Page 639 - Handbook Of Integral Equations

Hence,
m

arg D(t) (t – z k ) ν k =0, j =1, ... , m.
k=1 L j
Let us calculate the increment of the argument of the function D(t) m (t – z k ) ν k with respect to
k=1
the contour L 0 :
m m m

1 1 1
arg D(t) (t – z k ) ν k = arg D(t) + [ν k arg(t – z k )] L 0 = ν 0 + ν k = ν.
2π 2π L 0 2π
k=1 L 0 k=1 k=1
+
Since the origin belongs to the domain Ω , it follows that
=0, k =1, ... , m, =2π.
[arg t] L k [arg t] L 0
Therefore,
m

–ν
ν k
arg t (t – z k ) D(t) = 0. (55)
k=1 L
◦
1 . The Homogeneous Problem. Let us rewrite the boundary condition
+
–
Φ (t)= D(t)Φ (t) (56)
in the form m
t ν –ν –
+
ν k
Φ (t)= t (t – z k ) D(t) Φ (t). (57)
m
k=1
(t – z k ) ν k
k=1
The function t –ν m (t – z k ) D(t) has zero index on each of the contours L k (k =1, ... , m),
ν k
k=1
and hence it can be expressed as the ratio
+
m e G (t)
t –ν (t – z k ) D(t)= , (58)
ν k
e G – (t)
k=1
where
m
1 –ν dτ

ν k
G(z)= ln τ (τ – z k ) D(τ) . (59)
2πi L τ – z
k=1
The canonical function of the problem is given by the formulas
m
+
–
–ν k G (z) – –ν G (z)
+
X (z)= (z – z k ) e , X (z)= z e . (60)
k=1
Now the boundary condition (57) can be rewritten in the form
+
–
Φ (t) Φ (t)
= .
–
+
X (t) X (t)
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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