Page 160 - Handbook Of Integral Equations

x n

19. y(x)+ A k sinh[λ k (x – t)] y(t) dt = f(x).
a
k=1
◦
1 . This equation can be reduced to an equation of the form 2.2.19 with the aid of the formula
z
sinh z = 1 e – e –z . Therefore, the original integral equation can be reduced to a linear
2
nonhomogeneous ordinary differential equation of order 2n with constant coefficients.
2 . Let us find the roots z k of the algebraic equation
◦
n
λ k A k

+ 1 = 0. (1)
z – λ 2
k=1 k
By reducing it to a common denominator, we arrive at the problem of determining the roots
of an nth-degree characteristic polynomial.
Assume that all z k are real, different, and nonzero. Let us divide the roots into two groups
z 1 >0, z 2 >0, ... , z s > 0 (positive roots);
z s+1 <0, z s+2 <0, ... , z n < 0 (negative roots).

Then the solution of the integral equation can be written in the form
s n
x

y(x)=f(x)+ B k sinh µ k (x–t) + C k sin µ k (x–t) f(t) dt, µ k = |z k |. (2)
a
k=1 k=s+1
The coefficients B k and C k are determined from the following system of linear algebraic
equations:
s n
B k µ k C k µ k

+ + 1=0, µ k = |z k |, m =1, ... , n. (3)
2
λ – µ 2 λ + µ 2
2
k=0 m k k=s+1 m k
In the case of a nonzero root z s = 0, we can introduce the new constant D = B s µ s and
proceed to the limit µ s → 0. As a result, the term D(x – t) appears in solution (2) instead of
–2
B s sinh µ s (x – t) and the corresponding terms Dλ appear in system (3).
m
x sinh(λx)
20. y(x) – A y(t) dt = f(x).
a sinh(λt)
Solution:
x
sinh(λx)
y(x)= f(x)+ A e A(x–t) f(t) dt.
a sinh(λt)
x
sinh(λt)
21. y(x) – A y(t) dt = f(x).
a sinh(λx)
Solution:
x sinh(λt)
y(x)= f(x)+ A e A(x–t) f(t) dt.
a sinh(λx)
x

k
m
22. y(x) – A sinh (λx) sinh (µt)y(t) dt = f(x).
a
k m
This is a special case of equation 2.9.2 with g(x)= A sinh (λx) and h(t) = sinh (µt).

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 139