we show that the mapping M admits a fixed pointv in the closed bounded convex subset K(A), thenv will be our solution. Let us first show that the mappingMmaps the setK(A) toK(A).

Letv=W +S, such thatW solves

2W

∂t2 − 1 x

∂x x∂W

∂x

=Iu, (x, t)∈QT, (4.8) W(x,0) = 0, Wt(x,0) = 0, x∈(0,1), (4.9) Wx(1, t) = 0, W(1, t) = 0, t∈(0, T), (4.10) andS solves

2S

∂t2 −1 x

∂x x∂S

∂x

=f(x, t), (x, t)∈QT, (4.11) S(x,0) =Z1(x), St(x,0) =Z2(x), x∈(0,1) (4.12) Sx(1, t) = 0, S(1, t) = 0, t∈(0, T) (4.13) We now consider the following inner products inL2(Qy),y∈(0, T)

(LW, M1W)L2(Qy)= (Iu, M1W)L2(Qy), (4.14) (LS, M2S)L2(Qy)= (f, M2S)L2(Qy), (4.15) where

M1W = 2x2Wt+x2Wx+ 2xWt, M2S = 2x2St+x2Sx+ 2xSt.

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By integrating by parts in (4.14) and evoking conditions (4.9) and (4.10), we have kWt(x, y)k2L2

σ((0,1))+kWx(x, y)k2L2

σ((0,1))+kWt(x, y)k2L2 ρ((0,1))

+kWx(x, y)k2L2

ρ((0,1))+kWtk2L2 ρ(Qy)

=− Z 1

0

x2Wx(x, y)Wt(x, y)dx−2(Wt, Wx)L2

ρ(Qy)+ 2(Wt,Iu)L2 σ(Qy)

+ (Wx,Iu)L2σ(Qy)+ 2(Wt,Iu)L2ρ(Qy).

(4.16)

Thanks to Cauchyε-inequality, (4.16) reduces to kWt(x, y)k2L2

σ((0,1))+kWx(x, y)k2L2

σ((0,1))+1

2kWt(x, y)k2L2 ρ((0,1))

+1

2kWx(x, y)k2L2 ρ((0,1))

≤ kWtk2L2

ρ(Qy)+kWxk2L2

ρ(Qy)+kWtk2L2

σ(Qy)+1

2kWxk2L2 σ(Qy)

+kIuk2L2

ρ(Qy)+3 2kIuk2L2

σ(Qy).

(4.17)

Consider the following two elementary inequalities kW(x, y)k2L2

σ((0,1))≤ kWk2L2

σ(Qy)+kWtk2L2

σ(Qy), (4.18) kW(x, y)k2L2

ρ((0,1))≤ kWk2L2

ρ(Qy)+kWtk2L2

ρ(Qy). (4.19) Inequalities (4.17)–(4.19), yield

kW(x, y)k2L2

γ((0,1))+kWt(x, y)k2L2

γ((0,1))+kWx(x, y)k2L2 γ((0,1))

≤3 kWk2L2

γ(Qy)+kWtk2L2

γ(Qy)+kWxk2L2

γ(Qy)+kIuk2L2 γ(Qy)

.

(4.20) Application of Gronwall’s lemma to (4.20) gives

kW(x, y)k2L2

γ((0,1))+kWt(x, y)k2L2

γ((0,1))+kWx(x, y)k2L2 γ((0,1))

≤3e3TkIuk2L2

γ(QT). (4.21)

Integrating of both sides of (4.21) over (0, T) gives kW(x, t)k2L2(0,T;W1,1

2,γ(0,1)) ≤3T e3TkIuk2L2

γ(QT). (4.22) We now consider (4.15) and use conditions (4.12) and (4.13), to obtain

kSt(x, y)k2L2

σ((0,1))+kSx(x, y)k2L2

σ((0,1))+kSt(x, y)k2L2 ρ((0,1))

+kSt(x, y)k2L2

ρ((0,1))+kStk2L2 ρ(Qy)

=− Z 1

0

x2Sx(x, y)St(x, y)dx+ 2(St, Sx)L2

ρ(Qy)+ 2(St, f)L2 σ(Qy)

+ 2(Sx, f)L2

σ(Qy)+ 2(St, f)L2

ρ(Qy)+k∂Z1

∂x k2L2 γ((0,1))

+k(Z2)2k2L2

γ((0,1))+ Z 1

0

x2∂Z1

∂x Z2dx.

(4.23)

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By using the Cauchyε-inequality, we can transform (4.23) into kSt(x, y)k2L2

γ((0,1))+kSx(x, y)k2L2 γ((0,1))

≤4 kStk2L2

γ(Qy)+kSxk2L2 γ(Qy)

+ 4

k(Z1)2xk2L2

γ((0,1))+kZ2k2L2

γ((0,1))+kfk2L2 γ(Qy)

.

(4.24)

Consider the elementary inequality kS(x, y)k2L2

γ((0,1))≤ kSk2L2

γ(Qy)+kStk2L2

γ(Qy)+kZ1k2L2

γ((0,1)). (4.25) Combination of (4.24) and (4.25) leads to

kS(x, y)k2W1,1 2,γ(0,1)

≤4

kSk2W1,1

2,γ(Qy)+kZ1k2W1

2,γ((0,1))+kZ2k2L2

γ((0,1))+kfk2L2 γ(Qy)

.

(4.26) Applying of Gronwall’s lemma to inequality (4.26) and then integrating over (0, T) gives

kS(x, y)k2L2(0,T;W1,1 2,γ(0,1))

≤4T e4T kZ1k2W1

2,γ((0,1))+kZ2k2L2

γ((0,1))+kfk2L2 γ(Qy)

.

(4.27) It is obvious that

kIuk2L2

γ(QT)≤C12, (4.28)

whereC1>0.

Inequalities (4.22), (4.27) and (4.28) yield kvk2L2(0,T;W1,1

2,γ(0,1))≤2kWk2L2(0,T;W1,1

2,γ(0,1))+ 2kSk2L2(0,T;W1,1 2,γ(0,1))

≤6T e3TC12+ 8T e4TC22.

(4.29) and

kvtk2L2

γ(QT)≤2kWtk2L2

γ(QT)+ 2kStk2L2

γ(QT)≤6T e3TC12+ 8T e4TC22 (4.30) If we take A≥6T e3TC12+ 8T e4TC22, we then deduce from the inequalities (4.29) and (4.30) that

kvk2L2(0,T;W1,1

2,γ(0,1))≤A, kvtk2L2

γ(QT)≤A. (4.31) Hencev∈K(A) and consequently MmapsK(A) into itself.

We will now show that the mappingM:K(A)→K(A) is continuous. Letv1, v2∈K(A) and letV1=M(v1), andV2=M(v1).

We observe thatV =V1−V2satisfies

2V

∂t2 −1 x

∂x x∂V

∂x = d

dtmax(Λx(ξv1),0)− d

dtmax(Λx(ξv2),0), (4.32) V(1, t) = 0, Vx(1, t) = 0, (4.33) V(x,0) = 0, Vt(x,0) = 0 (4.34) Define the function

℘(x, t) = Z t

0

V(x, τ)dτ, (4.35)

then it follows from (4.32)–(4.35) that℘(x, t) satisfies

2

∂t2 − 1 x

∂x x∂℘

∂x

= max(Λx(ξv1),0)−max(Λx(ξv2),0), (4.36)

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℘(1, t) = 0, ℘x(1, t) = 0, (4.37)

℘(x,0) = 0, ℘t(x,0) = 0. (4.38) It is obvious that

kH(x, t)k2L2

γ(QT)≤ kv1−v2k2L2

γ(QT), (4.39)

where

H(x, t) = max(Λx(ξv1),0)−max(Λx(ξv2),0) (4.40) Then we have

kVk2L2

γ(QT)≤dkv1−v2k2L2

γ(QT), (4.41)

kV1−V2kL2γ(QT)=kM(v1)− M(v1)kL2γ(QT)≤√

dkv1−v2kL2γ(QT). (4.42) Consequently he mapping M : K(A) → K(A) is continuous. The set K(A) is

compact, because of the following result.

Theorem 4.2. Let B0 ⊂ B ⊂ B1 with compact embedding (reflexive Banach spaces) (see [13, 17]). Assume thatα, λ∈(0,∞)andT >0. Then

W ={θ:θ∈Lα(0, T;B), θt∈Lλ(0, T;B1)}

is compactly embedded inLα(0, T;B), that is the bounded sets are relatively compact inLα(0, T;B).

Observe thatL2(0, T;L2γ(0,1)) =L2γ(QT),M(K(A))⊂K(A)⊂L2γ(QT). Then by Schauder fixed point theorem the mappingMadmits a fixed pointv∈K(A).

5. A priori bound for the solution

We now obtain a priori bound for the solution of problem (1.1)–(1.3). This a priori bound can be used to establish some regularity results. We may expect the solution of (1.1)–(1.3) to be in the function spaceL2(0, T;Wq,γ1,1(0,1)) withq≤ ∞.

Theorem 5.1. Let u∈L2(0, T;W2,γ1,1(0,1)), be a solution of problem (1.1)–(1.3), then following a priori bound holds

kuk2L2(0,T;W1,1 2,γ(0,1))

≤8T e8T kZ1k2W1

2,γ((0,1))+kZ2k2L2

γ((0,1))+kfk2L2 γ(QT)

(5.1)

Proof. Note that Z s

0

d dt

Z 1

0

x2u2tdx

= 2 Z s

0

Z 1

0

x2ututtdx dt= 2 Z s

0

Z 1

0

x2ut

h1 x

∂x(x∂u

∂x) +Iui dx dt

= 2 Z s

0

x2(ut(1, t)ux(1, t)−x2ut(0, t)ux(0, t))dt−2 Z s

0

Z 1

0

x2utxuxdx dt

−2 Z s

0

Z 1

0

xutuxdx dt+ 2 Z s

0

Z 1

0

x2utIu dx dt+ 2 Z s

0

Z 1

0

x2utf dx dt (5.2)

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By using boundary and initial conditions (3.2) and (3.3), then from (5.2) one has Z 1

0

x2u2t(x, s)dx+

1

Z

0

x2u2x(x, s)dx

= Z 1

0

x2Z22(x)dx+ Z 1

0

x2(∂Z1

∂x )2dx−2 Z s

0

Z 1

0

xutuxdx dt

+ 2 Z s

0

Z 1

0

x2utIu dx dt+ 2 Z s

0

Z 1

1

x2utf dx dt.

(5.3)

On the other hand, Z s

0

Z 1

0

h

x2uxutt− ∂

∂x(x∂u

∂x) x∂u

∂x i

dx dt

= Z 1

0

x2ux(x, s)ut(x, s)dx+ Z s

0

Z 1

0

xu2tdx dt

= Z s

0

Z 1

0

x2uxIu dx dt+ Z s

0

Z 0

1

x2uxf dx dt,

(5.4)

and

2 Z s

)

Z 1

0

h

xututt− ∂

∂x(x∂u

∂x)uti dx

= Z 1

0

xu2t(x, T)dx+ Z 1

0

xu2x(x, T)dx

= 2 Z s

0

Z 1

0

xutIu dx dt+ 2 Z s

0

Z 1

0

xutf dx dt+ Z 1

0

xZ22(x)dx +

Z 1

0

x ∂Z1

∂x 2

dx.

(5.5)

Equalities (5.3)–(5.5) and the elementary equality Z 1

0

(x+x2)u2(x, s)dx

≤ Z s

0

Z 1

0

(x+x2)u2dx dt+ Z s

0

Z 1

0

(x+x2)u2tdx dt +

Z 1

0

(x+x2)Z12(x)dx,

(5.6)

lead to Z 1

0

x2u2t(x, s)dx+ Z 1

0

x2u2x(x, s)dx+ Z s

0

Z 1

0

xu2tdx dt

+ Z 1

0

xu2t(x, s)dx+ Z 1

0

xu2x(x, s)dx

= Z 1

0

(x2+x)Z22(x)dx+ Z 1

0

(x2+x) ∂Z1

∂x 2

dx−2 Z s

0

Z 1

0

xutuxdx dt

+ 2 Z s

0

Z 1

0

x2utIu dx dt+ 2 Z s

0

Z 1

0

x2utf dx dt+ Z 1

0

(x+x2)Z12(x)dx

(10)

+ Z s

0

Z 1

0

x2uxIu dx dt+ Z s

0

Z 1

0

x2uxf dx dt+ Z 1

0

x2ux(x, s)ut(x, s)dx + 2

Z s

0

Z 1

0

xutIu dx dt+ 2 Z s

0

Z 1

0

xutf dx dt.

Young’s inequality and a Poincar´e type of inequality [17], transforms the above inequality into

Z 1

0

(x+x2)u2(x, s)dx+ Z 1

0

(x+x2)u2t(x, s)dx+ Z 1

0

(x+x2)u2x(x, s)dx

≤8Z s 0

Z 1

0

(x+x2)u2dx dt+ Z s

0

Z 1

0

(x+x2)u2tdx dt+ Z s

0

Z 1

0

(x+x2)u2xdx dt + 8Z 1

0

(x2+x)Z22(x)dx+ Z 1

0

(x+x2)

Z12(x) +∂Z1

∂x 2

dx

+ Z s

0

Z 1

0

(x+x2)f2dx dt .

Using Gronwall’s lemma [14, Lemma 4.1] this implies ku(x, s)k2W1,1

2,γ(0,1))≤8T e8T kZ1k2W1

2,γ((0,1))+kZ2k2L2

γ((0,1))+kfk2L2 γ(QT)

Integrating with respect tosover the interval [0, T] gives the a priori bound (5.1).

Conclusion. The well posedness of a one point initial boundary value problem for a nonlinear hyperbolic singular integro-differential equation is proved. The Schauder fixed point theorem is the main tool used to establish the existence of solution. The solvability of the problem is proved when the given data are related to a weighted Sobolev space. A priori bound for the solution may allow to study the regularity of the solution is obtained.

Acknowledgements. The authors would like to extend their sincere appreciation to the Deanship of Scientific Research at King Saud University for its funding this Research group NO (RG-1435-043).

References

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theory revisited, Ann. l’Inst. H. Poincar´e. Anal. Non Lin´eaire., 25, No. 3 (2008), 567–585.

[5] S. Berrimi, S. A. Messaoudi;Existence and decay of solutions of a viscoelastic equation with a nonlinear source, Nonlinear Anal. 64 (2006) 2314–2331.

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Math. Anal. Appl. 316, (2006), 189-209.

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Said Mesloub

King Saud University, College of sciences Department of Mathematics, P.O. Box 2455, Riyadh 11451, Saudi Arabia

E-mail address:[email protected]

Imed Bachar

King Saud University, College of sciences Department of Mathematics, P.O. Box 2455, Riyadh 11451, Saudi Arabia

E-mail address:[email protected]

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