Page 634 - Handbook Of Integral Equations

Thus, we obtained results similar to those for a finite contour.
Indeed, for ν ≥ 0, the homogeneous and nonhomogeneous Riemann boundary value problems
for the half-plane are unconditionally solvable, and their solution linearly depends on ν + 1 arbitrary
constants. For ν < 0, the homogeneous problem is unsolvable. For ν < 0, the nonhomogeneous
problem is uniquely solvable; moreover, in the case ν = –1 the problem is unconditionally solvable,
and in the case ν < –1, it is solvable under –ν – 1 solvability conditions (40) only.
Let us also discuss the case of solutions vanishing at infinity (see also Subsection 10.3-4). On
+
–
substituting the relation Φ (∞)= Φ (∞) = 0 into the boundary condition, we obtain H(∞)=0.
Hence, for a Riemann problem to have a solution that vanishes at infinity, the right-hand side of
the boundary condition must vanish at infinity. Assume that this condition is satisfied. To obtain a
solution for the case under consideration, we must replace the expression P ν (z) in (38) by P ν–1 (z)
and equate the constant C in (39) with zero. Thus,

P ν–1 (z)
Φ(z)= X(z) Ψ(z)+ . (41)
(z + i) ν
For ν ≤ 0, we must set P ν–1 (z) ≡ 0 in this formula. We must add another condition to the solvability
conditions (40), namely, Ψ(–i) = 0, and finally we obtain the following solvability conditions:

∞
H(x) dx

=0, k =1, 2, ... , –ν. (42)
+
X (x) (x + i) k
–∞
Now, for ν > 0 we have a solution that depends on ν arbitrary constants. For ν ≤ 0, a solution is
unique, and for ν < 0, a solution exists if and only if –ν conditions hold.
12.3-9. Exceptional Cases of the Riemann Problem
In the statement of the Riemann boundary value problem it was required that the coefficient D(t)
satisfies the H¨ older condition (this prevents infinite values of this coefficient) and vanishes nowhere.
As can be observed from the solution (the use of ln D(t)), these restrictions are essential. Now we
assume that D(t) vanishes or tends to infinity, with an integer order, at some points of the contour.
We assume that the contour L consists of a single closed curve.
Consider the homogeneous problem. We rewrite the boundary condition of the homogeneous
Riemann problem in the form

µ

(t – α k ) m k
–
+
Φ (t)= k=1 D 1 (t)Φ (t). (43)
κ

(t – β j ) p j
j=1
Here α k (k =1, ... , µ) and β j (j =1, ... , κ) are some points of the contour, m k and p j are positive
integers, and D 1 (t) is a function that is everywhere nonzero and satisfies the H¨ older condition. The
points α k are zeros of the function D(t). The points β j will be called the poles of this function. The
use of the term “pole” is not completely rigorous because the function D(t) is not analytic. We shall
use this term for brevity for a point at which a function (not analytic) tends to infinity with some
integer order. We write

κ µ

Ind D 1 (t)= ν, p j = p, m k = m.
j=1 k=1

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 617