Page 526 - Handbook Of Integral Equations

◦
1 . Let the order η of the coefficient of the boundary value problem at infinity satisfy the condition
η ≥ 0, i.e., let D(u) have a zero of order η at infinity. It follows from (53) that n – ν ≥ m – h.On
equating the left- and right-hand sides of relation (55) with a polynomial P ν–n–1 (z), we obtain the
solution of the boundary value problem in the form
r
R – (z) +
+
Y (z)= (z – a i ) α i e G (z) P ν–n–1 (z),
Q – (z)
i=1
s (56)
Q + (z) –
–
Y (z)= (z – b j ) β j e G (z) P ν–n–1 (z).
R + (z)
j=1
This problem has ν – n linearly independent solutions for ν – n > 0 and only the trivial zero
solution for ν – n ≤ 0.
◦
2 . Let η < 0, i.e., let D(u) have a pole of order –η at infinity. In this case, m – h > n – ν, and we
can obtain the general solution from (56) by replacing P ν–n–1 (z)by P h–m–1 (z) in this expression.
In this case, the problem has h – m solutions for h – m > 0 and only the trivial zero solution for
h – m ≤ 0.
According to (53), we have
h – m = ν – n + η. (57)
Thus, in both cases under consideration, the number of linearly independent solutions is equal to
the index minus the total number of the poles (including the pole at infinity) of the coefficient D(u).
Hence, we have the following law: the number of linearly independent solutions of a homogeneous
Riemann problem is not affected by the number of zeros of the coefficient and is reduced by the
total number of its poles.

10.4-7. Exceptional Cases. The Nonhomogeneous Problem
Assume that the right-hand side has the same poles as the coefficient. The boundary condition can
be rewritten as follows:
r

(u – a i ) R + (u)R – (u)
α i
H 1 (u)
i=1
–
+
Y (u)= D 2 (u)Y (u)+ , (58)
s s

(u – b j ) Q + (u)Q – (u) (u – b j ) β j
β j
j=1 j=1
where D 2 (u) and H 1 (u) satisfy the H¨ older condition and some additional differentiability conditions
near the points a i , b j , and ∞.
◦
1 . Assume that the order η at infinity of the coefficient of the boundary value problem satisfies
the condition η ≥ 0. Since the first two terms of relation (58) vanish at infinity, it follows that the
minimal possible order of H 1 (u)atinfinity is equal to 1 – n. Just as in the homogeneous problem,
we replace D 2 (u) by the ratio of two functions (54) and write out the boundary condition in the
following form (under the braces, the orders of the functions at infinity are indicated):
s r
+
(u – b j ) Q – (u)Y (u) (u – a i ) R + (u)Y (u)
–
β j
α i
j=1 i=1 H 1 (u)Q – (u)
= +
R – (u)e G + (u) Q + (u)e G – (u) R – (u)e G + (u)

1–n–h 1–m–ν 1–n–h
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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