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By Myroslav L. Gorbachuk

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Additional info for Boundary Value Problems for Operator Differential Equations

Example text

Denoting by ζe (s) and ζo (s) the contributions of He and Ho to the zeta function, we compute     2 i 1 |λj | |λj | ζe (s) = dλ λ−s − +  λ(λ2j − λ) λ λ2 − λ  4 j 2π j γ = 1 4 |λj |−2s j i 2π dτ −τ −s−1 (1 − τ )−1 + τ −s−1 (1 − τ )−1/2 . γ The sum over j leads to the zeta function of A2 , |λj |−2s , ζ(s; A2 ) = j with a meromophic structure known from the considerations of a manifold without boundary. For (−t) < s < 0, the τ -integrals are determined using [235] Ft (s) := i 2π dτ τ −s−1 (1 − τ )−t γ ∞ i = e−iπ(s+1) − eiπ(s+1) 2π du u−s−1 (1 + u)−t 0 1 Γ(−s)Γ(s + t) = sin(π(s + 1)) π Γ(t) Γ(s + t) = .

P − λ) = j=−∞ If we write the symbol for HλB (x, x ) again in the form h−2−j homogeneous of degree −2 − j, eq. 12) reads ∞ j=0 h−2−j , with ∞ σ (P − λ) h−2−j (y, r, ω, τ, λ) = 0. , we find 0 = a(2) (y, r, ω, Dr , λ) h−2−j (y, r, ω, τ, λ) 1 + Dωα a(k) (y, r, ω, Dr , λ) α! 7) has been used. Note this equation is not a purely algebraic equation anymore, but instead an ordinary differential equation in the variable r. Supplemented by suitable boundary conditions, arising from eq. 13), this, under suitable assumptions, will provide uniquely defined symbols h−2−j .

The integral alone has poles at s = k ∈ IN which is seen by writing [(zν/a)2 − m2 ]−s = (−1)j (Γ(1 − s)/Γ(j + 1 − s))(dj /d(m2 )j )[(zν/a)2 − m2 ]−s+j . ) For this reason Z(s) gives no contribution to the residues of ζ(s) in that range. Furthermore, for s = −k, k ∈ IN0 , k < (−1 + N )/2, the prefactor guarantees Z(s) = 0 and thus no contributions to the values of the zeta function at these points arise. This result means that the heat kernel coefficients are just determined by the asymptotic terms Ai (s) with ∞ (2l + 1)Aνi (s).

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