Page 464 - Handbook Of Integral Equations

8.4-2. The Case in Which the Transform of the Solution is a Rational Function

Consider the important special case in which the transform (4) of the solution is a rational function
of the form
˜
f(p) R(p)
˜ y(p)= ≡ ,
˜
K(p) Q(p)
where Q(p) and R(p) are polynomials in the variable p and the degree of Q(p) exceeds that of R(p).
If the zeros of the denominator Q(p) are simple, i.e.,
Q(p) ≡ const (p – λ 1 )(p – λ 2 ) ... (p – λ n ),

and λ i ≠ λ j for i ≠ j, then the solution has the form

n
R(λ k )
y(x)= exp(λ k x),
Q (λ k )

k=1
where the prime stands for the derivatives.
Example 1. Consider the Volterra integral equation of the first kind
x
e –a(x–t) y(t) dt = A sinh(bx).
0
We apply the Laplace transform to this equation and obtain (see Supplement 4)
1 Ab
˜ y(p)= .
2
p + a p – b 2
This implies
Ab(p + a) Ab(p + a)
˜ y(p)= = .
2
p – b 2 (p – b)(p + b)
We have Q(p)=(p – b)(p + b), R(p)= Ab(p + a), λ 1 = b, and λ 2 = –b. Therefore, the solution of the integral equation has
the form
y(x)= 1 A(b + a)e bx + 1 A(b – a)e –bx = Aa sinh(bx)+ Ab cosh(bx).
2 2

8.4-3. Convolution Representation of a Solution
In solving Volterra integral equations of the first kind with difference kernel K(x – t) by means of
the Laplace transform, it is sometimes useful to apply the following approach.
Let us represent the transform (4) of a solution in the form

˜
˜
˜
˜
˜ y(p)= N(p)M(p)f(p), N(p) ≡ 1 . (6)
˜
˜
K(p)M(p)
˜
If we can find a function M(p) for which the inverse transforms
–1
–1
˜
˜
L M(p) = M(x), L N(p) = N(x) (7)
exist and can be found in a closed form, then the solution can be written as the convolution
x t

y(x)= N(x – t)F(t) dt, F(t)= M(t – s)f(s) ds. (8)
0 0
Example 2. Consider the equation
x √

sin k x – t y(t) dt = f(x), f(0)=0. (9)
0
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 446