Page 686 - Handbook Of Integral Equations

In this case Eq. (1) becomes

m b

y(x)= g k (x) h k (t)Φ t, y(t) dt. (3)
a
k=1
We write
b

A k = h k (t)Φ t, y(t) dt, k =1, ... , m, (4)
a
where the constants A k are yet unknown. Then it follows from (3) that
m

y(x)= A k g k (x). (5)
k=1
On substituting the expression (5) for y(x) into relations (4), we obtain (in the general case)
m transcendental equations of the form

A k = Ψ k (A 1 , ... , A m ), k =1, ... , m, (6)

which contain m unknown numbers A 1 , ... , A m .
For the case in which Φ(t, y) is a polynomial in y, i.e.,
n
Φ(t, y)= p 0 (t)+ p 1 (t)y + ··· + p n (t)y , (7)
where p 0 (t), ... , p n (t) are, for instance, continuous functions of t on the interval [a, b], system (6)
becomes a system of nonlinear algebraic equations for A 1 , ... , A m .
The number of solutions of the integral equation (3) is equal to the number of solutions of
system (6). Each solution of system (6) generates a solution (5) of the integral equation.
◦
2 . Consider the Urysohn equation of the second kind with the simplified degenerate kernel of the
following form:
n
b

y(x)+ g k (x)f k t, y(t) dt = h(x). (8)
a
k=1
Its solution has the form
n

y(x)= h(x)+ λ k g k (x), (9)
k=1
where the constants λ k can be defined by solving the algebraic (or transcendental) system of
equations
b n

λ m + f m t, h(t)+ λ k g k (t) dt =0, m =1, ... , n. (10)
a
k=1
To different roots of this system, there are different corresponding solutions of the nonlinear integral
equation. It may happen that (real) solutions are absent.
A solution of an Urysohn equation of the second kind with degenerate kernel in the general form
n
b
f x, y(x) + g k x, y(x) h k t, y(t) dt = 0. (11)
a
k=1
can be represented in the implicit form
n

f x, y(x) + λ k g k x, y(x) = 0, (12)
k=1

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 669