Page 640 - Handbook Of Integral Equations

As usual, by applying the theorem on analytic continuation and the generalized Liouville theorem
(see Subsection 12.3-1), we obtain
m
–
+
–ν k G (z) – –ν G (z)
+
Φ (z)= (z – z k ) e P ν (z), Φ (z)= z e P ν (z). (61)
k=1
We can see that this solution differs from the above solution of the problem for a simply
+
connected domain only in that the function Φ (z) has the factor m (z – z k ) –ν k . Under the additional
k=1
–
condition Φ (∞) = 0, in formulas (61) we must take the polynomial P ν–1 (z).
Applying the Sokhotski–Plemelj formulas, we obtain
–ν
1
±
G (t)= ± ln[t Π(t)D(t)] + G(t),
2
where G(t) is the Cauchy principal value of the integral (59) and
m

Π(t)= (t – z k ) .
ν k
k=1
On passing to the limit as z → t in formulas (60) we obtain

D(t) G(t) – 1 G(t)
+
X (t)= e , X (t)= √ e . (62)
ν
ν
t Π(t) t Π(t)D(t)
–ν
The sign of the root is determined by the (arbitrary) choice of a branch of the function ln[t Π(t)D(t)].
◦
2 . The Nonhomogeneous Problem. By the same reasoning as above, we represent the boundary
condition
+
–
Φ (t)= D(t)Φ (t)+ H(t) (63)
in the form
–
+
Φ (t) + Φ (t) –
– Ψ (t)= – Ψ (t),
+
–
X (t) X (t)
where Ψ(z)isdefined by the formula
1 H(τ) dτ
Ψ(z)= .
2πi L X (τ) τ – z
+
This gives the general solution
Φ(z)= X(z)[Ψ(z)+ P ν (z)] (64)
or
Φ(z)= X(z)[Ψ(z)+ P ν–1 (z)], (65)
–
if the solution satisfies the condition Φ (∞)=0.
For ν < 0, the nonhomogeneous problem is solvable if and only if the following conditions are
satisfied:

H(t) k–1
t dt = 0, (66)
+
X (t)
L
where k ranges from 1 to –ν – 1 if we seek solutions bounded at infinity and from 1 to –ν if we
–
assume that Φ (∞)=0.
Under conditions (66), the solution can also be found from formulas (64) or (65) by setting
P ν ≡ 0.
+
If the external contour L 0 is absent and the domain Ω is the plane with holes, then the main
difference from the preceding case is that here the zero index with respect to all contours L k
–ν
(k =1, ... , m) is attained by the function m (t – z k ) D(t) that does not involve the factor t .
ν k
k=1
Therefore, to obtain a solution to the problem, it suffices to repeat the above reasoning on omitting
this factor.
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 623