A dynamic model for viscoelastic materials with prescribed growing cracks
aa r X i v : . [ m a t h . A P ] N ov A DYNAMIC MODEL FOR VISCOELASTIC MATERIALS WITH PRESCRIBEDGROWING CRACKS
MAICOL CAPONI AND FRANCESCO SAPIO
Abstract.
In this paper, we prove the existence of solutions for a class of viscoelastic dynamic systems ontime–dependent cracked domains, with possibly degenerate viscosity coefficients. Under stronger regularityassumptions we also show a uniqueness result. Finally, we exhibit an example where the energy–dissipationbalance is not satisfied, showing there is an additional dissipation due to the crack growth.
Keywords : linear second order hyperbolic systems, dynamic fracture mechanics, elastodynamics, viscoelas-ticity, cracking domains.
MSC 2010 : 35L53, 35A01, 35Q74, 74H20, 74R10, 74D05.1.
Introduction
In the theory of Dynamic Fracture, the deformation of an elastic material evolves according to the elasto-dynamics system, while the evolution of the crack follows Griffith’s dynamic criterion, see [13]. This principle,originally formulated in [11] for the quasi–static setting, states that there is an exact balance between theenergy released during the evolution and the energy used to increase the crack, which is postulated to beproportional to the area increment of the crack itself.For an antiplane displacement, the elastodynamics system leads to the following wave equation¨ u ( t, x ) − ∆ u ( t, x ) = f ( t, x ) t ∈ [0 , T ] , x ∈ Ω \ Γ t , (1.1)with some prescribed boundary and initial conditions. Here, Ω ⊂ R d is an open bounded set with Lipschitzboundary, which represents the cross–section of the material, the closed set Γ t ⊂ Ω models the crack at time t in the reference configuration, u ( t ) : Ω \ Γ t → R is the antiplane displacement, and f is a forcing term. Inthis case, Griffith’s dynamic criterion reads E ( t ) + H d − (Γ t \ Γ ) = E (0) + work of external forces , where E ( t ) is the total energy at time t , given by the sum of kinetic and elastic energy, and H d − is the( d − u ( t, x ) − ∆ u ( t, x ) − ∆ ˙ u ( t, x ) = f ( t, x ) ( t, x ) ∈ (0 , T ) × Ω . (1.2)As it is well known, the solutions to (1.2) satisfy the energy–dissipation balance E ( t ) + Z t Z Ω |∇ ˙ u | d x d s = E (0) + work of external forces . (1.3)When we consider a crack in a viscoelastic material, Griffith’s dynamic criterion becomes E ( t ) + H d − (Γ t \ Γ ) + Z t Z Ω |∇ ˙ u | d x d s = E (0) + work of external forces . (1.4)For a prescribed crack evolution, this model was already considered by [3] in the antiplane case, and morein general by [18] for the vector–valued case. As proved in the quoted papers, the solutions to (1.2) on adomain with a prescribed time–dependent crack, i.e., with Ω replaced by Ω \ Γ t , satisfy (1.3) for every time. Preprint SISSA 12/2019/MATE.
This equality implies that (1.4) cannot be satisfied unless Γ t = Γ for every t . This phenomenon was alreadywell known in mechanics as the viscoelastic paradox, see for instance [17, Chapter 7].To overcome this problem, we modify Kelvin–Voigt’s model by considering a possibly degenerate viscosityterm depending on t and x . More precisely, we study the following equation¨ u ( t, x ) − ∆ u ( t, x ) − div(Ψ ( t, x ) ∇ ˙ u ( t, x )) = f ( t, x ) t ∈ [0 , T ] , x ∈ Ω \ Γ t . (1.5)On the function Ψ : (0 , T ) × Ω → R we only require some regularity assumptions (see (2.7)); a particularlyinteresting case is when Ψ assumes the value zero on some points of Ω, which means that the material hasno longer viscoelastic properties in such a zone.The main result of this paper is Theorem 3.1, in which we show the existence of a weak solution to (1.5).This is done in the more general case of linear elasticity, that is when the displacement is vector–valued andthe elastic energy depends only on the symmetric part of its gradient. To this aim, we first perform a timediscretization in the same spirit of [3], and then we pass to the limit as the time step goes to zero by relyingon energy estimates; as a byproduct, we obtain the energy–dissipation inequality (4.4). By using the changeof variables method implemented in [14, 7], we also prove a uniqueness result, but only in dimension d = 2and when Ψ( t ) vanishes on a neighborhood of the tip of Γ t .We complete our work by providing an example in d = 2 of a weak solution to (1.5) for which the fracturecan grow while balancing the energy. More precisely, when the cracks Γ t move with constant speed alongthe x –axis and Ψ( t ) is zero in a neighborhood of the crack tip, we construct a function u which solves (1.5)and satisfies E ( t ) + Z t Z Ω | Ψ ∇ ˙ u | d x d s + H (Γ t \ Γ ) = E (0) + work of external forces . (1.6)Notice that this is the natural extension of Griffith’s dynamic criterion (1.4) to this setting.The paper is organized as follows. In Section 2 we fix the notation adopted throughout the paper, welist the standard assumptions on the family of cracks { Γ t } t ∈ [0 ,T ] and on the function Ψ, and we specify thenotion of weak solution to problem (1.5). In Section 3 we state our main existence result (Theorem 3.1),and we implement the time discretization method. We conclude the proof of Theorem 3.1 in Section 4,where we show the validity of the initial conditions and the energy–dissipation inequality (4.4). Section 5deals with uniqueness: under stronger regularity assumptions on the cracks sets, in Theorem 5.5 we provethe uniqueness of a weak solution, but only when the space dimension is d = 2. To this aim, we assumealso that the function Ψ is zero in a neighborhood of the crack tip. We conclude with Section 6, where indimension d = 2 we show an example of a moving crack that satisfies Griffith’s dynamic energy–dissipationbalance (1.6). 2. Notation and Preliminary Results
The space of m × d matrices with real entries is denoted by R m × d ; in case m = d , the subspace ofsymmetric matrices is denoted by R d × dsym . Given two vectors v , v ∈ R d , their Euclidean scalar product isdenoted by v · v ∈ R and their tensor product is denoted by v ⊗ v ∈ R d × d ; we use v ⊙ v ∈ R d × dsym todenote the symmetric part of v ⊗ v , namely v ⊙ v := ( v ⊗ v + v ⊗ v ). Given A ∈ R m × d , we use A T to denote its transpose; we use A · A ∈ R to denote the Euclidean scalar product of two matrices A , A ∈ R d × d .The partial derivatives with respect to the variable x i are denoted by ∂ i . Given a function f : R d → R m ,we denote its Jacobian matrix by ∇ f , whose components are ( ∇ f ) ij := ∂ j f i , i = 1 , . . . , m , j = 1 , . . . , d .For a tensor field F : R d → R m × d , by div F we mean the divergence of F with respect to rows, namely(div F ) i := P dj =1 ∂ j F ij , for i = 1 , . . . , m .The d –dimensional Lebesgue measure is denoted by L d and the ( d − H d − . We adopted standard notations for Lebesgue and Sobolev spaces on open subsets of R d ; given anopen set Ω ⊆ R d we use k·k ∞ to denote the norm of L ∞ (Ω; R m ). The boundary values of a Sobolev functionare always intended in the sense of traces. Given a bounded open set Ω with Lipschitz boundary, we denoteby ν the outer unit normal vector to ∂ Ω, which is defined H d − –a.e. on the boundary.Given a Banach space X , its norm is denoted by k · k X ; if X is an Hilbert space, we use ( · , · ) X to denoteits scalar product. The dual space of X is denoted by X ′ , and we use h· , ·i X ′ to denote the duality productbetween X ′ and X . Given two Banach spaces X and X , the space of linear and continuous maps from X to X is denoted by L ( X ; X ); given A ∈ L ( X ; X ) and u ∈ X , we write A u ∈ X to denote the imageof u under A .Given an open interval ( a, b ) ⊆ R , L p ( a, b ; X ) is the space of L p functions from ( a, b ) to X . Given u ∈ L p ( a, b ; X ), we denote by ˙ u ∈ D ′ ( a, b ; X ) its distributional derivative. The set of continuous functionsfrom [ a, b ] to X is denoted by C ([ a, b ]; X ). Given a reflexive Banach space X , C w ([ a, b ]; X ) is the set ofweakly continuous functions from [ a, b ] to X , namely C w ([ a, b ]; X ) := { u : [ a, b ] → X : t
7→ h x ′ , u ( t ) i X ′ is continuous from [ a, b ] to R for every x ′ ∈ X ′ } . Let T be a positive real number and let Ω ⊂ R d be a bounded open set with Lipschitz boundary. Let ∂ D Ω be a (possibly empty) Borel subset of ∂ Ω and let ∂ N Ω be its complement. We assume the followinghypotheses on the geometry of the cracks:(E1) Γ ⊂ Ω is a closed set with L d (Γ) = 0 and H d − (Γ ∩ ∂ Ω) = 0;(E2) for every x ∈ Γ there exists an open neighborhood U of x in R d such that ( U ∩ Ω) \ Γ is the unionof two disjoint open sets U + and U − with Lipschitz boundary;(E3) { Γ t } t ∈ [0 ,T ] is a family of closed subsets of Γ satisfying Γ s ⊂ Γ t for every 0 ≤ s ≤ t ≤ T .Thanks (E1)–(E3) the space L (Ω \ Γ t ; R m ) coincides with L (Ω; R m ) for every t ∈ [0 , T ] and m ∈ N . Inparticular, we can extend a function u ∈ L (Ω \ Γ t ; R m ) to a function in L (Ω; R m ) by setting u = 0 onΓ t . Moreover, the trace of u ∈ H (Ω \ Γ) is well defined on ∂ Ω. Indeed, we may find a finite number ofopen sets with Lipschitz boundary U j ⊂ Ω \ Γ, j = 1 , . . . m , such that ∂ Ω \ (Γ ∩ ∂ Ω) ⊂ ∪ mj =1 ∂U j . Since H d − (Γ ∩ ∂ Ω) = 0, there exists a constant
C >
0, depending only on Ω and Γ, such that k u k L ( ∂ Ω) ≤ C k u k H (Ω \ Γ) for every u ∈ H (Ω \ Γ; R d ) . (2.1)Similarly, we can find a finite number of open sets U j ⊂ Ω \ Γ, j = 1 , . . . m , with Lipschitz boundary, suchthat Ω \ Γ = ∪ mj =1 U j . By using second Korn’s inequality in each U j (see, e.g., [15, Theorem 2.4]) and takingthe sum over j we can find a constant C K , depending only on Ω and Γ, such that k∇ u k L (Ω; R d × d ) ≤ C K (cid:16) k u k L (Ω; R d ) + k Eu k L (Ω; R d × dsym ) (cid:17) for every u ∈ H (Ω \ Γ; R d ) , (2.2)where Eu is the symmetric part of ∇ u , i.e., Eu := ( ∇ u + ∇ u T ).For every t ∈ [0 , T ] we define V t := { u ∈ L (Ω \ Γ t ; R d ) : Eu ∈ L (Ω \ Γ t ; R d × dsym ) } . Notice that in the definition of V t we are considering only the distributional gradient of u in Ω \ Γ t and notthe one in Ω. The set V t is a Hilbert space with respect to the following norm k u k V t := ( k u k H + k Eu k H ) for every u ∈ V t . To simplify our exposition, for every m ∈ N we set H := L (Ω; R m ) and H N := L ( ∂ N Ω; R m ); we alwaysidentify the dual of H by H itself and L (0 , T ; L (Ω; R m )) by L ((0 , T ) × Ω; R m ).Thanks to (2.2), the space V t coincides with the usual Sobolev space H (Ω \ Γ t ; R d ). Therefore, by (2.1),it makes sense to consider for every t ∈ [0 , T ] the set V Dt := { u ∈ V t : u = 0 on ∂ D Ω } , which is a Hilbert space with respect to k·k V t . Moreover, by combining (2.2) with (2.1), we derive also theexistence of a constant C tr > k u k H N ≤ C tr k u k V T for every u ∈ V T . (2.3)Let C , B : Ω → L ( R d × dsym ; R d × dsym ) be two fourth-order tensors satisfying: C ijhk , B ijhk ∈ L ∞ (Ω) for every i, j, h, k = 1 , . . . , d, (2.4) C ( x ) η · η = η · C ( x ) η , B ( x ) η · η = η · B ( x ) η for a.e. x ∈ Ω and for every η , η ∈ R d × dsym , (2.5) C ( x ) η · η ≥ λ | η | , B ( x ) η · η ≥ λ | η | for a.e. x ∈ Ω and for every η ∈ R d × dsym , (2.6)for two positive constants λ , λ independent of x . Consider a function Ψ : (0 , T ) × Ω → R satisfyingΨ ∈ L ∞ ((0 , T ) × Ω) , ∇ Ψ ∈ L ∞ ((0 , T ) × Ω; R d ) . (2.7) M. CAPONI AND F. SAPIO
Given f ∈ L (0 , T ; H ), w ∈ H (0 , T ; H ) ∩ H (0 , T ; V ), g ∈ H (0 , T ; H N ), u ∈ V with u − w (0) ∈ V D ,and u ∈ H , we want to find a solution to the viscoelastic dynamic system¨ u ( t ) − div( C Eu ( t )) − div(Ψ ( t ) B E ˙ u ( t )) = f ( t ) in Ω \ Γ t , t ∈ (0 , T ) , (2.8)satisfying the following boundary and initial conditions u ( t ) = w ( t ) on ∂ D Ω, t ∈ (0 , T ) , (2.9)( C Eu ( t ) + Ψ ( t ) B E ˙ u ( t )) ν = g ( t ) on ∂ N Ω, t ∈ (0 , T ) , (2.10)( C Eu ( t ) + Ψ ( t ) B E ˙ u ( t )) ν = 0 on Γ t , t ∈ (0 , T ) , (2.11) u (0) = u , ˙ u (0) = u . (2.12)As usual, the Neumann boundary conditions are only formal, and their meaning will be specified in Defini-tion 2.4.Throughout the paper we always assume that the family { Γ t } t ∈ [0 ,T ] satisfies (E1)–(E3), as well as C , B ,Ψ, f , w , g , u , and u the previous hypotheses. Let us define the following functional spaces: V := { ϕ ∈ L (0 , T ; V T ) : ˙ ϕ ∈ L (0 , T ; H ) , ϕ ( t ) ∈ V t for a.e. t ∈ (0 , T ) } , V D := { ϕ ∈ V : ϕ ( t ) ∈ V Dt for a.e. t ∈ (0 , T ) } , W := { u ∈ V : Ψ ˙ u ∈ L (0 , T ; V T ) , Ψ( t ) ˙ u ( t ) ∈ V t for a.e. t ∈ (0 , T ) } . Remark 2.1.
In the classical viscoelastic case, namely when Ψ is identically equal to 1, the solution u tosystem (2.8) has derivative ˙ u ( t ) ∈ V t for a.e. t ∈ (0 , T ) with E ˙ u ∈ L (0 , T ; H ). For a generic Ψ we expect tohave Ψ E ˙ u ∈ L (0 , T ; H ). Therefore W is the natural setting where looking for a solution to (2.8). Indeed,from a distributional point of view we haveΨ( t ) E ˙ u ( t ) = E (Ψ( t ) ˙ u ( t )) − ∇ Ψ( t ) ⊙ ˙ u ( t ) in D ′ (Ω \ Γ t ; R d × dsym ) for a.e. t ∈ (0 , T ) , and E (Ψ ˙ u ) , ∇ Ψ ⊙ ˙ u ∈ L (0 , T ; H ) if u ∈ W , thanks to (2.7). Remark 2.2.
The set W coincides with the space of functions u ∈ H (0 , T ; H ) such that u ( t ) ∈ V t andΨ( t ) ˙ u ( t ) ∈ V t for a.e. t ∈ (0 , T ), and satisfying Z T k u ( t ) k V t + k Ψ( t ) ˙ u ( t ) k V t d t < ∞ . (2.13)This is a consequence of the strong measurability of the maps t u ( t ) and t Ψ( t ) ˙ u ( t ) from (0 , T ) into V T , which gives that (2.13) is well defined and u, Ψ ˙ u ∈ L (0 , T ; V T ). To prove the strong measurability ofthese two maps, it is enough to observe that V T is a separable Hilbert space and that the maps t ˙ u ( t ) and t Ψ( t ) ˙ u ( t ) from (0 , T ) into V T are weakly measurable. Indeed, for every ϕ ∈ C ∞ c (Ω \ Γ T ) the maps t Z Ω \ Γ T Eu ( t, x ) ϕ ( x ) d x = − Z Ω \ Γ T u ( t, x ) ⊙ ∇ ϕ ( x ) d x,t Z Ω \ Γ T E (Ψ( t, x ) ˙ u ( t, x )) ϕ ( x ) d x = − Z Ω \ Γ T Ψ( t, x ) ˙ u ( t, x ) ⊙ ∇ ϕ ( x ) d x are measurable from (0 , T ) into R , and C ∞ c (Ω \ Γ T ) is dense in L (Ω). Lemma 2.3.
The spaces V and W are Hilbert spaces with respect to the following norms: k ϕ k V := ( k ϕ k L (0 ,T ; V T ) + k ˙ ϕ k L (0 ,T ; H ) ) for every ϕ ∈ V , k u k W := ( k u k V + k Ψ ˙ u k L (0 ,T ; V T ) ) for every u ∈ W . Moreover, V D is a closed subspace of V .Proof. It is clear that k·k V and k·k W are norms on V and W induced by scalar products. We just have tocheck the completeness of such spaces with respect to these norms.Let { ϕ k } k ⊂ V be a Cauchy sequence. Then, { ϕ k } k and { ˙ ϕ k } k are Cauchy sequences, respectively, in L (0 , T ; V T ) and L (0 , T ; H ), which are complete Hilbert spaces. Thus there exists ϕ ∈ L (0 , T ; V T ) with˙ ϕ ∈ L (0 , T ; H ) such that ϕ k → ϕ in L (0 , T ; V T ) and ˙ ϕ k → ˙ ϕ in L (0 , T ; H ). In particular there exists asubsequence { ϕ k j } j such that ϕ k j ( t ) → ϕ ( t ) in V T for a.e. t ∈ (0 , T ). Since ϕ k j ( t ) ∈ V t for a.e. t ∈ (0 , T ) we DYNAMIC MODEL FOR VISCOELASTIC MATERIALS WITH PRESCRIBED GROWING CRACKS 5 deduce that ϕ ( t ) ∈ V t for a.e. t ∈ (0 , T ). Hence ϕ ∈ V and ϕ k → ϕ in V . With a similar argument, we canprove that V D ⊂ V is a closed subspace.Let us now consider a Cauchy sequence { u k } k ⊂ W . We have that { u k } k and { Ψ ˙ u k } k are Cauchysequences, respectively, in V and L (0 , T ; V T ), which are complete Hilbert spaces. Thus there exist twofunctions u ∈ V and z ∈ L (0 , T ; V T ) such that u k → u in V and Ψ ˙ u k → z in L (0 , T ; V T ). Since ˙ u k → ˙ u in L (0 , T ; H ) and Ψ ∈ L ∞ ((0 , T ) × Ω), we also have that Ψ ˙ u k → Ψ ˙ u in L (0 , T ; H ), which gives that z = Ψ ˙ u .Finally let us prove that Ψ( t ) ˙ u ( t ) ∈ V t for a.e. t ∈ (0 , T ). By the fact that Ψ ˙ u k → Ψ ˙ u in L (0 , T ; V T ), thereexists a subsequence { Ψ ˙ u k j } j such that Ψ( t ) ˙ u k j ( t ) → Ψ( t ) ˙ u ( t ) in V T for a.e. t ∈ (0 , T ). Since Ψ( t ) ˙ u k j ( t ) ∈ V t for a.e. t ∈ (0 , T ) we deduce that Ψ( t ) ˙ u ( t ) ∈ V t for a.e. t ∈ (0 , T ). Hence u ∈ W and u k → u in W . (cid:3) We are now in position to define a weak solution to (2.8)–(2.11).
Definition 2.4 (Weak solution) . We say that u ∈ W is a weak solution to system (2.8) with boundaryconditions (2.9)–(2.11) if u − w ∈ V D and − Z T ( ˙ u ( t ) , ˙ ϕ ( t )) H d t + Z T ( C Eu ( t ) , Eϕ ( t )) H d t + Z T ( B E (Ψ( t ) ˙ u ( t )) , Ψ( t ) Eϕ ( t )) H d t − Z T ( B ∇ Ψ( t ) ⊙ ˙ u ( t ) , Ψ( t ) Eϕ ( t )) H d t = Z T ( f ( t ) , ϕ ( t )) H d t + Z T ( g ( t ) , ϕ ( t )) H N d t (2.14)for every ϕ ∈ V D such that ϕ (0) = ϕ ( T ) = 0.Notice that the Neumann boundary conditions (2.10) and (2.11) can be obtained from (2.14), by usingintegration by parts in space, only when u ( t ) and Γ t are sufficiently regular. Remark 2.5.
If ˙ u is regular enough (for example ˙ u ∈ L (0 , T ; V T ) with ˙ u ( t ) ∈ V t for a.e. t ∈ (0 , T )), then wehave Ψ E ˙ u = E (Ψ ˙ u ) − ∇ Ψ ⊙ ˙ u . Therefore (2.14) is coherent with the strong formulation (2.8). In particular,for a function u ∈ W we can defineΨ E ˙ u := E (Ψ ˙ u ) − ∇ Ψ ⊙ ˙ u ∈ L (0 , T ; H ) , (2.15)so that equation (2.14) can be rephrased as − Z T ( ˙ u ( t ) , ˙ ϕ ( t )) H d t + Z T ( C Eu ( t ) , Eϕ ( t )) H d t + Z T ( B Ψ( t ) E ˙ u ( t ) , Ψ( t ) Eϕ ( t )) H d t = Z T ( f ( t ) , ϕ ( t )) H d t + Z T ( g ( t ) , ϕ ( t )) H N d t for every ϕ ∈ V D such that ϕ (0) = ϕ ( T ) = 0. Definition 2.6 (Initial conditions) . We say that u ∈ W satisfies the initial conditions (2.12) iflim h → + h Z h ( k u ( t ) − u k V t + k ˙ u ( t ) − u k H ) d t = 0 . (2.16)3. Existence
We now state our main existence result, whose proof will be given at the end of Section 4.
Theorem 3.1.
There exists a weak solution u ∈ W to (2.8) – (2.11) satisfying the initial conditions u (0) = u and ˙ u (0) = u in the sense of (2.16) . Moreover u ∈ C w ([0 , T ]; V T ) , ˙ u ∈ C w ([0 , T ]; H ) ∩ H (0 , T ; ( V D ) ′ ) , and lim t → + u ( t ) = u in V T , lim t → + ˙ u ( t ) = u in H. To prove the existence of a weak solution to (2.8)–(2.11), we use a time discretization scheme in the samespirit of [3]. Let us fix n ∈ N and set τ n := Tn , u n := u , u − n := u − τ n u . We define V kn := V Dkτ n , g kn := g ( kτ n ) , w kn := w ( kτ n ) for k = 0 , . . . , n, M. CAPONI AND F. SAPIO f kn := 1 τ n Z kτ n ( k − τ n f ( s ) d s, Ψ kn := 1 τ n Z kτ n ( k − τ n Ψ( s ) d s, δg kn := g kn − g k − n τ n for k = 1 , . . . , n,δw n := ˙ w (0) , δw kn := w kn − w k − n τ n , δ w kn := δw kn − δw k − n τ n for k = 1 , . . . , n. For every k = 1 , . . . , n let u kn ∈ V T , with u kn − w kn ∈ V kn , be the solution to( δ u kn , v ) H + ( C Eu kn , Ev ) H + ( B Ψ kn Eδu kn , Ψ kn Ev ) H = ( f kn , v ) H + ( g kn , v ) H N for every v ∈ V kn , (3.1)where δu kn := u kn − u k − n τ n for k = 0 , . . . , n, δ u kn := δu kn − δu k − n τ n for k = 1 , . . . , n. The existence of a unique solution u kn to (3.1) is an easy application of Lax–Milgram’s theorem. Remark 3.2.
Since δu kn ∈ V ( k − τ n , then Ψ kn Eδu kn = E (Ψ kn u kn ) −∇ Ψ kn ⊙ u kn , so that the discrete equation (3.1)is coherent with the weak formulation given in (2.14).In the next lemma, we show a uniform estimate for the family { u kn } nk =1 with respect to n ∈ N that willbe used later to pass to the limit in the discrete equation (3.1). Lemma 3.3.
There exists a constant
C > , independent of n ∈ N , such that max i =1 ,..,n k δu in k H + max i =1 ,..,n k Eu in k H + n X i =1 τ n k Ψ in Eδu in k H ≤ C. (3.2) Proof.
We fix n ∈ N . To simplify the notation we set a ( u, v ) := ( C Eu, Ev ) H , b kn ( u, v ) := ( B Ψ kn Eu, Ψ kn Ev ) H for every u, v ∈ V T . By taking as test function v = τ n ( δu kn − δw kn ) ∈ V kn in (3.1), for k = 1 , . . . , n we obtain k δu kn k H − ( δu k − n , δu kn ) H + a ( u kn , u kn ) − a ( u kn , u k − n ) + τ n b kn ( δu kn , δu kn ) = τ n L kn , where L kn := ( f kn , δu kn − δw kn ) H + ( g kn , δu kn − δw kn ) H N + ( δ u kn , δw kn ) H + a ( u kn , δw kn ) + b kn ( δu kn , δw kn ) . Thanks to the following identities k δu kn k H − ( δu k − n , δu kn ) H = 12 k δu kn k H − k δu k − n k H + τ n k δ u kn k H ,a ( u kn , u kn ) − a ( u kn , u k − n ) = 12 a ( u kn , u kn ) − a ( u k − n , u k − n ) + τ n a ( δu kn , δu kn ) , and by omitting the terms with τ n , which are non negative, we derive12 k δu kn k H − k δu k − n k H + 12 a ( u kn , u kn ) − a ( u k − n , u k − n ) + τ n b kn ( δu kn , δu kn ) ≤ τ n L kn . We fix i ∈ { , . . . , n } and sum over k = 1 , . . . , i to obtain the following discrete energy inequality12 k δu in k H + 12 a ( u in , u in ) + i X k =1 τ n b kn ( δu kn , δu kn ) ≤ E + i X k =1 τ n L kn , (3.3)where E := k u k H + ( C Eu , Eu ) H . Let us now estimate the right–hand side in (3.3) from above.By (2.3) and (2.4) we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) i X k =1 τ n ( f kn , δu kn − δw kn ) H (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ k f k L (0 ,T ; H ) + 12 k ˙ w k L (0 ,T ; H ) + 12 i X k =1 τ n k δu kn k H , (3.4) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) i X k =1 τ n a ( u kn , δw kn ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ k C k ∞ k ˙ w k L (0 ,T ; V ) + k C k ∞ i X k =1 τ n k Eu kn k H , (3.5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) i X k =1 τ n ( g kn , δw kn ) H N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ k g k L (0 ,T ; H N ) + C tr k ˙ w k L (0 ,T ; V ) . (3.6) DYNAMIC MODEL FOR VISCOELASTIC MATERIALS WITH PRESCRIBED GROWING CRACKS 7
For the other term involving g kn , we perform the following discrete integration by parts i X k =1 τ n ( g kn , δu kn ) H N = ( g in , u in ) H N − ( g (0) , u ) H N − i X k =1 τ n ( δg kn , u k − n ) H N . (3.7)Hence for every ǫ ∈ (0 , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) i X k =1 τ n ( g kn , δu kn ) H N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ǫ k u in k H N + 12 ǫ k g k L ∞ (0 ,T ; H N ) + k g (0) k H N k u k H N + i X k =1 τ n k δg kn k H N k u k − n k H N ≤ C ǫ + ǫC tr k u in k V T + C tr i X k =1 τ n k u kn k V T , (3.8)where C ǫ is a positive constant depending on ǫ . Thanks to Jensen’s inequality we can write k u ln k V T ≤ k Eu ln k H + k u k H + l X j =1 τ n k δu jn k H ! ≤ k Eu ln k H + 2 k u k H + 2 T l X j =1 τ n k δu jn k H , so that (3.8) can be further estimated as (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) i X k =1 τ n ( g kn , δu kn ) H N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C ǫ + ǫC tr k Eu in k H + 2 k u k H + 2 T i X j =1 τ n k δu jn k H ! + C tr i X k =1 τ n k Eu kn k H + 2 k u k H + 2 T k X j =1 τ n k δu jn k H ! ≤ ˜ C ǫ + ǫC tr k Eu in k H + ˜ C i X k =1 τ n (cid:0) k δu kn k H + k Eu kn k H (cid:1) , (3.9)for some positive constants ˜ C ǫ and ˜ C , with ˜ C ǫ depending on ǫ . Similarly to (3.7), we can say i X k =1 τ n ( δ u kn , δw kn ) H = ( δu in , δw in ) H − ( δu n , δw n ) H − i X k =1 τ n ( δu k − n , δ w kn ) H , (3.10)from which we deduce that for every ǫ > (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) i X k =1 τ n ( δ u kn , δw kn ) H (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤k δu in k H k δw in k H + k u k H k ˙ w (0) k H + i X k =1 τ n k δu k − n k H k δ w kn k H ≤ ǫ k δw in k H + ǫ k δu in k H + k u k H k ˙ w (0) k H + 12 i X k =1 τ n k δu k − n k H + 12 i X k =1 τ n k δ w kn k H ≤ ¯ C ǫ + ǫ k δu in k H + 12 i X k =1 τ n k δu kn k H , (3.11)where ¯ C ǫ is a positive constant depending on ǫ . We estimate from above the last term in right-hand sideof (3.3) in the following way i X k =1 τ n b kn ( δu kn , δw kn ) ≤ i X k =1 τ n ( b kn ( δu kn , δu kn )) ( b kn ( δw kn , δw kn )) ≤ i X k =1 τ n b kn ( δu kn , δu kn ) + 12 k B k ∞ k Ψ k ∞ k ˙ w k L (0 ,T ; V ) . (3.12)By considering (3.3)–(3.12) and using (2.6) we obtain Å − ǫ ã k δu in k H + λ − ǫC tr k Eu in k H + 12 i X k =1 τ n b kn ( δu kn , δu kn ) ≤ ˆ C ǫ + ˆ C i X k =1 τ n (cid:0) k δu kn k H + k Eu kn k H (cid:1) M. CAPONI AND F. SAPIO for two positive constants ˆ C ǫ and ˆ C , with ˆ C ǫ depending on ǫ . We can now choose ǫ < min n , λ C tr o toderive the following estimate14 k δu in k H + 14 k Eu in k H + 12 i X k =1 τ n b kn ( δu kn , δu kn ) ≤ C + C i X k =1 τ n (cid:0) k δu kn k H + k Eu kn k H (cid:1) , (3.13)where C and C are two positive constants depending only on u , u , f , g , and w . Thanks to a discreteversion of Gronwall’s lemma (see, e.g., [1, Lemma 3.2.4]) we deduce the existence of a constant C > i and n , such that k δu in k H + k Eu in k H ≤ C for every i = 1 , . . . , n and for every n ∈ N . By combining this last estimate with (3.13) and (2.6) we finally get (3.2) and we conclude. (cid:3)
We now want to pass to the limit into the discrete equation (3.1) to obtain a weak solution to (2.8)–(2.11).We start by defining the following approximating sequences of our limit solution u n ( t ) := u kn + ( t − kτ n ) δu kn , ˜ u n ( t ) := δu kn + ( t − kτ n ) δ u kn t ∈ [( k − τ n , kτ n ] , k = 1 , . . . , n,u + n ( t ) := u kn , ˜ u + n ( t ) := δu kn t ∈ (( k − τ n , kτ n ] , k = 1 , . . . , n,u − n ( t ) := u k − n , ˜ u − n ( t ) := δu k − n t ∈ [( k − τ n , kτ n ) , k = 1 , . . . , n. Notice that u n ∈ H (0 , T ; H ) with ˙ u n ( t ) = δu kn = ˜ u + n ( t ) for t ∈ (( k − τ n , kτ n ) and k = 1 , . . . , n . Let usapproximate Ψ and w byΨ + n ( t ) := Ψ kn , w + n ( t ) := w kn t ∈ (( k − τ n , kτ n ] , k = 1 , . . . , n, Ψ − n ( t ) := Ψ k − n , w − n ( t ) := w k − n t ∈ [( k − τ n , kτ n ) , k = 1 , . . . , n. Lemma 3.4.
There exists a function u ∈ W , with u − w ∈ V D , such that, up to a not relabeled subsequence u n H (0 ,T ; H ) −−−−−−− ⇀ n →∞ u, u ± n L (0 ,T ; V T ) −−−−−−− ⇀ n →∞ u, ˜ u ± n L (0 ,T ; H ) −−−−−−− ⇀ n →∞ ˙ u, (3.14) ∇ Ψ ± n ⊙ ˜ u ± n L (0 ,T ; H ) −−−−−−− ⇀ n →∞ ∇ Ψ ⊙ ˙ u, E (Ψ ± n ˜ u ± n ) L (0 ,T ; H ) −−−−−−− ⇀ n →∞ E (Ψ ˙ u ) . (3.15) Proof.
Thanks to Lemma 3.3 the sequences { u n } n ⊂ H (0 , T ; H ) ∩ L ∞ (0 , T ; V T ), { u ± n } n ⊂ L ∞ (0 , T ; V T ), and { ˜ u ± n } n ⊂ L ∞ (0 , T ; H ) are uniformly bounded. By Banach-Alaoglu’s theorem there exist u ∈ H (0 , T ; H )and v ∈ L (0 , T ; V T ) such that, up to a not relabeled subsequence u n L (0 ,T ; V T ) −−−−−−− ⇀ n →∞ u, ˙ u n L (0 ,T ; H ) −−−−−−− ⇀ n →∞ ˙ u, u + n L (0 ,T ; V T ) −−−−−−− ⇀ n →∞ v. Since there exists a constant
C > k u n − u + n k L ∞ (0 ,T ; H ) ≤ Cτ n −−−−→ n →∞ , we can conclude that u = v . Moreover, given that u − n ( t ) = u + n ( t − τ n ) for t ∈ ( τ n , T ), ˜ u + n ( t ) = ˙ u n ( t ) for a.e. t ∈ (0 , T ), and ˜ u − n ( t ) = ˜ u + n ( t − τ n ) for t ∈ ( τ n , T ), we deduce u − n L (0 ,T ; V T ) −−−−−−− ⇀ n →∞ u, ˜ u ± n L (0 ,T ; H ) −−−−−−− ⇀ n →∞ ˙ u. By (3.2) we derive that the sequences { E (Ψ + n ˜ u + n ) } n ⊂ L (0 , T ; H ) and {∇ Ψ + n ⊙ ˜ u + n } n ⊂ L (0 , T ; H ) areuniformly bounded. Indeed there exists a constant C > n such that k∇ Ψ + n ⊙ ˜ u + n k L (0 ,T ; H ) = n X k =1 Z kτ n ( k − τ n k∇ Ψ kn ⊙ δu kn k H d t ≤ k∇ Ψ k ∞ n X k =1 τ n k δu kn k H ≤ C, k E (Ψ + n ˜ u + n ) k L (0 ,T ; H ) = n X k =1 Z kτ n ( k − τ n k E (Ψ kn δu kn ) k H d t = n X k =1 τ n k Ψ kn Eδu kn + ∇ Ψ kn ⊙ δu kn k H ≤ n X k =1 τ n k Ψ kn Eδu kn k H + 2 n X k =1 τ n k∇ Ψ kn ⊙ δu kn k H ≤ C. DYNAMIC MODEL FOR VISCOELASTIC MATERIALS WITH PRESCRIBED GROWING CRACKS 9
Therefore, there exists w , w ∈ L (0 , T ; H ) such that, up to a further not relabeled subsequence ∇ Ψ + n ⊙ ˜ u + n L (0 ,T ; H ) −−−−−−− ⇀ n →∞ w , E (Ψ + n ˜ u + n ) L (0 ,T ; H ) −−−−−−− ⇀ n →∞ w . We want to identify the limit functions w and w . Consider ϕ ∈ L (0 , T ; H ), then Z T ( ∇ Ψ + n ⊙ ˜ u + n , ϕ ) H d t = 12 Z T (˜ u + n , ϕ ∇ Ψ + n ) H d t + 12 Z T (˜ u + n , ϕ T ∇ Ψ + n ) H d t = Z T (˜ u + n , ϕ sym ∇ Ψ + n ) H d t, where ϕ sym := ϕ + ϕ T . Since ˜ u + n L (0 ,T ; H ) −−−−−−− ⇀ n →∞ ˙ u and ϕ sym ∇ Ψ + n L (0 ,T ; H ) −−−−−−−→ n →∞ ϕ sym ∇ Ψ by dominated convergencetheorem, we obtain Z T ( ∇ Ψ + n ⊙ ˜ u + n , ϕ ) H d t −−−−→ n →∞ Z T ( ˙ u, ϕ sym ∇ Ψ) H d t = Z T ( ∇ Ψ ⊙ ˙ u, ϕ ) H d t, and so w = ∇ Ψ ⊙ ˙ u . Moreover for φ ∈ L (0 , T ; H ) we have Z T (Ψ + n ˜ u + n , φ ) H d t = Z T (˜ u + n , φ Ψ + n ) H d t −−−−→ n →∞ Z T ( ˙ u, Ψ φ ) H d t = Z T (Ψ ˙ u, φ ) H d t, thanks to ˜ u + n L (0 ,T ; H ) −−−−−−− ⇀ n →∞ ˙ u and Ψ + n φ L (0 ,T ; H ) −−−−−−−→ n →∞ Ψ φ , again implied by dominated convergence theorem.Therefore Ψ + n ˜ u + n L (0 ,T ; H ) −−−−−−− ⇀ n →∞ Ψ ˙ u , from which E (Ψ + n ˜ u + n ) D ′ (0 ,T ; H ) −−−−−−→ n →∞ E (Ψ ˙ u ), that gives w = E (Ψ ˙ u ). Inparticular we have Ψ ˙ u ∈ L (0 , T ; V T ). By arguing in a similar way we also obtain ∇ Ψ − n ⊙ ˜ u − n L (0 ,T ; H ) −−−−−−− ⇀ n →∞ ∇ Ψ ⊙ ˙ u, E (Ψ − n ˜ u − n ) L (0 ,T ; H ) −−−−−−− ⇀ n →∞ E (Ψ ˙ u ) . Let us check that u ∈ W . To this aim, let us consider the following set F := { v ∈ L (0 , T ; V T ) : v ( t ) ∈ V t for a.e. t ∈ (0 , T ) } ⊂ L (0 , T ; V T ) . We have that F is a (strong) closed convex subset of L (0 , T ; V T ), and so by Hahn-Banach’s theorem the set F is weakly closed. Notice that { u − n } n , { Ψ − n ˜ u − n } n ⊂ F , indeed u − n ( t ) = u k − n ∈ V ( k − τ n ⊂ V t for t ∈ [( k − τ n , kτ n ), k = 1 , . . . , n, Ψ − n ( t )˜ u − n ( t ) = Ψ k − n δu k − n ∈ V ( k − τ n ⊆ V t for t ∈ [( k − τ n , kτ n ), k = 1 , . . . , n. Since u − n L (0 ,T ; V T ) −−−−−−− ⇀ n →∞ u and Ψ − n ˜ u − n L (0 ,T ; V T ) −−−−−−− ⇀ n →∞ Ψ ˙ u , we conclude that u, Ψ ˙ u ∈ F . Finally, to show that u − w ∈ V D we observe u − n ( t ) − w − n ( t ) = u k − n − w k − n ∈ V k − n ⊆ V Dt for t ∈ [( k − τ n , kτ n ), k = 1 , . . . , n. Therefore { u − n − w − n } n ⊂ { v ∈ L (0 , T ; V T ) : v ( t ) ∈ V Dt for a.e. t ∈ (0 , T ) } , which is a (strong) closed convexsubset of L (0 , T ; V T ), and so it is weakly closed. Since u − n L (0 ,T ; V T ) −−−−−−− ⇀ n →∞ u and w − n L (0 ,T ; V ) −−−−−−−→ n →∞ w , we get that u ( t ) − w ( t ) ∈ V Dt for a.e. t ∈ (0 , T ), which implies u − w ∈ V D . (cid:3) We now use Lemma 3.4 to pass to the limit in the discrete equation (3.1).
Lemma 3.5.
The limit function u ∈ W of Lemma 3.4 is a weak solution to (2.8) – (2.11) .Proof. We only need to prove that u ∈ W satisfies (2.14). We fix n ∈ N , ϕ ∈ C c (0 , T ; V T ) such that ϕ ( t ) ∈ V Dt for every t ∈ (0 , T ), and we consider ϕ kn := ϕ ( kτ n ) for k = 0 , . . . , n , δϕ kn := ϕ kn − ϕ k − n τ n for k = 1 , . . . , n, and the approximating sequences ϕ + n ( t ) := ϕ kn , ˜ ϕ + n ( t ) := δϕ kn t ∈ (( k − τ n , kτ n ] , k = 1 , . . . , n. If we use τ n ϕ kn ∈ V kn as test function in (3.1), after summing over k = 1 , ..., n , we get n X k =1 τ n ( δ u kn , ϕ kn ) H + n X k =1 τ n ( C Eu kn , Eϕ kn ) H + n X k =1 τ n ( B Ψ kn Eδu kn , Ψ kn Eϕ kn ) H = n X k =1 τ n ( f kn , ϕ kn ) H + n X k =1 τ n ( g kn , ϕ kn ) H N . (3.16)By these identities n X k =1 τ n ( δ u kn , ϕ kn ) H = − n X k =1 τ n ( δu k − n , δϕ kn ) H = − Z T (˜ u − n ( t ) , ˜ ϕ + n ( t )) H d t, from (3.16) we deduce − Z T (˜ u − n , ˜ ϕ + n ) H d t + Z T ( C Eu + n , Eϕ + n ) H d t − Z T ( B ∇ Ψ + n ⊙ ˜ u + n , Eϕ + n ) H d t + Z T ( B E (Ψ + n ˜ u + n ) , Eϕ + n ) H d t = Z T ( f + n , ϕ + n ) H d t + Z T ( g + n , ϕ + n ) H N d t. (3.17)Thanks to (3.14), (3.15), and the following convergences ϕ + n L (0 ,T ; V T ) −−−−−−−→ n →∞ ϕ, ˜ ϕ + n L (0 ,T ; H ) −−−−−−−→ n →∞ ˙ ϕ, f + n L (0 ,T ; H ) −−−−−−−→ n →∞ f, g + n L (0 ,T ; H N ) −−−−−−−−→ n →∞ g, we can pass to the limit in (3.17), and we get that u ∈ W satisfies (2.14) for every ϕ ∈ C c (0 , T ; V T ) suchthat ϕ ( t ) ∈ V Dt for every t ∈ (0 , T ). Finally, by using a density argument (see [8, Remark 2.9]), we concludethat u ∈ W is a weak solution to (2.8)–(2.11). (cid:3) Initial Conditions and Energy–Dissipation Inequality
To complete our existence result, it remains to prove that the function u ∈ W given by Lemma 3.5 satisfiesthe initial conditions (2.12) in the sense of (2.16). Let us start by showing that the second distributionalderivative ¨ u belongs to L (0 , T ; ( V D ) ′ ). If we consider the discrete equation (3.1), for every v ∈ V D ⊆ V kn ,with k v k V ≤
1, we have | ( δ u kn , v ) H | ≤k C k ∞ k Eu kn k H + k B k ∞ k Ψ k ∞ k Ψ kn Eδu kn k H + k f kn k H + C tr k g kn k H N . Therefore, taking the supremum over v ∈ V D with k v k V ≤
1, we obtain the existence of a positive constant C such that k δ u kn k V D ) ′ ≤ C ( k Eu kn k H + k Ψ kn Eδu kn k H + k f kn k H + k g kn k H N ) . If we multiply this inequality by τ n and we sum over k = 1 , . . . , n , we get n X k =1 τ n k δ u kn k V D ) ′ ≤ C n X k =1 τ n k Eu kn k H + n X k =1 τ n k Ψ kn Eδu kn k H + k f k L (0 ,T ; H ) + k g k L (0 ,T ; H N ) ! . (4.1)Thanks to (4.1) and Lemma 3.3 we conclude that P nk =1 τ n k δ u kn k V D ) ′ ≤ ˜ C for every n ∈ N for a positiveconstant ˜ C independent on n ∈ N . In particular the sequence { ˜ u n } n ⊂ H (0 , T ; ( V D ) ′ ) is uniformly bounded(notice that ˙˜ u n ( t ) = δ u kn for t ∈ (( k − τ n , kτ n ) and k = 1 , . . . , n ). Hence, up to extract a further (notrelabeled) subsequence from the one of Lemma 3.4, we get˜ u n H (0 ,T ;( V D ) ′ ) −−−−−−−−− ⇀ n →∞ w , (4.2)and by using the following estimate k ˜ u n − ˜ u + n k L (0 ,T ;( V D ) ′ ) ≤ τ n k ˙˜ u n k L (0 ,T ;( V D ) ′ ) ≤ ˜ Cτ n −−−−→ n →∞ w = ˙ u .Let us recall the following result, whose proof can be found for example in [9]. Lemma 4.1.
Let
X, Y be two reflexive Banach spaces such that
X ֒ → Y continuously. Then L ∞ (0 , T ; X ) ∩ C w ([0 , T ]; Y ) = C w ([0 , T ]; X ) . DYNAMIC MODEL FOR VISCOELASTIC MATERIALS WITH PRESCRIBED GROWING CRACKS 11
Since H (0 , T ; ( V D ) ′ ) ֒ → C ([0 , T ] , ( V D ) ′ ), by using Lemmas 3.4 and 4.1 we get that our weak solution u ∈ W satisfies u ∈ C w ([0 , T ]; V T ) , ˙ u ∈ C w ([0 , T ]; H ) , ¨ u ∈ L (0 , T ; ( V D ) ′ ) . By (3.14) and (4.2) we hence obtain u n ( t ) H −−−− ⇀ n →∞ u ( t ) , ˜ u n ( t ) ( V D ) ′ −−−− ⇀ n →∞ ˙ u ( t ) for every t ∈ [0 , T ] , (4.3)so that u (0) = u and ˙ u (0) = u , since u n (0) = u and ˜ u n (0) = u .To prove that lim h → + h Z h (cid:0) k u ( t ) − u k V t + k ˙ u ( t ) − u k H (cid:1) d t = 0we will actually show lim t → + u ( t ) = u in V T , lim t → + ˙ u ( t ) = u in H. This is a consequence of following energy–dissipation inequality which holds for the weak solution u ∈ W ofLemma 3.5. Let us define the total energy as E ( t ) := 12 k ˙ u ( t ) k H + 12 ( C Eu ( t ) , Eu ( t )) H t ∈ [0 , T ] . Notice that E ( t ) is well defined for every t ∈ [0 , T ] since u ∈ C w ([0 , T ]; V T ) and ˙ u ∈ C w ([0 , T ]; H ), and that E (0) = k u k H + ( C Eu , Eu ) H . Theorem 4.2.
The weak solution u ∈ W to (2.8) – (2.11) , given by Lemma 3.5, satisfies for every t ∈ [0 , T ] the following energy–dissipation inequality E ( t ) + Z t ( B Ψ E ˙ u, Ψ E ˙ u ) H d s ≤ E (0) + W tot ( t ) , (4.4) where Ψ E ˙ u is the function defined in (2.15) and W tot ( t ) is the total work on the solution u at time t ∈ [0 , T ] ,which is given by W tot ( t ) : = Z t [( f, ˙ u − ˙ w ) H + ( C Eu, E ˙ w ) H + ( B Ψ E ˙ u, Ψ E ˙ w ) H − ( ˙ u, ¨ w ) H − ( ˙ g, u − w ) H N ] d s + ( ˙ u ( t ) , ˙ w ( t )) H + ( g ( t ) , u ( t ) − w ( t )) H N − ( u , ˙ w (0)) H − ( g (0) , u − w (0)) H N . (4.5) Remark 4.3.
From the classical point of view, the total work on the solution u at time t ∈ [0 , T ] is givenby W tot ( t ) := W load ( t ) + W bdry ( t ) , (4.6)where W load ( t ) is the work on the solution u at time t ∈ [0 , T ] due to the loading term, which is defined as W load ( t ) := Z t ( f ( s ) , ˙ u ( s )) H d s, and W bdry ( t ) is the work on the solution u at time t ∈ [0 , T ] due to the varying boundary conditions, whichone expects to be equal to W bdry ( t ) := Z t ( g ( s ) , ˙ u ( s )) H N d s + Z t (( C Eu ( s ) + Ψ ( s ) B E ˙ u ( s )) ν, ˙ w ( s )) H D d s, being H D := L ( ∂ D Ω; R d ). Unfortunately, W bdry ( t ) is not well defined under our assumptions on u . Noticethat when Ψ ≡ U of the closure of ∂ N Ω, then every weak solution u to (2.8)–(2.11)satisfies u ∈ H (0 , T ; H ((Ω ∩ U ) \ Γ; R d )), which gives that u ∈ H (0 , T ; H N ) by our assumptions on Γ.Hence the first term of W bdry ( t ) makes sense and satisfies Z t ( g ( s ) , ˙ u ( s )) H N d s = ( g ( t ) , u ( t )) H N − ( g (0) , u (0)) H N − Z t ( ˙ g ( s ) , u ( s )) H N d s. The term involving the Dirichlet datum w is more difficult to handle since the trace of ( C Eu + Ψ B E ˙ u ) ν on ∂ D Ω is not well defined even when Ψ ≡ ∂ D Ω. If we assume that u ∈ H (0 , T ; H (Ω \ Γ; R d )) ∩ H (0 , T ; L (Ω; R d )) and that Γ is a smooth manifold, then we can integrate by part equation (2.14) to deduce that u satisfies (2.8). In this case, ( C Eu + Ψ B E ˙ u ) ν ∈ L (0 , T ; H D ) andby using (2.8), together with the divergence theorem and the integration by parts formula, we deduce Z t (( C Eu ( s ) + Ψ ( s ) B E ˙ u ( s )) ν, ˙ w ( s )) H D d s = Z t (cid:2) (div( C Eu ( s ) + Ψ ( s ) B E ˙ u ( s )) , ˙ w ( s )) H + ( C Eu ( s ) + Ψ ( s ) B E ˙ u ( s ) , E ˙ w ( s )) H − ( g ( s ) , ˙ w ( s )) H N (cid:3) d s = Z t (cid:2) (¨ u ( s ) , ˙ w ( s )) H − ( f ( s ) , ˙ w ( s )) H + ( C Eu ( s ) + Ψ ( s ) B E ˙ u ( s ) , E ˙ w ( s )) H − ( g ( s ) , ˙ w ( s )) H N (cid:3) d s = Z t [( C Eu ( s ) , E ˙ w ( s )) H + ( B Ψ( s ) E ˙ u ( s ) , Ψ( s ) E ˙ w ( s )) H − ( f ( s ) , ˙ w ( s )) H ] d s + Z t [( ˙ g ( s ) , w ( s )) H N − ( ˙ u ( s ) , ¨ w ( s )) H ] d s − ( g ( t ) , w ( t )) H N + ( ˙ u ( t ) , ˙ w ( t )) H + ( g (0) , w (0)) H N − ( u , ˙ w (0)) H . Hence, the definition of total work given in (4.5) is coherent with the classical one (4.6). Notice that if u is the solution to (2.8)–(2.11) given by Lemma 3.5, then (4.5) is well defined for every t ∈ [0 , T ], since g ∈ C ([0 , T ]; H N ), ˙ w ∈ C ([0 , T ]; H ), u ∈ C w ([0 , T ]; V T ), and ˙ u ∈ C w ([0 , T ]; H ). In particular, the function t
7→ W tot ( t ) from [0 , T ] to R is continuous. Proof.
Fixed t ∈ (0 , T ], for every n ∈ N there exists a unique j ∈ { , . . . , n } such that t ∈ (( j − τ n , jτ n ].After setting t n := jτ n , we can rewrite (3.3) as12 k ˜ u + n ( t ) k H + 12 ( C Eu + n ( t ) , Eu + n ( t )) H + Z t n ( B Ψ + n E ˜ u + n , Ψ + n E ˜ u + n ) H d s ≤ E (0) + W + n ( t ) , (4.7)where W + n ( t ) := Z t n (cid:2) ( f + n , ˜ u + n − ˜ w + n ) H + ( C Eu + n , E ˜ w + n ) H + ( B Ψ + n E ˜ u + n , Ψ + n E ˜ w + n ) H (cid:3) d s + Z t n (cid:2) (˜ u + n , ˜ w + n ) H + ( g + n , ˜ u + n − ˜ w + n ) H N (cid:3) d s. Thanks to (3.2), we have k u n ( t ) − u + n ( t ) k H = k u jn + ( t − jτ n ) δu jn − u jn k H ≤ τ n k δu jn k H ≤ Cτ n −−−−→ n →∞ , k ˜ u n ( t ) − ˜ u + n ( t ) k V D ) ′ = k δu jn + ( t − jτ n ) δ u jn − δu jn k V D ) ′ ≤ τ n k δ u jn k V D ) ′ ≤ Cτ n −−−−→ n →∞ . The last convergences and (4.3) imply u + n ( t ) H −−−− ⇀ n →∞ u ( t ) , ˜ u + n ( t ) ( V D ) ′ −−−− ⇀ n →∞ ˙ u ( t ) , and since k u + n ( t ) k V T + k ˜ u + n ( t ) k H ≤ C for every n ∈ N , we get u + n ( t ) V T −−−− ⇀ n →∞ u ( t ) , ˜ u + n ( t ) H −−−− ⇀ n →∞ ˙ u ( t ) . (4.8)By the lower semicontinuity properties of v
7→ k v k H and v ( C Ev, Ev ) H , we conclude k ˙ u ( t ) k H ≤ lim inf n →∞ k ˜ u + n ( t ) k H , (4.9)( C Eu ( t ) , Eu ( t )) H ≤ lim inf n →∞ ( C Eu + n ( t ) , Eu + n ( t )) H . (4.10)Thanks to Lemma 3.4 and (2.15), we obtainΨ + n E ˜ u + n = E (Ψ + n ˜ u + n ) − ∇ Ψ + n ⊙ ˜ u + n L (0 ,T ; H ) −−−−−−− ⇀ n →∞ E (Ψ ˙ u ) − ∇ Ψ ⊙ ˙ u = Ψ E ˙ u, so that Z t ( B Ψ E ˙ u, Ψ E ˙ u ) H d s ≤ lim inf n →∞ Z t ( B Ψ + n E ˜ u + n , Ψ + n E ˜ u + n ) H d s ≤ lim inf n →∞ Z t n ( B Ψ + n E ˜ u + n , Ψ + n E ˜ u + n ) H d s, (4.11) DYNAMIC MODEL FOR VISCOELASTIC MATERIALS WITH PRESCRIBED GROWING CRACKS 13 since t ≤ t n and v R t ( B v, v ) H d s is a non negative quadratic form on L (0 , T ; H ). Let us study theright–hand side of (4.7). Given that we have χ [0 ,t n ] f + n L (0 ,T ; H ) −−−−−−−→ n →∞ χ [0 ,t ] f, ˜ u + n − ˜ w + n L (0 ,T ; H ) −−−−−−− ⇀ n →∞ ˙ u − ˙ w, we can deduce Z t n ( f + n , ˜ u + n − ˜ w + n ) H d s −−−−→ n →∞ Z t ( f, ˙ u − ˙ w ) H d s. (4.12)In a similar way, we can prove Z t n ( C Eu + n , E ˜ w + n ) H d s −−−−→ n →∞ Z t ( C Eu, E ˙ w ) H d s, (4.13) Z t n ( B Ψ + n E ˜ u + n , Ψ + n E ˜ w + n ) H d s −−−−→ n →∞ Z t ( B Ψ E ˙ u, Ψ E ˙ w ) H d s, (4.14)since the following convergences hold χ [0 ,t n ] E ˜ w + n L (0 ,T ; H ) −−−−−−−→ n →∞ χ [0 ,t ] E ˙ w, C Eu + n L (0 ,T ; H ) −−−−−−− ⇀ n →∞ C Eu,χ [0 ,t n ] Ψ + n E ˜ w + n L (0 ,T ; H ) −−−−−−−→ n →∞ χ [0 ,t ] Ψ E ˙ w, Ψ + n E ˜ u + n L (0 ,T ; H ) −−−−−−− ⇀ n →∞ Ψ E ˙ u. It remains to study the behaviour as n → ∞ of the terms Z t n ( ˙˜ u n , ˜ w + n ) H d s, Z t n ( g + n , ˜ u + n − ˜ w + n ) H N d s. Thanks to formula (3.10) we have Z t n ( ˙˜ u n , ˜ w + n ) H d s = (˜ u + n ( t ) , ˜ w + n ( t )) H − ( u , ˙ w (0)) H − Z t n (˜ u − n , ˙˜ w n ) H d s. By arguing as before we hence deduce Z t n ( ˙˜ u n , ˜ w + n ) H d s −−−−→ n →∞ ( ˙ u ( t ) , ˙ w ( t )) H − ( u , ˙ w (0)) H − Z t ( ˙ u, ¨ w ) H d s, (4.15)thanks to (4.8) and by these convergences χ [0 ,t n ] ˙˜ w n L (0 ,T ; H ) −−−−−−−→ n →∞ χ [0 ,t ] ¨ w, ˜ u − n L (0 ,T ; H ) −−−−−−− ⇀ n →∞ ˙ u, k ˜ w + n ( t ) − ˙ w ( t ) k H = (cid:13)(cid:13)(cid:13)(cid:13) w ( jτ n ) − w (( j − τ n ) τ n − ˙ w ( t ) (cid:13)(cid:13)(cid:13)(cid:13) H = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) − Z jτ n ( j − τ n ( ˙ w ( s ) − ˙ w ( t )) d s (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) H ≤ − Z jτ n ( j − τ n k ˙ w ( s ) − ˙ w ( t ) k H d s −−−−→ n →∞ . Notice that in the last convergence we used the continuity of w from [0 , T ] in H . Similarly we have Z t n ( g + n , ˜ u + n − ˜ w + n ) H N d s = ( g + n ( t ) , u + n ( t ) − w + n ( t )) H N − ( g (0) , u − w (0)) H N − Z t n ( ˙ g n , u − n − w − n ) H N d s so that we get Z t n ( g + n , ˜ u + n − ˜ w + n ) H N d s −−−−→ n →∞ ( g ( t ) , u ( t ) − w ( t )) H N − ( g (0) , u − w (0)) H N − Z t ( ˙ g, u − w ) H N d s (4.16)thanks to (4.8), the continuity of s g ( s ) in H N , and the fact that χ [0 ,t n ] ˙ g n L (0 ,T ; H N ) −−−−−−−−→ n →∞ χ [0 ,t ] ˙ g, u − n − w − n L (0 ,T ; H N ) −−−−−−−− ⇀ n →∞ u − w. By combining (4.9)–(4.16), we deduce the energy–dissipation inequality (4.4) for every t ∈ (0 , T ]. Finally,for t = 0 the inequality trivially holds since u (0) = u and ˙ u (0) = u . (cid:3) We now are in position to prove the validity of the initial conditions.
Lemma 4.4.
The weak solution u ∈ W to (2.8) – (2.11) of Lemma 3.5 satisfies lim t → + u ( t ) = u in V T , lim t → + ˙ u ( t ) = u in H. (4.17) In particular u satisfies the initial conditions (2.12) in the sense of (2.16) .Proof. By sending t → + into the energy–dissipation inequality (4.4) and using that u ∈ C w ([0 , T ]; V T ) and˙ u ∈ C w ([0 , T ]; H ) we deduce E (0) ≤ lim inf t → + E ( t ) ≤ lim sup t → + E ( t ) ≤ E (0) , since the right–hand side of (4.4) is continuous in t , u (0) = u , and ˙ u (0) = u . Therefore there existslim t → + E ( t ) = E (0). By using the lower semicontinuity of t
7→ k ˙ u ( t ) k H and t ( C Eu ( t ) , Eu ( t )) H , wederive lim t → + k ˙ u ( t ) k H = k u k H , lim t → + ( C Eu ( t ) , Eu ( t )) H = ( C Eu , Eu ) H . Finally, since we have ˙ u ( t ) H −−−− ⇀ t → + u , Eu ( t ) H −−−− ⇀ t → + Eu , we deduce (4.17). In particular the functions u : [0 , T ] → V T and ˙ u : [0 , T ] → H are continuous at t = 0,which implies (2.16). (cid:3) We can finally prove Theorem 3.1.
Proof of Theorem 3.1.
It is enough to combine Lemmas 3.5 and 4.4. (cid:3)
Remark 4.5.
We have proved Theorem 3.1 for the d -dimensional linear elastic case, namely when thedisplacement u is a vector–valued function. The same result is true with identical proofs in the antiplanecase, that is when the displacement u is a scalar function and satisfies (1.5).5. Uniqueness
In this section we investigate the uniqueness properties of system (2.8) with boundary and initial con-ditions (2.9)–(2.12). To this aim, we need to assume stronger regularity assumptions on the crack sets { Γ t } t ∈ [0 ,T ] and on the function Ψ. Moreover, we have to restrict our problem to the dimensional case d = 2,since in our proof we need to construct a suitable family of diffeomorphisms which maps the time–dependentcrack Γ t into a fixed set, and this can be explicitly done only for d = 2 (see [7, Example 2.14]).We proceed in two steps; first, in Lemma 5.2 we prove a uniqueness result in every dimension d , butwhen the cracks are not increasing, that is Γ T = Γ . Next, in Theorem 5.5 we combine Lemma 5.2 with thefinite speed of propagation theorem of [5] and the uniqueness result of [8] to derive the uniqueness of a weaksolution to (2.8)–(2.12) in the case d = 2.Let us start with the following lemma, whose proof is similar to that one of [8, Proposition 2.10]. Lemma 5.1.
Let u ∈ W be a weak solution to (2.8) – (2.11) satisfying the initial condition ˙ u (0) = 0 in thefollowing sense lim h → + h Z h k ˙ u ( t ) k H = 0 . Then u satisfies − Z T ( ˙ u ( t ) , ˙ ϕ ( t )) H d t + Z T ( C Eu ( t ) , Eϕ ( t )) H d t + Z T ( B Ψ( t ) E ˙ u ( t ) , Ψ( t ) Eϕ ( t )) H d t = Z T ( f ( t ) , ϕ ( t )) H d t + Z T ( g ( t ) , ϕ ( t )) H N d t for every ϕ ∈ V D such that ϕ ( T ) = 0 , where Ψ E ˙ u is the function defined in (2.15) .Proof. We fix ϕ ∈ V D with ϕ ( T ) = 0 and for every ǫ > ϕ ǫ ( t ) := ® tǫ ϕ ( t ) t ∈ [0 , ǫ ] ,ϕ ( t ) t ∈ [ ǫ, T ] . DYNAMIC MODEL FOR VISCOELASTIC MATERIALS WITH PRESCRIBED GROWING CRACKS 15
We have that ϕ ǫ ∈ V D and ϕ ǫ (0) = ϕ ǫ ( T ) = 0, so we can use ϕ ǫ as test function in (2.14). By proceedingas in [8, Proposition 2.10] we obtainlim ǫ → + Z T ( ˙ u ( t ) , ˙ ϕ ǫ ( t )) H d t = Z T ( ˙ u ( t ) , ˙ ϕ ( t )) H d t, lim ǫ → + Z T ( C Eu ( t ) , Eϕ ǫ ( t )) H d t = Z T ( C Eu ( t ) , Eϕ ( t )) H d t, lim ǫ → + Z T ( f ( t ) , ϕ ǫ ( t )) H d t = Z T ( f ( t ) , ϕ ( t )) H d t. It remains to consider the terms involving B and g . We have Z T ( B Ψ( t ) E ˙ u ( t ) , Ψ( t ) Eϕ ǫ ( t )) H d t = Z ǫ ( B Ψ( t ) E ˙ u ( t ) , tǫ Ψ( t ) Eϕ ( t )) H d t + Z Tǫ ( B Ψ( t ) E ˙ u ( t ) , Ψ( t ) Eϕ ( t )) H d t, Z T ( g ( t ) , ϕ ǫ ( t )) H N d t = Z ǫ ( g ( t ) , tǫ ϕ ( t )) H N d t + Z Tǫ ( g ( t ) , ϕ ( t )) H N d t, hence by the dominated convergence theorem we get Z Tǫ ( B Ψ( t ) E ˙ u ( t ) , Ψ( t ) Eϕ ( t )) H d t −−−−→ ǫ → + Z T ( B Ψ( t ) E ˙ u ( t ) , Ψ( t ) Eϕ ( t )) H d t, (cid:12)(cid:12)(cid:12)(cid:12)Z ǫ ( B Ψ( t ) E ˙ u ( t ) , tǫ Ψ( t ) Eϕ ( t )) H d t (cid:12)(cid:12)(cid:12)(cid:12) ≤ k B k ∞ k Ψ k ∞ Z ǫ k Ψ( t ) E ˙ u ( t ) k H k Eϕ ( t ) k H d t −−−−→ ǫ → + , Z Tǫ ( g ( t ) , ϕ ( t )) H N d t −−−−→ ǫ → + Z T ( g ( t ) , ϕ ( t )) H N d t, (cid:12)(cid:12)(cid:12)(cid:12)Z ǫ ( g ( t ) , tǫ ϕ ( t )) H N d t (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z ǫ k g ( t ) k H N k ϕ ( t ) k H N d t −−−−→ ǫ → + . By combining together all the previous convergences we get the thesis. (cid:3)
We now state the uniqueness result in the case of a fixed domain, that is Γ T = Γ . We follow the sameideas of [12], and we need to assumeΨ ∈ Lip([0 , T ] × Ω) , ∇ ˙Ψ ∈ L ∞ ((0 , T ) × Ω; R d ) , (5.1)while on Γ we do not require any further hypotheses. Lemma 5.2 (Uniqueness in a fixed domain) . Assume (5.1) and Γ T = Γ . Then the viscoelastic dynamicsystem (2.8) with boundary and initial conditions (2.9) – (2.12) ( the latter in the sense of (2.16)) has a uniqueweak solution.Proof. Let u , u ∈ W be two weak solutions to (2.8)–(2.11) with initial conditions (2.12). The function u := u − u satisfies 1 h Z h ( k u ( t ) k V t + k ˙ u ( t ) k H ) d t −−−−→ h → + , (5.2)hence by Lemma 5.1 it solves − Z T ( ˙ u ( t ) , ˙ ϕ ( t )) H d t + Z T ( C Eu ( t ) , Eϕ ( t )) H d t + Z T ( B Ψ( t ) E ˙ u ( t ) , Ψ( t ) Eϕ ( t )) H d t = 0 (5.3)for every ϕ ∈ V D such that ϕ ( T ) = 0. We fix s ∈ (0 , T ] and consider the function ϕ s ( t ) := ® − R st u ( τ )d τ t ∈ [0 , s ] , t ∈ [ s, T ] . Since ϕ s ∈ V D and ϕ s ( T ) = 0, we can use it as test function in (5.3) to obtain − Z s ( ˙ u ( t ) , u ( t )) H d t + Z s ( C E ˙ ϕ s ( t ) , Eϕ s ( t )) H d t + Z s ( B Ψ( t ) E ˙ u ( t ) , Ψ( t ) Eϕ s ( t )) H d t = 0 . In particular we deduce − Z s dd t k u ( t ) k H d t + 12 Z s dd t ( C Eϕ s ( t ) , Eϕ s ( t )) H d t + Z s ( B Ψ( t ) E ˙ u ( t ) , Ψ( t ) Eϕ s ( t )) H d t = 0 , which implies 12 k u ( s ) k H + 12 ( C Eϕ s (0) , Eϕ s (0)) H = Z s ( B Ψ( t ) E ˙ u ( t ) , Ψ( t ) Eϕ s ( t )) H d t, (5.4)since u (0) = 0 = ϕ s ( s ). From the distributional point of view the following equality holdsdd t (Ψ Eu ) = ˙Ψ Eu + Ψ E ˙ u ∈ L (0 , T ; H ) , (5.5)indeed, for all v ∈ C ∞ c (0 , T ; H ) we have Z T ( dd t (Ψ( t ) Eu ( t )) , v ( t )) H d t = − Z T (Ψ( t ) Eu ( t ) , ˙ v ( t )) H d t = − Z T ( E (Ψ( t ) u ( t )) − ∇ Ψ( t ) ⊙ u ( t ) , ˙ v ( t )) H d t = Z T ( E ( ˙Ψ( t ) u ( t )) + E (Ψ( t ) ˙ u ( t )) , v ( t )) H d t − Z T ( ∇ ˙Ψ( t ) ⊙ u ( t ) + ∇ Ψ( t ) ⊙ ˙ u ( t ) , v ( t )) H d t = Z T ( ˙Ψ( t ) Eu ( t ) , v ( t )) H d t + Z T (Ψ( t ) E ˙ u ( t ) , v ( t )) H d t. In particular Ψ Eu ∈ H (0 , T ; H ) ⊂ C ([0 , T ] , H ), so that by (5.2) k Ψ(0) Eu (0) k H = lim h → h Z h k Ψ( t ) Eu ( t ) k H d t ≤ C lim h → h Z h k u ( t ) k V t d t = 0which yields Ψ(0) Eu (0) = 0. Thanks to (5.5) and to property Ψ u ∈ H (0 , T ; H ), we deducedd t ( B Ψ Eu, Ψ Eϕ s ) H = ( B ˙Ψ Eu, Ψ Eϕ s ) H + ( B Ψ E ˙ u, Ψ Eϕ s ) H + ( B Ψ Eu, ˙Ψ Eϕ s ) H + ( B Ψ Eu, Ψ E ˙ ϕ s ) H = 2( B Ψ Eu, ˙Ψ Eϕ s ) H + ( B Ψ E ˙ u, Ψ Eϕ s ) H + ( B Ψ Eu, Ψ E ˙ ϕ s ) H , and by integrating on [0 , s ] we get Z s ( B Ψ( t ) E ˙ u ( t ) , Ψ( t ) Eϕ s ( t )) H d t = Z s ï dd t ( B Ψ( t ) Eu ( t ) , Ψ( t ) Eϕ s ( t )) H − B Ψ( t ) Eu ( t ) , ˙Ψ( t ) Eϕ s ( t )) H − ( B Ψ( t ) E ˙ ϕ s ( t ) , Ψ( t ) E ˙ ϕ s ( t )) H ò d t ≤ ( B Ψ( s ) Eu ( s ) , Ψ( s ) Eϕ s ( s )) H − ( B Ψ(0) Eu (0) , Ψ(0) Eϕ s (0)) H + Z s h B Ψ( t ) Eu ( t ) , Ψ( t ) Eu ( t )) H ( B ˙Ψ( t ) Eϕ s ( t ) , ˙Ψ( t ) Eϕ s ( t )) H − ( B Ψ( t ) E ˙ ϕ s ( t ) , Ψ( t ) E ˙ ϕ s ( t )) H i d t ≤ Z s î ( B Ψ( t ) Eu ( t ) , Ψ( t ) Eu ( t )) H + ( B ˙Ψ( t ) Eϕ s ( t ) , ˙Ψ( t ) Eϕ s ( t )) H − ( B Ψ( t ) E ˙ ϕ s ( t ) , Ψ( t ) E ˙ ϕ s ( t )) H ó d t ≤ k B k ∞ k ˙Ψ k ∞ Z s k Eϕ s ( t ) k H d t, since Eϕ s ( s ) = 0 = Ψ(0) Eu (0) and E ˙ ϕ s = Eu in (0 , s ). By combining the previous inequality with (5.4)and using the coercivity of the tensor C , we derive λ k Eϕ s (0) k H + 12 k u ( s ) k H ≤
12 ( C Eϕ s (0) , Eϕ s (0)) H + 12 k u ( s ) k H ≤ k B k ∞ k ˙Ψ k ∞ Z s k Eϕ s ( t ) k H d t. Let us set ξ ( t ) := R t u ( τ )d τ , then k Eϕ s (0) k H = k Eξ ( s ) k H , k Eϕ s ( t ) k H = k Eξ ( t ) − Eξ ( s ) k H ≤ k Eξ ( t ) k H + 2 k Eξ ( s ) k H , DYNAMIC MODEL FOR VISCOELASTIC MATERIALS WITH PRESCRIBED GROWING CRACKS 17 from which we deduce λ k Eξ ( s ) k H + 12 k u ( s ) k H ≤ C Z s k Eξ ( t ) k H d t + Cs k Eξ ( s ) k H , (5.6)where C := 2 k B k ∞ k ˙Ψ k ∞ . Therefore, if we set s := λ C , for all s ≤ s we obtain λ k Eξ ( s ) k H ≤ Å λ − Cs ã k Eξ ( s ) k H ≤ C Z s k Eξ ( t ) k H d t. By Gronwall’s lemma the last inequality implies Eξ ( s ) = 0 for all s ≤ s . Hence, thanks to (5.6) we get k u ( s ) k H ≤ s ≤ s , which yields u ( s ) = 0 for all s ≤ s . Since s depends only on C , B , and Ψ, wecan repeat this argument starting from s , and with a finite number of steps we obtain u ≡ , T ]. (cid:3) In order to prove our uniqueness result in the case of a moving crack we need two auxiliary results, whichare [4, Theorem 6.1] and [8, Theorem 4.3]. For the sake of the readers, we rewrite below the statementswithout proof.The first one ([4, Theorem 6.1]) is a generalization of the well–known result of finite speed of propagationfor the wave equation. Given an open bounded set U ⊂ R d , we define by ∂ L U the Lipschitz part of theboundary ∂U , which is the collection of points x ∈ ∂U for which there exist an orthogonal coordinate system y , . . . , y d , a neighborhood V of x of the form A × I , with A open in R d − and I open interval in R , and aLipschitz function g : A → I , such that V ∩ U := { ( y , . . . , y d ) ∈ V : y d < g ( y , . . . , y d − ) } . Moreover, givena Borel set S ⊆ ∂ L U , we define H S ( U ; R d ) := { u ∈ H ( U ; R d ) : u = 0 on S } . Notice that H S ( U ; R d ) is a Hilbert space, and we denote its dual by H − S ( U ; R d ) . Theorem 5.3 (Finite speed of propagation) . Let U ⊂ R d be an open bounded set and let ∂ L U be the Lipschitzpart of ∂U . Let S and S be two Borel sets with S ⊆ S ⊆ ∂ L U , and let C : U → L ( R d × dsym ; R d × dsym ) be afourth-order tensor satisfying (2.4) – (2.6) . Let u ∈ L (0 , T ; H S ( U ; R d )) ∩ H (0 , T ; L ( U ; R d )) ∩ H (0 , T ; H − S ( U ; R d )) be a solution to h ¨ u ( t ) , ψ i H − S ( U ; R d ) + ( C Eu ( t ) , Eψ ) L ( U ; R d × dsym ) = 0 for every ψ ∈ H S ( U ; R d ) , with initial conditions u (0) = 0 and ˙ u (0) = 0 in the sense of L ( U ; R d ) and H − S ( U ; R d ) , respectively. Then u ( t ) = 0 a.e. in U t := { x ∈ U : dist( x, S \ S ) > t » k C k ∞ } for every t ∈ [0 , T ] .Proof. See [4, Theorem 6.1]. (cid:3)
The second one ([8, Theorem 4.3]) is a uniqueness result for the weak solutions of the wave equation ina moving domain. Let ˆ H be a separable Hilbert space, and let { ˆ V t } t ∈ [0 ,T ] be a family of separable Hilbertspaces with the following properties:( i ) for every t ∈ [0 , T ] the space ˆ V t is contained and dense in ˆ H with continuous embedding;( ii ) for every s, t ∈ [0 , T ], with s < t , ˆ V s ⊂ ˆ V t and the Hilbert space structure on ˆ V s is the one inducedby ˆ V t .Let a : ˆ V T × ˆ V T → R be a bilinear symmetric form satisfying the following conditions:( iii ) there exists M such that | a ( u, v ) | ≤ M k u k ˆ V T k v k ˆ V T for every u, v ∈ ˆ V T ;( iv ) there exist λ > ν ∈ R such that a ( u, u ) ≥ λ k u k V T − ν k u k H for every u ∈ ˆ V T . Assume that(U1) for every t ∈ [0 , T ] there exists a continuous and linear bijective operator Q t : ˆ V t → ˆ V , with contin-uous inverse R t : ˆ V → ˆ V t ; (U2) Q and R are the identity maps on ˆ V ;(U3) there exists a constant M independent of t such that k Q t u k ˆ H ≤ M k u k ˆ H for every u ∈ ˆ V t , k R t u k ˆ H ≤ M k u k ˆ H for every u ∈ ˆ V , k Q t u k ˆ V ≤ M k u k ˆ V t for every u ∈ ˆ V t , k R t u k ˆ V t ≤ M k u k ˆ V for every u ∈ ˆ V . Since ˆ V t is dense in ˆ H , (U3) implies that R t and Q t can be extended to continuous linear operators from ˆ H into itself, still denoted by Q t and R t . We also require(U4) for every v ∈ ˆ V the function t R t v from [0 , T ] into ˆ H has a derivative, denoted by ˙ R t v ;(U5) there exists η ∈ (0 ,
1) such that k ˙ R t Q t v k H ≤ λ (1 − η ) k v k V t for every v ∈ ˆ V t ;(U6) there exists a constant M such that k Q t v − Q s v k ˆ H ≤ M k v k ˆ V s ( t − s ) for every 0 ≤ s < t ≤ T and every v ∈ ˆ V s ;(U7) for very t ∈ [0 , T ) and for every v ∈ ˆ V t there exists an element of ˆ H , denoted by ˙ Q t v , such thatlim h → + Q t + h v − Q t vh = ˙ Q t v in ˆ H. For every t ∈ [0 , T ], define α ( t ) : ˆ V × ˆ V → R as α ( t )( u, v ) := a ( R t u, R t v ) for u, v ∈ ˆ V ,β ( t ) : ˆ V × ˆ V → R as β ( t )( u, v ) := ( ˙ R t u, ˙ R t v ) for u, v ∈ ˆ V ,γ ( t ) : ˆ V × ˆ H → R as γ ( t )( u, v ) := ( ˙ R t u, R t v ) for u ∈ ˆ V and v ∈ ˆ H,δ ( t ) : ˆ H × ˆ H → R as δ ( t )( u, v ) := ( R t u, R t v ) − ( u, v ) for u, v ∈ ˆ H. We assume that there exists a constant M such that(U8) the maps t α ( t )( u, v ), t β ( t )( u, v ), t γ ( t )( u, v ), and t δ ( t )( u, v ) are Lipschitz continuousand for a.e. t ∈ (0 , T ) their derivatives satisfy | ˙ α ( t )( u, v ) | ≤ M k u k ˆ V k v k ˆ V for u, v ∈ ˆ V , | ˙ β ( t )( u, v ) | ≤ M k u k ˆ V k v k ˆ V for u, v ∈ ˆ V , | ˙ γ ( t )( u, v ) | ≤ M k u k ˆ V k v k ˆ H for u ∈ ˆ V and v ∈ ˆ H, | ˙ δ ( t )( u, v ) | ≤ M k u k ˆ H k v k ˆ H for u, v ∈ ˆ H. Theorem 5.4 (Uniqueness for the wave equation) . Assume that ˆ H , { ˆ V t } t ∈ [0 ,T ] , and a satisfy ( i ) – ( iv ) andthat (U1)–(U8) hold. Given u ∈ ˆ V , u ∈ ˆ H , and f ∈ L (0 , T ; ˆ H ) , there exists a unique solution u ∈ ˆ V := { ϕ ∈ L (0 , T ; ˆ V T ) : ˙ u ∈ L (0 , T ; ˆ H ) , u ( t ) ∈ ˆ V t for a.e. t ∈ (0 , T ) } to the wave equation − Z T ( ˙ u ( t ) , ˙ ϕ ( t )) ˆ H d t + Z T a ( u ( t ) , ϕ ( t )) d t = Z T ( f ( t ) , ϕ ( t )) ˆ H d t for every ϕ ∈ ˆ V , satisfying the initial conditions u (0) = u and ˙ u (0) = u in the sense that lim h → + h Z h Ä k u ( t ) − u k V t + k ˙ u ( t ) − u k H ä d t = 0 . Proof.
See [8, Theorem 4.3]. (cid:3)
We now are in position to prove the uniqueness theorem in the case of a moving domain. We considerthe dimensional case d = 2, and we require the following assumptions:(H1) there exists a C , simple curve Γ ⊂ Ω ⊂ R , parametrized by arc–length γ : [0 , ℓ ] → Ω, such thatΓ ∩ ∂ Ω = γ (0) ∪ γ ( ℓ ) and Ω \ Γ is the union of two disjoint open sets with Lipschitz boundary;(H2) there exists a non decreasing function s : [0 , T ] → (0 , ℓ ) of class C , such that Γ t = γ ([0 , s ( t )]); DYNAMIC MODEL FOR VISCOELASTIC MATERIALS WITH PRESCRIBED GROWING CRACKS 19 (H3) | ˙ s ( t ) | < λ C K , where λ is the ellipticity constant of C and C K is the constant that appears in Korn’sinequality in (2.2).Notice that hypotheses (H1) and (H2) imply (E1)–(E3). We also assume that Ψ satisfies (5.1) and thereexists a constant ǫ > t ∈ [0 , T ]Ψ( t, x ) = 0 for every x ∈ { y ∈ Ω : | y − γ ( s ( t )) | < ǫ } . (5.7) Theorem 5.5.
Assume d = 2 and (H1)–(H3), (5.1) , and (5.7) . Then the system (2.8) with boundaryconditions (2.9) – (2.11) has a unique weak solution u ∈ W which satisfies u (0) = u and ˙ u (0) = u in thesense of (2.16) .Proof. As before let u , u ∈ W be two weak solutions to (2.8)–(2.11) with initial conditions (2.12). Then u := u − u satisfies (5.2) and (5.3) for every ϕ ∈ V D such that ϕ ( T ) = 0. Let us define t := sup { t ∈ [0 , T ] : u ( s ) = 0 for every s ∈ [0 , t ] } , and assume by contradiction that t < T . Consider first the case in which t >
0. By (H1), (H2), (5.1),and (5.7) we can find two open sets A and A , with A ⊂⊂ A ⊂⊂ Ω, and a number δ > t ∈ [ t − δ, t + δ ] we have γ ( s ( t )) ∈ A , Ψ( t, x ) = 0 for every x ∈ A , and ( A \ A ) \ Γ is the union oftwo disjoint open sets with Lipschitz boundary. Let us defineˆ V := { u ∈ H (( A \ A ) \ Γ t − δ ; R ) : u = 0 on ∂A ∪ ∂A } , ˆ H := L ( A \ A ; R ) . Since every function in ˆ V can be extended to a function in V Dt − δ , by classical results for linear hyperbolicequations (se, e.g., [9]), we deduce ¨ u ∈ L ( t − δ, t + δ ; ( ˆ V ) ′ ) and that u satisfies for a.e. t ∈ ( t − δ, t + δ ) h ¨ u ( t ) , φ i ( ˆ V ) ′ + ( C Eu ( t ) , Eφ ) ˆ H = 0 for every φ ∈ ˆ V . Moreover, we have u ( t ) = 0 as element of ˆ H and ˙ u ( t ) = 0 as element of ( ˆ V ) ′ , since u ( t ) ≡ t − δ, t ), u ∈ C ([ t − δ, t ]; ˆ H ), and ˙ u ∈ C ([ t − δ, t ]; ( ˆ V ) ′ ). We are now in position to apply the result of finitespeed of propagation of Theorem 5.3. This theorem ensures the existence of a third open set A , with A ⊂⊂ A ⊂⊂ A , such that, up to choose a smaller δ , we have u ( t ) = 0 on ∂A for every t ∈ [ t , t + δ ],and both (Ω \ A ) \ Γ and A \ Γ are union of two disjoint open sets with Lipschitz boundary.In Ω \ A the function u solves − Z t + δt − δ Z Ω \ A ˙ u ( t, x ) · ˙ ϕ ( t, x ) d x d t + Z t + δt − δ Z Ω \ A C ( x ) Eu ( t, x ) · Eϕ ( t, x ) d x d t + Z t + δt − δ Z Ω \ A B ( x )Ψ( t, x ) E ˙ u ( t, x ) · Ψ( t, x ) Eϕ ( t, x ) d x d t = 0for every ϕ ∈ L ( t − δ, t + δ ; ˆ V ) ∩ H ( t − δ, t + δ ; ˆ H ) such that ϕ ( t − δ ) = ϕ ( t + δ ) = 0, whereˆ V := { u ∈ H ((Ω \ A ) \ Γ t − δ ; R ) : u = 0 on ∂ D Ω ∪ ∂A } , ˆ H := L (Ω \ A ; R ) . Since u ( t ) = 0 on ∂ D Ω ∪ ∂A for every t ∈ [ t − δ, t + δ ] and u ( t − δ ) = ˙ u ( t − δ ) = 0 in the senseof (2.16) (recall that u ≡ t − δ, t )), we can apply Lemma 5.2 to deduce u ( t ) = 0 in Ω \ A for every t ∈ [ t − δ, t + δ ].On the other hand in A , by settingˆ V t := { u ∈ H ( A \ Γ t ; R ) : u = 0 on ∂A } , ˆ H := L ( A ; R ) , we get that the function u solves − Z t + δt − δ Z A ˙ u ( t, x ) · ˙ ϕ ( t, x ) d x d t + Z t + δt − δ Z A C ( x ) Eu ( t, x ) · Eϕ ( t, x ) d x d t = 0for every ϕ ∈ L ( t − δ, t + δ ; ˆ V t + δ ) ∩ H ( t − δ, t + δ ; ˆ H ) such that ϕ ( t ) ∈ ˆ V t for a.e. t ∈ ( t − δ, t + δ )and ϕ ( t − δ ) = ϕ ( t + δ ) = 0. Here we would like to apply the uniqueness result of Theorem 5.4 for thespaces { ˆ V t } t ∈ [ t − δ,t + δ ] and ˆ H , endowed with the usual norms, and for the bilinear form a ( u, v ) := Z A C ( x ) Eu ( x ) · Ev ( x )d x for every u, v ∈ ˆ V t + δ . As show in [7, Example 2.14] we can construct two maps Φ , Λ ∈ C , ([ t − δ, t + δ ] × A ; R ) such that forevery t ∈ [0 , T ] the function Φ( t, · ) : A → A is a diffeomorfism of A in itself with inverse Λ( t, · ) : A → A .Moreover, Φ(0 , y ) = y for every y ∈ A , Φ( t, Γ ∩ A ) = Γ ∩ A and Φ( t, Γ t − δ ∩ A ) = Γ t ∩ A for every t ∈ [ t − δ, t + δ ]. For every t ∈ [ t − δ, t + δ ], the maps ( Q t u )( y ) := u (Φ( t, y )), u ∈ ˆ V t and y ∈ A , and( R t v )( x ) := v (Λ( t, x )), v ∈ ˆ V t − δ and x ∈ A , provide a family of linear and continuous operators whichsatisfy the assumptions (U1)–(U8) of Theorem 5.4 (see [8, Example 4.2]). The only condition to check is(U5). The bilinear form a satisfies the following ellipticity condition a ( u, u ) ≥ λ k Eu k L ( A ; R × sym ) ≥ λ ˆ C k k u k V t δ − λ k u k H for every u ∈ ˆ V t + δ , (5.8)where ˆ C K is the constant in Korn’s inequality in ˆ V t + δ , namely k∇ u k L ( A ; R × ) ≤ ˆ C K ( k u k L ( A ; R ) + k Eu k L ( A ; R × sym ) ) for every u ∈ ˆ V t + δ . Notice that for t ∈ [ t − δ, t + δ ]( ˙ R t v )( x ) = ∇ v (Λ( t, x )) ˙Λ( t, x ) for a.e. x ∈ A , from which we obtain k ˙ R t Q t u k H ≤ Z A |∇ u ( x ) | | ˙Φ( t, Λ( t, x )) | d x. Hence, have to show the property | ˙Φ( t, y ) | < λ ˆ C K for every t ∈ [ t − δ, t + δ ] and y ∈ A . This is ensured by (H3). Indeed, as explained in [7, Example 3.1], we can construct the maps Φ and Λ insuch a way that | ˙Φ( t, y ) | < λ C K , since | ˙ s ( t ) | < λ C K . Moreover, every function in ˆ V t + δ can be extended to a function in H (Ω \ Γ; R d ). Hence,for Korn’s inequality in ˆ V t + δ , we can use the same constant C K of H (Ω \ Γ; R d ). This allows us to applyTheorem 5.4, which implies u ( t ) = 0 in A for every t ∈ [ t , t + δ ]. In the case t = 0, it is enough to argueas before in [0 , δ ], by exploiting (5.2). Therefore u ( t ) = 0 in Ω for every t ∈ [ t , t + δ ], which contradicts themaximality of t . Hence t = T , that yields u ( t ) = 0 in Ω for every t ∈ [0 , T ]. (cid:3) Remark 5.6.
Also Theorem 5.5 is true in the antiplane case, with essentially the same proof. Notice that,when the displacement is scalar, we do not need to use Korn’s inequality in (5.8) to get the coercivity inˆ V t + δ of the bilinear form a defined before. Therefore, in this case in (H3) it is enough to assume | ˙ s ( t ) | < λ .6. A Moving Crack Satisfying Griffith’s Dynamic Energy–Dissipation Balance
We conclude this paper with an example of a moving crack { Γ t } t ∈ [0 ,T ] and weak solution to (2.8)–(2.12)which satisfy the energy–dissipation balance of Griffith’s dynamic criterion, as happens in [4] for the purelyelastic case. In dimension d = 2 we consider an antiplane evolution, which means that the displacement u is scalar, and we take Ω := { x ∈ R : | x | < R } , with R >
0. We fix a constant 0 < c < cT < R ,and we set Γ t := { ( σ, ∈ Ω : σ ≤ ct } . Let us define the following function S ( x , x ) := Im ( √ x + ix ) = 1 √ x p | x | + x x ∈ R \ { ( σ,
0) : σ ≤ } , where Im denotes the imaginary part of a complex number. Notice that S ∈ H (Ω \ Γ ) \ H (Ω \ Γ ), andit is a weak solution to ® ∆ S = 0 in Ω \ Γ , ∇ S · ν = ∂ S = 0 on Γ . DYNAMIC MODEL FOR VISCOELASTIC MATERIALS WITH PRESCRIBED GROWING CRACKS 21
Let us consider the function u ( t, x ) := 2 √ π S Å x − ct √ − c , x ã t ∈ [0 , T ] , x ∈ Ω \ Γ t and let w ( t ) be its restriction to ∂ Ω. Since u ( t ) has a singularity only at the crack tip ( ct, w ( t ) can be seen as the trace on ∂ Ω of a function belonging to H (0 , T ; L (Ω)) ∩ H (0 , T ; H (Ω \ Γ )), stilldenoted by w ( t ). It is easy to see that u solves the wave equation¨ u ( t ) − ∆ u ( t ) = 0 in Ω \ Γ t , t ∈ (0 , T ) , with boundary conditions u ( t ) = w ( t ) on ∂ Ω, t ∈ (0 , T ) ,∂u∂ν ( t ) = ∇ u ( t ) · ν = 0 on Γ t , t ∈ (0 , T ) , and initial data u ( x , x ) := 2 √ π S Å x √ − c , x ã ∈ H (Ω \ Γ ) ,u ( x , x ) := − √ π c √ − c ∂ S Å x √ − c , x ã ∈ L (Ω) . Let us consider a function Ψ which satisfies the regularity assumptions (5.1) and condition (5.7), namelyΨ( t ) = 0 on B ǫ ( t ) := { x ∈ R : | x − ( ct, | < ǫ } for every t ∈ [0 , T ] , with 0 < ǫ < R − cT . In this case u is a weak solution, in the sense of Definition 2.4, to the damped waveequation ¨ u ( t ) − ∆ u ( t ) − div(Ψ ( t ) ∇ ˙ u ( t )) = f ( t ) in ∈ Ω \ Γ t , t ∈ (0 , T ) , with forcing term f given by f := − div(Ψ ∇ ˙ u ) = −∇ Ψ · ∇ ˙ u − Ψ ∆ ˙ u ∈ L (0 , T ; L (Ω)) , and boundary and initial conditions u ( t ) = w ( t ) on ∂ Ω, t ∈ (0 , T ) ,∂u∂ν ( t ) + Ψ ( t ) ∂ ˙ u∂ν ( t ) = 0 on Γ t , t ∈ (0 , T ) ,u (0) = u , ˙ u (0) = u . Notice that for the homogeneous Neumann boundary conditions on Γ t we used ∂ ˙ u∂ν ( t ) = ∇ ˙ u ( t ) · ν = ∂ ˙ u ( t ) = 0on Γ t . By the uniqueness result proved in the previous section, the function u coincides with that one foundin Theorem 3.1. Thanks to the computations done in [4, Section 4], we know that u satisfies for every t ∈ [0 , T ] the following energy–dissipation balance for the undamped equation, where ct coincides with thelength of Γ t \ Γ k ˙ u ( t ) k L (Ω) + 12 k∇ u ( t ) k L (Ω; R ) + ct = 12 k ˙ u (0) k L (Ω) + 12 k∇ u (0) k L (Ω; R ) + Z t ( ∂u∂ν ( s ) , ˙ w ( s )) L ( ∂ Ω) d s. (6.1)Moreover, we have Z t ( ∂u∂ν ( s ) , ˙ w ( s )) L ( ∂ Ω) d s = Z t ( ∇ u ( s ) , ∇ ˙ w ( s )) L (Ω; R ) d s − Z t ( ˙ u ( s ) , ¨ w ( s )) L (Ω) d s + ( ˙ u ( t ) , ˙ w ( t )) L (Ω) − ( ˙ u (0) , ˙ w (0)) L (Ω) . (6.2)For every t ∈ [0 , T ] we compute( f ( t ) , ˙ u ( t ) − ˙ w ( t )) L (Ω) = − Z (Ω \ B ǫ ( t )) \ Γ t div[Ψ ( t, x ) ∇ ˙ u ( t, x )]( ˙ u ( t, x ) − ˙ w ( t, x )) d x = − Z (Ω \ B ǫ ( t )) \ Γ t div[Ψ ( t, x ) ∇ ˙ u ( t, x )( ˙ u ( t, x ) − ˙ w ( t, x ))] d x + Z (Ω \ B ǫ ( t )) \ Γ t Ψ ( t, x ) ∇ ˙ u ( t, x ) · ( ∇ ˙ u ( t, x ) − ∇ ˙ w ( t, x )) d x. If we denote by ˙ u ⊕ ( t ) and ˙ w ⊕ ( t ) the traces of ˙ u ( t ) and ˙ w ( t ) on Γ t from above and by ˙ u ⊖ ( t ) and ˙ w ⊖ ( t ) thetrace from below, thanks to the divergence theorem we have Z (Ω \ B ǫ ( t )) \ Γ t div[Ψ ( t, x ) ∇ ˙ u ( t, x )( ˙ u ( t, x ) − ˙ w ( t, x ))] d x = Z ∂ Ω Ψ ( t, x ) ∂ ˙ u∂ν ( t, x )( ˙ u ( t, x ) − ˙ w ( t, x )) d x + Z ∂B ǫ ( t ) Ψ ( t, x ) ∂ ˙ u∂ν ( t, x )( ˙ u ( t, x ) − ˙ w ( t, x )) d x − Z (Ω \ B ǫ ( t )) ∩ Γ t Ψ ( t, x ) ∂ ˙ u ⊕ ( t, x )( ˙ u ⊕ ( t, x ) − ˙ w ⊕ ( t, x )) d H ( x )+ Z (Ω \ B ǫ ( t )) ∩ Γ t Ψ ( t, x ) ∂ ˙ u ⊖ ( t, x )( ˙ u ⊖ ( t, x ) − ˙ w ⊖ ( t, x )) d H ( x ) = 0 , since u ( t ) = w ( t ) on ∂ Ω, Ψ( t ) = 0 on ∂B ǫ ( t ), and ∂ ˙ u ( t ) = 0 on Γ t . Therefore for every t ∈ [0 , T ] we get( f ( t ) , ˙ u ( t ) − ˙ w ( t )) L (Ω) = k Ψ( t ) ∇ ˙ u ( t ) k L (Ω; R ) − (Ψ( t ) ∇ ˙ u ( t ) , Ψ( t ) ∇ ˙ w ( t )) L (Ω; R ) . (6.3)By combining (6.1)–(6.3) we deduce that u satisfies for every t ∈ [0 , T ] the following Griffith’s energy–dis-sipation balance for the viscoelastic dynamic equation12 k ˙ u ( t ) k L (Ω) + 12 k∇ u ( t ) k L (Ω; R ) + Z t k Ψ( s ) ∇ ˙ u ( s ) k L (Ω; R ) d s + ct = 12 k ˙ u (0) k L (Ω) + 12 k∇ u (0) k L (Ω; R ) + W tot ( t ) , (6.4)where in this case the total work takes the form W tot ( t ) := Z t (cid:2) ( f ( s ) , ˙ u ( s ) − ˙ w ( s )) L (Ω) + ( ∇ u ( s ) , ∇ ˙ w ( s )) L (Ω; R ) + (Ψ( s ) ∇ ˙ u ( s ) , Ψ( s ) ∇ ˙ w ( s )) L (Ω; R ) (cid:3) d s − Z t ( ˙ u ( s ) , ¨ w ( s )) L (Ω) d s + ( ˙ u ( t ) , ˙ w ( t )) L (Ω) − ( ˙ u (0) , ˙ w (0)) L (Ω) . Notice that equality (6.4) gives (1.6). This show that in this model Griffith’s dynamic energy–dissipationbalance can be satisfied by a moving crack, in contrast with the case Ψ = 1, which always leads to (1.3).
Acknowledgements.
The authors wish to thank Professors Gianni Dal Maso and Rodica Toader forhaving proposed the problem and for many helpful discussions on the topic. The authors are members of the
Gruppo Nazionale per l’Analisi Matematica, la Probabilit`a e le loro Applicazioni (GNAMPA) of the
IstitutoNazionale di Alta Matematica (INdAM).
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