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

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452 P. Ebenfelt and M. Viscardi CMFT by equation (10). By Cramer’s Rule, we may solve uniquely for the b k,j so that they satisfy the above system of equations, and therefore satisfy equation (6). Thus, we have found the desired b k,j and Q k (z), proving Lemma 2. We shall now use the results of Lemma 2 to construct the desired function u. Sum both sides of equation (6) as k goes from 1 to n to get p n∑ k p ∑ b k,j n∑ [f(z) − f(a k=1 j=1 k )] = ∑ k c k,j n∑ j (z − a k=1 j=1 k ) + Q j k (z) k=1 (14) = R(z) − ˜R(z) n∑ + Q k (z), where the second equality follows from equation (5). Define S(z) := ˜R(z) n∑ − Q k (z), ˆT (w) := n∑ k=1 j=1 k=1 p ∑ k k=1 b k,j [w − f(a k )] j . By Lemma 2 and the paragraph preceding it, S(z) ∈ O(Ω) and is rational. Also, since f is a bijective rational map from Ω to D and a k ∈ Ω for all k, we know that ˆT (w) is rational and ˆT (w) ∈ O(P 1 \D). Substituting these new functions into equation (14), we obtain i.e. ˆT (f(z)) = R(z) − S(z), (15) R(z) = S(z) + ˆT (f(z)). We now make one final substitution ( ) T (z) := ˆT 1 ∗ . f(z) Note that T is rational since ˆT and f are rational. Furthermore, since f : Ω → D and I : D → P 1 \D (where I denotes the inversion in the plane) are both bijective, T (z) ∈ O(Ω) since ˆT (z) ∈ O(P 1 \D). Finally, for z ∈ ∂Ω, we know that f(z) ∈ ∂D = {w : |w| 2 = ww = 1}, which gives ( ) ( ) T (z) = ˆT 1 ∗ = f(z) ˆT 1 = ˆT (f(z)), f(z) so that equation (15) becomes (16) R(z) = S(z) + T (z) for z ∈ ∂Ω.
5 (2005), No. 2 The Dirichlet Problem with Rational Holomorphic Data 453 We now define (17) u(z, z) := S(z) + T (z). Since S, T ∈ O(Ω), u is harmonic in Ω, and for z ∈ ∂Ω, equation (16) yields u(z, z) = R(z). Hence, u is the solution to the Dirichlet problem posed in Theorem 2. Since S and T are rational, equation (17) implies that u is rational. This completes the proof of Theorem 2. (ii) ⇒ (i). This is trivial. We shall also write the implication (i) ⇒ (v) as a theorem. Theorem 3. Let Ω be as in Theorem 1, and let a ∈ Ω. Suppose the solution u to the Dirichlet problem ⎧ ⎨∆u = 0 in Ω ⎩u(z, z) = 1 on ∂Ω z − a is rational. Then the Bergman kernel K(z, w) associated with Ω is also rational. Proof. Since u is harmonic in Ω, we may write u(z, z) = s(z) + t(z), where s, t ∈ O(Ω). Furthermore, since u is rational, s and t are also rational. Define R(z) := 1 − s(z) + t(z). z − a Note that R is rational. Since u(z, z) = s(z) + t(z) = 1/(z − a) on ∂Ω, we have R(z) = 1 − s(z) + t(z) = t(z) + t(z) ∈ R z − a on ∂Ω. Thus, we have constructed a rational function R such that R(∂Ω) ⊂ R. Now, since the boundary of Ω is a smooth real-analytic closed curve, it is well known that u can be extended as a harmonic function to an open neighborhood of Ω (this follows from a version of the Schwarz Reflection Principle; see e.g. [9, 14]). Thus, s and t can be extended as holomorphic functions past the boundary of Ω; in particular, they will be continuous on ∂Ω, so R will also be continuous on ∂Ω. Since Ω is bounded and simply connected, ∂Ω is a compact, connected subset of C. So by the above, R(∂Ω) is a compact, connected subset of R, i.e. a closed interval L = [z 1 , z 2 ] ⊂ R. Lemma 3. R maps Ω bijectively to P 1 \L ′ , where L ′ is some closed subinterval of L, i.e. L ′ = [z 3 , z 4 ] ⊆ [z 1 , z 2 ] with z 3 < z 4 .
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