On the Solution of the Dirichlet Problem with Rational Holomorphic ...

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450 P. Ebenfelt and M. Viscardi CMFT equation (7). We now Taylor expand the second term in the right-hand side of equation (7) about the point a k to obtain (8) b k,j [f(z) − f(a k )] j b k,j = f ′ (a k ) j (z − a k ) [1 + e 1,j(z − a k ) + e 2,j (z − a k ) 2 + · · · ] j = b [ k,j 1 f ′ (a k ) j (z − a k ) + e 1,j j (z − a k ) + · · · + e ] j−1,j + ψ j−1 k (z) , z − a k where the e n,j are constants in C and ψ k is rational (since we are assuming that f is rational). Now, b k,j /[f(z) − f(a k )] j clearly has a pole at a k , as well as at all other points z such that f(z) = f(a k ). However, since a k ∈ Ω and f is one-to-one inside Ω, a k is the only pole of the b k,j /[f(z) − f(a k )] j in Ω. Thus, a k is the only pole in Ω of the right-hand side of equation (8). But note that, by construction, ψ k does not have a pole at a k ; hence, ψ k only has poles outside of Ω. So ψ k is a rational function and holomorphic in Ω. We now sum both sides of equation (8) as j goes from 1 to p k to get (9) Define p k ∑ j=1 b k,j [f(z) − f(a k )] j = p ∑ k [ b k,j f j=1 ′ (a k ) j p ∑ k +ψ k (z) j=1 1 (z − a k ) + e 1,j j (z − a k ) + · · · + e ] j−1,j j−1 z − a k b k,j f ′ (a k ) j . Q k (z) := ψ k (z) p k ∑ j=1 b k,j f ′ (a k ) j . Since ψ k ∈ O(Ω) and is rational, we conclude that Q k ∈ O(Ω) and that Q k is rational. Thus, Q k satisfies the conditions of Lemma 2. For ease of notation, let us also define the following: Note that e 0,j := 1, (10) d j,j = d l,j := e (j−l),j f ′ (a k ) j , 1 ≤ l ≤ j, 1 ≤ j ≤ p k. 1 ≠ 0 for all j. f ′ (a k ) j
5 (2005), No. 2 The Dirichlet Problem with Rational Holomorphic Data 451 Equation (9) can now be written as (11) p k ∑ j=1 p b k,j [f(z) − f(a k )] = ∑ k j j=1 j∑ l=1 d l,j b k,j (z − a k ) l + Q k(z). Reversing the order of summation in the right-hand side of equation (11) yields (12) p k ∑ j=1 j=1 b k,j [f(z) − f(a k )] j = p k ∑ = p k ∑ l=1 j=l p ∑ k l=1 j=1 d l,j b k,j (z − a k ) l + Q k(z) ∑ pk j=l d l,jb k,j (z − a k ) l + Q k (z). We now swap the indices l and j in the right-hand side of equation (12) to get p k p ∑ b k ∑ k,j (13) [f(z) − f(a k )] = ∑ pk l=j d j,lb k,l + Q j (z − a k ) j k (z). Thus, by equation (6) we want to show that there exist b k,j ∈ C such that p ∑ k ∑ pk l=j d p j,lb k,l ∑ k c k,j + Q (z − a k ) j k (z) = (z − a k ) + Q k(z), j i.e. i.e. j=1 p k ∑ l=j p k ∑ j=1 ∑ pk l=j d j,lb k,l (z − a k ) j = j=1 p k ∑ j=1 c k,j (z − a k ) j , d j,l b k,l = c k,j for all j, 1 ≤ j ≤ p k . Writing this as a system of equations, we get d 1,1 b k,1 + d 1,2 b k,2 + · · · + d 1,pk b k,pk = c k,1 d 2,2 b k,2 + · · · + d 2,pk b k,pk = c k,2 . d pk ,p k b k,pk = c k,pk . The resulting coefficient matrix ⎛ ⎞ d 1,1 d 1,2 . . . d 1,pk 0 d A = ⎜ 2,2 . . . d 2,pk ⎝ . ⎟ . . .. . ⎠ 0 0 . . . d pk ,p k is triangular, so det A = d 1,1 d 2,2 . . . d pk ,p k ≠ 0
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