Page 430 - Handbook Of Integral Equations

b

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

y(x)+ e λ(x–t) f t, y(t) dt + e λ(t–x) f t, y(t) dt = g(x). (1)
a x
Differentiating (1) with respect to x twice yields
x b
2 λ(x–t) 2 λ(t–x)

y (x)+2λf x, y(x) +λ e f t, y(t) dt+λ e f t, y(t) dt = g (x). (2)
xx
xx
a 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 +2λf(x, y) – λ y = g (x) – λ g(x). (3)

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

y(a)+ e –λa e f t, y(t) dt = g(a),
a (4)
b

y(b)+ e λb e –λt f t, y(t) dt = g(b).
a
Let us solve equation (3) for f(x, y) and substitute the result into (4). Integrating by parts
yields
λb λa λa λb
e ϕ (b) – e ϕ (a)= λe ϕ(a)+ λe ϕ(b), ϕ(x)= y(x) – g(x);
x
x
e –λb ϕ (b) – e –λa ϕ (a)= λe –λa ϕ(a)+ λe –λb ϕ(b).

x
x
Hence, we obtain the boundary conditions for y(x):

ϕ (a)+ λϕ(a)=0, ϕ (b) – λϕ(b)=0; ϕ(x)= y(x) – g(x). (5)

x x
Equation (3) under the boundary conditions (5) determines the solution of the original
integral equation (there may be several solutions). Conditions (5) make it possible to calculate
the constants of integration that occur in solving the differential equation (3).
b

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

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

y (x)+2λf x, y(x) + λ 2 sinh[λ(x – t)]f t, y(t) dt

xx
a
b

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

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