Page 500 - Handbook Of Integral Equations

Example 2. Consider the equation
x 1 t
y(x)+ K y(t) dt = f(x). (21)
0 x x
In accordance with the method of model solutions, we consider the following auxiliary equation with power-law right-hand
side:
x 1 t
–s
y(x)+ K y(t) dt = x . (22)
0 x x
Its solution has the form (see Example 2 for λ = –s in Section 9.5)
1 1
–s
–s
Y (x, s)= x , B(s)= K(t)t dt. (23)
1+ B(s) 0
This, by means of formula (20), yields the solution of Eq. (21) for an arbitrary right-hand side:
1 c+i∞ ˆ f(s)
–s
y(x)= x ds, (24)
2πi c–i∞ 1+ B(s)
where ˆ f(s) is the Mellin transform (19) of the function f(x).

9.6-5. The Model Solution in the Case of a Sine-Shaped Right-Hand Side
Suppose that we have succeeded in finding a model solution Y = Y (x, u) that corresponds to the
sine on the right-hand side:
L [Y (x, u)] = sin(ux), λ = u. (25)
ˇ
Let f s (u) be the asymmetric sine Fourier transform of the function f(x):
∞

ˇ
f s (u)= F s {f(x)}, F s {f(x)}≡ f(x) sin(ux) dx. (26)
0
The solution of Eq. (1) for an arbitrary right-hand side f(x) can be expressed via the solution of the
simpler auxiliary equation with sine-shape right-hand side (25) by the formula

2 ∞
ˇ
y(x)= Y (x, u)f s (u) du. (27)
π 0

9.6-6. The Model Solution in the Case of a Cosine-Shaped Right-Hand Side
Suppose that we have succeeded in finding a model solution Y = Y (x, u) that corresponds to the
cosine on the right-hand side:

L [Y (x, u)] = cos(ux), λ = u. (28)
ˇ
Let f c (u) be the asymmetric Fourier cosine transform of the function f(x):

∞
ˇ
f c (u)= F c {f(x)}, F c {f(x)}≡ f(x) cos(ux) dx. (29)
0
The solution of Eq. (1) for an arbitrary right-hand side f(x) can be expressed via the solution of the
simpler auxiliary equation with cosine right-hand side (28) by the formula
2 ∞
ˇ
y(x)= Y (x, u)f c (u) du. (30)
π
0

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 482