Page 530 - Handbook Of Integral Equations

The Fourier transform of Eq. (5) results in the following boundary value problem:

K 2 (u) – F(u)
+
Y (u)= Y (u)+ , –∞ < u < ∞. (6)
K 1 (u) K 1 (u)
The coefficient of this problem is the ratio of functions that vanish at infinity, and hence, in contrast
to the preceding case, it can have a zero or a pole of some order at infinity.
µ
λ
Let K 1 (u)= T 1 (u)/u and K 2 (u)= T 2 (u)/u , where the functions T 1 (u) and T 2 (u) have zero
order at infinity. In the dependence of the sign of the difference η = µ – λ, two cases can occur. For
generality, we assume that there are exceptional points at finite distances as well. Let the functions
K 1 (u) and K 2 (u) have the representations
s p

K 1 (u)= (u – b j ) β j (u – c k ) K 11 (u),
γ k
j=1 k=1
r p

γ k
K 2 (u)= (u – a i ) α i (u – c k ) K 12 (u).
i=1 k=1
Along with the common zeros at points c k of multiplicity γ k , the functions K 1 (u) and K 2 (u)
have a common zero of order min(λ, µ)atinfinity.
The coefficient of the Riemann problem can be represented in the form

r

(u – a i ) R + (u)R – (u)
α i
i=1
D(u)= D 2 (u).
s

(u – b j ) Q + (u)Q – (u)
β j
j=1
It follows from (6) that this problem and the integral equation (5) are solvable if at any point c k that
is a common zero of the functions K 1 (u) and K 2 (u), the function F(u) has zero of order γ k , i.e.,
F(u) has the form
p

γ k
F(u)= (u – c k ) F 1 (u).
k=1
To this end, the following γ 1 + ··· + γ p = l conditions must hold:

(j k )
F u (u) =0, j k =0, 1, ... , γ k – 1, (7)
u=c k
or, which is the same, the conditions

∞

j k ic k x
f(x)x e dx = 0. (8)
–∞
For the case under consideration in which the equation is of the first kind, we must add other
d conditions, where
d = min(λ, µ)+1, (9)
that are imposed on the behavior of F(u)atinfinity because the functions K 1 (u) and K 2 (u)havea
common zero of order min(λ, µ)atinfinity. Hence, F(u) must satisfy the conditions (8) and have at
least the order d at infinity.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 512