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EQUATIONS OF ELASTIC HYPERSURFACES

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5. BOUNDARY VALUE PROBLEMS: EXAMPLES 153<br />

Proof: See Theorem 4.36, (4.180) for the mapping properties and (4.183a), (4.183b) for the<br />

Plemelji formulae.<br />

The concluding assertion about ellipticity also is standard and we quote [CW1, Du2, Fi1,<br />

KGBB1, Le1, Ma1] etc. for the proofs in case of domains in R n .<br />

Lemma 5.9 (The direct potential method). The boundary pseudodifferential equation<br />

(V −1 ψ)(s) = (N C (s, D)f)(s) − 1 2 g + (W 0(s, D)g)(s)<br />

= (D νΓ N C (s, D)f)(s) + (M ∗ −(s, D)g)(s) , (5.33)<br />

M − (s, D)u(s):=− 1 2 u(s) + W 0(s, D)u(s) , s ∈ Γ , (5.34)<br />

where ψ(s) := (D νΓ ϕ) + (s) is the unknown function, is equivalent to the Dirichlet BVP (5.16)<br />

via the representation formula (5.26), in which ϕ + = g and (D νΓ ϕ) + = ψ is the solution of<br />

(5.33).<br />

The boundary pseudodifferential equation<br />

(W +1 (s, D)ψ)(s) = −(D νΓ N C (s, D)f)(s) + 1 2 h + (W∗ 0(s, D)h)(s)<br />

= −(D νΓ N C (s, D)f)(s) + (M ∗ +(s, D)h)(s) , (5.35)<br />

M + (s, D)u(s):= 1 2 u(s) + W 0(s, D)u(s) , s ∈ Γ , (5.36)<br />

where ψ(s) := ϕ + (s) is the unknown function, is equivalent to the Neumann BVP (5.17) via<br />

the representation formula (5.26), in which (D νΓ ϕ) + = h and ϕ + = ψ is the solution of<br />

(5.35).<br />

Proof (cf. Theorem 4.42): By inserting the known data ϕ + = g (for the Dirichlet BVP)<br />

or (D νΓ ϕ) + = h (for the Neumann BVP) into the representation formulae (5.26) and by<br />

applying the appropriate Plemelji formulae (5.31) and get the equivalent boundary pseudodifferential<br />

equations (5.33) and (5.35), respectively.<br />

The next Lemma is only auxiliary one. The result is sharpened in Theorem 5.11.<br />

Lemma 5.10 The homogeneous equation<br />

V −1 (s, D)ϕ(s) = 0 , ϕ ∈ H θ p(Γ) , s ∈ Γ (5.37)<br />

has only a trivial solution ϕ = 0 for arbitrary s > 0 and 1 < p < ∞.<br />

Proof (cf. Lemma 4.44: From (5.27) and (5.25) follows that<br />

∮<br />

∮<br />

∆ C V Γ ψ(t)= ∆ C K ∆ (t, t − s)ψ(s) ds = δ(t − s)ψ(s) ds = 0 ∀ t ∉ Γ, (5.38)<br />

Γ<br />

Γ<br />

since δ(t − s) = 0 for all t ∉ Γ and all s ∈ Γ.

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