Mathematical Methods for Physics and Engineering - Matematica.NET

APPLICATIONS OF COMPLEX VARIABLESwhich there is no obvious solution for a general n(x). However, on the assumptionthat d 2 φ/dx 2 is small, an iterative solution can be found.As a first approximation φ ′′ is ignored, and the solutiondφdx ≈±k 0n(x)is obtained. From this, differentiation gives an approximate value ford 2 φdx 2 ≈±k dn0dx ,which can be substituted into equation (25.47) to give, as a second approximationfor dφ/dx, the expression[] 1/2dφdx ≈± k0n 2 2 dn(x) ± ik 0dx(= ± k 0 n 1 ± i dn···)2k 0 n 2 dx +≈±k 0 n + i dn2n dx .This can now be integrated to give an approximate expression for φ(x) as follows:∫ xφ(x) =± k 0 n(u) du + i ln[ n(x)], (25.48)x 02where the constant of integration has been formally incorporated into the lowerlimit x 0 of the integral. Now, noting that exp(i 1 2 i ln n) =n−1/2 , substitution of(25.48) into equation (25.46) givesy ± (x) =A [ ∫ x]n exp ± ik 1/2 0 n(u) du(25.49)x 0as two independent WKB solutions of the original equation (25.42). This result isessentially the same as that in (25.45) except that the amplitude has been dividedby √ n(x), i.e. by [ f(x)] 1/4 .Sincef(x) may be complex, this may introduce anadditional x-dependent phase into the solution as well as the more obvious changein amplitude.◮Find two independent WKB solutions of Stokes’ equation in the formd 2 y+ λxy =0, with λ real and > 0.dx2 The form of the equation is the same as that in (25.42) with f(x) =x, andthereforen(x) =x 1/2 . The WKB solutions can be read off immediately using (25.49), so long aswe remember that although f(x) is real, it has four fourth roots and that therefore theconstant appearing in a solution can be complex. Two independent WKB solutions arey ± (x) = A ±|x| 1/4 exp [± i √ λ∫ x√ udu]= A ±|x| 1/4 exp [± i 2√ λ3 x3/2 ].898(25.50)