Heat kernel for the elliptic system of linear elasticity with boundary conditions
aa r X i v : . [ m a t h . A P ] M a r Heat kernel for the elliptic system of linearelasticity with boundary conditions
Justin TaylorDepartment of MathematicsMurray State UniversityMurray, Kentucky 42071, USASeick Kim ∗ Department of Computational Science and EngineeringYonsei UniversitySeoul, 120-749, KoreaRussell Brown † Department of MathematicsUniversity of KentuckyLexington, Kentucky 40506, USAJune 26, 2018
Abstract
We consider the elliptic system of linear elasticity with bounded mea-surable coe ffi cients in a domain where the second Korn inequality holds.We construct heat kernel of the system subject to Dirichlet, Neumann, ormixed boundary condition under the assumption that weak solutions of theelliptic system are H¨older continuous in the interior. Moreover, we showthat if weak solutions of the mixed problem are H¨older continuous up tothe boundary, then the corresponding heat kernel has a Gaussian bound. Inparticular, if the domain is a two dimensional Lipschitz domain satisfyinga corkscrew or non-tangential accessibility condition on the set where wespecify Dirichlet boundary condition, then we show that the heat kernelhas a Gaussian bound. As an application, we construct Green’s functionfor elliptic mixed problem in such a domain. ∗ Seick Kim is supported by NRF Grant No. 2012-040411 and R31-10049 (WCU program). † This work was partially supported by a grant from the Simons Foundation ( Introduction
In a domain (i.e. a connected open set) Ω ⊂ R n ( n ≥ ff erential operator L u = ∂∂ x α A αβ ( x ) ∂ u ∂ x β ! , (1.1)where u is a column vector with components u , . . . , u n , A αβ ( x ) are n × n matrices whose elements a αβ ij ( x ) are bounded measurable functions whichsatisfy the symmetry condition a αβ ij ( x ) = a βα ji ( x ) = a i βα j ( x ) , (1.2)14 κ (cid:12)(cid:12)(cid:12) ξ + ξ T (cid:12)(cid:12)(cid:12) ≤ a αβ ij ( x ) ξ j β ξ i α ≤ κ (cid:12)(cid:12)(cid:12) ξ + ξ T (cid:12)(cid:12)(cid:12) , (1.3)where ξ is an arbitrary n × n matrix with real entries ξ α i , κ , κ >
0, and | ξ | = ( ξ , ξ ) / . Here and below, we will follow the convention that wesum over repeated indices and notation ( ξ , η ) = ξ i α η i α for n × n matrices ξ = ( ξ i α ) and η = ( η i α ). The operator L defined by (1.1) can also be written incoordinate form as follows.( L u ) i = n X α,β = n X j = ∂∂ x α a αβ ij ( x ) ∂ u j ∂ x β ! , i = , . . . , n . Our assumptions on the coe ffi cients a αβ ij will include the equations of linearelasticity and in this case, the coe ffi cients a αβ ij ( x ) are usually referred to as theelasticity tensor; see e.g. [2, 18]. In the classical theory of linear elasticity, theelasticity tensor for a homogeneous isotropic body is given by the formula a αβ ij = λδ i α δ j β + µ ( δ ij δ αβ + δ i β δ j α ) , where λ > µ > δ ij is the Kronecker symbol.In this case, the conditions (1.2) and (1.3) are satisfied with κ = µ and κ = µ + n λ . We define the traction τ = τ ( u ) by the formula τ ( u ) = ν α A αβ ( x ) ∂ u ∂ x β , (1.4)where ν = ( ν , . . . , ν n ) is the unit outward normal to ∂ Ω .We consider following boundary value problems, which are the onesmost frequently considered in the theory of linear elasticity.1. Dirichlet (displacement) problem L u = f in Ω , u = Φ on ∂ Ω . (DP)2. Neumann (traction) problem L u = f in Ω , τ ( u ) = ϕ on ∂ Ω . (NP) . Mixed problem L u = f in Ω , u = Φ on D , τ ( u ) = ϕ on N , (MP)where D and N are subsets of ∂ Ω such that D ∪ N = ∂ Ω and D ∩ N = ∅ .In the above, the equation as well as the boundary condition should beinterpreted in a weak sense; see Section 2 for a precise formulation.In this article, we are concerned with the heat kernel associated withthe mixed problem (MP). By allowing D = ∂ Ω , N = ∅ and D = ∅ , N = ∂ Ω in (MP), we may regard (DP) and (NP) as extreme cases of (MP) and thisis why we focus on (MP). By the heat kernel for (MP), we mean an n × n matrix valued function K ( x , y , t ) satisfying ∂∂ t K ( x , y , t ) − L x K ( x , y , t ) = Ω × (0 , ∞ ) , K ( x , y , t ) = D × (0 , ∞ ) , τ x ( K ( x , y , t )) = N × (0 , ∞ ) , K ( x , y , = δ y I on Ω , (1.5)where δ y ( · ) is Dirac delta function concentrated at y , I is the n × n identitymatrix, and the equation as well as the boundary condition should be inter-preted in some weak sense. The heat kernels for (DP) and (NP) are similarlydefined and they are frequently referred to as Dirichlet and Neumann heatkernels.We assume that Ω is a bounded ( ǫ, δ )-domain of Jones [15] and if D , ∅ and D , ∂ Ω , we assume further that it has a Lipschitz portion; if D = ∂ Ω ,then we require none of these conditions (see H1 in Section 2.5). We provethat if weak solutions of the system L u = L are H¨older continuous if n = ffi cients areuniformly continuous. We also prove that if the gradient of weak solutionsof the system L u = Dirichlet property (see H3 in Section 2.5), then the heat kernel has a Gaussian upper bound;see Theorem 3.10. The Dirichlet property is known to hold in the casewhen Ω is a Lipschitz domain in R and D satisfies the corkscrew condition (see [25]) or when the coe ffi cients and domains are su ffi ciently smooth and¯ D ∩ ¯ N = ∅ (see [9]). As an application, we construct Green’s function for theelliptic system from the heat kernel and in the presence of Dirichlet property,we show that the Green’s function has the usual bound of C | x − y | − n (orlogarithmic bound if n = ff erent method not involving the heat kernel. reviously, Taylor et al. [26] give a construction of the Green functionfor a class of mixed problems for the Laplacian in a Lipschitz domain indimensions two and higher.In recent years, there has been increasing interest in the study of ellipticequations under mixed boundary conditions from a variety of viewpoints.Haller-Dintelmann et al. [14] consider H¨older continuity of solutions ofa single equation with an interest in applications in control. Mazzucatoand Nistor [20] study the mixed problem for the elliptic system of elas-ticity in polyhedral domains under mixed boundary conditions. Finally,a recent monograph of Maz’ya and Rossmann [19] treats mixed problemsfor systems in polyhedral domains. We refer to the above works and theirreferences for background on the study of mixed problems for systems. Itwould be interesting to see if the well-posedness results in these works canbe used to obtain fundamental estimates H2 and H3 that we use to obtainfurther estimates for the heat kernel.The organization of the paper is as follows. In Section 2, we introducesome notation and definitions including the precise definition of the heatkernel for (MP). In Section 3, we state our main theorems (Theorems 3.1and 3.10), which we briefly described above. In Section 4, we construct, asan application, the Green’s function for (MP) and obtain the usual bounds.We give the proofs for our main results in Section 5 and some technicallemmas are proved in Appendix. Throughout the article, we let Ω denote a domain in R n and let D and N befixed subsets of ∂ Ω such that D ∪ N = ∂ Ω and D ∩ N = ∅ . We use X = ( x , t )to denote a point in R n + ; x = ( x , . . . , x n ) will always be a point in R n . Wealso write Y = ( y , s ), X = ( x , t ), and reserve notationˆ Y = ( y , = ( y , . . . , y n , . We define the parabolic distance in R n + by | X − Y | P = max( | x − y | , p | t − s | ) , where | · | denotes the usual Euclidean norm, and write | X | P = | X − | P .We use the following notation for basic cylinders in R n + . Q ( X , r ) = { Y ∈ R n + : | Y − X | P < r } , Q − ( X , r ) = { Y = ( y , s ) ∈ R n + : | Y − X | P < r , s < t } , Q + ( X , r ) = { Y = ( y , s ) ∈ R n + : | Y − X | P < r , s > t } . We also use B ( x , r ) = { y ∈ R n : | y − x | < r } to denote a ball in R n . For thevector valued function u = ( u , . . . , u n ) T , we denote by ε ( u ) (called the straintensor) the matrix whose elements are ε ij ( u ) = ∂ u i ∂ x j + ∂ u j ∂ x i ! . e denote by R the linear space of rigid displacements of R n ; i.e. R = { A x + b : A is real n × n skew-symmetric matrix and b ∈ R n } (2.1)Note that R is a real vector space of dimension N = n ( n + / h u , v i = Z Ω u · v dx = Z Ω v T u dx (2.2)defines an inner product on R . Let us fix an orthonormal basis { ω i } Ni = in R and define the projection operator π R : L ( Ω ) n → R by π R ( u ) = N X i = h u , ω i i ω i . (2.3)The above formula still makes sense for u = δ y e k , where e k is the k th unit(column) vector in R n . For y ∈ Ω , we denote by T y = T y ( x ) an n × n matrixvalued function such that T y e k = π R ( δ y e k ) if D = ∅ ,0 if D , ∅ . (2.4)Roughly speaking, T y is an orthogonal projection of δ y I on R if D = ∅ and0 otherwise. We set B ( u , v ) = a αβ ij ∂ u j ∂ x β ∂ v i ∂ x α . It follows from (1.2) the form B is symmetric (i.e., B ( u , v ) = B ( v , u )) andfrom (1.3) that κ − B ( u , u ) ≤ | ε ( u ) | ≤ κ − B ( u , u ) . (2.5)It is easy to verify that (1.2), (1.3) imply (see [22, Lemma 3.1, p. 30]) a αβ ij ξ j β η i α ≤ κ | ξ + ξ T | | η + η T | ≤ κ | ξ || η | . Finally, we denote d x = d ( x ) = dist( x , Ω c ) when Ω is clear from the contextand write a ∧ b = min( a , b ) and a ∨ b = max( a , b ) for a , b ∈ ¯ R . We use the notation in Gilbarg & Trudinger [13] for the standard functionsspaces defined on U ⊂ R n such as L p ( U ), W k , p ( U ), C k ,α ( ¯ U ), etc. For Γ ⊂ ∂ U ,let W , ( U ; Γ ) be the subspace obtained by taking the closure in W , ( U ) ofsmooth functions in ¯ U which vanish in a neighborhood of Γ . Note that wehave W , ( U ; ∂ U ) = W , ( U ). We shall denote˜ W , ( U ; Γ ) : = W , ( U ; Γ ) if Γ , ∅ n u ∈ W , ( U ): R U u dx = o if Γ = ∅ . For Ω and D as above, we define V : = W , ( Ω ; D ) n if D , ∅ n u ∈ W , ( Ω ) n : h u , v i = , ∀ v ∈ R o if D = ∅ , (2.6) here R and h u , v i are as in (2.1) and (2.2). Notice that V ⊂ ˜ W , ( Ω ; D ) n . For µ ∈ (0 , | u | µ ; U = [ u ] µ ; U + | u | U = sup x , y ∈ Ux , y | u ( x ) − u ( y ) || x − y | µ + sup x ∈ U | u ( x ) | . For spaces of functions defined on Q ⊂ R n + , we borrow notation mainlyfrom Ladyzhenskaya et al. [17]. To avoid confusion, spaces of functionsdefined on Q ⊂ R n + shall be always written in script letters . For q ≥
1, welet L q ( Q ) denote the Banach space consisting of measurable functions on Q that are q -integrable. For Q = Ω × ( a , b ), we denote by L q , r ( Q ) the Banachspace consisting of all measurable functions on Q with a finite norm k u k L q , r ( Q ) = Z ba Z Ω | u ( x , t ) | q dx ! r / q dt / r , where q ≥ r ≥
1. Thus L q , q ( Q ) is the space L q ( Q ). By C µ,µ/ ( ¯ Q )we denote the set of all bounded measurable functions u on Q for which | u | µ,µ/ Q is finite, where we define the parabolic H¨older norm as follows: | u | µ,µ/ Q = [ u ] µ,µ/ Q + | u | Q = sup X , Y ∈ QX , Y | u ( X ) − u ( Y ) || X − Y | µ P + sup X ∈ Q | u ( X ) | , µ ∈ (0 , . We write u ∈ C ∞ c ( Q ) (resp. C ∞ c ( ¯ Q )) if u is an infinitely di ff erentiable functionon R n + with compact support in Q (resp. ¯ Q ). We write D i u = ∂ u /∂ x i ( i = , . . . , n ) and u t = ∂ u /∂ t . We also write Du = D x u = ( D u , . . . , D n u ). Wewrite Q ( t ) for the set of all points ( x , t ) in Q and I ( Q ) for the set of all t suchthat Q ( t ) is nonempty. We denote ||| u ||| Q = Z Q | D x u | dx dt + ess sup t ∈ I ( Q ) Z Q ( t ) | u ( x , t ) | dx . The space W , q ( Q ) denotes the Banach space consisting of functions u ∈ L q ( Q ) with weak derivatives D i u ∈ L q ( Q ) ( i = , . . . , n ) with the norm k u k W , q ( Q ) = k u k L q ( Q ) + k D x u k L q ( Q ) and by W , q ( Q ) the Banach space with the norm k u k W , q ( Q ) = k u k L q ( Q ) + k D x u k L q ( Q ) + k u t k L q ( Q ) . In the case when Q has a finite height (i.e., Q ⊂ R n × ( − T , T ) for some T < ∞ ),we define V ( Q ) as the Banach space consisting of all elements of W , ( Q )having a finite norm k u k V ( Q ) : = ||| u ||| Q and the space V , ( Q ) is obtained bycompleting the set W , ( Q ) in the norm of V ( Q ). When Q does not havea finite height, we say that u ∈ V ( Q ) (resp. V , ( Q )) if u ∈ V ( Q T ) (resp. V , ( Q T )) for all T >
0, where Q T = Q ∩{| t | < T } , and ||| u ||| Q < ∞ . Note that thisdefinition allows that 1 ∈ V , ( Ω × (0 , ∞ )). Finally, we write u ∈ L q , loc ( Q ) if u ∈ L q ( Q ′ ) for all Q ′ ⋐ Q and similarly define W , q , loc ( Q ), etc. .3 Weak solutions For f , g α ∈ L ( U ) n , where α = , . . . , n , we say that u is a weak solution of L u = f + D α g α in U if u ∈ W , ( U ) n and for any v ∈ W , ( U ) n satisfies theidentity Z U B ( u , v ) = − Z U f · v + Z U g α · D α v . (2.7)Let Γ D , Γ N be disjoint subsets of ∂ U . Recall that the traction τ ( u ) is definedby the formula (1.4). We say that u is a weak solution of L u = f + D α g α in U , u = Γ D , τ ( u ) = g α ν α on Γ N if u ∈ W , ( U ; Γ D ) n and for any v ∈ W , ( U ; ∂ U \ Γ N ) n satisfies the identity(2.7). Let Ω , D , N be as above. For f ∈ L ( Ω ) n , we say that u is a weaksolution of the mixed problem L u = f in Ω , u = D , τ ( u ) = N if u ∈ W , ( Ω ; D ) n and satisfies for any v ∈ W , ( Ω ; D ) n the identity Z Ω B ( u , v ) = − Z Ω f · v . Let Q be a cylinder U × ( a , b ), where U ⊂ R n and −∞ < a < b < ∞ . For f ∈ L , ( Q ) n and g α ∈ L ( Q ) n , we say that u is a weak solution of u t − L u = f + D α g α if u ∈ V ( Q ) n and satisfies for all φ ∈ C ∞ c ( Q ) n the identity − Z Q u · φ t + Z Q B ( u , φ ) = Z Q f · φ − Z Q g α · D α φ . (2.8)Let Γ D , Γ N be disjoint subsets of ∂ U . We say that u is a weak solution of u t − L u = f + D α g α in U × ( a , b ) = : Q u = Γ D × ( a , b ) τ ( u ) = − g α ν α on Γ N × ( a , b ) = : S if u ∈ V ( Q ) n , u ( · , t ) ∈ W , ( U ; Γ D ) n for a.e. t ∈ ( a , b ), and it satisfies theidentity (2.8) for all φ ∈ C ∞ c ( Q ∪ S ) n . Next, denote Q = Ω × ( a , b ), and let f ∈ L , ( Q ) n , g α ∈ L ( Q ) n , and ψ ∈ L ( Ω ) n . By a weak solution in V ( Q ) n (resp. V , ( Q ) n ) of the problem u t − L u = f + D α g α in Ω × ( a , b ) u = D × ( a , b ) τ ( u ) = − g α ν α on N × ( a , b ) u ( · , a ) = ψ on Ω , (2.9)we mean u ( x , t ) in V ( Q ) n (resp. V , ( Q ) n ) such that u ( · , t ) ∈ W , ( Ω ; D ) n fora.e. t ∈ ( a , b ) and satisfying the identity − Z Q u · φ t + Z Q B ( u , φ ) − Z Q f · φ + Z Q g α · D α φ = Z Ω ψ ( x ) · φ ( x , a ) dx or any φ ( x , t ) ∈ C ∞ c ( ¯ Q ) n that vanishes on D × ( a , b ) and equals to zero for t = a . We may identify f ∈ L , ( Q ) n as an element in L ( a , b ; V ′ ) in the sense[ f ( · , t )]( v ) = Z Ω f ( x , t ) · v ( x ) dx , v ∈ V , and consider the problem u t − L u = f in Ω × ( a , b ) u = D × ( a , b ) τ ( u ) = N × ( a , b ) u ( · , a ) = ψ on Ω u ( · , t ) ∈ V for a.e. t ∈ ( a , b ) . (2.10)In the above, we impose the compatibility condition for ψ that h ψ , v i = v ∈ R in the case when D = ∅ . We shall say that u is a weak solutionof the problem (2.10) if u ∈ V , ( Q ) n , u ( · , t ) ∈ V for a.e. t ∈ ( a , b ), and satisfiesthe identity − Z Q u · φ t + Z Q B ( u , φ ) − Z Q f · φ = Z Ω ψ ( x ) · φ ( x , a ) dx for any φ ( x , t ) ∈ C ∞ c ( ¯ Q ) n that vanishes on D × ( a , b ), satisfies h φ ( · , t ) , v i = t ∈ [ a , b ] and any v ∈ R , and equals to zero for t = a . Note that if D , ∅ , then a weak solution of the problem (2.10) is also a weak solutionin V , ( Q ) n of the problem (2.9) with g α = α = , . . . , n , and viceversa. However, when D = ∅ , they are not the same in general; note that if f ( · , t ) − ˜ f ( · , t ) ∈ R , then they are the same as elements in L ( a , b ; V ′ ). We say that an n × n matrix valued function K ( x , y , t ), with measurableentries K ij : Ω × Ω × [0 , ∞ ) → ¯ R , is the heat kernel for (MP) if it satisfies thefollowing properties, where we denote Q = Ω × [0 , ∞ ).a) For all y ∈ Ω , elements of K ( · , y , · ) belong to W , , loc ( Q ) ∩ V ( Q \ Q + ( ˆ Y , r ))for any r > y ∈ Ω , K ( · , y , · ) is a generalized solution of the problem (1.5)in the sense that K ( · , y , t ) ∈ W , ( Ω ; D ) n for a.e. t > φ = ( φ , . . . , φ n ) T ∈ C ∞ c ( ¯ Q ) n that vanishes on N × [0 , ∞ ), we have theidentity − Z Q K ik ( x , y , t ) ∂∂ t φ i ( x , t ) dx dt + Z Q a αβ ij ∂∂ x β K jk ( x , y , t ) ∂∂ x α φ i ( x , t ) dx dt = φ k ( y , . (2.11)c) For any f = ( f , . . . , f n ) T ∈ C ∞ c ( ¯ Q ) n , the function u given by u ( x , t ) : = Z t Z Ω K ( y , x , t − s ) T f ( y , s ) dy ds s, for any T >
0, a unique weak solution in V , ( Ω × (0 , T )) n of theproblem u t − L u = f in Ω × (0 , T ) , u = D × (0 , T ) , τ ( u ) = N × (0 , T ) , u ( · , = Ω . (2.12)We note that part c) of the above definition gives the uniqueness of the heatkernel for (MP). H1.
We assume Ω ⊂ R n , n ≥ D = ∂ Ω , we donot make any further assumption. Otherwise, we assume the ( ǫ, δ )-condition of Jones [15] for some ǫ, δ >
0: For any x , y ∈ Ω such that | x − y | < δ , there is a rectifiable arc γ joining x to y and satisfying l ( γ ) ≤ ǫ | x − y | , d ( z ) ≥ ǫ | x − z || y − x || x − y | for all z on γ, where l ( γ ) denotes the arc length of γ and d ( z ) is the distance from zto the complement of Ω . If D , ∂ Ω and D , ∅ , we assume furtherthat D has a portion of Lipschitz boundary; i.e. there exist x ∈ D and a neighborhood V of x in R n and new orthogonal coordinates { y , . . . , y n } such that V is a hypercube in the new coordinates: V = { ( y , . . . , y n ) : − a j < y j < a j , ≤ j ≤ n } ;there exists a Lipschitz continuous function ϕ defined in V ′ = { ( y , . . . , y n − ) : − a j < y j < a j , ≤ j ≤ n − } ;and such that | ϕ ( y ′ ) | ≤ a n / y ′ = ( y , . . . , y n − ) ∈ V ′ , Ω ∩ V = { y = ( y ′ , y n ) ∈ V : y n < ϕ ( y ′ ) } , D ∩ V = { y = ( y ′ , y n ) ∈ V : y n = ϕ ( y ′ ) } . In other words, in a neighborhood V of x , Ω is below the graph of ϕ and D is the graph of ϕ .Basically, we introduce the assumption H1 is to guarantee the multiplicativeinequality (2.15) and the second Korn inequality (2.18) are available to us.We recall that the following multiplicative inequality holds for any u in W , ( R n ) with n ≥
1; see [17, Theorem 2.2, p. 62]. k u k L n + / n ( R n ) ≤ c ( n ) k Du k n / ( n + L ( R n ) k u k / ( n + L ( R n ) . (2.13)If we assume H1, then there is an extension operator E : W , ( Ω ) → W , ( R n )such that the following holds; see [24, Theorem 8]. k Eu k L ( R n ) ≤ C k u k L ( Ω ) , k Eu k W , ( R n ) ≤ C k u k W , ( Ω ) . (2.14) hen by combining (2.14) and (2.13), for any u ∈ ˜ W , ( Ω ; D ), we obtain k u k L n + / n ( Ω ) ≤ C k D ( Eu ) k n / ( n + L ( R n ) k Eu k / ( n + L ( R n ) ≤ C k u k n / ( n + W , ( Ω ) k u k / ( n + L ( Ω ) ≤ C k Du k n / ( n + L ( Ω ) k u k / ( n + L ( Ω ) , where in the last step we used H1 to apply the Friedrichs inequality (orPoincar´e’s inequality if D = ∅ ): k u k L ( Ω ) ≤ C k Du k L ( Ω ) , ∀ u ∈ ˜ W , ( Ω ; D ) . We have proved that H1 implies that there is γ = γ ( n , Ω , D ) such that forany u ∈ ˜ W , ( Ω ; D ), we have k u k L n + / n ( Ω ) ≤ γ k Du k n / ( n + L ( Ω ) k u k / ( n + L ( Ω ) . (2.15)If u ∈ V ( Ω × ( a , b )) is such that u ( · , t ) ∈ ˜ W , ( Ω ; D ) for a.e. t ∈ ( a , b ), where −∞ ≤ a < b ≤ ∞ , then by (2.15) we have (see [17, pp. 74–75]) k u k L n + / n ( Ω × ( a , b )) ≤ γ ||| u ||| Ω × ( a , b ) . (2.16)Another important consequence of the inequality (2.15) is the following:For any u ∈ V ( Ω × ( a , b )) with b − a < ∞ , we have k u k L n + / n ( Ω × ( a , b )) ≤ (cid:16) γ + ( b − a ) n / | Ω | − (cid:17) n + ||| u ||| Ω × ( a , b ) . (2.17)We refer to [17, Eq. (3.8), p. 77] for the proof of (2.17). Moreover, H1 impliesthe following second Korn inequality: (see [8] for the proof) k u k W , ( Ω ) ≤ C n k u k L ( Ω ) + k ε ( u ) k L ( Ω ) o . (2.18)In fact, if u ∈ W , ( Ω ; ∂ Ω ) n = W , ( Ω ) n , we have the first Korn inequality k D u k L ( Ω ) ≤ k ε ( u ) k L ( Ω ) . Also, we have the following inequalities for any u ∈ V : k u k W , ( Ω ) ≤ C k ε ( u ) k L ( Ω ) . (2.19)The inequality (2.19) is obtained by utilizing (2.18) in the proof of [22,Theorem 2.7, p. 21]. By (2.19) and (2.5), for any u ∈ V , we have Z Ω B ( u , u ) dx ≥ c Z Ω | D u | dx . (2.20) Lemma 2.21.
Assume H1 and let ψ ∈ L ( Ω ) n and f ∈ L , ( Q ) n , where Q =Ω × ( a , b ) and −∞ < a < b < ∞ . Then, there exists a unique weak solution u in V , ( Q ) n of the problem (2.9) . Moreover, if we assume that h ψ , v i = for all v ∈ R in the case when D = ∅ , then there also exists a unique weak solution of theproblem (2.10) . If k f k L n + / ( n + ( Q ) < ∞ , then the weak solution u of the problem (2.10) satisfies an energy inequality ||| u ||| Ω × ( a , b ) ≤ C n k f k L n + / ( n + ( Q ) + k ψ k L ( Ω ) o , (2.22) where C depends only on n , κ , κ and the constants appearing in (2.16) and (2.19) . roof. With the aid of the second Korn inequalities (2.18) or (2.19), it followsfrom the standard Galerkin’s method and the energy inequality. (cid:4)
H2.
There exist µ ∈ (0 ,
1] and A > u is a weak solution of L u = B = B ( x , r ), where x ∈ Ω and 0 < r ≤ d ( x ), then u is H¨oldercontinuous in B = B ( x , r /
2) with an estimate[ u ] µ ; B ≤ A r − µ ? B | u ( y ) | dy ! / . (2.23)Here, we use the notation > B u dx = | B | R B u dx .It follows from H2 that a weak solution of u t − L u is also locally H¨oldercontinuous. Lemma 2.24.
H2 implies that there exist µ ∈ (0 , µ ) and A > such thatwhenever u is a weak solution in V ( Q ) n of u t − L u = in Q = Q − ( X , r ) , whereX = ( x , t ) ∈ Q and < r ≤ d ( x ) , u is H¨older continuous in Q = Q − ( X , r / and we have an estimate [ u ] µ ,µ / Q ≤ A r − µ − ( n + / k u k L ( Q ) . Proof.
With the second Korn inequality available to us, the proof is essen-tially the same as that of [16, Theorem 3.3]. (cid:4)
Finally, we introduce a condition that was originally considered byAuscher and Tchamitchian [1] and is referred to as the
Dirichlet property . H3.
There exist µ ∈ (0 ,
1] and A > u is a weak solution of L u = B ∩ Ω , u = B ∩ D , τ ( u ) = B ∩ N , where B = B ( x , r ) with x ∈ Ω and 0 < r ≤ diam Ω , then for any0 < ρ ≤ r , we have Z B ( x ,ρ ) ∩ Ω | D u | dx ≤ A (cid:18) ρ r (cid:19) n − + µ Z B ( x , r ) ∩ Ω | D u | dx . (2.25)The following lemma says that H3 implies H2. Moreover, it shows that ifthere is β > x ∈ Ω and 0 < r ≤ diam Ω , we have | Ω ∩ B ( x , r ) | ≥ β r n , (2.26)then, weak solutions of u t − L u = ǫ, δ )-domains satisfy condition (2.26), so that domains that satisfy H1 also satisfy(2.26). Lemma 2.27.
Let Q = Ω × [0 , ∞ ) , D = D × [0 , ∞ ) , and N = N × [0 , ∞ ) . If Ω satisfies the condition (2.26) , then H3 implies that there exist µ ∈ (0 , µ ) andA > such that if u is a weak solution in V ( Q ∩ Q ) n of u t − L u = in Q ∩ Q , u = on Q ∩ D , τ ( u ) = on Q ∩ N , (2.28) here Q = Q − ( X , r ) with X = ( x , t ) ∈ Q and < r ≤ √ t ∧ diam Ω , then u isH¨older continuous in Q ∩ Q , where Q = Q − ( X , r / , and we have an estimater µ [ u ] µ ,µ / Q ∩Q + | u | Q ∩Q ≤ A r − ( n + / k u k L ( Q ∩Q ) . (2.29) Proof.
See Appendix 6.1. (cid:4)
Theorem 3.1.
Assume the conditions H1 and H2. Then there exists a uniqueheat kernel K ( x , y , t ) for (MP) . It satisfies the symmetry relation K ( x , y , t ) = K ( y , x , t ) T (3.2) and thus by (2.12) , for any f ∈ C ∞ c ( ¯ Ω × [0 , ∞ )) n , we have u ( x , t ) = Z t Z Ω K ( x , y , t − s ) f ( y , s ) dy ds (3.3) is, for any T > , a unique weak solution in V , ( Ω × (0 , T )) n of the problem (2.12) .Also, for ψ ∈ L ( Ω ) n , the function u given by u ( x , t ) = Z Ω K ( x , y , t ) ψ ( y ) dy (3.4) is, for any T > , a unique weak solution in V , ( Ω × (0 , T )) n of the problem u t − L u = in Ω × (0 , T ) , u = on D × (0 , T ) , τ ( u ) = on N × (0 , T ) , u ( · , = ψ on Ω (3.5) and if ψ is continuous at x ∈ Ω in addition, then lim ( x , t ) → ( x , x ∈ Ω , t > Z Ω K ( x , y , t ) ψ ( y ) dy = ψ ( x ) . (3.6) Moreover, the following estimates holds for all y ∈ Ω , where we use notation Q = Ω × [0 , ∞ ) , d y = d ( y ) , and ˆ Y = ( y , .1) k K ( · , y , · ) k L p ( Q + ( ˆ Y , r )) ≤ C p r − n + ( n + / p , ∀ r ∈ (0 , d y ] , ∀ p ∈ h , n + n (cid:17) .2) |{ ( x , t ) ∈ Q : | K ( x , y , t ) | > λ }| ≤ C λ − ( n + / n , ∀ λ > d − ny .3) k D x K ( · , y , · ) k L p ( Q + ( ˆ Y , r )) ≤ C p r − n − + ( n + / p , ∀ r ∈ (0 , d y ] , ∀ p ∈ h , n + n + (cid:17) .4) |{ ( x , t ) ∈ Q : | D x K ( x , y , t ) | > λ }| ≤ C λ − n + n + , ∀ λ > d − n − y .
5) For X = ( x , t ) ∈ Q satisfying | X − ˆ Y | P < d y / , we have | K ( x , y , t ) | ≤ C | X − ˆ Y | − n P . (3.7) ) For X = ( x , t ) and X ′ = ( x ′ , t ′ ) in Q satisfying | X ′ − X | P < | X − ˆ Y | P < d y / , we have | K ( x ′ , y , t ′ ) − K ( x , y , t ) | ≤ C | X ′ − X | µ P | X − ˆ Y | − n − µ P , (3.8) In the above, C are constants depending only on n , κ , κ , Ω , D , µ , A and C p depend on p in addition.Remark . It will be clear from the proof that besides the estimates 1) - 6)in Theorem 3.1, we also have k ˜ K ( · , y , · ) k L n + / n ( Q\ Q + ( ˆ Y , r )) ≤ Cr − n / , ∀ r ∈ (0 , d y ]. ||| ˜ K ( · , y , · ) ||| Q\ Q + ( ˆ Y , r ) ≤ Cr − n / , ∀ r ∈ (0 , d y ].Here, we use the notation˜ K ( x , y , t ) = K ( x , y , t ) if D , ∅ , K ( x , y , t ) − π R ( δ y I )( x ) if D = ∅ .where π R is as defined in (2.3); see also (5.28). Moreover, if Ω is such that itadmits a bounded linear trace operator from W , ( Ω ) to L ( ∂ Ω ), then it canbe shown that for f ∈ L q , r ( Ω × (0 , T )) n and g ∈ L q , r ( N × (0 , T )) n , where q k and r k ( k = ,
2) are subject to the conditions of [17, Theorem 5.1, p. 170],the function u defined by u ( x , t ) = Z t Z Ω K ( x , y , t − s ) f ( y , s ) dy ds + Z t Z Ω K ( x , y , t − s ) g ( y , s ) dS y ds is a unique weak solution in V , ( Ω × (0 , T )) n of the problem L u = f in Ω × (0 , T ) u = D × (0 , T ) τ ( u ) = g on N × (0 , T ) u ( · , = Ω . The proof is similar to that of the representation formula (3.3) and is omitted.
Theorem 3.10.
Assume H1 and H3. There exists the heat kernel K ( x , y , t ) for (MP) and it satisfies all the properties stated in Theorem 3.1. Moreover, for x , y ∈ Ω and t > , we have the Gaussian bound | K ( x , y , t ) | ≤ C (cid:16) √ t ∧ diam Ω (cid:17) n exp ( − ϑ | x − y | t ) , (3.11) where C = C ( n , κ , κ , Ω , D , µ , A ) and ϑ = ϑ ( κ , κ ) > . Furthermore, forX = ( x , t ) and X ′ = ( x ′ , t ′ ) in Q satisfying | X ′ − X | P < (cid:16) | X − ˆ Y | P ∧ diam Ω (cid:17) ,we have | K ( x ′ , y , t ′ ) − K ( x , y , t ) | ≤ C ( | X ′ − X | P | X − ˆ Y | P ∧ diam Ω ) µ × (cid:16) √ t ∧ diam Ω (cid:17) n exp ( − ϑ | x − y | t ) , (3.12) where µ is as in Lemma 2.27. Applications
1. If the coe ffi cients are constant, then it is well known that H2 holdswith µ = A = A ( n , κ , κ ). In fact, it is also known that ifthe coe ffi cients belong to the VMO class, then H2 holds with µ and A depending on the BMO modulus of the coe ffi cients as well as on n , κ , κ . Therefore, the conclusions of Theorem 3.1 are valid in thesecases.2. If the coe ffi cients belong to the VMO class, the domain Ω is of class C , and ¯ D ∩ ¯ N = ∅ , then it is known that H3 holds. Therefore, theconclusions of Theorem 3.10 are valid in this case.3. If n =
2, then it is well known that H2 holds with µ = µ ( κ , κ )and A = A ( κ , κ ). Therefore, the conclusions of Theorem 3.1 arevalid. In fact, if Ω is a Lipschitz domain and D is a (possibly empty)set satisfying the corkscrew condition , i.e., for each x ∈ ∂ D (where theboundary is taken with respect to ∂ Ω ) and r ∈ (0 , r ), we may find x r ∈ D so that | x − x r | ≤ r and dist( x r , ∂ Ω \ D ) ≥ M − r , where r > M > We say that an n × n matrix valued function G ( x , y ) is the Green’s functionof L for (MP) if it satisfies the following properties:i) G ( · , y ) ∈ W , loc ( Ω ) and G ( · , y ) ∈ W , ( Ω \ B ( y , r )) for all y ∈ Ω and r > D = ∅ , we require R Ω v ( x ) T G ( x , y ) dx = v ∈ R .ii) G ( · , y ) is a weak solution of − L G ( · , y ) = δ y I − T y in Ω , G ( · , y ) = D , τ ( G ( · , y )) = N in the sense that we have the identity Z Ω a αβ ij ∂∂ x β G jk ( · , y ) ∂φ i ∂ x α dx = φ k ( y ) . for any φ = ( φ , . . . , φ n ) T ∈ C ∞ ( ¯ Ω ) n ∩ V ; see (2.4) and (2.6) for thedefinition of T y and V .iii) For any f ∈ C ∞ c ( ¯ Ω ) n ∩ V , the function u defined by u ( x ) = Z Ω G ( y , x ) T f ( y ) dy (4.1)is the weak solution in V of the problem L u = f in Ω , u = D , τ ( u ) = N . (4.2) e note that with aid of the second Korn inequality (2.19), which is validfor any u ∈ V , the unique solvability of the problem (4.2) in the space V isan immediate consequence of Lax-Milgram lemma. We also note that theproperty iii) of the above definition together with the requirement Z Ω v ( x ) T G ( x , y ) dx = , ∀ v ∈ R gives the uniqueness of the Green’s function. Theorem 4.3.
Assume the conditions H1 and H2. Then, there exists a uniqueGreen’s function G ( x , y ) for (MP) . We have G ( y , x ) = G ( x , y ) T (4.4) and thus by (4.1) , for any f ∈ C ∞ ( ¯ Ω ) n ∩ V , we find u ( x ) = Z Ω G ( x , y ) f ( y ) dyis a unique weak solution in V of the problem (4.2) . If we assume H3 instead ofH2, then for any x , y ∈ Ω , we havei) n = | G ( x , y ) | ≤ C ( + ln diam Ω | x − y | !) , (4.5) ii) n ≥ | G ( x , y ) | ≤ C | x − y | − n , (4.6) and moreover, for any x , y ∈ Ω satisfying | x − x ′ | < | x − y | , we have | G ( x ′ , y ) − G ( x , y ) | ≤ C | x ′ − x | µ | x − y | − n − µ . (4.7) In the above, C is a constant depending on the prescribed parameters and diam Ω and µ ∈ (0 , is as in Lemma 2.27. Corollary 4.8.
Let n = and assume H1. Then, there exists the Green’s functionfor (MP) that satisfies (4.4) . Moreover, if Ω is a Lipschitz domain and D satisfiesthe corkscrew condition described in Section 4.1, then the estimates (4.5) , (4.7) hold, and µ and C are constants determined by κ , κ , Ω , and D .Proof. Follows from Example 3 in Section 4.1 and Theorem 4.3. (cid:4)
In the proof, we denote by C a constant depending on the prescribed pa-rameters n , κ , κ , µ , A as well as on Ω and D ; if it depends also on someother parameters such as p , it will be written as C p , etc.We fix a Φ ∈ C ∞ c ( R n ) such that Φ is supported in B (0 , ≤ Φ ≤
2, and R R n Φ =
1. Let y ∈ Ω be fixed but arbitrary. For ǫ >
0, we define Φ y ,ǫ ( x ) = ǫ − n Φ (( x − y ) /ǫ ) nd let v ǫ = v ǫ, y , k be a unique weak solution in V , ( Ω × (0 , T )) n of theproblem u t − L u = Ω × (0 , T ) , u = D × (0 , T ) , τ ( u ) = N × (0 , T ) , u ( · , = Φ y ,ǫ e k on Ω , (5.1)where e k is the k -th unit column vector in R n ; see Lemma 2.21. By theuniqueness, we find that v ǫ does not depend on a particular choice of T andthus by setting v ǫ ( x , t ) = t < T → ∞ , we may assumethe v ǫ is defined on the entire Ω × ( −∞ , ∞ ). We define the mollified heatkernel K ǫ ( x , y , t ) to be an n × n matrix valued function whose k -th column is v ǫ, y , k ( x , t ); i.e., K ǫ jk ( x , y , t ) = v j ǫ ( x , t ) = v j ǫ, y , k ( x , t ) . For f ∈ C ∞ c ( ¯ Ω × ( −∞ , ∞ )) n , fix a , b so that a < < b and supp f ⊂ ¯ Ω × ( a , b ).Let u be a weak solution in V , ( Ω × ( a , b )) n of the backward problem − u t − L u = f in Ω × ( a , b ) u = D × ( a , b ) τ ( u ) = N × ( a , b ) u ( · , b ) = Ω . Then, it is easy to see that we have Z Ω Φ y ,ǫ ( x ) u k ( x , dx = Z b Z Ω K ǫ ik ( x , y , t ) f i ( x , t ) dx dt . (5.2)Next, we define ˜ K ǫ ( x , y , t ) by˜ K ǫ ( x , y , t ) : = K ǫ ( x , y , t ) if D , ∅ , K ǫ ( x , y , t ) − [0 , ∞ ) ( t ) π R ( Φ y ,ǫ I )( x ) if D = ∅ , (5.3)where π R ( Φ y ,ǫ I )( x ) is an n × n matrix whose k -th column is π R ( Φ y ,ǫ e k )( x ).We set ˜ v ǫ = ˜ v ǫ, y , k to be the k -th column of ˜ K ǫ ( · , y , · ). It is easy to verify that˜ v ǫ ( · , t ) ∈ V for a.e. t >
0. Therefore, for any T >
0, it is the weak solution ofthe problem (see Section 2.3 and Lemma 2.21) u t − L u = Ω × (0 , T ) u = D × (0 , T ) τ ( u ) = N × (0 , T ) u ( · , = ψ ǫ, y , k on Ω u ( · , t ) ∈ V for a.e. t ∈ (0 , T ) , (5.4)where we denote ψ ǫ, y , k = Φ ǫ e k if D , ∅ , Φ y ,ǫ e k − π R ( Φ y ,ǫ e k ) if D = ∅ . By the energy inequality, we get (see Lemma 2.21) ||| ˜ v ǫ ||| Ω × ( −∞ , ∞ ) ≤ C k ψ ǫ, y , k k L ( Ω ) ≤ C ǫ − n / . (5.5) et f be a smooth function supported in Q + ( X , R ) ⊂ Ω × ( −∞ , ∞ ). Fix b > t + R and let ˜ u be the weak solution of the backward problem − u t − L u = f in Ω × ( a , b ) u = D × ( a , b ) τ ( u ) = N × ( a , b ) u ( · , b ) = Ω u ( · , t ) ∈ V for a.e. t ∈ ( a , b ) . (5.6)The unique solvability of the above problem is similar to Lemma 2.21 andby setting ˜ u ( x , t ) = t > b and letting a → −∞ , we may again assumethat ˜ u is defined on Ω × ( −∞ , ∞ ). Then, similar to (2.22), we have ||| ˜ u ||| Ω × ( −∞ , ∞ ) ≤ C k f k L n + / ( n + ( Q + ( X , R )) . (5.7)By using H¨older’s inequality, we derive from (5.7) that k ˜ u k L ( Q + ( X , R )) ≤ CR + n / k f k L ∞ ( Q + ( X , R )) . Then by Lemma 5.9 below and the above estimate, we obtain | ˜ u | Q + ( X , R / ≤ CR k f k L ∞ ( Q + ( X , R )) . (5.8) Lemma 5.9.
H2 implies that ˜ u is continuous in Q + ( X , R / and satisfies theestimate | ˜ u | Q ≤ C (cid:16) R − ( n + / k ˜ u k L ( Q ) + R k f k L ∞ ( Q ) (cid:17) , (5.10) where we denote α Q = Q + ( X , α R ) . In fact, the same conclusion is true if ˜ u is aweak solution in V ( Q ) n of − u t − L u = f (or u t − L u = f ) with f ∈ L ∞ ( Q ) n .Proof. See Appendix 6.2. (cid:4)
Note that similar to (5.2), we have the identity Z Ω Φ y ,ǫ ( x ) ˜ u k ( x , dx = Z ∞−∞ Z Ω ˜ K ǫ ik ( · , y , · ) f i dX . (5.11)If B ( y , ǫ ) × { } ⊂ Q + ( X , R / (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) " Q + ( X , R ) ˜ K ǫ ik ( · , y , · ) f i dX (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ | ˜ u | Q + ( X , R / ≤ CR k f k L ∞ ( Q + ( X , R )) . Therefore, by duality, it follows that we have k ˜ K ǫ ( · , y , · ) k L ( Q + ( X , R )) ≤ CR (5.12)provided 0 < R < d y and B ( y , ǫ ) × { } ⊂ Q + ( X , R / X ∈ Q such that0 < d : = | X − ˆ Y | P < d y /
6, if we set r = d / X = ( y , − d ), and R = d , thenit is easy to see that for ǫ < d /
3, we have B ( y , ǫ ) × { } ⊂ Q + ( X , R / , Q − ( X , r ) ⊂ Q + ( X , R ) , and also that ˜ v ǫ = ˜ v ǫ, y , k is a weak solution in V ( Q − ( X , r )) of u t − L u = emma 5.13. H2 implies that for any p > , we have | u | Q − ( X , r / ≤ C p r − ( n + / p k u k L p ( Q − ( X , r )) . Proof.
It follows from the estimate (5.10) in Lemma 5.9 together with astandard argument described in [12, pp. 80–82]. (cid:4)
Note that by Lemma 5.13 and (5.12), we obtain | ˜ v ǫ ( X ) | ≤ Cr − n − k ˜ v ǫ k L ( Q − ( X , r )) ≤ Cr − n − k ˜ v ǫ k L ( Q + ( X , R )) ≤ Cd − n , That is, for X = ( x , t ) ∈ Q satisfying 0 < | X − ˆ Y | P < d y /
6, we have | ˜ K ǫ ( x , y , t ) | ≤ C | X − ˆ Y | − n P , ∀ ǫ ≤ | X − ˆ Y | P . (5.14)Next, we claim that for 0 < R ≤ d y , we have ||| ˜ K ǫ ( · , y , · ) ||| Q\ Q + ( ˆ Y , R ) ≤ CR − n / , ∀ ǫ > . (5.15)To prove (5.15), we only need to consider the case when R > ǫ . Indeed, if R ≤ ǫ , then (5.5) yields ||| ˜ K ǫ ( · , y , · ) ||| Q\ Q + ( ˆ Y , R ) ≤ ||| ˜ K ǫ ( · , y , · ) ||| Ω × ( −∞ , ∞ ) ≤ C ǫ − n / ≤ CR − n / . Fix a cut-o ff function ζ ∈ C ∞ c ( Q ( ˆ Y , R )) such that ζ ≡ Q + ( ˆ Y , R / , ≤ ζ ≤ , | D x ζ | ≤ R − , | ζ t | ≤ R − . (5.16)By using the second Korn inequality (2.20) and (5.14), we derive from (5.4)that sup t ≥ Z Ω | (1 − ζ )˜ v ǫ ( x , t ) | dx + " Q | D x ((1 − ζ )˜ v ǫ ) | dxdt ≤ C " Q (cid:16) | D x ζ | + | (1 − ζ ) ζ t | (cid:17) | ˜ v ǫ | dxdt ≤ CR − " { R / < | X − ˆ Y | P < R } | X − ˆ Y | − n P dX ≤ CR − n , (5.17)which implies the desired estimate (5.15). In fact, we obtain from (5.17) that ||| (1 − ζ ) ˜ K ǫ ( · , y , · ) ||| Q ≤ CR − n / , ∀ ǫ > . (5.18)We claim that for 0 < R ≤ d y , we have k ˜ K ǫ ( · , y , · ) k L n + / n ( Q\ ¯ Q + ( ˆ Y , R )) ≤ CR − n / . (5.19)Indeed, set Q (1) : = Ω × ( R , ∞ ) and Q (2) : = ( Ω \ ¯ B ( y , R )) × (0 , R ) and note thatby (2.16) and (5.15) we have k ˜ K ǫ ( · , y , · ) k L n + / n ( Q (1) ) ≤ C γ ||| ˜ K ǫ ( · , y , · ) ||| Q (1) ≤ C γ R − n / and similarly, by (2.17) and (5.18), we have k ˜ K ǫ ( · , y , · ) k L n + / n ( Q (2) ) ≤ C (2 γ + n + γ R − n / . By combining the above two inequalities, we get (5.19). emma 5.20. For any y ∈ Ω and ǫ > , we have |{ X = ( x , t ) ∈ Q : | ˜ K ǫ ( x , y , t ) | > λ }| ≤ C λ − n + n , ∀ λ > d − ny , (5.21) |{ X = ( x , t ) ∈ Q : | D x ˜ K ǫ ( x , y , t ) | > λ }| ≤ C λ − n + n + , ∀ λ > d − ( n + y . (5.22) Also, for any y ∈ Ω , < R ≤ d y , and ǫ > , we have k ˜ K ǫ ( · , y , · ) k L p ( Q + ( ˆ Y , R )) ≤ C p R − n + ( n + / p , ∀ p ∈ [1 , n + n ) , (5.23) k D x ˜ K ǫ ( · , y , · ) k L p ( Q + ( ˆ Y , R )) ≤ C p R − n − + ( n + / p , ∀ p ∈ [1 , n + n + ) . (5.24) Proof.
We derive (5.23) and (5.24), respectively, from (5.21) and (5.22), whichin turn follow from (5.19) and (5.15), respectively; see [3, Lemmas 3.3 and3.4]. (cid:4)
Lemma 5.25.
Suppose { u k } ∞ k = is a sequence in V ( Q ) n such that sup k ||| u k ||| Q ≤ A < ∞ , then there exists u ∈ V ( Q ) n satisfying ||| u ||| Q ≤ A and a subsequence u k j that converges to u weakly in W , ( Ω × (0 , T )) n for any T > . Moreover, if each u k ( · , t ) ∈ V for a.e. t ∈ (0 , ∞ ) , then we also have u ( · , t ) ∈ V for a.e. t ∈ (0 , ∞ ) .Proof. See [3, Lemma A.1]. (cid:4)
The above two lemmas contain all the ingredients for the constructionof a function ˜ K ( · , y , · ) such that for a sequence ǫ µ tending to zero, we have˜ K ǫ µ ( · , y , · ) ⇀ ˜ K ( · , y , · ) weakly in W , q ( Q + ( ˆ Y , d y )) n , (1 − ζ ) ˜ K ǫ µ ( · , y , · ) ⇀ (1 − ζ ) ˜ K ( · , y , · ) weakly in W , ( Ω × (0 , T )) n , (5.26)where 1 < q < n + n + , ζ is as in (5.16) with R = ¯ d Y /
2, and T > K ( · , y , · ) satisfies the same estimates as in Lemma 5.20as well as (5.15) and (5.19); see [3, Section 4.2]. Note that by Lemma 5.25we have ˜ K ( · , t , y ) ∈ V for a.e. t >
0. We define K ( x , y , t ) by K ( x , y , t ) : = ˜ K ( x , y , t ) if D , ∅ ,˜ K ( x , y , t ) + [0 , ∞ ) ( t ) π R ( δ y I )( x ) if D = ∅ . (5.27)where π R ( δ y I ) is an n × n matrix valued function whose k -th column is π R ( δ y e k ) = N X i = ω ki ( y ) ω i , where ω i = ( ω i , . . . , ω ni ) T ∈ R and Z Ω ω i · ω j dx = δ ij . (5.28)Then, it is easy to see that K ( x , y , t ) satisfies the estimates 1) - 4) in Theorem3.1 because ˜ K ( x , y , t ) satisfies all of them as we noted above, and R is afinite dimensional vector space so that all norms over R are equivalent. Forexample, to see the estimate 3) in the theorem holds, observe that k D ω i k L p ( Q + ( ˆ Y , r )) ≤ r k ω i k W , ( B ( y , r )) | Q + ( ˆ Y , r ) | p − ≤ Cr − n / + ( n + / p ≤ C (diam Ω ) n / + r − n − + ( n + / p . ince π R ( δ y e k ) ∈ R , it is a weak solution of u t − L u =
0. Therefore, byrepeating the proof for (5.14), we find K ( x , y , t ) satisfies the estimate (3.7),while the estimate (3.8) is obtained from (3.7) and Lemma 2.24. We havethus shown that K ( x , y , t ) satisfies all the estimates 1) - 6) in Theorem 3.1.We now prove that K ( x , y , t ) satisfies all the properties stated in Sec-tion 2.4 so that it is indeed the heat kernel for (MP). First, note that theproperty a) is clear from 1), 3) in the theorem and 8) in Remark 3.9. Toverify the property b), first note that by (5.3) together with (5.26) and (5.27),we have K ǫ µ ( · , y , · ) ⇀ K ( · , y , · ) weakly in W , q ( Q + ( ˆ Y , d y )) n , (1 − ζ ) K ǫ µ ( · , y , · ) ⇀ (1 − ζ ) K ( · , y , · ) weakly in W , ( Ω × (0 , T )) n , (5.29)for any T >
0. Next, suppose φ = ( φ , . . . , φ n ) T is supported in ¯ Ω × (0 , T )and note that by (5.1) we have Z Ω Φ y ,ǫ ( x ) φ k ( x , dx = Z T Z Ω − K ǫ µ ik ( x , y , t ) ∂∂ t φ i ( x , t ) dx dt + Z T Z Ω a αβ ij ∂∂ x β K ǫ µ jk ( x , y , t ) ∂∂ x α φ i ( x , t ) dx dt . By writing φ = η φ + (1 − η ) φ , where η ∈ C ∞ c ( Q ( ˆ Y , d y )) satisfying η = Q ( ˆ Y , d y / µ → ∞ in the above, we getthe identity (2.11); see [3, p. 1662] for the details. To verify the propertyc), let us denote ˆ f ( x , t ) = f ( x , − t ) and let ˆ u be a unique weak solution in V , ( Ω × ( − T , n of the backward problem − u t − L u = ˆ f in Ω × ( − T , u = D × ( − T , τ ( u ) = N × ( − T , u ( · , = Ω . By letting T → ∞ , we may assume that ˆ u is defined on Ω × ( −∞ , t >
0, we have Z Ω Φ x ,ǫ ( y ) ˆ u k ( y , − t ) dy = Z − t Z Ω K ǫ ik ( y , x , s + t ) ˆ f i ( y , s ) dy ds . We note H2 implies, similar to Lemma 5.9, that ˆ u is continuous in Ω × ( −∞ , f = ζ f + (1 − ζ ) f and use (5.29) to getˆ u k ( x , − t ) = Z t Z Ω K ik ( y , x , t − s ) f i ( y , s ) dx ds If we set u ( x , t ) = ˆ u ( x , − t ), then it becomes be a weak solution in V ( Ω × (0 , T )) n of the problem (2.12), and thus by the uniqueness the property c)is confirmed. Therefore, we have shown that K ( x , y , t ) is indeed the heatkernel for (MP).Now, we prove the identity (3.2). Letˆ K δ il ( y , x , s ) = ˆ v i δ, x , l ( y , s ) = v i δ, x , l ( y , t − s ) = K δ il ( y , x , t − s ) (5.30) nd ˆ Φ x ,δ ( y , s ) = Φ x ,δ ( y , t − s ) , where v δ, x , l and Φ x ,δ are as above. Observe that ˆ v δ, x , l ( y , s ) is, for any − T < t ,a unique weak solution in V , ( Ω × ( − T , t )) n of the problem − v s − L v = Ω × ( − T , t ) v = D × ( − T , t ) τ ( v ) = N × ( − T , t ) v ( · , t ) = ˆ Φ x ,δ e l on Ω . (5.31)Then, similar to (5.2), we have Z Ω ˆ K δ kl ( · , x , Φ y ,ǫ = Z Ω K ǫ lk ( · , y , t ) ˆ Φ x ,δ . (5.32)By repeating the proof of [3, Lemma 3.5], we obtain (3.2) from (5.32) as wellas the the following representation of the mollified heat kernel: K ǫ ( x , y , t ) = Z Ω K ( z , x , t ) T Φ y ,ǫ ( z ) dz . In particular, by continuity of K ( · , x , t ) and (3.2), we havelim ǫ → K ǫ ( x , y , t ) = K ( x , y , t ) . (5.33)Now, we turn to the proof of the formula (3.4). Let u be the weaksolution in V , ( Ω × (0 , T )) n of the problem (3.5). Let X = ( x , t ) ∈ Ω × (0 , T )be fixed but arbitrary and let ˆ v δ = ˆ v δ, x , l be as in (5.30). Then, it follows fromthe equations (3.5) and (5.31) that for su ffi ciently small δ , we have Z Ω ψ i ( y ) ˆ v i δ ( y , dy = Z Ω u l ( y , t ) ˆ Φ δ, x ( y ) dy . Therefore, by using (5.30), we obtain Z Ω K ǫ µ il ( y , x , t ) ψ i ( y ) dy = Z Ω u l ( y , t ) ˆ Φ ǫ µ , x ( y ) dy . (5.34)By (5.15) and (5.33) with x in place of y , we find by the dominated conver-gence theorem thatlim µ →∞ Z Ω K ǫ µ il ( y , x , t ) ψ i ( y ) dy = Z Ω K il ( y , x , t ) ψ i ( y ) dy . By Lemma 5.9, we find that u is continuous at X = ( x , t ). Therefore, bytaking the limit µ → ∞ in (5.34) and using (3.2) we obtain (3.4).Finally, let u be a weak solution V , ( Ω × (0 , T )) n of the problem (3.5)and φ be a Lipschitz function on Ω satisfying |∇ φ | ≤ K a.e. for some K > I ( t ) : = Z Ω e φ | u ( x , t ) | dx . hen I ′ ( t ) satisfies for a.e. t > ff erential inequality I ′ ( t ) = − Z Ω ( e φ B ( u , u ) + e φ a αβ ij ∂ u j ∂ x β ∂φ∂ x α u i ) dx ≤ Z Ω n − κ e φ | ε ( u ) | + κ e ψ | ε ( u ) ||∇ φ || u | o dx ≤ Z Ω n − κ e φ | ε ( u ) | + κ e φ | ε ( u ) | + κ /κ ) K e φ | u | o dx ≤ κ /κ ) K I ( t ) . (5.35)Then, by repeating the argument in [3, Section 4.4], we obtain the formula(3.6). The theorem is proved. (cid:4) By Lemma 2.27 and a remark preceding it, we observe that the conditionsH1 and H3 imply the condition H2 and the condition (LB) in [4]. Then,by Theorem 3.1, the heat kernel K ( x , y , t ) exists and by using (5.35) andrepeating the proof of [4, Theorem 3.1] with R max = diam Ω , we obtain theGaussian bound (3.11). Also, by (2.29), for X = ( x , t ) and X ′ = ( x ′ , t ′ ) in Q satisfying 2 | X ′ − X | P < r ≤ r : = | X − ˆ Y | P ∧ diam Ω , we have | K ( x ′ , y , t ′ ) − K ( x , y , t ) | ≤ C | X ′ − X | µ P r − µ − ( n + / k K ( · , y , · ) k L ( Q − ( X , r ) ∩Q ) (5.36)Note that the estimate (3.11) implies that for 0 < s ≤ (diam Ω ) , we have | K ( z , y , s ) | ≤ C n | z − y | ∧ √ s o − n . (5.37)From the above the estimate, we obtain the estimate (3.12) by repeating theproof of [3, Theorem 3.7]. More precisely, we consider the following threepossible cases.i) Case | x − y | ≤ √ t < diam Ω : In this case, we have r = √ t = | X − ˆ Y | P and | x − y | / t ≤ . If | X ′ − X | P < r /
8, then we take r = r / | K ( x ′ , y , t ′ ) − K ( x , y , t ) | ≤ C | X ′ − X | µ P r − n − µ , which implies (3.12). If r / ≤ | X ′ − X | P ≤ r /
2, then we have | x ′ − y | ≤ r / r / ≤ √ t ′ ≤ r < diam Ω and thus, by (3.11) we get | K ( x ′ , y , t ′ ) − K ( x , y , t ) | ≤ | K ( x ′ , y , t ′ ) | + | K ( x , y , t ) | ≤ Cr − n , which also implies (3.12). i) Case √ t < | x − y | : In this case, r = | x − y | < diam Ω . Similar to [4,Eq. (5.22)], for all ( z , s ) ∈ Q − ( X , r / ∩ Q , we have | K ( z , y , s ) | ≤ Ct − n / exp n − ϑ | x − y | / t o . (5.38)If | X ′ − X | P < r /
4, then we take r = r / | K ( x ′ , y , t ′ ) − K ( x , y , t ) | ≤ C | X ′ − X | µ P r − µ t − n / exp n − ϑ | x − y | / t o . If r / ≤ | X ′ − X | P ≤ r /
2, then use (3.11) and (5.38) to get | K ( x ′ , y , t ′ ) − K ( x , y , t ) | ≤ Ct − n / exp n − ϑ | x − y | / t o . Therefore, we also obtain (3.12) in this case.iii) Case diam Ω ≤ √ t : In this case r = d : = diam Ω , and the desiredestimate (3.12) becomes | K ( x ′ , y , t ′ ) − K ( x , y , t ) | ≤ C | X ′ − X | µ P d − n − µ exp n − ϑ | x − y | / t o . (5.39)Since t ≥ d , for all ( z , s ) ∈ Q − ( X , r / ∩ Q , we haveexp n − ϑ | z − y | / s o ≤ e ϑ/ exp n − ϑ | x − y | / t o . (5.40)If | X ′ − X | P < r /
4, then we take r = r / r / ≤ | X ′ − X | P ≤ r /
2, then by (3.11) and (5.40) | K ( x ′ , y , t ′ ) − K ( x , y , t ) | ≤ Cd − n exp n − ϑ | x − y | / t o . Therefore, we also obtain (3.12) in this case.The theorem is proved. (cid:4)
Assuming H1 and H2, we construct the Green’s function G ( x , y ) for (MP)as follows. Note that (2.19) implies that there is a constant ̺ such that forany u ∈ V , we have k u k L ( Ω ) ≤ ̺ k ε ( u ) k L ( Ω ) . (5.41)By utilizing (5.41) and following the proof of [7, Lemma 3.2], we get thatfor x , y ∈ Ω with x , y , we have Z ∞ | ˜ K ( x , y , t ) | dt < ∞ , where ˜ K ( · , y , · ) is as in the proof of Theorem 3.1. We then define G ( x , y ) : = Z ∞ ˜ K ( x , y , t ) dt . (5.42)Then the symmetry relation (4.4) is an immediate consequence of (3.2) oncewe show that G ( x , y ) is the Green’s function. We shall prove below that G ( x , y ) indeed enjoys the properties stated in Section 4.2. Denoteˆ K ( x , y , t ) : = Z t ˜ K ( x , y , s ) ds so that we have G ( x , y ) = lim t →∞ ˆ K ( x , y , t ) . emma 5.43. The following holds uniformly for all t > and y ∈ Ω .i) k ˆ K ( · , y , t ) k L p ( B ( y , d y )) ≤ C p ( d y + ̺ ) d n / p − ny , ∀ p ∈ h , n + n (cid:17) .ii) k ˆ K ( · , y , t ) k L n + / n ( Ω \ B ( y , r )) ≤ C ( r + ̺ ) r − n ( n + n + , ∀ r ∈ (0 , d y ] .iii) k D ˆ K ( · , y , t ) k L p ( B ( y , d y )) ≤ C p ( d y + ̺ ) d − − n + n / py , ∀ p ∈ h , n + n + (cid:17) .iv) k D ˆ K ( · , y , t ) k L ( Ω \ B ( y , r )) ≤ C ( r + ̺ ) r − − n / , ∀ r ∈ (0 , d y ] .Proof. See [7, Lemma 3.23]. (cid:4)
By the above lemma, elements G ij ( x , y ) of G ( x , y ) satisfy G ij ( · , y ) ∈ W , ( Ω ) and G ij ( · , y ) ∈ W , ( Ω \ B ( y , r )) for any r > . Recall that columns of ˜ K ( · , y , t ) are members of V ; see Lemma 5.25. There-fore, in the case when D = ∅ , for any v ∈ R , we have Z Ω v ( x ) T ˜ K ( x , y , t ) dx = Z Ω v ( x ) T G ( x , y ) dx = . We have shown that G ( x , y ) satisfies the property i) in Section 4.2. For theproof of the property ii) in Section 4.2, we refer to [7, Section 3.2]. Finally, weshow that the property iii) in Section 4.2 also holds. Let f ∈ C ∞ ( ¯ Ω ) n ∩ V and u be defined by the formula (4.1). The integral (4.1) is absolutely convergentby the property i) of section 4.2. Similarly, Lemma 5.43 implies that v ( x , t ) : = Z Ω ˆ K ( x , y , t ) f ( y ) dy is well defined. Observe that v ( x , t ) = Z t Z Ω ˜ K ( x , y , s ) f ( y ) dy ds = Z t Z Ω ˜ K ( x , y , t − s ) f ( y ) dy ds . (5.44)Therefore, we have lim t →∞ v ( x , t ) = Z Ω G ( x , y ) f ( y ) dy = u ( x ) (5.45) v t ( x , t ) = Z Ω ˜ K ( x , y , t ) f ( y ) dy . (5.46)By (5.27), the assumption that f ∈ V , and (3.3), we find from (5.44) that v is,for any T >
0, the weak solution in V , ( Ω × (0 , T )) n of the problem v t − L v = f in Ω × (0 , T ) v = D × (0 , T ) τ ( v ) = N × (0 , T ) v ( · , = Ω . y a similar reasoning, (5.46) and the representation formula (3.4) impliesthat v t is, for any T >
0, the weak solution in V , ( Ω × (0 , T )) n of the problem(3.5) with ψ = f . Then, we have (see [7, Eq. (3.42)]) k v t ( · , t ) k L ( Ω ) ≤ Ce − κ ̺ − t k f k L ( Ω ) , ∀ t > . (5.47)Observe that by (5.44), (5.46), and the assumption that f ∈ V , we have v ( · , t ) ∈ V and v t ( · , t ) ∈ V for a.e. t > . Also, note that for any φ = ( φ , . . . , φ n ) T ∈ V and for a.e. t >
0, we have Z Ω a αβ ij ∂ v j ∂ x β ( · , t ) ∂φ i ∂ x α dx = Z Ω f i φ i dx − Z Ω v it ( · t ) φ i dx . (5.48)Then, by setting φ = v ( · , t ) in (5.48) and using (5.41), for a.e. t >
0, we have k ε ( v ( · , t )) k L ( Ω ) ≤ C (cid:16) k f k L ( Ω ) + k v t ( · , t ) k L ( Ω ) (cid:17) k v ( · , t ) k L ( Ω ) ≤ C k f k L ( Ω ) k ε ( v ( · , t )) k L ( Ω ) , where we have used (5.47). Therefore, by (2.19), for a.e. t >
0, we have k v ( · , t ) k W , ( Ω ) ≤ C k f k L ( Ω ) . Then, by the weak compactness of the space W , ( Ω ) n together with the factthat V is weakly closed in W , ( Ω ) n , we find that there is a sequence { t m } ∞ m = tending to infinity and ˜ u ∈ V such that v ( · , t m ) ⇀ ˜ u weakly in W , ( Ω ) n . By (5.45), we must have u = ˜ u ∈ V and thus, for all φ ∈ V , we getlim m →∞ Z Ω a αβ ij ∂ v j ∂ x β ( · , t m ) ∂φ i ∂ x α dx = Z Ω a αβ ij ∂ u j ∂ x β ∂φ i ∂ x α dx . (5.49)Then, by (5.49), (5.47), and (5.48), for any φ ∈ V , we obtain Z Ω a αβ ij ∂ u j ∂ x β ∂φ i ∂ x α dx = Z Ω f i φ i dx , which shows u is a weak solution in V of the problem (4.2); see the remarkthat appears above Theorem 4.3. Therefore, we verified that G ( x , y ) definedby the formula (5.42) also satisfies the property iii) in Section 4.2, and thusit is indeed the Green’s function for (MP).Next, we assume H3 instead of H2 and proceed to prove the secondpart of the theorem. In the rest of the proof, we shall denote d : = diam Ω . By Theorem 3.10, we have the Gaussian bound (3.11). In particular, for X = ( x , t ) ∈ Q satisfying √ t ≤ diam Ω , we have | ˜ K ( x , y , t ) | ≤ C | X − ˆ Y | − n P . (5.50) imilar to [4, Eq. (6.17)], we have | ˜ K ( x , y , t ) | ≤ Cr − n e − κ ̺ − ( t − r ) , t ≥ r , < r ≤ d . (5.51)We set r : = min( ̺, d ). If 0 < | x − y | ≤ r , then by (5.42), we have | G ( x , y ) | ≤ Z | x − y | + Z r | x − y | + Z ∞ r | ˜ K ( x , y , t ) | dt = : I + I + I . (5.52)It then follows from (5.50) and (5.51) that I ≤ C Z | x − y | | x − y | − n dt ≤ C | x − y | − n , I ≤ C Z r | x − y | t − n / dt ≤ C + C ln( r / | x − y | ) if n = C | x − y | − n if n ≥ I ≤ C Z ∞ r r − n e − κ ̺ − ( t − r ) dt ≤ C ̺ r − n . Combining all together we get that if 0 < | x − y | ≤ r , then | G ( x , y ) | ≤ C (cid:0) + ( ̺/ r ) + ln( r / d ) + ln( d / | x − y | ) (cid:1) if n = C (cid:0) + ( ̺/ r ) (cid:1) | x − y | − n if n ≥
3. (5.53)In the case when | x − y | ≥ r , we estimate by (5.50) and (5.51) that | G ( x , y ) | ≤ Z r | ˜ K ( x , y , t ) | dt + Z ∞ r | ˜ K ( x , y , t ) | dt ≤ C Z r r − n dt + C Z ∞ r r − n e − κ ̺ − ( t − r ) dt ≤ Cr − n + C ̺ r − n . (5.54)By (5.53) and (5.54), we get (4.5) and (4.6). Finally, we turn to the proof ofthe estimate (4.7). Because we assume H3, the conclusions of Theorem 3.10are valid. By (3.12) and the definition (5.27), if | X − ˆ Y | P ≤ d , then we have | ˜ K ( x ′ , y , t ) − ˜ K ( x , y , t ) | ≤ | K ( x ′ , y , t ) − K ( x , y , t ) | + C | x − x ′ |≤ C | x ′ − x | µ | X − ˆ Y | − n − µ P whenever | x − x ′ | < | x − y | . (5.55)We claim that for 0 < r ≤ d and t > r , we have | ˜ K ( x ′ , y , t ) − ˜ K ( x , y , t ) | ≤ C | x ′ − x | µ r − n − µ e − κ ̺ − ( t − r ) (5.56)whenever | x − x ′ | < | x − y | . Assume the claim (5.56) for the moment. Similarto (5.52), in the case when 0 < | x − y | ≤ r : = min( ̺, d ), we get | G ( x ′ , y ) − G ( x , y ) | ≤ Z | x − y | + Z r | x − y | + Z ∞ r | ˜ K ( x ′ , y , t ) − ˜ K ( x , y , t ) | dt = : I + I + I . t follows from (5.55) that I ≤ C | x ′ − x | µ Z | x − y | | x − y | − n − µ dt ≤ C | x ′ − x | µ | x − y | − n − µ , I ≤ C | x ′ − x | µ Z ∞| x − y | t − n / − µ / dt ≤ C | x ′ − x | µ | x − y | − n − µ . Also, by (5.56), we obtain I ≤ Cr − n − µ | x ′ − x | µ Z ∞ r e − κ ̺ − ( t − r ) dt ≤ C ̺ r − n − µ | x ′ − x | µ ≤ C (cid:18) ̺ r (cid:19) | x ′ − x | µ | x − y | − n − µ . Combining the above estimates together, we obtain (4.7) when 0 < | x − y | ≤ r .In the case when | x − y | ≥ r , by using (5.55) and (5.56), we estimate | G ( x ′ , y ) − G ( x , y ) | ≤ Z r + Z ∞ r | ˜ K ( x ′ , y , t ) − ˜ K ( x , y , t ) | dt ≤ C | x ′ − x | µ r − n − µ + C ̺ r − n − µ | x ′ − x | µ ≤ C (1 + ( ̺/ r ) ) | x ′ − x | µ ( r / d ) − n − µ d − n − µ ≤ C (1 + ( ̺/ r ) )( r / d ) − n − µ | x ′ − x | µ | x − y | − n − µ . Therefore, we also obtain (4.7) when | x − y | ≥ r . It only remains for us toprove the claim (5.56). The strategy is similar to the proof of (3.12). Notethat each column of ˜ K ( · , y , t ) is a weak solution in V ( Q ) of u t − L u = Q ⋐ Q \ { ˆ Y } . Therefore, similar to (5.36), for 0 < r ≤ d and t > r , we have | ˜ K ( x ′ , y , t ) − ˜ K ( x , y , t ) | ≤ C | x ′ − x | µ r − n / − − µ k ˜ K ( · , y , · ) k L ( Q − ( X , r ) ∩Q ) whenever | x − x ′ | < r /
2. Then by (5.51), we obtain (5.56). (cid:4)
We first show H3 implies H2. Suppose u is a weak solution of L u = B ( x , r ) ⊂ Ω . By a well-known theorem of Morrey [21, Theorem 3.5.2], wehave [ u ] µ ; B ( x , r / ≤ Cr − n − µ k D u k L ( B ( x , r / . By using the second Korn inequality, we get Caccioppoli’s inequality for u ,that is, for any λ ∈ R n and 0 < ρ ≤ r /
2, we have Z B ( x ,ρ ) | D u | dx ≤ C ρ − (cid:16) + (diam Ω ) (cid:17) Z B ( x , ρ ) | u − λ | dx . Then, we get the estimate (2.23) from (2.25). ext, we prove the estimate (2.29). We first establish a global versionof [16, Lemma 4.3]. For λ ≥
0, we denote by L ,λ ( U ) the linear space offunctions u ∈ L ( U ) such that k u k L ,λ ( U ) = sup x ∈ U < r < diam U r − λ Z B ( x , r ) ∩ U | u | dx / < ∞ . Lemma 6.1.
Let f ∈ L ,λ ( B ∩ Ω ) n , where λ ≥ and B = B ( x , R ) with x ∈ ¯ Ω and < R < diam Ω . Suppose u is a weak solution of L u = f in B ∩ Ω , u = on B ∩ D , τ ( u ) = on B ∩ N . If we assume H3, then, for ≤ γ < γ : = min( λ + , n + µ ) , we haver − γ Z B ( x , r ) ∩ Ω | D u | dx ≤ C Z B ∩ Ω | D u | dx + k f k L ,λ ( B ∩ Ω ) ! (6.2) uniformly for all x ∈ B ∩ ¯ Ω and < r ≤ R / . Here, B = B ( x , R / and C is aconstant depending only on n , κ , κ , µ , A , λ, γ , and Ω . We may take γ = γ in (6.2) if γ < n. Moreover, if γ < n, then u ∈ L ,γ ( B ∩ Ω ) and k u k L ,γ ( B ∩ Ω ) ≤ C (cid:16) k u k L ( B ∩ Ω ) + k D u k L ( B ∩ Ω ) + k f k L ,λ ( B ∩ Ω ) (cid:17) . (6.3) Proof.
Let x ∈ B ( x , R / ∩ ∂ Ω and 0 < r ≤ r ∧ ( R / r > Ω \ [ x ∈ ∂ Ω B ( x , r ) ⊃ B ( y , δ ) for some y ∈ Ω and δ > . (6.4)Denote ˜ N : = N ∩ B ( x , r ) , ˜ D : = ∂ ( B ( x , r ) ∩ Ω ) \ ˜ N and let v be a unique weak solution of the problem L v = f in B ( x , r ) ∩ Ω , v = D , τ ( v ) = N . By testing with v and using the Sobolev inequality, we get Z B ( x , r ) ∩ Ω | ε ( v ) | dy ≤ C Z B ( x , r ) ∩ Ω | f | q dy ! / q Z B ( x , r ) ∩ Ω | v | p dy ! / p , where 1 / p + / q = q = n / ( n +
2) if n ≥ q = / ( α +
1) if n = α ∈ [ µ , v to Ω by setting v = Ω \ B ( x , r ). Notethat v ∈ W , ( Ω ) n and D v = Ω \ B ( x , r ). By the Sobolev inequality andH¨older’s inequality, we then obtain k ε ( v ) k L ( Ω ) ≤ Cr α k f k L ( B ∩ Ω ) k v k W , ( Ω ) . By the assumption (6.4), we have c k D v k L ( Ω ) ≤ k ε ( v ) k L ( Ω ) , k v k W , ( Ω ) ≤ C k D v k L ( Ω ) , or some constants c , C > v , x , and r . Therefore, we have Z B ( x , r ) ∩ Ω | D v | dy ≤ Cr λ + α k f k L ,λ ( B ∩ Ω ) . Let w : = u − v . Then, w is a weak solution of L w = B ( x , r ) ∩ Ω . For0 < ρ < r , we have Z B ( x ,ρ ) ∩ Ω | D u | dy ≤ Z B ( x ,ρ ) ∩ Ω | D v | dy + Z B ( x ,ρ ) ∩ Ω | D w | dy ≤ Cr λ + α k f k L ,λ ( B ∩ Ω ) + C (cid:18) ρ r (cid:19) n − + µ Z B ( x , r ) ∩ Ω | D u | dy . Thus, by [11, Lemma 2.1, p. 86], we obtain (6.2) for x ∈ B ( x , R / ∩ ∂ Ω and 0 < r ≤ r ∧ ( R / (cid:4) Let u be a weak solution in V ( Q ∩ Q ) n of (2.28), where Q = Q − ( X , r )with X = ( x , t ) ∈ Q and 0 < r ≤ √ t ∧ diam Ω . Then, we have (see [16,Lemma 4.2]) ess sup t − ( r / ≤ t ≤ t Z B ( x , r / ∩ Ω | u t | ( x , t ) dx ≤ Cr − " Q ∩Q | u | dX , ess sup t − ( r / ≤ t ≤ t Z B ( x , r / ∩ Ω | D x u | ( x , t ) dx ≤ Cr − " Q ∩Q | u | dX . Also, we have (cf. [4, Lemma 8.6]) Z Q ∩Q | u − λ | dX ≤ Cr Z Q ∩Q | D x u | dX ; λ : = ? Q ∩Q u dX . By using Lemma 6.1 and the above inequalities, we repeat the proof of [16,Theorem 3.3] with obvious modifications to get[ u ] µ ,µ / Q ∩Q ≤ A r − µ − ( n + / k u k L ( Q ∩Q ) . (6.5)Finally, we show that the above estimate and the condition (2.26) implies | u | Q ∩Q ≤ A r − ( n + / k u k L ( Q ∩Q ) . (6.6)For Y ∈ Q ∩ Q and Z ∈ Q satisfying | Z − Y | P < r /
4, we have | u ( Y ) | ≤ − µ r µ [ u ] µ ,µ / , Q ∩Q + | u ( Z ) | . By taking the average over Z and using (6.5), we get | u ( Y ) | ≤ (cid:16) − µ A + β − n + (cid:17) r − n − k u k L ( Q ∩Q ) . By adopting a covering argument, we obtain (6.6), and thus (2.29). (cid:4) .2 Proof of Lemma 5.9 Recall that ˜ u is the weak solution of the problem (5.6), where f is understoodas an element of L (( a , b ); V ′ ). We define˜ f ( x , t ) : = f ( x , t ) − π R [ f ( · , t )]( x ) , ( x , t ) ∈ Q : = Ω × ( a , b ) . Note that f = ˜ f in L (( a , b ); V ′ ). Also, for all t ∈ ( a , b ), we have k π R f ( · , t ) k L ∞ ( Ω ) ≤ C k π R f ( · , t ) k L ( Ω ) ≤ C k f ( · , t ) k L ( Ω ) ≤ C | Ω | / k f k L ∞ ( Q ) , where we used the fact that all norms in R are equivalent and that π R is anorthogonal projection. Therefore, we have k ˜ f k L ∞ ( Q ) ≤ C k f k L ∞ ( Q ) . (6.7)Let u be the weak solution in V , ( Q ) n of the problem − u t − L u = ˜ f in Ω × ( a , b ) u = D × ( a , b ) τ ( u ) = N × ( a , b ) u ( · , b ) = Ω . Then it is easy to see that u ( · , t ) ∈ V for a.e. t ∈ ( a , b ) and thus u is a weaksolution of the problem (5.6). Therefore, by the uniqueness, we concludethat ˜ u = u . In particular, ˜ u is a weak solution in V ( Q ) n of − u t + L u = ˜ f .Because of the inequality (6.7), it is then enough to prove the second partof the lemma and we refer to [3, Section 3.2] for its proof. (cid:4) References [1] Auscher, P.; Tchamitchian, Ph.
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