Page 480 - Handbook Of Integral Equations

Taking into account the initial data (9), we find the dependence of the integration constants C 1
and C 2 on the parameter x. As a result, we obtain the solution of problem (8), (9):
t
ds
v = v 1 (x)Φ(x) 2 . (10)
x Φ(s)[v 1 (s)]
On substituting the expression (10) into formula (7) and eliminating the second derivative by means
of Eq. (8) we find the resolvent:
v 1 (x)Φ(x) t ds

R(x, t)= ϕ(t) + v 1 (x)Φ(x)[ϕ(t)v (t)+ ψ(t)v 1 (t)] .
t
v 1 (t)Φ(t) x Φ(s)[v 1 (s)] 2
Remark 2. The kernel of the integral equation under consideration can be rewritten in the form
K(x, t)= G 1 (t)+ xG 2 (t), where G 1 (t)= ϕ(t)+ tψ(t) and G 2 (t)= –ϕ(t).

n m–1
9.2-3. Equations With Kernel of the Form K(x, t)= ϕ m (x)(x – t)
m=1
To find the resolvent, we introduce an auxiliary function as follows:
x n–1
1 n–1 (x – t)

u(x, t)= R(s, t)(x – s) ds + ;
(n – 1)! (n – 1)!
t
at x = t, this function vanishes together with the first n – 2 derivatives with respect to x, and the
(n – 1)st derivative at x = t is equal to 1. Moreover,
n
d u(x, t)
(n)
(n)
R(x, t)= u (x, t), u = . (11)
x x n
dx
On substituting relation (11) into the resolvent equation (3) of Subsection 9.1-1, we see that
x
u (n) (x, t)= K(x, t)+ K(x, s)u (n) (s, t) ds. (12)
s
x
t
Integrating by parts the right-hand side in (12), we obtain
n–1
m
u (n) (x, t)= K(x, t)+ (–1) K (m) (x, s)u (n–m–1) (s, t) s=x . (13)
x s s s=t
m=0
On substituting the expressions for K(x, t) and u(x, t) into (13), we arrive at a linear homogeneous
ordinary differential equation of order n for the function u(x, t).
Thus, the resolvent R(x, t) of the Volterra integral equation with degenerate kernel of the above
form can be obtained by means of (11), where u(x, t) satisfies the following differential equation
and initial conditions:

u (n) – ϕ 1 (x)u (n–1) – ϕ 2 (x)u (n–2) – 2ϕ 3 (x)u (n–3) – ··· – (n – 1)! ϕ n (x)u =0,
x x x x

u = u = ··· = u (n–2) =0, u (n–1) =1.
x=t x x=t x x=t x x=t
The parameter t occurs only in the initial conditions, and the equation itself is independent of t
explicitly.

Remark 3. A kernel of the form K(x, t)= n φ m (x)t m–1 can be reduced to a kernel of the
m=1
above type by elementary transformations.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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