Page 633 - Handbook Of Integral Equations

increases by 2π as x ranges over the real axis in the positive direction. Thus,

x – i
Ind =1.
x + i
If Ind D(x)= ν, then the function
–ν
x – i

D(x)
x + i
has zero index. Its logarithm is single-valued on the real axis.
We construct the canonical function for which the point –i is the exceptional point as follows:
–ν
–
+ – z – i G (z)
G (z)
+
X (z)= e , X (z)= e , (36)
z + i
where –ν
1 ∞ τ – i dτ
G(z)= ln D(τ) .
2πi τ + i τ – z
–∞
Using the limit values of this function, we transform the boundary condition (35) to the form
+
–
Φ (x) Φ (x) H(x)
= + .
+
X (x) X (x) X (x)
+
–
Next, introducing the analytic function
1 ∞ H(τ) dτ
Ψ(z)= , (37)
+
2πi X (τ) τ – z
–∞
we represent the boundary condition in the form
–
+
Φ (x) + Φ (x) –
– Ψ (x)= – Ψ (x).
–
+
X (x) X (x)
–
Note that, in contrast with the case of a finite contour, here we have Ψ (∞) ≠ 0 in general. On
applying the theorem on analytic continuation and taking into account the fact that the only possible
singularity of the function under consideration is a pole at the point z = –i of order ≤ ν (for ν > 0),
on the basis of the generalized Liouville theorem we obtain (see Subsection 12.3-1)
–
+
Φ (z) + Φ (z) – P ν (z)
– Ψ (z)= – Ψ (z)= , ν ≥ 0,
–
+
X (z) X (z) (z + i) ν
where P ν (z) is a polynomial of degree ≤ ν with arbitrary coefficients. This gives the general solution
of the problem:

P ν (z)
Φ(z)= X(z) Ψ(z)+ for ν ≥ 0, (38)
(z + i) ν
(39)
Φ(z)= X(z)[Ψ(z)+ C] for ν <0,
where C is an arbitrary constant. For ν < 0, the function X(z) has a pole of order –ν at the point
–
z = –i, and therefore for the solvability of the problem we must set C = –Ψ (–i). For ν < –1, the
following conditions must additionally hold:
∞
H(x) dx

=0, k =2, 3, ... , –ν. (40)
X (x) (x + i) k
+
–∞
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 616