Weak solution of non-Newtonian polytropic variational inequality in fresh agricultural product supply chain problem

Tao Wu

doi:10.1515/math-2022-0590

Artikel Open Access

Weak solution of non-Newtonian polytropic variational inequality in fresh agricultural product supply chain problem

Tao Wu

Veröffentlicht/Copyright: 26. Juni 2023

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Abstract

In this article, we study a class of variational inequality problems with non-Newtonian polytropic parabolic operators. We introduce a mapping with an adjustable parameter to control the polytropic term, which exactly meets the conditions of Leray-Schauder fixed point theory. At the same time, we construct a penalty function to transform the variational inequality into a regular parabolic initial boundary value problem. Thus, the existence is treated with a Leray-Schauder fixed point theory as well as a suitable version of Aubin-Lions lemma. Then, the uniqueness and stability of the solution are analyzed.

Keywords: variational inequality problem; generalized solution; non-Newtonian polytropic operator; existence; uniqueness and stability

MSC 2010: 35K99; 97M30

1 Introduction

Let p ≥ 2 , m ∈ ( 0 , p − 1 ] and assume that Ω ⊂ R N ( N ≥ 2 ) be a bounded domain and Ω T = Ω × ( 0 , T ] with T > 0 . The author of this study considered a kind of variational inequality problem

(1) min { L u , u − u 0 } = 0 , ( x , t ) ∈ Ω T ,

with the non-Newtonian polytropic operator

(2) L u = ∂ t u − ∇ ( ∣ ∇ u m ∣ p − 2 ∇ u m ) + γ u , γ ≥ 0

and the Dirichlet initial-boundary value condition

(3) u ( 0 , x ) = u 0 ( x ) , x ∈ Ω ; u ( t , x ) = 0 , ( x , t ) ∈ ∂ Ω × ( 0 , T ) .

Variational inequality has a good application in the value analysis of financial products with early implementation clauses, for details see [1,2]. Recent years, much attention has been paid to the study of variational inequality with linear, quasi-linear, and degenerate parabolic operator [3–5]. In [3], Li and Bi considered two-dimensional variational inequality systems

min { L i u i − f i ( x , t , u 1 , u 2 ) , u i − u i , 0 } = 0 , ( x , t ) ∈ Ω T , u ( 0 , x ) = u 0 ( x ) , x ∈ Ω , u ( t , x ) = 0 , ( x , t ) ∈ ∂ Ω × ( 0 , T )

with degenerate parabolic operator

L i u i = ∂ t u i − div ( ∣ ∇ u i ∣ p i − 2 ∇ u i ) , i = 1 , 2 .

The existence of weak solution is studied by a limit process. In [4], Sun and Wu considered a kind of variational inequality problem with a double degenerate operator

L u = ∂ t u − u div ( a ( u ) ∣ ∇ u ∣ p ( x ) − 2 ∇ u ) − γ u p ( x ) − f ( x , t ) .

Similar methods to those presented in [3] have been used, and the existence and uniqueness of the solutions in the weak sense are proved.

The structure of parabolic initial boundary value problem is simpler than that of variational inequality. When m = 0 , initial boundary value problems with parabolic operator L u have been extensively studied in the last few years, see for details [6–13]. Some articles are focused on the existence of generalized solutions related to this article [6–8]. Some results for uniqueness of generalized solutions can be found in [9–11]. There are other arguments worth studying, such as stability of boundary output feedback [12,13].

In this article, we extend the corresponding results in [3,6,9] to study a class of variational inequality problems with non-Newtonian polytropic parabolic operators with the Dirichlet initial-boundary value condition. Since m and p are coupled in L u and L u is degenerate, we plan to solve this problem with Leray-Schauder fixed point theory by constructing a map. In order to overcome the difficulty of establishing generalized solutions on variational inequalities, we turn the variational inequalities into regular problems through penalty functions. Some estimates of regular problems and continuity, boundedness and compactness of Leray-Schauder map are given by the inequality technique as well as a suitable version of Aubin-Lions lemma. In what follows, we prove the existence, uniqueness, and stability of the solution under the proper setting of the parameters in (1).

2 Statement of the problem and the main results

Our consideration in this article is motivated by an application model about fresh agricultural product supply chain. Here we consider a fresh agricultural product supply chain formed by a supplier and a retailer in which retailers face uncertain market demand. Assuming that the current time is 0, the time agreed in the contract for retailers to purchase agricultural products is T , and the retail price of agricultural products agreed in the contract meets

d P ( t ) = μ P ( t ) d t + σ P ( t ) d W t , P ( 0 ) = P 0 ,

where μ and σ denote the expected rate and the volatility of return on the retail price of agricultural products, respectively. { W t , t ≥ 0 } is a Winner process, which drives the random noise of the market. P 0 represents the market price at time 0.

Since fresh agricultural products are easy to deteriorate and decay, retailers will have no residual value for their remaining agricultural products. Therefore, the order of agricultural products must be placed before the sales season T . Retailers can buy a call option contract in which they have the right to purchase a certain amount of fresh agricultural products at the agreed price of K from 0 to T . Of course, retailers need to pay a certain premium of C to obtain such rights. If the retailer finds a more suitable source of goods, it will give up the option contract and lose the option premium C . This means that retailers can decide whether to exercise or hold options based on their own earnings. According to the literature [1–3], the value of options meets

(4) min { LC , C − max ( P − K , 0 ) } = 0 , ( P , t ) ∈ [ 0 , B ] × [ 0 , T ] , C ( 0 , P ) = max ( P − K , 0 ) , P ∈ [ 0 , B ] , C ( t , B ) = P − exp { − r t } , C ( t , 0 ) = 0

where B represents the artificially set price ceiling for agricultural products,

LC = ∂ C ∂ t + 1 2 σ 2 P 2 ∂ 2 C ∂ P 2 + r P ∂ C ∂ P − r C .

In addition, if the loss of agricultural products during transportation is considered, then the expression of LC is more complicated that σ in LC is a function of C and ∂ C ∂ P , for details see [14].

More complex case than (5) is considered in this article. In doing so, we give a class of maximal monotone maps defined in [11,13]

(5) G ( x ) = 0 , x > 0 , − M 0 , x = 0 ,

where M 0 is a positive constant which can be chosen later. Since m and p are coupled in L u and L u is degenerate, problem (1) do not have a classical solution. So we consider its generalized case as follows.

Definition 1

A pair ( u , ξ ) is said to be a generalized solution of variational inequality (1), if ( u , ξ ) satisfies u ∈ L ∞ ( 0 , T , W 1 , p ( Ω ) ) , ∂ t u ∈ L 2 ( Ω T ) ,

u ( x , t ) ≥ u 0 ( x ) , u ( x , 0 ) = u 0 ( x ) for any ( x , t ) ∈ Ω T ,
ξ ∈ G ( u − u 0 ) for any ( x , t ) ∈ Ω T ,
for every test function φ ∈ C ∞ ( Ω T ) , there admits the equality
∫ ∫ Ω T ∂ t u ⋅ φ + ∣ ∇ u m ∣ p − 2 ∇ u m ∇ φ d x d t + γ ∫ ∫ Ω T u φ d x d t = ∫ ∫ Ω T ξ ⋅ φ d x d t .

Because of coupling in L u , we cannot prove the existence of solution of problem (1) by using a common limit method. Here, we plan to use the Leray-Schauder fixed point theory as well as a suitable version of Aubin-Lions lemma based on the map

(6) M : L ∞ ( 0 , T ; W 0 1 , p ( Ω ) ) × [ 0 , 1 ] → L ∞ ( 0 , T ; W 0 1 , p ( Ω ) ) ,

such that for every function ω ∈ L ∞ ( 0 , T ; W 0 1 , p ( Ω ) ) and θ ∈ [ 0 , 1 ] , u ε = M ( ω , θ ) is a solution of the equation

(7) L ε θ , ω u ε = − θ β ε ( u ε − u 0 ) , ( x , t ) ∈ Ω T , u ε ( x , 0 ) = u 0 ε ( x ) = u 0 + ε , x ∈ Ω , u ε ( x , t ) = ε , ( x , t ) ∈ ∂ Ω T ,

with an operator

(8) L ε θ , ω u ε = ∂ t u ε − div ( ∣ ∇ A θ ( u ε ) ∣ 2 + ε ) p − 2 2 ∇ A θ ( u ε ) + γ ω .

Here u ε m ∈ L ∞ ( 0 , T ; W 0 1 , p ( Ω ) ) , A θ ( u ε ) = θ u ε m + ( 1 − θ ) u ε , the penalty map β ε : R + → R − satisfies

(9) ε ∈ ( 0 , 1 ) , β ε ( ⋅ ) ∈ C 2 ( R ) , β ε ( x ) ≤ 0 , β ε ′ ( x ) ≥ 0 , β ε ″ ( x ) ≤ 0 , β ε ( x ) = 0 x ≥ ε , − M 0 x = 0 , lim ε → 0 + β ( x ) = 0 , x > 0 , − M 0 , x = 0 .

Then the existence of problem (1) is equivalent to u ε = M ( u ε , 1 ) . With a similar method to that in [6,7], we may prove that the regularized problem (7) admits a weak solution as follows:

Definition 2

A function u ε is said to be a solution of problem (7) if u ε satisfies u ε m ∈ L ( 0 , T ; W 1 , p ( Ω ) ) ,

(10) ∫ Ω ∂ t u ε ⋅ φ + ( ∣ ∇ A θ ( u ε ) ∣ 2 + ε ) p − 2 2 ∇ A θ ( u ε ) ∇ φ d x + γ ∫ Ω ω φ d x = ∫ Ω β ε ( u ε − u 0 ) ⋅ φ d x , ∀ t ∈ ( 0 , T )

for any φ ∈ C ∞ ( Ω T ) .

Because of the denseness of C ∞ ( Ω T ) in L ( 0 , T ; W 1 , p ( Ω ) ) , one can assert that the identities in Definition 2.1 and Definition 2.2 hold for any L ( 0 , T ; W 1 , p ( Ω ) ) . Furthermore, one can obtain the following inequalities [6,7]:

(11) u 0 ≤ u ε ≤ ∣ u 0 ∣ ∞ + ε , u ε 1 ≤ u ε 2 for ε 1 ≤ ε 2 .

It is clear that the constructed generalized solution in this article is non-negative. Then we give the restriction in (6) that ω ≥ 0 .

3 Some preliminaries

To discuss the existence of weak solution of equation (1), we give certain useful estimates.

Lemma 1

For any fixed t ∈ [ 0 , T ] and γ ≥ 0 ,

(12) ‖ u ε ‖ L p ( Ω ) p ≤ C ,

(13) ∫ Ω ( ∣ θ ∇ u ε m + ( 1 − θ ) ∇ u ε ∣ 2 + ε ) p − 2 2 ∣ θ ∇ u ε m + ( 1 − θ ) ∇ u ε ∣ 2 d x ≤ C ,

(14) ∫ Ω ∣ θ ∇ u ε m + ( 1 − θ ) ∇ u ε ∣ p d x ≤ C .

Proof

Since u ε m ∈ L ( 0 , T ; W 1 , p ( Ω ) ) , combining with (11) gives θ u ε m + ( 1 − θ ) u ε ∈ L ( 0 , T ; W 1 , p ( Ω ) ) . So choosing a test function φ as θ u ε m + ( 1 − θ ) u ε in (10), it can be easily verified that

(15) ∫ ∫ Ω T ∂ τ u ε ⋅ ( θ u ε m + ( 1 − θ ) u ε ) d x d t + Π 1 = − θ ∫ ∫ Ω β ε ( u ε − u 0 ) ( θ u ε m + ( 1 − θ ) u ε ) d x d t − γ ∫ ∫ Ω ω ( θ u ε m + ( 1 − θ ) u ε ) d x d t ,

where

Π 1 = ∫ ∫ Ω ( ∣ θ ∇ u ε m + ( 1 − θ ) ∇ u ε ∣ 2 + ε ) p − 2 2 ∣ θ ∇ u ε m + ( 1 − θ ) ∇ u ε ∣ 2 d x d t .

From the first part of (11), u ε − u 0 ≥ 0 for any ( x , t ) ∈ Ω T . Combining with (9) leads to ∣ β ε ( u ε − u 0 ) ∣ ≤ M 0 , such that ∫ ∫ Ω β ε ( u ε − u 0 ) ( θ u ε m + ( 1 − θ ) u ε ) d x d t can be estimated as

(16) − θ ∫ ∫ Ω β ε ( u ε − u 0 ) ( θ u ε m + ( 1 − θ ) u ε ) d x d t ≤ M 0 T ( max { ∣ u 0 , ε ∣ ∞ , ∣ u 0 , ε m ∣ ∞ } + 1 ) .

Since ω ≥ 0 , θ ∈ ( 0 , 1 ) , it is easy from the first part of (12) to see that

(17) ∫ ∫ Ω ω ⋅ ( θ u ε m + ( 1 − θ ) u ε ) d x d t ≥ 0 .

Using differential transformation method gives (recall that m ∈ ( 0 , p − 1 ] )

(18) ∫ ∫ Ω ∂ τ u ε ⋅ ( θ u ε m + ( 1 − θ ) u ε ) d x d t = ∫ ∫ Ω θ m + 1 ∂ τ u ε m + 1 + 1 2 ( 1 − θ ) ∂ τ u ε 2 d x d t = θ m + 1 ( ‖ u ε ‖ L m + 1 ( Ω ) m + 1 − ‖ u 0 , ε ‖ L m + 1 ( Ω ) m + 1 ) + 1 2 ( 1 − θ ) ( ‖ u ε ‖ L 2 ( Ω ) 2 − ‖ u 0 , ε ‖ L 2 ( Ω ) 2 ) .

Since A 1 ≥ 0 , we drop Π 1 to arrive at

(19) θ m + 1 ‖ u ε ‖ L p ( Ω ) p + 1 2 ( 1 − θ ) ‖ u ε ‖ L 2 ( Ω ) 2 ≤ θ m + 1 ‖ u 0 , ε ‖ L m + 1 ( Ω ) m + 1 + 1 2 ( 1 − θ ) ‖ u 0 , ε ‖ L 2 ( Ω ) 2 .

This implies (12) follows.

On the contrary, if we drop non-negative term ∫ ∫ Ω ω ( θ u ε m + ( 1 − θ ) u ε ) d x d t and non-positive term − θ ∫ ∫ Ω β ε ( u ε − u 0 ) ( θ u ε m + ( 1 − θ ) u ε ) d x d t in (14),

(20) Π 1 ≤ 2 m + 1 ‖ u 0 , ε ‖ L p ( Ω ) p + ‖ u 0 , ε ‖ L 2 ( Ω ) 2 .

Thus (13) follows. If ε = 0 ,

∫ ∫ Ω ∣ θ ∇ u ε m + ( 1 − θ ) ∇ u ε ∣ p d x d t ≤ ∫ ∫ Ω ( ∣ θ ∇ u ε m + ( 1 − θ ) ∇ u ε ∣ 2 + ε ) p − 2 2 ∣ θ ∇ u ε m + ( 1 − θ ) ∇ u ε ∣ 2 d x d t ,

such that estimate (14) is an immediate result of (13).□

Lemma 2

Assume that γ ≥ 0 , m > 0 , and p ≥ 2 . Then there exists a constant C, independent of ε , such that

(21) ∫ Ω ∣ ∂ t ( θ u ε m + ( 1 − θ ) u ε ) ∣ 2 d x ≤ C ( p , M 0 , ∣ Ω ∣ , ∣ u 0 , ε ∣ ∞ ) ,

(22) ∥ ∂ t u ε m ∥ L 2 ( 0 , T ; Ω ) ≤ C ( p , T , ∣ Ω ∣ ) ,

(23) ∥ ∂ t u ε ∥ L 2 ( 0 , T ; Ω ) ≤ C ( p , T , ∣ Ω ∣ ) .

Proof

Since θ ∈ ( 0 , 1 ) . It follows from the first part of (11) that

(24) ∂ t u ε ∂ t u ε m ≤ m ( ∣ u 0 ∣ ∞ m − 1 + 1 ) ∣ ∂ t u ε ∣ 2 .

Multiplying the first line of (7) by ∂ t ( θ u ε m + ( 1 − θ ) u ε ) and integrating both sides of the equality over Ω T , we have

(25) ρ ∫ ∫ Ω ∣ ∂ t ( θ u ε m + ( 1 − θ ) u ε ) ∣ 2 d x d t ≤ − Π 2 − Π 3 + Π 4 ,

where the constant π depends only upon m and ∣ u 0 ∣ ∞ ,

Π 2 = ∫ ∫ Ω ( ∣ θ ∇ u ε m + ( 1 − θ ) ∇ u ε ∣ 2 + ε ) p − 2 2 ( θ ∇ u ε m + ( 1 − θ ) ∇ u ε ) ∂ t ( θ ∇ u ε m + ( 1 − θ ) ∇ u ε ) d x d t , Π 3 = γ ∫ ∫ Ω ω ⋅ ∂ t ( θ u ε m + ( 1 − θ ) u ε ) d x d t , Π 4 = − θ ∫ ∫ Ω β ε ( u ε − u 0 ) ⋅ ∂ t ( θ u ε m + ( 1 − θ ) u ε ) d x d t .

Using some differential transformation techniques,

(26) − Π 2 = − 1 p ∫ 0 t ∫ Ω ∂ t ( ∣ θ ∇ u ε m + ( 1 − θ ) ∇ u ε ∣ 2 + ε ) p 2 d x d τ ≤ 1 p ∫ Ω ( ∣ θ ∇ u ε m ( , ⋅ 0 ) + ( 1 − θ ) ∇ u ε ( , ⋅ 0 ) ∣ 2 + ε ) p 2 d x .

It follows by Holder and Cauchy inequalities that

(27) ∣ Π 3 ∣ ≤ 2 π − 1 γ 2 ∫ Ω ω 2 d x + 1 8 π ∫ Ω ∣ ∂ t ( θ u ε m + ( 1 − θ ) u ε ) ∣ 2 d x ,

(28) ∣ Π 4 ∣ = 2 π − 1 M 0 2 ∣ Ω ∣ + 1 8 π ∫ Ω ∣ ∂ t ( θ u ε m + ( 1 − θ ) u ε ) ∣ 2 d x .

Since ω ∈ L ∞ ( 0 , T ; W 0 1 , p ( Ω ) ) , combining (25), (26), (27), and (28), it is easy to verify that

(29) 3 4 ρ ∫ Ω ∣ ∂ t ( θ u ε m + ( 1 − θ ) u ε ) ∣ 2 d x ≤ C ( p , M 0 , ∣ Ω ∣ , ∣ u 0 , ε ∣ ∞ ) .

This implies that (21) follows. Using (24) again,

(30) θ 2 ∫ Ω ∣ ∂ t u ε m ∣ 2 d x + ρ ∫ Ω ∣ ∂ t u ε ∣ 2 d x ≤ ∫ Ω ∣ ∂ t ( θ u ε m + ( 1 − θ ) u ε ) ∣ 2 d x .

(22) and (23) are immediate results of (29) and (30).□

Next, we analyze the continuity of map M . To this end, assume that θ k → θ and f k → f as k → ∞ and define u ε , k = M ( θ k , f k ) , such that u ε , k is the solution of (9) with degenerate parabolic operator

(31) L ε θ k , ω k u ε , k = ∂ t u ε , k − div ( ∣ ∇ A θ k ( u ε , k ) ∣ 2 + ε ) p − 2 2 ∇ A θ k ( u ε , k ) + γ ω k .

Lemma 3

The map M : L ∞ ( 0 , T ; W 0 1 , p ( Ω ) ) × [ 0 , 1 ] → L ∞ ( 0 , T ; W 0 1 , p ( Ω ) ) is continuous for any θ ∈ ( 0 , 1 ) .

Proof

From the first part of (11), it can be seen that { u ε , k , k = 1 , 2 , 3 , … } is bounded, which together with the uniform estimates in k allow one to extract from the sequence { u ε , k , k = 1 , 2 , 3 , … } a subsequence (for the sake of simplicity, we assume that it merely coincides with the whole of the sequence) and a function u ε such that

(32) u ε , k → u ε a.e . Ω T as k → ∞ .

So that we remain to prove that u ε = M ( θ , f ) . From (12) and (13), one can infer that for any fix ε ∈ ( 0 , 1 ) ,

(33) ( ∣ ∇ A θ k ( u ε , k ) ∣ 2 + ε ) p − 2 2 ∇ A θ k ( u ε , k ) w → χ ε in W 0 1 , p p − 1 ( Ω ) as k → ∞ ,

such that combining (10) and (33) gives

(34) ∫ Ω ∂ t u ε ⋅ φ + χ ε ∇ φ d x + γ ∫ Ω ω φ d x = − ∫ Ω β ε ( u ε − u 0 ) ⋅ φ d x .

Hence, it remains to prove that

(35) ∫ Ω χ ε ∇ φ d x = ∫ Ω ξ ⋅ ( ∣ ∇ A θ ( u ε ) ∣ 2 + ε ) p − 2 2 ∇ A θ ( u ε ) d x .

Choosing φ = u ε , k − u ε , multiplying u ε , k = M ( θ k , f k ) or L ε θ k , ω k u ε , k = β ε ( u ε − u 0 ) by φ , and integrating over Ω , we have that

(36) ∫ Ω ∂ t u ε , k ⋅ φ + ( ∣ ∇ A θ , k ( u ε , k ) ∣ 2 + ε ) p − 2 2 ∇ A θ , k ( u ε , k ) ∇ φ d x + γ ∫ Ω ω k φ d x = − ∫ Ω β ε ( u ε , k − u 0 ) ⋅ φ d x .

Combining (34) and (36), and integrating over [ 0 , T ] , one can have that

(37) ∫ ∫ Ω T ( ∂ t u k , ε − ∂ t u k , ε ) ⋅ φ + ( ∣ ∇ A θ , k ( u ε ) ∣ 2 + ε ) p − 2 2 ∇ A θ , k ( u ε ) − χ ε ∇ φ d x d t + γ ∫ ∫ Ω T ( ω k − ω ) φ d x d t = − ∫ 0 t ∫ Ω [ β ε ( u k , ε ) − β ε ( u ε ) ] ⋅ φ d x d t .

From (32), we infer that

(38) lim k → ∞ ∫ ∫ Ω T ( ω k − ω ) φ d x d t = lim k → ∞ ∫ 0 t ∫ Ω [ β ε ( u k , ε ) − β ε ( u ε ) ] ⋅ φ d x d t = 0 .

Since u k , ε ( x , 0 ) = u ε ( x , 0 ) for any x ∈ Ω ,

∫ ∫ Ω T ( ∂ t u k , ε − ∂ t u ε ) ⋅ φ d x d t = 1 2 ∫ Ω ( u k , ε − u ε ) 2 d x ≥ 0 ,

so we drop the non-negative term on the left-hand side and pass the limit k → ∞ to arrive at

(39) lim k → ∞ ∫ ∫ Ω T ( ∣ ∇ A θ , k ( u ε ) ∣ 2 + ε ) p − 2 2 ∇ A θ , k ( u ε ) − χ ε ∇ φ d x d t ≤ 0 ,

( ∣ ∇ A θ k ( u ε ) ∣ 2 + ε ) p − 2 2 ∇ A θ k ( u ε ) − ( ∣ ∇ A θ ( u ε ) ∣ 2 + ε ) p − 2 2 ∇ A θ ( u ε ) ( ∇ A θ k ( u ε ) − ∇ A θ ( u ε ) ) ≥ 0 .

Since ∇ A θ k ( u ε ) − ∇ A θ ( u ε ) and ∇ φ have the same sign,

(40) ( ∣ ∇ A θ k ( u ε ) ∣ 2 + ε ) p − 2 2 ∇ A θ k ( u ε ) − ( ∣ ∇ A θ ( u ε ) ∣ 2 + ε ) p − 2 2 ∇ A θ ( u ε ) ∇ φ ≥ 0 .

Subtracting (39) and (40), we infer that

(41) lim k → ∞ ∫ ∫ Ω T ( ∣ ∇ A θ ( u ε ) ∣ 2 + ε ) p − 2 2 ∇ A θ ( u ε ) − χ ε ∇ φ d x d t ≤ 0 .

Obviously, if we swap u ε , k and u ε , it is easy to obtain another inequality

(42) lim k → ∞ ∫ ∫ Ω T χ ε − ( ∣ ∇ A θ ( u ε ) ∣ 2 + ε ) p − 2 2 ∇ A θ ( u ε ) ∇ φ d x d t ≤ 0 .

(41) and (42) imply that Lemma 3.3 follows.□

By virtue of Lemmas 3.1 and 3.2, and the Ascoli-Arzela lemma [7], M maps any bounded set [ 0 , 1 ] × Q into a compact set of Q ⊂ L ∞ ( 0 , T ; W 0 1 , p ( Ω ) ) . So we have the following result.

Lemma 4

The map M : L ∞ ( 0 , T ; W 0 1 , p ( Ω ) ) × [ 0 , 1 ] → L ∞ ( 0 , T ; W 0 1 , p ( Ω ) ) is compact.

Let y ∈ L ∞ ( 0 , T ; W 0 1 , p ( Ω ) ) . By classical results (see, e.g., [8]), problem u ε = M ( 0 , f ) + y , or

(43) ∂ t u ε − ∇ ( ∣ ∇ u ε ∣ 2 + ε ) p − 2 2 ∇ u ε + γ u ε + y = 0

with the same initial boundary condition in (7), admits a unique solution in L ∞ ( 0 , T ; W 0 1 , p ( Ω ) ) . Combining Lemmas 3.1–3.4 and the results related to (43), the following initial boundary problem

(44) L u ε = − β ε ( u ε − u 0 ) , ( x , t ) ∈ Q T , u ε ( x , 0 ) = u 0 ε ( x ) , x ∈ Ω , u ε ( x , t ) = ε , ( x , t ) ∈ ∂ Q T ,

by using the properties of the Leray-Schauder degree, has a solution u ε ∈ L ∞ ( 0 , T ; W 1 , p ( Ω ) ) that satisfies ∂ t u ε ∈ L ∞ ( 0 , T ; L 2 ( Ω ) ) ,

(45) ∫ Ω ( ∂ t u ε ⋅ φ + ( ∣ ∇ u ε m ∣ 2 + ε ) p − 2 2 ∇ u ε m ∇ φ + γ u ε φ ) d x = − ∫ Ω β ε ( u ε − u 0 ) φ d x

with φ ∈ C 1 ( Ω ¯ T ) . And, the solution of (44) satisfies (11), that is,

(46) u 0 ε ≤ u ε ≤ ∣ u 0 ∣ ∞ + ε , u ε 1 ≤ u ε 2 for ε 1 ≤ ε 2 .

Based on Lemmas 3.1 and 3.2, we conclude that

(47) ‖ u ε ‖ L p ( Ω ) p ≤ C ,

(48) ∫ Ω ( ∣ ∇ u ε m ∣ 2 + ε ) p − 2 2 ∣ ∇ u ε m ∣ 2 d x ≤ C ,

(49) ‖ ∇ u ε m ‖ L p ( Ω ) p ≤ C ,

(50) ∥ ∂ t u ε m ∥ L 2 ( 0 , T ; Ω ) ≤ C ( p , T , ∣ Ω ∣ ) ,

(51) ∥ ∂ t u ε ∥ L 2 ( 0 , T ; Ω ) ≤ C ( p , T , ∣ Ω ∣ ) .

Indeed, choosing u ε m as a test function in (45) and repeating the proof of Lemma 3.1, (47)–(49) follow. As for (50) and (51), it is a consequence of taking ∂ t u ε m as a test function.

4 Existence

In this section, we consider the existence of generalized solution to (1). Using what we mentioned in (46)–(51), the sequence { u ε , ε ≥ 0 } contains a subsequence denoted again by itself weakly convergent to a function u ∈ L ∞ ( 0 , T ; W 0 1 , p ( Ω ) ) , that is,

(52) u ε → u a.e. weakly in Ω T as ε → 0 ,

(53) u ε m → ∗ u m weakly ∗ in L ∞ ( 0 , T ; W 0 1 , p ( Ω ) ) as ε → 0 ,

(54) ∂ t u ε → ∂ t u weakly in L 2 ( Ω T ) as ε → 0 .

From (46), one can infer that u ε ≤ u , ∀ ( x , t ) ≤ Ω T .

Lemma 5

Assume that γ ≥ 0 , m > 0 , and p ≥ 2 . Then for any fixed t ∈ ( 0 , T ) ,

∫ Ω ∣ u ε m − u m ∣ p d x → 0 a s ε → 0 .

Proof

Taking a test function ( u ε m − ε − u m ) in (45), we infer that

(55) ∫ Ω ∂ t u ε ⋅ ( u ε m − ε − u m ) d x + ∫ Ω ( ∣ ∇ u ε m ∣ 2 + ε ) p − 2 2 ∇ u ε m ∇ ( u ε m − u m ) d x = − γ ∫ Ω u ε ⋅ ( u ε m − ε − u m ) d x − ∫ Ω β ε ( u ε − u 0 ) ⋅ ( u ε m − ε − u m ) d x .

From (9) and (46), u ε m − ε − u m ≤ 0 , such that − ∫ Ω β ε ( u ε − u 0 ) ⋅ ( u ε m − ε − u m ) d x is non-positive and we drop it to arrive at

(56) ∫ Ω ∂ t u ε ⋅ ( u ε − ε − u ) d x + γ ∫ Ω u ε ⋅ ( u ε m − ε − u m ) d x ≤ − ∫ Ω ( ∣ ∇ u ε m ∣ 2 + ε ) p − 2 2 ∇ u ε m ∇ ( u ε m − u m ) d x .

This by adding ∫ Ω ( ∣ ∇ u ε m ∣ 2 + ε ) p − 2 2 ∇ u ε m − ∣ ∇ u m ∣ p − 2 ∇ u m ∇ ( u ε m − u m ) d x both hand sides of (50), leads to

(57) ∫ Ω ∂ t u ε ⋅ ( u ε m − ε − u m ) d x + γ ∫ Ω u ε ⋅ ( u ε m − ε − u m ) d x + ∫ Ω ( ∣ ∇ u ε m ∣ 2 + ε ) p − 2 2 ∇ u ε m − ∣ ∇ u m ∣ p − 2 ∇ u m ∇ ( u ε m − u m ) d x ≤ − ∫ Ω ∣ ∇ u m ∣ p − 2 ∇ u m ∇ ( u ε m − u m ) d x .

Applying Holder’s inequality and joining with Lemmas 3.1 and 3.2, (52), and (53), one can pass to the limit ε → 0 to arrive at

(58) ∫ Ω ∂ t u ε ⋅ ( u ε m − ε − u m ) d x ≤ C ∫ Ω ∣ ∂ t u ε ∣ 2 d x ⋅ ∫ Ω ( u ε m − ε − u m ) 2 d x → 0 ,

(59) ∫ Ω u ε ⋅ ( u ε m − ε − u m ) d x → 0 ,

(60) ∫ Ω ∣ ∇ u m ∣ p − 2 ∇ u m ∇ ( u ε m − u m ) d x → 0 .

Combining (57), (58), (59), and (60) and passing to the limit, one can infer that

(61) lim ε → 0 ∫ Ω ( ∣ ∇ u ε m ∣ 2 + ε ) p − 2 2 ∇ u ε m − ∣ ∇ u m ∣ p − 2 ∇ u m ∇ ( u ε m − u m ) d x ≤ 0 .

Since 0 ≤ ε ≤ 1 , it follows from [6] that

(62) ( ∣ ∇ u ε m ∣ 2 + ε ) p − 2 2 ∇ u ε m − ∣ ∇ u m ∣ p − 2 ∇ u m ∇ ( u ε m − u m ) ≥ ( ∣ ∇ u ε m ∣ p − 2 ∇ u ε m − ∣ ∇ u m ∣ p − 2 ∇ u m ) ∇ ( u ε m − u m ) ≥ C ( p ) ∣ ∇ u ε m − ∇ u m ∣ p ≥ 0 .

Combining (61) and (62), the proof of Lemma 4.1 is completed.□

Using Trigonometric inequality,

∫ Ω ∣ ∣ ∇ u ε m ∣ p − 2 ∇ u ε m − ∣ ∇ u m ∣ p − 2 ∇ u m ∣ d x ≤ ∫ Ω ∣ ∣ ∇ u ε m ∣ p − 2 − ∣ ∇ u m ∣ p − 2 ∣ ⋅ ∣ ∇ u ε m ∣ d x + ∫ Ω ∣ ∇ u m ∣ p − 2 ∣ ∇ u ε m − ∇ u m ∣ d x .

If p ∈ [ 2 , 3 ] , applying ∣ a r − b r ∣ ≤ ∣ a − b ∣ r , a > 0 , b > 0 , r ∈ [ 0 , 1 ] , Holder’s inequality as well as Lemma 4.1,

∫ Ω ∣ ∣ ∇ u ε m ∣ p − 2 − ∣ ∇ u m ∣ p − 2 ∣ ∣ ∇ u ε m ∣ d x ≤ ∫ Ω ∣ ∇ u ε m − ∇ u m ∣ p − 2 ∣ ∇ u ε m ∣ d x → 0 .

If p > 3 , one from Lemma 4.1 can obtain

∫ Ω ∣ ∣ ∇ u ε m ∣ p − 2 − ∣ ∇ u m ∣ p − 2 ∣ ∣ ∇ u ∣ d x ≤ ∫ Ω ( ∣ ∇ u ε m ∣ + ∣ ∇ u m ∣ ) p − 2 ∣ ∇ u ε m − ∇ u m ∣ d x → 0 .

Besides that using Lemma 4.1 also gives

∫ Ω ∣ ∇ u ε m ∣ p − 2 ∣ ∇ u ε m − ∇ u m ∣ d x → 0 .

Combining all the above, we have the following result.

Lemma 6

As ε → 0 , ∣ ∇ u ε m ∣ p − 2 ∇ u ε m converges to ∣ ∇ u m ∣ p − 2 ∇ u m with norm L 1 ( Ω ) .

Furthermore, we focus on the limit of β ε ( u ε − u 0 ) . Since β ε ( ⋅ ) ∈ C 2 ( R ) in (9), (46), and (47) , for all ( x , t ) ∈ Ω T , give

β ε ( u ε − u 0 ) → ξ as ε → 0 .

We consider ξ in two cases: u ε ≥ u 0 + ε and 0 < u ε < u 0 + ε . If u ε ≥ u 0 + ε , β ε ( u ε − u 0 ) = 0 , such that

(63) ξ ( x , t ) = 0 ⇔ u > u 0 .

If u 0 ≤ u ε < u 0 + ε , β ε ( 0 ) = − M 0 , and β ε ( ⋅ ) ∈ C 2 ( R ) imply that

(64) ξ ( x , t ) = − M 0 ⇔ u = u 0 .

Combining (63) and (64), it can be easily verified that

(65) − β ε ( u ε − u 0 ) → ξ ∈ G ( u − u 0 ) as ε → 0

for all ( x , t ) ∈ Ω T . Furthermore, passing the limit ε → 0 in the second line of (44) and the second part of (46),

(66) u ( x , 0 ) = u 0 ( x ) in Ω , u ( x , t ) ≥ u 0 ( x ) in Ω T .

Combining Lemmas 4.1, 4.2, (65), and (66), we infer that ( u , ξ ) satisfies the conditions of Definition 2.1, such that ( u , ξ ) is a generalized solution of (1).

Theorem 1

Assume that u 0 ∈ W 1 , p ( Ω ) , γ ≥ 0 . Then (1) admits a solution within the class of Definition 2.1.

5 Stability and uniqueness

In this section, we study the stability and uniqueness of generalized solution for all γ ≥ 0 , m > 0 , and p ≥ 2 . Consider generalized solutions ( u i , ξ i ) of (1) with initial conditions

(67) u ( 0 , x ) = u 0 , i ( x ) , x ∈ Ω , i = 1 , 2 .

Since u 1 , u 2 ∈ L ∞ ( 0 , T ; W 0 1 , p ( Ω ) ) , u 1 − u 2 ∈ L ∞ ( 0 , T ; W 0 1 , p ( Ω ) ) . So choosing φ = u 1 − u 2 in Definition 2.1, one can infer that

(68) ∫ ∫ Ω ∂ t φ ⋅ φ + ( ∣ ∇ u 1 m ∣ p − 2 ∇ u 1 m − ∣ ∇ u 2 m ∣ p − 2 ∇ u 2 m ) ∇ φ d x d t + γ ∫ ∫ Ω φ 2 d x d t = ∫ ∫ Ω ( ξ 1 − ξ 2 ) ⋅ φ d x d t .

As what we mentioned in (62),

( ∣ ∇ u 1 m ∣ p − 2 ∇ u 1 m − ∣ ∇ u 2 m ∣ p − 2 ∇ u 2 m ) ( ∇ u 1 m − ∇ u 2 m ) ≥ 0 .

f ( u ) = u and g ( u ) = u m are both increasing functions with u , sgn ( ∇ u 1 m − ∇ u 2 m ) = sgn ( ∇ u 1 − ∇ u 2 ) , so

(69) ( ∣ ∇ u 1 m ∣ p − 2 ∇ u 1 m − ∣ ∇ u 2 m ∣ p − 2 ∇ u 2 m ) ∇ φ ≥ 0 .

As for ∫ ∫ Ω ( ξ 1 − ξ 2 ) ⋅ φ d x d t , we first give the estimate

(70) ∫ ∫ Ω ( ξ 1 − ξ 2 ) ⋅ φ d x d t ≤ 0 , ∀ t ∈ [ 0 , T ] .

Indeed, if u 1 ( x , t ) > u 2 ( x , t ) , then u 1 ( x , t ) > u 1 , 0 ( x ) . (5) and (65) imply ξ 1 = 0 ≤ ξ 2 , such that

(71) ( ξ 1 − ξ 2 ) ⋅ φ = ( ξ 1 − ξ 2 ) ⋅ ( u 1 − u 2 ) ≤ 0 .

On the contrary, if u 1 ( x , t ) < u 2 ( x , t ) , we swap u 1 ( x , t ) and u 2 ( x , t ) , it is obvious to obtain

(72) ( ξ 1 − ξ 2 ) ⋅ φ = ( ξ 2 − ξ 1 ) ⋅ ( u 2 − u 1 ) ≤ 0 .

Furthermore, we remove the non-negative term (69) and γ ∫ ∫ Ω φ 2 d x d t and also drop the non-positive term (70) in (68) to arrive at

(73) ∫ ∫ Ω ∂ t φ ⋅ φ d x d t ≤ 0 ⇔ ∫ Ω ∣ u 1 − u 2 ∣ 2 d x ≤ ∫ Ω ∣ u 1 , 0 − u 2 , 0 ∣ 2 d x .

Uniqueness of solution can be found easily if u 1 , 0 = u 2 , 0 . Thus, we state our result as follows.

Theorem 2

Let ( u i , ξ i ) be a generalized solution of (1) with

u ( 0 , x ) = u 0 , i ( x ) , x ∈ Ω , i = 1 , 2 .

Then

‖ u 1 − u 2 ‖ L ∞ ( 0 , T ; L 2 ( Ω ) ) ≤ ‖ u 0 , 1 ( ⋅ ) − u 0 , 2 ( ⋅ ) ‖ L 2 ( Ω ) .

Moreover, problem (1) has a unique solution in the sense of Definition 2.1 with γ ≥ 0 , m > 0 , and p ≥ 2 .

6 Conclusion

In this work, the existence, uniqueness, and stability of generalized solution to a kind of variational inequality problem with non-Newtonian polytropic parabolic operators

L u = ∂ t u − ∇ ( ∣ ∇ u m ∣ p − 2 ∇ u m ) + γ u , γ ≥ 0 .

Since m and p are coupled in L u and L u is degenerate, we construct a map (6) based on the Leray-Schauder lemma’s conditions and a penalty map function used to deal with inequality constraint that regular problem (7) is established. Some estimates of problem (7) and continuity, boundedness and compactness of map (6) are proved a Leray-Schauder fixed point theory as well as a suitable version of Aubin-Lions lemma. The existence, uniqueness, and stability of the solution are analyzed.

In this article, we limit that γ ≥ 0 . In the proof of Lemma 3.1, formula (17) holds only when γ is greater than 0. The Holder inequality is used in Lemma 3.2, where p needs to satisfy p ≥ 2 . In addition, if parameter p depends on x or t , (26) in Lemma 3.2 does not work, Lemma 3.2 is difficult to be proved in this case. We will study those focuses in the future.

wutao@gznc.edu.cn

Acknowledgement

The author sincerely thanks the editors and anonymous reviewers for their insightful comments and constructive suggestions, which greatly improved the quality of the article.

Funding information: This work was supported by the National Social Science Fund of China in 2020 (No. 20BMZ043) and the Doctoral Project of Guizhou Education University (No. 2021BS037).
Author contributions: This is a single author article. The author read and approved the final manuscript.
Conflict of interest: The author states no conflict of interest.
Data availability statement: Not applicable.

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Received: 2023-01-15

Revised: 2023-03-30

Accepted: 2023-05-11

Published Online: 2023-06-26

This work is licensed under the Creative Commons Attribution 4.0 International License.

Artikel in diesem Heft

https://doi.org/10.1515/math-2022-0590

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variational inequality problem; generalized solution; non-Newtonian polytropic operator; existence; uniqueness and stability

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