Mathematical Methods for Physics and Engineering - Matematica.NET

INTEGRAL EQUATIONSSubstituting (23.31) into (23.30) we findy(x) =f(x)+˜f(x)+y(−x),but on changing x to −x and substituting back in for y(−x), this givesy(x) =f(x)+˜f(x)+f(−x)+˜f(−x)+y(x).Thus, in order for a solution to exist, we require that the function f(x) obeysf(x)+˜f(x)+f(−x)+˜f(−x) =0.This is satisfied if f(x) =−˜f(x), i.e. if the functional form of f(x) is minus the form of itsFourier transform. We may repeat this analysis for the case λ = −1/ √ 2π, and, in a similarway, we find that this time we require f(x) =˜f(x).In our case f(x) = exp(−x 2 /2), for which, as we mentioned above, f(x) = ˜f(x).Therefore, (23.28) possesses no solution when λ =+1/ √ 2π but has many solutions whenλ = −1/ √ 2π. ◭A similar approach to the above may be taken to solve equations with kernelsof the form K(x, y) =cosxy or sin xy, either by considering the integral over y ineach case as the real or imaginary part of the corresponding Fourier transformor by using Fourier cosine or sine transforms directly.23.4.3 DifferentiationA closed-form solution to a Volterra equation may sometimes be obtained bydifferentiating the equation to obtain the corresponding differential equation,which may be easier to solve.◮Solve the integral equationy(x) =x −∫ x0xz 2 y(z) dz. (23.32)Dividing through by x, weobtain∫y(x)xx=1− z 2 y(z) dz,0which may be differentiated with respect to x to give[ ][ ]d y(x)y(x)= −x 2 y(x) =−x 3 .dx xxThis equation may be integrated straightforwardly, and we find[ ] y(x)ln = − x4x 4 + c,where c is a constant of integration. Thus the solution to (23.32) has the form)y(x) =Ax exp(− x4, (23.33)4where A is an arbitrary constant.Since the original integral equation (23.32) contains no arbitrary constants, neithershould its solution. We may calculate the value of the constant, A, by substituting thesolution (23.33) back into (23.32), from which we find A =1.◭812