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EQUATIONS OF ELASTIC HYPERSURFACES

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2. OUTLINE <strong>OF</strong> DIFFERENTIAL GEOMETRY 35<br />

For arbitrary ϕ, ψ ∈ W 1 2(Ω) the Green formulae (2.124) and (2.124) follow by approximation<br />

ϕ j → ϕ, ψ j → ψ, ϕ j , ψ j ∈ C 2 (Ω).<br />

Stoke’s derivatives are concrete examples of weakly tangential operators<br />

M S := [M jk ] n×n<br />

, M jk := ν j ∂ k − ν k ∂ j = ∂ mj,k . (2.127)<br />

These derivatives are directional with respect to a tangential vector fields to S (cf. Definition<br />

2.4). In fact, the directing vector m jk (X ) = ν j (X )e k − ν k (X )e j of M jk , where {e j } n j=1 is<br />

the canonical frame in R n , is tangential to S :<br />

ν(X ) · m jk (X ) = ν j (X )ν k (X ) − ν k (X )ν j (X ) ≡ 0, X ∈ S . (2.128)<br />

Therefore the Stoke’s derivative M jk operator can be applied to functions defined on the<br />

surface S only.<br />

Corollary 2.34 Let Ω, S = ∂Ω and ν(τ) = (ν 1 (τ), . . . , ν n (τ)) ⊤ be as in Lemma 2.30.<br />

The following Stoke’s formulae<br />

∮<br />

(M jk f)(τ) dS = 0 (2.129)<br />

S<br />

holds for j, k = 1, . . . , n and for all f ∈ W 1 1(S ).<br />

The Stokes derivatives M j,k are skew-symmetric:<br />

∮<br />

∮<br />

(M jk ψ)(τ)ϕ(τ) dS = − ψ(τ)(M jk ϕ)(τ) dS (2.130)<br />

S<br />

S<br />

for j, k = 1, . . . , n and for arbitrary pair ϕ, ψ ∈ W 2 2(S ).<br />

Proof: We assume temporarily that f ∈ C 1 (S ) and extend this function into the domain<br />

F ∈ C 1 (Ω) ⋂ C 2 (Ω) with the trace on the boundary F ∣ S<br />

= f. Such extension is possible<br />

since the boundary is a Lipschitz hypersurface. It is possible to construct a direct extension<br />

by means of function theory (cf. E. Stein [St1]). But we consider here the following indirect<br />

construction: consider the Dirichlet problem for the Laplace operator ∆F = 0 in Ω with<br />

a boundary condition F ∣ S<br />

= f. It is well known that the solution exists and, moreover,<br />

F ∈ C ∞ (Ω) (cf., e.g., [Le1]). Herewith we have found the extension.<br />

Now apply the Gauß formula (2.112) to a function ∂ j ∂ k f = ∂ k ∂ j f twice:<br />

∫<br />

∮<br />

(∂ j ∂ k F )(y) dy = ν j (τ)(∂ k f)(τ) dS ,<br />

∫<br />

Ω<br />

Ω<br />

∮<br />

(∂ k ∂ j F )(y) dy =<br />

By taking the difference we get (2.129) immediately.<br />

S<br />

S<br />

ν k (τ)(∂ j f)(τ) dS .

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