Page 117 - Handbook Of Integral Equations

1.9-2. Equations With Difference Kernel: K(x, t)= K(x – t)

x
16. K(x – t)y(t) dt = f(x).
a
1 . Let K(0) = 1 and f(a) = 0. Differentiating the equation with respect to x yields a Volterra
◦
equation of the second kind:
x

y(x)+ K (x – t)y(t) dt = f (x).
x
x
a
The solution of this equation can be represented in the form
x

y(x)= f (x)+ R(x – t)f (t) dt. (1)
x t
a
Here the resolvent R(x) is related to the kernel K(x) of the original equation by

1
–1
˜
R(x)= L – 1 , K(p)= L K(x) ,
˜
pK(p)
–1
where L and L are the operators of the direct and inverse Laplace transforms, respectively.
∞ 1 c+i∞

˜
px ˜
˜

K(p)= L K(x) = e –px K(x) dx, R(x)= L –1 R(p) = e R(p) dp.
0 2πi c–i∞
◦
2 . Let K(x) have an integrable power-law singularity at x = 0. Denote by w = w(x) the
solution of the simpler auxiliary equation (compared with the original equation) with a =0
and constant right-hand side f ≡ 1,
x

K(x – t)w(t) dt = 1. (2)
0
Then the solution of the original integral equation with arbitrary right-hand side is expressed
in terms of w as follows:
d x x
y(x)= w(x – t)f(t) dt = f(a)w(x – a)+ w(x – t)f (t) dt.

t
dx a a
x

n
17. K(x – t)y(t) dt = Ax , n = 0,1,2, ...
–∞
This is a special case of equation 1.9.19 with λ =0.
◦
1 . Solution with n =0:
A ∞
y(x)= , B = K(z) dz.
B 0
◦
2 . Solution with n =1:
A AC ∞ ∞
y(x)= x + , B = K(z) dz, C = zK(z) dz.
B B 2 0 0
3 . Solution with n =2:
◦
A 2 AC AC 2 AD
y 2 (x)= x +2 x +2 – ,
B B 2 B 3 B 2
∞ ∞ ∞

2
B = K(z) dz, C = zK(z) dz, D = z K(z) dz.
0 0 0
4 . Solution with n =3, 4, ... is given by:
◦
∞
∂ n e λx –λz
y n (x)= A , B(λ)= K(z)e dz.
∂λ n B(λ)
λ=0 0
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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