Page 645 - Handbook Of Integral Equations

12.4-5. Equations of the First Kind With Hilbert Kernel

1 . Consider the simplest singular integral equation of the first kind with Hilbert kernel
◦
1 2π ξ – x
cot ϕ(ξ) dξ = f(x), 0 ≤ x ≤ 2π, (21)
2π 0 2
under the additional assumption
2π
ϕ(x) dx = 0. (22)
0
Equation (21) can have a solution only if a solvability condition is satisfied. This condition is
obtained by integrating Eq. (21) with respect to x from zero to 2π and, with regard for the relation

2π
ξ – x
cot dx =0,
2
0
becomes
2π

f(x) dx = 0. (23)
0
To construct a solution of Eq. (21), we apply the solution of the simplest singular integral
equation of the first kind with Cauchy kernel by assuming that the contour L is the circle of unit
radius centered at the origin (see Subsection 12.4-1). We rewrite the equation with Cauchy kernel
and its solution in the form
1 ϕ 1 (τ)
dτ = f 1 (t), (24)
π L τ – t
1 f 1 (τ)
ϕ 1 (t)= – dτ, (25)
π L τ – t
–1
which is obtained by substituting the function ϕ 1 (t) instead of ϕ(t) and the function f 1 (t)i instead
of f(t) into the relations of 12.4-1.
iξ
We set t = e ix and τ = e and find the relationship between the Cauchy kernel and the Hilbert
kernel:

dτ 1 ξ – x i
= cot dξ + dξ. (26)
τ – t 2 2 2
On substituting relation (26) into Eq. (24) and into the solution (25), with regard to the change of
variables ϕ(x)= ϕ 1 (t) and f(x)= f 1 (t) we obtain

1 2π ξ – x i 2π
cot ϕ(ξ) dξ + ϕ(ξ) dξ = f(x), (27)
2π 0 2 2π 0
2π 2π
1 ξ – x i

ϕ(x)= – cot f(ξ) dξ – f(ξ) dξ. (28)
2π 0 2 2π 0
Equation (21), under the additional assumption (22), coincides with Eq. (27), and hence its
solution is given by the expression (28). Taking into account the solvability conditions (23), on the
basis of (28) we rewrite a solution of Eq. (21) in the form

1 2π ξ – x
ϕ(x)= – cot f(ξ) dξ. (29)
2π 0 2
Formulas (21) and (29), together with conditions (22) and (23), are called the Hilbert inversion
formula.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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