Page 143 - Handbook Of Integral Equations

where

n–1 ν ν m–1
λ Γ (m/n) (νm/n)–1 b
R(x)= x + ε µ exp ε µ bx
Γ(νm/n) m
ν=1 µ=0
n–1 ν ν m–1 x
b λ Γ (m/n) (νm/n)–1
+ ε µ exp ε µ bx t exp –ε µ bt dt ,
m Γ(νm/n) 0
ν=1 µ=0
2πµi 2

n/m n/m
b = λ Γ (m/n), ε µ =exp , i = –1, µ =0, 1, ... , m – 1.
m
2 . Solution with any α from 0 < α <1:
◦
n

x ∞ 1–α
λΓ(1 – α)x
y(x)= f(x)+ R(x – t)f(t) dt, where R(x)= .
0 xΓ n(1 – α)
n=1
•
References: H. Brakhage, K. Nickel, and P. Rieder (1965), V. I. Smirnov (1974).
λ x y(t) dt
61. y(x) – = f(x), 0 < α ≤ 1.
x α 0 (x – t) 1–α
1 . The solution of the homogeneous equation (f ≡ 0) is
◦
y(x)= Cx β (β > –1, λ > 0). (1)
Here C is an arbitrary constant, and β = β(λ) is determined by the transcendental equation
λB(α, β + 1) = 1, (2)

where B(p, q)= 0 1 z p–1 (1 – z) q–1 dz is the beta function.
◦
2 . For a polynomial right-hand side,
N
n
f(x)= A n x
n=0
the solution bounded at zero is given by

N

A n

 n
 x for λ < α,
1 – (λ/λ n )



n=0
y(x)=
 N
 A n n β
x + Cx for λ > α and λ ≠ λ n ,



 1 – (λ/λ n )
n=0
(α) n+1
λ n = , (α) n+1 = α(α +1) ... (α + n).
n!
Here C is an arbitrary constant, and β = β(λ) is determined by the transcendental equation (2).
For special λ = λ n (n =1, 2, ... ), the solution differs in one term and has the form
n–1 N
A m m A m m λ n n n
¯
y(x)= x + x – A n x ln x + Cx ,
1 – (λ n /λ m ) 1 – (λ n /λ m ) λ n
m=0 m=n+1
1 –1

¯
z ln zdz
where λ n = (1 – z) α–1 n .
0
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 122