Existence of nonnegative nontrivial periodic solutions to a doubly degenerate parabolic equation with variable exponent

  • Zhongqing Li1Email author and

    Affiliated with

    • Wenjie Gao1

      Affiliated with

      Boundary Value Problems20142014:77

      DOI: 10.1186/1687-2770-2014-77

      Received: 19 October 2013

      Accepted: 20 March 2014

      Published: 2 April 2014

      Abstract

      The authors investigate a degenerate parabolic equation with delay and nonlocal term, which describes slow diffusive processes in physics or biology. The existence of a nonnegative nontrivial periodic solution is obtained through the use of the Leray-Schauder degree method.

      MSC:35D05, 35K55.

      Keywords

      degenerate parabolic equation periodic solution variable exponent topological degree De Giorgi iteration

      1 Introduction

      In this paper, we are interested in the following evolutional p ( x ) -Laplacian equation:
      { u t div ( | u m | p ( x ) 2 u m ) = [ a ( x , t ) Ω K ( ξ , t ) u 2 ( ξ , t τ ) d ξ ] u , ( x , t ) Q T , u ( x , t ) = 0 , ( x , t ) Γ T , u ( x , 0 ) = u ( x , T ) , x Ω .
      (1.1)

      Here, Ω is a bounded simply connected domain with smooth boundary Ω in R N , Q T = Ω × ( 0 , T ) , Γ T = Ω × ( 0 , T ) , T > 0 , and τ ( 0 , + ) . We assume m > 1 , p C 1 , α ( Ω ¯ ) , with p + : = max Ω ¯ p ( x ) , p : = min Ω ¯ p ( x ) , p + p > 2 and that a L ( Q T ) and K L ( Q T ) can be extended as T-periodic functions to Ω × R . Furthermore, we assume that K 0 for a.e. ( x , t ) Q T .

      Equation (1.1) is a doubly degenerate parabolic equation with delay and nonlocal term, which models diffusive periodic phenomena in physics and mathematical biology. In biology, it arises from population model, where u ( x , t ) denotes the density of population at time t located at x Ω , a is the natural growth rate of the population, the nonlocal term Ω K ( ξ , t ) u 2 ( ξ , t τ ) d ξ evaluates a weighted fraction of individual, and the delayed density u at time t τ appearing in the nonlocal term represents the time needed to an individual to become adult. In physics, problem (1.1) is proposed based on some evolution phenomena in electrorheological fluids [1]. It describes the ability of a conductor to undergo significant changes when an electric field is imposed on. This model has been employed for some technological applications, such as medical rehabilitation equipment and shock wave absorber.

      When p ( x ) is a constant and m > 1 , p > 2 , the model describes the slow diffusion process in physics, which has been extensively investigated; see [27]. For example, in [5], the authors studied the following doubly degenerate parabolic equation with logistic periodic sources:
      u t div ( | u m | p 2 u m ) = u α ( a b u β ) .

      They proved the existence of a nontrivial nonnegative periodic solution via monotonicity method. Using a Moser iterative method (see [811]), they also obtained some a priori bounds and asymptotic behaviors for the solutions.

      Recently, the variable exponent Sobolev space and its applications have attracted considerable interest; see [1, 1214] and the references therein. When p + p > 2 and m > 1 , the doubly degenerate parabolic equation (1.1) is a more realistic model which describes the rather slow diffusion process. In our models, the principal term div ( | u m | p ( x ) 2 u m ) , in place of the usual term u m or div ( | u m | p 2 u m ) , represents nonhomogeneous diffusion that depends on the position x Ω and thus gives a better description of nonhomogeneous character of the process.

      There are many differences between Sobolev spaces with constant exponent and those with variable exponent; many powerful tools applicable in constant exponent spaces are not available for variable exponent spaces. For instance, the variable exponent spaces are no longer translation invariant and Young’s inequality f g p ( ) C f p ( ) g 1 holds if and only if p is constant (see monograph [12]). As we all know, the frequently used Hölder’s inequality, Poincaré’s inequality, etc., will be presented in new forms for variable exponent spaces.

      The presence of the nonlocal term and p ( x ) -Laplacian term makes the sup-solution and sub-solution method (as in [5]) in vain. In our paper, we adopt the topological degree method (as in [810]) to show the existence of nontrivial periodic solutions to problem (1.1). However, the method employed in the variable exponent case [13] or in the constant exponent case [811] cannot be directly used to derive the uniform upper bound for solutions, which is a crucial step in applying the topological degree method. We apply a modified De Giorgi iteration to establish the crucial uniform bound. We believe that the modified De Giorgi iteration used in this paper can be employed to other types equations with nonstandard growth conditions.

      We now discuss the main plan of the paper. In Section 2, we review some preliminaries concerning the variable exponent Sobolev spaces and introduce a family of regularized problems for problem (1.1). We regularize the degenerate part through replacing the term div ( | u m | p ( x ) 2 u m ) by
      div { ( | ( σ u m + ϵ u ) | 2 + η ) p ( x ) 2 2 ( σ u m + ϵ u ) } , ϵ , η > 0 .

      In Section 3, in order to apply the topological degree method, we combine these regularized problems with a relatively simpler equation and derive some a priori estimates. By virtue of the De Giorgi iteration technique, we deduce an a priori L bound for solutions to the regularized problems in Proposition 3.2; and the uniform lower bound estimate is obtained in Proposition 3.5. In Section 4, we establish the existence of nonnegative nontrivial solution of (1.1) through the limit process as ϵ and η tend to zero. Finally, in the Appendix, we give a proof of the iteration lemma (Lemma 3.1) for the sake of readability.

      2 Preliminaries and the regularized problems of (1.1)

      First of all, for the reader’s convenience, we recall some preliminary results concerning the variable exponent Sobolev spaces. One may find these standard results in monographs [1, 12].

      Let p be a continuous function defined in Ω ¯ , p ( x ) > 1 , for any x Ω ¯ .
      1. 1.
        L p ( x ) ( Ω ) space: We have
        L p ( x ) ( Ω ) : = { u : u  is measurable in  Ω  and  Ω | u ( x ) | p ( x ) d x < } ,
        equipped with the following Luxemburg norm:
        | u | L p ( x ) ( Ω ) : = inf { λ > 0 : Ω | u ( x ) λ | p ( x ) d x 1 } .

        The space ( L p ( x ) ( Ω ) , | | L p ( x ) ( Ω ) ) is a separable, uniformly convex Banach space.

         
      2. 2.
        W 1 , p ( x ) ( Ω ) space: We have
        W 1 , p ( x ) ( Ω ) : = { u L p ( x ) ( Ω ) : | u | L p ( x ) ( Ω ) } ,

        endowed with the norm | u | W 1 , p ( x ) ( Ω ) : = | u | L p ( x ) ( Ω ) + | u | L p ( x ) ( Ω ) . We denote by W 0 1 , p ( x ) ( Ω ) the closure of C 0 ( Ω ) in W 1 , p ( x ) ( Ω ) . In fact, the norm | u | L p ( x ) ( Ω ) and | u | W 1 , p ( x ) ( Ω ) are equivalent norms in W 0 1 , p ( x ) ( Ω ) . W 1 , p ( x ) ( Ω ) and W 0 1 , p ( x ) ( Ω ) are separable and reflexive Banach spaces with the above norms.

         
      3. 3.
        Frequently used relationships in the estimate:
        min { ( Ω | u ( x ) | p ( x ) d x ) 1 p + , ( Ω | u ( x ) | p ( x ) d x ) 1 p } | u | L p ( x ) ( Ω ) max { ( Ω | u ( x ) | p ( x ) d x ) 1 p + , ( Ω | u ( x ) | p ( x ) d x ) 1 p } .
         
      4. 4.

        p ( x ) -Hölder’s inequality:

        For any u L p ( x ) ( Ω ) and v L q ( x ) ( Ω ) , with 1 p ( x ) + 1 q ( x ) = 1 , we have
        | Ω u v d x | ( 1 p + 1 q ) | u | L p ( x ) ( Ω ) | v | L q ( x ) ( Ω ) .
         
      5. 5.

        Embedding relationships:

        If p 1 and p 2 are in C ( Ω ¯ ) , and 1 p 1 ( x ) p 2 ( x ) , for any x Ω ¯ , then there exists a positive constant C p 1 ( x ) , p 2 ( x ) such that
        | u | L p 1 ( x ) ( Ω ) C p 1 ( x ) , p 2 ( x ) | u | L p 2 ( x ) ( Ω ) ,

        i.e. the embedding L p 2 ( x ) ( Ω ) L p 1 ( x ) ( Ω ) is continuous.

        If q C ( Ω ¯ ) and 1 q ( x ) < p ( x ) , for any x Ω ¯ , then the embedding W 0 1 , p ( x ) ( Ω ) L q ( x ) ( Ω ) is continuous and compact. Here
        p ( x ) : = { N p ( x ) N p ( x ) , p ( x ) < N , + , p ( x ) N .
         
      6. 6.

        p ( x ) -Poincaré’s inequality:

         

      There exists a positive constant C p such that | u | L p ( x ) ( Ω ) C p | u | L p ( x ) ( Ω ) , for any u W 0 1 , p ( x ) ( Ω ) .

      We next define the weak solutions to problem (1.1).

      Definition 2.1 u is said to be a weak periodic solution to (1.1) provided that u m L p ( 0 , T ; W 0 1 , p ( x ) ( Ω ) ) with | u m | L p ( x ) ( Q T ) , u C ( Q ¯ T ) and u satisfies
      0 = Q T { u φ t + | u m | p ( x ) 2 u m φ a u φ + u φ Ω K ( ξ , t ) u 2 ( ξ , t τ ) d ξ } d x d t ,
      (2.1)

      for all φ C 1 ( Q ¯ T ) satisfying φ ( x , T ) = φ ( x , 0 ) for x Ω and φ ( , t ) | Ω = 0 for t [ 0 , T ] .

      As in [7], we introduce the following regularized problem:
      { u t div { ( | ( σ u m + ϵ u ) | 2 + η ) p ( x ) 2 2 ( σ u m + ϵ u ) } = [ a Ω K ( ξ , t ) u 2 ( ξ , t τ ) d ξ ] u , a.e.  ( x , t ) Q T , u ( , t ) | Ω = 0 , a.e.  t ( 0 , T ) , u ( , 0 ) = u ( , T ) ,
      (2.2)

      where 0 < ϵ < 1 2 , 0 < η < ( 1 2 ) p + 2 p 2 and σ [ 0 , 1 ] are given constants.

      Definition 2.2 We say that u ϵ η is a weak periodic solution of (2.2), if u ϵ η m L p ( 0 , T ; W 0 1 , p ( x ) ( Ω ) ) with | u ϵ η m | L p ( x ) ( Q T ) , u ϵ η C ( Q ¯ T ) , and u ϵ η solves
      0 = Q T { u ϵ η φ t + ( | ( σ u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( σ u ϵ η m + ϵ u ϵ η ) φ a u ϵ η φ + u ϵ η φ Ω K ( ξ , t ) u ϵ η 2 ( ξ , t τ ) d ξ } d x d t ,
      (2.3)

      for all φ C 1 ( Q ¯ T ) satisfying φ ( x , T ) = φ ( x , 0 ) for x Ω and φ ( , t ) | Ω = 0 for t [ 0 , T ] .

      Remark 2.3 For any p C 1 , α ( Ω ¯ ) , the set { φ C 1 ( Q ¯ T ) : φ ( , T ) = φ ( , 0 ) } is dense in { φ C ( Q ¯ T ) L p ( 0 , T ; W 0 1 , p ( x ) ( Ω ) ) : | u m | L p ( x ) ( Q T ) , φ ( , T ) = φ ( , 0 ) } , thus in the sense of the definition of weak solution above, u ϵ η can be chosen as test function.

      We investigate problem (2.2) extensively before studying the limit process as ϵ , η 0 . Define a map G ϵ η : [ 0 , 1 ] × L ( Q T ) L ( Q T ) as follows:
      ( σ , f ) u ϵ η = G ϵ η ( σ , f ) ,
      where u ϵ η is a weak periodic solution of the problem:
      { u t div { ( | ( σ u m + ϵ u ) | 2 + η ) p ( x ) 2 2 ( σ u m + ϵ u ) } = f , a.e.  ( x , t ) Q T , u ( , t ) | Ω = 0 , a.e.  t ( 0 , T ) , u ( , 0 ) = u ( , T ) .
      (2.4)
      Given α L ( Q T ) , let f = f ( α ) L ( Q T ) be defined by
      f ( x , t ) = f ( α ) ( x , t ) = [ a ( x , t ) Ω K ( ξ , t ) α 2 ( ξ , t τ ) d ξ ] α for  ( x , t ) Q T .

      Therefore, if a nonnegative function u ϵ η L ( Q T ) satisfies u ϵ η = G ϵ η ( 1 , f ( u ϵ η ) ) , then u ϵ η is a weak solution of (2.2).

      Define
      T ϵ η ( σ , u ) = G ϵ η ( σ , f ( u ) ) , ( σ , u ) [ 0 , 1 ] × L ( Q T ) .

      Then, according to [3] or the classical regularity results from [4], one obtains the following lemma.

      Lemma 2.4 Assume that 0 < ϵ < 1 2 , 0 < η < ( 1 2 ) p + 2 p 2 and λ L ( Q T ) . Then T ϵ η is a continuous compact operator from [ 0 , 1 ] × L ( Q T ) to L ( Q T ) . Furthermore, u ϵ η = T ϵ η ( σ , λ ) C ( Q ¯ T ) .

      3 A priori estimates to the regularized problem

      First of all, the following modified De Giorgi iteration lemma will be useful (we give a proof in the Appendix).

      Lemma 3.1 (Iteration lemma)

      Suppose φ ( t ) is a nonnegative and nonincreasing function on [ K 0 , + ) , it satisfies
      φ ( h ) ( M h k ) α [ φ β ( k ) + φ γ ( k ) ] ,
      (3.1)
      for any h > k k 0 , and for some constants M > 0 , α > 0 , β > 1 , γ > 1 . Then
      φ ( k 0 + d ) = 0 ,

      where d = M 2 δ δ 1 ( φ β 1 ( k 0 ) + φ γ 1 ( k 0 ) ) 1 α , and δ = min { β , γ } .

      Next, we prove a crucial a priori L bound for u ϵ η via a De Giorgi iteration technique as in [15].

      Proposition 3.2 Let K 1 > 0 and assume that u ϵ η is a nonnegative T-periodic continuous function such that
      u t div { ( | ( u m + ϵ u ) | 2 + η ) p ( x ) 2 2 ( u m + ϵ u ) } K 1 u ,
      (3.2)
      u ( , t ) | Ω = 0 .
      (3.3)

      Then there exists a constant R > 0 , such that u ϵ η L ( Q T ) < R , where R is independent of ϵ and η.

      Proof Step 1. Multiplying (3.2) by u ϵ η m q , with any q > 1 . Integrating over Ω and noticing that u ϵ η ( , t ) | Ω = 0 , we have
      1 m q + 1 d d t Ω u ϵ η m q + 1 d x + Ω ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( u ϵ η m + ϵ u ϵ η ) u ϵ η m q d x K 1 Ω u ϵ η m q + 1 d x .
      (3.4)
      Since | u ϵ η m | p ( x ) | u ϵ η m | p 1 , we deal with the second term on the left-hand side of (3.4) as follows.
      Ω ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( u ϵ η m + ϵ u ϵ η ) u ϵ η m q d x q Ω u ϵ η m ( q 1 ) | u ϵ η m | p ( x ) d x q Ω u ϵ η m ( q 1 ) | u ϵ η m | p d x q Ω u ϵ η m ( q 1 ) d x = q ( p p + q 1 ) p Ω | u ϵ η m ( p + q 1 ) p | p d x q Ω u ϵ η m ( q 1 ) d x .
      (3.5)
      Combining (3.4) and (3.5), we have
      1 m q + 1 d d t Ω u ϵ η m q + 1 d x + q ( p p + q 1 ) p Ω | u ϵ η m ( p + q 1 ) p | p d x K 1 Ω u ϵ η m q + 1 d x + q Ω u ϵ η m ( q 1 ) d x .
      (3.6)
      We estimate the right-hand side of (3.6) by Hölder’s inequality, the embedding theorem and Young’s inequality with ϵ to deduce
      K 1 Ω u ϵ η m q + 1 d x + q Ω u ϵ η m ( q 1 ) d x K 1 ( Ω u ϵ η m ( p + q 1 ) d x ) m q + 1 m ( p + q 1 ) | Ω | 1 m q + 1 m ( p + q 1 ) + q ( Ω u ϵ η m ( p + q 1 ) d x ) q 1 p + q 1 | Ω | 1 q 1 p + q 1 C ( Ω | u ϵ η m ( p + q 1 ) p | p d x ) m q + 1 m ( p + q 1 ) + C ( Ω | u ϵ η m ( p + q 1 ) p | p d x ) q 1 p + q 1 ϵ 1 Ω | u ϵ η m ( p + q 1 ) p | p d x + C ( ϵ 1 ) C m ( p + q 1 ) m p m 1 + ϵ 2 Ω | u ϵ η m ( p + q 1 ) p | p d x + C ( ϵ 2 ) C p + q 1 p .
      (3.7)
      Choosing ϵ 1 and ϵ 2 appropriately, we have from (3.6) and (3.7)
      d d t Ω u ϵ η m q + 1 ( x , t ) d x C 1 ,
      (3.8)

      for any q > 1 , where C 1 depends on q, p , m, and Ω.

      Integrating (3.6) over [ τ , τ + T ] and using the T-periodicity of u ϵ η , we have
      q ( p p + q 1 ) p Q T | u ϵ η m ( p + q 1 ) p | p d x d t K 1 Q T u ϵ η m q + 1 d x d t + q Q T u ϵ η m ( q 1 ) d x d t .
      (3.9)
      Similarly to (3.7), we obtain
      Q T | u ϵ η m ( p + q 1 ) p | p d x d t C 2 ,
      (3.10)
      where C 2 depends on q, p , m, T and Ω. By Poincaré’s inequality, we have
      Q T u ϵ η m ( p + q 1 ) d x d t Q T | u ϵ η m ( p + q 1 ) p | p d x d t C 2 .
      (3.11)
      Recall our assumption that p > 2 , m > 1 , and thus m ( p + q 1 ) > m q + 1 . Consequently, considering (3.11), we obtain
      Q T u ϵ η m q + 1 ( x , t ) d x d t C 2 ,
      (3.12)
      which implies that there exists a t 0 ( τ , τ + T ) such that
      Ω u ϵ η m q + 1 ( x , t 0 ) d x C 2 .
      (3.13)
      From (3.8) and (3.13), we conclude
      Ω u ϵ η m q + 1 ( x , t ) d x C 2 + C 1 ( t t 0 ) ,
      (3.14)
      for any t t 0 . In view of the T-periodicity of u ϵ η , (3.14) shows
      Ω u ϵ η m q + 1 ( x , τ ) d x = Ω u ϵ η m q + 1 ( x , τ + T ) d x C 2 + C 1 T .
      We finally arrive at
      sup t ( τ , τ + T ) Ω u ϵ η m q + 1 ( x , t ) d x C ,
      (3.15)

      for any q > 1 , where C depends on q, p , m, T and Ω.

      Step 2. Let
      A k ( t ) = { x Ω ; u ϵ η ( x , t ) > k } , μ k = sup t ( τ , τ + T ) | A k ( t ) | ,
      where | A k ( t ) | is the Lebesgue measure of the set A k ( t ) . Multiplying (3.2) by ( u ϵ η k ) + m χ [ t 1 , t 2 ] ( t ) on both sides, where χ [ t 1 , t 2 ] ( t ) represents the characteristic function of the interval [ t 1 , t 2 ] , and integrating over Q T , we have
      1 m + 1 t 1 t 2 d d t Ω ( u ϵ η k ) + m + 1 d x d t + t 1 t 2 Ω ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( u ϵ η m + ϵ u ϵ η ) ( u ϵ η k ) + m d x d t K 1 t 1 t 2 Ω u ϵ η ( u ϵ η k ) + m d x d t .
      Let I k ( t ) : = Ω ( u ϵ η k ) + m + 1 d x . We assume that the absolutely continuous function I k ( t ) attains its maximum at ϱ [ τ , τ + T ] . Take t 1 = ϱ θ , t 2 = ϱ and θ small enough so that t 1 > τ . (In fact, this is always possible because of the periodicity of u ϵ η ; for example, if ϱ = τ , we take ϱ = τ + T , then ϱ > τ and t 1 > τ .) Then we have I k ( ϱ ) I k ( ϱ θ ) and
      1 θ ϱ θ ϱ Ω ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( u ϵ η m + ϵ u ϵ η ) ( u ϵ η k ) + m d x d t 1 θ K 1 ϱ θ ϱ Ω u ϵ η ( u ϵ η k ) + m d x d t .
      (3.16)
      Letting θ 0 + yields
      Ω ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( u ϵ η m + ϵ u ϵ η ) ( u ϵ η k ) + m d x K 1 Ω u ϵ η ( u ϵ η k ) + m d x .
      (3.17)
      After a direct computation, we obtain an estimate for the left-hand side of (3.17) as follows:
      Ω ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( u ϵ η m + ϵ u ϵ η ) ( u ϵ η k ) + m d x Ω | m ( u ϵ η k ) + m 1 ( u ϵ η k ) + | p ( x ) d x = Ω | ( u ϵ η k ) + m | p ( x ) d x A k ( ϱ ) | ( u ϵ η k ) m | p d x | A k ( ϱ ) | .
      (3.18)
      Substituting (3.18) into (3.17), we have
      A k ( ϱ ) | ( u ϵ η k ) m | p d x K 1 A k ( ϱ ) u ϵ η ( u ϵ η k ) m d x + μ k .
      (3.19)
      We now deal with (3.19). On one hand, by the embedding theorem
      ( A k ( ϱ ) ( u ϵ η k ) m r d x ) p r S p A k ( ϱ ) | ( u ϵ η k ) m | p d x ,
      (3.20)
      where S is the Sobolev embedding constant, and
      r = { N p N p , if  p < N , 2 p ( N + p ) N , if  p N .
      On the other hand, from (3.15), where we may fix a special q, using Hölder’s inequality, we obtain
      K 1 A k ( ϱ ) u ϵ η ( u ϵ η k ) m d x K 1 ( A k ( ϱ ) u ϵ η N + p p d x ) p N + p ( A k ( ϱ ) ( u ϵ η k ) m N + p N d x ) N N + p C ( A k ( ϱ ) ( u ϵ η k ) m N + p N d x ) N N + p C | A k ( ϱ ) | N r N p r ( N + p ) ( A k ( ϱ ) ( u ϵ η k ) m r d x ) 1 r .
      (3.21)
      Let J k ( ϱ ) = A k ( ϱ ) ( u ϵ η k ) m r d x . Then (3.19), (3.20), and (3.21) imply
      1 S p [ J k ( ϱ ) ] p r C μ k N r N p r ( N + p ) [ J k ( ϱ ) ] 1 r + μ k .
      (3.22)
      Utilizing Young’s inequality with ϵ, we obtain from (3.22)
      J k 2 r p ( S r C r p μ k N r N p p ( N + p ) J k 1 p + S r μ k r p ) C μ k N r N p p ( N + p ) J k 1 p + C μ k r p C ( ϵ J k + C ( ϵ ) μ k N r N p ( p 1 ) ( N + p ) ) + C μ k r p .
      Upon choosing ϵ appropriately, one obtains
      J k ( ϱ ) C ( μ k N r N p ( p 1 ) ( N + p ) + μ k r p ) .
      (3.23)
      For any h > k > 0 , it is easy to see
      J k ( ϱ ) | A h ( ϱ ) | ( h k ) r m .
      (3.24)
      The relationships (3.23) and (3.24) above imply that
      μ h ( M h k ) r m ( μ k 1 + N r p ( N + p ) ( p 1 ) ( N + p ) + μ k r p ) .
      (3.25)
      Noticing that N r p ( N + p ) ( p 1 ) ( N + p ) > 0 and r p > 1 , by the iteration Lemma 3.1, we obtain μ R = 0 and thus u ϵ η L ( Q T ) < R , where
      R = M 2 Λ Λ 1 ( | Q T | N r p ( N + p ) ( p 1 ) ( N + p ) + | Q T | r p 1 ) 1 r m ,
      with
      Λ = min { N r N p ( p 1 ) ( N + p ) , r p } .

       □

      Theorem 3.3 Assume K 0 , for a.e. ( x , t ) Q T . Then there exists a positive constant R such that
      deg ( u T ϵ η ( 1 , u + ) , B R , 0 ) = 1 ,

      where u + = max { u , 0 } .

      Proof From Proposition 3.2, we take K 1 = a L ( Q T ) , it implies that there exists a positive constant R > 0 independent of ϵ and η, such that u ϵ η G ϵ η ( 1 , ρ f ( u ϵ η + ) ) , for any u ϵ η B R , ρ [ 0 , 1 ] . Hence the topological degree deg ( u G ϵ η ( 1 , ρ f ( u + ) ) , B R , 0 ) is well defined in B R . Thanks to the homotopy invariance property of the Leray-Schauder degree, we have
      deg ( u G ϵ η ( 1 , f ( u + ) ) , B R , 0 ) = deg ( u G ϵ η ( 1 , 0 ) , B R , 0 ) .
      (3.26)
      Using the fact that G ϵ η ( 1 , 0 ) = 0 , one has
      deg ( u G ϵ η ( 1 , 0 ) , B R , 0 ) = deg ( I , B R , 0 ) = 1 .
      (3.27)

      From (3.26) and (3.27), we get deg ( u T ϵ η ( 1 , u + ) , B R , 0 ) = 1 . □

      Using the standard method, similar to that in [3] or [13], one can prove the following.

      Proposition 3.4 Assume that a L ( Q T ) , K L ( Q T ) . If u ϵ η solves u = G ϵ η ( σ , ρ f ( u + ) + 1 σ ) , for some σ [ 0 , 1 ] and ρ [ 0 , 1 ] , then u ϵ η 0 for any ( x , t ) Q T . Moreover, if u ϵ η 0 , then u ϵ η > 0 in Q T .

      In what follows, we prove a lower bound for the regularized problem.

      Proposition 3.5 Let μ 1 be the first eigenvalue of
      { Δ u = μ u , x Ω , u = 0 , x Ω ,
      and let e 1 be the associated positive eigenfunction such that e 1 L 2 ( Ω ) = 1 . Assume that 1 T Q T a e 1 2 d x d t > μ 1 , 0 < ϵ < 1 2 and 0 < η < ( 1 2 ) p + 2 p 2 . If u ϵ η 0 satisfies u ϵ η = G ϵ η ( σ , f ( u ϵ η + ) + 1 σ ) for some σ [ 0 , 1 ] , then u ϵ η L ( Q T ) r 0 , where
      r 0 = min { ( 1 2 m ) 1 m 1 , ( Q T a e 1 2 d x d t μ 1 T M ) γ p + } , γ p ( x ) = p ( x ) p ( x ) 2 , γ p + = max Ω ¯ γ p ( x ) = p p 2 , γ p = min Ω ¯ γ p ( x ) = p + p + 2 , M = K L 1 ( Q T ) + 2 p + 2 ( 1 γ p + 2 p ) C p + , p ( x ) max Ω ¯ | e 1 | 2 | Ω | 2 p + × max { ( a L 1 ( Q T ) + | Ω | T ) 1 γ p ± T γ p ± 1 γ p ± } ,

      C p + , p ( x ) is the embedding constant of L p + ( Ω ) into L p ( x ) ( Ω ) , and | Ω | is the Lebesgue measure of the domain Ω.

      Proof We argue by contradiction. If not, then for each σ [ 0 , 1 ] and r ( 0 , r 0 ) , there exists a u ϵ η 0 such that u ε η = G ϵ η ( σ , f ( u ε η + ) + 1 σ ) , with u ϵ η L ( Q T ) r . For clarity, we divide the proof into four steps.

      Step 1. Note that, by Proposition 3.4, u ϵ η > 0 in Q T . Taking ϕ C 0 ( Ω ) and multiplying
      u ϵ η t div { ( | ( σ u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( σ u ϵ η m + ϵ u ϵ η ) } = [ a Ω K ( ξ , t ) u ϵ η 2 ( ξ , t τ ) d ξ ] u ϵ η + 1 σ
      (3.28)
      by ϕ 2 u ϵ η , integrating over Q T and using the T-periodicity of u ϵ η , we obtain
      Q T [ a Ω K ( ξ , t ) u ϵ η 2 ( ξ , t τ ) d ξ ] ϕ 2 d x d t + Q T ( 1 σ ) ϕ 2 u ϵ η d x d t = Q T ϕ 2 u ϵ η div { ( | ( σ u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( σ u ϵ η m + ϵ u ϵ η ) } d x d t : = ( R ) .
      (3.29)
      Step 2. Using u ϵ η ( ϕ 2 u ϵ η ) = | ϕ | 2 u ϵ η 2 | ( ϕ u ϵ η ) | 2 , we have
      ( R ) = Q T ( | ( σ u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( σ u ϵ η m + ϵ u ϵ η ) ( ϕ 2 u ϵ η ) d x d t = Q T ( | ( σ u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( σ m u ϵ η m 1 + ϵ ) u ϵ η ( ϕ 2 u ϵ η ) d x d t = Q T ( | ( σ u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( σ m u ϵ η m 1 + ϵ ) | ϕ | 2 d x d t Q T ( | ( σ u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( σ m u ϵ η m 1 + ϵ ) u ϵ η 2 | ( ϕ u ϵ η ) | 2 d x d t Q T ( | ( σ u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( σ m u ϵ η m 1 + ϵ ) | ϕ | 2 d x d t .
      Since r < r 0 ( 1 2 m ) 1 m 1 and 0 < ϵ < 1 2 , we have σ m u ϵ η m 1 + ϵ m u ϵ η m 1 + ϵ < 1 2 + ϵ < 1 . Hence
      ( R ) Q T ( | ( σ u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 | ϕ | 2 d x d t 2 p + 2 2 Q T ( | ( σ u ϵ η m + ϵ u ϵ η ) | p ( x ) 2 + η p 2 2 ) | ϕ | 2 d x d t = 2 p + 2 2 Q T | ( σ u ϵ η m + ϵ u ϵ η ) | p ( x ) 2 | ϕ | 2 d x d t + 2 p + 2 2 η p 2 2 Q T | ϕ | 2 d x d t .
      (3.30)
      Thanks to the p ( x ) -Hölder’s inequality in variable exponent space, we have
      Ω | ( σ u ϵ η m + ϵ u ϵ η ) | p ( x ) 2 | ϕ | 2 d x ( 1 γ p + 2 p ) | | ( σ u ϵ η m + ϵ u ϵ η ) | p ( x ) 2 | L γ p ( x ) ( Ω ) | | ϕ | 2 | L p ( x ) 2 ( Ω ) ( 1 γ p + 2 p ) C p + , p ( x ) | ϕ | 2 L p + 2 ( Ω ) × max { ( Ω | ( σ u ϵ η m + ϵ u ϵ η ) | p ( x ) d x ) 1 γ p ± } .
      (3.31)
      Noting that 1 γ p ± < 1 and using Hölder’s inequality, we have
      0 T ( Ω | ( σ u ϵ η m + ϵ u ϵ η ) | p ( x ) d x ) 1 γ p ± d t T 1 1 γ p ± ( Q T | ( σ u ϵ η m + ϵ u ϵ η ) | p ( x ) d x d t ) 1 γ p ± .
      (3.32)
      Integrating (3.31) over [ 0 , T ] and noting (3.32), we get
      Q T | ( σ u ϵ η m + ϵ u ϵ η ) | p ( x ) 2 | ϕ | 2 d x d t ( 1 γ p + 2 p ) C p + , p ( x ) | ϕ | 2 L p + 2 ( Ω ) × max { ( Q T | ( σ u ϵ η m + ϵ u ϵ η ) | p ( x ) d x d t ) 1 γ p ± T γ p ± 1 γ p ± } .
      (3.33)
      Step 3. Multiplying (3.28) by σ u ϵ η m + ϵ u ϵ η , integrating over Q T , noticing the T-periodicity of u ϵ η and 0 < r < r 0 ( 1 2 m ) 1 m 1 < 1 , we deduce
      Q T | ( σ u ϵ η m + ϵ u ϵ η ) | p ( x ) d x d t Q T ( | ( σ u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 | ( σ u ϵ η m + ϵ u ϵ η ) | 2 d x d t Q T a u ϵ η ( u ϵ η m + u ϵ η ) d x d t + ( 1 σ ) Q T ( u ϵ η m + u ϵ η ) d x d t ( u ϵ η L ( Q T ) m + 1 + u ϵ η L ( Q T ) 2 ) a L 1 ( Q T ) + ( 1 σ ) ( u ϵ η L ( Q T ) m + u ϵ η L ( Q T ) ) | Ω | T ( r m + 1 + r 2 ) a L 1 ( Q T ) + ( r m + r ) | Ω | T 2 r ( a L 1 ( Q T ) + | Ω | T ) .
      Substituting this inequality into (3.33), we have
      Q T | ( σ u ϵ η m + ϵ u ϵ η ) | p ( x ) 2 | ϕ | 2 d x d t ( 1 γ p + 2 p ) C p + , p ( x ) | ϕ | 2 L p + 2 ( Ω ) × max { ( 2 r ( a L 1 ( Q T ) + | Ω | T ) ) 1 γ p ± T γ p ± 1 γ p ± } ( 1 γ p + 2 p ) C p + , p ( x ) | ϕ | 2 L p + 2 ( Ω ) 2 1 γ p r 1 γ p + × max { ( a L 1 ( Q T ) + | Ω | T ) 1 γ p ± T γ p ± 1 γ p ± } .
      (3.34)
      Substituting (3.34) into (3.30) and noticing that 2 p + 2 2 2 1 γ p 2 p + 2 , we get
      ( R ) 2 p + 2 ( 1 γ p + 2 p ) r 1 γ p + C p + , p ( x ) | ϕ | 2 L p + 2 ( Ω ) × max { ( a L 1 ( Q T ) + | Ω | T ) 1 γ p ± T γ p ± 1 γ p ± } + 2 p + 2 2 η p 2 2 Q T | ϕ | 2 d x d t .
      (3.35)
      Considering that 0 < η < ( 1 2 ) p + 2 p 2 , from (3.29) and (3.35), we have
      Q T [ a Ω K ( ξ , t ) u ϵ η 2 ( ξ , t τ ) d ξ ] ϕ 2 d x d t Q T | ϕ | 2 d x d t + 2 p + 2 ( 1 γ p + 2 p ) r 1 γ p + C p + , p ( x ) | ϕ | 2 L p + 2 ( Ω ) × max { ( a L 1 ( Q T ) + | Ω | T ) 1 γ p ± T γ p ± 1 γ p ± } .
      (3.36)
      Step 4. We claim
      Q T a e 1 2 d x d t μ 1 T r 1 γ p + M ,
      (3.37)
      from which we will derive a contradiction. First, to show (3.37), let ϕ = e 1 in (3.36). Using the fact that e 1 ( C 1 ( Ω ¯ ) ) N and noting e 1 L 2 ( Ω ) = 1 and Ω | e 1 | 2 d x = μ 1 , we get
      Q T a e 1 2 d x d t μ 1 T 2 p + 2 ( 1 γ p + 2 p ) r 1 γ p + C p + , p ( x ) | e 1 | 2 L p + 2 ( Ω ) × max { ( a L 1 ( Q T ) + | Ω | T ) 1 γ p ± T γ p ± 1 γ p ± } + 0 T Ω K ( ξ , t ) u ϵ η 2 ( ξ , t τ ) d ξ Ω e 1 2 d x d t 2 p + 2 ( 1 γ p + 2 p ) r 1 γ p + C p + , p ( x ) max Ω ¯ | e 1 | 2 | Ω | 2 p + × max { ( a L 1 ( Q T ) + | Ω | T ) 1 γ p ± T γ p ± 1 γ p ± } + K L 1 ( Q T ) r 2 r 1 γ p + [ 2 p + 2 ( 1 γ p + 2 p ) C p + , p ( x ) max Ω ¯ | e 1 | 2 | Ω | 2 p + × max { ( a L 1 ( Q T ) + | Ω | T ) 1 γ p ± T γ p ± 1 γ p ± } + K L 1 ( Q T ) ] = r 1 γ p + M .
      Now the definition of r 0 and (3.37) yield
      r 0 ( Q T a e 1 2 d x d t μ 1 T M ) γ p + r ,
      (3.38)

      which is clearly a contradiction to the assumption that r ( 0 , r 0 ) . This completes the proof. □

      Theorem 3.6 Let r 0 be as given in Proposition  3.5. Then deg ( u T ϵ η ( 1 , u + ) , B r , 0 ) = 0 for all 0 < r < r 0 .

      Proof In view of Proposition 3.5, for any fixed r ( 0 , r 0 ) , we have proved that u G ϵ η ( σ , f ( u + ) + 1 σ ) for all u B r , σ [ 0 , 1 ] . So the Leray-Schauder topological degree deg ( u G ϵ η ( σ , f ( u + ) + 1 σ ) , B r , 0 ) is well defined for all σ [ 0 , 1 ] . Thanks to the homotopy invariance of the topological degree, we have
      deg ( u G ϵ η ( 1 , f ( u + ) ) , B r , 0 ) = deg ( u G ϵ η ( 0 , f ( u + ) + 1 ) , B r , 0 ) .
      (3.39)

      Also, from Proposition 3.5, we infer that u = G ϵ η ( 0 , f ( u + ) + 1 ) admits no nontrivial solution in B r . Moreover, u ϵ η = 0 is not a solution of u = G ϵ η ( 0 , f ( u + ) + 1 ) . So deg ( u G ϵ η ( 0 , f ( u + ) + 1 ) , B r , 0 ) = 0 . Together with (3.39), we have deg ( u T ϵ η ( 1 , u + ) , B r , 0 ) = 0 . □

      4 Existence of nontrivial nonnegative solution to (1.1)

      Theorem 4.1 Assume K ( x , t ) 0 for a.e. ( x , t ) Q T and 1 T Q T a e 1 2 d x d t > μ 1 . Then problem (1.1) has a nontrivial nonnegative periodic solution.

      Proof We consider the regularized problem (2.2). By Theorem 3.3 and Theorem 3.6, we conclude that there exist R and r, independent of ϵ and η, with R > r > 0 , such that
      deg ( u G ϵ η ( 1 , f ( u + ) ) , B R B ¯ r , 0 ) = 1

      for 0 < ϵ < 1 2 and 0 < η < ( 1 2 ) p + 2 p 2 . Using the solvability of the Leray-Schauder degree, we conclude that the regularized problem (2.2) admits a nontrivial nonnegative solution u ϵ η in B R B ¯ r .

      We prove that u ϵ η m L p ( 0 , T ; W 0 1 , p ( x ) ( Ω ) ) with | u ϵ η m | L p ( x ) ( Q T ) and that a solution to problem (1.1) is obtained as a limit of u ϵ η as ϵ , η 0 . We proceed in several steps.

      Step 1. In view of K ( x , t ) 0 , choosing C = a L ( Q T ) , we have
      u ϵ η t div { ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( u ϵ η m + ϵ u ϵ η ) } C u ϵ η .
      (4.1)
      Multiplying (4.1) by u ϵ η m + ϵ u ϵ η , integrating over Q T and noting the T-periodicity of u ϵ η and the boundedness of u ϵ η , we have
      Q T | u ϵ η m | p ( x ) d x d t Q T ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 | ( u ϵ η m + ϵ u ϵ η ) | 2 d x d t C Q T ( u ϵ η m + 1 + ϵ u ϵ η 2 ) d x d t M ,
      (4.2)
      where M is a positive constant independent of ϵ and η. Moreover,
      0 T | u ϵ η m | L p ( x ) ( Ω ) p d t = [ 0 , T ] { t : | u ϵ η m | L p ( x ) ( Ω ) > 1 } | u ϵ η m | L p ( x ) ( Ω ) p d t + [ 0 , T ] { t : | u ϵ η m | L p ( x ) ( Ω ) 1 } | u ϵ η m | L p ( x ) ( Ω ) p d t [ 0 , T ] { t : | u ϵ η m | L p ( x ) ( Ω ) > 1 } Ω | u ϵ η m | p ( x ) d x d t + T Q T | u ϵ η m | p ( x ) d x d t + T M + T .
      (4.3)

      So u ϵ η m L p ( 0 , T ; W 0 1 , p ( x ) ( Ω ) ) and u ϵ η m is uniformly bounded in the space L p ( 0 , T ; W 0 1 , p ( x ) ( Ω ) ) . Thus, up to subsequence if necessary, we may assume that u ϵ η m u m L p ( 0 , T ; W 0 1 , p ( x ) ( Ω ) ) . In what follows, our main goal is to prove that u is a weak solution of problem (1.1).

      Step 2. The following relation is obvious:
      Q T ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 | ( u ϵ η m + ϵ u ϵ η ) | 2 d x d t Q T | ( u ϵ η m + ϵ u ϵ η ) | p ( x ) d x d t .
      (4.4)
      From (4.2) and (4.4), we have
      Q T | ( u ϵ η m + ϵ u ϵ η ) | p ( x ) d x d t C .
      (4.5)
      Owing to the embedding results in the variable exponent space, one has
      ( u ϵ η m + ϵ u ϵ η ) L 2 ( Ω ) C 2 , p ( x ) | ( u ϵ η m + ϵ u ϵ η ) | L p ( x ) ( Ω ) C 2 , p ( x ) max { ( Ω | ( u ϵ η m + ϵ u ϵ η ) | p ( x ) d x ) 1 p ± } .
      (4.6)
      Integrating (4.6) over [ 0 , T ] and using Hölder’s inequality, we have
      0 T ( u ϵ η m + ϵ u ϵ η ) L 2 ( Ω ) d t C 2 , p ( x ) max { ( Q T | ( u ϵ η m + ϵ u ϵ η ) | p ( x ) d x d t ) 1 p ± T 1 1 p ± } .
      (4.7)
      From (4.5) and (4.7), there exists a positive constant C independent of ϵ and η, such that
      Q T | ( u ϵ η m + ϵ u ϵ η ) | 2 d x d t C .
      (4.8)
      In the following, we prove
      Q T | ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( u ϵ η m + ϵ u ϵ η ) | p ( x ) p ( x ) 1 d x d t C .
      (4.9)
      First, denote
      K 1 p ( x ) = p ( x ) p ( x ) 1 , K 1 p = min Ω ¯ K 1 p ( x ) , K 1 p + = max Ω ¯ K 1 p ( x ) , K 2 p ( x ) = 2 ( p ( x ) 1 ) p ( x ) , K 2 p = min Ω ¯ K 2 p ( x ) , K 2 p + = max Ω ¯ K 2 p ( x ) , K 2 p ( x ) = 2 ( p ( x ) 1 ) p ( x ) 2 , K 2 p = min Ω ¯ K 2 p ( x ) , K 2 p + = max Ω ¯ K 2 p ( x ) .
      A straightforward computation shows that
      Q T | ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( u ϵ η m + ϵ u ϵ η ) | p ( x ) p ( x ) 1 d x d t Q T | 2 p + 2 2 ( | ( u ϵ η m + ϵ u ϵ η ) | p ( x ) 2 + η p 2 2 ) ( u ϵ η m + ϵ u ϵ η ) | p ( x ) p ( x ) 1 d x d t 2 p + 2 2 K 1 p + Q T | | ( u ϵ η m + ϵ u ϵ η ) | p ( x ) 2 ( u ϵ η m + ϵ u ϵ η ) + ( u ϵ η m + ϵ u ϵ η ) | p ( x ) p ( x ) 1 d x d t 2 p + 2 2 K 1 p + 2 K 1 p + Q T ( | ( u ϵ η m + ϵ u ϵ η ) | p ( x ) + | ( u ϵ η m + ϵ u ϵ η ) | p ( x ) p ( x ) 1 ) d x d t M p { Q T | ( u ϵ η m + ϵ u ϵ η ) | p ( x ) d x d t + Q T | ( u ϵ η m + ϵ u ϵ η ) | p ( x ) p ( x ) 1 d x d t } .
      (4.10)
      By the p ( x ) -Hölder’s inequality, we have
      Ω | ( u ϵ η m + ϵ u ϵ η ) | p ( x ) p ( x ) 1 d x ( 1 K 2 p + 1 K 2 p ) | | ( u ϵ η m + ϵ u ϵ η ) | p ( x ) p ( x ) 1 | L K 2 p ( x ) ( Ω ) | 1 | L K 2 p ( x ) ( Ω ) ( 1 K 2 p + 1 K 2 p ) max { | Ω | 1 K 2 p + , | Ω | 1 K 2 p } × max { ( Ω | ( u ϵ η m + ϵ u ϵ η ) | 2 d x ) 1 K 2 p ± } .
      (4.11)
      Integrating (4.11) over [ 0 , T ] , using the p ( x ) -Hölder’s inequality again, we get
      Q T | ( u ϵ η m + ϵ u ϵ η ) | p ( x ) p ( x ) 1 d x d t ( 1 K 2 p + 1 K 2 p ) max { | Ω | 1 K 2 p + , | Ω | 1 K 2 p } max { T 1 K 2 p + , T 1 K 2 p } × max { ( Q T | ( u ϵ η m + ϵ u ϵ η ) | 2 d x d t ) 1 K 2 p ± } .
      (4.12)
      Substituting (4.5), (4.8), and (4.12) into (4.10), we derive (4.9). Therefore, there exists a H ( L p ( x ) p ( x ) 1 ( Q T ) ) N such that
      ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( u ϵ η m + ϵ u ϵ η ) H ,
      (4.13)

      weakly in ( L p ( x ) p ( x ) 1 ( Q T ) ) N as ϵ , η 0 .

      Step 3. Using a method analogous to [7], we get u ϵ η m t L 2 ( Q T ) C , where C is independent of ϵ and η. Since u ϵ η m is uniformly bounded in L p ( 0 , T ; W 0 1 , p ( x ) ( Ω ) ) , and W 0 1 , p ( x ) ( Ω ) compact L p ( x ) ( Ω ) L 1 ( Ω ) , by compactness theorem (Corollary 4 in [16]), it follows that u ϵ η m u m in L p ( 0 , T ; L p ( x ) ( Ω ) ) . Thus, we have
      0 = Q T { u φ t + H φ a u φ + u φ Ω K ( ξ , t ) u 2 ( ξ , t τ ) d ξ } d x d t
      (4.14)

      for any φ C 1 ( Q ¯ T ) satisfying φ ( x , T ) = φ ( x , 0 ) for x Ω and φ ( , t ) | Ω = 0 for t [ 0 , T ] (and hence, by density, for any φ C ( Q ¯ T ) L p ( 0 , T ; W 0 1 , p ( x ) ( Ω ) ) with | φ | L p ( x ) ( Q T ) and T-periodicity). The continuity of u follows from similar Hölder estimates in [17].

      Step 4. It remains to verify for any φ C 1 ( Q ¯ T ) ,
      Q T | u m | p ( x ) 2 u m φ d x d t = Q T H φ d x d t .
      (4.15)
      We consider matrix function Π ( Y ) = ( | Y | 2 + η ) p ( x ) 2 2 Y . Then Π ( Y ) = ( | Y | 2 + η ) p ( x ) 2 2 I + ( p ( x ) 2 ) ( | Y | 2 + η ) p ( x ) 4 2 Y Y T is a positive definite matrix. Choosing v L p ( 0 , T ; W 0 1 , p ( x ) ( Ω ) ) with | v | L p ( x ) ( Q T ) , by mean value theorem, there exists a matrix Y such that
      Π ( ( u ϵ η m + ϵ u ϵ η ) ) Π ( v ) , ( u ϵ η m + ϵ u ϵ η ) v = Π ( Y ) ( ( u ϵ η m + ϵ u ϵ η ) v ) , ( u ϵ η m + ϵ u ϵ η ) v 0 ,
      (4.16)
      which gives
      0 Q T { ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( u ϵ η m + ϵ u ϵ η ) ( | v | 2 + η ) p ( x ) 2 2 v } [ ( u ϵ η m + ϵ u ϵ η ) v ] d x d t = Q T ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 | ( u ϵ η m + ϵ u ϵ η ) | 2 d x d t Q T ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( u ϵ η m + ϵ u ϵ η ) v d x d t Q T ( | v | 2 + η ) p ( x ) 2 2 v [ ( u ϵ η m + ϵ u ϵ η ) v ] d x d t .
      (4.17)
      Multiplying the equation
      u ϵ η t div { ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( u ϵ η m + ϵ u ϵ η ) } = [ a Ω K ( ξ , t ) u ϵ η 2 ( ξ , t τ ) d ξ ] u ϵ η
      by u ϵ η m + ϵ u ϵ η , integrating over Q T and using T-periodicity of u ϵ η , one has
      Q T ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 | ( u ϵ η m + ϵ u ϵ η ) | 2 d x d t = Q T [ a Ω K ( ξ , t ) u ϵ η 2 ( ξ , t τ ) d ξ ] ( u ϵ η m + 1 + ϵ u ϵ η 2 ) d x d t .
      (4.18)
      Thus, (4.17) and (4.18) imply
      Q T ( | ( u ϵ η m + ϵ u ϵ η ) | 2 + η ) p ( x ) 2 2 ( u ϵ η m + ϵ u ϵ η ) v d x d t + Q T ( | v | 2 + η ) p ( x ) 2 2 v [ ( u ϵ η m + ϵ u ϵ η ) v ] d x d t Q T [ a Ω K ( ξ , t ) u ϵ η 2 ( ξ , t τ ) d ξ ] ( u ϵ η m + 1 + ϵ u ϵ η 2 ) d x d t .
      Letting ϵ , η 0 , by (4.13), we have
      Q T H v d x d t + Q T | v | p ( x ) 2 v ( u m v ) d x d t Q T [ a Ω K ( ξ , t ) u 2 ( ξ , t τ ) d ξ ] u m + 1 d x d t .
      (4.19)
      Let φ = u m in (4.14) and, by the T-periodicity of u, we get
      Q T H u m d x d t = Q T [ a Ω K ( ξ , t ) u 2 ( ξ , t τ ) d ξ ] u m + 1 d x d t .
      (4.20)
      Combining (4.19) with (4.20), we obtain
      0 Q T ( H | v | p ( x ) 2 v ) ( u m v ) d x d t .
      (4.21)
      Taking v = u m λ φ , with λ > 0 and φ C 1 ( Q ¯ T ) , we get
      0 Q T ( H | ( u m λ φ ) | p ( x ) 2 ( u m λ φ ) ) φ d x d t .
      (4.22)
      Letting λ 0 in (4.22) yields
      0 Q T ( H | u m | p ( x ) 2 u m ) φ d x d t .
      (4.23)
      On the other hand, if we take v = u m + λ φ , with λ > 0 and φ C 1 ( Q ¯ T ) and let λ 0 , we get
      0 Q T ( H | u m | p ( x ) 2 u m ) φ d x d t .
      (4.24)

      From (4.23) and (4.24) we have (4.15). This completes the proof of Theorem 4.1. □

      Appendix

      In this appendix, we prove Lemma 3.1 for the reader’s convenience.

      Proof of Lemma 3.1 Define the following sequence:
      k s = k 0 + d d 2 s , s = 0 , 1 , 2 , ,
      where d is to be determined later. Then (3.1) implies the recursive relationship
      φ ( k s + 1 ) M α 2 ( s + 1 ) α d α [ φ β ( k s ) + φ γ ( k s ) ] , s = 0 , 1 , 2 , .
      (5.1)
      By induction, one has
      φ ( k s ) φ ( k 0 ) r s , s = 0 , 1 , 2 , ,
      (5.2)
      where r > 1 is to be chosen. In fact, if (5.2) is right, then
      φ ( k s + 1 ) M α 2 ( s + 1 ) α d α [ φ β ( k 0 ) r s β + φ γ ( k 0 ) r s γ ] M α 2 ( s + 1 ) α d α φ β ( k 0 ) + φ γ ( k 0 ) r s δ = φ ( k 0 ) r s + 1 [ M α 2 ( s + 1 ) α d α φ β 1 ( k 0 ) + φ γ 1 ( k 0 ) r [ s ( δ 1 ) 1 ] ] .

      We choose r = 2 α δ 1 and M α 2 α δ δ 1 d α ( φ β 1 ( k 0 ) + φ γ 1 ( k 0 ) ) 1 . Consequently, these choices guarantee φ ( k s + 1 ) φ ( k 0 ) r s + 1 . From (5.2) and φ ( t ) is nonnegative and nonincreasing, we have deduced the result. □

      Declarations

      Acknowledgements

      The authors would like to thank the anonymous referees for their valuable comments on and suggestions regarding the original manuscript. This work was supported by National Science Foundation of China (11271154), by Key Lab of Symbolic Computation and Knowledge Engineering of Ministry of Education and by the 985 program of Jilin University.

      Authors’ Affiliations

      (1)
      College of Mathematics, Jilin University

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