Page 625 - Handbook Of Integral Equations

into (3) and assume that ξ and η are differentiable. Then we obtain

1 ξdη – ηdξ 1 l ξ(s)η (s) – η(s)ξ (s)

s
s
ν = = ds. (6)
2
2
2
2π Γ ξ + η 2 2π 0 ξ (s)+ η (s)
n
Example. Let us calculate the index of D(t)= t with respect to an arbitrary contour L surrounding the origin.
n
n
First method. The function t is the boundary value of the function z , which has precisely one zero of order n inside
the contour. Hence
n
ν = Ind t = n.
n
Second method. If the argument of t is ϕ, then the argument of t is nϕ. As the point t traverses the contour L and
returns to the original value, the argument ϕ obtains the increment 2π. Hence,
n
Ind t = n.
The index can also be found numerically. Since the index is integer-valued, an approximate
value whose error is less than 1 can be rounded off to the nearest integer to obtain the exact value.
2
12.3-4. Statement of the Riemann Problem
Let L be a simple smooth closed contour which divides the complex plane into the interior domain Ω +
–
and the exterior domain Ω , and let two functions of points of the contours D(t) and H(t) satisfying
the H¨ older condition (see Subsection 12.2-2) be given; moreover, suppose that D(t) does not vanish.
The Riemann Problem. Find two functions (or a single piecewise analytic function), namely, a
–
–
+
+
function Φ (z) analytic in Ω and a function Φ (z) analytic in the domain Ω including z = ∞,so
that the following linear relation is satisfied on the contour L:
+
–
Φ (t)= D(t)Φ (t) (the homogeneous problem) (7)
or
+
–
Φ (t)= D(t)Φ (t)+ H(t) (the nonhomogeneous problem). (8)
The function D(t) is called the coefficient of the Riemann problem, and the function H(t)is
called the right-hand side.
We first consider a Riemann problem of special form that is called the jump problem. Let a
function ϕ(t)defined on a closed contour L satisfy the H¨ older condition. The problem is to find
+
+
–
–
a piecewise analytic function Φ(z)(Φ(z)= Φ (z) for z ∈ Ω and Φ(z)= Φ (z) for z ∈ Ω ) that
vanishes at infinity and has a jump of magnitude ϕ(t)on L, i.e., such that
–
+
Φ (t) – Φ (t)= ϕ(t).
It follows from the Sokhotski–Plemelj formulas (see Subsection 12.2-5) that the function
1 ϕ(τ)
Φ(z)= dτ
2πi τ – z
L
is the unique solution to the above problem.
Thus, an arbitrary function ϕ(t) given on the closed contour and satisfying the H¨ older condition
+
–
can be uniquely represented as the difference of functions Φ (t) and Φ (t) that are the boundary
–
–
+
values of analytic functions Φ (z) and Φ (z) under the additional condition Φ (∞)=0.
–
If we neglect the additional condition Φ (∞) = 0, then the solution will be given by the formula
1 ϕ(τ)
Φ(z)= dτ + const . (9)
2πi τ – z
L
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 608