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Korn coercivity with a partial Dirichlet segment

Statement

Assume the geometry of Partial-trace displacement space on a polygon and let V={v∈H1(Ω;R2):TDv=0}. Then there is cK>0, depending on Ω and ΓD, such that ∥v∥H1(Ω)≤cK∥ε(v)∥L2(Ω)(v∈V).

Facts & Assumptions

Given: A nonempty bounded connected Lipschitz extension polygon Ω⊂R2, a straight Dirichlet segment ΓD of length L>0, and V as above.

[A1]

Countable Choice is used by the bounded-Lipschitz divergence right inverse and the closure-defined H01 approximation. (The Axiom of Countable Choice (ACω))

[A2]

The Axiom of Choice is used through the bounded C1 trace theorem and Poincare-Wirtinger on the bounded connected extension domain. (The Axiom of Choice, The Lp trace operator on a bounded C1 domain, Poincare-Wirtinger on bounded connected extension domains by Rellich compactness)

[F0]

A nonempty open bounded set has finite positive Lebesgue measure: it contains a ball and is contained in a ball. (Euclidean balls have positive finite Lebesgue measure)

[F1]

There is a bounded linear BΩ:L02(Ω)→H01(Ω;R2) with div⁡BΩq=q and ∥∇BΩq∥2≤Cdiv(Ω)∥q∥2. (Divergence right inverse on bounded connected Lipschitz domains)

[F2]

For z∈H1(Ω) of zero mean, ∥z∥2≤CP∥∇z∥2; apply this componentwise. (Poincare-Wirtinger on bounded connected extension domains by Rellich compactness)

[F3]

TD is bounded and V=ker⁡TD; in particular TDv=0 for v∈V. (Partial-trace displacement space on a polygon, The partial-trace displacement space is a Hilbert space)

[F4]

Every planar rigid motion r(x)=a+ωJx satisfies ∥r∥H1(Ω)≤CD∥TDr∥L2(ΓD). (A planar rigid motion is controlled by its trace on a segment)

[F5]

H01(Ω) is the H1 closure of Cc∞(Ω). (Zero-boundary Sobolev space as a norm closure)

[F6]

ε(v)=(∇v+(∇v)T)/2, and J=(0−110); adding a rigid motion a+ωJx does not change ε(v). (Small-strain elasticity with a frictionless contact gap, A planar rigid motion is controlled by its trace on a segment)

[F7]

For finite-measure Ω, every L2(Ω) function is integrable and ∥g∥L1≤∣Ω∣1/2∥g∥L2. (Cauchy-Schwarz inequality for L2)

Proof

technique · direct

Given: The polygon, trace segment and space V of the statement, with the Axiom of Choice and Countable Choice available through [A1]–[A2].

1.1A1givenF0F6F7constructalgebra

Since Ω is nonempty, open and bounded, [F0] gives 0<∣Ω∣<∞. Hence 1∈L2(Ω), and [F7] with the Cauchy–Schwarz inequality makes every L2 component of v and its weak gradient integrable; boundedness of Ω also makes x and Jx integrable. Thus the means in ω=(2∣Ω∣)−1∫Ωcurl⁡v and a=∣Ω∣−1∫Ω(v−ωJx) exist. Put r(x)=a+ωJx and z=v−r. Then ∫Ωz=0, ∫Ωcurl⁡z=0, and ε(z)=ε(v) by [F6].

2.1A1F1F5step 1.1

Let q=curl⁡z∈L02(Ω) and choose w=BΩq by [F1]. Define the matrix R(w) rowwise by R(w)i1=∂2wi and R(w)i2=−∂1wi. For w∈Cc∞(Ω;R2) each row of R(w) has zero divergence, so distributional integration by parts gives (∇z,R(w))L2=0. Approximate the actual w∈H01 by compactly supported smooth fields using [F5]; since R(wm)→R(w) in L2, the same identity holds for w.

3.1A1A2F1F2F6step 1.1step 2.1algebra

Set G=∇z−R(w). In two dimensions J:G=G21−G12=curl⁡z−div⁡w=q−q=0, so the off-diagonal entries of G agree and G is symmetric. Write S=(∇z−(∇z)T)/2, which is skew-symmetric. Symmetric and skew-symmetric matrices are orthogonal in the Frobenius inner product, so (S,G)L2=0 and (∇z,G)=(ε(z),G). By step 2.1, (∇z,G)=∥∇z∥22−(∇z,R(w))=∥∇z∥22. The entries of R(w) are the entries of ∇w permuted and sign-changed, so ∥R(w)∥2=∥∇w∥2. Also ∣q∣2=∣∂1z2−∂2z1∣2≤2(∣∂1z2∣2+∣∂2z1∣2)≤2∣∇z∣F2 almost everywhere; integration gives ∥q∥2≤2∥∇z∥2. Thus [F1] and the triangle inequality imply ∥G∥2≤∥∇z∥2+∥∇w∥2≤(1+2Cdiv)∥∇z∥2. Therefore ∥∇z∥22≤∥ε(z)∥2∥G∥2≤(1+2Cdiv)∥ε(z)∥2∥∇z∥2. If ∇z=0 the gradient estimate is immediate; otherwise divide by ∥∇z∥2. Then [F2] and ∫Ωz=0 give ∥z∥2≤CP∥∇z∥2, hence ∥z∥H1≤1+CP2∥∇z∥2≤C0∥ε(v)∥2.

4.1A2F3F4step 3.1givenalgebra∎

For v∈V, [F3] gives TDr=−TDz. Apply [F4] and the trace bound [F3] to obtain ∥r∥H1≤CD∥TDz∥L2(ΓD)≤CDCT∥z∥H1. Since v=z+r, step 3.1 yields ∥v∥H1≤∥z∥H1+∥r∥H1≤(1+CDCT)C0∥ε(v)∥2, proving the assertion.

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