Electronic Journal of Qualitative Theory of Differential Equations 2012, No. 50, 1-16; http://www.math.u-szeged.hu/ejqtde/
ON A HIGHER ORDER TWO DIMENSIONAL
THERMOELASTIC SYSTEM COMBINING A LOCAL
AND NONLOCAL BOUNDARY CONDITIONS
SAIDMESLOUB
Abstrat. Due to theirimportane and numerousappliations,
evolutionmixed problemswithnonloalonstraintsin thebound-
ary onditions havebeenextensively studied during the two last
deades. Inthispaper,weonsideraninitialboundaryvalueprob-
lemforahigherorderthermoelastisystemarisingin linearther-
moelastiitywhihombinessomeDirihletandweightedintegral
boundary onditions. The studied systemmodelizes in general a
Kirho plate. We prove the well posedness of the given prob-
lem. Ourproofsaremainlybasedonsomeaprioriboundsinsome
Sobolevtypespaefuntionsandonsomedensityarguments.
1. Introdution
During the few last deades, many researhers have studied linear
and nonlinear systems of thermoelasti equations and many results
havebeenpublished. Mostof theseresultsweredealingwiththestudy
of existene, asymptoti behavior, regularity, ontrollability, propa-
gation of singularities and blow up of solutions. For example Assila
[1℄, has studied global existene and asymptoti behavior of solutions
for a purely linear multidimensional system of nonhomogeneous and
anisotropi thermoelastiity, assoiated with nonlinear boundary on-
ditions. Dafermos and Hsiao [4℄, Hrusa and Messaoudi [6℄, Munoz-
Rivera [15℄, Rake [16℄, Rake and Shibata [17℄ and Slemrod[19℄ have
studiedandobtained someresultsaboutthe existene, regularity,on-
trollabilityandlong-timebehaviorofsomesystemsofthermoelastiity.
Also Rake and Wang [18℄ have onsidered some linear and semilin-
ear Cauhy problems and desribed the propagation of singularities.
Mixedproblems forthermoelastisystemsbeomeveryhardtohandle
in ase of the presene of nonloal onstraints in the boundary (suh
Key words and phrases. Thermoelasti System, integral ondition, Energy In-
equality,Existene,Uniqueness,Weaksolution
2000MathematisSubjetClassiation: 35L70,35L20,35K,35R.
ondition). Thereare two types ofnonloalproblems, spatialnonloal
problems and time nonloalproblems. These type of mixed problems,
arise mainly when the data on the boundary annot be measured di-
retly,butitanbereplaedbyanonloalonditionsuhasanintegral
ondition. Physially, this kind of ondition an represent a mean, a
total energy ora total mass. Boundary value problems with nonloal
onstraintshave many importantappliationssuhasinunderground-
water ow, population dynamis, hemial diusion, thermoelastiity,
heatondution proesses,ertainbiologialproesses, nulearreator
dynamis,ontroltheory, medialsiene, biohemistry,and transmis-
sion theory. Nonloal problems were rst investigated by using the
method of separation of variables and the orresponding eigenvalues
and eigenfuntions were onsidered. Later on, other methods suh as
the funtional analysis method,the energeti method and the method
ofsingularintegralequations wereappliedtomixednonloalproblems
but with great diulties. Forsome nonloalmixed problems for par-
aboli and hyperboli equations the reader should refer to Mesloub
[7,8,9,10℄, Mesloub and Mesloub.F [12℄, Mesloub and Messaoudi [13℄
and Mesloub and Bouziani [11℄.
Motivated by the previous studies, we onsider the following ini-
tialboundaryvalue problemfor a fourth order two dimensional linear
thermoelastisystemwithDirihletandnonloalonstraintsofintegral
type:
L 1 u = ∂ ∂t 2 2 u + △ 2 u − α△θ + c 1 u + c 2 u t = f(x
y
t)
(x
y
t) ∈ Q, L 2 θ = β ∂θ ∂t − η△θ + σθ + α△u t = g(x
y
t)
(x
y
t) ∈ Q, u(x
y
0) = u ◦ (x, y), u t (x
y
0) = u 1 (x
y), θ(x
y
0) = θ ◦ (x
y),
u(0
y
t) = 0
u(a
y
t) = 0
0 < y < b
0 < t < T, u(x
0
t) = 0
u(x
b
t) = 0
0 < x < a
0 < t < T, R a
0 x k udx = 0R b
0 y k udy = 0, R a
0 x k θdx = 0R b
0 y k θdy = 0, k = 0, 1.
(1)
where
Q = Ω × [0
T ]
withΩ = (0
a) × (0
b)
T < ∞
a < ∞
andb < ∞
.The given data satisfy the onsisteny onditions
u ◦ (0
y) = u ◦ (a
y) = u ◦ (x
0) = u ◦ (x
b) = 0,
(2)u 1 (0
y) = u 1 (a
y) = u 1 (x
0) = u 1 (x
b) = 0,
Z a 0
x k u ◦ dx = 0
Z b
0
y k u ◦ dy = 0
Z a
0
x k u 1 dx = 0
(3)Z b 0
y k u 1 dy = 0
Z a
0
x k θ ◦ dx = 0
Z b
0
y k θ ◦ dy = 0, k = 0, 1
.Problem
(1)
arises inlinearthermoelastiity. It modelizesaKirho plate, whereu
is the displaementθ
is the thermal damping. Theintegralonditionsmaybeinterpretedastheaverageandthe weighted
average ofthe displaementand the thermaldamping and
α
β
η, σ, c 1 and c 2 are positiveonstants.
Onthebasisofsomeaprioribounds,energyestimatesandsomeden-
sity arguments,weprovethe existene, uniqueness and the ontinuous
dependene of the solution on the data of the given problem
(1) .
Thepaperisorganizedas follows: Insetion 1,westart byanintrodution
about previous and related results onerning the subjet. In setion
2, we introdue some notations, funtion spaes and reformulation of
the studied problem. Setion 3 is devoted to the study of uniqueness
of the solution of the stated problem. In the fourth and last setion,
weestablishtheexisteneofthe weaksolutionofour problemand give
some remarks. At the end of the paper, we give a list of some used
referenes.
2. Notations, funtional frame, some auxiliary inequalities
and reformulation of the problem
Wedenoteby
L 2 (Q)
theusualsquareintegrablefuntionsspae,
andby
W 2 1,0 (Q)
the Sobolev spaehaving the inner produt(u
υ) W 1 , 0
(Q) = (u υ) L 2 (Q) + (u x
υ x ) L 2 (Q) .
Let
B 2 m (0, a)
(See [2,3℄) be the spae onstituted of funtionsu ∈ L 2 (0, a),
ifm = 0
and of funtionsu
suh thatℑ m x u ∈ L 2 (0, a),
ifm ≥ 1,
whereℑ m x V = 1 (m − 1)!
x
Z
0
(x−ξ) m−1 V (ξ, t)dξ =
x
Z
0 ξ 1
Z
0
...
ξ m− 1
Z
0
V (η, t)dηdξ m−1 ...dξ 1 ,
withinnerprodut
(u, v) B m 2 (0,a) =
a
Z
0
ℑ m x u.ℑ m x vdx,
andassoiatednormkuk B m
2 (0,a) = kℑ m x uk L 2 (0,a) , for m ≥ 1.
Wealsousethefuntionspaes
C(I, B 2 m (Ω)), C (I, B 2 1,x (Ω)), C(I, B 2 1,y (Ω))
and
C(I, B 2 1,x,y (Ω))
of ontinuous mappings fromI = [0, T ]
onto theHilbert spaes
B 2 m (Ω) , B 2 1,x (Ω) , B 2 1,y (Ω),
andB 2 1,x,y (Ω)
respetively, and with inner produts (respetively) given by(u, v) B 2 m (Ω) = Z
Ω
ℑ m x u.ℑ m x vdxdy, (u, v) B 1 ,x
2 (Ω) =
Z
Ω
ℑ x u.ℑ x vdxdy,
(u, v) B 1 ,y
2 (Ω) =
Z
Ω
ℑ y u.ℑ y vdxdy, (u, v) B 1 ,x,y
2 (Ω) =
Z
Ω
ℑ xy u.ℑ xy vdydx.
The followinginequalitiesare used inour paper:
A) For every
u ∈ L 2 (Λ)
(Λ
is either(0, d)
orΩ
) and allm ∈ IN ∗,
we have the inequalities
khk 2 B m
2 (0,d) ≤ d 2
2 khk 2 B m− 1
2 (0,d)
(1*)and
khk 2 B m 2 (0,d) ≤
d 2 2
m
. khk 2 L 2 (0,d) ,
(2*)kℑ x uk 2 L 2 (Ω) ≤ a 2
2 kuk 2 L 2 (Ω) (3*)
khk 2 B 1 ,x,y
2 (Ω) = kℑ xy hk 2 L 2 (Ω) ≤ (ab) 2
4 khk 2 L 2 (Ω) ,
(4*)B) Growall's Lemma [5, Lemma 7.1℄. If
f 1 (s)
f 2 (s)
andf 3 (s)
arenonnegativefuntions on
(0
T )
f 1 (s)
andf 2 (s)
integrablefuntions,and
f 3 (s)
is nondereasing on (0
T )
then ifR s
0 f 1 (t) dt + f 2 (s) ≤ c R s
0 f 2 (t) dt + f 3 (s) ,then R s
0 f 1 (t) dt + f 2 (s) ≤ exp (cs) .f 3 (s).
C)Cauhy
ε−
inequality: Forallε > 0
, and for arbitrarya
b
inR
we have
|ab| ≤ ε
2 |a| 2 + 1 2ε |b| 2 .
Let us now formulate problem
(1)
in its operator form. Problem(1)
an be viewed as the problem of solving the operator equation:JU = G,
withU = (u
θ)
JU = (J 1 u
J 2 θ)
andG= (G 1 G 2 ) =
({f
u ◦ u 1 } , {g
θ ◦ }) ,
where
u 1 } , {g
θ ◦ }) ,
whereJ 1 u = {L 1 u
ℓ 1 u
ℓ 2 u}
J 2 θ = {L 2 θ
ℓ 3 θ}
.The unbounded operator
A
is onsidered on the Banah spaeB
intoa Hilbert spae
H
with domainD(J)
dened byD (J) =
U = (u
θ) ∈ (L 2 (Q)) 2 suh that u t θ t u tt ∂ i u
θ t u tt ∂ i u
∂ i u
∂x i
∂ i u
∂y i
u tx
u txx u ty u tyy ∈ L 2 (Q) , i = 1.4
u tyy ∈ L 2 (Q) , i = 1.4
(4)
and the funtions
U = (u
θ)
satisfy boundary onditions in(1).
HereB
istheBanah spaeobtained byenlosingD (J )
withrespet tothenite norm
kU k B =
ku t (.
.
τ )k 2 C(I,B 1 ,x,y
2 (Ω)) + ku(. y
τ)k 2 C(I,B 1 ,x
2 (Ω))
+ ku(x
.
τ)k 2 C(I,B 1 ,y
2 (Ω)) + kθ(. .
τ)k 2 C(I,B 1 ,x,y
2 (Ω))
1/2
,
(5)where
ℑ x u = Z x
0
u(ξ, y, t)dξ, ℑ xy u = Z x
0 y
Z
0
u(ξ, η, t)dηdξ.
Theelements
U = (u
θ) ∈ B
arethe setof ontinuous funtionsu
andθ
onI = [0, T ],
wherefuntionsu
have valuesinB 2 1,x (Ω) , B 2 1,y (Ω)
andhavederivatives
u twhihareontinuousonI
withvalues inB 2 1,x,y (Ω)
and
θ
have values inB 2 1,x,y (Ω) .
Let
H = H 1 ×H 2betheHilbertspae{L 2 (Q) × W 2 1 (Ω) × L 2 (Ω)}×
{L 2 (Q) × L 2 (Ω)}
havingthe nite normkGk 2 H = kf k 2 L 2 (Q) + ku ◦ k 2 W 1
2 (Ω) + ku 1 k 2 L 2 (Ω) + kg k 2 L 2 (Q) + kθ ◦ k 2 L 2 (Ω) . (6)
3. Uniqueness of solution
We now state the rst result onerning the uniqueness of solution
of problem
(1)
.Theorem 1. For any funtion
U = (u
θ) ∈ D (J)
there exists apositiveonstant
C > 0
independent ofU
suh thatkU k 2 B − C kJU k 2 H ≤ 0
(7)Proof. Taking the inner produt in
L 2 (Q τ )
of equationsL 1 u = f
,L 2 u = g
and the intgrodierential operatorsℑ 2 xy u t and ℑ 2 xy θ,
respetively, where
Q τ = (0
τ) × Ω
with0 ≤ τ ≤ T,
we havec 1 u
ℑ 2 xy u t
L 2 (Q τ ) + c 2 u t
ℑ 2 xy u t
L 2 (Q τ ) + u tt
ℑ 2 xy u t
L 2 (Q τ )
+ ∂ 4 u
∂x 4 ℑ 2 xy u t
L 2 (Q τ )
+ 2
∂ 4 u
∂x 2 ∂y 2 ℑ 2 xy u t
L 2 (Q τ )
+ ∂ 4 u
∂y 4 ℑ 2 xy u t
L 2 (Q)
−α ∂ 2 θ
∂x 2 ℑ 2 xy u t
L 2 (Q τ )
− α ∂ 2 θ
∂y 2 ℑ 2 xy u t
L 2 (Q τ )
+ β θ t ℑ 2 xy θ
L 2 (Q τ )
−η ∂ 2 θ
∂x 2 ℑ 2 xy θ
L 2 (Q)
− η ∂ 2 θ
∂y 2 ℑ 2 xy θ
L 2 (Q)
+σ θ
ℑ 2 xy θ
L 2 (Q τ ) + α
∂ 3 u
∂x 2 ∂t
ℑ 2 xy θ
L 2 (Q τ )
+ α
∂ 3 u
∂y 2 ∂t
ℑ 2 x θ
L 2 (Q τ )
= f
ℑ 2 xy u t
L 2 (Q τ ) + g ℑ 2 xy θ
L 2 (Q τ ) . (8)
By suessive integration by parts of eah term of
(8),
and usingboundary onditions in
(1),
we derivethe following equalities:c 1 u
ℑ 2 xy u t
L 2 (Q τ ) = c 1
2 kℑ xy u(.
.
τ)k 2 L 2 (Ω) − c 2
2 kℑ xy u 0 k 2 L 2 (Ω) ,
(9)c 2 u t ℑ 2 xy u t
L 2 (Q τ ) = c 2 kℑ xy u t k 2 L 2 (Q τ ) , (10)
u tt
ℑ 2 xy u t
L 2 (Q τ ) = 1
2 kℑ xy u t (.
.
τ )k 2 L 2 (Ω) − 1
2 kℑ xy u 1 k 2 L 2 (Ω), (11)
∂ 4 u
∂x 4 ℑ 2 xy u t
L 2 (Q τ )
= 1
2 kℑ y u x (x
.
τ )k 2 L 2 (Ω) − 1 2
ℑ y
∂u ◦
∂x
2 L 2 (Ω)
,
(12)2
∂ 4 u
∂x 2 ∂y 2 ℑ 2 xy u t
L 2 (Q τ )
= ku(x
y
τ )k 2 L 2 (Ω) − ku ◦ k 2 L 2 (Ω) (13)
∂ 4 u
∂y 4 ℑ 2 xy u t
L 2 (Q τ )
= 1
2 kℑ x u y (.
y
τ)k 2 L 2 (Ω) − 1 2
ℑ x ∂u ◦
∂y
2 L 2 (Ω)
(14)
−α ∂ 2 θ
∂x 2 ℑ 2 xy u t
L 2 (Q τ )
= α (ℑ y θ
ℑ y u t ) L 2 (Q τ ) (15)
−α ∂ 2 θ
∂y 2 ℑ 2 xy u t
L 2 (Q τ )
= α (ℑ x θ
ℑ x u t ) L 2 (Q τ ) (16)
β θ t ℑ 2 xy θ
L 2 (Q τ ) = β
2 kℑ xy θ (.
.
τ)k 2 L 2 (Ω) − β
2 kℑ xy θ ◦ k 2 L 2 (Ω) (17)
−η ∂ 2 θ
∂x 2 ℑ 2 xy θ
L 2 (Q τ )
= η kℑ y θk 2 L 2 (Q τ ) (18)
−η ∂ 2 θ
∂y 2 ℑ 2 xy θ
L 2 (Q τ )
= η kℑ x θk 2 L 2 (Q τ ) (19)
σ θ
ℑ 2 xy θ
L 2 (Q τ ) = σ kℑ xy θk 2 L 2 (Q τ )
(20)α
∂ 3 u
∂x 2 ∂t
ℑ 2 xy θ
L 2 (Q τ )
= −α (ℑ y u t ℑ y θ) L 2 (Q τ ) (21)
α
∂ 3 u
∂y 2 ∂t
ℑ 2 xy θ
L 2 (Q τ )
= −α (ℑ x u t ℑ x θ) L 2 (Q τ ) (22)
f
ℑ 2 xy u t
L 2 (Q τ ) = (ℑ xy f ℑ xy u t ) L 2 (Q τ )
(23)
g
ℑ 2 xy θ
L 2 (Q τ ) = (ℑ xy g ℑ xy θ) L 2 (Q τ )
. (24)
Ifweuse Cauhy-
ε
-inequalitygiven inC),Poinare'inequalityoftype(4*)
,
and equalities(9) − (24),
then equation(8)
redues tokℑ xy u(.
.
τ)k 2 L 2 (Ω) + kℑ xy u t (.
.
τ )k 2 L 2 (Ω) + kℑ x u (.
y
τ )k 2 L 2 (Ω) + kℑ x u y (.
y
τ )k 2 L 2 (Ω) + kℑ y u (x
.
τ )k 2 L 2 (Ω) + kℑ xy u t k 2 L 2 (Q τ )
+ kℑ y u x (x
.
τ)k 2 L 2 (Ω) + kℑ xy θ (.
.
τ)k 2 L 2 (Ω)
≤ C 1
ku 1 k 2 L 2 (Ω) + ku ◦ k 2 L 2 (Ω) + ∂u ◦
∂x
2
L 2 (Ω) + kℑ xy u 0 k 2 L 2 (Ω) +
∂u ◦
∂y
2
L 2 (Ω) + kθ ◦ k 2 L 2 (Ω) + kf k 2 L 2 (Q τ )
+ kgk 2 L 2 (Q τ ) + kℑ xy u t k 2 L 2 (Q τ ) + kℑ xy θk 2 L 2 (Q τ )
(25)
where
C 1 =
max n
1
a 4 2
b 4 2
,(ab) 8 2
,β(ab) 8 2 , c 2 1 o min a 2
4
b 2
4 , c 2 2 .
Applying the Gronwall's lemma given in B) to inequality
(25)
anddisarding the fourth, sixth and seventh term of its left-hand, we
obtain
kℑ xy u(.
.
τ)k 2 L 2 (Ω) + kℑ xy u t (.
.
τ )k 2 L 2 (Ω) + kℑ x u (.
y
τ )k 2 L 2 (Ω) + kℑ y u (x
.
τ)k 2 L 2 (Ω) + kℑ xy θ (.
.
τ)k 2 L 2 (Ω)
≤ C 1 e C 1 T ku 1 k 2 L 2 (Ω) + ku ◦ k 2 W 1
2 (Ω) + kθ ◦ k 2 L 2 (Ω) + kf k 2 L 2 (Q) + kgk 2 L 2 (Q)
!
.
(26)Eah term on the left-hand side of
(26)
is bounded and sine theright-handoftheaboveinequality
(26)
isindependentofτ
,weantakethe least upper bound of eah term of the left-hand side with respet
to
τ
over[0
T ]
,weget thedesiredestimate(7)
withC = C 1 e C 1 T. This
ompletes the proof of Theorem 1.
Itan beprovedinastandard way thatthe operator
J : B −→ H = H 1 × H 2 is losable.
Proposition 2. The operator
J : B → H = H 1 × H 2 has alosure
See
[15].
Let
J
be the losure ofJ
andD(J)
its domain of denition.
Wedenethe stronggeneralizedsolution ofproblem
(1)
as the solutionofthe operator equation
JU = (J 1 u
J 2 θ),
withJ 1 u = {L 1 u
ℓ 1 u
ℓ 2 u}
J 2 θ = {L 2 θ
ℓ 3 θ} , (J 1 u
J 2 θ) ∈ H.
If we pass to the limit in(7)
wehavethe result
Corollary 3. There exists a positive onstant
C
suh thatkU k 2 B − C
JU
2
H ≤ 0, ∀U ∈ D J
.
(27)Wededue fromtheaprioriestimate
(27)
thatastronggeneralizedso- lutionof(1)
ifitexists isuniqueanddependsontinuouslyonG= (G 1
G 2 ) = ({f
u ◦ u 1 } , {g
θ ◦ }) ∈ H
, and that the range R J
u 1 } , {g
θ ◦ }) ∈ H
, and that the rangeR J
of
J
islosed in
H
andR J
= R (J ).
Existene of the solution of the stated problem.
Theorem 4. Problem
(1)
has aunique strong solutionverifyingu ∈ C(I, B 2 1,x (Ω)), u ∈ C(I, B 2 1,y (Ω)), θ ∈ C(I, B 2 1,x,y (Ω)), ∂u
∂t ∈ C(I, B 2 1,x,y (Ω)).
Moreover, the funtions
ℑ x u, ℑ y u, , ℑ xy θ, ℑ xy θ
depend ontinuously on the free termsf ∈ L 2 (Q)
,g ∈ L 2 (Q) ,
and on the initial datau ◦ ∈ W 2 1 (Ω)
u 1 ∈ L 2 (Ω)
θ ◦ ∈ L 2 (Ω)
that isku(.
.
τ )k 2 C(I,B 1 ,x,y
2 (Ω)) ≤ C kJU k H ,
ku(x
.
τ)k C(I,B 1 ,y
2 (Ω)) ≤ C kJU k H ,
ku(.
y
τ )k C(I,B 1 ,x
2 (Ω)) ≤ C kJU k H ,
ku t (.
.
τ )k C(I,B 1 ,x,y
2 (Ω)) ≤ C kJU k H ,
kθ(.
.
τ )k 2 C(I,B 1 ,x,y
2 (Ω)) ≤ C kJU k H . (28)
Proof. To establish the existene of the solutionof problem
(1)
, itis suientto prove that the imageof the operator
J
is dense inH.
General ase for density: Sine
H
is a Hilbert spae,R (J) = H
isequivalenttotheorthogonalityofavetor
Φ = (G 1 G 2 ) = ({σ 1 σ 2 σ 3 }
{σ 4 σ 5 }) ∈ H
tothe range R (J ) ,
that is the equality
σ 2 σ 3 }
{σ 4 σ 5 }) ∈ H
tothe range R (J ) ,
that is the equality
σ 5 }) ∈ H
tothe rangeR (J ) ,
that is the equality+ (ℓ 2 u
σ 4 ) L 2 (Ω) + (ℓ 3 θ
σ 5 ) L 2 (Ω)
+ (L 1 u
σ 1 ) L 2 (Q) + (L 2 θ
σ 2 ) L 2 (Q) + (ℓ 1 u
σ 3 ) W 1
2 (Ω)
= 0,
(29)for all
U = (u
θ) ∈ D (J) ,
implies thatΦ = 0.
LetD 0 (J)
bea subsetof
D(J )
for whihℓ 1 u = 0
ℓ 2 u = 0, ℓ 3 θ = 0.
IfU ∈ D 0 (J),
then(29)
redues to
(L 1 u
σ 1 ) L 2 (Q) + (L 2 θ
σ 2 ) L 2 (Q) = 0.
(30)We have to prove that
Ψ = (σ 1 σ 2 ) = 0
everywhere in Q.
Thus we
must prove the followingspeial ase (ofdensity) and then gobak to
the general ase.
Proposition5. If,forsomefuntion
Ψ = (σ 1 σ 2 ) ∈ (L 2 (Q)) 2 and
for allelements
U ∈ D ◦ (J )
we have(L 1 u
σ 1 ) L 2 (Q) + (L 2 θ
σ 2 ) L 2 (Q) = 0
(31)then
Ψ
vanishesalmost everywhere inQ.
Proof. Sinerelation
(31)
holdsforany elementofD ◦ (J)
wethentake anelement
U = (u
θ)
with a speial formgiven byU =
( (0
0)
0 ≤ t ≤ s, R t
s (τ − t) u τ τ dτ R t s θ τ dτ
s ≤ t ≤ T
,(32)
suh that
(u tt θ t )
isa solutionof the system
ℑ 2 xy u tt = E 1 (r
t) =
Z T t
σ 1 (r
τ) dτ
ℑ 2 xy θ t = E 2 (r
t) = Z T
t
σ 2 (r
τ) dτ
(33)
where
E 1 (x
t) = R T
t σ 1 (r τ ) dτ
and E 2 (x
t) = R T
t σ 2 (r τ) dτ.
It
is lear that
σ 1 = −ℑ 2 xy u ttt , σ 2 = −ℑ 2 xy θ tt .
(34)Proposition 6. The funtion
Ψ = (σ 1 σ 2 ) ∈ (L 2 (Q)) 2 dened in
(34)
is in(L 2 (Q)) 2 .
(34)
is in(L 2 (Q)) 2 .
Proof. it an bearried out as in[10℄.
Now replaing the funtions
σ 1and σ 2 given by (34)
in (31),
we
(34)
in(31),
weobtain
−c 1 u
ℑ 2 xy u ttt
L 2 (Q) − c 2 u t
ℑ 2 xy u ttt
L 2 (Q) − u tt
ℑ 2 xy u ttt
L 2 (Q)
− ∂ 4 u
∂x 4 ℑ 2 xy u ttt
L 2 (Q)
− 2
∂ 4 u
∂x 2 ∂y 2 ℑ 2 xy u ttt
L 2 (Q)
− ∂ 4 u
∂y 4 ℑ 2 xy u ttt
L 2 (Q)
+α ∂ 2 θ
∂x 2 ℑ 2 xy u ttt
L 2 (Q)
+ α ∂ 2 θ
∂y 2 ℑ 2 xy u ttt
L 2 (Q)
−β θ t ℑ 2 xy θ tt
L 2 (Q) + η ∂ 2 θ
∂x 2 ℑ 2 xy θ tt
L 2 (Q)
+η ∂ 2 θ
∂y 2 ℑ 2 xy θ tt
L 2 (Q)
+ −σ θ
ℑ 2 xy θ tt
L 2 (Q)
−α
∂ 3 u
∂x 2 ∂t
ℑ 2 xy θ tt
L 2 (Q)
− α
∂ 3 u
∂y 2 ∂t
ℑ 2 xy θ tt
L 2 (Q)
= 0
(35)Taking into aount the speial form of
U
given by(32)
and(33)
using boundary onditions in
(1)
and integrating by parts eah terms of(35)
gives−c 1 u
ℑ 2 xy u ttt
L 2 (Q) = c 1 Z
Q s
ℑ x u.ℑ xyy u ttt dxdydt
= −c 1
Z
Q s
ℑ xy u.ℑ xy u ttt dxdydt
= c 1
2 kℑ xy u t (.
.
T )k 2 L 2 (Ω) ,
(36)−c 2 u t ℑ 2 xy u ttt
L 2 (Q) = c 2 Z
Q s
ℑ x u t .ℑ xyy u ttt dxdydt
= −c 2
Z
Q s
ℑ xy u t .ℑ xy u ttt dxdydt
= c 2 kℑ xy u tt k 2 L 2 (Ω) ,
(37)− u tt ℑ 2 xy u ttt
L 2 (Q) = Z
Q s
ℑ x u tt .ℑ xyy u ttt dxdydt
= − Z
Q s
ℑ xy u tt .ℑ xy u ttt dxdydt
= 1
2 kℑ xy u tt (.
.
s)k 2 L 2 (Ω) (38)
− ∂ 4 u
∂x 4 ℑ 2 xy u ttt
L 2 (Q)
= Z
Q s
∂ 3 u
∂x 3 .ℑ xyy u ttt dxdydt
= − Z
Q s
∂ 2 u
∂x 2 .ℑ 2 y u ttt dxdydt
= Z
Q s
u x .ℑ 2 y u tttx dxdydt
= − Z
Q s
ℑ y u x .ℑ y u tttx dxdydt
= Z
Q s
ℑ y u tx .ℑ y u ttx dxdydt
= 1
2 kℑ y u xt (.
.
T )k 2 L 2 (Ω) (39)
−2
∂ 4 u
∂x 2 ∂y 2 ℑ 2 xy u ttt
L 2 (Q)
= 2 Z
Q s
∂ 3 u
∂x∂y 2 .ℑ xyy u ttt dxdydt
= −2 Z
Q s
∂ 2 u
∂y 2 .ℑ 2 y u ttt dxdydt
= 2 Z
Q s
u y .ℑ y u ttt dxdydt
= −2 Z
Q s
u.u ttt dxdydt
= 2 Z
Q s
u t .u tt dxdydt
= ku t (x
y
T )k 2 L 2 (Ω) (40)
− ∂ 4 u
∂y 4 ℑ 2 xy u ttt
L 2 (Q)
= Z
Q s
∂ 3 u
∂y 3 .ℑ xxy u ttt dxdydt
= Z
Q s
u y .ℑ 2 x u ttty dxdydt
= − Z
Q s
ℑ x u y .ℑ x u ttty dxdydt
= Z
Q s
ℑ x u ty .ℑ x u tty dxdydt
= 1
2 kℑ x u ty (.
y
T )k 2 L 2 (Ω) (41)
α ∂ 2 θ
∂x 2 ℑ 2 xy u ttt
L 2 (Q)
= −α Z
Q s
θ x .ℑ xyy u ttt dxdydt
= α Z
Q s
θ.ℑ 2 y u ttt dxdydt
= −α Z
Q s
ℑ y θ.ℑ y u ttt dxdydt
= α (ℑ y θ t ℑ y u tt ) L 2 (Q s ) (42)
α ∂ 2 θ
∂y 2 ℑ 2 xy u ttt
L 2 (Q)
= −α Z
Q s
θ y .ℑ xxy u ttt dxdydt
= α Z
Q s
θ.ℑ 2 x u ttt dxdydt
= −α Z
Q s
ℑ x θ.ℑ x u ttt dxdydt
= α (ℑ x θ t ℑ x u tt ) L 2 (Q
s )
(43)−β θ t ℑ 2 xy θ tt
L 2 (Q) = β Z
Q s
ℑ x θ t .ℑ xyy θ tt dxdydt
= −β Z
Q s
ℑ xy θ t .ℑ xy θ tt dxdydt
= β
2 kℑ xy θ t (.
.
s)k 2 L 2 (Ω) (44)
η ∂ 2 θ
∂x 2 ℑ 2 xy θ tt
L 2 (Q)
= −η Z
Q s
θ x .ℑ xyy θ tt dxdydt
= η Z
Q s
θ.ℑ 2 y θ tt dxdydt
= −η Z
Q s
ℑ y θ.ℑ y θ tt dxdydt
= η kℑ y θ t k 2 L 2 (Q
s )
(45)η ∂ 2 θ
∂y 2 ℑ 2 xy θ tt
L 2 (Q)
= η kℑ x θ t k 2 L 2 (Q
s )
(46)−σ θ
ℑ 2 xy θ tt
L 2 (Q) = σ Z
Q s
ℑ x θ.ℑ xyy θ tt dxdydt
= −σ Z
Q s
ℑ xy θ.ℑ xy θ tt dxdydt
= σ kℑ xy θ t k 2 L 2 (Q s ) (47)
−α
∂ 3 u
∂x 2 ∂t
ℑ 2 xy θ tt
L 2 (Q)
= α Z
Q s
∂ 2 u
∂x∂t .ℑ xyy θ tt dxdydt
= −α Z
Q s
u t .ℑ 2 y θ tt dxdydt
= α Z
Q s
ℑ y u t .ℑ y θ tt dxdydt
= −α (ℑ y u tt ℑ y θ t ) L 2 (Q s ) (48)
−α
∂ 3 u
∂y 2 ∂t
ℑ 2 xy θ tt
L 2 (Q)
= α Z
Q s
∂ 2 u
∂y∂t .ℑ xxy θ tt dxdydt
= −α Z
Q s
u t .ℑ 2 x θ tt dxdydt
= α Z
Q s
ℑ x u t .ℑ x θ tt dxdydt
= −α (ℑ x u tt ℑ x θ t ) L 2 (Q s ) (49)
Combiningequalities
(36) − (49)
and(35)
we obtain1
2 kℑ xy u tt (.
.
s)k 2 L 2 (Ω) + 1
2 kℑ y u xt (.
.
T )k 2 L 2 (Ω) + ku t (x
y
T )k 2 L 2 (Ω) + 1
2 kℑ x u ty (.
y
T )k 2 L 2 (Ω) + β
2 kℑ xy θ t (.
.
s)k 2 L 2 (Ω) + η kℑ y θ t k 2 L 2 (Q s ) +η kℑ x θ t k 2 L 2 (Q
s ) + σ kℑ xy θ t k 2 L 2 (Q
s ) + c 1
2 kℑ xy u t (.
.
T )k 2 L 2 (Ω) +c 2 kℑ xy u tt k 2 L 2 (Ω)
= 0
(50)where
Q s = Ω × (s
T ) .
Equality
(50)
impliesthatℑ xy u tt (ζ
ς
s) = 0
onΩ
andℑ xy θ t = 0
on
Q s hene we dedue that Ψ = (σ 1 σ 2 ) = (0
0)
almost everywhere
σ 2 ) = (0
0)
almost everywherein
Q s .
Proeeding in this way step by step, we prove that
Ψ = 0
almostevery where in
Q.
Now bak tothe generalase: Sine
Ψ = (σ 1 σ 2 ) = 0
everywhere in
Q
, thereforeequality (29)
beomes
(ℓ 1 u
σ 3 ) W 1
2 (Ω) + (ℓ 2 u σ 4 ) L 2 (Ω) + (ℓ 3 θ
σ 5 ) L 2 (Ω) = 0.
(51)
Sine the three quantities in
(51)
vanish independently and sine the rangesof thetraeoperatorsℓ 1 ℓ 2 andℓ 3 are respetively everywhere
dense in the spaes W 2 1 (Ω)
L 2 (Ω)
and L 2 (Ω)
therefore it follows,
ℓ 3 are respetively everywhere
dense in the spaes W 2 1 (Ω)
L 2 (Ω)
and L 2 (Ω)
therefore it follows,
from
(51)
thatσ 3 = σ 4 = σ 5 = 0.
HeneR (J) = H.
This ahievesthe proof of Theorem 4.
Remark: The following larger lass of problems may handled by
using the previous same tehniques
L 1 u = ∂ ∂t 2 u 2 + △ 2 u − α△θ + (c 1 u + c 2 u t ) = f(x
y
t
,θ, u)
(x
y
t) ∈ Q, L 2 θ = β ∂θ ∂t − η△θ + σθ + α△u t − c 3 ∆u = g(x
y
t
,θ, u)
(x
y
t) ∈ Q,
u(x
y
0) = u ◦ (x, y), u t (x
y
0) = u 1 (x
y), θ(x
y
0) = θ ◦ (x
y), u(0
y
t) = 0
u(a
y
t) = 0
0 < y < b
0 < t < T, u(x
0
t) = 0
u(x
b
t) = 0
0 < x < a
0 < t < T, R a
0 x k udx = 0R b
0 y k udy = 0, R a
0 x k θdx = 0R b
0 y k θdy = 0, k = 0, 1,
(47)
where the funtions
f
andg
satisfy the onditions|f(x
y
t
,θ 1 , u 1 ) − f (x
y
t
,θ 2 , u 2 )| ≤ µ (|θ 1 − θ 2 | + |u 1 − u 2 |) ,
|g(x
y
t
,θ 1 , u 1 ) − g(x
y
t
,θ 2 , u 2 )| ≤ µ (|θ 1 − θ 2 | + |u 1 − u 2 |) .
and
α
β
η
,σ, c 1 and c 2 are positive onstants.
We rst deal with the assoiated linear problem,that is when
f (x
y
t
,θ, u) = f (x
y
t )
andg(x
y
t
,θ, u) = g(x
y
t )
and then,on the basis of the obtained results of the linear problem, we apply
an iterative proess to establish the existene and uniqueness of the
nonlinear problem.
Aknowledgment: The author extends his appreiation to the
Deanship of Sienti Researh at King Saud University for funding
the work through the researh groupprojetNo: RGP-VPP-117.
Referenes
[1℄ M. Assila, Nonlinear boundary stabilization of an inhomogeneous and
anisotropithermoelastiitysystem,AppliedMathLetters.13 (2000),71-76.
[2℄ A.Bouziani,OnaninitialboundaryvalueproblemwithDirihletintegralon-
ditionsforahyperboliequationwiththeBesseloperator,JournalofApplied
Mathematis,10(2003)487-502.
paraboliequation,J.Appl.Math.StohastiAnal.9(1996),no3,323-330.
[4℄ C. M. Dafermos, L. Hsiao, Development of singularities in solutions of the
equations on nonlinear thermoelastiity system, Q. Appl. Math 44 (1986),
463-474.
[5℄ L.Garding,Cauhy'sProblemforHyperboliEquations.LetureNotes.Uni-
versityofChiago: Chiago,1957.
[6℄ W.J.Hrusa,S.A.Messaoudi,Onformationofsingularitiesonone-dimensional
nonlinearthermoelastiity,Arh.RationalMeh.Anal3(1990),135-151.
[7℄ S.Mesloub,Anonlinearnonloalmixedproblemforaseondorderparaboli
equation,J.Math.Anal.Appl.316(2006)189-209.
[8℄ S.Mesloub,Onasingulartwodimensionalnonlinearevolutionequationwith
nonloalonditions,NonlinearAnalysis68(2008)2594-2607.
[9℄ S.Mesloub,Onanonlinearsingularhyperboliequation,MathematialMeth-
odsintheAppliedSienes,Vol33,Issue1(2010)57-70.
[10℄ S. Mesloub, On anon loal problem for apluriparaboliequation, Ata Si.
Math.(Szeged)67(2001),203-219.
[11℄ S. Mesloub, A. Bouziani, On a lass of singular hyperboli equation with a
weighted integral ondition, Internat. J. Math. &Math. Si. Vol 22. N 511-
519,(1999).
[12℄ S. Mesloub, and F. Mesloub, Solvability of amixed non loal problem for a
nonlinearSingularVisoelastiequation,Ata.Appl.Mathematiae,Vol110,
Number1,109-129.
[13℄ S.Mesloub,S.Messaoudi,GlobalExistene, Deay,andBlowupofSolutions
ofaSingularNonloalVisoelastiProblem,AtaAppliandaeMathematiae,
Volume: 110Issue: 2(2010),705-724.
[14℄ S. Mesloub,N. Lekrine, Onanon loal hyperbolimixed problem, Ata Si.
Math.(Szeged),70(2004),65-75.
[15℄ J.E. Munoz Rivera, R. K.Barreto,Existene andexponentialdeayin non-
linearthermoelastiity,NonlinearAnalysis31No.1/2(1998),149-162.
[16℄ R. Rake, Blow up in nonlinear three dimensional thermoelastiity, Math.
MethodsAppl.Si. 12No3(1990),273-276.
[17℄ R. Rake, Y. Shibata, Global smooth solutions and asymptoti stability in
one-dimensional nonlinearthermoelastiity, Arh. rational. Meh. Anal. 116
(1991),1-34.
[18℄ R. Rake, Y. G.Wang, Propagationof singularities in one-dimensionalther-
moelastiity,J.Math.Anal. Appl,223(1998),216-247.
[19℄ M.Slemrod,Globalexistene,uniqueness,andasymptotistabilityoflassial
solutionsinone-dimensionalthermoelastiity, Arh. rational.Meh.Anal. 76
(1981),97-133.
(Reeived September 5,2011)
King Saud University, College of sienes Department of Mathe-
matis,P.O.Box2455, Riyadh11451, SaudiArabia
E-mailaddress: mesloubsyahoo.om