Mathematical Methods for Physics and Engineering - Matematica.NET

23.4 CLOSED-FORM SOLUTIONSso we can writey(x) =f(x)+ √ 2πλỹ(x). (23.25)If we now take the Fourier transform of (23.25) but continue to denote theindependent variable by x (i.e. rather than k, for example), we obtainỹ(x) =˜f(x)+ √ 2πλy(−x). (23.26)Substituting (23.26) into (23.25) we findy(x) =f(x)+ √ √ ]2πλ[˜f(x)+ 2πλy(−x) ,but on making the change x →−x and substituting back in for y(−x), this givesy(x) =f(x)+ √ [ 2πλ˜f(x)+2πλ 2 f(−x)+ √ ]2πλ˜f(−x)+2πλ 2 y(x) .Thus the solution to (23.24) is given byy(x) =1[1 − (2π) 2 λ 4 f(x)+(2π) 1/2 λ˜f(x)+2πλ 2 f(−x)+(2π) 3/2 λ 3˜f(−x) ].(23.27)Clearly, (23.24) possesses a unique solution provided λ ≠ ±1/ √ 2π or ±i/ √ 2π;these are easily shown to be the eigenvalues of the corresponding homogeneousequation (for which f(x) ≡ 0).◮Solve the integral equation) ∫ ∞y(x) =exp(− x2+ λ exp(−ixz) y(z) dz, (23.28)2−∞where λ is a real constant. Show that the solution is unique unless λ has one of two particularvalues. Does a solution exist for either of these two values of λ?Following the argument given above, the solution to (23.28) is given by (23.27) withf(x) =exp(−x 2 /2). In order to write the solution explicitly, however, we must calculatethe Fourier transform of f(x). Using equation (13.7), we find ˜f(k) =exp(−k 2 /2), fromwhich we note that f(x) has the special property that its functional form is identical tothat of its Fourier transform. Thus, the solution to (23.28) is given byy(x) =11 − (2π) 2 λ 4 [1+(2π) 1/2 λ +2πλ 2 +(2π) 3/2 λ 3] exp(− x22).(23.29)Since λ is restricted to be real, the solution to (23.28) will be unique unless λ = ±1/ √ 2π,at which points (23.29) becomes infinite. In order to find whether solutions exist for eitherof these values of λ we must return to equations (23.25) and (23.26).Let us first consider the case λ =+1/ √ 2π. Putting this value into (23.25) and (23.26),we obtainy(x) =f(x)+ỹ(x), (23.30)ỹ(x) =˜f(x)+y(−x). (23.31)811