Short time heat diffusion in compact domains with discontinuous transmission boundary conditions
aa r X i v : . [ m a t h . A P ] S e p Short time heat diffusion in compact domains withdiscontinuous transmission boundary conditions
CLAUDE BARDOS ∗ , DENIS GREBENKOV † ,ANNA ROZANOVA-PIERRAT ‡ August 8, 2018
Abstract
We consider a heat problem with discontinuous diffusion coefficients and dis-continuous transmission boundary conditions with a resistance coefficient. For allcompact ( ǫ, δ ) -domains Ω ⊂ R n with a d -set boundary (for instance, a self-similarfractal), we find the first term of the small-time asymptotic expansion of the heatcontent in the complement of Ω , and also the second-order term in the case of aregular boundary. The asymptotic expansion is different for the cases of finite andinfinite resistance of the boundary. The derived formulas relate the heat content tothe volume of the interior Minkowski sausage and present a mathematical justifica-tion to the de Gennes’ approach. The accuracy of the analytical results is illustratedby solving the heat problem on prefractal domains by a finite elements method. Keywords: heat content; discontinuous transmission condition; Minkowski sausage.
We consider a compact domain Ω ⊂ R n with boundary ∂ Ω that splits R n into “hot”and “cold” media, Ω + = Ω and Ω − = R n \ Ω , characterized by (distinct) heat diffusioncoefficients D + and D − (Fig. 1). On the boundary ∂ Ω is also defined a function ≤ λ ( x ) ≤ ∞ which describes the resistivity to heat exchange through the boundary.We are interested in propagation of the heat content associated with the following ∗ Laboratory Jacques Louis Lions, University of Paris 6, Pierre et Marie Curie, 4 place Jussieu, Paris,France, [email protected] † Laboratoire de Physique de la Matière Condensée, CNRS – Ecole Polytechnique, Palaiseau, France,[email protected] ‡ Laboratory Applied Mathematics and Systems, CentraleSupélec Paris, Grande Voie des Vignes,Châtenay-Malabry, France, [email protected] ∂ t u ± − D ± ∆ u ± = 0 x ∈ Ω ± , t > , (1) u + | t =0 = 1 , u − | t =0 = 0 , (2) D − ∂u − ∂n | ∂ Ω = λ ( x )( u − − u + ) | ∂ Ω , (3) D + ∂u + ∂n | ∂ Ω = D − ∂u − ∂n | ∂ Ω , (4)where ∂/∂n is the normal derivative directed outside the domain Ω .A rigorous analysis of the problem (1)–(4) for irregular boundaries requires its vari-ational formulation in appropriate functional spaces (see Section 2). The variationalproblem is shown to have a unique weak solution with the desired trace properties onthe boundary ∂ Ω (see Section 2). The variational problem is equivalent to the prob-lem (1)–(4) for a piecewise Lipschitz ∂ Ω according to the classical trace theorem. Inturn, extensions of the trace theorem have to be used for fractal boundaries or, moreprecisely, d -sets (see Subsection 2.2).Once a unique solution u ± of the problem (1)–(4) is established, we study the asymp-totic expansion of the heat content as t → N ( t ) = Z R n \ Ω u − ( x, t )d x = Vol(Ω) − Z Ω u + ( x, t )d x. (5)PSfrag replacements Ω + = Ω “hot” Ω − = R n \ Ω “cold” ∂ Ω Figure 1: Illustration of the heat content problem for a planar domain Ω with prefractalboundary ∂ Ω presented by the third generation of the Minkowski fractal (of fractaldimension / ). This boundary splits the plane into two complementary regions. Attime t = 0 , the inner region Ω + = Ω is “hot” (functions on Ω + are denoted withsubscript + ), while the outer region Ω − = R n \ Ω is “cold” (functions on Ω − are denotedwith subscript − ).Eqs. (1)–(4) describe heat exchange between two media prepared initially at differ-ent temperatures and separated by a partially isolating boundary[1, 2]. In fact, u ( x, t ) λ ), or the shape of the exchange boundary. It is therefore crucial to understand how theshape of the boundary influences heat exchange. In particular, would an irregular (e.g.,fractal) boundary with a very large exchange area significantly speed up cooling?Similar equations can describe molecular diffusion between two media across semi-permeable membranes [3, 4]. In that case, u ( x, t ) represents the (normalized) concen-tration of molecules, while Eqs. (1)–(4) can model the leakage of molecules from a cell( Ω + ) to the extracellular space ( Ω − ) or, more generally, the diffusive exchange betweentwo compartments (e.g., oxygen or carbon dioxide exchange between air and blood acrossthe alveolar membrane in the lungs). The resistance λ is related to the cellular mem-brane permeability. As for heat exchange, one may need to enhance or to slow down themolecular leakage, and the shape of the boundary may play an important role.The discontinuity of the initial condition, of the diffusion coefficient, and of the solution u ( x, t ) across the boundary between two domains constitutes one of the mathematical dif-ficulties to be treated. From a physical point of view, such discontinuities might appearunrealistic. For instance, the diffusive flux at the boundary at time t = 0 is infinite.For any physical setting of heat or molecular diffusion, there would be an intermediatelayer between two media in which the material properties would change rapidly but con-tinuously. When the thickness of this intermediate layer is much smaller than the size ofthe domain, the physical problem with continuously varying parameters can be approx-imated by the heat problem (1)–(4). Such an approximation is applicable starting froma small cut-off time while understanding the heat exchange at smaller time scales wouldneed either restituting an intermediate layer, or introducing nonlinear terms into the heatequation. Throughout this paper, we focus on the mathematical problem (1)–(4).The physical properties of the two media Ω + and Ω − are supposed to be different: D + = D − . This implies the discontinuity of the metric on ∂ Ω . The case of continuousmetric ( g − | ∂ Ω = g + | ∂ Ω ) on smooth compact n -dimensional Riemannian manifolds witha smooth boundary ∂ Ω was considered in Ref. [5]. The case of continuous transmissionboundary conditions for the expansion of the heat kernel on the diagonal was treatedin Ref. [6] (see also Ref. [7] for a survey of results on asymptotic expansion of the heatkernel for different boundary conditions). The heat content asymptotic expansion withDirichlet boundary condition was found • up to the third-order term for a compact connected domain Ω ⊂ R n with a regularboundary ∂ Ω ∈ C (Refs. [8, 9]); • up to an exponentially small error for a compact connected domain Ω ⊂ R with apolygonal ∂ Ω (Ref. [10]) and for Ω ⊂ R with ∂ Ω given by the triadic Von Koch3nowflake (Ref. [11]); • up to the second-order term for the general case of self-similar fractal compactconnected domains in R n (Ref. [12]).In general, the boundary between two media can have some resistance to heat ex-change, described by the function λ ( x ) ≥ ( x ∈ ∂ Ω ) that may account for partialthermal isolation. We outline three cases of boundary conditions according to λ :1. If < λ ( x ) < ∞ for all x ∈ ∂ Ω , u is discontinuous on ∂ Ω and we have: (cid:18) λ ( x ) u − − D − ∂u − ∂n (cid:19) | ∂ Ω = λ ( x ) u + | ∂ Ω , D + ∂u + ∂n | ∂ Ω = D − ∂u − ∂n | ∂ Ω .
2. If λ = + ∞ for all x ∈ ∂ Ω , u is continuous on ∂ Ω due to the transmissioncondition and in this case u + | ∂ Ω = u − | ∂ Ω , D + ∂u + ∂n | ∂ Ω = D − ∂u − ∂n | ∂ Ω .
3. If λ = 0 for all x ∈ ∂ Ω , we have the Neumann boundary condition ∂u − ∂n | ∂ Ω = ∂u + ∂n | ∂ Ω = 0 that models the complete thermal isolation of ∂ Ω and implies the trivial solutiongiven by u − ( x, t ) = 0 and u + ( x, t ) = 1 for all time t ≥ .The main goal of the article is to develop the preliminary study[13] and especiallyto formalize the seminal approach by de Gennes[14]. In the case λ = + ∞ , de Gennesargued that as t → +0 , N ( t ) is proportional to the volume µ ( ∂ Ω , √ D + t ) of the interiorMinkowski sausage of ∂ Ω of the width equal to the diffusion length √ D + t : µ ( ∂ Ω , ℓ ) = Vol (cid:0) { x ∈ Ω | dist( x, ∂ Ω) < ℓ } (cid:1) (see also Ref. [12]). In particular, • for a regular boundary ∂ Ω , N ( t ) is proportional to Vol( ∂ Ω) √ D + t ; • for a fractal boundary ∂ Ω of the Hausdorff dimension d , N ( t ) is proportional to ( D + t ) n − d .The de Gennes scaling argument was further investigated in Ref. [13], both experimentallyand numerically. It was shown that irregularly shaped passive coolers rapidly dissipateat short times, but their efficiency decreases with time. The de Gennes scaling argumentwas shown to be only a large scale approximation, which is not sufficient to describeadequately the temperature distribution close to the irregular frontier.In the present paper, we provide a mathematical foundation and further understandingfor the de Gennes approach. We obtain three results valid for all compact ( ǫ, δ ) -domains Ω in R n with connected boundary ∂ Ω , presented by a closed d -set (see Section 2.2 for4he definitions of ( ǫ, δ ) -domains and d -sets): the well-posedness of the problem (1)–(4),the continuity of the solution on λ (see Section 2), and the asymptotic expansion of theheat content (5). In particular, these results hold for domains with a self-similar fractalboundary.We show in Theorem 5 that the heat content N ( t ) is approximated by the volume ofthe interior Minkowski sausage of ∂ Ω of the radius √ D + t : N ( t ) = τ λ h C λ ( ∂ Ω) µ (cid:16) ∂ Ω , p D + t (cid:17) + O (cid:16) µ (cid:16) ∂ Ω , p D + t (cid:17)(cid:17)i , (6)where τ λ is equal to if λ = ∞ and √ t if λ > is finite. Here C λ ( ∂ Ω) is a constantdepending only on the shape of ∂ Ω and finiteness of λ (see Theorem 5 for the exactformulas). Formula (6) is the first approximation of Eqs (80), (82) given in Theorem 5,which allows to find N ( t ) up to terms of the order τ λ O (cid:0) √ t µ (cid:0) ∂ Ω , √ D + t (cid:1)(cid:1) .Moreover, the asymptotic relation (6) remains valid even for mixed boundary condi-tions for three disjoint boundary parts, i.e. when λ = ∞ on one part of the boundary, λ = 0 on another part, and < λ < ∞ on the remaining boundary (see Theorem 3).However, changes of the type of the boundary condition should be continuous (see The-orem 2) such that u remains a continuous function of λ . In this more general case, thecoefficient C λ ( ∂ Ω) in Eq. (6) is given either by Eq. (83) for < λ < ∞ , or by Eq. (84)for λ = ∞ , or is equal to for λ = 0 (the boundary with λ = 0 does not contributeto the short-time asymptotics of the heat content). Finding the asymptotics for mixedboundary conditions with a discontinuous jump from a finite λ to λ = ∞ is still an openproblem.As expected, the resistivity of the boundary to heat transfer makes heat diffusion slower due to the presence of the coefficient τ λ = √ t .For a fractal boundary we replace µ (cid:0) ∂ Ω , √ D + t (cid:1) by the volume of the interiorMinkowski sausage which scales as (4 D + t ) ( n − d ) / , where d is the fractal dimension[12].In the fractal case the integral over ∂ Ω should be understood by using the Hausdorffmeasure (see Ref. [15, 16, 17]).The comparison between the asymptotic formula (6) and a numerical solution of theproblem (1)–(4) for the unit square and a prefractal domain is shown in Fig. 2 for a finite λ and in Fig. 3 for λ = + ∞ . The numerical solution was obtained in FreeFem++by a finite elements method with the implicit θ -schema, also known as Crank-Nicolsonschema, for the time discretization with θ = and ∆ t = 10 − . The domain Ω wascentered in a ball B of diameter (at least) twice bigger than the diameter of Ω . TheNeumann boundary condition was imposed on the boundary of the ball. According to theprinciple “not feeling the boundary” [11] (see also Section 3), the heat content propagationin R with a prescribed boundary ∂ Ω can be very accurately approximated at small timesby the heat content propagation computed in B . The accuracy of this approximationcan also be checked by changing the diameter of the ball. In the case of the squaredomain Ω , the ball was replaced by a square with four times bigger edge. Each pre-fractal edge was discretized with 27 space points while 57 points were used in the externalboundary of the ball. The mesh size was varied to check the accuracy of the presentednumerical solutions. For the case of the discontinuous solution on the boundary (when < λ < ∞ ) we apply the domain decomposition method and match the boundary5 −4 −3 t N ( t ) (a) numericalasymptotic −5 −4 −3 −3 −2 −1 t N ( t ) (b) numericalasymptoticfractal Figure 2: Comparison between the asymptotic formula (6) (solid line) and a FreeFem++numerical solution of the problem (1)–(4) (circles) for two domains: (a) the unit square(
Vol( ∂ Ω) = 4 ) and (b) the third generation of the Minkowski fractal (
Vol( ∂ Ω) = 2 · ), with D + = 1 / , D − = 1 , and λ = 17 . Since the Hausdorff dimension of theboundaries of these domains is (even for the prefractal case), Eq. (6) for a constant λ is reduced, according to Theorem 5, to N ( t ) = 2 √ tC λµ ( ∂ Ω , √ D + t ) + O ( t ) with µ ( ∂ Ω , √ D + t ) ≃ √ D + t Vol( ∂ Ω) and C given by Eq. (97). For plot (b), dashedline shows the fractal asymptotic (that would be exact for the infinite generation of thefractal) with de Gennes approximation of µ (cid:0) ∂ Ω , √ D + t (cid:1) in Eq. (6) by (4 D + t ) . Thisapproximation is valid for intermediate times. −3 t N ( t ) (a) numericalasymptotic −5 −4 −3 −2 −1 t N ( t ) (b) numericalasymptoticfractal Figure 3: Comparison between the asymptotic formula (6) (solid line) and a FreeFem++numerical solution of the problem (1)–(4) (circles) for two domains: (a) the unit square(
Vol( ∂ Ω) = 4 ), and (b) the third generation of the Minkowski fractal (
Vol( ∂ Ω) = 2 · ),with D + = 0 . , D − = 1 , and λ = ∞ . Since the prefractal boundary ∂ Ω has theHausdorff dimension , Eq. (6) is reduced to Eq. (95), i.e., N ( t ) ∝ √ t . In turn, dashedline shows the fractal asymptotic (that would be exact for the infinite generation of thefractal) with de Gennes approximation of µ (cid:0) ∂ Ω , √ D + t (cid:1) in Eq. (6) by . D + t ) . Thisapproximation is valid for intermediate times.values of the respective solutions on ∂ Ω by a Picard fixed point method. We considertherefore the numerical solution of heat propagation for small times as a reference, towhich asymptotic formulas are compared with. In particular, deviations between thenumerical solution and the asymptotic formulas observed at longer times illustrate therange of validity of the short-time expansion.6or the regular case ∂ Ω ∈ C , we obtain the heat content approximation up to thethird-order term. The formulas are given in Theorem 6. For the case λ < ∞ , thecoefficient in front of the second-order term ( t ) in the asymptotic expansion dependson the mean curvature. In turn, for λ = ∞ , the second-order term (here, t ) in theasymptotic expansion vanishes: N ( t ) = 2 1 − e − √ π √ D − D + √ D + + √ D − Vol( ∂ Ω) √ t + O ( t ) . (7)The rest of the paper is organized as follows. In Section 2, we describe the class ofirregular boundaries and prove the well-posedness of the model relying on the variationalformulation of the problem. The boundary conditions are treated in the weak sense bygeneralizing the trace operator and the Green formula to fractals using fractal Besovspaces, B , β ( ∂ Ω) and B , − β ( ∂ Ω) ( β = 1 − n − d > for a d -dimensional ∂ Ω ) definedin A. In Section 2 we also establish the continuity of u as a function of λ . In Section 3we prove that the problem to find N ( t ) can be replaced by a heat problem localizedin O ( √ t ) -interior Minkowski sausage of the boundary by a variant of the principle “notfeeling the boundary” [11] in the general case in R n . This allows, due to the continuity of u on λ , to establish Theorem 3 for a mixed boundary condition including zero, finite, orinfinite values of λ . Considering a regular ∂ Ω (at least in C ) and using the localizationproperties from Section 3, we rewrite in Section 4 the formula for N ( t ) in the terms ofthe local coordinates. Section 5 gives the approximation of the heat problem solutionthrough the solution of one-dimensional constant coefficient problem. The heat contentis calculated in terms of the volume of the interior Minkowski sausage of the boundary inSection 6. Firstly, to illustrate the technique of the proof on a simple case, we give theproof for the case of continuous diffusion coefficients D + = D − , just with discontinuity ofthe initial condition. In this case, all formulas given in Section 6 are valid for all types ofthe boundary introduced in Subsection 2.2. The calculation relies on the Green functionof the problem with constant coefficients for Ω being a half-space (see B). We also obtainthe Green function used in Section 7 for the proof of the asymptotic heat expansion upto the third-order term for a regular ∂ Ω ∈ C . Let Ω be an open connected bounded subset of R n such that ∂ Ω is closed with Vol(Ω) < ∞ . We denote by Ω + = Ω and Ω − = R n \ Ω (Fig. 1).We are looking for the solution of the problem (1)–(4), where D + = D − , D + > and D − > , λ ( x ) ≥ for all x ∈ ∂ Ω . The boundary ∂ Ω is divided into two disjointparts: Γ ∞ = { x ∈ ∂ Ω | λ ( x ) = + ∞} and ∂ Ω \ Γ ∞ = { x ∈ ∂ Ω | ≤ λ ( x ) < + ∞} . Eachof the parts can be the empty set. We thus assume that λ ∈ L ∞ ( ∂ Ω \ Γ ∞ ) . Firstly, we consider the case when ∂ Ω is regular (at least piecewise Lipschitz) and Γ ∞ is the empty set. 7o prove the existence, the uniqueness, and the stability of a solution of the prob-lem (1)–(4), we proceed with its variational formulation.We introduce the space H = L ( R n ) and the space V = { f ∈ H | f + = f | Ω + ∈ H (Ω + ) , and f − = f | Ω − ∈ H (Ω − ) } of functions f = f + Ω + + f − Ω − defined on Ω + ∪ Ω − such that their restrictions f + = f | Ω + and f − = f | Ω − belong to H . We equip V with the norm: k u k V = D + Z Ω + |∇ u + | d x + D − Z Ω − |∇ u − | d x + Z Ω + ∪ Ω − | u | d x. We notice that V is a Hilbert space, V ⊂ L (Ω) , and V is dense in L (Ω) . In addition, V ⊂ L ( R n ) ⊂ V ′ , where V ′ is the dual space to V . Finally, since ∂ Ω is regular, theinclusion V ⊂ L ( R n ) is compact.Applying the usual trace theorem under the assumptions that Ω is bounded and ∂ Ω is at least piecewise Lipschitz, the bilinear form a ( u, v ) = D + Z Ω + ∇ u + ∇ v + + D − Z Ω − ∇ u − ∇ v − + Z ∂ Ω + λ ( x )( u + − u − )( v + − v − ) dσ (8)is continuous, | a ( u, v ) | ≤ C ( k λ k L ∞ ( ∂ Ω) , D + , D − , Ω + ) k u k V k v k V ( for a constant C > , and coercive on V × V , i.e., a ( u, u ) = D + Z Ω + |∇ u + | d x + D − Z Ω − |∇ u − | d x + Z ∂ Ω + λ ( x ) | u + − u − | dσ ≥ k u k V − k u k L ( R n ) > . Thus we conclude[18] that the bilinear form a ( u, v ) defines an operator A : V → V ′ by a ( u, v ) = h Au, v i . Moreover, − A | L ( R n ) with D ( A ) = { u ∈ V | Au ∈ L ( R n ) } generates an analytical semigroup. Remark 1
When Γ ∞ is not empty, the variational form (8) is well adaptable to the casewhere u is continuous across the part Γ ∞ ⊂ ∂ Ω of the interface. By convention we puton this part λ ( x ) = ∞ which implies u + = u − on Γ ∞ (see also Theorem 2).For Γ ∞ = ∅ , we introduce V as the space of functions u ∈ L ( R n ) such that u + = u | Ω + ∈ H (Ω + ) , u − = u − | Ω − ∈ H (Ω − ) , u + | Γ ∞ = u − | Γ ∞ , and, therefore, we consider the bilinear continuous and coercive form on V × Va ( u, v ) = D + Z Ω + ∇ u + ∇ v + + D − Z Ω − ∇ u − ∇ v − + Z ∂ Ω \ Γ ∞ λ ( x )( u + − u − )( v + − v − ) dσ. (9) In particular, for Γ ∞ = ∂ Ω , we get V = H (Ω + ∪ Ω − ) and a ( u, v ) = D + Z Ω + ∇ u + ∇ v + + D − Z Ω − ∇ u − ∇ v − . .2 Extension to d -sets (fractal case for d > n − ) Let us define a class of fractal domains to be considered. We will see that the existence anduniqueness results of a weak solution of the problem (1)–(4) hold for a class of bounded ( ǫ, δ ) -domains[20, 21, 22] Ω + such that ∂ Ω is a d -set[21]: Definition 1 ( d -set[21, 22, 23]) Let Γ be a closed subset of R n and < d ≤ n .A positive Borel measure m d with support Γ is called a d -measure of Γ if, for somepositive constants c , c > , c r d ≤ m d (Γ ∩ U r ( x )) ≤ c r d , for ∀ x ∈ Γ , < r ≤ , where U r ( x ) ⊂ R n denotes the Euclidean ball centered at x and of radius r .The set Γ is a d -set if there exists a d -measure on Γ . As it is known from Ref. [[24], p.30], any two d -measures on Γ are equivalent. Definition 2 ( ( ǫ, δ ) -domain[20, 21, 22]) An open connected subset Ω of R n is an ( ǫ, δ ) -domain, ǫ > , < δ ≤ ∞ , if whenever x, y ∈ Ω and | x − y | < δ , there is arectifiable arc γ ⊂ Ω with length ℓ ( γ ) joining x to y and satisfying1. ℓ ( γ ) ≤ | x − y | ǫ and2. d ( z, ∂ Ω) ≥ ǫ | x − z | | y − z || x − y | for z ∈ γ . In particular, a Lipschitz domain Ω is an ( ǫ, δ ) -domain and also a n -set[22] (i.e., a d -setwith d = n ). Self-similar fractals (e.g., von Koch’s snowflake domain) are examples of ( ǫ, δ ) -domains with the d -set boundary[19, 22], d > n − .In order to describe irregular boundaries of fractal dimension d > n − , we definesets preserving Markov’s inequality (Ref. [21] Ch. II): Definition 3
A closed subset V in R n preserves Markov’s inequality if for every fixedpositive integer k , there exists a constant c = c ( V, n, k ) > , such that max V ∩ U r ( x ) |∇ P | ≤ cr max V ∩ U r ( x ) | P | for all polynomials P ∈ P k and all closed balls U r ( x ) , x ∈ V and < r ≤ . Examples of sets that preserves Markov’s inequality are d -sets in R n , where d > n − ,and self-similar sets that are not a subset of any ( n − -dimensional subspace of R n (seeRefs. [22, 25]).To extend the variational formulation introduced in Subsection 2.1 to fractal bound-aries of the type of d -sets, we use the existence of the d -dimensional Hausdorff measure m d on ∂ Ω (the d -measure from Definition 1) and the theorem which generalizes theusual trace theorem and the Green formula.For example, for d = n − and a Lipschitz ∂ Ω , we know[18, 26] that the traceoperator is linear continuous and surjective from H (Ω) onto H ( ∂ Ω) , and the formula Z Ω v ∆ u d x = h ∂u∂ν , Tr v i (( H ( ∂ Ω)) ′ ,H ( ∂ Ω)) − Z Ω ∇ v ∇ u d x, u ∈ H (Ω) such that ∆ u ∈ L (Ω) and v ∈ H (Ω) .To generalize the trace operator and the Green formula to fractal boundaries, oneintroduces the Besov space B , β ( ∂ Ω) with β = 1 − n − d > (see A). Note that for d = n − , one has β = and B , ( ∂ Ω) = H ( ∂ Ω) , i.e., one recovers the above relations. In general,1. For an arbitrary open set Ω of R n , the trace operator Tr is defined[21, 25, 27] for u ∈ L loc (Ω) by Tr u ( x ) = lim r → m (Ω ∩ U r ( x )) Z Ω ∩ U r ( x ) u ( y ) dy, (10)where m denotes the Lebesgue measure. The trace operator Tr is considered forall x ∈ Ω for which the limit exists.2. If Ω is a bounded ( ǫ, δ ) -domain in R n such that its boundary ∂ Ω is a closed d -setpreserving Markov’s inequality, then[21, 22](a) the trace operator Tr : H (Ω) → B , β ( ∂ Ω) is linear continuous and surjective;(b) the Green formula holds (see also Refs. [27, 28] for the von Koch case in R ): Z Ω v ∆ u d x = h ∂u∂ν , Tr v i (( B , β ( ∂ Ω)) ′ ,B , β ( ∂ Ω)) − Z Ω ∇ v ∇ u d x, (11)where the dual Besov space ( B , β ( ∂ Ω)) ′ = B , − β ( ∂ Ω) is introduced in Ref. [23](see A).Let us also notice that the Green’s formula (11) still holds whatever u ∈ H (Ω) suchthat ∆ u ∈ L (Ω) and v ∈ H (Ω) . The above preliminaries allow us to prove the following Proposition:
Proposition 1
1. Let
Ω = Ω + be a bounded domain in R n with a closed piecewiseLipschitz boundary ∂ Ω and < λ ( x ) ≤ + ∞ be a given function defined on ∂ Ω .By Γ ∞ is denoted the part of ∂ Ω such that ∀ x ∈ Γ ∞ λ ( x ) = + ∞ , in the such way that λ ∈ L ∞ ( ∂ Ω \ Γ ∞ ) . Then the bilinear form a ( u, v ) = D + Z Ω + ∇ u + ∇ v + + D − Z Ω − ∇ u − ∇ v − + Z ∂ Ω \ Γ ∞ λ ( x )( u + − u − )( v + − v − ) dσ is continuous and coercive on V × V with V = { u ∈ L ( R n ) | u + = u | Ω + ∈ H (Ω + ) , u − = u | Ω − ∈ H (Ω − ) ,u + = u − on Γ ∞ } . (12)10 . Let Ω = Ω + be a bounded ( ǫ, δ ) -domain in R n with a closed d -set boundary ∂ Ω and λ ∈ C ( ∂ Ω) be a positive continuous function defined on ∂ Ω . By m d is denotedthe d -measure on ∂ Ω (see Definition 1). Then the bilinear form a ( u, v ) = D + Z Ω + ∇ u + ∇ v + + D − Z Ω − ∇ u − ∇ v − + Z ∂ Ω λ ( x )Tr( u + − u − )Tr( v + − v − ) dm d is continuous and coercive on V × V ( V is defined in Eq. (12)).3. Let Ω = Ω + be a bounded ( ǫ, δ ) -domain in R n with a closed d -set boundary ∂ Ω and λ ( x ) = + ∞ for all x ∈ ∂ Ω . Then the bilinear form a ( u, v ) = D + Z Ω + ∇ u + ∇ v + + D − Z Ω − ∇ u − ∇ v − is continuous and coercive on V × V with V = H ( R n ) . Consequently, we obtain the following theorem:
Theorem 1 (Well-posedness)
In all cases from Proposition 1 for all u ∈ H = L ( R n ) there exists a unique solution u ∈ C ( R + t , L ( R n )) ∩ L ( R + t , V ) of the variational problem ∀ v ∈ V ddt h u, v i H + a ( u, v ) = 0 , u ( x,
0) = u ∈ L ( R n ) , (13) where by h· , ·i H is denoted the inner product in H . In addition, this solution verifies theenergy equality: Z R n | u ( t ) | d x + Z t a ( u, u )d s = 12 Z R n | u ( x ) | d x. (14) Remark 2
On one hand, any “smooth enough” solution of the problem (1)–(4) givesthe solution of Theorem 1. On the other hand, any solution from Theorem 1 satisfiesthe relations (1)–(2) and, in a weak sense (in the sense of the duality presented above),satisfies the relations (3)–(4).
Finally, we prove
Theorem 2 (Continuity of u λ on λ and the case λ = ∞ ) Let ( λ k ) k ∈ N be a positivesequence converging to λ ∗ in L ∞ ( ∂ Ω) . Then the corresponding sequence of the solutions ( u λ k ) k ∈ N of the system (1)–(4) converges strongly to u ∗ λ in C ( R + t , L ( R n )) ∩ L ( R + t , V ) , i.e. , u λ is continuous as a function of λ .If λ k → ∞ in L ∞ ( ∂ Ω) , then u λ k → u ∞ in C ( R + t , L ( R n )) ∩ L ( R + t , V ) with ( u ∞ ) + =( u ∞ ) − on ∂ Ω . In this case, u ∞ ∈ C ( R + t , L ( R n )) ∩ L ( R + t , H ( R n )) solves ∀ v ∈ H ( R n ) Z R n ∂ t u ∞ v d x + Z R n D ( x ) ∇ u ∞ ∇ v d x = 0 , u ∞ ( x,
0) = u ∈ L ( R n ) , (15) with D ( x ) = Ω + D + + Ω − D − . roof. Firstly we suppose that λ ∗ is a finite bounded function on ∂ Ω ( k λ ∗ k L ∞ ( ∂ Ω) < ∞ ).Since u (0) = u does not depend on λ , the equality (14) implies that the sequence ( u λ k ) is bounded in C ( R + t , L ( R n )) ∩ L ( R + t , V ) .Therefore, due to the unicity of the solution for λ ∗ and the unicity of the weak limit,the convergence λ k → λ ∗ in L ∞ ( ∂ Ω) implies u λ k ⇀ u λ ∗ . Since u k | ∂ Ω ∈ B , β ( ∂ Ω) with β = 1 − n − d > , with the help of (13) and the coercive behavior of a ( u, u ) , a ( u, u ) > α k u k V , for α > , we conclude that u λ k → u λ ∗ in C ( R + t , L ( R n )) ∩ L ( R + t , V ) .In the case k λ k k L ∞ ( ∂ Ω) → + ∞ , we find from (14) that ∀ k ∈ N Z ∂ Ω λ k ( x )Tr(( u k ) + − ( u k ) − ) dm d < ∞ , (cid:18)Z ∂ Ω Tr(( u k ) + − ( u k ) − ) dm d (cid:19) ≤ p k λ k k L ∞ ( ∂ Ω) Z R n | u ( x ) | d x. Therefore, we obtain in this case that ( u ∞ ) + = ( u ∞ ) − on ∂ Ω , where by u ∞ we denotethe limit of u k as λ → + ∞ . In addition, u ∞ ( t, · ) ∈ H ( R n ) and it is the solution of (15). (cid:3) s the initial condition is zero in R n \ Ω , we have N ( t ) = Z Ω (1 − u ( x, t ))d x = Vol(Ω) − Z Ω u ( x, t )d x, (16)or equivalently, in terms of the Green function of the problem (1)–(4), N ( t ) = Vol(Ω) − Z Ω Z Ω G ( x, y, t )d y d x. Let us show that it is sufficient to integrate only on a small neighborhood of the boundary ∂ Ω to obtain the desired heat content with an exponentially small error: Lemma 1
Let F ⊂ Ω be a non-empty open bounded set in R n , such that dist( F, ∂
Ω) = ǫ > . Then for t → +0 and u = u + Ω + u − R n \ Ω the solution of (1)–(4), associatedwith the Green function G ( x, y, t ) ,1. it holds Z F (1 − u + ( x, t ))d x = Z F (cid:18) − Z Ω G ( x, y, t )d y (cid:19) d x = O (cid:18) ǫ √ D + t (cid:19) n − e − ǫ / (4 D + t ) ! . (17)12 . for ǫ > √ D + t such that ǫ = O ( √ t ) , there exists δ > (a constant independenton time) such that the heat content N ( t ) can be expressed as N ( t ) = Z R n \ Ω u − ( x, t )d x = Z Ω ǫ (cid:18) − Z Ω ǫ G ( x, y, t )d y (cid:19) d x + O (cid:16) e − tδ (cid:17) , (18) where Ω ǫ is the ǫ -neighborhood of ∂ Ω . Proof.
As it was shown, the problem (1)–(4) has a unique solution u = u + Ω + u − R n \ Ω . Let G ( x, y, t ) be the Green function so that u ( x, t ) = Z Ω G ( x, y, t )d y. Thus, using the properties of G such as G ≥ for all ( x, y, t ) ∈ R n × R n × R + and R R n G ( x, y, t )d y = 1 , we easily see that ≤ Z Ω G ( x, y, t )d y = u ( x, t ) ≤ Z R n G ( x, y, t )d y = 1 . We notice that, by the assumption, λ ( x ) > is a regular function on ∂ Ω and allother coefficients are constant. By definition u + is the solution of the system ( ∂ t − D + ∆) u + = 0 , x ∈ Ω ⊂ R n ,u + | t =0 = 1 ,u + | ∂ Ω = (cid:18) u − − D − λ ∂u − ∂n (cid:19) | ∂ Ω , λ > , which can be reformulated for ˆ v = 1 − u + ( ∂ t − D + ∆) ˆ v = 0 , x ∈ Ω ⊂ R n , ˆ v | t =0 = 0 , (1 − ˆ v ) | ∂ Ω = (cid:18) u − − D − λ ∂u − ∂n (cid:19) | ∂ Ω , where ≤ u − ≤ for all t . Moreover, as ≤ ˆ v ≤ , it follows that ≤ (cid:18) u − − D − λ ∂u − ∂n (cid:19) | ∂ Ω ≤ and, as u − is increasing in time on ∂ Ω , then ˆ v is decreasing in time on ∂ Ω . Therefore, ˆ v ≤ v , where v is the solution of the following problem: ( ∂ t − D + ∆) v = 0 , x ∈ Ω ⊂ R n ,v | t =0 = 0 ,v | ∂ Ω = 1 , n = 2 , but now in R n ( n ≥ ), we find thatfor the ball Ω = U r ( z ) centered at z and of radius r , the solution satisfies as t → +0 v ( z, t ) ≤ C (cid:18) r √ D + t (cid:19) n − exp (cid:18) − r D + t (cid:19) , with a constant C > depending only on n ( C can be explicitly obtained by theintegration by parts in the generalized spherical coordinates in R n , where the coefficient (cid:18) r √ D + t (cid:19) n − corresponding to the leading term as t → +0 , appears from the integral R + ∞ r √ Dt e − w w n − dw ). Consequently (see Ref. [29] Corollary 12.8 p.232), for z ∈ int { Ω } and t → +0 we find v ( z, t ) ≤ C (cid:18) dist( z, ∂ Ω) √ D + t (cid:19) n − exp (cid:18) − dist( z, ∂ Ω) D + t (cid:19) . Then we immediately obtain Eq. (17) by integration.For n = 2 we obtain directly the exponential decay in Eq. (17) for all ǫ > . If n > , we still have the exponential decay for a small constant α > depending only on ǫ : O (cid:18) ǫ √ D + t (cid:19) n − e − ǫ / (4 D + t ) ! = O ( e − α/t ) . Note that O (cid:16) e − ǫ / (4 D + t ) (cid:17) gives an exponentially small remaining term iff ǫ = 2 √ D + t − δ for a constant δ > . For small enough δ we have ǫ = O ( √ D + t ) , also knowing that ǫ > √ D + t .So, for this ǫ , we split Ω in two parts: Ω ǫ , the neighborhood of ∂ Ω such that dist( x, ∂ Ω) ≤ ǫ , and Ω \ Ω ǫ . For all F ⊆ Ω \ Ω ǫ , dist( F, ∂ Ω) > ǫ > √ D + t , we have Z F (cid:18) − Z Ω G ( x, y, t )d y (cid:19) d x = O (cid:16) e − c ( F ) /t δ ( F ) (cid:17) , where c ( F ) and δ ( F ) are positive constants depending only on the distance between F and ∂ Ω and the dimension n .To complete the proof of the second statement, we first find that N ( t ) = Z R n \ Ω u − ( x, t )d x = Z R n \ Ω Z Ω G ( x, y, t )d y d x = Z R n Z Ω G ( x, y, t )d y d x − Z Ω Z Ω G ( x, y, t )d y d x = Vol(Ω) − Z Ω Z Ω G ( x, y, t )d y d x. Ω = Ω ǫ ∪ (Ω \ Ω ǫ ) we can write Z Ω Z Ω G ( x, y, t )d y d x = (cid:18)Z Ω ǫ Z Ω + Z Ω \ Ω ǫ Z Ω (cid:19) G ( x, y, t )d y d x = Vol(Ω \ Ω ǫ ) + Z Ω ǫ Z Ω G ( x, y, t )d y d x + O (cid:16) e − tδ (cid:17) = Vol(Ω) − Vol(Ω ǫ ) + Z Ω ǫ Z Ω ǫ G ( x, y, t )d y d x + Z Ω ǫ Z Ω \ Ω ǫ G ( x, y, t )d y d x + O (cid:16) e − tδ (cid:17) . Moreover, Z Ω ǫ Z Ω \ Ω ǫ G ( x, y, t )d y d x = Z Ω Z Ω \ Ω ǫ G ( x, y, t )d y d x − Z Ω \ Ω ǫ Z Ω \ Ω ǫ G ( x, y, t )d y d x = Vol(Ω \ Ω ǫ ) + O (cid:16) e − tδ (cid:17) − Z Ω \ Ω ǫ Z Ω \ Ω ǫ G ( x, y, t )d y d x and since Z Ω \ Ω ǫ Z Ω \ Ω ǫ G ( x, y, t )d y d x − Vol(Ω \ Ω ǫ ) ≤ Z Ω \ Ω ǫ Z Ω G ( x, y, t )d y d x − Vol(Ω \ Ω ǫ ) = O (cid:16) e − tδ (cid:17) , we conclude that − Z Ω ǫ Z Ω \ Ω ǫ G ( x, y, t )d y d x = O (cid:16) e − tδ (cid:17) and finally N ( t ) = Vol(Ω ǫ ) − Z Ω ǫ Z Ω ǫ G ( x, y, t )d y d x + O (cid:16) e − tδ (cid:17) , that completes the proof. (cid:3) A variant of Lemma 1 for n = 2 can be found in Ref. [11], where the heat localizationnear the boundary is also called by the principle of “not feeling the boundary”. In addition,we can consider the case of the distinct parts of the boundary: Corollary 1
Let X and Y be different closed parts of ∂ Ω such that dist( X, Y ) > ǫ ,where ǫ = O ( √ t ) > √ D + t . Let U r ( X ) = { x ∈ R n | d ( x, X ) < r } be the openneighborhood of X of size r > . Consider u + and ˆ u + as the respective solutions of thefollowing systems: ∂ t u + − D + △ u + = 0 , x ∈ Ω ⊂ R n ,u + | t =0 = 1 ,u + | ∂ Ω = (cid:18) u − − D − λ ∂u − ∂n (cid:19) | ∂ Ω , λ > , t ˆ u + − D + △ ˆ u + = 0 , x ∈ Ω ⊂ R n , ˆ u + | t =0 = 1 , ˆ u + | ∂ Ω ∩ U ( X ) = (cid:18) u − − D − λ ∂u − ∂n (cid:19) | ∂ Ω ∩ U ( X ) , λ > u + | ∂ Ω \ ( ∂ Ω ∩ U ( X )) = 1 , where U ( X ) is an open neighborhood of X of a radius strictly greater than ǫ : U ǫ ( X ) ⊂ U ( X ) .Then there exists δ > such that Z U ǫ ( X ) | u + − ˆ u + | d x = O (cid:16) e − tδ (cid:17) . Moreover, if ˜ u + is the solution of the system: ∂ t ˜ u + − D + △ ˜ u + = 0 , x ∈ Ω ⊂ R n , ˜ u + | t =0 = 1 , ˜ u + | Y = (cid:18) u − − D − λ ∂u − ∂n (cid:19) | Y , λ > u + | ∂ Ω \ Y = 1 , then Z U ǫ ( X ) (1 − ˜ u + )d x = Z Ω \ U ǫ ( Y ) (1 − ˜ u + )d x = O (cid:16) e − tδ (cid:17) . The proof of Corollary 1 follows from the proof of the first statement of Lemma 1.Note that the continuity of u on λ (see Theorem 2) and the localization of the heatcontent near the boundary allow one to consider mixed boundary conditions: Theorem 3
Let Ω be a bounded ( ǫ, δ ) -domain (see Section 2) with a closed connected d -set boundary ∂ Ω = Γ ⊔ Γ λ ⊔ Γ ∞ . Let λ ∈ C (Γ ⊔ Γ λ ) such that λ ( x ) = , x ∈ Γ , < f ( x ) < ∞ , x ∈ Γ λ , + ∞ , x ∈ Γ ∞ and ǫ = O ( √ t ) > √ D + t . We assume that the connection between different types ofboundary is performed in the continuous way (see Theorem 2) such that the solution u remains continuous as a function of λ .We split the ǫ -interior Minkowski sausage of ∂ Ω into disjoint subsets Ω ǫ = Ω Γ ǫ ⊔ Ω Γ λ ǫ ⊔ Ω Γ ∞ ǫ such that each subset Ω Γ ǫ is contained in the ǫ -interior Minkowski sausage of Γ ( Γ ⊂ ∂ Ω ). Then, for δ > from Lemma 1, the heat content of the problem (1)–(4), N ( t ) = Z Ω (1 − u ( x, t ))d x = Z Ω ǫ (1 − u ( x, t ))d x + O ( e − tδ ) , an be found as a sum of two heat contents: N ( t ) = Z Ω Γ λǫ (1 − u ( x, t ))d x + Z Ω Γ ∞ ǫ (1 − u ( x, t ))d x + O ( e − tδ ) . In order to locally approximate the solution of the problem (1)–(4) by considering theproblem with coefficients frozen on a fixed boundary point, according to Corollary 1, wealso obtain the following proposition:
Proposition 2
Let σ be a fixed point of the boundary ∂ Ω and let define B lǫ,ǫ = U lǫ ( σ ) ∩ (Ω ǫ ∪ Ω − ǫ ) for l ∈ N , (19) where U lǫ ( σ ) ⊂ R n is a ball of radius lǫ centered at σ , ǫ is defined in Corollary 1. Let φ σ ∈ ˚ C ∞ ( B ǫ,ǫ ( σ )) be a smooth cut-off function with a compact support on B ǫ,ǫ ( σ ) : φ σ ( x ) = x ∈ B ǫ,ǫ ( σ ) , a smooth function ≤ η < x ∈ B ǫ,ǫ ( σ ) \ B ǫ,ǫ ( σ ) , x ∈ Ω \ B ǫ,ǫ ( σ ) (20) If u is the solution of the problem (1)–(4), then φ σ u is the solution of the followingproblem: ∂ t ( φ σ u ± ) − D ± △ ( φ σ u ± ) = (cid:26) − (1 − u ± ) D ± △ φ σ ) x ∈ B ǫ,ǫ ( σ ) \ B ǫ,ǫ ( σ ) , elsewhere in Ω , (21) ( φ σ u ± ) | t =0 = Ω ( x ) φ σ ( x ) , (22) D − ∂ ( φ σ u − ) ∂n | ∂ Ω = λ ( x ) φ σ ( x )( u − − u + ) | ∂ Ω , (23) D + ∂ ( φ σ u + ) ∂n | ∂ Ω = D − ∂ ( φ σ u − ) ∂n | ∂ Ω . (24) Therefore, there exists δ > such that Z B ǫ,ǫ ( σ ) | u − φ σ u | d x = O (cid:16) e − tδ (cid:17) , and if φ σ u σ is the solution of the problem (21)–(24) with frozen coefficients in the bound-ary point σ , then Z Ω \ B ǫ,ǫ ( σ ) φ σ (1 − u σ )d x = O (cid:16) e − tδ (cid:17) . (25) ∂ Ω ∈ C In order to prove Eq. (6) for a large class of ( ǫ, δ ) -compact connected domains Ω in R n , we first prove it for the case of domains with regular boundary ∂ Ω ∈ C ∞ or atleast in C . As Ω is compact, for all types of connected ∂ Ω , the volume of Ω is finiteand, therefore, the volume of the ǫ -neighborhood of ∂ Ω in Ω is also finite and can be17pproximated by a sequence of volumes of Minkowski sausages with regular boundaries(the same argument was used in Ref. [11] p.378).Let us consider the regular boundary ∂ Ω ∈ C .Given a positive ǫ > provided in Lemma 1, we denote by Ω ǫ and Ω − ǫ the open ǫ -neighborhoods of ∂ Ω in Ω and in R n \ Ω , respectively.According to Eq. (18) and the regularity of the boundary ∂ Ω , we can decompose Ω ǫ ∪ ∂ Ω ∪ Ω − ǫ = F Ii =1 B i,ǫ ( I is a finite integer because Ω + ∪ Ω − ǫ is a compact domain)in such way that on each B i,ǫ it is possible to introduce the local coordinates. In addition,we assume that for all i = 1 , . . . , I there exists σ i ∈ ∂ Ω ∩ B i,ǫ such that B i,ǫ ⊂ B ǫ,ǫ ( σ i ) (see Eq. (19) for the definition). Due to Proposition 2, the last assumption ensures that Z B i,ǫ (1 − u )d x = Z B i,ǫ φ σ i (1 − u )d x + O (cid:16) e − tδ (cid:17) . For all i we perform the change of the space variables ( x , . . . , x n ) ∈ B i,ǫ to the localcoordinates ( θ , . . . , θ n − , s ) by the formula x = ˆ x ( θ , . . . , θ n − ) − sn ( θ , . . . , θ n − ) (cid:26) < s < ǫ for x ∈ B i,ǫ ∩ Ω ǫ − ǫ < s < for x ∈ B i,ǫ ∩ Ω − ǫ , (26)where ˆ x ( θ , . . . , θ n − ) ∈ ∂ Ω and x , ˆ x and n are the vectors in R n such that (cid:26) ∂ ˆ x∂θ , . . . , ∂ ˆ x∂θ n − , n (cid:27) is an orthonormal basis in R n .In what follows we denote B i,ǫ ∩ Ω ǫ by Ω i, + ǫ and B i,ǫ ∩ Ω − ǫ by Ω i, − ǫ respec-tively. In each of two regions, Ω i, + ǫ and Ω i, − ǫ , the change of variables ( x , . . . , x n ) ( θ , . . . , θ n − , s ) is a local C -diffeomorphism.In local coordinates ∂ Ω is described by s = 0 .Thus, Eq. (18) becomes N ( t ) = I X i =1 Z Ω i, + ǫ (1 − u ( x, t ))d x + O ( e − tδ ) . (27)Denoting θ = ( θ , . . . , θ n − ) , the integration domain Ω i, + ǫ in (27) becomes Ω i, + ǫ = { < s < ǫ, θ ∈ ∂ Ω ∩ Ω i, + ǫ } , which is actually a parallelepiped neighborhood ( ∂ Ω ∩ Ω i, + ǫ ) × ]0 , ǫ [ . For this change of variables we have |∇ x s | = 1 , ∇ x s ∇ x θ i = 0 , ∇ x θ j ∇ x θ i = δ ij (1 − sk i ) for i, j = 1 , . . . , n − , ∇ u ± ∇ φ ± = ∂u ± ∂s ∂φ ± ∂s + n − X i ∂u ± ∂θ i ∂φ ± ∂θ i − sk i ) , | J ( s, θ ) | = n − Y i =1 (1 − sk i ) (28)for the Jacobian and k i = k i ( θ , . . . , θ n − ) of the principal curvatures for ∂ Ω curvingaway the outward normal n to ∂ Ω like in the case of the sphere, we find that for all testfunctions φ = ( φ + , φ − ) ∈ V | B i,ǫ Z B i,ǫ ∂ t u | J ( s, θ ) | φ d sdθ · · · dθ n − − Z Ω i, + ǫ " ∂∂s (cid:18) D + | J ( s, θ ) | ∂u + ∂s (cid:19) + n − X i =1 ∂∂θ i (cid:18) D + | J ( s, θ ) | (1 − sk i ) ∂u + ∂θ i (cid:19) φ + d sdθ · · · dθ n − − Z Ω i, − ǫ " ∂∂s (cid:18) D − | J ( s, θ ) | ∂u − ∂s (cid:19) + n − X i =1 ∂∂θ i (cid:18) D − | J ( s, θ ) | (1 − sk i ) ∂u − ∂θ i (cid:19) φ − d sdθ · · · dθ n − + Z s =0 λ ( θ )( u + − u − )( φ + − φ − ) dθ = 0 . The regularity of the boundary ensures that the principal curvatures k i ( θ ) are at least in C ( ∂ Ω ∩ ∂B i,ǫ ) . Therefore, the problem (1)–(4) locally becomes ∂∂t u + − D + ∂ ∂s + n − X i =1 ∂ ∂θ i ! u + = D + n − X i =1 sk i ( θ )1 − sk i ( θ ) (cid:18) − sk i ( θ ) (cid:19) ∂ u + ∂θ − D + n − X i =1 k i ( θ ) + s n − X i =1 k i ( θ )1 − sk i ( θ ) ! ∂u + ∂s + D + | J ( s, θ ) | n − X i =1 ∂∂θ i (cid:18) | J ( s, θ ) | (1 − sk i ( θ )) (cid:19) ∂u + ∂θ i , < s < ǫ, θ ∈ ( ∂ Ω ∩ Ω i, + ǫ ) (29) ∂∂t u − − D − ∂ ∂s + n − X i =1 ∂ ∂θ i ! u − = D − n − X i =1 sk i ( θ )1 − sk i ( θ ) (cid:18) − sk i ( θ ) (cid:19) ∂ u − ∂θ − D − n − X i =1 k i ( θ ) + s n − X i =1 k i ( θ )1 − sk i ( θ ) ! ∂u − ∂s + D − | J ( s, θ ) | n − X i =1 ∂∂θ i (cid:18) | J ( s, θ ) | (1 − sk i ( θ )) (cid:19) ∂u − ∂θ i , − ǫ < s < , θ ∈ ( ∂ Ω ∩ Ω i, + ǫ ) , (30) u + | t =0 = 1 , u − | t =0 = 0 , (31) D − ∂u − ∂s | s = − = λ ( θ )( u − − u + ) | s =0 , (32) D + ∂u + ∂s | s =+0 = D − ∂u − ∂s | s = − . (33)19e emphasize that the problem (29)–(33) should be considered as the trace of Eqs. (1)–(4)on B i,ǫ in the sense of the problem (21)–(24) with φ θ i ≡ on B i,ǫ . Therefore, we can rewrite (27) in new coordinates and use the parallelepiped propertyof Ω i, + ǫ in the space of variables ( s, θ ) : N ( t ) = I X i =1 Z Ω i, + ǫ (1 − u ( s, θ, t )) | J ( s, θ ) | dsdθ + O ( e − tδ )= I X i =1 Z ∂ Ω ∩ Ω i, + ǫ dθ Z [0 ,ǫ ] ds (1 − u ( s, θ, t )) | J ( s, θ ) | + O ( e − tδ ) . Since this local representation holds for all i (the form of the problem (29)–(33) is thesame for all i ) and P Ii =1 R ∂ Ω ∩ Ω i, + ǫ dθ = R ∂ Ω dθ , we can formally write N ( t ) = Z ∂ Ω dθ Z [0 ,ǫ ] ds (1 − u ( s, θ, t )) | J ( s, θ ) | + O ( e − tδ ) , (34)where u is the solution of (29)–(33) in ] − ǫ, ǫ [ × ∂ Ω in the local sense, as explainedpreviously. We denote by ˜ G ( s , θ , s , θ , t ) the Green function of the problem (29)–(33) in ∂ Ω × ] − ǫ, ǫ [ . Let us fix a boundary point (0 , θ ) .We denote by G θ the Green function corresponding to the following constant coeffi-cient problem, considered as a local trace problem, i.e. in the sense of the problem (21)–(24) with φ θ i ≡ on B i,ǫ : ∂∂t u + − D + ∂ ∂s + n − X i =1 ∂ ∂θ i ! u + = 0 , < s < ǫ (35) ∂∂t u − − D − ∂ ∂s + n − X i =1 ∂ ∂θ i ! u − = 0 , − ǫ < s < (36) u + | t =0 = 1 , u − | t =0 = 0 , (37) D − ∂u − ∂s | s = − = λ ( θ )( u − − u + ) | s =0 ,D + ∂u + ∂s | s =+0 = D − ∂u − ∂s | s = − . (38)Next, let G θ R n ( s , θ , s , θ , t ) = { s > } G θ ++ ( s , θ , s , θ , t ) + { s < } G θ − + ( s , θ , s , θ , t )
20e the Green function of the constant coefficient problem in the half space, explicitlyobtained in B. Then, according to Ref. [30] p.48–49, due to Varadhan’s bound propertyof Green functions, in U ǫ (0 , θ ) the difference between the Green function φ θ G θ of theproblem (35)–(38) and the analogous Green function in R n , G θ R n , is exponentially small: | ( φ θ G θ − G θ R n ) | U ǫ (0 ,θ ) × U ǫ (0 ,θ ) | = O (cid:16) e − tδ (cid:17) . Therefore, following the ideas of McKean and Singer[30] (p.49), we approximate ˜ G bythe Green function G θ with the frozen coefficients on (0 , θ ) , whose replacement by G θ R n yields only an exponentially small error.For an abstract operator Cauchy problem ∂∂t u − Au = R u, (39) u | t =0 = u the solution u can be found by the Duhamel formula u ( t ) = e − tA u + Z t e − ( t − τ ) A R u ( τ )d τ. (40)Therefore, by the Duhamel formula, locally, we have the following infinite expansion u + ( s, θ , t ) = Z Ω ǫ dθ ds G θ ++ ( s, θ , s , θ , t )+ Z t d τ Z Ω ǫ dθ ds G θ ++ ( s, θ , s , θ , t − τ ) ·· Z Ω ǫ dθ ds R G θ ++ ( s , θ , s , θ , τ )+ Z t d τ Z Ω ǫ dθ ds G θ ++ ( s, θ , s , θ , t − τ ) ·· Z τ d τ Z Ω ǫ dθ ds R G θ ++ ( s , θ , s , θ , τ − τ ) ·· Z Ω ǫ dθ ds R G θ ++ ( s , θ , s , θ , τ ) + . . . + O (cid:16) e − tδ (cid:17) (41)where the operator R is defined by R = R s ( s , θ ) + R θ ( s , θ ) , (42) R s ( s, θ ) = R ( s, θ ) ∂∂s = − D + n − X i =1 k i ( θ ) + s n − X i =1 k i ( θ )1 − sk i ( θ ) ! ∂∂s , (43) R θ ( s, θ ) = n − X i =1 D + sk i ( θ )1 − sk i ( θ ) (cid:18) − sk i ( θ ) (cid:19) ∂ ∂θ i + D + | J ( s, θ ) | n − X i =1 ∂∂θ i (cid:18) | J ( s, θ ) | (1 − sk i ( θ )) (cid:19) ∂∂θ i . (44)We substitute Eq. (41) into Eq. (34) with θ = θ and prove the following theorem:21 heorem 4 Let ˆ u = (cid:26) ˆ u + , < s < ǫ ˆ u − , − ǫ < s < be the solution of the one-dimensional problem ∂∂t ˆ u − D ± ∂ ∂s ˆ u = R s ( s, θ )ˆ u − ǫ < s < ǫ, θ ≡ θ , (45) ˆ u | t =0 = ,s > } G θ ++ ( s , θ , s , θ , t )+ { s < ,s > } G θ − + ( s , θ , s , θ , t ) . Due to Eq. (34), we need to know only G θ ++ G θ ++ ( s , θ , s , θ , t ) = (cid:0) h θ + ( s , s , t ) − f θ + ( s , s , t ) (cid:1) K ( θ , θ , D + t ) , h θ + ( s , s , t ) = 1 √ πD + t (cid:18) exp (cid:18) − ( s − s ) D + t (cid:19) + a ( λ, , θ ) exp (cid:18) − ( s + s ) D + t (cid:19)(cid:19) , (53) f θ + ( s , s , t ) = b ( λ, , θ ) λ ( θ ) D + exp (cid:18) λ ( θ ) α √ D + ( s + s ) + λ ( θ ) α t (cid:19) · Erfc (cid:18) s + s √ D + t + λ ( θ ) α √ t (cid:19) , (54)where a ( λ, , θ ) = , λ ( θ ) < ∞ , √ D + − √ D − √ D + + √ D − , λ ( θ ) = ∞ ,b ( λ, , θ ) = (cid:26) , λ ( θ ) < ∞ , , λ ( θ ) = ∞ , and K ( θ , θ , D ± t ) is the heat kernel in R n − : K ( θ , θ , D ± t ) = 1(4 πD ± t ) n − exp (cid:18) − | θ − θ | D ± t (cid:19) . (55)Since N ( t ) = Z Ω ǫ (1 − u ǫ ( s, θ, t )) | J ( s, θ ) | d sdθ + O ( e − tδ )= Vol(Ω ǫ ) − Z Ω ǫ Z Ω ǫ G ( s, θ, s , θ , t ) | J ( s, θ ) | d sdθ d s dθ + O ( e − tδ ) , (56)in what follows we use P ( t ) for the notation of the principal part of N ( t ) : P ( t ) = Z Ω ǫ Z Ω ǫ G ( s, θ, s , θ , t ) | J ( s, θ ) | d sdθ d s dθ . (57)To prove Theorem 4 we need the following Lemma: Lemma 2
The principal part P ( t ) of the heat content for the solution of the system (29)–(33), defined in Eq. (57), is given by P ( t ) = Z Ω ǫ Z Ω ǫ G θ ++ ( s, θ , s , θ , t ) | J ( s, θ ) | d sdθ dθ ds + G θ ++ ♯ ( R θ s + R θ θ ) G θ ++ + G θ ++ ♯ ( R θ s + R θ θ ) G θ ++ ♯ ( R θ s + R θ θ ) u + + O ( e − tδ ) , (58) with notation G θ ++ ♯ ( R θ s + R θ θ ) G θ ++ = Z t d τ Z Ω ǫ d sdθ | J ( s, θ ) | Z Ω ǫ G θ ++ ( s, θ , s , θ , t − τ ) ·· Z Ω ǫ ( R θ s ( s , θ ) + R θ θ ( s , θ )) G θ ++ ( s , θ , s , θ , τ ) dθ ds dθ ds . (59)23 oreover, the following identities hold M ( t ) = Z Ω ǫ Z Ω ǫ G θ ++ ( s, θ , s , θ , t ) | J ( s, θ ) | d sdθ dθ ds = Z ∂ Ω dθ Z [0 ,ǫ ] ds d s ( h θ + ( s, s , t ) − f θ + ( s, s , t )) | J ( s, θ ) | , (60) G θ ++ ♯ R θ θ G θ ++ = G θ ++ ♯ R θ θ G θ ++ ♯ R θ θ G θ ++ = . . . = 0 , (61) Z ∂ Ω dθ K ( θ , θ , D + t )= Z ∂ Ω dθ Z ∂ Ω dθ K ( θ , θ , D + ( t − τ )) K ( θ , θ , D + τ )= . . . = ∂ Ω ( θ ) . (62) Proof.
Formula (58) is the direct corollary of the Duhamel formula (see (40) and (41)).Let us start to prove (60).Indeed, we find that M ( t ) = Z Ω ǫ Z Ω ǫ ( h θ + ( s , s , t ) − f θ + ( s, s , t )) K ( θ , θ , D + ( θ ) t ) | J ( s, θ ) | ds d sdθ dθ = Z R n − Z R n − dθ dθ πD + t ) n − exp (cid:18) − | θ − θ | D + t (cid:19) ∂ Ω ( θ ) ∂ Ω ( θ )Φ( θ , t ) , where Φ( θ , t ) = Z [0 ,ǫ ] dsds ( h θ + ( s, s , t ) − f θ + ( s, s , t )) | J ( s, θ ) | . (63)With the change of variables θ v = θ − θ √ D + t , M ( t ) becomes M ( t ) = Z R n − Z R n − e −| v | π n − ∂ Ω ( θ ) ∂ Ω+ √ D + tv ( θ )Φ( θ , t ) dvdθ . By our construction, θ ∈ ∂ Ω and θ = θ − p D + tv ∈ ∂ Ω , that implies ∂ Ω ( θ ) − ∂ Ω ( θ ) ∂ Ω+ √ D + tv ( θ ) ≡ . It can be interpreted in the following way: if we take a point on the boundary and moveit along the boundary, we obtain another point which is still a boundary point.Consequently, we find (60) M ( t ) = Z R n − Z R n − e − v π n − ∂ Ω ( θ )Φ( θ , t ) dvdθ = Z ∂ Ω Φ( θ , t ) dθ , Z ∂ Ω dθ K ( θ , θ , D + t ) = ∂ Ω ( θ ) . Let us now prove that in the computation of P ( t ) all terms containing the derivativesover the transversal variable θ vanish.For all terms in (29) containing a derivative over θ , we calculate (see (44)) R θ K ( θ , θ , D + t )= n − X i =1 D + s k i ( θ )1 − s k i ( θ ) (cid:18) − s k i ( θ ) (cid:19) · D + t (cid:18) ( θ i − θ i ) D + t − (cid:19) K ( θ , θ , D + t ) − | J ( s , θ ) | n − X i =1 ∂∂θ i (cid:18) D + | J ( s , θ ) | (1 − s k i ( θ )) (cid:19) ( θ i − θ i )2 D + t K ( θ , θ , D + t ) . Let us prove Eq. (61), noting that G θ ++ ♯ R θ G θ ++ = Z t d τ Z Ω ǫ d sdθ | J ( s, θ ) | Z Ω ǫ G θ ++ ( s, θ , s , θ , t − τ ) ·· Z Ω ǫ R θ G θ ++ ( s , θ , s , θ , τ ) dθ ds dθ ds . We can schematically rewrite G θ ++ ♯ R θ G θ ++ in the following form: G θ ++ ♯ R θ G θ ++ == Z t d τ Z R n − dθ Z R n − dθ Z R n − dθ K ( θ , θ , D + ( t − τ )) R θ K ( θ , θ , D + τ ) · ∂ Ω ( θ ) ∂ Ω ( θ ) ∂ Ω ( θ ) Z [0 ,ǫ ] d s d s d s φ ( s, s , s , t, τ, θ ) . With the change of variables involving θ : ˜ θ = θ − θ p D + ( t − τ ) , θ = θ − p D + ( t − τ )˜ θ , (64) ˜ θ = θ − θ √ D + τ , θ = θ − p D + τ ˜ θ , and so θ = θ − p D + ( t − τ )˜ θ − p D + τ ˜ θ , (65)and since for all θ ∈ ∂ Ω ∂ Ω ( θ ) − ∂ Ω ( θ ) ∂ Ω+2 √ D + ( t − τ )˜ θ ( θ ) ·· ∂ Ω+2 √ D + ( t − τ )˜ θ +2 √ D + τ ˜ θ ( θ ) = 0 ,
25e obtain the separation of variables on ˜ θ from ( θ , s , ˜ θ ) : G θ ++ ♯ R θ G θ ++ = Z t d τ Z ∂ Ω dθ Z [0 ,ǫ ] d s d s d s φ ( s, s , s , t, τ, θ ) · n − Y i =1 "Z R d ˜ θ i e − (˜ θ i ) √ π Z R d ˜ θ i e − (˜ θ i ) √ π (cid:16) C i (2(˜ θ i ) − − C i ˜ θ i (cid:17) , where C i and C i are the functions of s , θ , ˜ θ , but not of ˜ θ , and consequently G θ ++ ♯ R θ G θ ++ = 0 . By the same reason we have Eq. (61). Changing variables θ i to ˜ θ i from (64)–(65), wealso obtain the last part of (62). (cid:3) Let us know prove Theorem 4.
Proof.
To find Eq. (48), we study Eq. (58) using proved relations (60)–(62). Forinstance, we have G θ ++ ♯ ( R θ s + R θ θ ) G θ ++ = G θ ++ ♯ R θ s G θ ++ = Z t d τ Z Ω ǫ d sdθ | J ( s, θ ) | Z Ω ǫ G θ ++ ( s, θ , s , θ , t − τ ) ·· Z Ω ǫ R θ s ( s , θ ) G θ ++ ( s , θ , s , θ , τ ) dθ ds dθ ds . As G θ ++ ( s, θ , s , θ , t ) = ( h θ + ( s, s , t ) − f θ + ( s, s , t )) K ( θ , θ , D + ( θ ) t ) , we have G θ ++ ♯ R θ s G θ ++ = ( h θ + − f θ + ) K♯ R θ s ( h θ + − f θ + ) K. Now we perform the change of variables (64)–(65). Since locally k i ( θ ) ∈ C , then forall θ ∈ ∂ Ω , for t → +0 we can develop k i ( θ − p D + ( t − τ )˜ θ ) = k i ( θ ) − ∇ k i ( θ )2 p D + ( t − τ )˜ θ + O ( t − τ ) . Consequently, by definition of R s ( s , θ ) in (43), which is a composition of the operatorof the first derivative by s and of a multiplication by a function of the class C on θ (locally, in the sense of local variables), we also have for t → +0 R s ( s , θ ) = R s ( s , θ )[1 + O ( t − τ )] − ∇ θ R s ( s , θ )2 p D + ( t − τ )˜ θ . As p D + ( t − τ ) R R d ˜ θ i e − (˜ θ i ) ˜ θ i = 0 , we obtain G θ ++ ♯ R θ s G θ ++ = Z t d τ Z ∂ Ω dθ Z [0 ,ǫ ] d s | J ( s, θ ) | Z [0 ,ǫ ] d s Z [0 ,ǫ ] d s · ( h θ + − f θ + )( s, s , t − τ ) R s ( s , θ ) [1 + O ( t − τ )] ( h θ + − f θ + )( s , s , τ ) · n − Y i =1 "Z R d ˜ θ i e − (˜ θ i ) √ π Z R d ˜ θ i e − (˜ θ i ) √ π = Z t d τ Z ∂ Ω dθ Z [0 ,ǫ ] d s | J ( s, θ ) | Z [0 ,ǫ ] d s · Z [0 ,ǫ ] d s ( h θ + − f θ + )( s, s , t − τ ) R s ( s , θ ) [1 + O ( t − τ )] ( h θ + − f θ + )( s , s , τ ) , P ( t ) = Z ∂ Ω dθ Z [0 ,ǫ ] d s Z [0 ,ǫ ] d s ( h θ + ( s, s , t ) − f θ + ( s, s , t )) | J ( s, θ ) | + [1 + O ( t )] Z t d τ Z ∂ Ω dθ Z [0 ,ǫ ] d s | J ( s, θ ) | Z [0 ,ǫ ] d s ( h θ + − f θ + )( s, s , t − τ ) · Z [0 ,ǫ ] d s R s ( s , θ )( h θ + − f θ + )( s , s , τ )+ [1 + O ( t )] Z ∂ Ω dθ ( h θ + − f θ + ) ♯ R s ( s, θ )( h θ + − f θ + ) ♯ R s ( s, θ ) u + ǫ . (66)We notice that the solution ˆ u ( s, θ , t ) of the one-dimensional system (45)–(47) is givenby ˆ u ( s, θ , t ) = Z [0 ,ǫ ] d s ( h θ + ( s, s , t ) − f θ + ( s, s , t ))+ Z t d τ Z [0 ,ǫ ] d s ( h θ + − f θ + )( s, s , t − τ ) Z [0 ,ǫ ] d s R s ( s , θ )( h θ + − f θ + )( s , s , τ )+( h θ + − f θ + ) ♯ R s ( s, θ )( h θ + − f θ + ) ♯ R s ( s, θ )ˆ u. To obtain (48) of Theorem 4 from formula (66), we estimate
N N ( t ) = O ( t ) Z t d τ Z ∂ Ω dθ Z [0 ,ǫ ] d s | J ( s, θ ) |· Z [0 ,ǫ ] d s ( h θ + − f θ + )( s, s , t − τ ) Z [0 ,ǫ ] d s R s ( s , θ )( h θ + − f θ + )( s , s , τ ) . (67)In fact, from (52), proven in what follows, it holds (see (68) for the definition of N N ( t ) ) N N ( t ) = O ( t ) N N ( t ) = O ( t ) (cid:26) O ( t ) , < λ < ∞ O ( t ) , λ = ∞ = (cid:26) O ( t ) , < λ < ∞ O ( t ) , λ = ∞ . To conclude, we note that if all principal curvatures k j ( θ ) on Ω ǫ are constant, thenfor all θ ∈ ∂ Ω R s ( s, θ ) ≡ R s ( s, θ ) , and thus N ( t ) = Z ∂ Ω dθ Z [0 ,ǫ ] d s (1 − ˆ u ( s, θ , t )) | J ( s, θ ) | + O ( e − tδ ) . To show (52), we need to estimate
N N j ( t ) = j X l =1 Γ θ ++ ♯ R s Γ θ ++ ♯ . . . ♯ R s Γ θ ++ ( l − fold ) , (68)where Γ θ ++ = ( h θ + − f θ + ) . j ≥ | N N j ( t ) | ≤ C ( t j µ ( ∂ Ω , √ D + t ) , < λ < ∞ t j µ ( ∂ Ω , √ D + t ) , λ = ∞ . (69)Due to Lemma 2, we start with (see (66)) N N ( t ) = Z t d τ Z ∂ Ω dθ Z [0 ,ǫ ] d s | J ( s, θ ) |· Z [0 ,ǫ ] d s ( h θ + − f θ + )( s, s , t − τ ) Z [0 ,ǫ ] d s R s ( s , θ )( h θ + − f θ + )( s , s , τ ) . Therefore, we have to estimate four terms:
N N ( t ) = X j =1 M M j ( t ) , where M M ( t ) = Z t d τ Z ∂ Ω dθ Z [0 ,ǫ ] d s | J ( s, θ ) |· Z [0 ,ǫ ] d s h θ + ( s, s , t − τ ) Z [0 ,ǫ ] d s R s ( s , θ ) h θ + ( s , s , τ ) ,M M ( t ) = − Z t d τ Z ∂ Ω dθ Z [0 ,ǫ ] d s | J ( s, θ ) |· Z [0 ,ǫ ] d s f θ + ( s, s , t − τ ) Z [0 ,ǫ ] d s R s ( s , θ ) h θ + ( s , s , τ ) ,M M ( t ) = − Z t d τ Z ∂ Ω dθ Z [0 ,ǫ ] d s | J ( s, θ ) |· Z [0 ,ǫ ] d s h θ + ( s, s , t − τ ) Z [0 ,ǫ ] d s R s ( s , θ ) f θ + ( s , s , τ ) ,M M ( t ) = Z t d τ Z ∂ Ω dθ Z [0 ,ǫ ] d s | J ( s, θ ) |· Z [0 ,ǫ ] d s f θ + ( s, s , t − τ ) Z [0 ,ǫ ] d s R s ( s , θ ) f θ + ( s , s , τ ) . We aim to approximate R s ( s , θ ) = R ( s , θ ) ∂ s from Eq. (43) near the point ( s, θ ) .For t → +0 and < s < ǫ = O ( √ t ) , we find that − s k i ( θ ) = 1 + s k i ( θ ) + O ( s ) , which gives R ( s , θ ) = − D + n − X i =1 k i ( θ ) + s n − X i =1 k i ( θ ) + O ( s ) ! . C ± = s ± s , I s ± s ( τ ) = exp (cid:18) − ( s ± s ) D + τ (cid:19) , c = 18 πD (0 , θ ) p ( t − τ ) τ , we find h θ + ( s, s , t − τ ) R s ( s , θ ) h θ + ( s , s , τ ) = − cR ( s , θ ) ( C − [ I s − s ( t − τ ) I s − s ( τ )+ a ( λ, , θ ) I s + s ( t − τ ) I s − s ( τ )]+ a ( λ, , θ ) C + [ I s − s ( t − τ ) I s + s ( τ ) + a ( λ, , θ ) I s + s ( t − τ ) I s + s ( τ )]) . We now change s to z and s to z by the following change of variables: • for P s ∓ s ( τ ) : z = s ∓ s √ D + τ and s = ± s ∓ √ D + τ z , • for P s ∓ s ( t − τ ) : z = s ∓ s √ D + ( t − τ ) and s = ± s ∓ p D + ( t − τ ) z .Let us notice that t is a constant parameter and, as τ takes its values between and t ,hence, z and z are in R + or R . But at the same time p D + ( t − τ ) z = s ± s and √ D + τ z = s ± s are bounded to the interval [ − ǫ, ǫ ] and hence are of the order of O ( √ t ) . In what follows, we suppose that τ and t − τ have the same order of smallnessas t : O ( t ) = O ( τ ) = O ( t − τ ) . Therefore, for < s = ± s ∓ p D + ( t − τ ) z < ǫ we have R s ( s , θ ) = [ φ ( θ ) ∓ ψ ( s, z , θ )] ∂∂s , where φ ( θ ) = − D + n − X i =1 k i ( θ ) ,ψ ( s, z , θ ) = ( s − p D + ( t − τ ) z ) D + n − X i =1 k i ( θ ) + O ( t ) . If we develop R in the neighborhood of (0 , θ ) , we find R ( s , θ ) = R (0 , θ ) + O ( √ t ) = φ ( θ ) + O ( √ t ) . (70)For λ = ∞ on ∂ Ω , we simply have M M ( t ) = M M ( t ) = M M ( t ) = 0 , and | N N ( t ) | = | M M ( t ) | = | h θ + ♯ R s h θ + |≤ C (cid:12)(cid:12)(cid:12)(cid:12)Z t d τ √ τ Z ∂ Ω dθ Z ǫ d s | J ( s, θ ) | (cid:12)(cid:12)(cid:12)(cid:12) ≤ C √ tµ ( ∂ Ω , p D + t ) .
29y iteration of the proof, we show that | N N j ( t ) | ≤ Ct j µ ( ∂ Ω , √ D + t ) for j ≥ .Now, for λ < ∞ , M M ( t ) = h θ + ♯ R s h θ + = − Z t d τ √ τ Z ∂ Ω dθ π √ D + Z R d s | J ( s, θ ) | Z R d z e − z · (cid:20)Z R d z φ ( θ ) z e − z (cid:21) X i =1 χ i ( s, z , z ) − Z t d τ √ τ Z ∂ Ω dθ π √ D + Z R d s | J ( s, θ ) |· Z R d z e − z · (cid:20)Z R d z ψ ( s, z , θ ) z e − z (cid:21) · X i =1 χ i ( s, z , z ) − X i =3 χ i ( s, z , z ) ! . Here for v + ( z ) = 2 p D + ( t − τ ) z R + ( z ) , v ( z ) = 2 p D + ( t − τ ) z , w + ( z ) = 2 √ D + τ z R + ( z ) and w ( z ) = 2 √ D + τ z χ ( s, z , z ) = [0 ,ǫ ] ( s ) [0 ,ǫ ] ( s − v ( z )) [0 ,ǫ ] ( s − v ( z ) − w ( z )) ,χ ( s, z , z ) = [0 ,ǫ ] ( s ) [0 ,ǫ ] ( s − v ( z )) [0 ,ǫ ] ( − s + v ( z ) + w + ( z )) ,χ ( s, z , z ) = [0 ,ǫ ] ( s ) [0 ,ǫ ] ( − s + v + ( z )) [0 ,ǫ ] ( − s + v + ( z ) − w ( z )) ,χ ( s, z , z ) = [0 ,ǫ ] ( s ) [0 ,ǫ ] ( − s + v + ( z )) [0 ,ǫ ] ( s − v + ( z ) + w + ( z )) . Considering two formulas: η = [0 ,ǫ ] ( s ) − [0 ,ǫ ] ( s − v ) [0 ,ǫ ] ( s ) , and ζ = [0 ,ǫ ] ( s ) [0 ,ǫ ] ( − s + v + ) , we find that η = 0 ⇐⇒ (cid:26) < v < ǫ < s < v − ǫ < v < ǫ + v < s < ǫ (71) ζ = 0 ⇐⇒ (cid:26) < v < ǫ < s < vǫ < v < ǫ v − ǫ < s < ǫ (72)It means that for < v = v + < ǫ , it holds η ( s ) = ζ ( s ) = [0 ,v + ] ( s ) and for − ǫ < v < and ǫ < v + = v + 2 ǫ < ǫ it holds η ( s ) = ζ ( s ) = [ v + − ǫ,ǫ ] ( s ) = [ ǫ + v,ǫ ] ( s ) . Consequently, we found the formula [0 ,ǫ ] ( s ) − [0 ,ǫ ] ( s − v ) [0 ,ǫ ] ( s ) = [0 ,ǫ ] ( s ) [0 ,ǫ ] ( − s + v + ) , (73)from which it follows χ = [0 ,ǫ ] ( s ) [0 ,ǫ ] ( s − v ) − χ , χ = [0 ,ǫ ] ( s ) [0 ,ǫ ] ( − s + v + ) − χ . Therefore, we have X i =1 χ i ( s, z , z ) = [0 ,ǫ ] ( s ) and χ + χ − χ − χ = [0 ,ǫ ] ( s ) − · [0 ,v + ( z )] ( s ) , z . Since Z R z e − z dz = 0 and Z R (2 z − e − z dz = 0 , we obtain exactly | M M ( t ) | = | h♯ R s h | = 0 . For
M M we find in completely analogous way M M ( t ) = 2 √ π Z t d τ √ t − τ √ τ Z ∂ Ω dθ λ ( θ ) D + Z R d s | J ( s, θ ) |· Z R + d z e λ ( θ ) αz √ t − τ + λ ( θ ) α ( t − τ ) Erfc( z + λ ( θ ) α √ t − τ ) · Z R d z h φ ( θ ) z e − z i ( χ ( s, z , z ) + χ ( s, z , z ))+ 2 √ π Z t d τ √ t − τ √ τ Z ∂ Ω dθ λ ( θ ) D + Z R d s | J ( s, θ ) | Z R + d z · e λ ( θ ) αz √ t − τ + λ ( θ ) α ( t − τ ) Erfc( z + λ ( θ ) α √ t − τ ) · Z R d z h ψ ( s, z , θ ) z e − z i ( χ ( s, z , z ) − χ ( s, z , z )) . Since χ ( s, z , z ) + χ ( s, z , z ) = [0 ,ǫ ] ( s ) [0 ,ǫ ] ( − s + v + ( z )) = [0 ,v + ( z )] ( s ) ,χ ( s, z , z ) − χ ( s, z , z ) = 2 χ ( s, z , z ) − [0 ,ǫ ] ( s ) [0 ,ǫ ] ( − s + v + ( z )) , the parts of M M , which contain the integration over s on [0 , p D + ( t − τ ) z ] , areequal to zero. In addition, for ℓ = 0 , (cid:12)(cid:12)(cid:12)(cid:12)Z R + d z z ℓ e λ ( θ ) αz √ t − τ + λ ( θ ) α ( t − τ ) Erfc( z + λ ( θ ) α √ t − τ ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C. As ψ is of the order O ( √ t ) and linear on z , and ǫ = O ( √ t ) , we directly obtain | M M ( t ) | = (cid:12)(cid:12)(cid:12)(cid:12) √ π Z t d τ √ t − τ √ τ Z ∂ Ω dθ λ ( θ ) D + Z R d s | J ( s, θ ) | Z R + d z · e λ ( θ ) αz √ t − τ + λ ( θ ) α ( t − τ ) Erfc( z + λ ( θ ) α √ t − τ ) · Z R d z ψ ( s, z , θ ) z e − z χ ( s, z , z ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C Z t d τ √ τ Z ∂ Ω dθ Z ǫ d s | J ( s, θ ) | ≤ Ct µ ( ∂ Ω , p D + t ) . Since µ ( ∂ Ω , √ D + t ) = C √ t for a regular boundary, then | M M ( t ) | ≤ Ct .
31o estimate
M M we find ∂ s f θ + ( s , s , τ ) = λ ( θ ) α √ D + f θ + ( s , s , τ ) − λ ( θ ) D + √ πD + τ exp (cid:18) − ( s + s ) D + τ (cid:19) . In our notations, using (70), we have h θ + ( s, s , t − τ ) R s ( s , θ ) f θ + ( s , s , τ ) = P s − s + P s + s p πD + ( t − τ ) ·· ( φ ( θ ) + O ( √ t )) (cid:26) λ ( θ ) α √ D + f θ + ( s , s , τ ) − λ ( θ ) D + √ πD + τ P s + s (cid:27) . Changing variables s to z and s to z , we obtain χ ± χ for the area of s , whichgives intervals (linearly) depending on the values of z and z . Thus, we majorate s by ǫ and estimate M M : | M M ( t ) | ≤ C | Z t d τ √ τ Z ∂ Ω dθ Z ǫ d s | J ( s, θ ) |· Z R d z e − z Z R + d z ( φ ( θ ) + O ( √ t )) f ( z , τ ) | , where f ( z , τ ) = λ ( θ ) α √ D + exp (cid:0) λ ( θ ) αz √ τ + λ ( θ ) α τ (cid:1) · Erfc( z + λ ( θ ) α √ τ ) − √ πD + τ e − z . We see that √ τ (cid:12)(cid:12)(cid:12)(cid:12)Z R + d z ( φ ( θ ) + O ( √ t )) f ( z , τ ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C. Therefore, we have | M M ( t ) | ≤ Ctµ ( ∂ Ω , p D + t ) . In the same way, since χ depends on z and z at the same time, we have | M M ( t ) | ≤ C (cid:12)(cid:12)(cid:12)(cid:12)Z t d τ p τ ( t − τ ) Z ∂ Ω dθ Z ǫ d s | J ( s, θ ) |· Z + ∞ d z e λ ( θ ) αz √ t − τ + λ ( θ ) α ( t − τ ) Erfc( z + λ ( θ ) α √ t − τ ) · Z + ∞ d z ( φ ( θ ) + O ( √ t )) f ( z , τ ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Ct µ ( ∂ Ω , p D + t ) . By iteration of the proof, we show for j ≥ that | N N j ( t ) | ≤ Ct j µ ( ∂ Ω , p D + t ) . (cid:3) Relation of the heat content expansion with the in-terior Minkowski sausage
Let us start with a heat problem with just a discontinuous initial condition. D + = D − = const Lemma 3
Let Ω ⊂ R n be a compact connected bounded domain with a connected bound-ary ∂ Ω of the Hausdorff dimension d and u is the solution of the following problem: ∂ t u − D △ u = 0 x ∈ R n , t > , (74) u | t =0 = Ω , (75) Then for t → +0 we have N ( t ) = Z e − z √ π µ ( ∂ Ω , √ Dtz )d z + o (cid:16) t n − d (cid:17) . (76) Moreover, it can be approximated by N ( t ) = β n − d µ ( ∂ Ω , √ Dt ) + o (cid:16) t n − d (cid:17) , (77) with the prefactor β x ≡ Z z x e − z √ π d z = 12 √ π γ (cid:18) x + 12 , (cid:19) (78) is expressed through the incomplete Gamma function. Proof.
Let us prove formula (76). By definition N ( t ) = Z R n \ Ω Z R n G ( x, y, t ) Ω d x d y, where this time G is the heat kernel in R n G ( x, y, t ) = (4 Dπt ) − n exp (cid:18) − | x − y | Dt (cid:19) . Therefore, we have N ( t ) = Vol(Ω) − Z R n Z R n πDt ) n e − | x − y | Dt Ω ( x ) Ω ( y )d x d y = Vol(Ω) − Z R n π n e −| v | (cid:18)Z R n Ω ( x ) Ω ( x + 2 √ Dtv )d x (cid:19) d v = Z R n π n e −| v | (cid:20)Z Ω (cid:0) Ω ( x ) − Ω − √ Dtv ( x ) (cid:1) d x (cid:21) d v, where Ω − √ Dtv ( x ) = Ω ( x + 2 √ Dtv ) and the notation Ω − √ Dtv means that Ω isshifted by the vector − √ Dtv ∈ R n . 33et us firstly suppose that ∂ Ω is regular, i.e of the class C . We see that for all points x ∈ Ω for which d ( x, ∂ Ω) ≥ √ Dt k v k , it holds ( x + 2 √ Dtv ) ∈ Ω . Thus, it follows thatfor ǫ = 2 √ Dt k v k , Ω ( x ) (cid:0) Ω ( x ) − Ω − √ Dtv ( x ) (cid:1) = 0 for all x ∈ Ω \ Ω ǫ . Therefore, only x belonging to Ω ǫ with k v k < ǫ √ Dt contribute to N ( t ) and we canwrite: N ( t ) = Z R n π n e −| v | (cid:20)Z Ω ǫ (cid:0) Ω ǫ ( x ) − Ω ǫ − √ Dtv ( x ) (cid:1) d x (cid:21) d v + O (cid:16) e − tδ (cid:17) , where the exponentially small error with a δ > is defined by the integral Z k v k > ǫ √ Dt π n e −| v | d v. Since ∂ Ω is regular, we introduce (see Section 4) the local coordinates x = ( θ, s ) and thus have ˆ x ( θ ) ∈ ∂ Ω and x ∈ Ω ǫ iff < s < ǫ . In this case, χ √ Dt,v ( x ) = Ω ǫ ( x ) − Ω ǫ − √ Dtv ( x ) = 0 iff x ∈ Ω ǫ and ˆ x ( θ ) − sn ( θ ) + 2 √ Dtv / ∈ Ω . Moreover, withthe notation ( v, n ) for the Euclidean inner product of two vectors in R n , (ˆ x ( θ ) − sn ( θ ) + 2 √ Dtv ) · n ( θ ) = − s + 2 √ Dt ( v, n ) . We deduce that χ √ Dt,v ( x ) = 0 iff s − √ Dt ( v, n ) < . Consequently, if ( v, n ) < , as s > , it is not possible to have s − √ Dt ( v, n ) < . Inturn, if < ( v, n ) then s ∈ ]0 , √ Dt ( v, n )[ . Considering only ( v, n ) > , we can define ǫ = 2 √ Dt ( v, n ) and, since v = x − y √ Dt and x, y ∈ Ω √ Dt ( v,n ) , we have < ( v, n ) < .Thus, the vector v can be locally decomposed in two parts: v = (( v, n ) , ( v, ˆ x )) = ( v n , v ˆ x ) .Thus, returning to N ( t ) , we obtain with the error O ( t ) which comes from the Jacobianapproximation (see | J ( s, θ ) | in Section 4) N ( t ) = Z R n − π n − e −| v ˆ x | dv ˆ x Z √ π e −| v n | (cid:18)Z Ω ǫ χ √ Dt,v n ( x )d x (cid:19) dv n + O ( t )= Z e − z √ π µ ( ∂ Ω , √ Dtz ) dz + o ( t n − d ) . If ∂ Ω is regular, then d = n − and o ( t n − d ) = o ( √ t ) , which, as it was mentioned, isactually O ( t ) . The last formula that depends only on a volume of the interior Minkowskisausage, holds for all types of connected boundaries described in Subsection 2.2.The formula (77) follows from Eq. (76) and the relation µ ( ∂ Ω , ǫz ) = z n − d µ ( ∂ Ω , ǫ ) + O ( ǫ n − d ) ) , (79)which, for a fixed z and ǫ → +0 , is evident for the regular case and can be proved byapproximating the fractal volume by a converging sequence of the volumes for smoothboundaries. For d = n − in Eq. (77), one has β = − e − √ π ≈ . . (cid:3) A comparison between the asymptotic formula (77) and a numerical solution of theproblem (74)–(75) is illustrated in Fig. 4 (for a square and a prefractal domain).34 −5 −4 −3 −2 −3 −2 −1 t N ( t ) (a) numericalasymptotic −5 −4 −3 −2 −1 t N ( t ) (b) numericalasymptotic Figure 4: Comparison between the asymptotic formula (77) (solid line) and a FreeFem++numerical solution of the problem (74)–(75) (circles) for two domains: (a) the unitsquare (with
Vol( ∂ Ω) = 4 ) and (b) the second generation of the Minkowski fractal,with
Vol( ∂ Ω) = 2 · . We set D + = D − = D = 1 . Let us come back to the problem (1)–(4).According to Theorem 4 (Eq. (52)), the heat content can be found up to the termseither t , or t (depending on values of λ ), by integrating over all boundary points θ ofthe solution ˆ u hom of the homogeneous problem (49)–(51) with constant coefficients takenat a boundary point (0 , θ ) . Obviously, Eq. (52) is valid only for regular boundaries.Let us reformulate it to allow an explicit calculation of the heat content for all types ofboundaries mentioned in Section 2.For this purpose, given ǫ = O ( √ t ) , ǫ > √ D + t , we divide ∂ Ω (which is still supposedto be regular) into J disjoint parts B j ( j = 1 , . . . , J ) of the size δ n − with < δ ≤ ǫ such that ∂ Ω = ⊔ Jj =1 B j .For t → +0 , δ → and thus, due to regularity of ∂ Ω on each B j × ]0 , ǫ [ thelocal change of variables from Section 4 is a C -diffeomorphism. In addition, since u continuously depends on λ (see Theorem 2), ˆ u hom , considered as a function of θ , by thecontinuity of λ , is continuous on θ . Therefore, by the mean value theorem and due tothe positivity of | J ( s, θ ) | , we deduce that for all j = 1 , . . . J there exists θ j ∈ B j suchthat Z B j dθ Z [0 ,ǫ ] d s (1 − ˆ u hom ( s, θ, t )) | J ( s, θ ) | = Z [0 ,ǫ ] d s (1 − ˆ u hom ( s, θ j , t )) Z B j dθ | J ( s, θ ) | . From Eq. (69), Eq. (52) becomes N ( t ) − J X j =1 Z [0 ,ǫ ] d s (1 − ˆ u hom ( s, θ j , t )) Z B j dθ | J ( s, θ ) | = (cid:26) O ( t µ ( ∂ Ω , √ t )) , < λ < ∞ O ( √ t µ ( ∂ Ω , √ t )) , λ = ∞ . Hence we prove 35 heorem 5
The heat content for the solution of the problem (1)–(4) can be explicitlyfound for all types of boundaries ∂ Ω (a connected boundary of a compact domain describedin Subsection 2.2) using the following expressions:1. for λ < ∞ on ∂ Ω : N ( t ) = 2 √ t √ D + Vol( ∂ Ω) (cid:20) µ ( ∂ Ω , p D + t ) Z ∂ Ω dσλ ( σ ) Z d zf ( σ, z, t ) − Z d zµ ( ∂ Ω , p D + t ( z − Z ∂ Ω dσλ ( σ ) f ( σ, z, t ) − Z d zµ ( ∂ Ω , p D + tz ) Z ∂ Ω dσλ ( σ ) f ( σ, z, t ) (cid:21) + O ( tµ ( ∂ Ω , √ t )) , (80) where α = √ D − + √ D + and f ( σ, z, t ) = exp (cid:16) λ ( σ ) α √ tz + λ ( σ ) α t (cid:17) Erfc( z + λ ( σ ) α √ t ) . (81)
2. for λ = ∞ on ∂ Ω : N ( t ) = 2 √ D − √ D − + √ D + Z e − z √ π µ ( ∂ Ω , p D + tz )d z + O ( √ t µ ( ∂ Ω , √ t )) . (82) Formulas (80) and (82) can be approximated by1. for λ < ∞ on ∂ Ω : N ( t ) = 2 √ t µ ( ∂ Ω , √ D + t ) √ D + Vol( ∂ Ω) (cid:20)Z ∂ Ω dσλ ( σ ) Z d zf ( σ, z, t ) − Z d z ( z − n − d Z ∂ Ω dσλ ( σ ) f ( σ, z, t ) − Z d zz n − d Z ∂ Ω dσλ ( σ ) f ( σ, z, t ) (cid:21) + O ( √ t µ ( ∂ Ω , √ t ) ) , (83)
2. for λ = ∞ on ∂ Ω : N ( t ) = 2 √ D − β n − d √ D − + √ D + µ ( ∂ Ω , p D + t ) + O ( µ ( ∂ Ω , √ t ) ) , (84) where β x was defined in Eq. (78). Proof.
Using Eqs. (56)–(60), N ( t ) becomes N ( t ) − µ ( ∂ Ω , ǫ ) + J X j =1 Z [0 ,ǫ ] ds d s ( h θ j + ( s, s , t ) − f θ j + ( s, s , t )) Z B j dθ | J ( s, θ ) | = (cid:26) O ( t µ ( ∂ Ω , √ D + t )) , < λ < ∞ O ( √ t µ ( ∂ Ω , √ D + t )) , λ = ∞ . N h j ( t ) = Z [0 ,ǫ ] ds d s h θ j + ( s, s , t ) Z B j dθ | J ( s, θ ) | . Changing variables as in the proof of Theorem 4,
N h j ( t ) becomes N h j ( t ) = Z B j dθ Z R e − z √ π (cid:18)Z R [0 ,ǫ ] ( s ) [0 ,ǫ ] ( s − p D + tz ) | J ( s, θ ) | d s (cid:19) d z + a ( λ, , θ j ) Z B j dθ Z R e − z √ π (cid:18)Z R [0 ,ǫ ] ( s ) [0 ,ǫ ] ( − s + p D + tz ) | J ( s, θ ) | d s (cid:19) d z. Therefore, we obtain
N h j ( t ) = Z B j dθ (cid:20)Z [0 ,ǫ ] | J ( s, θ ) | d s − Z R e − z √ π (cid:18)Z [0 ,ǫ ] ( [0 ,ǫ ] ( s ) − [0 ,ǫ ]+ √ D + tz ( s )) | J ( s, θ ) | d s (cid:19) d z + a ( λ, , θ j ) Z R e − z √ π (cid:18)Z R [0 ,ǫ ] ( s ) [ − ǫ, √ D + tz ( s ) | J ( s, θ ) | d s (cid:19) d z . Applying formula (73) with v = √ D + t z (see also Subsection 6.1), we find N h j ( t ) = µ ( B j , ǫ ) − (1 − a ( λ, , θ j )) Z e − z √ π µ ( B j , p D + tz )d z. Thus, for λ < ∞ N h j ( t ) = µ ( B j , ǫ ) since a = 1 .We treat the second part in the same way, N f j ( t ) = − Z [0 ,ǫ ] ds d s f θ j + ( s, s , t ) Z B j dθ | J ( s, θ ) | , which is equal to zero for λ = ∞ . For f ( θ j , z, t ) from Eq. (81), we find that N f j ( t ) = − λ ( θ j ) √ t √ D + Z R d s d z [0 ,ǫ ] ( s ) [0 ,ǫ ] ( − s + 2 p D + tz ) f ( z, t ) Z B j | J ( s, θ ) | dθ = − λ ( θ j ) √ t √ D + "Z d zf ( θ j , z, t ) Z B j Z √ D + t ( z − √ D + t | J ( s, θ ) | d sdθ + Z d zf ( θ j , z, t ) Z B j Z √ D + tz | J ( s, θ ) | d sdθ = − λ ( θ j ) √ t √ D + (cid:20) µ ( B j , p D + t ) Z f ( θ j , z, t )d z − Z f ( θ j , z, t ) µ ( B j , p D + t ( z − z + Z f ( θ j , z, t ) µ ( B j , p D + tz )d z (cid:21) . Putting two results together, we obtain the following approximations for N ( t ) :37. for λ < ∞ on ∂ Ω : N ( t ) = J X j =1 µ ( B j , p D + t ) 2 λ ( θ j ) √ t √ D + Z f ( θ j , z, t )d z − J X j =1 λ ( θ j ) √ t √ D + (cid:20)Z f ( θ j , z, t ) µ ( B j , p D + t ( z − z − Z f ( θ j , z, t ) µ ( B j , p D + tz )d z (cid:21) + O ( t µ ( ∂ Ω , √ t )) , (85)2. for λ = ∞ on ∂ Ω : N ( t ) = 2 √ D − √ D − + √ D + J X j =1 Z e − z √ π µ ( B j , p D + t z )d z + O ( √ t µ ( ∂ Ω , √ t )) . (86)It means that if the formulas for µ ( B j , δ ) are known, we get the approximation of N ( t ) up to terms of the order of t n − d +22 for λ < ∞ , and of the order of t n − d for λ = ∞ .Moreover, this approximation, depending only on the volume of ∂ Ω , holds for all typesof boundaries, even fractals (see Subsection 2.2 and p. 378 of Ref. [11] for a similarconclusion).Let us now change the sum over j with the integral over z and make J → + ∞ : lim J → + ∞ J X j =1 C ( z, t, θ j ) µ ( B j , p D + tz ) = Z ∂ Ω C ( z, t, σ ) dist( σ, p D + tz ) dσ, where dσ is understood in the sense of the Hausdorff measure ( d -measure) defined on ∂ Ω . Thus, again with the help of the mean value theorem, we have Z ∂ Ω C ( z, t, σ ) dist( σ, p D + tz ) dσ = µ ( ∂ Ω , √ D + t )Vol( ∂ Ω) Z ∂ Ω C ( z, t, σ ) dσ, from which Eqs. (80) and (82) follow. We use Eq. (79) to obtain formulas (83) and (84). (cid:3) In the case of a regular boundary we provide the asymptotic expansion of the heat contentup to the third-order term.In this case, we can approximate the solution of the system (29)–(33) by the solution38 of the following problem (instead of (35)–(38), as previously) ∂∂t u + − D + ∂ ∂s + n − X i =1 ∂ ∂θ i ! u + + D + n − X i =1 k i ( θ ) ∂u + ∂s = 0 , < s < ǫ (87) ∂∂t u − − D − ∂ ∂s + n − X i =1 ∂ ∂θ i ! u − + D − n − X i =1 k i ( θ ) ∂u − ∂s = 0 , − ǫ < s < (88) u + | t =0 = 1 , u − | t =0 = 0 , (89) D − ∂u − ∂s | s = − = λ ( θ )( u − − u + ) | s =0 ,D + ∂u + ∂s | s =+0 = D − ∂u − ∂s | s = − . (90)In this approximation the remainder terms of the system (29)–(33) contain only thecoefficients of the order √ t (to compare with (70)): R ( s , θ ) = s D + n − X i =1 k i ( θ ) + O ( s ) , that gives R ( s , θ ) = ( s ∓ p D + ( t − τ ) z ) D + n − X i =1 k i ( θ ) + O ( t ) = O ( √ t ) . (91)The basis of the parametrix is the Green function given by (see Section B.1) h θ + ( s , s , t ) = 1 √ πD + t (cid:18) exp (cid:18) − ( s − s − tD + γ ( θ )) D + t (cid:19) + a ( λ, , θ ) exp (cid:18) − ( s + s − tD + γ ( θ )) D + t (cid:19)(cid:19) , (92) f θ + ( s , s , t ) = b ( λ, , θ ) λ ( θ ) D + ·· exp (cid:18) λ ( θ ) α √ D + ( s + s − tD + γ ( θ )) + λ ( θ ) α t (cid:19) ·· Erfc (cid:18) s + s − tD + γ ( θ )2 √ D + t + λ ( θ ) α √ t (cid:19) , (93)where γ ( θ ) = P n − i =1 k i ( θ ) . The estimate (69) becomes N ( t ) − Z ∂ Ω dθ Z [0 ,ǫ ] d s (1 − ˆ u homǫ ( s, θ , t )) | J ( s, θ ) | = (cid:26) O ( t ) , < λ < ∞ O ( t ) , λ = ∞ (94)Consequently, for the regular case we have 39 heorem 6 Let Ω be a compact domain of R n with a connected boundary ∂ Ω ∈ C ∞ ( R n ) .Then for λ = ∞ we have N ( t ) = 2 1 − e − √ π √ D + D − √ D + + √ D − Vol( ∂ Ω) √ t + O ( t ) . (95) In the case of < λ < ∞ , we have N ( t ) = 4 C t Z ∂ Ω λ ( σ ) dσ − C t (cid:20) (cid:18) √ D + + 1 √ D − (cid:19) Z ∂ Ω λ ( σ ) dσ − p D + ( n − Z ∂ Ω λ ( σ ) H ( σ ) dσ (cid:21) + O ( t ) , (96) where H is the mean curvature, and C = 1 + 32 erf(1) −
94 erf(2) + 1 √ π (cid:18) e − e (cid:19) ≈ . , (97) C = 1 √ π − e − − e − √ π − ≈ . . (98) Proof.
Let us consider the case λ = ∞ . Using the Green function given in Eqs. (92)and (93), we obtain N ( t ) = J X j =1 β Z − √ tD + γ ( θj e − z √ π µ ( B j , p D + tz + tD + γ ( θ j )) dz + O ( t ) , (99)where β = √ D − √ D − + √ D + . In Eq. (99) the remainder term also contains the integrals R − √ tD + γ ( θj dz . From Eq. (99) we find N ( t ) = J X j =1 β Z − √ tD + γ ( xj )2 e − z √ π Z B j dθ Z √ D + tz + γ ( x j ) D + t ds (1 − s ( n − H ( θ ))+ O ( t ) . Therefore, we have N ( t ) = √ t C p D + J X j =1 Vol( B j ) ! − t ( n − J X j =1 ξ " Vol( B j ) H ( x j ) − Z B j H ( σ ) dσ + O ( t ) , (100)where C = 1 − e − √ π √ D − √ D + + √ D − ,ξ = (cid:18) e − √ π − erf(2) (cid:19) D + √ D − √ D + + √ D − .
40n addition, for all σ ∈ B j , the distance between x j (which also belongs to B j ) and σ goes to as J → + ∞ . Thus, since | H ( x j ) − H ( σ ) | ≤ H ′ ( σ ) | x j − σ | ≤ C Vol( B j ) , we have lim J → + ∞ J X j =1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Vol( B j ) H ( x j ) − Z B j H ( σ ) dσ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 . Hence, from Eq. (100) we obtain Eq. (95).The case < λ < ∞ can be treated in the similar way using in Eq. (85) the expansionof the f ( σ, t, z ) : f ( σ, t, z ) = exp (cid:16) λ ( σ ) α √ tz + λ ( σ ) α ( σ ) t (cid:17) Erfc( z + λ ( σ ) α √ t )= Erfc( z ) − λ ( σ ) α √ t (cid:18) √ π e − z − z Erfc( z ) (cid:19) + O ( t ) . (cid:3) Acknowledgment
We thank Frédéric Abergel and François Golse for helpful discussions and Jean-BaptisteApoung Kamga for advises on FreeFem++ simulations.
A Definitions of Besov spaces on fractals
Let us define the Besov space B , β ( ∂ Ω) on a d -set ∂ Ω (see Ref. [21] p.135 andRef. [22]).There are many equivalent definitions[23, 31] of Besov spaces. To give one of them,we introduce[21, 23] a net N with mesh − ν , ν ∈ N , i.e. a division of R n into half-opennon-overlapping cubes W with edges of length − ν , obtained by intersecting R n withhyperplanes orthogonal to the axes. In addition, we denote by P k ( N ) the set of functionswhich on each cube W in the net N coincide with a polynomial of degree at most k . Definition 4 (Besov space B p,qβ (Γ) , β > , see Ref. [21]) Let Γ be a closed subsetof R n which is a d -set preserving Markov’s inequality for < d ≤ n and let m d be afixed d -measure on Γ . We say that f ∈ B p,qβ (Γ) , β > , ≤ p, q ≤ + ∞ , if f ∈ L p ( m d ) and there is a sequence B = ( B ν ) ν ∈ N ∈ ℓ q such that for every net N with mesh − ν , ν ∈ N there exists a function s ( N ) ∈ P [ β ] ( N ) (by [ β ] is denoted the integer part of β )satisfying k f − s ( N ) k L p ( m d ) ≤ − νβ B ν . The norm of f in B p,qβ (Γ) is given by the formula k f k B p,qβ (Γ) = k f k L p ( m d ) + inf B k B k ℓ q , where the infimum is over all such sequences B . ( B , β ( ∂ Ω)) ′ = B , − β ( ∂ Ω) is introduced in Ref. [23]. To give thedefinition of the Besov space B , − β ( ∂ Ω) we need to define the atoms: Definition 5 (Atom[23])
Let β > , ≤ p ≤ ∞ , and let W with W ∩ Γ = ∅ bea cube with edge length − ν , ν ∈ N . A function a = a W ∈ L p ( m d ) is a ( − β, p ) -atomassociated with W if1. supp a ⊂ W , where W is the cube obtained by expanding W twice from itscenter,2. R x γ a ( x ) dm d = 0 for | γ | ≤ [ β ] if ν > ,3. k a k L p ( m d ) ≤ νβ . Let N ν (Γ) = { W ∈ N ν | W ∩ Γ = ∅ } with the notation N ν of the net with mesh − ν such that the origin is a corner of some cube in the net. Then we can define the Besovspace with a negative parameter − β , B , − β ( ∂ Ω) , which is actually[23] the dual Besovspace of B , β ( ∂ Ω) : Definition 6 (Besov space B p,q − β (Γ) , β > , see Ref. [21]) The space B p,q − β (Γ) , β > , ≤ p, q ≤ ∞ consists of functions f ∈ D ′ ( R n ) which are given by ∀ φ ∈ D ( R n ) h f, φ i = X ν ∈ N X W ∈N ν (Γ) s W Z a W φdm d , where a W are ( − β, p ) -atoms and s W are numbers such that S = ( S ν ) ν ∈ N ∈ ℓ q and S ν is defined by S ν = X W ∈N ν (Γ) | s W | p p . The norm of f is defined by k f k B p,q − β (Γ) = inf k S k ℓ q , where the infimum is taken over all possible atomic decompositions of f : f = X ν ∈ N X W ∈N ν (Γ) s W a W . B Explicit computations for half space problem withconstant coefficients .1 Case λ = ∞ The Green function of the one-dimensional problem (49)–(51) with λ = ∞ and s ∈ R was treated in Ref. [6, 1] and it is given by Γ( s, s , t ) = { s> , s > } Γ ++ ( s, s , t ) + { s< , s > } Γ − + ( s, s , t ) with Γ ++ ( s, s , t ) = 1 √ πD + t (cid:18) exp (cid:18) − ( s − s ) D + t (cid:19) + A exp (cid:18) − ( s + s ) D + t (cid:19)(cid:19) , (101) Γ − + ( s, s , t ) = B √ πD + t exp − (cid:16) s − s q D − D + (cid:17) D − t , (102)where A = √ D + − √ D − √ D + + √ D − and B = √ D + √ D + + √ D − .Let us use this result to find the Green function Γ reg ( s, s , t ) of the following one-dimensional problem ∂∂t u + − D + ∂ ∂s u + + D + k ∂∂s u + = 0 , s > (103) ∂∂t u − − D − ∂ ∂s u − + D − k ∂∂s u − = 0 , s < (104) u + | t =0 = 1 , u − | t =0 = 0 , (105) u + | s =+0 = u − | s = − , D + ∂∂s u + | s =+0 = D − ∂∂s u − | s = − , (106)The constant coefficient problem ∂∂t u − D ∂ ∂s u + Dk ∂∂s u = 0 , s ∈ R , (107) u | t =0 = u , (108)has the Green function of the form K ( s, s , t ) = 1 √ πDt e − ( s − y − tDk )24 Dt , that means that the change of variables s − tDk = X transforms (107) to ∂∂t u − D ∂ ∂X u = 0 with the Green function K ( X, Y, t ) = 14 πDt e − ( X − Y )24 Dt .
43n addition[6], we know (see (101)–(102)) the Green function for the constant coefficientproblem ∂∂t u + − D + ∂ ∂X u + = 0 X > ,∂∂t u − − D − ∂ ∂X u − = 0 X < ,u + | t =0 = 1 , u − | t =0 = 0 ,u + | X =+0 = u − | X = − , D + ∂∂X u + | X =+0 = D − ∂∂X u − | X = − . Consequently, we perform the following change of variables in Eqs. (103)–(106): X = s> ( s ) ( s + tD + k ) + s< ( s ) (cid:18) D − D + s − tD − k (cid:19) and obtain for z = D + D − X that ∂∂t u + − D + ∂ ∂X u + = 0 X > tD + k,∂∂t u − − D − ∂ ∂z u − = 0 z < tD + k,u + | t =0 = 1 , u − | t =0 = 0 ,u + | X =+ tD + k = u − | z = − tD + k , D + ∂∂X u + | X =+ tD + k = D − ∂∂z u − | z = − tD + k . Thus, Γ reg ++ ( s, s , t ) = 1 √ πD + t (cid:18) exp (cid:18) − ( s − s − tD + k ) D + t (cid:19) + A exp (cid:18) − ( s + s − tD + k ) D + t (cid:19)(cid:19) for s, s > , and Γ reg − + ( s, s , t ) = 1 √ πD + t B exp − ([ D + /D − ] s − s p D + /D − + tD + k ) D − t ! for s < , s > . Now, to obtain the Green function of the multidimensional problem ∂∂t u + − D + ∂ ∂s + n − X i =1 ∂ ∂θ i ! u + + D + k ∂∂s u + = 0 , s > , θ i ∈ R ,∂∂t u − − D − ∂ ∂s + n − X i =1 ∂ ∂θ i ! u − + D − k ∂∂s u − = 0 , s < , θ i ∈ R u + | t =0 = 1 , u − | t =0 = 0 ,u + | s =+0 = u − | s = − , D + ∂∂s u + | s =+0 = D − ∂∂s u − | s = − ,
44e apply the Fourier transform in s i variables and, due to the boundary conditionsdepending only on s , we obtain that ˆ G ± + , the Fourier transform of the Green function G ± + , can be found by the formula ˆ G ± + ( s, s , ξ, t ) = e − D ± | ξ | t Γ ± + ( s, s , t ) , where Γ ± + ( s, s , t ) is the Green function of the corresponding one-dimensional problem.This implies Γ reg ++ ( s, θ, s , θ , t ) = 1(4 πD + t ) n (cid:18) exp (cid:18) − ( s − s − tD + k ) D + t (cid:19) + A exp (cid:18) − ( s + s − tD + k ) D + t (cid:19)(cid:19) exp (cid:18) − | θ − θ | D + t (cid:19) for s, s > , θ, θ ∈ R n − , Γ reg − + ( s, θ, s , θ , t ) = B (4 πD − t ) n − √ πD + t exp − ( D + D − s − s q D + D − + tD + k ) D − t · exp (cid:18) − | θ − θ | D − t (cid:19) for s < , s > , θ, θ ∈ R n − . For Γ and Γ reg we also have Varadhan’s bounds[30] for s = s lim t → t ln Γ ++ ( s, s , t ) = lim t → t ln Γ reg ++ ( s, s , t ) = − d ( s, s ) D + , lim t → t ln Γ − + ( s, s , t ) = lim t → t ln Γ reg − + ( s, s , t ) = − d (cid:0) s, s q D − D + (cid:1) D − , where d ( s, s ) is the Riemannian distance between s and s , which is equal here to theEuclidean distance, since D + and D − are constant in Ω + and Ω − respectively. B.2 Case < λ < ∞ Let us consider the one-dimensional problem (49)–(51) with λ ≡ λ ( θ ) and s ∈ R . Theassociated problem for the heat kernel is then given by (cid:0) ∂ t − D ± ∂ s (cid:1) G ( s, s , t ) = 0 ,G | t =0 = δ ( s, s ) for s > ,D − ∂∂s G ( − , s , t ) = λ ( G ( − , s , t ) − G (+0 , s , t )) , (109) D + ∂∂s G (+0 , s , t ) = D − ∂∂s G ( − , s , t ) . (110)We search the explicit solution of the problem[6] with G ( s, s , t ) = (cid:26) G − + , s < , s > G ++ , s > , s > .
45e seek for G − + and G ++ in terms of free heat kernel K ( s, s , D ± t ) (see Eq. (55)) andsingle layer heat potentials for s > : G ++ ( s, s , t ) = K ( s, s , D + t ) + D + Z t K ( s, , D + ( t − τ )) α + ( s , τ )d τ ( s > ,G − + ( s, s , t ) = D − Z t K ( s, , D − ( t − τ )) α − ( s , τ )d τ ( s < , where α ± ( s , τ ) are unknown densities to be determined. Considering the boundaryconditions (109)–(110) and the jumps of the first derivatives of G ± + , ∂∂s G ++ | s =+0 = − α + ( s , t ) + ∂∂s K (0 , s , D + t ) ,∂∂s G − + | s = − = − α − ( s , t ) , we obtain two relations D − α − ( s , t ) = − D + α + ( s , t ) + 2 D + ∂∂s K (0 , s , D + t ) ,D − α − ( s , t ) = 2 λK (0 , s , D + t ) + λ √ D + √ π Z t α + ( s , τ ) √ t − τ d τ − λ √ D − √ π Z t α − ( s , τ ) √ t − τ d τ. Following the method from Ref. [6], we solve the system corresponding to α − ( s , t ) and α + ( s , t ) : D − α − ( s , t ) + D + α + ( s , t ) = 2 D + ∂∂s K (0 , s , D + t ) ,D − α − ( s , t ) + λ √ D − √ π s D − D + ! Z t α − ( s , τ ) √ t − τ d τ = 4 λK (0 , s , D + t ) . We obtain therefore the Abel integral equation of the second kind for α − ( s , t ) α − ( s , t ) + γ Z t α − ( s , τ ) √ t − τ d τ = 4 λD − K (0 , s , D + t ) , where γ = λ √ πD − (cid:16) q D − D + (cid:17) . Consequently, α − ( s , t ) = 4 λD − K (0 , s , D + t ) − γ λD − Z t K (0 , s , D + τ ) √ t − τ d τ + πγ λD − Z t e πγ ( t − τ ) (cid:18) K (0 , s , D + τ ) − γ Z τ K (0 , s , D + s ) √ τ − s ds (cid:19) d τ. G ++ ( s, s , t ) = 1 √ πD + t (cid:18) exp (cid:18) − ( s − s ) D + t (cid:19) + exp (cid:18) − ( s + s ) D + t (cid:19)(cid:19) − λD + exp (cid:18) λα √ D + ( s + s ) + λ α t (cid:19) Erfc (cid:18) s + s √ D + t + λα √ t (cid:19) , where α = √ D − + √ D + . By the same way, G − + ( s, s , t ) = λ √ D − D + exp λα √ D − − s + s s D − D + ! + λ α t ! · Erfc − s + s q D − D + √ D − t + λα √ t . We see that the Green function G ++ for λ = 0 becomes the Green function of theproblem with the Neumann boundary conditions and in this case N ( t ) = 0 , as u − ≡ .This property, N ( t ) = 0 , can be also directly found using the Green function.In R n for x = ( s, θ ) and y = ( s , θ ) ∈ R × R n − we have G ++ ( s, θ, s , θ , t ) R n = G ++ ( s, s , t ) R K ( θ, θ , D + t ) R n − ,G − + ( s, θ, s , θ , t ) R n = G − + ( s, s , t ) R K ( θ, θ , D − t ) R n − . Therefore in R n for Varadhan’s bounds with x = y we have lim t → t ln G ++ ( x, y, t ) R n = − d ( x, y ) D + , lim t → t ln G − + ( x, y, t ) R n = − d (cid:0) s, s q D − D + (cid:1) + d ( θ, θ ) D − . Remark 3
Applying this framework to the same system but with the transmittal boundary ondition for < λ < ∞ , we obtain G ++ ( s, θ, s , θ , t ) = 1(4 πD + t ) n (cid:18) exp (cid:18) − ( s − s − tD + k ) D + t (cid:19) + exp (cid:18) − ( s + s − tD + k ) D + t (cid:19)(cid:19) exp (cid:18) − d ( θ, θ ) D + t (cid:19) − πD + t ) n − λD + exp (cid:18) λα √ D + ( s + s − tD + k ) + λ α t (cid:19) ·· Erfc (cid:18) s + s − tD + k √ D + t + λα √ t (cid:19) exp (cid:18) − d ( θ, θ ) D + t (cid:19) ,G − + ( s, θ, s , θ , t ) = 1(4 πD − t ) n − λ √ D − D + · exp λα √ D − − D + D − s + s s D + D − + tD + k ! + λ α t ! · Erfc − D + D − s + s q D + D − + tD + k √ D − t + λα √ t exp (cid:18) − d ( θ, θ ) D − t (cid:19) . We also notice that for a fixed t > for λ → + ∞ we obtain G ++ ( s, s , t ) → Γ ++ ( s, s , t ) and G − + ( s, s , t ) → Γ − + ( s, s , t ) . References [1] H. S. Carslaw and J. C. Jaeger,
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