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–ν
On representing the function t D(t) with zero index as the ratio of boundary values of analytic
functions,
+
–ν
e G (t) 1 ln[τ D(τ)]
–ν
t D(t)= , G(z)= dτ, (16)
e G – (t) 2πi τ – z
L
we obtain the following expression for the canonical function:
+
–ν G (z)
–
+
X (z)= e G (z) , X (z)= z e – . (17)
–
+
Since X (t)= D(t)X (t), it follows that the coefficient of the Riemann problem can be represented
as the ratio of canonical functions:
+
X (t)
D(t)= . (18)
–
X (t)
The representation (18) is often called a factorization.
For ν ≥ 0, the canonical function, which has a zero of order ν at infinity, is a particular solution
of the boundary value problem (7). For ν < 0, the canonical function has a pole of order |ν| at infinity
and is not a solution, but in this case it is still used as an auxiliary function in the solution of the
nonhomogeneous problem.
12.3-5. The Solution of the Homogeneous Problem
Let ν = Ind D(t) be an arbitrary integer. On representing D(t) by formula (18), we reduce the
boundary condition (7) to the form
–
+
Φ (t) Φ (t)
= .
–
X (t) X (t)
+
The left-hand side of the last relation contains the boundary value of a function that is analytic
+
in Ω , and the right-hand side contains the boundary value of a function that has at least the order –ν
at infinity. By the principle of continuity (see Subsection 12.3-1), the functions on the left-hand
side and on the right-hand side are analytic continuations of each other to the entire plane possibly
except for the point at infinity at which, in the case ν > 0, a pole of order ≤ ν can occur. Hence, for
ν > 0, by the generalized Liouville theorem (see Subsection 12.3-1), this single analytic function is a
polynomial of degree ≤ ν with arbitrary coefficients. For ν < 0, it follows from the Liouville theorem
that this function is constant. However, since this function must vanish at infinity, it follows that it
is identically zero. Hence, for ν < 0, the homogeneous problem has only the trivial solution (which
is identically zero). A problem that has no nontrivial solutions is said to be unsolvable. Thus, for a
negative index, the homogeneous problem (7) is unsolvable.
Let ν > 0. Let P ν (z) stand for a polynomial of degree ν with arbitrary coefficients. In this case,
we obtain a solution in the form
Φ(z)= P ν (z)X(z),

or
+
–
–
–ν
+
Φ (z)= P ν (z)e G (z) , Φ (z)= z P ν (z)e G (z) , (19)
where G(z) is determined by formula (16).
Thus, if the index ν of the Riemann boundary value problem is nonnegative, then the homoge-
neous problem (7) has ν + 1 linearly independent solutions
–
k G (z)
+
–
Φ (z)= z e + , Φ (z)= z k–ν G (z) (k =0, 1, ... , ν). (20)
e
k k
The general solution contains ν + 1 arbitrary constants and is given by formula (19). For a negative
index, problem (7) is unsolvable.
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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