Chapter 40

Korn’s InequalityA fundamental inequality used in elasticity to obtain coercivity and then apply the LaxMilgram theorem or some other theorem is Korn’s inequality. The proof given here of thisfundamental result follows [101] and [46].

40.1 A Fundamental InequalityThe proof of Korn’s inequality depends on a fundamental inequality involving negativeSobolev space norms. The theorem to be proved is the following.

Theorem 40.1.1 Let f ∈ L2 (Ω) where Ω is a bounded Lipschitz domain. Then there existconstants, C1 and C2 such that

C1 || f ||0,2,Ω ≤

(|| f ||−1,2,Ω +

n

∑i=1

∣∣∣∣∣∣∣∣ ∂ f∂xi

∣∣∣∣∣∣∣∣−1,2,Ω

)≤C2 || f ||0,2,Ω ,

where here ||·||0,2,Ω represents the L2 norm and ||·||−1,2,Ω represents the norm in the dualspace of H1

0 (Ω) , denoted by H−1 (Ω) .

Similar conventions will apply for any domain in place of Ω. The proof of this theoremwill proceed through the use of several lemmas.

Lemma 40.1.2 Let U− denote the set,

{(x,xn) ∈ Rn : xn < g(x)}

where g : Rn−1→ R is Lipschitz and denote by U+ the set

{(x,xn) ∈ Rn : xn > g(x)} .

Let f ∈ L2 (U−) and extend f to all of Rn in the following way.

f (x,xn)≡−3 f (x,2g(x)− xn)+4 f (x,3g(x)−2xn) .

Then there is a constant, Cg, depending on g such that

|| f ||−1,2,Rn +n

∑i=1

∣∣∣∣∣∣∣∣ ∂ f∂xi

∣∣∣∣∣∣∣∣−1,2,Rn

≤Cg

(|| f ||−1,2,U− +

n

∑i=1

∣∣∣∣∣∣∣∣ ∂ f∂xi

∣∣∣∣∣∣∣∣−1,2,U−

).

Proof: Let φ ∈C∞c (Rn) . Then,∫

Rnf

∂φ

∂xndx =

∫U+

∂φ

∂xn[−3 f (x,2g(x)− xn)+4 f (x,3g(x)−2xn)]dx

+∫

U−f

∂φ

∂xndx. (40.1.1)

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