Page 345 - Handbook Of Integral Equations

b

38. y(x)+ sinh(λ|x – t|)g(t)y(t) dt = f(x), a ≤ x ≤ b.
a
◦
1 . Let us remove the modulus in the integrand:

x b
y(x)+ sinh[λ(x – t)]g(t)y(t) dt + sinh[λ(t – x)]g(t)y(t) dt = f(x). (1)
a x
Differentiating (1) with respect to x twice yields

x

y (x)+2λg(x)y(x)+ λ 2 sinh[λ(x – t)]g(t)y(t) dt
xx
a
b

+ λ 2 sinh[λ(t – x)]g(t)y(t) dt = f (x). (2)

xx
x
Eliminating the integral terms from (1) and (2), we arrive at a second-order ordinary
differential equation for y = y(x),
2
2
y +2λg(x)y – λ y = f (x) – λ f(x). (3)

xx xx
◦
2 . Let us derive the boundary conditions for equation (3). We assume that the limits of
integration satisfy the conditions –∞ < a < b < ∞. By setting x = a and x = b in (1), we
obtain two corollaries
b

y(a)+ sinh[λ(t – a)]g(t)y(t) dt = f(a),
a
(4)
b
y(b)+ sinh[λ(b – t)]g(t)y(t) dt = f(b).
a

Let us express g(x)y from (3) via y and f xx and substitute the result into (4). Integrating
xx
by parts yields the desired boundary conditions for y(x),

sinh[λ(b – a)]ϕ (b) – λ cosh[λ(b – a)]ϕ(b)= λϕ(a),
x
(5)

sinh[λ(b – a)]ϕ (a)+ λ cosh[λ(b – a)]ϕ(a)= –λϕ(b); ϕ(x)= y(x) – f(x).
x
Equation (3) under the boundary conditions (5) determines the solution of the original
integral equation. Conditions (5) make it possible to calculate the constants of integration
that occur in solving the differential equation (3).

b
39. y(x)+ sin(λ|x – t|)g(t)y(t) dt = f(x), a ≤ x ≤ b.
a
1 . Let us remove the modulus in the integrand:
◦
x b

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

y (x)+2λg(x)y(x) – λ 2 sin[λ(x – t)]g(t)y(t) dt
xx
a

b

– λ 2 sin[λ(t – x)]g(t)y(t) dt = f (x). (2)
xx
x
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 324