Page 308 - Handbook Of Integral Equations

2 . With the 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 and taking into account (3), we find that

n n
2

y (x)+ σ n y(x) – A k λ I k = f (x), σ n =2 A k λ k . (5)
xx
k
xx
k=1 k=1
Eliminating the integral I n from (4) and (5) yields
n–1
2 2 2
2

y (x)+(σ n + λ )y(x)+ A k (λ – λ )I k = f (x)+ λ f(x). (6)
xx
n
n
xx
n
k
k=1
Differentiating (6) with respect to x twice and eliminating I n–1 from the resulting equation
with the aid of (6), we obtain a similar equation whose left-hand side is a second-order linear
n–2

differential operator (acting on y) with constant coefficients plus the sum B k I k .If we
k=1
successively eliminate I n–2 , I n–3 , ... , with the aid of double differentiation, then we finally
arrive at a linear nonhomogeneous ordinary differential equation of order 2n with constant
coefficients.
◦
3 . The boundary conditions for y(x) can be found by setting x = a in the integral equation
and all its derivatives. (Alternatively, these conditions can be found by setting x = a and x = b
in the integral equation and all its derivatives obtained by means of double differentiation.)

∞ sin(x – t)
33. y(x) – λ y(t) dt = f(x).
x – t
–∞
Solution:
∞
λ sin(x – t) 2
y(x)= f(x)+ √ f(t) dt, λ ≠ .
2π – πλ –∞ x – t π
•
Reference: F. D. Gakhov and Yu. I. Cherskii (1978).
4.5-3. Kernels Containing Tangent
b

34. y(x) – λ tan(βx)y(t) dt = f(x).
a
This is a special case of equation 4.9.1 with g(x) = tan(βx) and h(t)=1.
b
35. y(x) – λ tan(βt)y(t) dt = f(x).
a
This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = tan(βt).

b
36. y(x) – λ [A tan(βx)+ B tan(βt)]y(t) dt = f(x).
a
This is a special case of equation 4.9.4 with g(x) = tan(βx).

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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