Page 632 - Handbook Of Integral Equations

◦
3 .We have
2
p +(t)= t, p –(t)=1, q +(t)= t – 1, q –(t)=1, m + =1, n + =2, ν = –1,
+

 z for z ∈ Ω ,

1 τ 1 1/[τ(τ – 1)]
Ψ(z)= dτ + dτ = 1 –
2πi L τ – z 2πi L τ – z  – z(z – 1) for z ∈ Ω .
The solution of the problem exists only under the solvability conditions (34) or, for the case in question, under the single
condition

q –(τ)
H(τ) dτ =0.
L p –(τ)
On calculating this integral, we obtain
τ – τ +1 dτ dτ
3 2
dτ = τdτ + – = 0+2πi – 2πi =0.
2
L τ – τ L L τ – 1 L τ
Thus, the solvability condition holds, and the unique solution of the problem is
z +1
–
+
Φ (z)= z, Φ (z)= – .
z 2
◦
4 .We have
2
p +(t)=1, p –(t)= t, q +(t)= t – 1, q –(t)=1, ν = m + – n + = –2<0.
For the solvability of the problem, the following two conditions are necessary:

q –(τ)
H(τ)τ k–1 dτ =0, k =1, 2.
L p –(τ)
On calculating the last integral for k = 1, we obtain
τ – τ +1 1 1 1
3 2
dτ = 1 – – + dτ =2πi ≠ 0.
2
L τ(τ – τ) L τ τ 2 τ – 1
Thus, the solvability condition fails, and hence the problem has no solution.
Note that if we formally calculate the function Φ(z), then it has a pole at infinity, and hence cannot be a solution of the
problem.
12.3-8. The Riemann Problem for a Half-Plane
Let the contour L be the real axis. Just as above, the Riemann problem is to find two bounded analytic
+
–
functions Φ (z) and Φ (z) in the upper and the lower half-plane, respectively (or a single piecewise
analytic function Φ(z) on the plane), whose limit values on the contour satisfy the boundary condition
+
–
Φ (x)= D(x)Φ (x)+ H(x), –∞ < x < ∞. (35)
The given functions D(x) and H(x) satisfy the H¨ older condition both at the endpoints and in a
neighborhood of the point at infinity on the contour. We also assume that D(x) ≠ 0.
The main difference from the above case of a finite curve is that here the point at infinity and the
origin belong to the contour itself, and therefore cannot be taken as exceptional points at which the
canonical function can have a nonzero order. Instead of the auxiliary function t which was used in
the above discussion (and has the unit index with respect to L), we use the linear-fractional function
on the real axis with the same property:
x – i
.
x + i
The argument of this function

x – i (x – i) 2
arg =arg = 2 arg(x – i)
2
x + i x + i

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 615