Page 461 - Handbook Of Integral Equations

The substitution
x
h 2 (t)
w(x)= u(t) dt (4)
a h 1 (t) t
reduces Eq. (3) to the first-order linear ordinary differential equation

h 2 (x) h 2 (x)

[g 1 (x)h 1 (x)+ g 2 (x)h 2 (x)]w – g 2 (x)h 1 (x) w = f(x)h 1 (x) . (5)
x
h 1 (x) h 1 (x)
x x
1 . In the case g 1 (x)h 1 (x)+ g 2 (x)h 2 (x) /≡ 0, the solution of equation (5) satisfying the condition
◦
w(a) = 0 (this condition is a consequence of the substitution (4)) has the form
x
h 2 (t) f(t)h 1 (t) dt
w(x)= Φ(x) , (6)
a h 1 (t) t Φ(t)[g 1 (t)h 1 (t)+ g 2 (t)h 2 (t)]
x
h 2 (t) g 2 (t)h 1 (t) dt
Φ(x)=exp . (7)
a h 1 (t) t g 1 (t)h 1 (t)+ g 2 (t)h 2 (t)
Let us differentiate relation (4) and substitute the function (6) into the resulting expression. After
integrating by parts with regard to the relations f(a)=0 and w(a) = 0, for f /≡ const g 2 we obtain
x

g 2 (x)h 1 (x)Φ(x) f(t) dt
u(x)= .
g 1 (x)h 1 (x)+ g 2 (x)h 2 (x) g 2 (t) Φ(t)
a t
Using formula (2), we find a solution of the original equation in the form
x
1 d g 2 (x)h 1 (x)Φ(x) f(t) dt
y(x)= , (8)
h 1 (x) dx g 1 (x)h 1 (x)+ g 2 (x)h 2 (x) a g 2 (t) t Φ(t)
where the function Φ(x) is given by (7).
If f(x) ≡ const g 2 (x), the solution is given by formulas (8) and (7) in which the subscript 1 must
be changed by 2 and vice versa.

2 . In the case g 1 (x)h 1 (x)+ g 2 (x)h 2 (x) ≡ 0, the solution has the form
◦

1 d (f/g 2 ) x 1 d (f/g 2 ) x
y(x)= = – .
h 1 dx (g 1 /g 2 ) h 1 dx (h 2 /h 1 )
x x
8.2-2. Equations With General Degenerate Kernel
A Volterra equation of the first kind with general degenerate kernel has the form

n x

g m (x) h m (t)y(t) dt = f(x). (9)
a
m=1
Using the notation
x
w m (x)= h m (t)y(t) dt, m =1, ... , n, (10)
a
we can rewrite Eq. (9) as follows:
n

g m (x)w m (x)= f(x). (11)
m=1

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 443