Page 234 - Handbook Of Integral Equations

a

2. cosh(βx) – cosh(µt) y(t) dt = f(x), β >0, µ >0.

0
This is a special case of equation 3.8.4 with g(x) = cosh(βx) and λ = µ/β.
b

k k
3. cosh x – cosh t| y(t) dt = f(x), 0 < k <1.
a
k
This is a special case of equation 3.8.3 with g(x) = cosh x.
Solution:
1 d f (x)

x
y(x)= .
2k dx sinh x cosh k–1 x
The right-hand side f(x) of the integral equation must satisfy certain relations (see item 2 ◦
of equation 3.8.3).
b
y(t)
4. dt = f(x), 0 < k <1.
a |cosh(λx) – cosh(λt)| k
This is a special case of equation 3.8.7 with g(x) = cosh(λx)+ β, where β is an arbitrary
number.

3.3-2. Kernels Containing Hyperbolic Sine
b

5. sinh λ|x – t| y(t) dt = f(x), –∞ < a < b < ∞.
a
1 . Let us remove the modulus in the integrand:
◦
x b
sinh[λ(x – t)]y(t) dt + sinh[λ(t – x)]y(t) dt = f(x). (1)
a x
Differentiating (1) with respect to x twice yields
x b

2λy(x)+ λ 2 sinh[λ(x – t)]y(t) dt + λ 2 sinh[λ(t – x)]y(t) dt = f (x). (2)
xx
a x
Eliminating the integral terms from (1) and (2), we obtain the solution
2

y(x)= 1 f (x) – λ f(x) . (3)

xx
2λ
2 . The right-hand side f(x) of the integral equation must satisfy certain relations. By setting
◦
x = a and x = b in (1), we obtain two corollaries
b b

sinh[λ(t – a)]y(t) dt = f(a), sinh[λ(b – t)]y(t) dt = f(b). (4)
a a
Substituting solution (3) into (4) and integrating by parts yields the desired conditions for f(x):

sinh[λ(b – a)]f (b) – λ cosh[λ(b – a)]f(b)= λf(a),
x
(5)

sinh[λ(b – a)]f (a)+ λ cosh[λ(b – a)]f(a)= –λf(b).
x
The general form of the right-hand side is given by
f(x)= F(x)+ Ax + B, (6)
where F(x) is an arbitrary bounded twice differentiable function, and the coefficients A and B

are expressed in terms of F(a), F(b), F (a), and F (b) and can be determined by substituting
x
x
formula (6) into conditions (5).
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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