Page 151 - Handbook Of Integral Equations

Differentiating (4) with respect to x and taking account of (3), we obtain

n n

y (x)+ σ n y(x)+ A k λ k I k = f (x), σ n = A k . (5)
x
x
k=1 k=1
Eliminating the integral I n from (4) and (5), we find that
n–1

y (x)+ σ n – λ n )y(x)+ A k (λ k – λ n )I k = f (x) – λ n f(x). (6)
x x
k=1
Differentiating (6) with respect to x and eliminating I n–1 from the resulting equation with
the aid of (6), we obtain a similar equation whose left-hand side is a second-order linear
n–2
1
differential operator (acting on y) with constant coefficients plus the sum A I k .If we
k
k=1
proceed with successively eliminating I n–2 , I n–3 , ... , I 1 with the aid of differentiation and
formula (3), then we will finally arrive at an nth-order linear nonhomogeneous ordinary
differential equation with constant coefficients.
The initial conditions for y(x) can be obtained by setting x = a in the integral equation
and all its derivative equations.
2 . The solution of the equation can be represented in the form
◦
n
x
y(x)= f(x)+ B k e µ k (x–t) f(t) dt. (7)
a
k=1
The unknown constants µ k are the roots of the algebraic equation

n
A k

+ 1 = 0, (8)
z – λ k
k=1
which is reduced (by separating the numerator) to the problem of finding the roots of an
nth-order characteristic polynomial.
After the µ k have been calculated, the coefficients B k can be found from the following
linear system of algebraic equations:
n
B k

+1 = 0, m =1, ... , n. (9)
λ m – µ k
k=1
◦
Another way of determining the B k is presented in item 3 below.
If all the roots µ k of equation (8) are real and different, then the solution of the original
integral equation can be calculated by formula (7).
To a pair of complex conjugate roots µ k,k+1 = α ± iβ of the characteristic polynomial (8)
there corresponds a pair of complex conjugate coefficients B k,k+1 in equation (9). In this case,
the corresponding terms B k e µ k (x–t) + B k+1 e µ k+1 (x–t) in solution (7) can be written in the form

B k e α(x–t) cos β(x – t) + B k+1 e α(x–t) sin β(x – t) , where B k and B k+1 are real coefficients.
◦
3 .For a = 0, the solution of the original integral equation is given by
x

–1
y(x)= f(x) – R(x – t)f(t) dt, R(x)= L R(p) , (10)
0

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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