Page 572 - Handbook Of Integral Equations

11.9-2. An Integral Equation of the Second Kind With Two Kernels
Consider an integral equation of convolution type of the second kind with two kernels of the form
0
1 ∞ 1
y(x)+ √ K 1 (x – t)y(t) dt + √ K 2 (x – t)y(t) dt = f(x), –∞ < x < ∞. (16)
2π 0 2π –∞
Note that each of the kernels K 1 (x) and K 2 (x)isdefined on the entire real axis. On representing the
desired function as the difference of one-sided functions,
y(x)= y + (x) – y – (x), (17)
we rewrite the equation in the form

1 ∞ 1 ∞
y + (x)+ √ K 1 (x – t)y + (t) dt – y – (x) – √ K 2 (x – t)y – (t) dt = f(x). (18)
2π –∞ 2π –∞
Applying the Fourier integral transform (see Subsection 7.4-3), we obtain
+
–
[1 + K 1 (u)]Y (u) – [1 + K 2 (u)]Y (u)= F(u). (19)
This implies the relation
1+ K 2 (u) – F(u)
+
Y (u)= Y (u)+ . (20)
1+ K 1 (u) 1+ K 1 (u)
Here K 1 (u), K 2 (u), and F(u) stand for the Fourier integrals of known functions. The unknown
–
+
transforms Y (u) and Y (u) are the boundary values of functions that are analytic on the upper and
lower half-planes, respectively. Thus, we have obtained a Riemann boundary value problem.
1 . Assume that the normality conditions are satisfied, i.e.,
◦
1+ K 1 (u) ≠ 0, 1 + K 2 (u) ≠ 0,
then we can rewrite the Riemann problem in the usual form (see Subsection 10.4-4):
+
–
Y (u)= D(u)Y (u)+ H(u), –∞ < u < ∞, (21)
where
1+ K 2 (u) F(u)
D(u)= , H(u)= . (22)
1+ K 1 (u) 1+ K 1 (u)
The Riemann problem (21), (22) is equivalent to Eq. (16): these problems are solvable or
unsolvable simultaneously, and have the same number of arbitrary constants in their general solutions.
If the index
1+ K 2 (u)
ν = Ind (23)
1+ K 1 (u)
is positive, then the homogeneous equation (16) (f(x) ≡ 0) has precisely ν linearly independent
solutions, and the nonhomogeneous equation is unconditionally solvable; moreover, the solution of
this equation depends on ν arbitrary complex constants.
In the case ν ≤ 0, the homogeneous equation has no nonzero solutions. The nonhomogeneous
equation is unconditionally solvable for ν = 0, and the solution is unique. For the case in which the
index ν is negative, the conditions

F(u) du
∞
=0, k =1, 2, ... , –ν, (24)
+
X (u)[1 + K 1 (u)](u + i) k
–∞
are necessary and sufficient for the solvability of the nonhomogeneous equation.
In all cases for which the solution of Eq. (16) exists, this solution can be found by the formula
∞
1 + – –iux
y(x)= √ [Y (u) – Y (u)]e du, –∞ < x < ∞, (25)
2π –∞
+
–
where Y (u), Y (u) is the solution of the Riemann problem (21), (22) constructed with respect to
the scheme of Subsection 10.4-4 (see Fig. 3).
Thus, the solution of Eq. (16) is equivalent to the solution of a Riemann boundary value problem
and is reduced to the calculation of finitely many Fourier integrals.
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 555