A mixed problem with only integral boundary conditions for a hyperbolic equation

PII. S0161171204204240 http://ijmms.hindawi.com © Hindawi Publishing Corp.

A MIXED PROBLEM WITH ONLY INTEGRAL BOUNDARY

CONDITIONS FOR A HYPERBOLIC EQUATION

ABDELFATAH BOUZIANI

Received 29 April 2002

We investigate an initial boundary value problem for a second-order hyperbolic equation with only integral conditions. We show the existence, uniqueness, and continuous depen-dence of a strongly generalized solution. The proof is based on an energy inequality estab-lished in a nonclassical function space, and on the density of the range of the operator asso-ciated to the abstract formulation of the studied problem by introducing special smoothing operators.

2000 Mathematics Subject Classification: 35L20, 35B45, 35D05, 35B30.

1. Introduction. In recent years, new attention has been devoted to mixed problems for hyperbolic equations with integral conditions. Such conditions appear in case, where for instance, direct measurement quantities are impossible and their mean values are known. Such situations take place in studying, for example, the dynamics of ground waters [12,13]. The first investigation of this type of problems goes back to [3] in 1996, in which the author proved the existence, uniqueness, and continuous dependence of the solution upon the data of certain hyperbolic problems with only integral boundary conditions. The proof used in [3] is based, on the one hand, on the method derived by Ladyženskaya [11] to show the existence of the solution, and, on the other hand, on a priori estimates to prove the uniqueness and continuous dependence; these estimates are established by taking the scalar product inL2_{-space of the considered equations}

and integrodifferential operators constructed for each problem. Later, mixed problems for hyperbolic equations with integral condition(s) were treated in [2,4,6,7,13,14].

In this note, we prove the existence, uniqueness, and continuous dependence of the solution of a mixed problem with only integral boundary conditions for a second-order hyperbolic equation. To this end, we reformulate the stated problem as a problem of solving an operator equation, then we show that the operator generated by the problem is bijective. To prove the injection, we have established an a priori estimate inBm

2

respect tot. These operators can be used to solve a large class of evolution problems possessing second-order derivative in time.

2. Statement of the problem and notations. In the rectangular domainQ=(a, b)× (0, T ), we consider the following problem: find the functionv=v(x, t)satisfying

ᏸv=∂_∂t2v2− ∂ ∂x

p(x, t)∂v ∂x

+q(x, t)∂v

∂t +r (x, t) ∂v ∂x +s(x, t)v=f(x, t), (x, t)∈Q,

0v=v(x,0)=v0(x), x∈(a, b),

1v=∂v(x,_∂t 0)=v1(x), x∈(a, b),

b

a

v(x, t)dx=E(t), t∈(0, T ),

b

axv(x, t)dx=M(t), t∈(0, T ),

(2.1)

with

b

a

v0(x)dx=E(0),

b

a

xv0(x)dx=M(0),

b

a

v1(x)dx=E(0),

b

a

xv1(x)dx=M(0),

(2.2)

wherev0,v1,E,M, f,p,q,r, andsare known functions andT >0,a, andbare given constants.

Assumption2.1. We will assume, for(x, t)∈Q, that

0< c0≤p(x, t)≤c1, ∂p ∂t

≤c2, ∂p

∂x

≤c3, q(x, t)≥0,

∂q_∂x≤c4, r (x, t)≤c5, ∂r ∂x

≤c6, s(x, t)≤c7.

(2.3)

InAssumption 2.1and throughout,ciare positive constants.

Since integral boundary conditions are inhomogeneous, it is convenient to convert problem (2.1) to an equivalent problem with homogeneous integral conditions. For this, we introduce a new functionu(x, t)representing the deviation of the functionv(x, t)

from the function

U (x, t)=12(x−a)(3x−a−2b) M(t) (b−a)4

−18(x−a)2_{−12(x−a)(b−a)−(b−a)}2 E(t) (b−a)3.

Then, our problem becomes as follows: find a functionu=u(x, t)satisfying

ᏸu=f(x, t)−ᏸU=f (x, t), (x, t)∈Q, (2.5)

0u=u(x,0)=u0(x), x∈(a, b), (2.6)

1u=∂u(x,0)

∂t =u1(x), x∈(a, b), (2.7)

b

a

u(x, t)dx=0, t∈(0, T ), (2.8) b

a

xu(x, t)dx=0, t∈(0, T ), (2.9)

with

b

aui(x)dx=0,

b

axui(x)dx=0 (i=0,1). (2.10)

We introduce the appropriate function spaces that will be used in the rest of the note. LetHbe a Hilbert space with a norm·H.

Definition_2.2. _{(i) Denote byL}2_{(0, T;H)the set of all measurable abstract functions} u(·, t)from(0, T )intoHsuch that

uL2(0,T;H)=

T

0

u(·, t)2_Hdt 1/2

<∞. (2.11)

(ii) LetC(0, T;H)be the set of all continuous functionsu(·, t):(0, T )→Hwith

uC(0,T;H)= max 0≤t≤T

_{u(·, t)H<∞. (2.12)}

Definition2.3. Denote byB₂m(a, b)the space equipped with the scalar product

(u, w)B₂m(a,b)=

b

a m

xu·mxv dx (2.13)

and the associated norm

uBm 2(a,b)=

b

a

m xu

2 dx

1/2

, (2.14)

where

m xu=

1

(m−1)! x

a

(x−ξ)m−1_{u(ξ,·)dξ, m≥1. (2.15)}

Lemma_2.4. _Form≥_{1, the following inequality holds:}

u2 B₂m(a,b)≤

(b−a)2

2 u

2

Bm₂−1(a,b). (2.16)

Definition2.5. LetBn,w₂ (0, T )be theweightedB₂n-spaceequipped with the norm

uBn,w 2 (0,T )=

T

0

w(t)n tu

2 dt

1/2

. (2.17)

In this note,w(t)=e−ct_.

Definition_2.6. _WriteBn_{2(0, T;H)(resp.,B2}n,w_{(0, T;H)) for the space of functions}

from(0, T )intoH, which is aB2n-space (resp., aB n,w

2 -space, ) for the measuredt. It is

a Hilbert space for the norm

uBn

2(0,T;H)=

T

0

n

tu(·, τ)H

2 dt

1/2

, (2.18)

respectively,

uBn,w

2 (0,T;H)=

T

0

w(t)n

tu(·, τ)H

2 dt

1/2

. (2.19)

Problem (2.5), (2.6), (2.7), (2.8), and (2.9) can be written in the following abstract form:

Lu=f , u0, u1, (2.20)

whereL=(ᏸ, 0, 1). The operatorL, with domainD(L), acts fromBtoF, whereD(L)

is the set of all functionsu∈L2_{(0, T;B}1

2(a, b))for which∂iu/∂ti,∂iu/∂xi(i=1,2),

and∂3_u/∂x∂t2_{belong toL}2_{(0, T;B}1

2(a, b))andusatisfies conditions (2.8) and (2.9),B

is the Banach space obtained by the closure ofD(L)in the norm

uB=

u2

C(0,T;L2_(a,b))+

∂u_∂t2_C(0,T;B1 2(a,b))

1/2

, (2.21)

andF is the Hilbert spaceL2_{(0, T;B}1

2(a, b))×L2(a, b)×B21(a, b).

LetLbe the closure ofLwith domainD(L).

Definition2.7. A solution of the abstract equation

Lu=f , u0, u1 (2.22)

is called astrongly generalized solutionof problem (2.5), (2.6), (2.7), (2.8), and (2.9).

3. Uniqueness and continuous dependence. We first establish an energy inequality; the uniqueness and continuous dependence of the solution with respect to the data are immediate consequences.

Theorem_3.1. _{UnderAssumption 2.1, the following inequality holds for any function} u∈D(L):

uB≤cLuF, (3.1)

Proof. Taking the scalar product, inB₂1(a, b), of (2.5) and∂u/∂tand integrating by

parts, we get

2 b

aq

x∂u_∂t

2 dx+_∂t∂

∂u(_∂t·, t)

2

B₂1(a,b)+

b

apu 2_dx

=2

f (·, t),∂u(·, t) ∂t

B1₂(a,b)+

b

a ∂p

∂tu 2_dx−2

b

a ∂p ∂xux

∂u ∂tdx −2 b a ∂q ∂xx ∂u ∂t 2 x ∂u ∂tdx−2

b

a

r ux∂u ∂tdx−2

b a ∂r ∂xu 2 x ∂u ∂tdx −2 b

ax(su)x ∂u

∂tdx.

(3.2)

According to theε-inequality, the right-hand side of (3.2) is bounded by

_{f (·, t)}2 B₂1(a,b)+

b a ∂p ∂t + _∂p ∂x 2 +r2+

_∂r

∂x

2

u2dx+4∂u(·, t)

∂t

2_B1

2(a,b)

+2∂u(·, t)

∂t

2_B2

2(a,b) + b a _∂q ∂x 2 x∂u_∂t

2 dx+

b

a

x(su)2dx.

(3.3)

Substituting (3.3) into (3.2) and integrating the result over(0, τ), with 0≤τ≤T, we obtain, by using inequality (2.16) form=1 andm=2,

2 τ 0 b a q x∂u ∂t 2 dx dt+

b

a

p(x, τ)u2_{(x, τ)dx+}∂u(·, τ) ∂t

2_B1

2(a,b)

≤

τ

0

f (·, t)2_B1

2(a,b)dt+

b

a

p(·,0)u2

0dx+u1 2 B₂1(a,b)

+ τ 0 b a ∂p ∂t + _∂p ∂x 2 +r2+

_∂r

∂x

2

+(b−a)2

2 s

2 _u2_{dx dt}

+4+(b−a)2 τ

0

∂u(·, t)_∂t 2_B1 2(a,b) dt+ τ 0 b a _∂q ∂x 2 x∂u ∂t 2 dx dt. (3.4)

By virtue ofAssumption 2.1, we have

u(·, τ)2_L2_(a,b)+

∂u(_∂t·, τ)2_B1 2(a,b)

≤c8

τ

0

_{f (·, t)}2

B1₂(a,b)dt+u0 2

L2(a,b)+u1 2 B₂1(a,b)

+c9

τ

0

u(·, t)2_L2_(a,b)+

∂u(_∂t·, t)2_B1 2(a,b)

dt,

where

c8=max

1, c1

min1, c0,

c9=max

c2+c32+c52+c62+

(b−a)2_/2c2

7,4+(b−a)2+c24

min1, c0 .

(3.6)

The application of Gronwall’s lemma implies

_{u(·, τ)}2 L2(a,b)+

∂u(_∂t·, τ)2_B1 2(a,b) ≤c8expc9Tf2

L2(0,T;B₂1(a,b))+u0 2

L2_(a,b)+u1 2 B₂1(a,b)

.

(3.7)

The right-hand side here is independent ofτ; thus taking the upper bound for 0≤τ≤T

in the left-hand side, we obtain estimate (3.1), wherec=c₈1/2exp(c8T /2).

As we have no information concerningR(L)expect thatR(L)⊂F, we must extendL, so that estimate (3.1) holds for the extension and its range is the whole space. We first state the following result.

Proposition3.2. Under the hypotheses ofTheorem 3.1, the operatorLacting from BintoF has a closure.

Proof. The proof is analogous to that of [5, Proposition 1].

The next corollary follows fromTheorem 3.1andProposition 3.2.

Corollary_3.3. _{Under the assumptions ofTheorem 3.1, the operatorLhas onR(L)} a continuous inverseL−1, that is, there exists a constantc >0such that

uB≤cLu_F, u∈DL. (3.8)

Proof. Inequality (3.8) can be obtained by passing to the limit in (3.1).

Corollary3.4. If problem (2.5), (2.6), (2.7), (2.8), and (2.9) has a strongly generalized solution, then this solution is unique and depends continuously on(f , u0, u1).

Corollary_3.5. _{The range of the operatorLequals the closure of the range of the} operatorL, that is,R(L)=R(L).

Proof. The proof is the same as that of [5, Corollary 2.5].

4. Existence of the solution. Now, we are able to state and prove our main result.

Theorem_4.1. _{Assume that the hypothesis ofTheorem 3.1holds. Moreover, it is} as-sumed that the coefficients ∂2_p/∂t2_{, ∂}3_p/∂t2_{∂x are bounded. Then, for f ∈L}2_{(0, T;} B1

that satisfies

uB≤cfL2_(0,T;B1

2(a,b))+u0L2(a,b)+u1B12(a,b)

, (4.1)

wherecis a positive constant independent ofu.

Proof. _{By virtue of Corollary 3.5, we conclude that it is sufficient to prove that} R(L)=F. To this end, we will first establish the density for a particular case in which

Lis reduced toL0andu∈D0(L0), whereL0=(ᏸ0, 1, 2),ᏸ0is the principal part of

ᏸ, that is,ᏸ0=∂2/∂t2−(∂/∂x)(p(x, t)(∂/∂x)),D0(L0)=D0(L), andD0(L)is the set

of all functionsu∈D(L)for whichiu=0(i=0,1).

Proposition4.2. Under the hypotheses ofTheorem 4.1, if

ᏸ0u, ω_L2_(0,T;B1

2(a,b))=0 (4.2)

for arbitraryu∈D0(L0)and someω∈L2_{(0, T;B}1

2(a, b)), thenωvanishes almost ev-erywhere inQ.

Suppose for a moment thatProposition 4.2has been established and return to the proof ofTheorem 4.1. Let the element(f , u0, u1)ofF=L2_{(0, T;B}1

2(a, b))×L2(a, b)× B21(a, b)be orthogonal toR(L0), that is,

ᏸ0u, f_L2_(0,T;B1 2(a,b))+

0u, u0L2(a,b)+

1u, u1_B1

2(a,b)=0, u∈D

L0. (4.3)

Ifu∈D0(L0), it follows fromProposition 4.2thatfvanishes almost everywhere inQ. Hence

0u, u0_L2_(a,b)+

1u, u1_B1

2(a,b)=0, u∈D

L0. (4.4)

ButR(1)and R(2)are everywhere dense in L2_{(a, b)and B}1

2(a, b), respectively. So, ω1=ω2=0, from which we conclude thatR(L0)=F.

We turn back to the general case. The operator L−L0 maps continuously B into

F; thenTheorem 4.1can be proved by the method of continuation with respect to the parameter. We will not describe it here; however we refer the reader, for instance, to [8]. To complete the proof ofTheorem 4.1, it remains to establish the proof ofProposition 4.2.

Proof ofProposition4.2. We need to introduce the family of smoothing

opera-tors with respect tot:

−ε2θ=

1

√ ε

t

0

sin√1

ε(t−τ)θ(x, τ)dτ, ε >0, (4.5)

ε−2

∗_θ=√1

ε

T

t sin

1

√

These operators provide the solutions of the problems

ε∂ 2−2

ε θ(x, t) ∂t2 +−

2

ε θ(x, t)=θ(x, t),

−2

ε θ(x,0)=0,

∂−2 ε θ(x,0)

∂t =0,

ε∂ 2−2

ε

∗

θ(x, t)

∂t2 +

−ε2

∗_{(x, t)=θ(x, t),}

−ε2

∗

θ(x, T )=0, ∂

−2 ε

∗_{θ(x, T )}

∂t =0,

(4.7)

respectively. They have the following properties.

Lemma4.3. Ifθ∈L2(0, T ), then

(1) −2

ε θ∈H2(0, T )and−ε2θ(x,0)=0,∂−ε2θ(x,0)/∂t=0,

(2) (−2

ε )∗θ∈H2(0, T )and(−ε2)∗θ(x, T )=0,∂(−ε2)∗θ(x, T )/∂t=0. Lemma_4.4. _{Ifθandgare inL}2_{(0, T ), then}

T

0

−2

ε θ·g dt=

T

0 θ·

−ε2

∗

g dt. (4.8)

Lemmas4.3and4.4are proved directly by using the definitions of the operators−2 ε

and(−2 ε )∗.

Lemma_4.5. _{For allθ}∈_L2_{(0, T )such that∂}2_θ/∂t2_{is inL}2_{(0, T ), the following identity} holds:

−ε2 ∂2_θ ∂τ2=

∂2−2 ε

θ ∂t2 −

1

εθ(x,0)cos t √

ε−

1

√ ε

∂θ(x,0) ∂t

t

0sin s √

εds. (4.9)

To prove this lemma, it suffices to integrate by parts the expression−2

ε (∂2θ/∂τ2). Lemma_4.6. _{For allθ}∈_L2_{(0, T;H),}

(1) 0T−ε2θ2dt≤

T

0θ2Hdtand

T

0ε−2θ−θH2dt→0, whenε→0,

(2) 0T(−ε2)∗2Hdt≤

T

0θ2Hdtand

T

0(−ε2)∗−θ2Hdt→0, whenε→0.

The proof ofLemma 4.6is similar to that of [1, Lemma 2.18].

Lemma_4.7. _{SetP (t)}=_{(∂/∂x)(p(x, t)(∂/∂x)); then the following relation holds:}

P (t)−2

ε =−ε2P (τ)−ε2+−ε2P(τ)−ε2, (4.10)

whereP(τ)=(∂/∂x)((∂2_{p(x, t)/∂t}2_{)(∂/∂x)).}

Lemma 4.5is proved directly by integrating by parts−2 ε P (τ).

Note that similar operators related to equations of order one in time and to opera-tional equations are established in [9,10].

From (4.2), we have _∂2_u

∂t2, ω

L2_(0,T;B1 2(a,b))

=(P u, ω)_L2_(0,T;B1

We replaceuby−2

ε uin (4.11) and applyLemma 4.5to the left-hand side andLemma 4.7

to the right-hand side; we get

−2 ε

∂2_u ∂τ2, ω

L2_(0,T;B1 2(a,b))

=−2

ε P ε−2u+εε−2P−ε2u, ω

L2(0,T;B1₂(a,b)). (4.12)

According toLemma 4.4, we obtain

_∂2_u

∂t2,

ε−2

∗_ω

L2_(0,T;B1 2(a,b))

=P −ε2u+εPε−2u,

−ε2

∗_ω

L2_(0,T;B1 2(a,b))

, (4.13)

from which we have

Q u∂

2−2 ε

∗_∗

x

ξω ∂t2 dx dt

=

Q

P (t)−ε2u+εP(t)−ε2u

−ε2

∗∗

x

ξωdx dt.

(4.14)

The operatorP (t)with conditions (2.8) and (2.9) has, onL2_{(0, T ), a continuous inverse}

defined by

P−1_(t)g=

x a dξ p(ξ, t) ξ a

g(ζ, t)dζ+ 1

(b−a)ab

(a−x)(b−x)/p(x, t)dx

×

b

a

b2_−x2 p(x, t) dx

b

a

(b−x) p(x, t)dx

x

ag(ξ, t)dξ

−

b

a (b−x) p(x, t)dx

b

a

b2_−x2 p(x, t) dx

x

a

g(ξ, t)dξ

−(b−a)

b

a

(a−x)(b−x) p(x, t) dx

x

ag(ξ, t)dξ x

a dξ p(ξ, t) .

(4.15)

Hence, we can writeP(t)−2

ε uas follows:

P(t)−ε2u=P(t)P−1(t)P (t)ε−2u=Π(t)P (t)−ε2u=Π(t)g, (4.16)

where

Π(t)g=

∂3_p ∂t2_∂x

1

p− ∂2_p

∂t2 1 p2 ∂p ∂x ×  x

a

g(ξ, t)dξ−

b

a

(a−x)(b−x)/p(x, t)dxaxg(ξ, t)dξ

b

a

(a−x)(b−x)/p(x, t)dx

 ₊∂2p

∂t2

1

pg.

Therefore, (4.14) becomes

Qu ∂2−2

ε

∗_∗

x

ξω

∂t2 dx dt

=

Q

P (t)ε−2u+εΠ(t)P (t)−ε2u

−ε2

∗_∗

x

ξωdx dt

=

Q

P (t)−2

ε u·Λε−ε2

∗∗

x

ξωdx dt,

(4.18)

where

Λε−ε2

∗_=I+εΠ∗−2 ε

∗_,

Π∗−2 ε

∗_∗

x

ξω=

b

x

_∂3_p

∂ξ∂t2

1

p− ∂2_p ∂t2 ∂p ∂ξ 1 p2

−ε2

∗_∗

ξ

ζωdξ

−

b x

(a−ξ)(b−ξ)/p(ξ, t)dξ

b a

(a−x)(b−x)/p(x, t)dx

×

b

a

_∂3_p

∂x∂t2

1

p− ∂2_p

∂t2 ∂p ∂x 1 p2 −2 ε ∗∗ x

ξωdx

+∂_∂t2p2

1

p

−ε2

∗∗

x

ξω.

(4.19)

Since the left-hand side of (4.18) is a continuous linear functional ofu, then the function

Λεhas∂Λε/∂xand∂2Λε/∂x2belonging toL2(Q), and such that

Λε|x=a=Λε|x=b=0, ∂Λε

∂x|x=a= ∂Λε

∂x|x=b=0.

(4.20)

For sufficiently smallε, we haveεΠ∗L2_(Q)<1, from which we conclude that the op-eratorΛεpossesses a continuous inverse onL2(Q), that is,∗x(ξω)∈L2(Q). We

dif-ferentiateΛεwith respect tox:

∂Λε ∂x −2 ε ∗∗ x

ξω= −I+εΠ∗−2 ε

∗_xω+ε∂Π∗

∂x −2 ε ∗∗ x

ξω, (4.21)

where

∂Π∗

∂x

−ε2

∗∗

x

ξω=b(a−x)(b−x)/p(x, t) a

(a−x)(b−x)/p(x, t)

×

b

a

_∂3_p

∂x∂t2

1

p− ∂2_p

∂t2 ∂p ∂x 1 p2

ε−2

∗∗

x

ξω.

(4.22)

Relation (4.23) implies that∂Π∗_{(t)/∂xis bounded onL}2_{(Q); thus according to (4.22),}

we deduce thatxω∈L2_{(Q)for sufficiently smallε; in other words,ω∈L}2_{(0, T;B}1 2(a, b))for sufficiently smallε. Similarly, using the boundedness of∂2_Π∗_(t)/∂x2_onL2_(Q),

According to (4.19) and (4.20), we have

I+ε1 p

∂2_p ∂t2

−2 ε

∗∗

x

ξω_x=a=0, (4.23)

I+ε1 p

∂2_p ∂t2

−ε2

∗∗

x

ξωx=b=0, (4.24)

I+ε1 p

∂2_p ∂t2

−ε2

∗_xω

x=a=0, (4.25)

I+ε1 p

∂2_p ∂t2

−2 ε

∗_xω

x=b=0. (4.26)

Analogously, for fixedx∈[a, b]and sufficiently smallε, the norm of(ε(1/p)(∂2_p/∂t2₎₎

onL2_{(0, T )is smaller than 1. Therefore, the operator(I+ε(1/p)(∂}2_p/∂t2_{))has a}

con-tinuous inverse onL2_{(0, T ). Then, from (4.23), (4.24), (4.25), and (4.26), we get}

∗

x

ξω_x=a=0, (4.27)

xωx=b=0. (4.28)

We will now construct the functionω. For this, we introduce the function

v(x, t)=ect

cω+∂ω ∂t

(4.29)

with

ω(x,0)=0, (4.30)

where

c∈

 c2+

c22+8c23

c0 ,+∞



. (4.31)

Solving the differential equation (4.29) with respect tot, by taking into account (4.30), we obtain

ω(x, t)= −e−ct_{tv. (4.32)}

It follows from relations (4.27), (4.28), and (4.32) that

bv=0, (4.33)

2

bv=0. (4.34)

Substituting (4.32) into (4.11) yields

−∂2u ∂t2, e−

ct_tv

L2(0,T;B1 2(a,b))

= −P u, e−cttv_L2_(0,T;B1

2(a,b)). (4.35)

In identity (4.35), we set

u= 2

from which we have

−v, e−ct_tv L2_(0,T;B1

2(a,b))= −

P2

tv, e−cttv

L2_(0,T;B1

2(a,b)). (4.37) Integrating by parts both sides of (4.37), we get

e−cTTv2_B1

2(a,b)+cv 2

B1,w₂ (0,T;B₂1(a,b))

= −

b

ae

−cT_{p(x, T )Tv}2 dx−

Qe

−ct_cp−∂p ∂t

tv2dx dt

−2

Q e−ct∂p

∂x 2 tvx

tvdx dt.

(4.38)

The last integral on the right-hand side of (4.38) is dominated by

c

2v

2

B₂1,w(0,T;B1₂(a,b))+

2

c

Qe

−ct∂p ∂x

2

tv2dx dt, (4.39)

where the first term is absorbed in the left-hand side. It then follows, by omitting the first term of each side of (4.38) and by usingAssumption 2.1, that

c2

2v

2

B1,w₂ (0,T;B₂1(a,b))≤ −

c2c0−cc2−2c32

v2

B₂2(0,T;L2_(a,b)). (4.40)

According to (4.31), we deduce thatvand thusωvanish almost everywhere inQ. This achieves the proof ofProposition 4.2.

This completes the proof ofTheorem 4.1.

Acknowledgment. _{This work has been done in the Abdus Salam International}

Centre of Theoretical Physics, Trieste, Italy. This work was supported by Laboratoire de matériaux et structure des systémes électromécaniques et leurs fiabilités, Centre Universitaire Larbi Ben M’Hidi, Oum El Bouagui, Algeria.

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