Mathematical Methods for Physics and Engineering - Matematica.NET

23.4 CLOSED-FORM SOLUTIONS23.4.2 Integral transform methodsIf the kernel of an integral equation can be written as a function of the differencex − z of its two arguments, then it is called a displacement kernel. An integralequation having such a kernel, and which also has the integration limits −∞ to∞, may be solved by the use of Fourier transforms (chapter 13).If we consider the following integral equation with a displacement kernel,y(x) =f(x)+λ∫ ∞−∞K(x − z)y(z) dz, (23.17)the integral over z clearly takes the form of a convolution (see chapter 13).Therefore, Fourier-transforming (23.17) and using the convolution theorem, weobtainỹ(k) =˜f(k)+ √ 2πλ ˜K(k)ỹ(k),which may be rearranged to give˜f(k)ỹ(k) =1 − √ 2πλ ˜K(k) . (23.18)Taking the inverse Fourier transform, the solution to (23.17) is given byy(x) = √ 1 ∫ ∞ ˜f(k) exp(ikx)2π −∞ 1 − √ 2πλ ˜K(k) dk.If we can perform this inverse Fourier transformation then the solution can befound explicitly; otherwise it must be left in the form of an integral.◮Find the Fourier transform of the function{1 if |x| ≤a,g(x) =0 if |x| >a.Hence find an explicit expression for the solution of the integral equation∫ ∞sin(x − z)y(x) =f(x)+λy(z) dz. (23.19)−∞ x − zFind the solution for the special case f(x) =(sinx)/x.The Fourier transform of g(x) is given directly by˜g(k) = √ 1 ∫ a[ ] a√1 exp(−ikx) 2 sin kaexp(−ikx) dx = √ = .2π −a2π (−ik)−aπ k(23.20)The kernel of the integral equation (23.19) is K(x − z) = [sin(x − z)]/(x − z). Using(23.20), it is straightforward to show that the Fourier transform of the kernel is˜K(k) ={√π/2 if |k| ≤1,0 if |k| > 1.(23.21)809