Mathematical Methods for Physics and Engineering - Matematica.NET

APPLICATIONS OF COMPLEX VARIABLESexp[± iφ(z) ] will be purely real. What is more, one such factor, known as thedominant term, will be exponentially large, whilst the other (the subdominant term)will be exponentially small. A Stokes line is precisely where this happens.We can now see how the change takes place without an observable discontinuityoccurring. Suppose that y 1 (z) is very large and y 2 (z) is very small on a Stokes line.Then a finite change in A 2 will have a negligible effect on Y (z); in fact, Stokesshowed, for some particular cases, that the change is less than the uncertaintyin y 1 (z) arising from the approximations made in deriving it. Since the solutionwith any particular asymptotic form is determined in a region bounded by twoStokes lines to within an overall multiplicative constant and the original equationis linear, the change in A 2 when one of the Stokes lines is crossed must beproportional to A 1 ,i.e.A 2 changes to A 2 + SA 1 ,whereS is a constant (the Stokesconstant) characteristic of the particular line but independent of A 1 and A 2 .Itshould be emphasised that, at a Stokes line, if the dominant term is not presentin a solution, then the multiplicative constant in the subdominant term cannotchange as the line is crossed.As an example, consider the Bessel function J 0 (z) ofzeroorder.Itissinglevalued,differentiable everywhere, and can be written as a series in powers of z 2 .Itis therefore an integral even function of z. However, its asymptotic approximationsfor two regions of the z-plane, Re z>0andz real and negative, are given byJ 0 (z) ∼ √ 1 1 √z(e iz e −iπ/4 + e −iz e iπ/4) , | arg(z)| < 12π2 π, | arg(z−1/2 )| < 1 4 π,J 0 (z) ∼ 1 √2π1 √z(e iz e 3iπ/4 + e −iz e iπ/4) , arg(z) =π, arg(z −1/2 )=− 1 2 π.We note in passing that neither of these expressions is naturally single-valued,and a prescription for taking the square root has to be given. Equally, neither isan even function of z. For our present purpose the important point to note isthat, for both expressions, on the line arg z = π/2 both z-dependent exponentsbecome real. For large |z| the second term in each expression is large; this is thedominant term, and its multiplying constant e iπ/4 is the same in both expressions.Contrarywise, the first term in each expression is small, and its multiplyingconstant does change, from e −iπ/4 to e 3iπ/4 ,asargz passes through π/2 whilstincreasing from 0 to π. It is straightforward to calculate the Stokes constant forthis Stokes line as follows:S = A 2(new) − A 2 (old)A 1= e3iπ/4 − e −iπ/4e iπ/4= e iπ/2 − e −iπ/2 =2i.If we had moved (in the negative sense) from arg z = 0 to arg z = −π, the relevantStokes line would have been arg z = −π/2. There the first term in each expressionis dominant, and it would have been the constant e iπ/4 in the second term thatwould have changed. The final argument of z −1/2 would have been +π/2.904