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where a 0 , a 1 , ... , a n are the coefficients of the polynomial U n (z), and the a –k are the coefficients
of the expansion of the function Ψ(z), which are given by the obvious formula

κ k–1
1 H(τ)τ

a –k = – (τ – β j ) p j dτ.
+
2πi X (τ)
L j=1
The solvability conditions acquire the form
a n = a n–1 = ··· = a n–p+ν+2 =0.

–
If a solution must satisfy the additional condition Φ (∞) = 0, then, for ν –p > 0, in formulas (53)
we must take the polynomial P ν–p–1 (z), and for ν – p <0, p – ν conditions must be satisfied.

12.3-10. The Riemann Problem for a Multiply Connected Domain
Let L = L 0 + L 1 + ··· + L m be a collection of m + 1 disjoint contours, and let the interior of the
+
contour L 0 contain the other contours. By Ω we denote the (m + 1)-connected domain interior
+
–
for L 0 and exterior for L 1 , ... , L m .By Ω we denote the complement of Ω + L in the entire
+
complex plane. To be definite, we assume that the origin lies in Ω . The positive direction of the
+
contour L is that for which the domain Ω remains to the left, i.e., the contour L 0 must be traversed
counterclockwise and the contours L 1 , ... , L m , clockwise.
We first note that the jump problem
–
+
Φ (t) – Φ (t)= H(t)
is solved by the same formula
1 H(τ) dτ
Φ(z)=
2πi L τ – z
as in the case of a simply connected domain. This follows from the Sokhotski–Plemelj formulas,
which have the same form for a multiply connected domain as for a simply connected domain.
The Riemann problem (homogeneous and nonhomogeneous) can be posed in the same way as
for a simply connected domain.
We write ν k = 1 (all contours are passed in the positive direction). By the index
2π [arg D(t)] L k
of the problem we mean the number
m

ν = ν k . (54)
k=0
If ν k (k =1, ... , m) are zero for the inner contours, then the solution of the problem has just the
same form as for a simply connected domain.
To reduce the general case to the simplest one, we introduce the function
m

(t – z k ) ,
ν k
k=1
where the z k are some points inside the contours L k (k =1, ... , m). Taking into account the fact
= –2π, we obtain
that [arg(t – z k )] L j = 0 for k ≠ j and [arg(t – z j )] L j
m
1 1

arg (t – z k ) ν k = arg(t – z j ) ν j = –ν j , j =1, ... , m.
2π 2π L j
k=1 L j

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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