Page 172 - Handbook Of Integral Equations

Case 2.If z 1 < 0 and z 2 < 0, then the solution of the integral equation has the form

x

y(x)= f(x)+ {B 1 sin[µ 1 (x – t)] + B 2 sin µ 2 (x – t) f(t) dt, µ i = |z i |,
a
where B 1 and B 2 are determined from the system

B 1 µ 1 B 2 µ 2 B 1 µ 1 B 2 µ 2
+ – 1=0, + – 1=0.
2
2
2
2
λ – µ 2 λ – µ 2 λ – µ 2 λ – µ 2
1 1 1 2 2 1 2 2
Case 3.If z 1 > 0 and z 2 < 0, then the solution of the integral equation has the form
x

y(x)= f(x)+ {B 1 sinh[µ 1 (x – t)] + B 2 sin µ 2 (x – t) f(t) dt, µ i = |z i |,
a
where B 1 and B 2 are determined from the system
B 1 µ 1 B 2 µ 2 B 1 µ 1 B 2 µ 2
+ – 1=0, + – 1=0.
2
2
2
2
λ + µ 2 λ – µ 2 λ + µ 2 λ – µ 2
1 1 1 2 2 1 2 2
Remark. The solution of the original integral equation can be obtained from the solution
of equation 2.3.18 by performing the following change of parameters:
2
λ k → iλ k , µ k → iµ k , A k → –iA k , B k → –iB k , i = –1(k = 1, 2).
x n

19. y(x)+ A k sin[λ k (x – t)] y(t) dt = f(x).
a k=1
1 . This integral equation can be reduced to a linear nonhomogeneous ordinary differential
◦
equation of order 2n with constant coefficients. Set
x

I k (x)= sin[λ k (x – t)]y(t) dt. (1)
a
Differentiating (1) with respect to x twice yields

x x
2
k
k
I = λ k cos[λ k (x – t)]y(t) dt, I = λ k y(x) – λ k sin[λ k (x – t)]y(t) dt, (2)
a a
where the primes stand for differentiation with respect to x. Comparing (1) and (2), we see
that
2

I = λ k y(x) – λ I k , I k = I k (x). (3)
k k
With aid of (1), the integral equation can be rewritten in the form
n

y(x)+ A k I k = f(x). (4)
k=1
Differentiating (4) with respect to x twice taking into account (3) yields

n n
2

y (x)+ σ n y(x) – A k λ I k = f (x), σ n = A k λ k . (5)
k
xx
xx
k=1 k=1
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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