
7.4 Inhomogeneous Wiener–Hopf Integral Equation of the Second Kind 269
if and only if
1
2πi
−C
2
ζ
j
Y
+
(ζ )
ˆ
f
−
(ζ )dζ = 0, j = 0, ...,
|
ν
|
− 1. (7.4.54)
If this condition is satisfied, we get from the j =
|
ν
|
term onward,
ˆ
φ
−
(k) → C/k
1+
|
ν
|
as
k
→∞, (7.4.55)
so that
ˆ
φ
−
(k)canbeinvertedforφ(x), which is the unique solution to the inhomo-
geneous problem. To understand this solvability condition (7.4.54), we first recall
the Parseval identity,
+∞
−∞
dk
ˆ
h(k)
ˆ
g(−k) = 2π
+∞
−∞
h(y)g(y)dy, (7.4.56)
where
ˆ
h(k)and
ˆ
g(k) are the Fourier transforms of h(y)andg(y), respectively. Then
we consider the homogeneous adjoint problem. Recall that for a real kernel,
K
adj
(x, y) = K(y, x).
Thus, corresponding to the original homogeneous problem,
φ(x) = λ
+∞
0
K(x − y)φ(y)dy,
there exists the homogeneous adjoint problem,
φ
adj
(x) = λ
+∞
0
K(y − x)φ
adj
(y)dy, (7.4.57)
whose translation kernel is related to the original one by
K
adj
(ξ) = K(−ξ). (7.4.58)
Now, when we take the Fourier transform of the homogeneous adjoint problem,
we find
1 −λ
ˆ
K(−k)
ˆ
φ
adj
−
(k) =−
ˆ
ψ
adj
+
(k), (7.4.59)
where the only difference from the original equation is the sign of k inside
ˆ
K(−k).
However, since 1 − λ
ˆ
K(−k)isjustthereflectionof1−λ
ˆ
K(k) through the origin, a
zero of 1 −λ
ˆ
K(k) in the upper half plane corresponds to a zero of 1 −λ
ˆ
K(−k)in
the lower half plane, etc. Thus, when the original 1 −λ
ˆ
K(k) has a negative index
ν<0 with respect to a line, Im k = k
2
= A, the homogeneous adjoint problem
1 − λ
ˆ
K(−k) has a positive index
|
ν
|
relative to the line Im k = k
2
=−A.So,in