LAPLACE EQUATION IN A DOMAIN WITH A RECTILINEAR CRACK: HIGHER ORDER DERIVATIVES OF THE ENERGY WITH RESPECT TO THE CRACK LENGTH

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1 LAPLACE EQUATION IN A DOMAIN WITH A RECTILINEAR CRACK: HIGHER ORDER DERIVATIVES OF THE ENERGY WITH RESPECT TO THE CRACK LENGTH GIANNI DAL MASO, GIANLUCA ORLANDO, AND RODICA TOADER Abtract We conider the weak olution of the Lalace equation in a lanar domain with a traight crack, recribing a homogeneou Neumann condition on the crack and a nonhomogeneou Dirichlet condition on the ret of the boundary For every k we exre the k -th derivative of the energy with reect to the crack length in term of a finite number of coefficient of the aymtotic exanion of the olution near the crack ti and of a finite number of other arameter, which only deend on the hae of the domain Keyword: cracked domain, energy releae rate, higher order derivative, aymtotic exanion of olution MSC 010: 35J0, 35C0, 74R10 Content 1 Introduction 1 Regularity of the olution with reect to crack length 3 3 Exanion near the crack ti 7 4 Comutation of the derivative of the energy 1 5 The main theorem 16 Reference 19 1 Introduction Motivated by roblem that arie in the tudy of fracture mechanic for brittle material (ee, eg, ] for a recent urvey on variational method alied to thi field, in thi work we deal with the olution u H 1 ( \ Γ of the following roblem: u = 0 in \ Γ u (11 = g on \ Γ u = 0 on Γ, ν where i a bounded connected oen et of the lane containing the origin, and Γ := {(x 1, 0 x 1 } For imlicity, the olution u 0 to roblem (11 with = 0 will be denoted by u The energy aociated to the olution u i defined by (1 E( := 1 dx \Γ u (x Prerint SISSA 1/014/MATE 1

2 G DAL MASO, G ORLANDO, AND R TOADER It i eay to rove that E( i C (ee Corollary 3 The main reult of thi aer i an algorithm to comute d (0 k Similar reult in the cae of lane elaticity have been obtained in 5] and 1] The tarting oint of our analyi i the well known aymtotic exanion of u near it crack ti: (13 u(ρ, θ = (14 + n=0 v an+1ρ n+1 in ( n+1 θ + b nρ n co (nθ ], where (ρ, θ are olar coordinate, with ρ > 0 and π < θ < π An intereting reult, obtained for the firt time by Irwin in 4], give an exlicit relation between the energy releae rate de (0 and the coefficient a1 of the exanion (13 of the olution u, called d tre intenity factor (ee, eg, 3]: de d (0 = π 4 a 1 Thi equality how that the firt derivative of the energy uniquely deend on the local behaviour of the olution near the crack ti and doe not deend on the hae of the domain Thi i no longer true for higher order derivative, a noticed by 5] and 1] in the cae of lane elaticity For k we hall ee that dk E (0 deend alo on the olution v d k H1 ( \ Γ 0, 1 j k 1, of the roblem = 0 in \ Γ0 v v j+1 = ρ in ( j+1 θ on \ Γ 0 = 0 on Γ0 ν Thee function have the following aymtotic exanion near the origin: (15 v For every k 1, let = + n=0 n+1 c n+1 (ρ in ( n+1 θ + d n (ρn co (nθ ] (16 λ k ( = (c n+1 ( n+j, conidered a an element of R k( We are now in a oition to tate the main reult of the aer Let A be the collection of the admiible oen et, whoe recie definition will be given at the beginning of Section Theorem 11 For every k 1 there exit a function Ψ k : R k R k( R uch that for every admiible et A and for every boundary condition g H 1 ( \ Γ 0 we have d k (0 = Ψ k(α k, λ k (, where λ k ( are defined in (16 and α k = (a 1, a 3,, a, a i being the coefficient of the aymtotic exanion (13 Moreover, for all λ R k(, the function α Ψ k (α, λ i a homogeneou olynomial of degree, and for every α R k the function α Ψ k (α, λ i a olynomial of degree k 1 The roof rovide alo an iterative algorithm for comuting Ψ k An eential ingredient i a careful analyi of the harmonic function u on \ Γ 0, defined a the derivative of the olution with reect to the crack length: u u u 0 := lim 0, u u (j 1 := lim 0 u (j 1 0, for j

3 HIGHER ORDER DERIVATIVES OF THE ENERGY 3 A crucial te i the roof of the formula j u = a m+1 m+1 ρ in ( m+1 θ + v (m ], which connect the function u m=1 to the hae function v defined in (14 Regularity of the olution with reect to crack length Let be a bounded connected oen ubet of R with Lichitz boundary Suoe that contain the origin 0 R To decribe the crack lying on the traight line R {0}, we fix m < 0 < M, and for every ( m, M we et Γ := {(x 1, 0 m x 1 } We aume that Γ i contained in excet for the end-oint ( m, 0, which belong to Moreover, we uoe that there exit two oen triangle with one vertex at ( m, 0, contained in \ Γ 0 and lying above and below Γ 0 Henceforth, the cla of domain atifying thee roertie will be denoted by A It i eay to ee that for every A and for every ( m, M, the oen et \ Γ can be written a the union of two domain with Lichitz boundary Therefore we can define a trace oerator from H 1 ( \ Γ into L ( \ Γ and emloy the Poincaré inequality in \ Γ, by conidering earately thee Lichitz ubdomain Γ We now fix a function g H 1 ( \ Γ 0 In order to make recie the notion of olution of roblem (11, we introduce the ace of tet function: (1 H := {ψ H 1 ( \ Γ ψ = 0 on \ Γ } For every ( m, M, we ay that u i a olution of (11 if u H 1 ( \ Γ, u = g on \ Γ and ( u ψ dx = 0 for every ψ H \Γ The olution of roblem (11 correonding to = 0 will be imly denoted by u Here we focu our attention on the deendence of the olution u on the crack length and we tudy it regularity with reect to the arameter To do thi we reformulate ( a an equation over a fixed domain, by uing uitable -deendent diffeomorhim To thi aim, we fix η Cc ( uch that η 1 in a neighbourhood of the crack ti 0 and conider the ma F : R R defined by (3 F (x 1, x = (x 1 + η(x 1, x, x It i eay to ee that there exit δ 0 > 0 uch that for every ( δ 0, δ 0 the ma F i a diffeomorhim and it atifie the following roertie: F (Γ 0 = Γ and F coincide with the identity near We may aume that m < δ 0 and δ 0 < M In addition, we uoe that η i radial and atifie η(ρ = 1 for ρ R 0/ and η(ρ = 0 for ρ R 0, for ome 0 < R 0 < dit(0, Let U be the olution u of roblem (11 in the new coordinate, ie, (4 U := u F H 1 ( \ Γ 0

4 4 G DAL MASO, G ORLANDO, AND R TOADER Remark 1 Since F doe not modify the boundary of, urely U g H 0, where H 0 i the ace defined in (1 for = 0 Moreover, by alying the change of coordinate in (, it i traightforward to check that U olve (5 T C ξ dx = 0 for all ξ H 0, \Γ 0 where the matrix C i defined by (6 C (x := DF 1 (F (x(df 1 (F (x T det DF (x, which i a ymmetric matrix, mooth with reect to the variable (, x and atifie the uniform elliticity condition, ie, ζ T C ζ λ ζ for ome λ > 0, for all x \ Γ 0, for all ( δ 0, δ 0 and for all ζ R An exlicit exreion of the matrix C will be ueful From the definition (6 we can comute ( (D η(x D η(x(1 + D 1η(x (7 C (x = 1 + D 1η(x D η(x(1 + D 1η(x (1 + D 1η(x Let C be the j -th derivative of thi matrix with reect to the arameter ; a uual we et C (0 = C Before facing the roblem of the regularity of u with reect to, which i quite a delicate iue, we invetigate the regularity of U Theorem The function ( δ 0, δ 0 U H 1 ( \ Γ 0 i of cla C Proof The theorem i a conequence of the Imlicit Function Theorem on Banach ace In fact, let H 0 be the dual ace of H 0, and for ( δ 0, δ 0, let A : H 0 H 0 be the oerator defined by A V, ξ := V T C ξ dx for every V, ξ H 0 \Γ 0 For every ( δ 0, δ 0, the function V := U g H 0 where the ma L : ( δ 0, δ 0 H 0 H 0 L(, V = 0 in H 0, i defined by L(, V := A V + A g i the unique olution of the roblem It i oible to deduce from the moothne of the matrix C that the ma L i mooth Moreover, for every 0 ( δ 0, δ 0, it derivative with reect to V comuted at ( 0, 0 i given by L V (0, 0 = A 0 L(H 0, H 0, and the oerator A 0 i invertible by the Lax-Milgram Theorem Hence, by the Imlicit Function Theorem, there exit δ > 0 uch that the locu defined by L(, V = 0 i the grah of a mooth function ( 0 δ, 0 + δ H 0 The next corollary deal with the regularity of the energy Corollary 3 The function E( introduced in (1 i C Proof Uing the change of variable (3 we obtain that (8 E( := 1 T C U dx \Γ 0 The concluion follow from Theorem and from the fact that C deend moothly on The following theorem how that all artial derivative of U deend moothly on Theorem 4 Let ω be an oen et with ω \ Γ 0 Then the function ( δ 0, δ 0 U H m (ω i of cla C for every integer m 1

5 HIGHER ORDER DERIVATIVES OF THE ENERGY 5 Proof Conider an oen et ω uch that ω ω \ Γ 0 and let ζ be a cut-off function comactly uorted in ω and uch that ζ 1 on ω After ome comutation done integrating by art, one can ee that the function D j(ζu olve a roblem of the form A D j(ζu + G = 0 in H 1 (ω where A : H0 1 (ω H 1 (ω i defined by A V, ψ := ( V T C ψ dx for all V, ψ H0 1 (ω, ω and where G i a uitable element of H 1 (ω, deending moothly on Thank to the moothne of the matrix C and to the regularity of the function U H 1 ( \ Γ 0 obtained in Theorem, we deduce that the ma L : ( δ 0, δ 0 H 1 0 (ω H 1 (ω defined by L(, V := A V + G i mooth A in the roof of Theorem, it follow from the Imlicit Function Theorem that ( δ 0, δ 0 D j(ζu H0 1 (ω i C Thi how that the function ( δ 0, δ 0 U H (ω i C Arguing by induction, one can rove that for every multi-index α, the function D α (ζu belong to C (( δ 0, δ 0; H0 1 (ω Thi how that for every integer m 1 the function U belong to C (( δ 0, δ 0; H m (ω The next corollary eaily follow from Theorem 4 and from the Sobolev Embedding Theorem For every bounded oen et ω R, the ace C k (ω i endowed with the uual toology of uniform convergence of the function and all their derivative u to order k Corollary 5 Let ω be an oen et with ω \ Γ 0 Then the function ( δ 0, δ 0 U C k (ω i of cla C for every integer k 1 We can extend the regularity reult of Corollary 5 alo for ome oen et ω \ Γ 0 whoe boundary touche Γ 0 Theorem 6 Let ω be an oen ubet of \ Γ 0 of the form ω := B r(x 0 ± = {(x 1, x B r(x 0 ± x > 0}, where x 0 Γ 0 Aume that 0 / ω and ω Then the function ( δ 0, δ 0 U C k (ω i of cla C k Proof We give the roof only for B r(x 0 + Under our aumtion there exit ome r > r uch that ω := B r (x 0 + and 0 / ω Conider the function Ũ H1 (B r (x 0 defined by { U (x 1, x if x 0 Ũ (x 1, x := U (x 1, x if x 0 Let c ij be the coefficient of the matrix C Since η i radial, from (7 we ee that c 11 and c are even in x, while c 1 = c 1 i odd in x Therefore, from (5 it follow that Ũ olve the roblem (9 ( ŨT C ψ dx = 0 for all ψ H0 1 (B r (x 0 B r (x 0 We conclude now a in the roof of Theorem 4 and Corollary 5 In view of Theorem, we are allowed to define the derivative of U with reect to by imly taking the limit (10 U := lim h 0 U +h U h in the trong toology of H 1 ( \ Γ 0 Moreover, for j, we can define further derivative of U by the recurive formula (11 U U (j 1 +h := lim U (j 1, h 0 h

6 6 G DAL MASO, G ORLANDO, AND R TOADER where the limit i taken in the trong toology of H 1 ( \ Γ 0 Oberve that, for j 1, the function U i the limit of function in H 0, and hence it belong to H 0 The convergence in (10 and (11 take lace in C k (ω for every oen et ω \ Γ 0 and alo for every half-dik ω a in Theorem 6 A uual we et U (0 = U and we adot the notation U for U comuted at = 0 We can now deal with the regularity of the olution u with reect to the arameter Theorem 7 Let 0 ( δ 0, δ 0 and let ω be an oen et with ω Aume that either ω \ Γ 0 or that ω = B r(x 0 ± = {(x 1, x B r(x 0 ±x > 0} with x 0 Γ 0 and ( 0, 0 / ω Then for every integer k 0 there exit δ > 0 uch that the function ( 0 δ, 0 + δ u C k (ω i of cla C k Proof The reult follow from Corollary 5 and Theorem 6, by noticing that u i the comoition of U with the change of coordinate F 1 The reviou reult allow u to define the derivative of u with reect to the arameter Define (1 u := lim h 0 u +h u h and for every j, by the recurive formula, (13 u u (j 1 +h := lim u(j 1 h 0 h The convergence in (1 and (13 take lace in C k (ω for every ω a in Theorem 7 A uual we et u (0 = u and we adot the notation u for u comuted at = 0 Prooition 8 For every j 1, the function u i harmonic on \ Γ 0 and atifie the Neumann condition u ν u (14 lim x x 0 ν x \Γ 0 = 0 on Γ 0 \ {0}, in the ene that (x = 0 for every x0 Γ0 \ {0} Proof The function u i harmonic, becaue by definition (1 it i the uniform limit on comact et of harmonic function It alo atifie the Neumann condition u = 0 on ν Γ0 \{0}, ince u h u atifie the Neumann condition on Γ 0 and the limit in (1 take lace in C k (ω for the half ball ω conidered in Theorem 7 By induction, it follow from the ame reaon that u i harmonic and atifie the Neumann condition u = 0 on Γ ν 0 \ {0} The following lemma how the relationhi between the derivative of U with reect to and the derivative of u with reect to and x 1 Lemma 9 For every j 0 it hold (15 U (x = j ( j D =0 1u (j (F (xη (x, where η i the cut-off function involved in the definition (3 of the change of coordinate F denote the derivative of order in the direction x1 D 1 Proof The imle roof can be done by induction and it i omitted and Remark 10 Since we choe the cut-off function η in uch a way that it vanihe outide the ball B R0, from formula (15 we ee that U and u coincide out of B R0 In articular u i H 1 far from the crack ti and it trace on \ Γ 0 vanihe Moreover, ince u i mooth in \ Γ 0 and can be moothly extended to both ide of Γ 0 \{0}, we conclude that u belong to H 1 ((B R \ B ε \ Γ 0, for every 0 < ε < R < dit(0, The following lemma how that the function U equation are weak olution of uitable differential

7 HIGHER ORDER DERIVATIVES OF THE ENERGY 7 Lemma 11 For all j 0 we have j ( (16 j (j T C ( ξ dx = 0 =0 \Γ 0 for all ξ H 0 Proof The roof roceed by induction on j For j = 0, equation (16 i exactly equation (5, which hold true Let j 1 and aume that the tatement i true for j 1 Let u rove that it i true for j Equation (16 for j 1 read j 1 ( j 1 (j 1 T C ( ξ dx = 0 for all ξ H 0 =0 \Γ 0 Thank to Theorem, we can derive thi equation with reect to We obtain j 1 ( 0 = j 1 (j T C ( ξ + (j 1 T C (+1 ξ ] dx =0 \Γ 0 j 1 = T ( C ξ dx + j 1 =1 \Γ 0 \Γ 0 j ( + j 1 (j 1 T C (+1 = =0 which i what we wanted to rove (j T C ( ξ dx+ ξ dx + T C ξ dx =0 \Γ 0 \Γ 0 j ( j (j T C ( ξ dx, \Γ 0 3 Exanion near the crack ti In thi ection we find the aymtotic exanion for the harmonic function u and u near the crack ti, which coincide with the origin We tart by recalling the claical reult for u, which can be obtained by elementary method of comlex analyi Prooition 31 Let 0 < R < dit(0, Then (13 hold in the cracked ball B R \ Γ 0 centred at 0, and the erie in (13 converge uniformly on every cracked ball B r \ Γ 0, with 0 < r < R Thi exanion i a articular cae of a more general reult concerning the u roved in Prooition 3 (ee Remark 33 The new difficulty about the u i that, in general, they do not belong to H 1 ( \ Γ 0, ince they exhibit a tronger ingularity at the origin Prooition 3 Let j 0 and 0 < R < dit(0, Then (31 u (ρ, θ = + n= a n+1 ρ n+1 in ( n+1 θ + b n ρn co (nθ ] + c log ρ in the cracked ball B R \ Γ 0 centred at 0, and the erie in (31 converge uniformly on every et of the form (B r \ B r \ Γ 0, with 0 < r < r < R Proof We oen the crack by uing the bi-holomorhic change of coordinate Φ : B R \ Γ 0 B + R := {(x 1, x B R x1 > 0} z z, where we identify (x 1, x with the comlex number z = x 1 + ix Notice that the change of coordinate Φ tranform the art of the crack Γ 0 B R into the egment S := {(x 1, x R x 1 = 0, R x R}

8 8 G DAL MASO, G ORLANDO, AND R TOADER Uing the fact that Φ i bi-holomorhic, it i eay to how that for every ε > 0 the function v := u Φ 1 belong to H 1 ((B + R \ B ε \ Γ 0 and by Prooition 8 it olve the roblem v = 0 in (B + R \ B ε \ Γ 0 v = 0 on S \ Bε ν By reflecting the function v, we can define a harmonic function on the whole annulu B R \ Bε : { v(x 1, x if x 1 0 w(x 1, x = v( x 1, x if x 1 < 0 Thi contruction can be reeated for every ε > 0, hence we can extend the function w to a harmonic function on the unctured dik B R \ {0} Therefore there exit a contant c uch that the function w(x c log x i the imaginary art of a holomorhic function f on the unctured dik B R \ {0} The function f can be exanded in a Laurent erie (3 f(z = + n= (a n + ib n z n, a n, b n R, which uniformly converge on every annulu centred at 0 and trictly contained in the unctured dik B R \ {0} By taking the imaginary art in (3, we obtain an exanion for the function w and hence for the function v in olar coordinate in the half dik B + R v(ρ, θ = + n= a n ρ n in (nθ + b n ρ n co (nθ ] + c log ρ By alying the holomorhic change of coordinate Φ to thi exanion, we get exactly the exanion (31 in the tatement of the rooition, ince the coefficient a n for n even and b n for n odd mut vanih becaue the Neumann condition u θ = 0 i atified θ=±π Remark 33 In the cae j = 0, the function u belong to H 1 (B R \ Γ 0, hence v H 1 (B + R Thi imlie that w H 1 (B R A a conequence c(0 = 0 and the function f i holomorhic in the whole dik B R, and therefore (3 reduce to a Taylor exanion, ie a(0 n = 0, b (0 n = 0 for n < 0 Thi lead to the claical exanion (13 for u Corollary 34 Let u fix j 0, k 0, let a n+1, b n v k (ρ, θ := + n=k Then v k H k+1 (B r \ Γ 0 for every 0 < r < R The roof i baed on the following two lemma be a in Prooition 3, and let a n+1 ρ n+1 in ( n+1 θ + b n ρn co (nθ ] Lemma 35 Let D be a domain in C, and let f : D C be a holomorhic function Let u = Ref and v = Imf Aume that f, f,, f (k L (D Then u, v H k (D Proof The thei follow from the fact that for every h 1 and for every 0 α 1, α, β 1, β h with α 1 β 1 = 1, α β = 1, and α 1 + α = β 1 + β = h, we have f (h = σd α 1 1 D α v + id β 1 1 D β v] for a uitable contant σ {±1, ±i} Thi can be eaily roved by induction on h, uing the fact that for every holomorhic function g we have g = D 1g = id g Lemma 36 Let f : B R C be a holomorhic function and let ϕ : BR \ Γ0 C be defined by ϕ(z := f( z Aume that f(z = z k g(z for ome k 0 and ome holomorhic function g : B R C Then ϕ, ϕ,, ϕ (k+1 L (B r \ Γ 0 for every r < R

9 HIGHER ORDER DERIVATIVES OF THE ENERGY 9 Proof We oberve that ϕ(z = z k ψ(z, where ψ(z = g( z We can eaily rove by induction that for every h 1 we have ψ (h (z = h c h m g (m ( z z 1 m h, m=1 for uitable contant c h m Let u fix 0 < r < R Since g i holomorhic in B R, the function g (m are bounded in B r Hence there exit a contant C h uch that ψ (h (z C h z 1 h for every z B r, for every 1 h k + 1 Since ϕ(z = z k ψ(z, for every 1 h k, by Leibniz rule we have the etimate ϕ (h (z h C m,h,k z k h+m z 1 m + C 0,h,k z k h C z k h for every z B r, m=1 which how that ϕ (h i bounded on B r for 1 h k A for h = k + 1, by Leibniz rule we have ϕ (k+1 (z k+1 m=1 which how that ϕ (k+1 L (B r C m,k z 1+m z 1 m C z 1 for every z B r, Proof of Corollary 34 Since the function f introduced in the roof of Prooition 3 ha the exanion (3, we have that the function (33 f k (z = + n=k (a n + ib n z n i holomorhic in B R A the coefficient a n for n even and b n for n odd vanih (ee again the roof of Prooition 3, the function v k i the imaginary art of f k ( z The concluion follow now from Lemma 35 and Lemma 36 The next te i to how that indeed the erie in (31 ha finitely many nonzero term for n negative To do thi, we rove an etimate on u which how that it belong to the dual of a uitable ubace of H j 1 (B R \ Γ 0 Let B ± R := {(x1, x BR ± x > 0} and let T be the cla of tet function ϕ C (B R \ Γ 0 C (B + R C (B R, which vanih on a neighbourhood of BR \ Γ0 and on a neighbourhood of the crack ti 0 For every ϕ T and for every k 0 we et ( D k ϕ L (B R \Γ 0 := D α ϕ 1 L (B R \Γ 0 α =k Lemma 37 Let R > 0 with η = 1 on B R For every j 1 there exit a oitive contant C j > 0 uch that the etimate (34 ϕ dx C j D j 1 ϕ L (B R \Γ 0 hold for every ϕ T B R \Γ 0 u Proof Let u rove the claim by induction on j For the cae j = 1, imly oberve that by Lemma 9 we have that u = U D 1u L (B R \ Γ 0 and therefore etimate (34 hold Let j and uoe that the claim i true for j with 1 j 1 From Lemma 9 we deduce that j u = U ( j D 1u (j in B R \ Γ 0 =1

10 10 G DAL MASO, G ORLANDO, AND R TOADER Fix a tet function ϕ T Since U L (B R \ Γ 0, by alying Hölder inequality and Poincaré inequality to all the derivative of ϕ of order le or equal than j, we get the etimate ϕ dx U L (B R \Γ 0 ϕ L (B R \Γ 0 U L (B R \Γ 0 D j 1 ϕ L (B R \Γ 0 B R \Γ 0 U Let u etimate the other term of the um, integrating by art with reect to the variable x 1 and uing the induction hyothei: (D1u (j ϕ dx = u (j D 1ϕ dx C j D j 1 D 1ϕ L (B R \Γ 0 B R \Γ 0 B R \Γ 0 Thi conclude the roof C j D j 1 ϕ L (B R \Γ 0 Prooition 38 Let 0 < R < dit(0, Then (35 u (ρ, θ = + n= j a n+1 ρ n+1 in ( n+1 θ + b n ρn co (nθ ] + c log ρ in the cracked ball B R \ Γ 0, and the erie in (35 converge uniformly on every et of the form (B r \ B r \ Γ 0, with 0 < r < r < R Proof From Prooition 3 we know that (31 hold Let u rove that for every n j we have a (n+1 = 0 In order to do thi, fix 0 < r < R < R uch that η = 1 on B R and let ϕ T be a tet function of the form ϕ(ρ, θ = ψ(ρ in ( n+1 θ, where ψ i a nonzero mooth function, with uψ (r, R and ψa (n+1 0 Define the recaled function ϕ ε(ρ, θ := ϕ ( ρ = ψ ( ( ρ ε ε in n+1 θ Etimate (34 hold for ϕ ε, and therefore C j D j 1 ϕ ε L (B R \Γ 0 εr π εr π u (ρ, θ ψ ( ( ρ ε in n+1 θ ρ dθ dρ By the uniform convergence of the erie in (31 and by the orthogonality of the trigonometric function, from the reviou inequality we obtain On the other hand and hence we have that (36 C j D j 1 ϕ ε L (B R \Γ 0 π εr εr ψ ( ρ n+1 ε a (n+1 ρ + a n+1 ρ n+1 ] ρ dρ R = πε ψ(ρ a r R + πε ψ(ρ a r n+1 (n+1 ρ ε n+1 n+1 ρ n+1 ε n+1 ] ρ dρ D j 1 ϕ ε L (B R \Γ 0 = ε j D j 1 ϕ L (B R \Γ 0 C j D j 1 ϕ L (B R \Γ 0 πε R n+1 ( j + πε ψ(ρ a r R n+1 + ( j ψ(ρ a r ] ρ dρ n+1 (n+1 ρ n+1 ρ n+1 ] ρ dρ ] ρ dρ

11 HIGHER ORDER DERIVATIVES OF THE ENERGY 11 If a (n+1 were different from zero, the right-hand ide in (36 would diverge to + for ε 0+ Therefore a (n+1 = 0 Arguing in the ame way and uing a tet function of the form we obtain that b n = 0 for n > j Theorem 39 Let 0 < R < dit(0, Then (37 u (ρ, θ = + n= j ϕ(ρ, θ = ψ(ρ co (nθ, a n+1 ρ n+1 in ( n+1 θ + + b n ρn co (nθ in the cracked ball B R \ Γ 0, and the erie in (37 converge uniformly on every et of the form (B r \ B r \ Γ 0, with 0 < r < r < R Proof Beide the binomial coefficient, we hall ue the Pochhammer ymbol defined by (38 (x := x(x 1 (x + 1 for every x R and every integer 1 We et alo (x 0 = 1 Fix 0 < r < R < R where R i uch that η = 1 on B R Uing Prooition 38 and Corollary 34, it i traightforward to check that for every 1 j the function D 1 u(j ha the following exanion in B r \ Γ 0 : D 1u (j = + 1 n= j+ 1 n= j+ ( n+1 a (j n+1 ρ n+1 n=0 in (( n+1 θ + (n b (j ρ n co ((n θ ( 1 ( 1! c (j ρ co (θ + v j,, n where v j, H 1 (B r \ Γ 0 In the formula above c (0 = 0 by Remark 33 By Lemma 9, on B r \ Γ 0 we have 1 j U ( j = j D 1u (j ( = j ( n+1 a (j n+1 ρ n+1 in (( n+1 θ + + =0 j 1 =0 n= j+ + c log ρ + w j, =0 n= j+ ( j (n b (j n j 1 ρ n co ((n θ ( 1 ( 1! ( j c (j ρ co (θ + where w j H 1 (B r \ Γ 0 We et V := U w j From the reviou formula, we obtain that the function j V := α n ρ n+1 in ( n+1 θ j + β n ρ n co (nθ + c log ρ, where (39 (310 n=1 β n := n 1 =0 α n := j =0 n=1 =1 ( j ( ( n+1 a (j ( n+1, ( j ( n b (j ( n ( 1n (n 1! ( j n c (j n Since U H 1 ( \ Γ 0 by Theorem, we conclude that V H 1 (B r \ Γ 0 Let u rove that β j = 0 To thi aim we conider D ρv, which can be written a (311 D ρv = jβ j ρ j 1 (co (jθ + h(ρ, θ where h(ρ, θ 0 a ρ 0, uniformly with reect to θ Since D ρv L (B r \ Γ 0, it follow that β j = 0 Then we can rove in a imilar way that α j = 0, β j 1 = 0,, α 1 = 0, and

12 1 G DAL MASO, G ORLANDO, AND R TOADER c = 0 Therefore, by (310, for all 1 n j we have (31 n 1 ( j ( n b (j ( n = 0 =0 From (31 it i eay to ee by induction on j that b n concluion follow from (35 Remark 310 A a byroduct of the roof of Theorem 39, we obtain that j =0 ( j ( ( n+1 a (j ( n+1 = 0 for all 1 n j, a a conequence of (39 and from the equality α n ueful in the roof of the main theorem = 0 for all for all 1 n j The = 0 for 1 n j Thee relation will be 4 Comutation of the derivative of the energy In thi ection we exre the derivative dk E d k (0 of the energy E( defined in (1 in term of the coefficient of the aymtotic exanion (13 of the olution u for = 0 and in term of the coefficient of the correonding exanion (37 of the j -th derivative u of u with reect to We already know that E i C by Corollary 3 Prooition 41 For every k 1 we have (41 for every ( δ 0, δ 0 d k ( = 1 k =1 ( k \Γ 0 (k T C ( U dx Proof In order to rove formula (41, we roceed by induction on k For the cae k = 1, imly differentiate formula (8 with reect to : de d ( = T C U dx + 1 T C U dx \Γ \Γ Since U belong to the ace H 0 introduced in (1, we can ue it a a tet function for roblem (5 Since C i ymmetric, thi imlie that de d ( = 1 T C U dx, \Γ which i formula (41 for k = 1 Let u uoe by induction that the tatement i true for k 1, and let u rove that it i true for k By differentiating with reect to the following formula d E d ( = 1 =1 ( \Γ 0 ( T C ( U dx we obtain d k ( = 1 =1 ( \Γ 0 (k T C ( U + ( T C (+1 U + ( T C ( U ] dx

13 HIGHER ORDER DERIVATIVES OF THE ENERGY 13 We now ue U a a tet function for roblem (16 olved by U ( function for the roblem olved by U, and we obtain that 1 ( ( T C ( U dx = 1 =1 \Γ 0 \Γ 0 = 1 \Γ 0 and then U ( ( T C U dx ( T C U dx, a a tet which, ubtituted in the exreion of dk E d k, yield d k ( = 1 =1 Thi conclude the roof ( \Γ 0 (k ] T C ( U + ( T C (+1 U dx + 1 ( T C U dx \Γ 0 = k 1 ( T C U dx + 1 ( = \Γ 0 \Γ ( (k 1 T C ( U dx + 1 = \Γ 0 \Γ ( T C U dx \Γ 0 = 1 k ( k (k T C ( U dx =1 \Γ 0 (k T C ( U dx + T C (k U dx + In order to exand formula (41, we comute the derivative of the matrix C with reect to Lemma 4 The following equalitie hold ( C 0 D1η D η = D η D 1η for every k, C (k 0 = ( ( 1 k k!(d 1η k η Proof From the exreion of C written in (7, we have that C = DF 1 (DF 1 T 1 det DF = 1 + D M, 1η where ( 1 + D η D η(1 + D 1η M = D η(1 + D 1η (1 + D 1η Hence, by differentiating with reect to we obtain that (4 C = D 1η (1 + D 1η M D M 1η which comuted in = 0 give the formula we wanted to rove for the firt derivative Intead of roving the tatement of the lemma, it i convenient to how by induction a tronger reult, ie, that for every k C (k = ( 1k k!(d 1η k (1 + D 1η k+1 M + ( 1 k!(d 1η M (1 + D 1η k + ( 1k k!(d 1η k (1 + D 1η M The bae cae k = i obtained by imly differentiating formula (4 with reect to The inductive te follow eaily from the fact that the matrix M doe not actually deend on, ince the entrie of M are olynomial of degree with reect to

14 14 G DAL MASO, G ORLANDO, AND R TOADER Prooition 43 For every k 1, we have d k (0 = k j=0 ( j η (ρη(ρ j R 0 0 π π D j+1 1 u (k j 1 D ρu + 1 ρ DDj 1u (k j 1 D θ u ] ρ dθ dρ where R 0 i the radiu of the ball which uort the cutoff function η Proof Let u tart by comuting dk E at 0 uing formula (41 and emloying the exreion of the d k derivative of C found in Lemma 4: d k (0 = 1 = k + 1 k =1 \Γ 0 ( k (k T C ( 0 u dx \Γ 0 D1ηD 1U ( D 1u D ηd 1U ( D u + D ηd U ( D 1u + D 1ηD U ( D ] u dx + k ( k ( 1!(D 1η η D 1U (k D 1u dx \Γ 0 = By uing Lemma 9 and exanding the derivative, we get d k (0 = k + 1 j=0 ( j k k = j=0 \Γ 0 D1ηD 1uD j+1 1 u ( j η j j(d 1η D 1uD j 1u ( j η j 1 D ηd ud j+1 1 u ( j η j jd 1ηD ηd ud j 1u ( j η j 1 D ηd 1uD D j 1u ( j η j j(d η D 1uD j 1u ( j η j D 1ηD ud D j 1u ( j η j + jd 1ηD ηd ud j 1u ( j η j 1] dx + (D1η η D 1uD j+1 1 u (k j η j + ( k k ( j ( 1! \Γ 0 + j(d 1η 1 η D 1uD j 1u (k j η j 1] dx After ome algebraic maniulation, we obtain d k (0 = k j=0 ( j \Γ 0 D1ηD 1uD j+1 1 u ( j η j D ηd ud j+1 1 u ( j η j D ηd 1uD D j 1u ( j η j + D 1ηD ud D j 1u ( j η j] dx + k ( j η D j 1uD j 1u ( j η j 1 dx + j=1 \Γ k k ( k k ( j ( 1! (D 1η η D 1uD j+1 1 u (k j η j dx + = j=0 \Γ k ( k k ( j ( 1! j(d 1η 1 η D 1uD j 1u (k j η j 1 dx = j=1 \Γ 0 = I + II + III + IV

15 HIGHER ORDER DERIVATIVES OF THE ENERGY 15 where I, II, III, IV are the four um aearing in the formula above Let u conider the term of the um III correonding to = : 1 k j=0 = k = k ( k k ( j! η D 1uD j+1 1 u (k j η j dx = \Γ 0 k ( j+1 (j + 1 η D 1uD j+1 1 u (k j η j dx j=0 \Γ 0 ( j j η D 1uD j 1u ( j η j 1 dx = II j=1 \Γ 0 Let u now conider all the other term of the um III : k 1 k ( k k ( j ( 1! (D 1η η D 1uD j+1 1 u (k j η j dx = =3 j=0 \Γ 0 = 1 k 1 ( k ( k 1 +1 j ( 1 +1 ( + 1! (D 1η 1 η D 1uD j+1 1 u (k 1 j η j dx = j=0 \Γ 0 = 1 k ( k ( k 1 +1 j 1 ( 1 +1 ( + 1! (D 1η 1 η D 1uD j 1u (k j η j 1 dx = j=1 \Γ 0 = 1 k ( k k ( j ( 1!j 1 η D 1uD j 1u (k j η j 1 dx = IV It follow that = j=1 d k (0 = k j=0 ( j \Γ 0 \Γ 0 (D 1η Now we integrate in olar coordinate to deduce D1ηD 1uD j+1 1 u ( j η j D ηd ud j+1 1 u ( j η j D ηd 1uD D j 1u ( j η j + D 1ηD ud D j 1u ( j η j] dx d k (0 = k = k j=0 j=0 ( j ( j Thi conclude the roof R 0 π 0 π R 0 η (ρη(ρ j D 1u co θd j+1 1 u ( j D u in θd j+1 1 u ( j D 1u in θd D j 1u ( j + D u co θd D j 1u ( j] ρ dθ dρ η (ρη(ρ j 0 π π D j+1 1 u (k j 1 D ρu + 1 ρ DDj 1u (k j 1 D θ u ] ρ dθ dρ The following theorem allow u to exre the k -th derivative of the energy in term of the following coefficient of the exanion of the olution u and of it derivative u with reect to the crack length: a ( k+3 a ( k+5 a ( 1 a (k k+5 a (1 1 a 1 a (k k+7 a (1 1 a 3 a (k 1 a (1 k 5 a k 3 a ( 1 a (k 3 a (1 k 3 a

16 16 G DAL MASO, G ORLANDO, AND R TOADER Theorem 44 For every k 1 we have d k E (43 d (0 = π j k j=0 n= k+j+1 ( k j+1 ( n+1 j+ a (k j 1 n+1 a (j n+1 d Proof Formula (43 eaily follow from the exreion of k E (0 found in Prooition 43, uing d k the uniform convergence of the exanion (31 of the function u to jutify the integration term by term, and emloying imle trigonometric identitie to integrate the ingle term 5 The main theorem Formula (43 alo hold for k = 1 Hence the firt derivative of the energy i given by de (51 d (0 = π 4 a 1 Thi i a well known reult, which how that the firt derivative of the energy uniquely deend on the local behaviour of the olution near the crack ti We now tudy the cae k We hall ee that the higher order derivative of the energy deend not only on the local behaviour of the olution near the crack ti, but alo on the hae of \ Γ 0 Indeed, we hall how that thee derivative can be exreed in term of a finite number of coefficient of the aymtotic exanion of the olution and of a finite number of other arameter, which only deend on the hae of the domain In order to do thi, we need to introduce ome technical tool Definition 51 For every j 1 let v H1 ( \ Γ 0 be the weak olution of the roblem = 0 in \ Γ0 v v v ν j+1 = ρ in ( j+1 θ on \ Γ 0 = 0 on Γ0 Remark 5 Notice that the function v deend only on the hae of the domain \ Γ0, and doe not deend on the boundary value g recribed in roblem (11 In view of Prooition 31, the function v (5 v = + n=0 can be exanded in a erie near the crack ti: n+1 c n+1 (ρ in ( n+1 θ + d n (ρn co (nθ ] The following rooition rovide ome equalitie which will be ued to rove the main theorem, combined with thoe found in Remark 310 Prooition 53 For every j 1 and for every n 0 we have a n+1 = j m=1 a m+1 c(m n+1 ( Proof Define on \ Γ 0 the function j (53 w := u a m+1 ρ m+1 in ( m+1 θ =: u + σ m=1 We now how that w i the variational olution of a uitable boundary value roblem Firt of all, the function w belong to H 1 ( \ Γ 0 Indeed, the function u i in H 1 away from the crack ti by Remark 10, and the um σ i alo mooth away from the crack ti Moreover, from (37, we deduce that w ha the following exanion w = + n=0 a n+1 ρ n+1 in ( n+1 θ + b n ρn co (nθ ], which belong to H 1 near the crack ti, by Corollary 34 We conclude that w H 1 ( \ Γ 0 We oberve that w i harmonic in \ Γ 0, ince by Prooition 8 the function u i harmonic and by direct check the um σ i harmonic too Let u ee which boundary condition are atified by

17 HIGHER ORDER DERIVATIVES OF THE ENERGY 17 w Both u and the um σ atify the Neumann condition on Γ 0 \ {0}, by Prooition 8 and by direct check reectively Let u conider now the trace of w on \ Γ 0 By Remark 10 the trace of u vanihe on \ Γ 0, hence the trace of w on \ Γ 0 i σ In concluion, w H 1 ( \ Γ 0 i a weak olution of the roblem w = 0 in \ Γ 0 j w = a m+1 ρ m+1 in ( m+1 θ on \ Γ 0 w ν m=1 = 0 on Γ 0 By the uniquene of the olution of thi roblem, we have that and hence by (53 u = j m=1 w = j m=1 a m+1 v(m a m+1 m+1 ρ in ( m+1 θ + v (m ] Comaring the exanion of both ide of the lat equation, we get the thei For every k, let (54 λ k ( = (c n+1 ( n+j, conidered a an element of R k( Written in a convenient way, the entrie of λ k ( make u the following triangular matrix (55 c (1 1 c (1 3 ( c( 1 ( c( 3 c (1 k 5 ( c (1 k 3 ( ( c(k 1 ( c ( 1 ( ( c(k 3 ( c( k 5 ( We are now ready to rove the main reult of the aer Proof of Theorem 11 Fix A and g H 1 ( \ Γ 0 In view of (51 it uffice to rove the theorem for k In formula (43 the k -th derivative of the energy wa exreed a a linear combination of the following term: (56 a ( k+3 a a ( k+5 a k 3 A k = a (k k+5 a a (1 1 a a 1a a (k k+7 a k 3 a (1 1 a k 3 a 3a k 3 a ( 1 a 3 a (k 1 a 3 a (1 k 5 a3 a k 3a 3 a ( 1 a 1 a (k 3 a 1 a (1 k 3 a1 a a 1 We collect the term different from a 1,, a in the matrix a ( k+3 a (k k+5 a ( 3 a (1 1 a ( k+5 a ( 1 a (k k+7 a ( 1 a (1 1 a (k 1 a ( k 7 a (1 k 5 a ( 1 a (k 3 a ( k 5 a (1 k 3 R k (

18 18 G DAL MASO, G ORLANDO, AND R TOADER With thi notation, formula (43 for the k -th derivative of the energy can be written in the following comact way (57 d k (0 = E k(a k, α k + F k (α k, α k, where E k : R k ( R k R and F k : R k R k R are uitable bilinear ma We now rove that we can exre the entrie of A k in term of α := (a 1,, a k 3 and λ k ( Indeed, we how by induction that for every k there exit a ma uch that Λ k : R R k( R k ( A k = Λ k (α, λ k ( Moreover we hall ee that for every λ R k(, the function α Λ k (α, λ i linear, and for every α R, the function λ Λ k (α, λ i a olynomial of degree k 1 To do thi, we will make ue of the relation found in Remark 310 and in Prooition 53 (58 (59 a ( n+1 = ( ( ( n+1 a ( ( n+1, for all 1 n k 1, =1 a ( n+1 = m=1 a ( m+1 c(m n+1 (, for all n 0 Bae cae: Let u define Λ : R R R 1 From equation (58 and (59 with k =, we deduce that (510 a (1 1 = 1 a1 and a(1 1 = a (1 1 c(1 1 ( = 1 a1c(1 1 ( Hence A = Λ (a 1, c (1 1 (, where Λ (α, λ := ( 1 α 1 αλ The ma Λ i linear with reect to α and it i a olynomial of degree 1 with reect to λ Inductive te: Suoe that there exit a ma uch that Λ : R k R ((k R ( (k A = Λ (α k, λ (, and that for every λ R ((k, the function α Λ (α, λ i linear, and for every α R k, the function λ Λ (α, λ i a olynomial of degree k We want to define Λ k The matrix A k can be written a a block matrix ( βk A where β k = a ( k+3 a ( k+5 a ( 1 A k = a ( 1 γ T k R, γ k = a (k 3 a ( k 5 a (1 k 3 R k Thank to formula (58, we can exre all the entrie of β k in term of the entrie of A and of α We can therefore define a linear ma β k : R ( (k R R uch that β k = β k (A, α Then we can ue (59 to exre all the element of γ k in term of the entrie of A and of the element of λ k ( In articular, we can define a bilinear ma γ k : R ( (k R k( R uch that γ k = γ k (A, λ k (

19 HIGHER ORDER DERIVATIVES OF THE ENERGY 19 Finally, we ue again formula (59 to exre a ( 1 in term of the element of β k and of λ k ( Hence there i a bilinear function ã k : R ( R k( R uch that a ( 1 = ã k (β k, λ k ( In concluion, alying the inductive hyothei: ( β k (A, α A A k = ã k ( β k (A, α, λ k ( γ k (A, λ k ( T ( β k (Λ (α k, λ (, α Λ (α k, λ ( = ã k ( β k (Λ (α k, λ (, α, λ k ( γ k (Λ (α k, λ (, λ k ( T =: Λ k (α, λ k ( Notice that for every λ R k(, the function α Λ k (α, λ i linear, and for every α R the function λ Λ k (α, λ i a olynomial of degree k 1 Eventually, we can aly what we roved to formula (57, concluding that d k (0 = E k(a k, α k + F k (α k, α k = E k (Λ k (α, λ k (, α k + F k (α k, α k =: Ψ k (α k, λ k ( The ma Ψ k : R k R k( R defined a above atifie the requeted roertie Remark 54 The roof of Theorem 11 i contructive In articular it allow u to obtain Λ k from Λ uing only elementary comutation In thi way for every k we can find an exlicit exreion for dk E (0 in term of a d 1,, a k and of the coefficient in (55 We write here the lit of the firt three derivative of the energy: de d (0 = π 4 a 1, d E d (0 = π 4 a 1c (1 1 ( 3 4 πa1a3, d 3 E d (0 = 9 ( 3 8 πa1a3c(1 1 ( + 3 ( c (1 1 8 ( c(1 3 ( c( 1 ( πa πa1a5 9 8 πa 3 Acknowledgement Thi material i baed on work uorted by the Italian Minitry of Education, Univerity, and Reearch under the Project Calculu of Variation (PRIN and by the Euroean Reearch Council under Grant No Quaitatic and Dynamic Evolution Problem in Platicity and Fracture The firt and third author are member of the Gruo Nazionale er l Analii Matematica, la Probabilità e le loro Alicazioni (GNAMPA of the Itituto Nazionale di Alta Matematica (INdAM Reference 1 M Ametoy and J-B Leblond, Crack ath in lane ituation II: Detailed form of the exanion of the tre intenity factor, International Journal of Solid and Structure 9 (199, B Bourdin, GA Francfort, and J-J Marigo, The variational aroach to fracture, J Elaticity 91 (008, P Grivard, Singularitie in boundary value roblem, Maon Editeur, GR Irwin, Analyi of tree and train near the end of a crack travering a late, Tran ASME J Al Mech 4 (1957, J-B Leblond, Crack ath in lane ituation I general form of the exanion of the tre intenity factor, International Journal of Solid and Structure 5 (1989,

20 0 G DAL MASO, G ORLANDO, AND R TOADER (Gianni Dal Mao SISSA, Via Bonomea 65, Triete, Italy addre, Gianni Dal Mao: dalmao@iait (Gianluca Orlando SISSA, Via Bonomea 65, Triete, Italy addre, Gianluca Orlando: gianlucaorlando@iait (Rodica Toader DIMI, Univerità di Udine, Via delle Scienze 06, Udine, Italy addre, Rodica Toader: rodicatoader@uniudit