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Finally,γ→0finishes the proof of the lower bound.

condition has the same scaling as Rr →1.

As the constant for Korn’s inequality is invariant under dilations of the domain, we will only consider the caseR= 1, r <1.

The technique to prove an upper bound for the constant is a covering argument which is classical for proving Korn’s inequality for thin domains. We show first a uniform bound on the constant for special subsets of the annulus.

Lemma 5.A.1. For r <1 define the set Ur={t(sinθ,−cosθ) :r <1<1,|θ|< 1−r2 }. There exist r0<1 and a constantC >0 such that for every r0< r <1 and every u∈H1(Ur;R2)there exists a skew-symmetric matrixW such that

ˆ

Ur

|∇u−W|2dx≤C ˆ

Ur

|(∇u)sym|2dx.

Proof. Let us define the square U˜r = r−12 ,1−r2

×(0,1−r) and the function fr : ˜Ur → Ur by fr(θ, t) = (1−t)(sinθ,−cosθ), see figure 5.6. Clearly,fris a diffeomorphism and its derivative is

∇fr(t, θ) = (1−t) cosθ (1−t) sinθ

−sinθ cosθ

! .

It holds|∇fr(t, θ)−Id| ≤ O(r)where O(r)→0as r→1.

Now, letu∈H1(Ur;R2)such that ´

Ur(∇u)skewdx= 0. Defineu˜: ˜Ur→R2 by ˜u(θ, t) =u(fr(θ, t)).

As∇fris almost the identity matrix, the symmetric part of∇u˜is approximately(∇u)sym◦fr. Indeed, by the chain rule we find

|(∇˜u)sym−(∇u)sym◦fr| ≤ O(r)|((∇u)◦fr)sym| ≤ O(r)|(∇u)◦fr|. (5.36) By Korn’s inequality applied tou, there exists a skew-symmetric matrix˜ W and a constantK ≥1

such that ˆ

U˜r

|∇˜u−W|2dx≤K ˆ

U˜r

|(∇u)˜ sym|2dx.

Note that by scaling the constant does not depend onr. By (5.36), we find further ˆ

U˜r

|(∇u)˜ sym|2dx≤2 ˆ

U˜r

|(∇u)sym◦fr|2+O(r)2|(∇u)◦fr|2dx

= 2 ˆ

Ur

(|(∇u)sym|2+O(r)2|∇u|2)|det∇fr−1|.

On the other hand, one proves similarly that ˆ

U˜r

|∇˜u−W|2dx≥ 1 2 ˆ

Ur

(|∇u|2− O(r)2|∇u|2)|det∇fr−1|dx, where one uses that from´

Ur(∇u)skewdx= 0it follows´

Ur|∇u−W|2dx≥´

Ur|∇u|2dx.

Notice that for r close to 1 the gradient ∇fr is uniformly close to the identity. The same holds for∇fr−1. Chooser0 so close to 1 such that for all smaller r it holds that |det∇fr−1−1| ≤ 12 and

1−r fr

r 1−r Ur

Figure 5.6: Sketch of the situation in Lemma 5.A.1.

O(r)≤ 1

4

K. Then, one obtains, by combining the previous inequalities, ˆ

Ur

|∇u|2dx≤12K ˆ

Ur

|(∇u)sym|2dx+13 16

ˆ

|∇u|2dx.

Absorbing the very right term to the left hand side ends the proof.

Remark 5.A.1. The same holds true for any rotated version of Ur with the same constant since Korn’s constant does not depend on the choice of coordinates.

Remark 5.A.2. For an explicit bound on the constructed constant in the previous lemma, note that an upper bound for the optimal constant for the square was computed in [49], namely8 + 4√

2. In [48], it is conjectured that the optimal constant for a square is7.

In the following lemma, we state an upper bound on Korn’s constant for the annulusB1(0)\Br(0) forcurl-free fields.

Lemma 5.A.2. There existsr0<1such that for allr0< r <1Korn’s constantK(r)forB1(0)\Br(0) is less or equal than C(1−r)−2 where C >0 is a universal constant. Precisely, for every function β ∈ L2(B1(0)\Br(0);R2×2) satisfying curlβ = 0 there exists a skew-symmetric matrix W ∈ R2×2

such that ˆ

B1(0)\Br(0)

|β−W|2dx≤C(1−r)−2 ˆ

B1(0)\Br(0)

sym|2dx.

Proof. Letr0as in Lemma 5.A.1 and0< r < r0. In analogy to the previous lemma, we define the sets Urk =

t(cosθ,sinθ) :r < t <1 and k−1

2 (1−r)< θ < k+ 1 2 (1−r)

wherek= 1, . . . , 4π

1−r

=Lr.

As b

1−rc+1

2 (1 − r) > 2π, it follows B1(0)\ Br(0) ⊂ SLr

k=1Urk. Moreover, notice that for all k= 1, . . . , Lr−1 it holds |Urk|

|Urk∩Urk+1| = 2andPLr

k=11Ukr ≤2.

Now, let β ∈ L2(B1(0)\Br(0);R2×2) such that curlβ = 0. In particular, β can be written as a gradient on each Ukr. For each k = 1, . . . , Lr, we apply Lemma 5.A.1 and Remark 5.A.1 on Urk to obtain a skew-symmetric matrixWk such that

ˆ

Urk

|β−Wk|2dx≤C ˆ

Urk

sym|2dx. (5.37)

Note that C >0 does not depend onk, rnorβ.

For every k= 1, . . . , Lr−1, the distance betweenWk andWk+1 can be estimated as follows

|Wk−Wk+1|2≤2

Urk∩Urk+1

|Wk−β|2+|Wk+1−β|2dx≤ 4C

|Urk∩Urk+1| ˆ

Urk∪Urk+1

sym|2dx.

x+εu(x) b=εe1

Figure 5.7: Sketch of the almost optimal displacement constructed in the proof of Lemma 5.A.3 for ξ=e1,ε= 14 andr=45.

Consequently, we obtain for1≤k < l≤Lr

|Wk−Wl|2=

l−1

X

i=k

Wi−Wi+1

2

≤(l−k)

l−1

X

i=k

|Wi−Wi+1|2

≤Lr Lr−1

X

i=1

4C

|Uri∩Uri+1| ˆ

Uri∪Uri+1

sym|2dx. (5.38)

We defineW =W1. Then, we derive from (5.37) and (5.38) the following chain of inequalities ˆ

B1(0)\Br(0)

|β−W|2≤2

Lr

X

k=1

ˆ

Urk

|β−Wk|2+|W −Wk|2dx

≤2

Lr

X

k=1

C ˆ

Urk

sym|2dx+Lr Lr−1

X

i=1

4C |Uri|

|Uri∩Uri+1| ˆ

Uri∪Uri+1

sym|2dx

!

≤4C ˆ

B1(0)\Br(0)

sym|2dx+ 16C L2r

Lr−1

X

i=1

ˆ

Uri∪Uri+1

sym|2dx

≤C(4 + 64L2r) ˆ

B1(0)\Br(0)

sym|2dx.

Note that by definitionL2r(1−r)16π22 andC is a universal constant.

Finally, we construct for eachξ ∈R2 a function which is curl-free, whose circulation is exactlyξ, and whose elastic energy is optimal in scaling, see Figure 5.7.

Lemma 5.A.3. Let ξ ∈R2. There exists 0< r0 <1 such that for allr0 < r <1 andξ ∈R2 there exists a functionβ :B1(0)\Br(0)→R2×2 such thatcurlβ= 0 and´

∂B1β·τ dH1=ξand min

W∈Skew(2)

ˆ

B1(0)\Br(0)

|β−W|2dx≥c(1−r)−2 ˆ

B1(0)\Br(0)

sym|2dx. (5.39) The constantcdoes not depend on r, ξorβ.

Proof. Letξ= (ξ1, ξ2)∈R2. Moreover, we writeeρ(θ) = (cosθ,sinθ)andeθ(θ) = (−sinθ,cosθ).

We define the functionu:B1(0)\Br(0)→R2 in polar coordinates by

u(ρeρ(θ)) = 1 π

ˆ θ 0

1cosϕ+ξ2sinϕ)eρ(ϕ)dϕ+ (1−ρ)(ξ1cosθ+ξ2sinθ)eθ(θ)

! .

For a visualization, see Figure 5.7.

The function u has a jump on the line {θ = 0} of height ξ. Hence, the absolutely continuous part of the derivative of u, β =∇u, satisfies the circulation condition´

B1β·τ dH1=ξand curlβ = 0in B1(0)\Br(0). In addition, we can compute explicitly

β=1 π

1

ρ(ξ1cosθ+ξ2sinθ)eθ(θ)⊗eρ(θ)−(ξ1cosθ+ξ2sinθ)eρ(θ)⊗eθ(θ)

−1−ρ

ρ (ξ1cosθ+ξ2sinθ)eθ(θ)⊗eρ(θ) +1−ρ

ρ (−ξ1sinθ+ξ2cosθ)eθ(θ)⊗eθ(θ)

=1 π

−(ξ1cosθ+ξ2sinθ)eρ(θ)⊗eθ(θ) + (ξ1cosθ+ξ2sinθ)eθ(θ)⊗eρ(θ)

+ 1 π

1−ρ

ρ (−ξ1sinθ+ξ2cosθ)eθ(θ)⊗eθ(θ)

skewsym. One sees directly that ´

B1(0)\Br(0)βskewdx = 0. Hence, minW∈Skew(2)´

B1(0)\Br(0)|β −W|2dx =

´

B1(0)\Br(0)|β|2dx. In addition, a straightforward computation show that asr→1we obtain ˆ

B1(0)\Br(0)

skew|2dx= (1−r2)1

π|ξ|2∼2(1−r)1 π|ξ|2 and

ˆ

B1(0)\Br(0)

sym|2dx=

−log(r)−2(1−r) +1

2(1−r2) 1

π|ξ|2∼1

3(1−r)3 1 π|ξ|2, which proves the claim for rclose enough to1 independently ofξ.

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