Page 465 - Handbook Of Integral Equations

Applying the Laplace transform, we obtain (see Supplement 4)
2
2
˜ y(p)= √ p 3/2 exp(α/p) ˜ f(p), α = 1 4 k . (10)
πk
Let us rewrite the right-hand side of (10) in the equivalent form
2 2
–1/2 1 2
˜ y(p)= √ p p exp(α/p) ˜ f(p), α = k , (11)
πk 4
2
where the factor in the square brackets corresponds to ˜ M(p) in formula (6) and ˜ N(p) = const p .
By applying the Laplace inversion formula according to the above scheme to formula (11) with regard to the relation
(see Supplement 5)
d 2
1 √
2
L –1 p ˜ϕ(p) = ϕ(x), L –1 p –1/2 exp(α/p) = √ cosh k x ,

dx 2 πx
we find the solution √
2 d 2 x cosh k x – t
y(x)= √ f(t) dt.
πk dx 2 0 x – t
8.4-4. Application of an Auxiliary Equation
Consider the equation
x

K(x – t)y(t) dt = f(x), (12)
a
where the kernel K(x) has an integrable singularity at x =0.
Let w = w(x) be the solution of the simpler auxiliary equation with f(x) ≡ 1 and a =0,
x
K(x – t)w(t) dt = 1. (13)
0
Then the solution of the original equation (12) with arbitrary right-hand side can be expressed as
follows via the solution of the auxiliary equation (13):
d x x

y(x)= w(x – t)f(t) dt = f(a)w(x – a)+ w(x – t)f (t) dt. (14)
t
dx
a a
Example 3. Consider the generalized Abel equation
x y(t) dt

= f(x), 0 < µ <1. (15)
a (x – t) µ
We seek a solution of the corresponding auxiliary equation
x w(t) dt
=1, 0 < µ <1, (16)
0 (x – t) µ
by the method of indeterminate coefficients in the form
β
w(x)= Ax . (17)
Let us substitute (17) into (15) and then perform the change of variable t = xξ in the integral. Taking into account the
relationship
1 Γ(p)Γ(q)
B(p, q)= ξ p–1 (1 – ξ) 1–q dξ =
0 Γ(p + q)
between the beta and gamma functions, we obtain
Γ(β +1)Γ(1 – µ)
A x β+1–µ =1.
Γ(2 + β – µ)
From this relation we find the coefficients A and β:
1 sin(πµ)
β = µ – 1, A = = . (18)
Γ(µ)Γ(1 – µ) π
Formulas (17) and (18) define the solution of the auxiliary equation (16) and make it possible to find the solution of the
generalized Abel equation (15) by means of formula (14) as follows:

sin(πµ) d x f(t) dt sin(πµ) f(a) x f (t) dt
y(x)= = + t . (19)
π dx a (x – t) 1–µ π (x – a) 1–µ a (x – t) 1–µ

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 447