Page 531 - Handbook Of Integral Equations

If these conditions are satisfied, then the boundary value problem (6) becomes

r

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

(u – b j ) Q + (u)Q – (u) (u – b j ) β j
β j
j=1 j=1
The solution was given above in Subsection 10.4-7. For the case in which η ≥ 0(µ ≥ λ), this solution
can be rewritten in the form
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 (10)
Q + (z) G (z)
–
–
–
Y (z)= V (z)+ (z – b j ) β j e P ν–n+1 (z).
R + (z)
j=1
For the case in which η <0 (µ < λ), this solution becomes
r
R – (z) +
+
+
Y (z)= V (z)+ (z – a i ) α i e G (z) P h–m–1 (z),
1
Q – (z)
i=1
s (11)
Q + (z) –
–
–
Y (z)= V (z)+ (z – b j ) β j e G (z) P h–m–1 (z).
1
R + (z)
j=1
In both cases, the solution of the original integral equation can be obtained by substituting the
expressions (10) and (11) into the formula

1 ∞ + – –iux
y(x)= √ [Y (u) – Y (u)]e du. (12)
2π –∞
Example. Consider the following equation of the first kind:
1 ∞ 1 0
√ K 1 (x – t)y(t) dt + √ K 2 (x – t)y(t) dt = f(x),
2π 0 2π –∞
where
√
0 for x >0, – 2πie –2x for x >0, 0 for x >0,
K 1 (x)= √ 3x 2x K 2 (x)= f(x)= √ 3x 2x (13)
2π (e – e ) for x <0, 0 for x <0, 2π (e – e ) for x <0.
Applying the Fourier transform to the functions in (13), we obtain
1 1 1
K 1 (u)= , K 2 (u)= , F(u)= .
(u – 2i)(u – 3i) u +2i (u – 2i)(u – 3i)
Here the boundary value problem (6) becomes
(u – 2i)(u – 3i)
–
+
Y (u)= Y (u)+1.
u +2i
The coefficient D(u) has a first-order pole at infinity (ν = –1). In this case
m + =2, n + =0, ν = m + – n + = 2, min(λ, µ)=1, d =2.
The function F(u) has second-order zero at infinity, and hence the necessary condition for the solvability is satisfied.
In the class of functions that vanish at infinity, the homogeneous problem
(u – 2i)(u – 3i)
–
+
Y (u)= Y (u)
u +2i
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 513