Page 493 - Handbook Of Integral Equations

9.5-2. A Generating Function of Exponential Form

Consider a linear equation with exponential right-hand side

λx
L [y]= e . (5)
Suppose that the solution is known and is given by formula (2). In Table 4 we present solutions
of the equation L [y]= f(x) with various right-hand sides; these solutions are expressed via the
solution of Eq. (5).

Remark 1. When applying the formulas indicated in the table, we need not know the left-hand
side of the linear equation (5) (the equation can be integral, differential, etc.) provided that a particular
solution of this equation for exponential right-hand side is known. It is only of importance that the
left-hand side of the equation is independent of the parameter λ.
Remark 2. When applying formulas indicated in the table, the convergence of the integrals
occurring in the resulting solution must be verified.
Example 1. We seek a solution of the equation with exponential right-hand side

∞

y(x)+ K(x – t)y(t) dt = e λx (6)
x
in the form y(x, λ)= ke λx by the method of indeterminate coefficients. Then we obtain
1 ∞
y(x, λ)= e λx , B(λ)=1 + K(–z)e λz dz. (7)
B(λ) 0
It follows from row 3 of Table 4 that the solution of the equation

∞

y(x)+ K(x – t)y(t) dt = Ax (8)
x
has the form
A AC
y(x)= x – ,
D D 2
∞ ∞

D =1 + K(–z) dz, C = zK(–z) dz.
0 0
For such a solution to exist, it is necessary that the improper integrals of the functions K(–z) and zK(–z) exist. This
holds if the function K(–z) decreases more rapidly than z –2 as z →∞. Otherwise a solution can be nonexistent. It is of
interest that for functions K(–z) with power-law growth as z →∞ in the case λ < 0, the solution of Eq. (6) exists and is
given by formula (7), whereas Eq. (8) does not have a solution. Therefore, we must be careful when using formulas from
Table 4 and verify the convergence of the integrals occurring in the solution.
It follows from row 15 of Table 4 that the solution of the equation
∞

y(x)+ K(x – t)y(t) dt = A sin(λx) (9)
x
is given by the formula

A
y(x)= B c sin(λx) – B s cos(λx) ,
2 2
B c + B s

∞ ∞
B c =1 + K(–z) cos(λz) dz, B s = K(–z) sin(λz) dz.
0 0

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 475