Initial boundary value problems for second order parabolic systems in cylinders with polyhedral base

  • Vu Trong Luong1Email author and

    Affiliated with

    • Do Van Loi2

      Affiliated with

      Boundary Value Problems20112011:56

      DOI: 10.1186/1687-2770-2011-56

      Received: 18 September 2011

      Accepted: 24 December 2011

      Published: 24 December 2011

      Abstract

      The purpose of this article is to establish the well posedness and the regularity of the solution of the initial boundary value problem with Dirichlet boundary conditions for second-order parabolic systems in cylinders with polyhedral base.

      1 Introduction

      Boundary value problems for partial differential equations and systems in nonsmooth domains have been attracted attentions of many mathematicians for more than last 50 years. We are concerned with initial boundary value problems (IBVP) for parabolic equations and systems in nonsmooth domains. These problems in cylinders with bases containing conical points have been investigated in [1, 2] in which the regularity and the asymptotic behaviour near conical points of the solutions are established. Parabolic equations with discontinuous coefficients in Lipschitz domains have also been studied (see [3] and references therein).

      In this study, we consider IBVP with Dirichlet boundary conditions for second-order parabolic systems in both cases of finite cylinders and infinite cylinders whose bases are polyhedral domains. Firstly, we prove the well posedness of this problem by Galerkin's approximating method. Next, by this method we obtain the regularity in time of the solution. Finally, we apply the results for elliptic boundary value problems in polyhedral domains given in [4, 5] and former our results to deal with the global regularity of the solution.

      Let Ω be an open polyhedral domain in ℝ n (n = 2, 3), and 0 < T ≤ ∞. Set Q T = Ω × (0, T), S T = ∂Ω × (0, T). For a vector-valued function u = (u1, u2, ..., u s ) and p = (p1, p2, ..., p n ) ∈ ℕ n we use the notation D p u = ( x p u 1 , , x p u s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq1_HTML.gif.

      Let m, k be non negative integers. We denote by H m (Ω), H ˚ m ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq2_HTML.gif the usual Sobolev spaces as in [6]. By the notation (., .) we mean the inner product in L2(Ω).

      We denote by Hm,k(Q T , γ) (γ ∈ ℝ) the weighed Sobolev space of vector-valued functions u defined in Q T with the norm
      u H m , k ( Q T , γ ) = Q T 0 p m D p u 2 + j = 1 k u t j 2 e - γ t d x d t 1 2 < + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equa_HTML.gif

      Let us note that if T < +∞, then Hm,k(Q T , γ) ≡ Hm,k(Q T ).

      The space H ˚ m , k ( Q T , γ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq3_HTML.gif is the closure in Hm,k(Q T , γ) of the set consisting of all vector-valued functions uC(Q T ) which vanish near S T .

      Let ∂ sing Ω be the set of all singular points of ∂Ω, namely, the set of vertexes of Ω for the case n = 2 and the union of all edges of Ω for the case n = 3. Let ρ(x) be the distance from a point x ∈ Ω to the set ∂ sing Ω. For a ∈ ℝ, we denote by H a m ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq4_HTML.gif the weighed Sobolev space of vector functions u defined on Ω with the norm
      u H a m ( Ω ) = Ω 0 p m ρ ( x ) 2 ( p - a ) D p u 2 d x 1 2 < + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equb_HTML.gif

      It is obvious from the definition that continuous imbeddings H m ( Ω ) H 0 m ( Ω ) H a - 1 m ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq5_HTML.gif hold for all a ≤ 1.

      The weighed Sobolev spaces H a m , k ( Q T , γ ) , H a m ( Q T , γ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq6_HTML.gif are defined as sets of all vector-valued functions defined in Q T with respect to the norms
      u H a m , k ( Q T , γ ) = Q T 0 p m ρ ( x ) 2 ( p - a ) D p u 2 + j = 1 k u t j 2 e - γ t d x d t 1 2 < + , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equc_HTML.gif
      and
      u H a m ( Q T , γ ) = Q T 0 p + k m ρ ( x ) 2 ( p + k - a ) D p u t k 2 e - γ t d x d t 1 2 < + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equd_HTML.gif
      Let
      L ( x , t ; D ) = - i , j = 1 n D i ( A i j ( x , t ) D j ) + i = 1 n B i ( x , t ) D i + C ( x , t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Eque_HTML.gif

      be a second-order partial differential operator, where D i = x i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq7_HTML.gif, and A ij , B i , C are s × s matrices of bounded functions with complex values from C ( Q T ¯ ) , A i j = A j i * , A j i * http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq8_HTML.gif is the transposed conjugate matrix of A ji .

      We assume that the operator L is uniformly strong elliptic, that is, there exists a constant C > 0 such that
      i , j = 1 n ( A i j ( x , t ) η η ̄ ) ξ i ξ j C ξ 2 η 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ1_HTML.gif
      (1)

      for all ξ ∈ ℝ n , η ∈ ℂ s and a.e. (x, t) ∈ Q T .

      In this article, we study the following problem:
      u t + L ( x , t ; D ) u = f in Q T , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ2_HTML.gif
      (2)
      u = 0 on S T , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ3_HTML.gif
      (3)
      u t = 0 = 0 in Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ4_HTML.gif
      (4)

      where f(x, t) is given.

      Let us introduce the following bilinear form
      B ( u , v ; t ) = Ω i , j = 1 n ( A i j ( x , t ) D j u D i v ¯ + i = 1 n B i ( x , t ) D i u v ̄ + C ( x , t ) u v ̄ d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equf_HTML.gif
      Then the following Green's formula
      ( L ( x , t ; D ) u , v ) = B ( u , v ; t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equg_HTML.gif

      is valid for all u , v C 0 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq9_HTML.gif and a.e. t ∈ [0, T).

      Definition 1.1. A function u H ˚ 1 , 1 ( Q T , γ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq10_HTML.gif is called a generalized solution of problem (2) -(4) if and only if u|t = 0= 0 and the equality
      ( u t , v ) + B ( u , v ; t ) = ( f , v ) , a . e . t [ 0 , T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ5_HTML.gif
      (5)

      holds for all v H ˚ 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq11_HTML.gif.

      From (1) it follows that there exist constants µ0 > 0, λ0 ≥ 0 such that
      Re B ( u , u ; t ) μ 0 u H 1 ( Ω ) 2 - λ 0 u L 2 ( Ω ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ6_HTML.gif
      (6)
      holds for all u H ˚ 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq12_HTML.gif and t ∈ [0, T). By substituting u = v e - λ 0 t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq13_HTML.gif into (2), we can assume for convenience that λ0 in (6) is zero. Hence, throughout the present paper we also suppose that B(., .; t) satisfies the following inequality
      Re B ( u , u ; t ) μ 0 u H 1 ( Ω ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ7_HTML.gif
      (7)

      for all u H ˚ 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq12_HTML.gif and t ∈ [0, T).

      Now, let us present the main results of this article. Firstly, we give a theorem on well posedness of the problem:

      Theorem 1.1. Let fL2(Q T , γ0), γ0 > 0, and suppose that the coefficients of the operator L satisfy
      sup { A i j , B i , C : i , j = 1 , , n ; ( x , t ) Q ̄ T } μ , μ = const . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equh_HTML.gif
      Then for each γ > γ0, problem (2) -(4) has a unique generalized solution u in the space H ˚ 1 , 1 ( Q T , γ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq14_HTML.gif and the following estimate holds
      u H 1 , 1 ( Q T , γ ) 2 C f L 2 ( Q T , γ 0 ) 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ8_HTML.gif
      (8)

      where C is a constant independent of u and f.

      Write A i j t k = k A i j t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq15_HTML.gif, B i t k = k B i t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq16_HTML.gif, C t k = k C t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq17_HTML.gif. Next, we give results on the smoothness of the solution:

      Theorem 1.2. Let m ∈ ℕ*, γ 0 = 2 μ n μ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq18_HTML.gif, σ = γ - γ0, γ k = (2k + 1)γ0. Assume that the coefficients of L satisfy
      sup { A i j t k , B i t k , C t k : i , j = 1 , , n ; ( x , t ) Q ̄ T , k m + 1 } μ , μ = const . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equi_HTML.gif
      Furthermore,
      f t k H m ( Q T , γ k ) , f o r k = 0 , 1 , 2 ; f t k ( x , 0 ) = 0 , f o r k = 0 , , m - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equj_HTML.gif
      Then there exists η > 0 such that u belongs to H a + 1 2 + m ( Q T , γ 2 + m + σ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq19_HTML.gif for any |a| < η, and
      u H a + 1 2 + m ( Q T , γ 2 + m + σ ) 2 C k = 0 2 f t k H m ( Q T , γ k ) 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ9_HTML.gif
      (9)

      where C is a constant independent of u and f.

      2 The proof of Theorem 1.1

      Firstly, we will prove the existence by Galerkin's approximating method. Let { ω k ( x ) } k = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq20_HTML.gif be an orthogonal basis of H ˚ 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq21_HTML.gif which is orthonormal in L2(Ω). Put
      u N ( x , t ) = k = 1 N C k N ( t ) ω k ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equk_HTML.gif
      where C k N ( t ) , k = 1 , , N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq22_HTML.gif, is the solution of the following ordinary differential system:
      ( u t N , ω k ) + B ( u N , ω k ; t ) = ( f , ω k ) , k = 1 , , N , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ10_HTML.gif
      (10)
      with the initial conditions
      C k N ( 0 ) = 0 , k = 1 , , N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ11_HTML.gif
      (11)
      Let us multiply (10) by C k N ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq23_HTML.gif, then take the sum with respect to k from 1 to N to arrive at
      ( u t N , u N ) + B ( u N , u N ; t ) = ( f , u N ) , t [ 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equl_HTML.gif
      Now adding this equality to its complex conjugate, we get
      d d t u N L 2 ( Ω ) 2 + 2 Re B ( u N , u N ; t ) = 2 Re ( f , u N ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ12_HTML.gif
      (12)
      Utilizing (7), we obtain
      Re B ( u N , u N ; t ) μ 0 u N H 1 ( Ω ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equm_HTML.gif
      By the Cauchy inequality, for an arbitrary positive number ε, we have
      2 ( f , u N ) 2 f L 2 ( Ω ) u N L 2 ( Ω ) C f L 2 ( Ω ) 2 + ε u N L 2 ( Ω ) 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equn_HTML.gif
      where C = C(ε) is a constant independent of u N , f and t. Combining the estimates above, we get from (12) that
      d d t u N ( . , t ) L 2 ( Ω ) 2 + 2 μ 0 u N ( . , t ) H 1 ( Ω ) 2 C f ( . , t ) L 2 ( Ω ) 2 + ε u N ( . , t ) L 2 ( Ω ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ13_HTML.gif
      (13)
      for a.e. t ∈ [0, T). Now write
      η ( t ) : = u N ( . , t ) L 2 ( Ω ) 2 ; ξ ( t ) : = f ( . , t ) L 2 ( Ω ) 2 , t [ 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equo_HTML.gif
      Then (13) implies
      η ( t ) ε . η ( t ) + C ξ ( t ) , for a .e . t [ 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equp_HTML.gif
      Thus the differential form of Gronwall-Belmann's inequality yields the estimate
      η ( t ) C e ε t 0 t ξ ( s ) d s , t [ 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ14_HTML.gif
      (14)
      We obtain from (14) the following estimate:
      u N ( . , t ) L 2 ( Ω ) 2 C e ( ε + γ 0 ) t 0 t e - γ 0 S f L 2 ( Ω ) 2 d s C e ( γ 0 + ε ) t f L 2 ( Q T , γ 0 ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equq_HTML.gif
      Now multiplying both sides of this inequality by e-γt, γ > γ0 + ε, then integrating them with respect to t from 0 to T, we obtain
      u N L 2 ( Q T , γ ) 2 C f L 2 ( Q T , γ 0 ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ15_HTML.gif
      (15)
      Multiplying both sides of (13) by e-γt, then integrating them with respect to t from 0 to τ, τ ∈ (0, T), we obtain
      0 τ e - γ t d d t u N L 2 ( Ω ) 2 d t + 2 μ 0 0 τ e - γ t u N H 1 ( Ω ) 2 d t C ( f L 2 ( Q T , γ 0 ) 2 + u N L 2 ( Q T , γ ) 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equr_HTML.gif
      Notice that
      0 τ e - γ t d d t u N L 2 ( Ω ) 2 d t = 0 τ d d t ( e - γ t u N L 2 ( Ω ) 2 ) d t + γ 0 τ e - γ t u N L 2 ( Ω ) 2 d t = e - γ τ u N ( x , τ ) L 2 ( Ω ) 2 + γ 0 τ e - γ t u N L 2 ( Ω ) 2 d t 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equs_HTML.gif
      We employ the inequalities above to find
      2 μ 0 τ e - γ t u N H 1 ( Ω ) 2 d t C f L 2 ( Q T , γ 0 ) 2 , τ ( 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ16_HTML.gif
      (16)
      Since the right-hand side of (16) is independent of τ, we get
      u N H 1 , 0 ( Q T , γ ) 2 C f L 2 ( Q T , γ 0 ) 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ17_HTML.gif
      (17)

      where C is a constant independent of u, f and N.

      Fix any v H ˚ 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq24_HTML.gif, with v H 1 ( Ω ) 2 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq25_HTML.gif and write v = v1 + v2, where v 1 span { ω k } k = 1 N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq26_HTML.gif and (v2, ω k ) = 0, k = 1, ..., N, ( v 2 span { ω k } k = 1 N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq27_HTML.gif. We have v 1 H 1 ( Ω ) v H 1 ( Ω ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq28_HTML.gif. Utilizing (10), we get
      ( u t N , v 1 ) + B ( u N , v 1 ; t ) = ( f , v 1 ) for a . e . t [ 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equt_HTML.gif
      From u N ( x , t ) = k = 1 N C k N ( t ) ω k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq29_HTML.gif, we can see that
      ( u t N , v ) = ( u t N , v 1 ) = ( f , v 1 ) - B ( u N , v 1 ; t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equu_HTML.gif
      Consequently,
      ( u t N , v ) C f L 2 ( Ω ) 2 + u N H 1 ( Ω ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equv_HTML.gif
      Since this inequality holds for all v H ˚ 1 ( Ω ) , v H 1 ( Ω ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq30_HTML.gif, the following inequality will be inferred
      u t N L 2 ( Ω ) 2 C f L 2 ( Ω ) 2 + u N H 1 ( Ω ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ18_HTML.gif
      (18)
      Multiplying (18) by e-γt, γ > γ0 + ε, then integrating them with respect to t from 0 to T, and by using (17), we obtain
      u t N L 2 ( Q T , γ ) 2 C f L 2 ( Q T , γ 0 ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ19_HTML.gif
      (19)
      Combining (17) and (19), we arrive at
      u N H 1 , 1 ( Q T , γ ) 2 C f L 2 ( Q T , γ 0 ) 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ20_HTML.gif
      (20)

      where C is a constant independent of f and N.

      From the inequality (20), by standard weakly convergent arguments, we can conclude that the sequence { u N } N = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq31_HTML.gif possesses a subsequence weakly converging to a function u H ˚ 1 , 1 ( Q T , γ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq32_HTML.gif, which is a generalized solution of problem (2) -(4). Moreover, it follows from (20) that estimate (8) holds.

      Finally, we will prove the uniqueness of the generalized solution. It suffices to check that problem (2)-(4) has only one generalized solution u ≡ 0 if f ≡ 0. By setting v = u(., t) in identity (5) (for f ≡ 0) and adding it to its complex conjugate, we get
      d d t ( u ( . , t ) 2 ) + 2 Re B ( u , u ; t ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equw_HTML.gif
      From (7), we have
      d d t ( u L 2 ( Ω ) 2 ) + 2 μ 0 u H 1 ( Ω ) 2 0 , for a .e . t [ 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equx_HTML.gif

      Since u|t = 0= 0, it follows from this inequality that u ≡ 0 on Q T . The proof is complete.

      3 The proof of Theorem 1.2

      Firstly, we establish the results on the smoothness of the solution with respect to time variable of the solution which claims that the smoothness depends on the smoothness of the coefficients and the right-hand side of the systems.

      To simplify notation, we write
      B t k ( u , v ; t ) = Ω i , j = 1 n ( A i j t k ( x , t ) D j u D i v ¯ + i = 1 n B i t k ( x , t ) D i u v ̄ + C t k ( x , t ) u v ̄ d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equy_HTML.gif

      Proposition 3.1. Let h ∈ ℕ*. Assume that there exists a positive constant µ such that

      (i) sup A i j t k , B i t k , C t k : i , j = 1 , , n ; ( x , t ) Q ̄ T , k h + 1 μ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq33_HTML.gif,

      (ii) f t k L 2 ( Q T , γ k ) , k h ; f t k ( x , 0 ) = 0 , 0 k h - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq34_HTML.gif.

      Then for an arbitrary real number γ satisfying γ > γ0, the generalized solution u H ˚ m , 1 ( Q T , γ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq35_HTML.gif of problem (2)-(4) has derivatives with respect to t up to order h with u t k H ˚ 1 , 1 ( Q T , γ k + σ ) , k = 0 , , h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq36_HTML.gif, and the estimate
      u t h H 1 , 1 ( Q T , γ h + σ ) 2 C j = 0 h f t j L 2 ( Q T , γ j ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ21_HTML.gif
      (21)

      holds, where C is a constant independent of u and f.

      Proof. From the assumptions on the coefficients of operator L and the function f, it implies that the solution { C k N } k = 1 N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq37_HTML.gif of problem (10)-(11) has derivatives with respect to t up to order h + 1. We will prove by induction that
      u t h N ( . , τ ) H 1 ( Ω ) 2 C e ( γ h + σ 2 ) τ j = 0 h f t j L 2 ( Q T , γ j ) 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ22_HTML.gif
      (22)
      and
      u t h N H 1 , 0 ( Q T , γ h + σ ) 2 C j = 0 h f t j L 2 ( Q T , γ j ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ23_HTML.gif
      (23)
      Firstly, we differentiate h times both sides of (10) with respect to t to find the following equality:
      ( u t h + 1 N , ω k ) + l = 0 h h l B t h - l ( u t l N , ω k ; t ) = ( f t h , ω k ) , k = 1 , , N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ24_HTML.gif
      (24)
      From the equalities above together with the initial condition (11) and assumption (ii), we can show by induction on h that
      u t k N t = 0 = 0 for k = 0 , , h . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ25_HTML.gif
      (25)
      Equality (24) is multiplied by d h + 1 C k N ( t ) d t h + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq38_HTML.gif and sum k = 1, ..., N, so as to discover
      ( u t h + 1 N , u t h + 1 N ) + j = 0 h h j B t h - j ( u t j N , u t h + 1 N ; t ) = ( f t h , u t h + 1 N ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equz_HTML.gif
      Adding this equality to its complex conjugate, we get
      2 u t h + 1 N ( . , t ) L 2 ( Ω ) 2 + 2 Re j = 0 h h j B t h - j ( u t j N , u t h + 1 N ; t ) = 2 Re ( f t h , u t h + 1 N ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ26_HTML.gif
      (26)
      Next, we show that inequalities (22) and (23) hold for h = 0. According to (26) (with h = 0), we have
      2 u t N ( . , t ) L 2 ( Ω ) 2 + 2 Re B ( u N , u t N ; t ) = 2 Re ( f , u t N ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equaa_HTML.gif
      Then the equality is rewritten in the form:
      2 u t N ( . , t ) L 2 ( Ω ) 2 + t B ( u N , u N ; t ) = B t ( u N , u N ; t ) + 2 Re ( f , u t N ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equab_HTML.gif
      Integrating both sides of this equality with respect to t from 0 to τ, τ ∈ (0, T), employing Garding inequality (7) and Cauchy inequality, and by simple calculations, we deduce that
      u N ( . , τ ) H 1 ( Ω ) 2 2 μ n μ 0 0 τ u N ( . , t ) H 1 ( Ω ) 2 d t + 0 τ f ( . , t ) L 2 ( Ω ) 2 d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equac_HTML.gif
      Thus Gronwall-Belmann's inequality yields the estimate
      u N ( . , τ ) H 1 ( Ω ) 2 C e γ 0 τ 0 τ e - γ 0 t f ( . , t ) L 2 ( Ω ) 2 d t C e γ 0 τ f L 2 ( Q T , γ 0 ) 2 , for all τ ( 0 , T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ27_HTML.gif
      (27)
      where γ 0 = 2 μ n μ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq39_HTML.gif. Multiplying both sides of (27) by e ( - γ 0 - σ ) t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq40_HTML.gif, then integrating them with respect to t from 0 to T, we arrive at
      u N H 1 , 0 ( Q T , γ 0 + σ ) 2 C f L 2 ( Q T , γ 0 ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ28_HTML.gif
      (28)

      From inequalities (27) and (28), it is obvious that (22) and (23) hold for h = 0.

      Assume that inequalities (22) and (23) are valid for k = h - 1, we need to prove that they are true for k = h. With regard to equality (26), the second term in left-hand side of (26) is written in the following form:
      2 Re j = 0 h h j B t h - j ( u t j N , u t h + 1 N ; t ) = 2 Re B ( u t h N , u t h + 1 N ; t ) + 2 Re j = 0 h - 1 h j B t h - j ( u t j N , u t h + 1 N ; t ) = t [ B ( u t h N , u t h N ; t ) ] - B t ( u t h N , u t h N ; t ) + 2 Re j = 0 h - 1 h j t B t h - j ( u t j N , u t h N ; t ) - B t h - j ( u t j + 1 N , u t h N ; t ) - B t h - j + 1 ( u t j N , u t h N ; t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equad_HTML.gif
      Hence, from (26) we have
      2 u t h + 1 L 2 ( Ω ) 2 + t [ B ( u t h N , u t h N ; t ) ] - B t ( u t h N , u t h N ; t ) + 2 Re j = 0 h - 1 h j t B t h - j ( u t j N , u t h N ; t ) - B t h - j ( u t j + 1 N , u t h N ; t ) - B t h - j + 1 ( u t j N , u t h N ; t ) = 2 Re ( f t h , u l h + 1 N ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ29_HTML.gif
      (29)
      Integrating both sides of (29) with respect to t from 0 to τ, 0 < τ < T, and using the integration by parts, we find
      2 u t h + 1 L 2 ( Q τ ) 2 + B ( u t h N , u t h N ; τ ) = 0 τ B t ( u t h N , u t h N ; t ) d t - 2 Re j = 0 h - 1 h j B t h - j ( u t j N , u t h N ; τ ) + 2 Re j = 0 h - 1 h j 0 τ B t h - j + 1 ( u t j N , u t h N ; t ) + 2 Re j = 0 h - 1 h j 0 τ B t h - j ( u t j + 1 N , u t h N ; t ) + 2 Re Q τ f t h u t h + 1 N ¯ d x d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ30_HTML.gif
      (30)
      For convenience, we abbreviate by I, II, III, IV, V the terms from the first to the fifth, respectively, of the right-hand side of (30). By using assumption (i) and the Cauchy inequality, we obtain the following estimates:
      ( I ) 2 μ n 0 τ u t h H 1 ( Ω ) 2 d t . ( II ) C ( ε ) j = 0 h - 1 u t j N H 1 ( Ω ) 2 + ε u t h N H 1 ( Ω ) 2 . ( III ) C ( ε ) j = 0 h - 1 0 τ u t j N H 1 ( Ω ) 2 + ε 0 τ u t h N ( x , t ) H 1 ( Ω ) 2 d t . ( IV ) C ( ε ) j = 1 h - 1 0 τ u t j N H 1 ( Ω ) 2 + ε 0 τ u t h N H 1 ( Ω ) 2 d t + 4 μ n h 0 τ u t h N H 1 ( Ω ) 2 d t . ( V ) C ( ε 1 ) Q τ f t h 2 d x d t + ε 1 Q τ u t h + 1 N 2 d x , ( ε 1 < 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equae_HTML.gif
      Employing the estimates above, we get from (30) that
      B ( u t h N , u t h N ; τ ) C Q τ f t h 2 d x d t + C 1 j = 0 h - 1 0 τ u t j N H 1 ( Ω ) 2 d t + 2 ( 2 h + 1 ) μ n 0 τ u t h N H 1 ( Ω ) 2 d t + ε 0 τ u t h N H 1 ( Ω ) 2 d t + ε u t h N H 1 ( Ω ) 2 + C 2 j = 0 h - 1 u t j N H 1 ( Ω ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ31_HTML.gif
      (31)
      By using (7) again, we obtain from (31) the estimate
      u t h N H 1 ( Ω ) 2 C Q τ f t h 2 d x d t + C 1 j = 0 h - 1 0 τ u t j N H 1 ( Ω ) 2 d t + C 2 j = 0 h - 1 u t j N H 1 ( Ω ) 2 + 2 ( 2 h + 1 ) μ n + ε μ 0 - ε 0 τ u t h N H 1 ( Ω ) 2 d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ32_HTML.gif
      (32)
      From (32) and the induction assumptions, we get
      u t h N H 1 ( Ω ) 2 C Q τ f t h 2 d x d t + C 1 j = 0 h - 1 e γ j τ 0 τ e - γ j τ u t j N H 1 ( Ω ) 2 d t + C 2 j = 0 h - 1 e ( γ j + σ 2 ) τ k = 0 j f t k L 2 ( Q T , γ k ) 2 + 2 ( 2 h + 1 ) μ n + ε μ 0 - ε 0 τ u t h N H 1 ( Ω ) 2 d t C Q τ f t h 2 d x d t + C 1 j = 0 h - 1 e γ j τ f t j L 2 ( Q T , γ j ) 2 + C 2 j = 0 h - 1 e ( γ j + σ 2 ) τ k = 0 j f t k L 2 ( Q T , γ k ) 2 + ( γ h + σ 2 ) 0 τ u t h N H 1 ( Ω ) 2 d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ33_HTML.gif
      (33)
      where ε > 0 is chosen such that
      2 ( 2 h + 1 ) μ n + ε μ 0 - ε < 2 ( 2 h + 1 ) μ n μ 0 + σ 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equaf_HTML.gif
      By the Gronwall-Bellmann inequality, we receive from (33) that
      u t h N ( . , τ ) H 1 ( Ω ) 2 C Q τ f t h 2 d x d t + j = 0 h - 1 e ( γ j + σ 2 ) τ f t j L 2 ( Ω T , γ j ) 2 + C e ( γ h + σ 2 ) τ 0 τ e - ( γ h + σ 2 ) t f t h L 2 ( Ω ) 2 + j = 0 h - 1 e ( γ j + σ 2 ) t f t j L 2 ( Q T , γ j ) 2 d t C e ( γ h + σ 2 ) τ j = 0 h f t j L 2 ( Q T , γ j ) 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equag_HTML.gif
      (γ h > γ j , for j = 0, ..., h - 1). Now multiplying both sides of this inequality by e ( - γ h - σ ) τ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq41_HTML.gif, then integrating them with respect to τ from 0 to T, we arrive at
      u t h N H 1 , 0 ( Q T , γ h + σ ) 2 C j = 0 h f t j L 2 ( Q T , γ j ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ34_HTML.gif
      (34)

      It means that the estimates (22) and (23) hold for k = h.

      By the similar arguments in the proof of Theorem 1.1, we obtain the estimate
      u t h N L 2 ( Q T , γ h + σ ) 2 C j = 0 h f t j L 2 ( Q T , γ j ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ35_HTML.gif
      (35)
      Then the combination between (34) and (35) produces the following inequality:
      u t h N H 1 , 1 ( Q T , γ h + σ ) 2 C j = 0 h f t j L 2 ( Q T , γ j ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ36_HTML.gif
      (36)

      Accordingly, by again standard weakly convergent arguments, we can conclude that the sequence { u t k N } N = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq42_HTML.gif possesses a subsequence weakly converging to a function u ( k ) H ˚ 1 , 1 ( Q T , γ k + σ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq43_HTML.gif. Moreover, u(k)is the k th generalized derivative in t of the generalized solution u of problem (2)-(4). Estimate (21) follows from (36) by passing the weak convergences.   □

      Next, by changing problem (2) -(4) into the Dirichlet problem for second order elliptic depending on time parameter, we can apply the results for this problem in polyhedral domains (cf. [4, 5]) and our previous ones to deal with the regularity with respect to both of time and spatial variables of the solution.

      Proposition 3.2. Let the assumptions of Theorem 3.1 be satisfied for a given positive integer h. Then there exists η > 0 such that u t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq44_HTML.gif belongs to H a + 1 2 , 0 ( Q T , γ k + σ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq45_HTML.gif for any |a| < η, k = 0, ..., h and
      k = 0 h u t k H a + 1 2 , 0 ( Q T , γ k + σ ) 2 C k = 0 h f t k L 2 ( Q T , γ k ) 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ37_HTML.gif
      (37)

      where C is a constant independent of u and f.

      Proof. We prove the assertion of the theorem by an induction on h. First, we consider the case h = 0. Equalities (2), (3) can be rewritten in the form:
      L ( x , t ; D ) u = f 1 : = f - u t in Q T , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ38_HTML.gif
      (38)
      u = 0 on S T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ39_HTML.gif
      (39)
      Since u satisfies
      B ( u , v ; t ) = ( f 1 , v ) , v H ˚ 1 ( Ω ) for a .e . t ( 0 , T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equah_HTML.gif
      it is clear that for a.e. t ∈ (0, T), u is the solution of the Dirichlet problem for system (38) with the right-hand side f 1 ( . , t ) = f ( . , t ) - u t ( . , t ) L 2 ( Ω ) H a - 1 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq46_HTML.gif for all a ≤ 1. From Theorem 4.2 in [5] (or Theorem 1.1. in [4]), it implies that there exists η > 0 such that u ( . , t ) H a + 1 2 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq47_HTML.gif for any |a| ≤ η. Furthermore, we have
      u ( . , t ) H a + 1 2 ( Ω ) 2 C f 1 ( . , t ) H a - 1 0 ( Ω ) 2 C f ( . , t ) L 2 ( Ω ) 2 + u ( . , t ) L 2 ( Ω ) 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ40_HTML.gif
      (40)
      where C is a constant independent of u, f and t. Now multiplying both sides of (40) with e - ( γ 0 + σ ) t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq48_HTML.gif, then integrating with respect to t from 0 to T and using estimates from Theorem 3.1, we obtain
      u H a + 1 2 , 1 ( Q T , γ 0 + σ ) 2 C f L 2 ( Q T , γ 0 ) 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equai_HTML.gif
      where C is a constant independent of u, f. Thus, the theorem is valid for h = 0. Suppose that the theorem is true for h - 1; we will prove that this also holds for h. By differentiating h times both sides of (38)-(39) with respect to t, we get
      L ( x , t ; D ) u t h = F : = f t h - u t h + 1 - k = 0 h - 1 h k L t h - k ( x , t ; D ) u t k , in Q T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ41_HTML.gif
      (41)
      u t h = 0 , on S T , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ42_HTML.gif
      (42)
      where
      L t k ( x , t ; D ) = - i , j = 1 n D i ( A i j t k ( x , t ) D j ) + i = 1 n B i t k ( x , t ) D i + C t k ( x , t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equaj_HTML.gif
      By the induction assumption, it implies that
      u t k H a + 1 2 , 1 ( Q T , γ k + σ ) L 2 ( Q T , γ k + σ ) , k = 0 , 1 , , h - 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equak_HTML.gif
      and
      f t h L 2 ( Q T , γ h + 1 ) L 2 ( Q T , γ h ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equal_HTML.gif
      Moreover,
      u t h + 1 L 2 ( Q T , γ h ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equam_HTML.gif
      by Theorem 3.1. Hence, for a.e. t ∈ (0, T), we have F ( . , t ) L 2 ( Ω ) H a - 1 0 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq49_HTML.gif and the estimate
      F ( . , t ) H a - 1 0 ( Ω ) 2 C f t h ( . , t ) L 2 ( Ω ) 2 + u t h + 1 ( . , t ) L 2 ( Ω ) 2 + k = 0 h - 1 u t k ( . , t ) L 2 ( Ω ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ43_HTML.gif
      (43)
      Applying Theorem 4.2 in [5] again, we conclude from (41)-(42) that u t h ( . , t ) H a + 1 2 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq50_HTML.gif and
      u t h ( . , t ) H a + 1 2 ( Ω ) 2 C F ( . , t ) H a - 1 0 ( Ω ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equan_HTML.gif
      From the inequality above and (43), it follows that
      u t h ( . , t ) H a + 1 2 ( Ω ) 2 C f t h ( . , t ) L 2 ( Ω ) 2 + u t h + 1 ( . , t ) L 2 ( Ω ) 2 + k = 0 h - 1 u t k ( . , t ) L 2 ( Ω ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ44_HTML.gif
      (44)
      Multiplying both sides of (44) with e - ( γ h + σ ) t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq51_HTML.gif, then integrating with respect to t from 0 to T and using Theorem 3.1 with a note that γ k < γ h for k = 0, 1, ..., h - 1, we obtain
      u t h H a + 1 2 , 0 ( Q T , γ h + σ ) 2 C k = 0 h f t k L 2 ( Q T , γ k ) 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equao_HTML.gif

      where C is the constant independent of u and f. The proof is completed.   □

      Proof of Theorem 1.2. We will prove the theorem by an induction on m. It is easy to see that
      u H a + 1 2 ( Q T , γ 2 + σ ) 2 k = 0 2 u t k H a + 1 2 - k , 0 ( Q T , γ k + σ ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equap_HTML.gif
      Hence, Proposition 3.2 implies that the theorem is valid for m = 0. Assume that the theorem is true for m - 1, we will prove that it also holds for m. It is only needed to show that
      u t s H a + 1 2 + m - s , 0 ( Q T , γ 2 + m - s + σ ) for s = m , m - 1 , 0 , and u t s H a + 1 2 + m - s ( Q T , γ 2 + m - s + σ ) 2 C k = 0 2 f t k H m ( Q T , γ k ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ45_HTML.gif
      (45)
      Suppose that (45) is true for s = m, m - 1, ..., j + 1, return one more to (41) (h=j), and set v = u t j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq52_HTML.gif, we obtain
      L v = F j , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ46_HTML.gif
      (46)
      where F j = ( f t j - v t ) - ( - 1 ) m k = 1 j j k L t j - k u t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq53_HTML.gif. By the inductive assumption with respect to s, we see that
      v t H a + 1 1 + m - j ( Ω ) H a - 1 m - j ( Ω ) for a .e . t ( 0 , T ) , f t j H m ( Ω ) H a - 1 m - j ( Ω ) , for a .e . t ( 0 , T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equaq_HTML.gif
      and
      L t j - k u t k H m - k ( Ω ) H a - 1 m - j ( Ω ) , k = 1 , , j , for a .e . t ( 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equar_HTML.gif

      Thus, the right-hand side of (46) belongs to H a - 1 m - j ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq54_HTML.gif. Applying Theorem 4.2 in [5] again, we get that v = u t j H a + 1 2 + m - j ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq55_HTML.gif for a.e. t ∈ (0, T). It means that v = u t j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq56_HTML.gif belongs to H a + 1 2 + m - j , 0 ( Q T , γ 2 + m - j + σ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq57_HTML.gif.

      Furthermore, we have
      v H a + 1 2 + m - j , 0 ( Q T , γ 2 + m - j + σ ) 2 C F j H a - 1 h - j , 0 ( Q T , γ 2 + m - j + σ ) C k = 0 2 f t k H m - j ( Q T , γ k ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ47_HTML.gif
      (47)
      Therefore,
      u t j H 1 + a 2 + m - j ( Q T , γ 2 + m - j + σ ) 2 u t j + 1 H 1 + a 2 + m - j - 1 ( Q T , γ 2 + m - j + σ ) 2 + u t j H a + 1 2 + m - j , 0 ( Q T , γ 2 + m - j + σ ) 2 C k = 0 2 f t k H m ( Q T , γ k ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equas_HTML.gif

      It implies that (45) holds for s = j. The proof is complete for j = 0.

      An example. In order to illustrate the results above, we show an example for the case L = -Δ, and Ω is a curvilinear polygonal domain in the plane.

      Denote by A1, A2, ..., A k the vertexes of Ω. Let α j be the opening of the angle at the vertex A j . Set
      K j = { ( x 1 , x 2 ) 2 : r > 0 , 0 < θ < α j } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equat_HTML.gif

      as the angle at vertex A j with sides γ j - : θ = 0 , γ j + : θ = α j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq58_HTML.gif. Here r, θ are the polar coordinates of the point x = (x1, x2), noting that r(x) = ρ(x) is the distance from a point xK j U to the set {A1, A2, .... A k }, where U is a small neighbourhood of A j .

      Let η j = π α j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq59_HTML.gif be the eigenvalue of the pencil U ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq60_HTML.gif (cf. [7]) arises from the Dirichlet problem for Laplace operator via the Mellin transformation rλ. Let η = min{η j }. We consider the Cauchy-Dirichlet problem for the classical heat equation
      u t - Δ u = f in Q T , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ48_HTML.gif
      (48)
      u = 0 on S T , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ49_HTML.gif
      (49)
      u t = 0 = 0 in Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ50_HTML.gif
      (50)

      where f : Q T → ℂ is given.

      Combining Theorem 1.2 and Theorem 4.4 in [5] we receive the following theorem.

      Theorem 3.1. Let Ω ⊂ ℝ2 be a bounded curvilinear polygonal domain in the plane. Furthermore,
      f t k L 2 ( Q T , γ k ) , f o r k = 0 , 1 , 2 ; f t k ( x , 0 ) = 0 , f o r k = 0 , 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equau_HTML.gif
      Then the generalized solution u of problem (48)-(50) belongs to H a + 1 2 ( Q T , γ 2 + σ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq61_HTML.gif for any |a| < η := min η j , as above, and u satisfies the following estimate
      u H a + 1 2 ( Q T , γ 2 + σ ) 2 C k = 0 2 f t k L 2 ( Q T , γ k ) 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_Equ51_HTML.gif
      (51)

      where C is a constant independent of u and f.

      Remark: Let us notice that f ( . , t ) L 2 ( Ω ) = V 0 0 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq62_HTML.gif, u H ˚ 1 ( Ω ) V 0 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq63_HTML.gif, the weighed Sobolev space V β m ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq64_HTML.gif is defined in [[7], p. 191]. Applying Theorem 6.1.4 in [[7], p. 205] with l2 = 2, β2 = 1 - a, l1 = 1, β1 = 0, n = 2 and the strip 0 < Reλ < a < η does not contain any eigenvalue of U ( λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq60_HTML.gif, we obtain u ( . , t ) V 1 - a 2 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq65_HTML.gif. It is easy to see that H a + 1 2 ( Ω ) V 1 - a 2 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-56/MediaObjects/13661_2011_Article_101_IEq66_HTML.gif. Hence, the regularity of the solution of problem (48)-(50) is better than the regularity result, which can obtain from helps of Theorem 6.1.4 in [[7], p. 205].

      Declarations

      Acknowledgements

      This study was supported by the Vietnam's National Foundation for Science and Technology Development (NAFOSTED: 101.01.58.09).

      Authors’ Affiliations

      (1)
      Department of Mathematics, Taybac University
      (2)
      Department of Mathematics, Hongduc University

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      © Luong and Loi; licensee Springer. 2011

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