Page 50 - Handbook Of Integral Equations

x

41. A 1 sinh[λ 1 (x – t)] + A 2 sinh[λ 2 (x – t)] y(t) dt = f(x), f(a)= f (a)=0.
x
a
1 . Introduce the notation
◦
x x

I 1 = sinh[λ 1 (x – t)]y(t) dt, I 2 = sinh[λ 2 (x – t)]y(t) dt,
a a
x x

J 1 = cosh[λ 1 (x – t)]y(t) dt, J 2 = cosh[λ 2 (x – t)]y(t) dt.
a a
Let us successively differentiate the integral equation four times. As a result, we have (the
first line is the original equation):
A 1 I 1 + A 2 I 2 = f, f = f(x), (1)

A 1 λ 1 J 1 + A 2 λ 2 J 2 = f , (2)
x
2 2

(A 1 λ 1 + A 2 λ 2 )y + A 1 λ I 1 + A 2 λ I 2 = f , (3)
1 2 xx
3
3
(A 1 λ 1 + A 2 λ 2 )y + A 1 λ J 1 + A 2 λ J 2 = f , (4)

x 1 2 xxx
4
4
3
3
(A 1 λ 1 + A 2 λ 2 )y +(A 1 λ + A 2 λ )y + A 1 λ I 1 + A 2 λ I 2 = f . (5)
xx 1 2 1 2 xxxx
Eliminating I 1 and I 2 from (1), (3), and (5), we arrive at the following second-order linear
ordinary differential equation with constant coefficients:
2
2
2 2

(A 1 λ 1 + A 2 λ 2 )y xx – λ 1 λ 2 (A 1 λ 2 + A 2 λ 1 )y = f xxxx – (λ + λ )f xx + λ λ f. (6)
1 2
1
2
The initial conditions can be obtained by substituting x = a into (3) and (4):

(A 1 λ 1 + A 2 λ 2 )y(a)= f (a), (A 1 λ 1 + A 2 λ 2 )y (a)= f (a). (7)
xx x xxx
Solving the differential equation (6) under conditions (7) allows us to find the solution of the
integral equation.
2 . Denote
◦
A 1 λ 2 + A 2 λ 1
∆ = λ 1 λ 2 .
A 1 λ 1 + A 2 λ 2
2.1. Solution for ∆ >0:
x

(A 1 λ 1 + A 2 λ 2 )y(x)= f (x)+ Bf(x)+ C sinh[k(x – t)]f(t) dt,
xx
a
√ 1
2
2
2
2
2
2 2
k = ∆, B = ∆ – λ – λ , C = √ ∆ – (λ + λ )∆ + λ λ .
1 2 1 2 1 2
∆
2.2. Solution for ∆ <0:
x

(A 1 λ 1 + A 2 λ 2 )y(x)= f (x)+ Bf(x)+ C sin[k(x – t)]f(t) dt,
xx
a
√ 1
2
2
2
2 2
2
2
k = –∆, B = ∆ – λ – λ , C = √ ∆ – (λ + λ )∆ + λ λ .
1 2 1 2 1 2
–∆
2.3. Solution for ∆ =0:
x

2 2
2
2

(A 1 λ 1 + A 2 λ 2 )y(x)= f (x) – (λ + λ )f(x)+ λ λ (x – t)f(t) dt.
xx 1 2 1 2
a
2.4. Solution for ∆ = ∞:
2 2
2
2

f xxxx – (λ + λ )f xx + λ λ f

2
1
1 2
y(x)= 3 3 , f = f(x).
A 1 λ + A 2 λ
1 2
In the last case, the relation A 1 λ 1 + A 2 λ 2 = 0 is valid, and the right-hand side of the
integral equation is assumed to satisfy the conditions f(a)= f (a)= f (a)= f xxx (a)=0.

x
xx
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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