A discrete crystal model in three dimensions: The line-tension limit for dislocations

Proposition A.1 (From [12]).

Let Ω ⊆ R 3 be a bounded connected Lipschitz set, q ∈ [ 1 , 3 2 ] . Then there is a constant c = c ⁢ ( Ω , q ) such that for any β ∈ L q ⁢ ( Ω ; R 3 × 3 ) there is S ∈ R skew 3 × 3 such that

Proof.

The important case is the critical exponent q = 3 2 , which was proven in [12]. The case q ∈ [ 1 , 3 2 ) follows from the same argument, and we briefly mention the changes using the notation of [12] and referring to formulas in that proof. For clarity we work in generic dimension n, proving the assertion for q ≤ 1 * = n - 1 n . One starts with the unit cube; [12, equation (8)] holds unchanged and gives by Hölder a bound for Y in L q ⁢ ( Q ) , Q := ( 0 , 1 ) n . Equation [12, equation (9)] becomes

which by Korn’s inequality implies the assertion if Ω = Q .

After scaling to Q r = ( 0 , r ) n , this leads to (cf. [12, equation (11), note that the right-hand side of that equation should contain β instead of Du])

The next part of the proof, up to [12, equation (16)], is unchanged, replacing s by q. Then [12, equation (17)] becomes

Since q ≤ 1 * , the factor | Q j | 1 - q 1 * is uniformly bounded and can be dropped. Therefore, as in [12, equation (17)],

This concludes the proof. ∎

Proof of Lemma 4.7.

We divide the proof into three steps.

Step 1. We show that, for every d ∈ ℕ , β ∈ L p ⁢ ( ω out ; ℝ d × 3 ) with curl ⁡ β = 0 in ω out , there is β ^ ∈ L p ⁢ ( ω ; ℝ d × 3 ) such that β ^ = β on ω out , curl ⁡ β ^ = 0 on ω ∖ ( { 0 } × ( 0 , ℓ ) ) , and

By scaling, we can assume ρ = 1 . Since the curl is computed row-wise, it suffices to consider the case that β is a vector, i.e., d = 1 , and we can identify column vectors with row vectors. For λ ∈ ( 3 2 , 2 ) chosen below, we set θ ⁢ ( t ) := λ + ( 1 - λ ) ⁢ t . We observe that θ is a bijective map from [ 0 , 1 ] onto [ 1 , λ ] and that θ ⁢ ( 1 ) = 1 . We define φ : ω in ∖ ( { 0 } × ( 0 , ℓ ) ) → ω out as

where x ′ := ( x 1 , x 2 , 0 ) denotes the projection of x onto the plane spanned by e 1 and e 2 . We compute

where Id 2 := e 1 ⊗ e 1 + e 2 ⊗ e 2 ∈ ℝ 3 × 3 is the identity matrix in the first 2 × 2 block. In particular, D ⁢ φ ⁢ ( x ) is symmetric and has eigenvalues

We define

In ω in , we compute

which componentwise reads

Since curl ⁡ β = 0 on ω out , we have that D ⁢ β is a symmetric matrix. Hence, D ⁢ β ^ is symmetric and curl ⁡ β ^ = 0 in ω in ∖ ( { 0 } × ℝ ) . Let x ∈ ∂ ⁡ ω in ∩ ∂ ⁡ ω out and let τ be a vector tangential to ∂ ⁡ ω in ∩ ∂ ⁡ ω out in x. Since φ ⁢ ( x ) = x on this set, we have D ⁢ φ ⁢ ( x ) ⁢ τ = τ . Therefore, the traces satisfy

This proves that curl ⁡ β ^ = 0 in ω ∖ ( { 0 } × ℝ ) .

It remains to estimate the norm. By (A.2) and λ ∈ ( 3 2 , 2 ) , we obtain

From the definition of θ we obtain | x ′ | = λ - θ ⁢ ( | x ′ | ) λ - 1 . We compute

where χ { | y ′ | < λ } = 1 if | y ′ | := | ( y 1 , y 2 ) | < λ and 0 otherwise. We set

and average over all possible choices of λ. Recalling that p ∈ [ 1 , 2 ) ,

Therefore, there is λ ∈ ( 3 2 , 2 ) such that f ⁢ ( λ ) ≤ 2 ⁢ c * ⁢ ∥ β ∥ L p ⁢ ( ω out ) p . Inserting in (A.3) concludes the proof of (A.1).

Step 2: Proof of (4.6). By scaling, we can work with ρ = 1 and ℓ ≥ 1 . Let N := 2 ⁢ ⌊ 2 ⁢ ℓ ⌋ , z j := j ⁢ ( ℓ - 1 2 ) / N , so that 0 = z 0 < z 1 < … < z N = ℓ - 1 2 and z i + 1 - z i = ( ℓ - 1 2 ) / ( 2 ⁢ ⌊ 2 ⁢ ℓ ⌋ ) ∈ [ 1 8 , 1 4 ) . Consider the sets ω i := B 2 ′ × ( z i , z i + 1 2 ) , ω i out := ( B 2 ′ ∖ B 1 ′ ) × ( z i , z i + 1 2 ) , ω i in := B 1 ′ × ( z i , z i + 1 2 ) . By Korn’s inequality, for every i there is R i ∈ ℝ skew 3 × 3 such that

Since z i + 1 ≤ z i + 1 4 , we obtain ℒ 3 ⁢ ( ω i out ∩ ω i + 1 out ) ≥ 3 4 ⁢ π and, therefore,

We define the affine isometries u i : ℝ 3 → ℝ 3 as

where d 0 = 0 and d i + 1 := d i + R i ⁢ e 3 ⁢ z i - R i + 1 ⁢ e 3 ⁢ z i , so that

We fix a partition of unity θ 0 , … , θ N ∈ C ∞ ⁢ ( ℝ ; [ 0 , 1 ] ) , with ∑ i θ i ⁢ ( t ) = 1 for all t ∈ ℝ and supp ⁡ θ 0 ⊆ ( - ∞ , 1 / 2 ) , supp ⁡ θ N ⊆ ( z N , ∞ ) , supp ⁡ θ i ⊆ ( z i , z i + 1 2 ) for 1 ≤ i ≤ N - 1 , and | θ i ′ | ⁢ ( t ) ≤ C for all i ∈ { 0 , … , N } and all t ∈ ℝ . We define u : ℝ 3 → ℝ 3 as

and observe that

where, for x ∈ ω j , only the terms with j - 4 ≤ i ≤ j + 4 are relevant. With (A.6), (A.5) and D ⁢ u j + D ⁢ u j T = 0 , we obtain

and similarly with (A.4)

We apply Step 1 to the function β - D ⁢ u and obtain a function β ^ ∈ L p ⁢ ( ω ; ℝ 3 × 3 ) with curl ⁡ β ^ concentrated on { 0 } × ( 0 , ℓ ) and

We set β ~ := D ⁢ u + β ^ , so that curl ⁡ β ~ = curl ⁡ β ^ in ω and β ~ = β in ω out , and use (A.7) and (A.8) to obtain

Step 3: Structure of curl ⁡ β , case θ = 0 . The extension provided above, together with Lemma 4.5, gives that there exists θ ∈ ℝ d such that curl ⁡ β ~ = θ ⊗ e 3 ⁢ ℋ 1 ⁢ ⌞ ⁢ ( { 0 } × ( 0 , ℓ ) ) . In the case θ = 0 , this implies that β ~ is a gradient in ω. Therefore, there a function u ∈ W 1 , p ⁢ ( ω out ; ℝ d ) such that D ⁢ u = β . By standard extension argument in Sobolev spaces, an extension u ~ of u in W 1 , p ⁢ ( ω ; ℝ d ) is obtained, and a redefinition β ~ := D ⁢ u ~ . Hence, Step 2 is unchanged for every p ∈ [ 1 , ∞ ) . ∎

Proof of Lemma 4.8.

We divide the proof into three steps.

Step 1. We show that for every p ∈ [ 1 , 2 ] , d ∈ ℕ , if β ∈ L p ⁢ ( ω out ; ℝ d × 3 ) with curl ⁡ β = 0 in ω out is given, then there is β ~ ∈ L p ⁢ ( ω ; ℝ d × 3 ) such that β ~ = β on ω out , curl ⁡ β ~ = 0 on ω := ω in ∪ ω out , and

The construction is similar that in Step 1 of the proof of Lemma 4.7. Also in this case, we work in the case ρ = 1 and d = 1 . We set θ ⁢ ( t ) := 2 - t and consider φ : B 1 ∖ { 0 } → B 2 ∖ B 1 defined as

We observe that φ ⁢ ( ( 0 , 1 ] ⁢ v i ) = [ 1 , 2 ) ⁢ v i , so that φ ⁢ ( ω in ∖ { 0 } ) ⊆ ω out . We compute

In particular, D ⁢ φ ⁢ ( x ) is symmetric and has eigenvalues

so that | det D ϕ | ( x ) = ( 2 | x | - 1 ) 2 ≥ 1 and

We define

By the same computation as in Lemma 4.7, curl ⁡ β ~ = 0 in B 2 ∖ ⋃ i [ 0 , 2 ) ⁢ v i .

We compute as in (A.3), and using (A.11), p ≤ 2 and | D ⁢ φ | ⁢ ( x ) ≥ 1 pointwise, we have

which concludes the proof of (A.9).

Step 2: Proof of (4.7). By scaling, it suffices to consider the case ρ = 1 . As p ≤ 3 2 , by the rigidity estimate of Proposition A.1 there is S ∈ ℝ skew 3 × 3 such that

By Step 1, we can obtain an extension β ~ of β - S such that

The map β ^ := S + β ~ satisfies | β ^ + β ^ T | = | β ~ + β ~ T | pointwise and, therefore, (4.7).

The proof of (4.8) is analogous, using Korn’s inequality instead of Proposition A.1.

Step 3: Structure of curl ⁡ β ^ . Since β ^ ∈ L 1 ⁢ ( B 2 ; ℝ 3 × 3 ) and curl ⁡ β ^ = 0 on B 2 ∖ ⋃ i [ 0 , 2 ) ⁢ v i , it follows from Lemma 4.5 that in ( 𝒟 ⁢ ( B 2 ∖ { 0 } ; ℝ 3 × 3 ) ) ′ we obtain curl ⁡ β ^ | B 2 ∖ { 0 } = ∑ i θ i ⊗ v i ⁢ ℋ 1 ⁢ ⌞ ⁢ ( 0 , 2 ) ⁢ v i . It remains to deal with the origin. Fix a test function η ∈ C c ∞ ⁢ ( B 2 ; ℝ 3 × 3 ) and, for any r ∈ ( 0 , 1 / 2 ) , select h r ∈ C c ∞ ⁢ ( B 2 ⁢ r ) such that h r = 1 on B r and | D ⁢ h r | ≤ 2 r . We write

and estimate the last term using

where C η depends on η but not on r. Taking the limit r → 0 in (A.12) leads to

for every η ∈ C c ∞ ⁢ ( B 2 ; ℝ 3 × 3 ) , as desired. ∎