Page 569 - Handbook Of Integral Equations

and the nonhomogeneous equation is unconditionally solvable and its solution depends on ν arbitrary
complex constants.
In the case ν ≤ 0, the homogeneous equation has no nonzero solutions. For ν = 0, the nonhomo-
geneous equation is unconditionally solvable, and the solution is unique. If the index ν is negative,
then the conditions

∞
F(u) du
=0, k =1, 2, ... , –ν, (7)
+
X (u)[1 + K(u)](u + i) k
–∞
are necessary and sufficient for the solvability of the nonhomogeneous equation (see Subsec-
tion 10.4-4).
For all cases in which the solution of Eq. (1) exists, it can be found by the formula
1 ∞ + –iux
y(x)= y + (x)= √ Y (u)e du, x > 0, (8)
2π –∞
+
where Y (u) is the solution of the Riemann problem (4) and (5) that is constructed by the scheme of
–
Subsection 10.4-4 (see Fig. 3). The last formula shows that the solution does not depend on Y (u),
i.e., is independent of the choice of the extension of the equation to the negative semiaxis.
◦
2 . Now let us study the exceptional case of the integral equation (1) in which the normality
condition for the Riemann problem (3) (see Subsections 10.4-6 and 10.4-7) is violated. In this case,
the coefficient D(u)= [1 + K(u)] –1 has no zeros, and its order at infinity is η = 0. The general
solution to the boundary value problem (3) can be obtained by formulas (63) of Subsection 10.4-7
for α i = 0. The solution of the original integral equation (1) can be determined from the solution of
the boundary value problem on applying formula (8).
Figure 4 depicts a scheme of solving the Wiener–Hopf equations (see also Subsection 10.5-1).
Example. Consider the equation

∞ –|x–t|
y(x)+ (a + b|x – t|)e y(t) dt = f(x), x > 0, (9)
0
where the constants a and b are real, and b ≠ 0. The kernel K(x – t) of Eq. (1) is given by the expression
√ –|x|
K(x)= 2π (a + b|x|)e .
Let us find the transform of the kernel,
2
∞ u (a – b)+ a + b
K(u)= (a + b|x|)e –|x|+iux dx =2 2 2 .
–∞ (u +1)
Hence,
P(u)
2
4
1+ K(u)= , P(z)= z +2(a – b +1)z +2a +2b +1.
2
(u +1) 2
On the basis of the normality condition, we assume that the constants a and b are such that the polynomial P(z) has no real
roots. Let α + iβ be a root of the biquadratic equation P(z) = 0 such that α > 0 and β > 0. Since the coefficients of the
equation are real, it is clear that (α – iβ), (–α + iβ), and (–α – iβ) are the other three roots. Since the function 1 + K(u)is
real as well, it follows that it has zero index, and hence Eq. (9) is uniquely solvable.
–
+
On factorizing, we obtain the relation 1 + K(u)= X (u)/X (u), where
(u + i) 2 (u – α – iβ)(u + α – iβ)
–
+
X (u)= , X (u)= .
(u + α + iβ)(u – α + iβ) (u – i) 2
Applying this result, we represent the boundary condition (4), (5) in the form
+
2
–
+
Y (u) (u – i) F (u) Y (u)
– = , –∞ < u < ∞. (10)
–
+
X (u) (u – α – iβ)(u + α – iβ) X (u)
It follows from the theorem on the analytic continuation and the generalized Liouville theorem (see Subsection 10.4-3) that
both sides of the above relation are equal to
C 1 C 2
+ ,
u – α – iβ u + α – iβ
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 552