Existence of solutions of the parabolic variational inequality with variable exponent of nonlinearity

J. Math. Anal. Appl. 377 (2011) 450–463

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Existence of solutions of the parabolic variational inequality with variable exponent of nonlinearity R.A. Mashiyev a,∗ , O.M. Buhrii b a b

Dicle University, Faculty of Science, Department of Mathematics, 21280-Diyarbakir, Turkey Department of Differential Equations, Faculty of Mechanics and Mathematics, Ivan Franko National University of Lviv, 79000-Lviv, Ukraine

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 10 May 2010 Available online 13 November 2010 Submitted by V. Radulescu

In this article, new properties of variable exponent Lebesgue and Sobolev spaces are examined. Using these properties we prove the existence of the solution of some parabolic variational inequality. © 2010 Elsevier Inc. All rights reserved.

Keywords: Variable exponent Lebesgue and Sobolev spaces Nonlinear parabolic variational inequality

1. Introduction In the recent years, much attention has been paid to the study of mathematical models of electro-rheological fluids. These models include parabolic or elliptic equations which are nonlinear with respect to the gradient of the solution and variable exponents of nonlinearity. In that context, we refer to [1,20,34] and the references therein. We also refer to [26] for a multiplicity result for a nonlinear degenerate problem arising in the theory of electro-rheological fluids and to [27,28] for eigenvalue problems involving variable exponent growth conditions. Let Ω ⊂ Rn be a bounded domain of C 0,1 class, T ∈ (0, ∞), Q t1 ,t2 = Ω × (t 1 , t 2 ), 0  t 1 < t 2  T , Ωτ = {(x, t ): x ∈ Ω, t = τ }, τ ∈ [0, T ]. We consider the parabolic variational inequality

τ



 v t (t ) + A (t )u (t ) − f (t ), v (t ) − u (t ) V dt +

0



1

 | v − u |2 dx −

2 Ωτ

1

τ



 ϕ v (t ) dt −

0

 | v − u 0 |2 dx

2

τ





ϕ u (t ) dt 0

(1.1)

Ω0

where v : Q 0, T → R is a test function, A (t ) : V → V ∗ , f (t ) ∈ V ∗ , t ∈ (0, T ), u 0 ∈ V , ϕ : V → (−∞, ∞], ϕ ≡ ∞ and τ ∈ (0, T ]. Here V is a suitably defined subspace of variable exponent Sobolev space W 1, p (x) (Ω) and V ∗ is its conjugate space (see Section 3). In the case when p is constant, there have been many results about the different kinds of the parabolic variational inequalities. We refer the reader to the bibliography given in [6,7,15,23,24,30,32,33]. Parabolic variational inequalities with variable exponent of nonlinearity are studied in [9–11].

*

Corresponding author. E-mail addresses: [email protected] (R.A. Mashiyev), [email protected] (O.M. Buhrii).

0022-247X/$ – see front matter doi:10.1016/j.jmaa.2010.11.006

©

2010 Elsevier Inc. All rights reserved.

R.A. Mashiyev, O.M. Buhrii / J. Math. Anal. Appl. 377 (2011) 450–463

451

The initial–boundary value problems for the parabolic equations with nonstandard growth conditions are also considered in [2–5,8,12,35,37]. In [2] Alkhutov, Antontsev and Zhikov considered the initial–boundary value problem for equation





ut − div |∇ u | p (x,t )−2 ∇ u = 0. They proved the existence of weak solutions of this problem, assuming that exponent p (x, t ) is a log-continuous function. In [3] Antontsev and Shmarev considered the problem

ut (t ) + A (t )u + ϕ u (t ) = f (t ),

t ∈ (0, T ),

u (0) = u 0 ,

(1.2) (1.3)



where ϕ = 0, A (t ) = A 1 (t ),  A 1 (t )u , v V = Ω |u |γ (x,t ) (∇ u , ∇ v ) dx. They proved the existence and uniqueness of weak solutions and established sufficient conditions on the problem data and the exponent γ (x, t ) for the existence of such properties as finite speed of propagation of disturbances, the waiting time effect, finite time vanishing of the solution. In the case when A (t ) = A 1 (t ), ϕ u = 0, this problem is also considered in [4,35]. In the present paper, applying some minimal assumptions on the regularity of variable exponent p (x), we prove the existence of the solution of the variational inequality (1.1) for which the authors have also proved the uniqueness of solutions in [14]. The outline of this paper is the following: in Section 2, we describe the physical process whose mathematical modelling leads to a problem which is under discussion. In Section 3, we show some new properties of variable exponent Lebesgue and Sobolev spaces. In Section 4, using the results obtained in Section 3, the main existence result is stated and proved in Theorem 4.5 and we extend this result to the case when the assumptions on the data are less restrictive. 2. Mathematical models of electromagnetic fields

= (0, 0, u (x, t )), where x = (x1 , x2 ) ∈ Consider an electromagnetic field (see also [34]) with vector of magnetic density B

= ( H 1 , H 2 , H 3 ) be a magnetic field intensity, J = ( J 1 , J 2 , J 3 ) be a current density, E = ( H 1 , H 2 , H 3 ) be an Ω ⊂ R2 . Let H electrostatic field intensity and r be a resistivity. Consider Maxwell’s equations − →

∂B − → − → + rot E = 0 , ∂t

J ≈ rot H

,

= λH

, B

(2.1) (2.2) (2.3)

= r J , E

(2.4) u

where λ > 0. By using (2.3), we have H 3 = λ . Therefore,

∂ H3 ∂ H2 1 ∂u − = , ∂ x2 ∂ x3 λ ∂ x2 ∂ H1 ∂ H3 1 ∂u J2 ≈ − =− , ∂ x3 ∂ x1 λ ∂ x1 ∂ H2 ∂ H1 J3 ≈ − = 0. ∂ x1 ∂ x2 J1 ≈

(2.5)

Now suppose that r = r0 | J |α , where r0 > 0 is a constant and geneous environment. Taking into account (2.5), we get

| J | =



J 12 + J 22 + J 32 =

1

λ

α = α (x1 , x2 ) is a function which depends on a nonhomo-

|∇ u |, r

where ∇ u = ( ∂∂xu , ∂∂xu ). If a(x1 , x2 ) = α(x1 ,0x2 )+1 , then we have 1 2 λ

= r J = a|∇ u |α E



∂u ∂u ,− ,0 , ∂ x2 ∂ x1

(2.6)

is as a generalization of Ohm’s law (2.4). Hence the third coordinate of the vector rot E

(rot E )3 =





  ∂ E2 ∂ E1 ∂ ∂ ∂u ∂u − = a|∇ u |α −a|∇ u |α − = − div a|∇ u |α ∇ u . ∂ x1 ∂ x2 ∂ x1 ∂ x1 ∂ x2 ∂ x2

Using (2.1), we obtain





ut − div a(x)|∇ u |α (x) ∇ u = 0,

(x, t ) ∈ Q 0,T = Ω × (0, T ).

(2.7)

452

R.A. Mashiyev, O.M. Buhrii / J. Math. Anal. Appl. 377 (2011) 450–463

We also consider the initial–boundary value conditions

−

 ∂ u  = Φ(u ), ∂ νa ∂Ω×[0,T ]

(2.8)

u |t =0 = u 0 ,

(2.9)

is a unit normal vector of ∂Ω , u 0 : Ω → R, and Φ : R → R are the given functions. where ∂∂νu = a|∇ u |α (x) (∇ u , ν ), ν a Thus we have initial–boundary value problem (2.7)–(2.9). Now we give the definition of the generalized solutions of this problem (see [15, Ch. 1, Sec. 3.2–3.3]). Assuming that u is a smooth solution of problem (2.7)–(2.9), and v is a some smooth function (see for instance [17, p. 296]) and using (2.7) we get









ut ( v − u ) − div a|∇ u |α (x) ∇ u ( v − u ) dx dt = 0,

τ ∈ (0, T ).

(2.10)

Q 0,τ

Integrating by parts, we have



 ut ( v − u ) dx dt = Q 0,τ



 v t ( v − u ) dx dt −

Q 0,τ

  − div a|∇ u |α (x) ∇ u ( v − u ) dx =

Ω

If ψ(η) =



t =τ  | v − u |2 dx ,

 −

∂u ( v − u ) dS + ∂ νa

∂Ω

Φ(η) dη is a convex function and ϕ ( w ) =

ψ( v ) − ψ(u ) +

t =0

Ωt







a|∇ u |α (x) ∇ u , ∇ v − ∇ u dx.

Ω

 ∂Ω

ψ( w (x)) dS, then

∂u ( v − u ) = ψ( v ) − ψ(u ) − ψ (u )( v − u )  0. ∂ νa

Thus, taking into account (2.8), (2.9), and (2.10), we get





 v t ( v − u ) + a(x)|∇ u |α (x) ∇ u , ∇ v − ∇ u dx dt + 

Q 0,τ



1

 | v − u |2 dx −

2 Ωτ

1



2

τ



 ϕ v (t ) dt −

0

   v (0) − u 0 2 dx

τ





ϕ u (t ) dt 0

(2.11)

Ω0

for all τ ∈ (0, T ) and for all test functions v. So, solution to problem (2.7)–(2.9) satisfies the parabolic variational inequality (2.11). This inequality is a definition of generalized solution to (2.7)–(2.9) (see [15]). In this paper we investigate some inequality of type (2.11). 3. Some properties of the variable exponent Lebesgue and Sobolev spaces First, we introduce several concepts and symbols which will be used later. Let  ·  B be a norm of some Banach space B, B k be a Cartesian product of B, where k ∈ N, B ∗ be a conjugate space of B, ·,· B be a scalar product between B ∗ and B. def

If u : (0, T ) → B, then u (t ) = u (·, t ). Notation B 1 → B 2 means that the space B 1 is continuously embedded in B 2 and _ B 1 → B 2 means that the space B 1 is continuously and densely embedded in B 2 . The variable exponent Lebesgue space was firstly introduced in [31]. For fundamental properties of these spaces we refer to [11,13,14,18,21,25,29,36,38] and references therein. Let Ω ⊂ Rn be a bounded domain of C 0,1 class,





∞ L∞ + (Ω) = h ∈ L (Ω): ess inf h (x) > 1 . x∈Ω

For any h ∈ L ∞ + (Ω) we denote

h1 := ess inf h(x), x∈Ω

h2 := ess sup h(x), x∈Ω

and 1/h(x) + 1/h (x) = 1, x ∈ Ω . Suppose that h, p , q ∈ L ∞ + (Ω). We define a modular by setting



ρh ( v , Ω) = Ω

   v (x)h(x) dx.

R.A. Mashiyev, O.M. Buhrii / J. Math. Anal. Appl. 377 (2011) 450–463

453

The space L h(x) (Ω) is called the variable exponent Lebesgue space if for all v ∈ L h(x) (Ω), the following conditions hold: (i) v : Ω → R is a measurable function; (ii) ρh ( v , Ω) < ∞. L h(x) (Ω) is a reflexive Banach space (see [36]) with respect to the norm

   v Lh(x) (Ω) :=  v h(x),Ω = inf δ > 0: ρh ( v /δ, Ω)  1

and [ L h(x) (Ω)]∗ = L h (x) (Ω). L r (x) (Ω) → L q(x) (Ω) if r (x)  q(x) (see [21, pp. 599–600]). Remark 3.1. Let



S h (s) =

sh 1 , sh 2 ,



s ∈ [0, 1], s > 1,

S 1/h (s) =

s1/h2 , s1/h1 ,

s ∈ [0, 1], s > 1.

From [16, Lem. 2.1] we have 1)  v h(x),Ω  S 1/h (ρh ( v , Ω)) if ρh ( v , Ω) < ∞; 2) ρh ( v , Ω)  S h ( v h(x),Ω ) if  v h(x),Ω < ∞. The set of all functions v ∈ L p (x) (Ω) such that v x1 , . . . , v xn ∈ L p (x) (Ω) is called the variable exponent Sobolev space and is denoted by W 1, p (x) (Ω). W 1, p (x) (Ω) is a reflexive (see [21, p. 604]) Banach space with respect to the norm

 v W 1, p(x) (Ω) :=  v 1, p (x),Ω =

n 

 v xi  p (x),Ω +  v  p (x),Ω .

i =1 1, p (x)

Let W 0

(Ω) be a closure of C 0∞ (Ω) with respect to the norm of W 1, p (x) (Ω).

Remark 3.2. In [38, Sec. 3] Zhikov proved that there exists a function p ∈ L ∞ (Ω) = { v ∈ W 1, p (x) (Ω): + (Ω) such that W 0 v |∂Ω = 0}, where ∂Ω ⊂ C ∞ . This simple example shows that every elementary property of classical Sobolev spaces needs verification for variable exponent Sobolev spaces. 1, p (x)

1, p (x)

Let X be a closed subspace of W 1, p (x) (Ω) such that W 0 X=

1, p (x) W0 (Ω)

(Ω) → X → W 1, p (x) (Ω). For example, we may put either or X = W 1, p (x) (Ω). By definition, put V = X ∩ L 2 (Ω) ∩ L q(x) (Ω) and





U ( Q 0, T ) = u: (0, T ) → V |u ∈ L 2 ( Q 0, T ) ∩ L q(x) ( Q 0, T ), u x1 , . . . , u xn ∈ L p (x) ( Q 0, T ) . We define the norms on V and U ( Q 0, T ) by the formulas

 z V =

n 

zxi  p (x),Ω + z2,Ω + zq(x),Ω ,

z∈V

i =1

and

u U ( Q 0,T ) =

n 

u xi  p (x), Q 0,T + u 2, Q 0,T + u q(x), Q 0,T ,

u ∈ U ( Q 0, T )

i =1

respectively. _ _

Remark 3.3. It is easily seen that V → L 2 (Ω) → V ∗ = X ∗ + L q (x) (Ω) + L 2 (Ω) and



∗ _ s  _ _ U ( Q 0, T ) → L 2 ( Q 0, T ) → U ( Q 0, T ) → L s−1 0, T ; V ∗ ,

where s = max{ p 2 , 2, q2 } (see [10]). Therefore, U ( Q 0, T ) ⊂ D ∗ (0, T ; V ∗ ) and [U ( Q 0, T )]∗ ⊂ D ∗ (0, T ; V ∗ ), where D ∗ (0, T ; V ∗ ) is the space of distributions (see [19]). In [14] it is shown that if g ∈ [U ( Q 0, T )]∗ , then for all u ∈ U ( Q 0, T )

T  g , u U ( Q 0,T ) = 0





g (t ), u (t )

V

dt

454

R.A. Mashiyev, O.M. Buhrii / J. Math. Anal. Appl. 377 (2011) 450–463



and there exist elements g 0 ∈ L 2 ( Q 0, T ) + L q (x) ( Q 0, T ), g 1 , . . . , gn ∈ L p (x) ( Q 0, T ) such that

  n

 g , u U ( Q 0,T ) =



g i u xi + g 0 u dx dt ,

∀u ∈ U ( Q 0,T ).

i =1

Q 0, T

We can define the space U ( Q 0,τ ) similarly to how we defined U ( Q 0, T ), where operators A (t ) : V → V ∗ , t ∈ (0, T ), A τ : U ( Q 0,τ ) → [U ( Q 0,τ )]∗ , τ ∈ (0, T ) such that



A (t )u , v

 V

  n

=

Ωt



 p (x)−2

ai (x, t )u xi (x)

τ ∈ (0, T ). Consider a family of the

u xi (x) v xi (x) + c (x, t )u (x) v (x)

i =1

  q(x)−2 + g (x, t )u (x) u (x) v (x) dx, τ

 A τ z, y ≡  A τ z, y U ( Q 0,τ ) =





A (t ) z(t ), y (t )

V

u , v ∈ V , t ∈ (0, T ),

dt ,

z, y ∈ U ( Q 0, T ),

(3.1)

τ ∈ (0, T ).

(3.2)

0

We will need the following assumption: (A1) 1) ai ∈ L ∞ ( Q 0, T ), ai (x, t )  a0 > 0 for almost every (x, t ) ∈ Q 0, T , i = 1, . . . , n; 2) c ∈ L ∞ ( Q 0, T ), c (x, t )  c 0 for almost every (x, t ) ∈ Q 0, T , where c 0 ∈ R; 3) g ∈ L ∞ ( Q 0, T ), g (x, t )  g 0 > 0 for almost every (x, t ) ∈ Q 0, T . Theorem 3.4. If t , τ ∈ (0, T ) and the condition (A1) holds, then 1) A (t ) is a bounded operator; 2) the operator A (t ) satisfies the estimate



A (t )u − A (t ) v , u − v

 V

 |u − v |2 dx,

 c0

u , v ∈ V , t ∈ (0, T );

(3.3)

Ωt

3) if c 0 > 0, then the operators A (t ) and A τ are coercive and they satisfy the estimates



A (t )u , u

 V

 α0 u αV1 , α2

 A τ z, z  α0 zU ( Q 0,τ ) ,

u ∈ V,

(3.4)

z ∈ U ( Q 0,τ ),

(3.5)

1 where α0 = min{a0 ( n+ ) p 2 , c 0 ( n+1 2 )2 , g 0 ( n+1 2 )q2 }, α1 > 1, α2 > 1; 2 4) the operators A (t ) and A τ are semicontinuous, that is the functions

  R  μ → A (t )( v 1 + μ v 2 ), v 3 ∈ R,

  R  μ → A τ ( z1 + μ z2 ), z3 ∈ R

are continuous functions for every v 1 , v 2 , v 3 ∈ V and z1 , z2 , z3 ∈ U ( Q 0, T ) respectively; 5) the operator A τ is pseudomonotone, that is for every sequence { w m }m∈N ⊂ U ( Q 0,τ ) such that

wm → w

as m → ∞ weakly in U ( Q 0,τ ),

lim sup A τ w m , w m − w  0,

(3.6)

m→∞

we have

∀ v ∈ U ( Q 0,τ ):

lim inf A τ w m , w m − v   A τ w , w − v .

(3.7)

m→∞

Proof. 1) The proof is trivial. 2) Using inequality (|r1 |s−2 r1 − |r2 |s−2 r2 )(r1 − r2 )  0, where r1 , r2 ∈ R, s > 1, we obtain



A (t )u − A (t ) v , u − v

 V

=

  n 



ai |u xi | p (x)−2 u xi − | v xi | p (x)−2 v xi (u xi − v xi ) + c |u − v |2

Ωt

i =1

   q(x)−2  q(x)−2 + g |u | u − |v | v (u − v ) dx  c 0 |u − v |2 dx, Ωt

u , v ∈ V , t ∈ (0, T ).

R.A. Mashiyev, O.M. Buhrii / J. Math. Anal. Appl. 377 (2011) 450–463

3) Consider u ∈ V , t ∈ (0, T ). Let

η = u  V =

ηi = u xi  p(x),Ω , i = 1, . . . , n,

n

i =1

455

ηi + η0,2 + η0,q , where

η0,2 = u 2,Ω ,

η0,q = u q(x),Ω .

η

If  η = max{η1 , . . . , ηn , η0,2 , η0,q }, then  η  n+2 . Suppose that  η = η j , where j ∈ {1, . . . , n}. Using Remark 3.1, we get



A (t )u , u



 a0

V

n 



ρ p (u xi , Ω) + c0 ρ2 (u , Ω) + g0 ρq (u , Ω)  a0 ρ p (u x j , Ω)  a0 min η pj 1 , η pj 2

i =1



 a0 min



p1

p2  η , . n+2 n+2

η

η

In the same way we obtain  A (t )u , u V  c 0 ( n+2 )2 if  η = η0,2 and



A (t )u , u





 g 0 min

V

q1

 η q2 , , n+2 n+2

η

if  η = η0,q . Therefore, we have (3.4), where

⎧ p2 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ p1 , α1 = 2, ⎪ ⎪ ⎪ q2 , ⎪ ⎪ ⎪ ⎩q ,

 η ∈ {η1 , . . . , ηn }, n+η 2  1,

 η ∈ {η1 , . . . , ηn }, n+η 2 > 1,  η = η0,2 ,  η = η0,q , n+η 2  1,  η = η0,q , n+η 2 > 1.

1

In the same way we can prove (3.5). 4) We consider only the operator A (t ). Let t ∈ (0, T ), v 1 , v 2 , v 3 ∈ V ,

v

m

≡ v 1 + μm v 2 → v 1 + μ0 v 2 ≡ v

Let N 0 : L q(x) (Ω) → L

q (x)

0

μ0 , μm ∈ R, μm → μ0 as m → ∞ in R. Then

as m → ∞ strongly in V .

(Ω), N 1 , . . . , Nn : L p (x) (Ω) → L p (x) (Ω) be the Nemytsky operators such that

N 0 ( v ) = g (t )| v |q(x)−2 v , N i ( z) = ai (t )| z| p (x)−2 z,

v ∈ L q(x) (Ω), z ∈ L p (x) (Ω).

In [21, Thm. 4.2], Kováˇcik and Rákosník proved that N 0 , N 1 , . . . , N n are continuous and bounded operators. Therefore, using generalized Hölder inequality (see Theorem 2.1 [21, p. 594]), we get

     A (t ) v m , v 3 − A (t ) v 0 , v 3     n   m  0    0   m   m 0  N i v xi − N i v xi v 3,xi + c (t ) v − v v 3 + N 0 v − N 0 v v 3 dx Ω

 C1



i =1 n    m   N i v − N i v 0  xi

xi

i =1

p (x),Ω

  + vm − v 0

2,Ω

     + N0 vm − N0 v 0 

q (x),Ω

  v 3 V

→ 0 as m → ∞.

5) Every bounded semicontinuous monotone operator is the pseudomonotone (see [24, Ch. 2, Sec. 2.4, Prop. 2.5]). Therefore statement (3.7) follows from (3.6). This completes the proof of Theorem 3.4. 2 Now we introduce the following concept. Consider any Banach space B, w : (0, T ) → B, u 0 ∈ B. A set of functions { w μ }μ∈N is called a smoothing sequence (see [24, Ch. 3, Sec. 6.3], [32, p. 59], [14]) for the function w if

1

μ



w μ (t )

t

+ w μ (t ) = w (t ),

t ∈ (0, T ),

w μ (0) = u 0 ,

(3.8)

for all μ ∈ N. The equalities (3.8) are the equalities in the space B. If w ∈ L 1 (0, T ; B ), then there exists the solution w μ ∈ C ([0, T ]; B ) of the problem (3.8) such that ( w μ )t ∈ L 1 (0, T ; B ). It is easy to see that

w μ (t ) = u 0 e

−μt

t +μ 0

e μ(s−t ) w (s) ds,

t ∈ (0, T ).

(3.9)

456

R.A. Mashiyev, O.M. Buhrii / J. Math. Anal. Appl. 377 (2011) 450–463

Notice that if w ∈ U ( Q 0, T ), u 0 ∈ L 2 (Ω), then the authors proved that (see [14]) w μ → w as U ( Q 0, T ) and strongly in L 2 ( Q 0, T ). Assume that ϕ is a properly convex function that is (F1) 1) ϕ : V → (−∞, ∞], ϕ ≡ ∞; 2) for every β ∈ [0, 1], u , v ∈ V we have

μ → ∞ weakly in the

ϕ (β u + (1 − β) v )  β ϕ (u ) + (1 − β)ϕ ( v ).

Lemma 3.5. If w μ is defined as in (3.9) and if ϕ satisfies the condition (F1), then 1) (see for comparison [32, p. 59], [6, p. 154])





ϕ w μ (t )  ϕ (u 0 )e−μt + μ

t





e μ(s−t ) ϕ w (s) ds,

t ∈ (0, T );

(3.10)

0

τ 2)



 1 ϕ w μ (t ) dt  ϕ (u 0 ) +

τ

μ

0





ϕ w (s) ds, τ ∈ (0, T ).

(3.11)

0

Proof. Suppose the conditions of Lemma 3.5 are satisfied. 1 1 1) Notice that s = t + μ ln r iff r = e μ(s−t ) . Substituting t + μ ln r for s in (3.10), we get

w μ (t ) = e

−μt



u0 + 1 − e

−μt





1

1

(1 − e −μt )

w t+ e −μt

1

μ

ln r dr ,

t ∈ (0, T ).

Since β = e −μt ∈ (0, 1], then by definition of a convex function and Jensen’s inequality, we have



   ϕ w μ (t )  e−μt ϕ (u 0 ) + 1 − e−μt ϕ    e −μt ϕ (u 0 ) + 1 − e −μt

(1 − e −μt )

(1 − e −μt )

e −μt

w t+ e −μt



1

1

ϕ w t+



1

1



1

= e −μt ϕ (u 0 ) +



ϕ w t+ e −μt

1

μ 1

μ

ln r dr



ln r



1

μ

ln r

dr ,

t ∈ (0, T ).

Taking r = e μ(s−t ) gives (3.10). 2) Taking into account (3.10), we obtain

τ



 ϕ w μ (t ) dt  ϕ (u 0 )

0

τ e

−μt

τ dt +

0

t dt μ

0





e μ(s−t ) ϕ w (s) ds

0

t =τ τ τ   −e −μt  = ϕ (u 0 ) + ds μ e μ(s−t ) ϕ w (s) dt μ  t =0

=

1

μ

s

0

  ϕ (u 0 ) 1 − e−μτ +

τ

  τ  μ (s−t ) ϕ w (s) μ e dt ds. 

s

0

Since

τ

μ

t =τ

e μ(s−t ) dt = −e μ(s−t ) t =s = 1 − e μ(s−τ )  1,

s

we have (3.11). This completes the proof of Lemma 3.5.

2





dr

R.A. Mashiyev, O.M. Buhrii / J. Math. Anal. Appl. 377 (2011) 450–463

457

4. Existence of solution of the variational inequality Assume the all conditions of Sections 1 and 3 are satisfied, and Ωτ = {(x, t ) ∈ Rn+1 | x ∈ Ω, t = τ }, consider the operator A (t ) : V → V ∗ , t ∈ (0, T ), such that (3.1) holds.

τ ∈ [0, T ]. Moreover,

Definition 4.1. The function u ∈ U ( Q 0, T ) ∩ C ([0, T ]; L 2 (Ω)) is called a solution of the parabolic variational inequality (1.1) if u satisfies (1.1) for all v ∈ U ( Q 0, T ) ∩ C ([0; T ]; L 2 (Ω)), v t ∈ [U ( Q 0, T )]∗ and for all τ ∈ (0, T ]. It is easy to prove that if u is a solution of (1.1), then

u (x, 0) = u 0 (x)

for almost every x ∈ Ω.

(4.1)

We will need the following assumption (F2) 1) if either um → u as m → ∞ weakly in U ( Q 0, T ) or um → u as m → ∞ strongly in L 2 ( Q 0, T ), then

τ



 ϕ um (t ) dt →

0

2)

τ





ϕ u (t ) dt as m → ∞, ∀τ ∈ (0, T ], 0

and it is uniformly convergent for any ϕ (0) = 0, ϕ ( v )  0 for all v ∈ V .

τ ∈ (0, T ].

By definition, put K ϕ = { v ∈ V | ϕ ( v ) < ∞}. In [14] it is shown that K ϕ is a convex set, K ϕ = ∅ and u (t ) ∈ K ϕ for almost every t ∈ (0, T ), where u is a solution of (1.1). Theorem 4.2. (See [14].) Suppose (A1), (F1) and (F2) hold. If u 0 ∈ K ϕ and f ∈ L 2 ( Q 0, T ), then the solution of the parabolic variational inequality (1.1) is unique. Remark 4.3. (See Theorem 2 [13, p. 51].) If u ∈ U ( Q 0, T ) ∩ L ∞ (0, T ; L 2 (Ω)), ut ∈ [U ( Q 0, T )]∗ , then for a.e. t 1 , t 2 ∈ (0, T ), t 1 < t 2 , we get

t2





ut (t ), u (t )

V



1

dt =

|u |2 dx −

2

t1

1

 |u |2 dx.

2

Ωt2

Ωt1

Lemma 4.4. If v , w ∈ U ( Q 0, T ) ∩ C ([0, T ]; L 2 (Ω)), v t , w t ∈ [U ( Q 0, T )]∗ , then for all t 1 , t 2 ∈ [0, T ], t 1 < t 2 , we obtain

t2



 v t (t ), w (t ) V dt =

t1



 v w dx −

Ωt2

Ωt1

t2 v w dx −





w t (t ), v (t )

V

dt .

(4.2)

t1

Proof. Since v , w ∈ C ([0, T ]; L 2 (Ω)), then we have formulas of Remark 4.3 for all t 1 , t 2 ∈ [0, T ], t 1 < t 2 . Taking u = v − w, we obtain (4.2) easily. This completes the proof of Lemma 4.4. 2 Consider the following conditions: (A2) a1 , . . . , an , c , g ∈ C ([0, T ]; L ∞ (Ω)); (F3) ϕ is differentiable at u ∈ V and the following conditions hold: 1) ϕ : V → V ∗ is a bounded semicontinuous monotone operator, 2) Ker ϕ ≡ { w ∈ V | ϕ ( w ) = 0} is a convex closed non-empty set, 0 ∈ Ker ϕ . For the sake of simplicity we can replace ϕ ( z) = ϕ z, where z ∈ V and ϕ ( z(·)) = ϕ z(·), where z : (0, T ) → V . Theorem 4.5. If the conditions (A1)–(A2) and (F1)–(F3) hold, u 0 ∈ V , f ∈ L 2 ( Q 0, T ), then the inequality (1.1) has solution u. Proof. First we consider the initial–boundary value problem (1.2), (1.3). The equalities (1.2) are the equalities in space of distributions D ∗ (0, T ; V ∗ ). The existence and uniqueness of solution u ∈ U ( Q 0, T ) ∩ C ([0, T ]; L 2 (Ω)), ut ∈ [U ( Q 0, T )]∗ , of the problem (1.2), (1.3) were proved by the second author of the present paper and Lavrenyuk in [12]. Since (F1) and (F3) hold, we have





ϕ ( z1 ) − ϕ ( z2 )  ϕ z2 , z1 − z2 V , ∀ z1 , z2 ∈ V .

(4.3)

458

R.A. Mashiyev, O.M. Buhrii / J. Math. Anal. Appl. 377 (2011) 450–463

Let u be a solution of (1.2), (1.3). Using (4.3), we get

τ



 ut (t ) + A (t )u (t ) − f (t ), v (t ) − u (t ) V dt +

0

 ϕ v (t ) dt − 

0

τ =

τ





ϕ v (t ) dt −

τ





ϕ u (t ) dt −

0

for all

τ

0



τ





ϕ u (t ) dt 0



ϕ u (t ), v (t ) − u (t ) V dt  0

(4.4)

0

τ ∈ (0, T ], v ∈ U ( Q 0,T ) ∩ C ([0, T ]; L 2 (Ω)). Let v t ∈ [U ( Q 0,T )]∗ . Since (see Remark 4.3) τ



τ



ut (t ), v (t ) − u (t )

V

dt =

0



τ



v t (t ), v (t ) − u (t )

V

dt −

0

τ =





v t (t ) − ut (t ), v (t ) − u (t )

V

dt

0





v t (t ), v (t ) − u (t )

V

dt −

1

 | v − u |2 dx +

2

0

1

 | v − u 0 |2 dx

2

Ωτ

Ω0

we obtain (1.1) from (4.4). This completes the proof of Theorem 4.5.

2

Remark 4.6. It is easy to prove that any solution of (1.1) satisfies (1.2) and (1.3) (see [15]). In the following we shall consider a weaker form of the condition (F3): (F4) the function ϕ satisfies conditions (F1), (F2) and there exists a sequence of functions ϕ1 , ϕ2 , . . . such that the following conditions are executed: 1) for every m ∈ N the function ϕm satisfies conditions (F1), (F2), (F3); 2) there exist a continuous function γ : R+ → R+ and the sequence {sm }m∈N ⊂ R+ such that sm → 0 as m → ∞ and for every τ ∈ (0, T ], m ∈ N and w ∈ U ( Q 0, T )

τ  τ           ϕm w (t ) dt − ϕ w (t ) dt   sm γ  w U ( Q 0,τ )   0

0

holds; 3) if w m → w weakly in U ( Q 0, T ) as m → ∞, then for every

τ





ϕm w m (t ) dt 

lim inf m→∞

0

Suppose that

τ



τ





ϕ w (t ) dt . 0

ϕ satisfies condition (F4) with the sequence {ϕm }m∈N . For every m ∈ N consider the problem



τ



v t (t ) + A (t )um (t ) − f (t ), v (t ) − um (t )

0



1 2

τ ∈ (0, T ] we get

   v − um 2 dx − 1

2

| v − u 0 | dx,





τ





ϕm um (t ) dt

ϕm v (t ) dt − 0



2

Ωτ

V

dt +

0

(4.5)

Ω0

where v ∈ U ( Q 0, T ) ∩ C ([0; T ]; L (Ω)), v t ∈ [U ( Q 0, T )]∗ , and τ ∈ (0, T ]. Theorem 4.5 implies the existence of the solutions u of the parabolic variational inequality (4.5) such that um ∈ U ( Q 0, T ) ∩ C ([0, T ]; L 2 (Ω)). If m = 1, 2, 3, . . . then we obtain the sequence u 1 , u 2 , u 3 , . . . of the solutions to (4.5). 2

Lemma 4.7 (Uniformly estimating of um ). Suppose the conditions (A1), (A2), (F1), (F2), and (F4) are satisfied. If u 0 ∈ V , f ∈ L 2 ( Q 0, T ), then the sequence {um }m∈N of the solutions to (4.5) is bounded in the space U ( Q 0, T ) ∩ L ∞ (0, T ; L 2 (Ω)) and there exists a constant C 0 > 0 such that

τ





ϕm um (t ) dt  C 0 , τ ∈ (0, T ], m ∈ N. 0

(4.6)

R.A. Mashiyev, O.M. Buhrii / J. Math. Anal. Appl. 377 (2011) 450–463

459

Proof. Take a number m ∈ N. The inequality (4.5) implies that

 





v t v − um +

n    p (x)−2 m     m  a i u m u xi v xi − u m v − um xi xi + cu i =1

Q 0,τ

 τ τ  m q(x)−2 m         m m   +g u u v −u − f v −u dx dt + ϕm v (t ) dt − ϕm um (t ) dt





1 2

0



   v − um 2 dx − 1

| v − u 0 | dx,

2

Ωτ

0

2

(4.7)

Ω0

where v ∈ U ( Q 0, T ) ∩ C ([0; T ]; L (Ω)), v t ∈ [U ( Q 0, T )]∗ and

τ ∈ (0, T ]. Taking v = 0 gives us     τ τ n  m  p (x)  m 2  m q(x)  m  1  m 2 u  dx +       a i u xi +c u +g u dx dt + ϕm u (t ) dt − ϕm (0) dt 2

2

Ωτ



1

i =1

Q 0,τ



|u 0 |2 dx +

2

0



0

f um dx dt .

(4.8)

Q 0,τ

Ω

It follows from (F4) 1), (F2) 2), and (F3) 2) that

ϕm (0) = 0.

ϕm (0) = 0,

(4.9)

Therefore inequality f um  12 | f |2 + 12 |um |2 and (4.8) imply that

1



2

 m 2 u  dx +

Ωτ





τ n   m  p (x)  m q(x)   1  m 2       a0 u xi + c0 − u + g0 u dx dt + ϕm um (t ) dt 0



1

|u 0 |2 dx +

2

2

i =1

Q 0,τ



1

 

| f |2 dx dt  C 1 ,

2

(4.10)

Q 0,τ

Ω

τ . Using (4.9) and (4.3), we obtain   ϕm (0) + ϕm (0), um − 0 V = 0.

where C 1 > 0 is independent of m,



ϕm u

m









By definition, put y (τ ) = Ω |um (x, τ )|2 dx, τ



y (τ )  2C 1 + 2|c 0 | + 1



τ ∈ [0, T ]. Therefore,

y (t ) dt 0

for all τ ∈ [0, T ]. Using Gronwall’s inequality (see [17]), we obtain y (τ )  C 2 for all we have

 sup

τ ∈[0, T ]

τ ∈ [0, T ]. Combining this with (4.10),

n   m        u (x, τ )2 dx + ρ p um , Q 0,T + ρ2 um , Q 0,T + ρq um , Q 0,T +

T

i =1

Ω





ϕm um (t ) dt  C 3 ,

xi

(4.11)

0

where C 3 > 0 is independent of m. This completes the proof of Lemma 4.7.

2

Theorem 4.8. Assume that the conditions (A1), (A2), (F1), (F2) 1) and (F4) are satisfied. If u 0 ∈ V and f ∈ L 2 ( Q 0, T ), then the inequality (1.1) has a solution u. Proof. Let {um }m∈N be a sequence from Lemma 4.7. By (4.11), we have the existence of the subsequence {um j } j ∈N ⊂ {um }m∈N such that

um j → u





as j → ∞ ∗-weakly in L ∞ 0, τ ; L 2 (Ω) and weakly in U ( Q 0,τ ),

A τ u m j → χτ



∗

as j → ∞ weakly in U ( Q 0,τ ) ,

τ ∈ (0, T ].

460

R.A. Mashiyev, O.M. Buhrii / J. Math. Anal. Appl. 377 (2011) 450–463

Take k, j ∈ N. Assume that umk and um j are the solutions to inequality (1.1) with ϕmk , ϕm j respectively. Substituting u = umk , v = um j and u = um j , v = umk in (4.4), we obtain two inequalities. Summing these inequalities, we get

τ



 m  + ut j (t ), umk (t ) − um j (t ) V dt



m

ut k (t ), um j (t ) − umk (t )

V

0

τ +





A (t )umk (t ), um j (t ) − umk (t )

V

  + A (t )um j (t ), umk (t ) − um j (t ) V dt

0

τ +



0

where G k, j (τ ) =



1 2



f (t ), um j (t ) − umk (t )

τ 0

V

  + f (t ), umk (t ) − um j (t ) V dt + G k, j (τ )  0,

(4.12)

[ϕmk (um j (t )) − ϕmk (umk (t )) + ϕm j (umk (t )) − ϕm j (um j (t ))] dt, k, j ∈ N, τ ∈ (0, T ]. Hence,

 m  u k − um j 2 dx +

τ





A (t )umk (t ) − A (t )um j (t ), umk (t ) − um j (t )

V

dt  G k, j (τ ).

(4.13)

0

Ωτ

Since

τ











ϕmk um j (t ) − ϕmk umk (t ) dt

0

τ



































ϕmk um j (t ) − ϕ um j (t ) + ϕ um j (t ) − ϕ u (t ) + ϕ u (t ) − ϕ umk (t ) + ϕ umk (t ) − ϕmk umk (t ) dt

= 0

   smk γ um j U ( Q

 0,τ )

 τ  τ            m      mj k +  ϕ u (t ) − ϕ u (t ) dt  +  ϕ u (t ) − ϕ u (t ) dt  + smk γ umk U ( Q ) , 0,τ     0

0

we have (see Remark 3.1, estimates (4.11) and (F2) 1))

τ lim



k, j →∞



ϕmk u

mj

τ     m        mj k (t ) − ϕmk u (t ) dt  2 lim smk γ (C 4 ) + lim  ϕ u (t ) − ϕ u (t ) dt   j →∞ k→∞

0

0

τ     m     + lim  ϕ u (t ) − ϕ u k (t ) dt  = 0  k→∞ 0

uniformly for any

τ lim

k, j →∞



τ ∈ [0, T ]. Similarly, 







ϕm j umk (t ) − ϕm j um j (t ) dt  0.

(4.14)

0

Thus G k, j (τ ) → 0 as k, j → ∞ uniformly for any Using (4.13), (3.3), we obtain

τ ∈ [0, T ].

y (τ )  2|c 0 | y (τ ) + 2G k, j (τ ),  where y (τ ) = Q |umk − um j |2 dx dt, τ ∈ (0, T ). From Lemma 1.1 [22, p. 152] we get 0,τ



y (τ )  e

2τ |c 0 |

τ y (0) + 2



G k, j (s) ds = 2e 0

2τ |c 0 |

τ G k, j (s) ds,

τ ∈ (0, T ).

0

This inequality means that um j → u as j → ∞ strongly in L 2 ( Q 0, T ). Taking into account (4.15) we have

 Ωτ

 m  u k − um j 2 dx = y (τ )  4|c 0 |e 2τ |c0 |

τ G k, j (s) ds + 2G k, j (τ ) → 0 0

as k, j → ∞

(4.15)

R.A. Mashiyev, O.M. Buhrii / J. Math. Anal. Appl. 377 (2011) 450–463

τ ∈ [0, T ]. Therefore,

uniformly for any

u

461

  → u as j → ∞ strongly in C [0, T ]; L 2 (Ω) .

mj

(4.16)

Let us get the estimate (3.7) for um j . Suppose that v ∈ U ( Q 0, T ) ∩ C ([0; T ]; L 2 (Ω)), v t ∈ [U ( Q 0, T )]∗ and ing (4.5) with m = m j , we have



1 2

     v − um j 2 dx + A τ um j , um j − u

Ωτ



τ ∈ (0, T ). Us-



1

  | v − u 0 |2 dx + A τ um j , um j − u

2 Ω0

τ +



τ



v t (t ) + A (t )um j (t ) − f (t ), v (t ) − um j (t )

V

dt +

0

=

τ



0



1





| v − u 0 |2 dx + A τ um j , v − u

2





ϕm j um j (t ) dt

ϕm j v (t ) dt − 0



Ω0

τ +



v t (t ) − f (t ), v (t ) − u

mj

 (t ) V dt +

0

Hence

τ

ϕm j



 v (t ) dt −

0



τ





ϕm j um j (t ) dt . 0



lim sup A τ um j , um j − u  H (τ , v ),

(4.17)

j →∞

where

H (τ , v ) =

1



τ 2

| v − u 0 | dx +

2



 v t (t ) + χτ (t ) − f (t ), v (t ) − u (t ) V dt +

0

Ω0

τ



 ϕ v (t ) dt −

0

τ





ϕ u (t ) dt . 0

Consider the smoothing sequence (see Section 3) for the function u:

1

μ where



u μ (t )

t

+ u μ (t ) = u (t ),

t ∈ (0, T ), u μ (0) = u 0 ,

μ ∈ N. Using (3.11), we get τ H (τ , u μ ) =



 (u μ )t (t ) + χτ (t ) − f (t ), u μ (t ) − u (t ) V dt +

0

τ 

τ



τ



ϕ u μ (t ) dt − 0





χτ (t ) − f (t ), u μ (t ) − u (t ) V dt +

1

μ





ϕ u (t ) dt 0

ϕ (u 0 ) → 0 as μ → ∞.

0

Therefore, lim sup j →∞  A τ um j , um j − u  0 and Theorem 3.4 imply that





lim inf A τ um j , um j − v   A τ u , u − v .

(4.18)

j →∞

Using (4.5) with m = m j , we have

τ 0



v t (t ) − f (t ), v (t ) − u

mj

 (t ) V dt +

τ

ϕm j 0



 v (t ) dt 

τ







ϕm j um j (t ) dt + A τ um j , um j − v 0

+

1



2

   v − um j 2 dx − 1

 | v − u 0 |2 dx.

2

Ωτ



Ω0

462

R.A. Mashiyev, O.M. Buhrii / J. Math. Anal. Appl. 377 (2011) 450–463

By (F4) 3), (4.18), and (4.16) we have

τ



τ



v t (t ) − f (t ), v (t ) − u (t )

0

 τ = lim

j →∞

V

dt +





ϕ v (t ) dt 0



v t (t ) − f (t ), v (t ) − u

mj

 (t ) V dt +

0

τ

ϕm j



 v (t ) dt

0

τ









ϕm j um j (t ) dt + lim inf A τ um j , um j − v + lim

 lim inf j →∞

j →∞

j →∞

0

τ 





1



2

   v − um j 2 dx − 1



ϕ u (t ) dt +  A τ u , u − v +

1 2

0

This completes the proof of Theorem 4.8.

| v − u |2 dx − Ωτ

1

| v − u 0 |2 dx

2

Ωτ



 Ω0

 | v − u 0 |2 dx.

2 Ω0

2

Acknowledgments The authors would like to thank the referees for their many helpful suggestions and corrections, which improve the presentation of this paper. The authors also wishes to express his sincerely thanks to Professor M.M. Bokalo for several helpful suggestions on this paper. This research project was supported by DUBAP -10-FF-15, Dicle University, Turkey.

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