Mathematical Methods for Physics and Engineering - Matematica.NET

25.6 STOKES’ EQUATION AND AIRY INTEGRALScontour integral (25.35) with the particular solution of Stokes’ equation thatdecays monotonically to zero for real z>0as|z| →∞. As discussed in subsection25.6.1, all solutions except the one called Ai(z) tendto±∞ as z (real) takes onincreasingly large positive values and so their asymptotic forms reflect this. In aworked example in subsection 25.8.2 we use the method of steepest descents (asaddle-point method) to show that the function defined by (25.39) has exactlythe characteristic asymptotic property expected of Ai(z) (see page 911). It followsthat it is the same function as Ai(z), up to a real multiplicative constant.The choice of definition (25.36) as the other named solution Bi(z) ofStokes’equation is a less obvious one. However, it is made on the basis of its behaviour fornegative real values of z. As discussed earlier, Ai(z) oscillates almost sinusoidallyin this region, except for a relatively slow increase in frequency and an evenslower decrease in amplitude as −z increases. The solution Bi(z) is chosen to bethe particular function that exhibits the same behaviour as Ai(z) exceptthatitisin quadrature with Ai, i.e. it is π/2 out of phase with it. Specifically, as x →−∞,Ai(x) ∼Bi(x) ∼(1 2|x|3/2√ sin + π ), (25.40)2πx1/4 3 4(1 2|x|3/2√ cos + π2πx1/4 3 4). (25.41)There is a close parallel between this choice and that of taking sine and cosinefunctions as the basic independent solutions of the simple harmonic oscillatorequation. Plots of Ai(z) andBi(z) forrealz are shown in figure 25.11.◮By choosing a suitable contour for C 1 in (25.35), express Ai(0) in terms of the gammafunction.With z set equal to zero, (25.35) takes the formAi(0) = 1 ∫exp(− 1 32πit3 ) dt.C 1We again use the freedom to choose the specific line of the contour so as to make theactual integration as simple as possible.HereweconsiderC 1 as made up of two straight-line segments: one along the linearg t =4π/3, starting at infinity in the correct sector and ending at the origin; the otherstarting at the origin and going to infinity along the line arg t =2π/3, thus ending in thecorrect final sector. On each, we set 1 3 t3 = s, wheres is real and positive on both lines.Then dt = e 4πi/3 (3s) −2/3 ds on the first segment and dt = e 2πi/3 (3s) −2/3 ds on the second.893