Page 661 - Handbook Of Integral Equations

We regard the polynomial U ρ (z) as an operator that maps the right-hand side f(t) of Eq. (1) to
the polynomial that interpolates the Cauchy type integral (37) as above. Let us denote this operator
by
1
2 T[f(t)] = U ρ (z). (38)
Here the coefficient 1 is taken for the convenience of the subsequent manipulations.
2
Furthermore, by analogy with the normal case, from (36) we can find
+
X (t) 1 f(t) 1 f(τ) dτ 1 1
+
Φ (t)= + – T[f(t)] – A 0 (t)P ν–p–1 (t) ,
η + +
2 s(t)X (t) 2πi L s(τ)X (τ) τ – t 2 2
(t – β j ) p j
j=1
–
X (t) 1 f(t) 1 f(τ) dτ 1 1
–
Φ (t)= – + – T[f(t)] – A 0 (t)P ν–p–1 (t) .
µ + +
2 s(t)X (t) 2πi L s(τ)X (τ) τ – t 2 2
(t – α k ) m k
k=1
We introduced the coefficient – 1 in the last summands of these formulas using the fact that the
2
coefficients of the polynomial P ν–p–1 (t) are arbitrary. Hence,

∆ 1 (t)f(t) 1 f(τ) dτ
–
+
ϕ(t)= Φ (t)–Φ (t)= +∆ 2 (t) –T[f(t)]–A 0 (t)P ν–p–1 (t) , (39)
+
s(t)X (t) πi s(τ)X (τ)(τ – t)
+
L
where
–
+
+
–
X (t) X (t) X (t) X (t)
∆ 1 (t)= + , ∆ 2 (t)= – .
η µ η µ

2 (t – β j ) p j 2 (t – α k ) m k 2 (t – β j ) p j 2 (t – α k ) m k
j=1 k=1 j=1 k=1
We write
+
–
Z(t)= s(t)X (t)= r(t)X (t), (40)
and, applying relation (34), represent formula (39) as follows:
1 b(t)Z(t) f(τ) dτ
ϕ(t)= a(t)f(t) – + b(t)Z(t)T[f(t)] + b(t)Z(t)P ν–p–1 (t).
A 0 (t) πi L Z(τ) τ – t
Let us introduce the operator R 1 [f(t)] by the formula
1 b(t)Z(t) f(τ) dτ
R 1 [f(t)] ≡ a(t)f(t) – + b(t)Z(t)T[f(t)] , (41)
A 0 (t) πi L Z(τ) τ – t

and finally obtain
ϕ(t)= R 1 [f(t)] + b(t)Z(t)P ν–p–1 (t). (42)

Formula (42) gives a solution of Eq. (1) for the exceptional case in which ν – p > 0. This
solution linearly depends on ν –p arbitrary constants. If ν –p < 0, then the solution exists only under
p – ν special solvability conditions imposed on f(t), which follow from the solvability conditions
for the Riemann problem (35) corresponding to this case.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC

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