RENORMALIZED ENERGY AND PEACH-KÖHLER FORCES FOR SCREW DISLOCATIONS WITH ANTIPLANE SHEAR

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1 RENORMALIZED ENERGY AND PEACH-KÖHLER FORCES FOR SCREW DISLOCATIONS WITH ANTIPLANE SHEAR TIMOTHY BLASS AND MARCO MORANDOTTI Abstract. We present a variational framework for studying screw dislocations subject to antiplane shear. Using a classical model developed by Cermelli & Gurtin [CG99], methods of Calculus of Variations are exploited to prove existence of solutions, and to derive a useful expression of the Peach-Köhler forces acting on a system of dislocation. This provides a setting for studying the dynamics of the dislocations, which is done in [BFLM14]. 1. Introduction Dislocations are one-dimensional defects in crystalline materials [Nab67]. Their modeling is of great interest in materials science because important material properties, such as rigidity and conductivity, can be strongly affected by the presence of dislocations. For example, large collections of dislocations can result in plastic deformations in solids under applied loads. In this note, we derive an expression for the renormalized energy associated to a system screw dislocations in cylindrical crystalline materials using a continuum model introduced by Cermelli and Gurtin [CG99]. We use the renormalized energy to derive an expression for the forces on the dislocations, called Peach-Köhler forces. These forces drive the dynamics of the system, which is studied in [BFLM14]. The proofs of some results that are used in [BFLM14] are contained in this note. Following [CG99], we consider here an elastic body B R 3 undergoing antiplane shear in the x 3 direction, with B = Ω R, where Ω R 2 is a bounded simply connected open set with Lipschitz boundary. The deformations Φ : B B are of the form Φx 1, x 2, x 3 := x 1, x 2, x 3 + ux 1, x 2, with u : Ω R, and the deformation gradient F is given by F = Φ := u = I + e u 2 u The assumption of antiplane shear allows us to reduce the three-dimensional problem to a two-dimensional problem. We will consider strain fields h that are defined on the cross-section Ω, taking values in R 2. When no dislocations are present, in this setting it turns out that h = u; however, the presence of dislocations causes the strain field to be singular at the sites of the dislocations. A screw dislocation is a line singularity in the strain field for the body B. In the antiplane shear setting, this line is parallel to the x 3 axis; in the cross-section Ω a screw dislocation is represented as a point singularity. A screw dislocation is characterized by a position z Ω and a vector b R 3, called the Burgers vector. The position z Ω is a point where the strain field fails to be the gradient of a smooth function; the Burgers vector measures the severity of this failure. More specifically, a strain field associated with a system of N screw dislocations at the positions Z := {z 1,..., z N }, with corresponding Burgers vectors satisfies the relation B := {b 1,..., b N } curl h = b i δ zi in Ω 1.2 1

2 in the sense of distributions. The notation curl h denotes the scalar curl, 1 h 2 2 h 1. Thus, in the antiplane shear setting, the Burgers vectors can be written as b i = b i e 3, the scalar b i is called the Burgers modulus for the dislocation at z i, and it is given by b i = h t ds, l i where l i is any counterclockwise loop surrounding the dislocation point z i and no other dislocation points, t is the tangent to l i, and ds is the line element. Since b i = b i e 3 for all i {1,..., N}, by abuse of notation from now on we will use the symbol B both for the set of Burgers vectors and for the set of Burgers moduli. When dislocations are present, the deformation gradient F cannot be any longer represented by the last expression in 1.1, which needs to be replaced with F = I + e 3 Our goal is to derive an energy associated to systems of screw dislocation and obtain an expression for the forces on the dislocations Peach-Köhler forces [PK50]. This lays the foundation needed to study the dynamics of the dislocations, which is investigated in [BFLM14], where the expression for the Peach-Köhler force is used along with an energy-dissipation criterion described in [CG99] to obtain an evolution equation for the system of dislocations. Our investigation of the energy associated to a system of dislocations will be undertaken in the context of linear elasticity for singular strains h. The energy density W is given by h 0 W h := 1 2 h Lh as functions of the strain h, where the elasticity tensor L is a symmetric, positive-definite matrix. In suitable coordinates, L is written in terms of the Lamé moduli λ, µ of the material µ 0 L := 0 µλ We require µ > 0, and the energy is isotropic if and only if λ = 1. The energy of a strain field h is given by Jh := W hx dx, and the equilibrium equation is Ω. div Lh = 0 in Ω. 1.4 Equations 1.2 and 1.4 provide a characterization of strain fields describing screw dislocation systems in linearly elastic materials. To be precise, we say that a strain field h L 2 Ω; R 2 corresponds to a system of dislocations at the positions Z with Burgers vectors B if h satisfies { N curl h = b iδ zi in Ω, 1.5 div Lh = 0 in the sense of distributions. In analogy to the theory of Ginzburg-Landau vortices [BBH94], no variational principle can be associated with 1.5 because the elastic energy of a system of screw dislocations is not finite see, e.g., [CL05, CG99, Nab67], so we cannot study 1.5 directly in terms of energy minimization. Indeed, the simultaneous requirements of finite energy and 1.2 are incompatible, since, if curl h = δ z0, z 0 Ω, and if B ε z 0 Ω, then h 2 dx = O log ε. Ω\B εz 0 In the engineering literature see, e.g., [CG99, Nab67], this problem is usually overcome by regularizing the energy. By removing small cores of size ε > 0 centered at the dislocations, we will replace J by J ε cf. Section 2 and obtain finite-energy strains, h ε, as minimizers of J ε. Upon taking ε 0 we will recover a 2

3 unique limiting strain h 0 = lim ε 0 h ε, satisfying 1.5. From this, we can compute a renormalized energy U associated with the limiting strain. In Section 3, we show that J ε h ε = C log ε + Uz 1,..., z N + Oε. 1.6 This type of asymptotic expansion was first proved by Bethuel, Brezis, and Hélein in [BBH93] for Ginzburg- Landau vortices, while the case of edge dislocations was studied in [CL05]. Note that 1.6 can also be obtained as a consequence of Γ-convergence, see, e.g., [AP14, SS03] and the references therein for Ginzburg- Landau vortices, [DLGP12, GPPS13, GLP10] for edge dislocations, and [ACP11, CG09, CGM11, FG07, GM05, GM06, SZ12] for other dislocations models. Finally, it is important to mention that we ignore here the core energy. We refer to [Nab67, TOP96, VKHLLO12] for more details. The renormalized energy U is a function only of the positions {z 1,..., z N }, and its gradient with respect to z i gives the opposite of the Peach-Köhler force on z i, denoted j i. In Section 4 see Theorem. 4.1, we show that j i = zi U = {W h 0 I h 0 Lh 0 } n ds, l i where l i is a suitably chosen loop around z i and n is the outer unit normal to the set bounded by l i containing z i. The quantity W h 0 I h 0 Lh 0 is the Eshelby stress tensor, see [Esh51, Gur95]. There are two different kinds of forces acting on a dislocation when other dislocations are present: the interactions with the other dislocations and the interactions with Ω. Therefore, the expression of j i contains two contributions, where the one coming from the boundary balances the tractions of the forces generated by all the dislocations. It is the gradient of the solution u 0 to an elliptic problem with Neumann boundary conditions Precisely, we show that j i has the form [ ] j i z 1,..., z N = b i JL k j z i ; z j + u 0 z i ; z 1,..., z N, j i where J is the rotation matrix of an angle π/2, and k j ; z j is the fundamental singular strain generated by the dislocation z j, see 2.4. It is important to notice that the force on the i-th dislocation is a function of the positions all the dislocations. This explicit formula is useful for computing j i, and is employed in [BFLM14] to study the motion of the dislocations. In Section 2 we show how to regularize the energy to use variational techniques to study the problem. In Section 3 we derive the renormalized energy, which we will use in Section 4 to derive the Peach-Köhler force. 2. Regularized Energies and Singular Strains We consider a system of dislocations at the positions Z = {z 1,..., z N } with Burgers vectors B = {b 1,..., b N }. We regularize the energy J by removing the singular points from the domain Ω. We define the sets E ε,i := E ε z i, where { } 2 E r z = x 1, x 2 R 2 : x 1 z 1 2 x2 z 2 + < r 2 λ is an ellipse centered at z for r > 0; the parameter λ is one of the the Lamé moduli of the material cf Let ε 0 > 0 be fixed depending on Ω, Z, and λ such that for all ε 0, ε 0 we have E ε,i Ω, and E ε,i E ε,j = for all i j. We define the sets N := Ω \ E ε,i for ε 0, ε The shape of the cores E ε,i is not crucial, but in the sequel we will find the ellipses E ε,i centered at z i to be useful for computations. We now define the energy functional at the level ε by setting J ε h := W h dx. 3

4 By removing cores around the singular set Z, we have regularized the energy, in the sense that it will not necessarily be infinite on strains satisfying 1.5. However, since we have effectively removed the dislocations from the problem, we account for their presence by a judicious choice of function space. We define and H curl 0, Z, B := H curl := {h L 2, R 2 : curl h L 2 } { } h H curl, curl h = 0, h t ds = b i, i = 1,..., N, 2.2 E ε,i where t is the unit tangent vector to E ε,i. The condition on h involving the Burgers moduli b i in 2.2 reintroduces the dislocations into the regularized problem, and it prevents the minimizers of J ε from being gradients of H 1 functions. In order to keep the notation shorter, we will write only H0 curl whenever it is possible to do so without confusion. We will denote by n the unit outward normal on. Lemma 2.1. Let h ε be a minimizer of J ε in H curl 0. Then it satisfies the Euler equations { divlhε = 0 in, Lh ε n = 0 on. Moreover, the solution to 2.3 is unique. Proof. Given that the functional J ε is quadratic, the result is achieved by computing the vanishing of its first variation. Let w H 1 ; then J ε h ε + t w J ε h ε δj ε h ε [w] = lim t 0 t 1 = lim t w Lh ε + 1 t 0 t 2 t2 w L w dx = w Lh ε dx = w divlh ε dx + w Lh ε n dsx. By setting δj ε h ε [w] = 0 for all w H 1, we get 2.3. To prove uniqueness, assume that h ε and h ε both solve system 2.3. Then the path integral of the difference h ε h ε over any loop in must vanish, and so h ε h ε = u for some function u H 1. Since u solves the weak Euler equation w L u dx = 0, for all w H 1, taking w = u we obtain J ε u = 0, and as L is positive definite, we conclude that u = Singular Strains and the Limit ε 0. We introduce the singular strains k i ; z i : R 2 \ {z i } R 2, i = 1,..., N, as b i λ x2 z k i x; z i = i,2 2πλ 2 x 1 z i,1 2 + x 2 z i, x 1 z i,1 We will often abbreviate k i ; z i as k i. Each k i enjoys several nice properties; in particular, each can be written as the gradient of a multi-valued function, precisely k i x; z i = b i 2π x2 z i,2 x arctan, λx 1 z i,1 and it is straightforward to compute directly that In particular, by 2.5c and 2.5b, L k i y; z i ny dsy = Ω 2.3 curl x k i x; z i = b i δ zi x in R 2, 2.5a div x Lk i x; z i = 0 in R 2 \ {z i }, 2.5b Lk i x; z i n = 0 on E ε,i. 2.5c L 4 k i y; z i ny dsy =

5 However, it is important to note that the integral in 2.6 is only well-defined when the dislocations are away from the boundary ε 0 > 0. Indeed, the integral is divergent when z i Ω, which would require ε 0 = 0. These singular strains k i are fundamental to the problem of screw dislocations in the sense that they are the building blocks of the singular part of the strain field h that represents the system of dislocations. Lemma 2.2. Let h H curl 0, Z, B. Then h = k i + u for some u H 1. Moreover, the minimization problem { } min J ε h h H0 curl, Z, B is equivalent to the minimization problem where min I ε u = W udx { } I ε u u H 1, 2.8 Ω ulk i n ds j i E ε,i ulk j n ds 2.9 here the last integral comes with a minus sign because n points outside E ε,i, by definition of outer normal. Minimizers u ε of 2.8 exist and are solutions of the Neumann problem div L u = 0 in, L u + N k i L u + j i k j n = 0 on Ω, n = 0 on E ε,i, i = 1, 2,..., N Proof. Let h H0 curl, Z, B. By 2.5a, l h N k i dx = 0 for any loop l h N k i = u for some u H 1. In turn and thus, N 1 J ε h = J ε k i + Lk i k j dx + I ε u 2.11 j=i+1 where I ε u is given by 2.9 and where in the last sum in the expression for I ε we omit the terms with i = j because Lk i n = 0 on each E ε,i see 2.5c. Hence, the minimization of J ε over h H0 curl, Z, B is achieved by minimizing I ε over u H 1. To show that minimizers solve the Neumann problem 2.10, we compute the first variation of I ε and apply Stokes s theorem to find that, given ϕ H 1, δi ε u[ϕ] = ϕdivl u dx + ϕl u + k i n ds ϕl u + k j n ds. Ω E ε,i j i By requiring that δi ε u[ϕ] = 0 for all ϕ H 1, we obtain that 2.10 is satisfied. The following two lemmas are slight adaptations of [CL05, Lemmas 4.2, 4.3], so we do not present the full proofs here. The key tool is an ε-independent Poincaré inequality for, [CL05, Proposition A.2]. Lemma 2.3. Let ε 0 > 0 be fixed as in 2.1. Assume that L is positive definite. Then there exist positive constants c 1 and c 2, depending only on L and ε 0 in particular, independent of ε, such that for all u H 1 subject to the constraint I ε u c 1 u 2 H 1 c 2 u H ux dx =

6 Moreover, for every ε 0, ε 0 the minimization problem 2.8 admits a unique solution u ε H 1 satisfying Each u ε satisfies u H1 M, 2.14 where M > 0 is a constant independent of ε. Sketch of Proof. Since L is positive definite, we have I ε u C u 2 dx sup Lk i x, z i x Ω Ω u ε ds sup Lk j x, z j u ε ds x E ε,i E ε,i j i Adapting the proof of [CL05, Proposition A.2], for which 2.13 is crucial, we can find a constant c 1 = c 1 λ, ε 0 such that u 2 dx c 1 u 2 H 1 Ω ε Moreover, in [CL05] it is proved that there exist constants C 1, C 2 independent of ε such that u ε ds C 1 u ε H1, and u ε ds C 2 u ε H Ω E ε,i From the definition of k i x, z i see 2.4, it is easy to see that there exist constants c = c λ, ε 0 and c = c λ, ε 0 such that sup Lk i x, z i < c, and sup Lk j x, z j < c, i j x Ω x E ε,i Estimates 2.15, 2.16, and 2.17 prove The uniqueness of the solution and the bound 2.14 are straightforward conclusions from the convexity of the functional I ε and the fact that I ε 0 = 0. Lemma 2.4. Assume that L is positive definite, and let u ε be the unique solution to 2.8 that satisfies Then, as ε 0, the sequence {u ε } converges strongly in Hloc 1 Ω \ Z to a solution u 0 of the problem { } min I 0 u u H 1 Ω, 2.18 where Moreover, I ε u ε I 0 u 0. I 0 u := Ω W u dx + Ω u Lk i n ds. Sketch of Proof. One can extend u ε to Ω and obtain an inequality u ε H1 Ω cm, with M as in 2.14 [CL05, Prop. A.7], which leads to E ε,i u ε Lk k n ds 0 as ε 0. Also, a subsequence not relabeled of {u ε } converges u ε u 0 weakly in H 1 Ω. Now, if we fix δ 0, ε 0 and consider ε < δ, from 2.9 we have I ε u ε W u ε dx + u ε Lk i n ds u ε Lk j n ds. Ω δ Ω j i E ε,i Taking ε 0 gives lim inf ε 0 I ε u ε Ω δ W u 0 dx + N Ω u 0Lk i n ds. Taking δ 0 gives lim inf ε 0 I ε u ε I 0 u 0. But I ε u ε I ε u 0, so lim sup ε 0 I ε u ε I 0 u 0, and I ε u ε I 0 u 0. Strong convergence of u ε u 0 in H 1 Ω \ Z follows from convergence of the energies, see [Eva90]. Remark 2.5. The solutions u 0 to 2.18 are also solutions of the Neumann problem { div L u = 0 in Ω, L u + N k i n = 0 on Ω, 2.19 and therefore u 0 can be represented in terms of a Green s function u 0 x; z 1,..., z N = Gx, yl k i y; z i ny dsy, 2.20 Ω 6

7 exhibiting the explicit dependence on the parameters z 1,..., z N. The function u 0 x; z 1,..., z N represents the elastic strain at the point x Ω due to the presence of Ω and the dislocations at z i with Burgers moduli b i. For this reason, we refer to u 0 x; z 1,..., z N as the boundary-response strain at x due to Z. Combining the results of Lemmas 2.2, 2.3, and 2.4, we conclude the following theorem, which characterizes the strain field associated to a system of dislocations. Theorem 2.6. Let Z and B be given, and let Ω R 2 be a bounded domain with Ω C 2. Then the minimization problem min W hdx h H0 curl,z,b admits a unique solution, h ε. Moreover, h ε h 0 strongly in L 2 loc Ω \ Z, where h 0 x = k i x; z i + u 0 x; z 1,..., z N 2.21 is a solution of { N curl h = b iδ zi in Ω, div Lh = 0 in the sense of distributions, and u 0 is a minimizer of 2.18 and solves the Neumann problem Alternative form of the fundamental singular strains. When computing integrals over the cores E R,i, we find some computations later in the paper are simplified by using eccentric anomaly centered at z i tan θi τ i := arctan. λ Using τ, the ellipse E i,r is parametrized by the curve ρτ i = z i + R cos τ i, λr sin τ i, so 1 λ cos τi n =, k λ2 cos 2 τ i + sin 2 τ i sin τ i R, τ i ; z i = b l λ sin τi i 2πλR cos τ i 3. The Renormalized Energy Recall the definition of ε 0 from the beginning of Section 2. Theorem 3.1. Let 0 < ε < ε 0 and let h ε be a solution of 2.7. Then 1 J ε h ε = 2 h µλb 2 i ε Lh ε dx = 4π log 1 ε + Uz 1,..., z N + Oε, 3.1 where and, using 2.21, for any ε < R < ε 0 Uz 1,..., z N := U S z 1,..., z N + U I z 1,..., z N + U E z 1,..., z N 3.2 U S z 1,..., z N := U I z 1,..., z N := N 1 U E z 1,..., z N := µλb 2 i 4π j=i+1 Ω N log R + Ω W u 0 dx + k j Lk i dx, Ω\E R,i W k i dx, 3.3 Ω u 0 Lk i n ds. 3.4 Remark 3.2. We refer to the energy U in 3.2 as the renormalized energy. U S is the self energy associated to the presence of a dislocation, U I is the energy associated to the interaction between dislocations, and U E is the energy associated to the elastic medium. Note that Theorem 3.1 asserts that the renormalized energy is independent of ε, and we will show that it can be written in terms of the limit shear h 0 as in Theorem 2.6. This fact will be used in identifying the force on a dislocation in Section 4. 7

8 Proof. If we expand J ε h ε as in 2.11, we see that the three terms on the right side of 2.11 correspond to the terms U S, U I, U E. We begin with N J εk i and fix R ε, ε 0. Each term in this sum can be written as J ε k i = W k i dx + W k i dx, \E R,i A i,r,ε where A i,r,ε := E R,i \ E ε,i. Using the representation for k i in 2.22, we have 1 A i,r,ε 2 k i Lk i dx = µλb2 i R 4π log, 3.5 ε and this accounts for the log 1 ε term in the energy 3.1 and the log R term in 3.3. To show that N 1 N 1 k j Lk i dx k j Lk i dx as ε 0, 3.6 j=i+1 j=i+1 we note that k i is integrable in E R,i it grows like r 1 i and Lk j is bounded on E R,i for j i, hence 3.6 holds by Lebesgue Dominated Convergence Theorem. From Lemma 2.4, we have that I ε u ε I 0 u 0 as ε 0, whence 3.4 follows. To show that U is independent of R, we need only show that U S is independent of R. If we take R R, without loss of generality we can assume R < R, then by 3.5 W k i dx W k i dx = W k i dx = µλb2 i Ω\E R,i Ω\E R,i A i,r,r 4π log R R, so that Ω\E R,i W k i dx + µλb2 i 4π log R = which shows that 3.3 is independent of the choice of R < ε 0. Ω W k i dx + µλb2 i Ω\E R,i 4π log R, The renormalized energy U will blow up like the log of the distance between dislocations, i.e. log z i z j. This is made precise in [BFLM14]. 4. The Force on a Dislocation In this section we determine the force j i on the dislocation at z i for a given a system of dislocations Z with Burgers vectors B, and show that j i = zi U. Following [CG99], the Peach-Köhler force on the dislocation at z i also called the net configurational traction is given by j i := lim Cn ds, 4.1 R 0 E R,i where the stress tensor is the Eshelby stress [Esh51, Gur95] Here I is the identity matrix and h 0 is defined in U C := W h 0 I h 0 Lh Theorem 4.1. Let h 0 be the limiting singular strain defined by 2.21 and let U the associated renormalized energy given in 3.2. Then for l {1,..., N} and any R 0, ε 0 zl Uz 1,..., z N = {W h 0 I h 0 Lh 0 } n ds, 4.3 and so the force on the dislocation at z l is given by Moreover, j l = zl U. 4.4 j l z 1,..., z N = b l JL u 0 z l ; z 1,..., z N + k i z l ; z i 0 1, where J = 1 0 i l 8, 4.5

9 and u 0 is the solution to Proof. Formula 4.3 is proved in the Appendix. From 4.3, we show 4.4 and 4.5 as follows. Recall that the renormalized energy is independent of R < ε 0 see the proof of Theorem 3.1, so zl U = Cn ds = lim E R 0 R,l Cn ds = j l, 4.6 establishing 4.4 in view of 4.1. The field h 0 has a singularity at z l which comes from the term k l see 2.21, and we decompose h 0 into the singular part at z l and the regular part at z l, h 0 x = k l x; z l + hx, where hx := u 0 x + i l k i x; z i. 4.7 Using 4.7, we write the Eshelby stress C from 4.2 as 1 C = 2 k l Lk l + k l L h h L h I k l Lk l k l L h h Lk l h L h. Since h is smooth and bounded on E R,l we have 1 lim R 0 2 h L h n ds = 0 and lim h L hn ds = 0. R 0 Using the fact that Lk l n = 0 on see 2.5c we have h Lkl n ds = 0, and k l Lk l n ds = 0, Using 2.22 we have k l Lk l = µb 2 l /4π2 R 2 on, and so 1 2 k l Lk l n ds = µb2 l 8π 2 R 2 n ds = 0, for all R < ε 0. Therefore the only contribution in 4.6 will come from k l L h I k l L h n = n k l L h k l n L h. R < R. Now, using 2.22, it is easy to see that b l n k l k l n = 2πλR λ 2 cos 2 τ + sin 2 τ 0 λ λ 0 and, since ds = R λ 2 cos 2 τ + sin 2 τ dτ, n k l k l n L h ds = b l 2π Since the integrand is smooth on E R,l, we conclude that lim n k l k l n L h ds = b l R 0 2π which, in view of 4.6, establishes π 0 2π 0 JL h dτ. JL hz l dτ = b l JL hz l, Remark 4.2. The formula 4.5 gives the force on the dislocation at z l, and it shows that, as a function of z l, the force j l is smooth in the interior of Ω \ {z 1,..., z l 1, z l+1,..., z N }. That is, provided z l is not colliding with another dislocation or with Ω, then the force is given by a smooth function. Of course, j l depends on the positions of all the dislocations, and the same reasoning applies to j l as a function of any z i. 9

10 Remark 4.3. We find agreement between 4.5 and equation 8.18 from [CG99], where the force on z l is given by b l times a π/2-rotation of the regular part of the strain at z l i.e., h. Since we have a formula for the regular part, we are able to write the Peach-Köhler force more explicitly in terms of the solution to We have also shown that assumption A3 from [CG99] holds for screw dislocations, validating the derivation of 8.18 in [CG99]. Acknowledgements The authors warmly thank the Center for Nonlinear Analysis NSF Grants No. DMS and DMS , where part of this research was carried out. T. Blass also acknowledges support of the National Science Foundation under the PIRE Grant No. OISE M. Morandotti also acknowledges support of the Fundação para a Ciência e a Tecnologia Portuguese Foundation for Science and Technology through the Carnegie Mellon Portugal Program under the grant FCT UTA/CMU/MAT/0005/ Appendix We present the proof of 4.3 along with some necessary lemmas. We begin by noting that Uz 1,..., z N = Ûz 1,..., z N + Uz 1,..., z N where Ûz 1,..., z N = W h 0 dx, N 1 Uz 1,..., z N = W k i dx + k j Lk i dx + m i + m=1 W u 0 dx + m=1 j=i+1 m=1 u 0 Lk i n ds, 5.1 which follows from a direct computation and integration by parts to eliminate the integral over Ω from U E. We introduce the notation vu for the derivative of a function u = ux; z 1,..., z N with respect to the l-th dislocation location in the direction v, v ux := d dξ ux; z 1,..., z l + ξv,..., z N. ξ=0 Lemma 5.1. The fields k i x; z i, u 0 x; z 1,..., z N, and h 0 x; z 1,..., z N are smooth with respect to z l for every l {1,..., N}. Moreover, D v l k ix = 0 if l i, D v l k l x = Dk l xv = k l x v 5.2 D v l h 0 x = wx, where wx = D v l u 0 x k l x v 5.3 Proof. The form of k in 2.4 shows that k i is smooth with respect to z l for all i, l = 1,..., N, and in particular that k i x = kx; z i is independent of z l if l i so vk i = 0. That form also shows that kx; z l + ξv = kx ξv; z l = k l x ξv so that vk lx = Dk l v, where Dk l is the derivative of k l with respect to x. Now because curl k l = 0, we have Dk l xv = k l x v, which establishes 5.2. Since u 0 solves the elliptic problem 2.19, it can be represented as in 2.20, in terms of the Green s function Gx, y. The smoothness of u 0 in z l follows from the smoothness of k i for each i, l = 1,..., N. Hence, h 0 is smooth in z l and vh 0 = v u 0 + vk l = vu 0 k l v, which establishes 5.3. We will take derivatives of the energy with respect to the dislocations positions. This will involve integrals over cores that are centered at z l + ξv whose integrands are evaluated on these shifted cores or on their complements in Ω. Thus, we will need to be able to take derivatives of integrals over sets that depend on ξ and whose integrands are functions that depend on ξ. Lemma 5.2. Let f = fx, ξ, g = gx, ξ, and r = rx, ξ be defined on E ε x 0 + ξv, E ε x 0 + ξv, and Ω \ E ε x 0 + ξv, respectively, for ξ a real parameter, ε > 0, v R 2. Then 10

11 d dξ E εx 0+ξv fx, ξ dx = ξ=0 = d gx, ξ ds = dξ E εx 0+ξv ξ=0 d rx, ξ dx = dξ Ω\E εx 0+ξv ξ=0 where D ξ f := ξ f + f v. Proof. We compute E εx 0 E εx 0 E εx 0 Ω\E εx 0 D ξ fx, 0 dx ξ fx, 0 dx + fx, 0v n ds, 5.4 E εx 0 D ξ gx, 0 ds, 5.5 ξ rx, 0 dx rx, 0v n ds, 5.6 E εx 0 d fx, ξdx = d fx + ξv, ξdx = ξ fx + ξv, ξ + fx + ξv, ξ vdx. dξ E εx 0+ξv dξ E εx 0 E εx 0 If we send ξ 0 and apply the divergence theorem we obtain 5.4. A similar computation gives 5.5 but the divergence theorem is not applied. If r is a smooth extension of r to Ω then d rx, ξdx = d dξ Ω\E εx 0+ξv dξ = Ω rx, ξdx d Ω dξ ξrx, ξdx E εx 0 E εx 0+ξv rx, ξdx ξrx + ξv, ξdx E εx 0 rx + ξv, ξv n ds. Setting ξ = 0 and combining the first two integrals on the right side yields 5.6. Remark 5.3. Lemma 5.2 applies to the vector-valued k i. When applying Lemma 5.2 to integrals of kx; z l + ξv over E ε z l + ξv we will get cancellations from D ξ kx; z l + ξv = ξ kx; z l + ξv + Dkx; z l + ξvv = D v l k l x + Dk l v = The last equality follows from 5.2. Proof of Equation 4.3. The log ε term in the energy is independent of the positions of the dislocations so it vanishes upon taking the derivative of the energy with respect to z l. To compute the derivative of U with respect to z l will split zl U into zl Û + zl U. To compute zl Û we apply 5.6 to get zl Û = v W h 0 dx = v h 0 Lh 0 dx W h 0 v n ds 5.8 Using 5.3, divlh 0 = 0 in Ω, and Lh 0 n = 0 on Ω, we have v h 0 Lh 0 dx = v u 0 k l v Lh 0 dx = v u 0 k l vlh 0 n ds = v u 0 k l vlh 0 n ds = wlh 0 n ds E ε,j E ε,j j=1 11 j=1 5.9

12 using the notation w = D v l u 0 k l v from 5.3. Combining 5.8 and 5.9, and adding and subtracting h 0 Lh 0 n v from the integrand, we have zl Û = v u 0 k l vlh 0 n ds W h 0 v n ds j=1 E ε,j = {W h 0 I h 0 Lh 0 } n v ds v u 0 k l vlh 0 n ds + j l E ε,j [ v u 0 k l vlh 0 n + h 0 Lh 0 n v] ds; also, h 0 Lh 0 n v = h 0 vlh 0 n = u 0 v + N k i vlh 0 n, so D v l u 0 k l vlh 0 n + h 0 Lh 0 n v = D v l u 0 k l v + u 0 v + = D ξ u 0 + k i v Lh 0 n i l k i v Lh 0 n where D ξ u 0 = vu 0 + u 0 v. Hence, zl Û = {W h 0 I h 0 Lh 0 } n v ds v u 0 k l vlh 0 n ds j l E ε,j D ξ u 0 + k i v Lh 0 n ds i l 5.10 We compute zl U in several steps. We split the first sum in the right side of 5.1 into the integral over E ε,l and the rest of the terms W k i dx = m i E ε,l W k m dx + i m W k i dx In the first of these, each k m does not vary as z l z l + ξv because m l. Hence we apply 5.4 directly with D ξ k m = ξ k m + vk m = k m v because ξ k m = 0. We have v W k m dx = D ξ k m Lk m dx = k m v Lk m dx E ε,l E ε,l E ε,l 5.12 = k m vlk m n ds, where we used divlk m = 0. The second term from 5.11 involves integrals over for m l, so these domains do not move as z l z l + ξv. Also, the terms W k i for i l vanish when we apply v, so v W k i dx = v k l Lk l dx = k l v Lk l dx i m = 5.13 k l vlk l n ds, where we used 5.2 and divlk l = 0. 12

13 The second sum from 5.1 is split into the integral over E ε,l and the rest of the terms N 1 m=1 j=i+1 k j Lk i dx = E ε,l i<j k j Lk i dx + Applying 5.4 to the first term on the right side yields k j Lk i dx = D ξ k j Lk i + D ξ k i Lk j dx D v l E ε,l i<j = = i l E ε,l i<j E ε,l i,j l,i<j j i i<j k j v Lk i + k i v Lk j dx E ε,l k j v Lk i dx = i l j i k j Lk i dx k j v Lk i n ds 5.15 Between the first and second lines we used D ξ k i = k i v for i l and D ξ k l = 0 by 5.7, and in the third line we used divlk i = 0. For the second sum of 5.14, using 5.2 from Lemma 5.1, we have v k j Lk i dx = v k l Lk i dx = k l v Lk i dx i<j i l i l = 5.16 k l vlk i n ds i l The third term comprising U in 5.1 is split as W u 0 dx = W u 0 dx + W u 0 dx E ε,l m=1 To compute the derivative of the first term on the right side of 5.17, we use 5.4, but integrate the D ξ term by parts directly. Using D ξ u 0 = vu 0 + u 0 v and divl u 0 = 0 we have v W u 0 dx = D ξ u 0 L u 0 dx = D ξ u 0 L u 0 n ds E ε,l E ε,l 5.18 = v u 0 + u 0 vl u 0 n ds. Computing the derivative of the second term on the right side of 5.17 is almost the same as in 5.18 except the domains do not depend on z l because m l. Hence v W u 0 dx = v u 0 L u 0 dx = v u 0 L u 0 n ds Turning to the the final term in 5.1, which we split as m=1 u 0 Lk i n ds = u 0 Lk i n ds + u 0 Lk i n ds, 5.20 we compute the derivative of the first term using 5.5 to get N v u 0 Lk i n ds = D ξ u 0 Lk i n ds + u 0 LD ξ k i n ds

14 From 5.7 we have D ξ k l = 0 and from 5.2 we have D ξ k i = k i v for i l. Hence, for i l we have u 0 LD ξ k i n ds = u 0 L k i v n ds = div u 0 L k i v dx E ε,l = u 0 L k i v dx + u 0 div L k i v dx 5.22 E ε,l E ε,l = k i v L u 0 dx = k i v L u 0 n ds. E ε,l E ε,l We used div L k i v = v divlk i = 0, which follows from curl k i = 0 and divlk i = 0. Combining 5.21 and 5.22 we get N v u 0 Lk i n ds = D ξ u 0 Lk i n ds + k i vl u 0 n ds 5.23 i l Finally, the derivative of the second term in 5.20 is computed similarly to the first, but is simpler because the domains of integration are independent of z l. Hence, v u 0 Lk i n ds = u 0 L v k l n ds + v u 0 Lk i n ds 5.24 because vk i = 0 when i l. Using div L k i v = 0, as we did to get 5.22, we have u 0 L v k l n ds = u 0 L k l v n ds = divu 0 L k l v dx E ε,m = u 0 L k l v dx = k l vl u 0 n ds Then 5.24 and 5.25 give N D v l m=1 u 0 Lk i n ds = 5.25 N v u 0 Lk i n k l vl u 0 n ds 5.26 Combining 5.12, 5.15, 5.18, and 5.23 we have i vlk i n + i lk k j v Lk i n + i l j i + D ξ u 0 L u 0 n + ξ u 0 Lk i n + D k i vl u 0 n ds i l = k i v L u 0 + Lk j + D ξ u 0 L u 0 + Lk j n ds i l j=1 j=1 = D ξ u 0 + k i v Lh 0 n ds i l 5.27 Combining 5.13, 5.16, 5.19, and 5.26 we have k l vlk l k l vlk i + v u 0 L u 0 + v u 0 Lk i k l vl u 0 n ds i l = v u 0 k l v Lh 0 n ds

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