Page 528 - Handbook Of Integral Equations

Adding V(z) to the above general solution of the homogeneous problem, we find the general
solution of the nonhomogeneous problem under consideration:
r
+
R – (z) G (z)
+
+
Y (z)= V (z)+ (z – a i ) α i e P ν–n–1 (z),
Q – (z)
i=1
s (63)
Q + (z) G (z)
–
–
–
Y (z)= V (z)+ (z – b j ) β j e P ν–n–1 (z).
R + (z)
j=1
For ν – n > 0, the problem has ν – n linearly independent solutions. In the case ν – n ≤ 0 we must set
P ν–n–1 (z) ≡ 0. For ν –n < 0, the canonical function V(z) has the order ν –n < 0 at infinity and hence
is no longer a solution of the nonhomogeneous problem. However, on subjecting the right-hand side
to n – ν conditions we can increase the order of the function V(u)atinfinity by n – ν and thus make
the canonical function V(z) be a solution of the nonhomogeneous problem again.
k
To make the above operations possible, it suffices to require that the functions u H 1 (u) and
D 2 (u) have derivatives of order ≤ n – ν at infinity, and these derivatives satisfy the H¨ older condition.
2 . Let η < 0. The least possible order at infinity of H 1 (u)is h – ν – m + 1. In this case, the function
◦
+
[H 1 (u)Q – (u)]/[R – (u)e G (u) ] in the boundary condition (58) has the order 1–m–ν at infinity. After
+
selecting the principal part of the expansion of [H 1 (u)Q – (u)]/[R – (u)e G (u) ] in a neighborhood of
the point at infinity for m + ν – 1 > 0, the boundary condition can be rewritten in the form
s r

+
–
(u – b j ) Q – (u)Y (u) (u – a i ) R + (u)Y (u)
β j
α i
j=1 + i=1 –
– W (u)= – W (u)+ S(u).
R – (u)e G + (u) Q + (u)e G – (u)
The canonical function of the nonhomogeneous problem can be expressed via the interpolation
polynomial as follows:
+
R – (z)e G (z) +
+
V (z)= s [W (z) – U p (z)],
1

(z – b j ) Q – (z)
β j
j=1
(64)
–
Q + (z)e G (z) –
–
V (z)= r [W (z) – S(z) – U p (z)].
1

(z – a i ) R + (z)
α i
i=1
The general solution of problem (58) becomes
r
+
R – (z) G (z)
+
+
Y (z)= V (z)+ (z – a i ) α i e P h–m–1 (z),
1 Q – (z)
i=1
s (65)
Q + (z) –
–
–
Y (z)= V (z)+ (z – b j ) β j e G (z) P h–m–1 (z).
1
R + (z)
j=1
For h – m > 0, the problem has h – m linearly independent solutions. In the case h – m ≤ 0,
we must set the polynomial P h–m–1 (z) to be identically zero and, for the case in which h – m <0,
impose m–h conditions of the same type as in the previous case on the right-hand side. Under these
conditions, the nonhomogeneous problem (58) has a unique solution.
Remark. In Section 10.5 we consider equations that can be reduced to the problem by applying
the convolution theorem for the Fourier transform. Equations to which the convolution theorems for
other integral transforms can be applied, for instance, for the Mellin transform, can be investigated
in a similar way.
•
References for Section 10.4: F. D. Gakhov and Yu. I. Cherskii (1978), S. G. Mikhlin and S. Pr¨ ossdorf (1986),
N. I. Muskhelishvili (1992).
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
Page 510