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We obtain the solution by formula (3) taking into account Eq. (5):
2 ∞ 2
y(x)= f(x)+ λ cos(xt)f(t) dt , λ ≠ ± . (6)
2 – πλ 2 0 π
◦
2 . Consider the equation
∞
y(x) – λ tJ ν (xt)y(t) dt = f(x), (7)
0
1
where J ν (x) is the Bessel function, Re ν > – .
2
Here the operator L coincides, up to a constant factor, with the Hankel transform:
∞
L [y]= tJ ν (xt)y(t) dt (8)
0
2
and acts by the rule L = 1 (see Subsection 7.6-1).
We obtain the solution by formula (3), for k = 1, taking into account Eq. (8):
∞
1
y(x)= f(x)+ λ tJ ν (xt)f(t) dt , λ ≠ ±1. (9)
1 – λ 2 0
•
Reference for Section 11.7: A. D. Polyanin and A. V. Manzhirov (1998).
11.8. Methods of Integral Transforms and Model
Solutions
11.8-1. Equation With Difference Kernel on the Entire Axis
Consider an integral equation of convolution type of the second kind with one kernel
∞
1
y(x)+ √ K(x – t)y(t) dt = f(x), –∞ < x < ∞, (1)
2π –∞
where f(x) and K(x) are the known right-hand side and the kernel of the integral equation and y(x)
is the unknown function. Let us apply the (alternative) Fourier transform to Eq. (1). In this case,
taking into account the convolution theorem (see Subsection 7.4-4), we obtain
Y(u)[1 + K(u)] = F(u). (2)
Thus, on applying the Fourier transform we reduce the solution of the original integral equation (1)
to the solution of the algebraic equation (2) for the transform of the unknown function. The solution
of Eq. (2) has the form
F(u)
Y(u)= . (3)
1+ K(u)
Formula (3) gives the transform of the solution of the original integral equation in terms of the
transforms of the known functions, namely, the kernel and the right-hand side of the equation. The
solution itself can be obtained by applying the Fourier inversion formula:
∞ ∞
1 –iux 1 F(u) –iux
y(x)= √ Y(u)e du = √ e du. (4)
2π –∞ 2π –∞ 1+ K(u)
In fact, formula (4) solves the problem; however, sometimes it is not convenient because it
requires the calculation of the transform F(u) for each right-hand side f(x). In many cases, the
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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