Page 635 - Handbook Of Integral Equations

We seek the solution in the class of functions bounded on the contour.
Let X(z) be the canonical function of the Riemann problem with coefficient D 1 (t). Let us
–
+
substitute the expression D 1 (t)= X (t)/X (t) into (43) and rewrite the boundary condition in the
form
+
–
Φ (t) Φ (t)
= . (44)
µ κ
–
+
X (t) (t – α k ) m k X (t) (t – β j ) p j
k=1 j=1
To the last relation we apply the theorem on analytic continuation and the generalized Liouville
theorem (see Subsection 12.3-1). The points α k and β j cannot be singular points of the same analytic
+
–
function because this would contradict the assumption that Φ (t)or Φ (t) be bounded. Hence, the
–
only possible singularity is the point at infinity. The order at infinity of X (z)is ν, and the order of
–
–
κ

(z – β j ) p j is equal to –p. Hence, the order at infinity of the function Φ (z)/ X (z) κ (z – β j ) p j
j=1 j=1
is –ν + p.For ν – p ≥ 0 it follows from the generalized Liouville theorem that
–
+
Φ (z) Φ (z)
= = P ν–p (z),
µ κ

+
–
X (z) (z – α k ) m k X (z) (z – β j ) p j
k=1 j=1
and hence
µ κ
– –
+
+
Φ (z)= X (z) (z – α k ) m k P ν–p (z), Φ (z)= X (z) (z – β j ) P ν–p (z). (45)
p j
k=1 j=1
If ν – p < 0, then we must set P ν–p (z) ≡ 0, and hence the problem has no solutions.
The boundary value problem with coefficient D 1 (t) is called the reduced problem. The index ν
of the reduced problem will be called the index of the original problem. Formulas (45) show that
the degree of the occurring polynomial is less by p than the index ν of the problem.
Hence, the number of solutions of problem (43) in the class of functions bounded on the contour
is independent of the number of zeros of the coefficient and is diminished by the total number of
all poles. In particular, if the index is less than the total order of the poles, then the problem is
unsolvable. If the problem is solvable, then its solution can be expressed by formulas (45) in which
the canonical function X(z) of the reduced problem can be found by formulas (16) and (17) after
–
replacing D(t)by D 1 (t) in these formulas. Under the additional condition Φ (∞) = 0, the number
of solutions is diminished by one, and the degree of the polynomial in (45) must be at most ν – p – 1.
+
Now let us extend the class of solutions by assuming that one of the desired functions Φ (z)
–
and Φ (z) can tend to infinity with integral order at some points of the contour, and at the same
time another function remains bounded at these points. We can readily see that this assumption
implies no modifications at nonexceptional points. Here the boundedness of one of the functions
automatically implies the boundedness of the other. This is not the case for the exceptional points.
Let us rewrite the boundary condition (43) in the form
κ µ
+
–
(t – β j ) Φ (t) (t – α k ) m k Φ (t)
p j
j=1 k=1
= . (46)
–
+
X (t) X (t)
Applying the above reasoning and taking into account the fact that the right-hand side has a pole of
order ν + m at infinity, we obtain the general solution in the form
µ κ
– –
+
+
Φ (z)= X (z) (z – α k ) –m k P ν+m (z), Φ (z)= X (z) (z – β j ) –p j P ν+m (z). (47)
k=1 j=1
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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