Chap . 5 Series Solutions of ODEs Special Functions This chapter introduces you to higher functions that are needed in physics and its engineering applications. Just as you could not do problems in calculus without knowing trigonometric and exponential functions, you could not solve important models involving ordinary or partial differential equations without knowing Bessel functions and Legendre polynomials, to mention just the two most prominent classes of these functions. Although these and the many other special functions of practical interest have quite different properties and serve very distinct purposes, it is most remarkable that these functions are accessible by the same instrument, namely, power series , (whose simplest examples you know from calculus), perhaps multiplied by a fractional power or a logarithm. You should see and learn which of the many functions and series are of importance in applications and what methods are available for investigating properties through series and for discovering relations between functions, in addition to the calculation of values by series and representing the results graphically by curves. Such an overview will help you to find your way through the vast number of books and articles in journals related to special functions when you need help in solving specific engineering or other problems. Your CAS knows all the functions that you will ever need, but it is your task to establish a general orientation in this wide field of relationships and formulas, in order to find and select what functions and relations you need for specific tasks and situations. Sec . 5.1 Power Series Method Problem Set 5.1. Page 170 5. Terminating power series . The ODE 2 + x y′ = y can be solved by separating variables, dy dx = , ln |y| = ln |x + 2 | + c, y = c̃ x + 2. y x + 2 The solution is a polynomial, a terminating power series, and you will see why, and how termination occurs. Substitute the power series for y and y′ into the given ODE to get 2 2 2 + xa1 + 2a2 x + 3a3 x + ⋯ = a0 + a1 x + a2 x + ⋯. Write the left side as a power series by multiplying out and ordering, 2 2a1 + 4a2 x + 6a3 x + ⋯ 2 3 + a1 x + 2a2 x + 3a3 x + ⋯ 2 = 2a1 + 4a2 + a1 x + 6a3 + 2a2 x + ⋯. Now compare powers on both sides: a0 a0 = 2a1 hence a1 = 2 a1 = 4a2 + a1 a2 = 0 a2 = 6a3 + 2a2 a3 = 0, etc. and you see how it happens that the power series terminates after two terms. a0 remains arbitrary, so that x you have obtained a general solution of your first-order ODE, namely, y = a0 1 + 2 . 13. Verhulst ODE , initial value . y′ = y − y2 = y1 − y has the general solution (see (9) on p. 31 in Sec. 1.5 with A = B = 1) y = 1/Ce−t + 1, Chap. 5 Series Solutions of ODEs. Special Functions 41 and the initial value 1/2 gives the solution (C = 1) y = 1/e−t + 1. 1 1 You see that you can introduce y0 = 2 into the power series, obtaining y = 2 + a1 x + ⋯, so that also 1 1 − y = 2 − a1 x − ⋯ in the other factor on the right. Hence by substitution, 2 3 4 a1 + 2a2 x + 3a3 x + 4a4 x + 5a5 x + ⋯ 1 2 3 1 2 3 = + a1 x + a2 x + a3 x + ⋯ a1 x a2 x a3 x ⋯ . 2 2 − − − − 0 1 1 1 1 Compare like powers on both sides. x gives a1 = 2 − 4 = 4 . Then x gives 1 1 1 2a2 = a0 a1 + a1 a0 a1 = + = 0, hence a2 = 0. − − 8 4 − 8 x2 gives 2 1 1 3a3 = a0 a2 + a + a0 1 a0 = , hence a3 = . − 1 − − 16 − 48 x3 gives 1 1 4a4 = a3 + a3 = 0, hence a4 = 0. − 2 2 x4 gives 1 1 1 5a5 = a4 a1 a3 + a2 a2 + a3 a1 + a4 = , 2 − − − − − 4 96 hence a5 = 1/480, as given on p. A13. The value s1 = 0.73125 (the sum of the coefficients just calculated!) is exact to 3D, and the value exact to 5D is 1/e−1 + 1 = 0.73106. This calculation shows that for a nonlinear ODE, working with power series is inconvenient . Indeed, for ODEs these series find their main application in connection with linear ODEs; and all the ODEs we shall now consider in this chapter will be linear. Sec . 5.2 Theory of the Power Series Method Problem Set 5.2. Page 176 7. Radius of convergence . A power series in powers of x may converge for all x (this is the best possible case) or within an interval with the center x0 as midpoint (in the complex plane: within a disk with center x0) or only at the center (the practically useless case). In the second case the interval of convergence has length 2R, where R is called the radius of convergence (it is a radius in the complex case, as has just been said) and is given by (7a) or (7b) of Sec. 5.2. Here it is assumed that the limits in these formulas exist. This will be the case in most applications. (For help when this is not the case, see Sec. 15.2.) The convergence radius is important whenever you want to use series for computing values, exploring properties of functions represented by series, or proving relations between functions, tasks of which you will gain a first impression in Secs. 5.3-5.7 and corresponding problems. In Prob. 7 you may set 2 m x − 1 = t. Then you have a power series in t with coefficients of absolute value |am| = 1/4 . Hence the root in (7a) is 1/4, so that the radius of convergence of the power series in t is 4. That is, the series converges for |t| < 4. This implies |x − 1| = |t| < 2. Hence the given series has the convergence radius 2. Confirm this by using (7b). The quotient in (7b) is |am+1/am| = 1/4. This leads to the same result as before. Note that the problem is special; in general, the sequences of those roots and quotients in (7) will not be constant, that is, the terms of such a sequence of quotients (or roots) will not be all the same. 42 Ordinary Differential Equations (ODEs) Part A 17. General solution . Whittaker functions . A general solution of a second-order ODE involves two arbitrary constants. In power series solutions these are two coefficients which are left arbitrary and to which the other coefficients are recursively related. Usually these are a0 and a1. In simpler cases the coefficients a2, a4, ⋯ are then related to a0, and the coefficients a3, a5, ⋯ of the odd powers are related to a1. For instance, this is the situation in the case of a power series solution of y′′ + y = 0 in which you obtain y = a0 cos x + a1 sin x. However, it may very well happen that each of the two functions of a basis of solutions involves both even and odd powers. This is the case for our present ODE y′′ − y′ + x2 y = 0 2 as you will see. If you ask your CAS, it will tell you that solutions of this ODE are Mk,mix and 2 x/2 Wk,mix , i = −1 , multiplied by e / x . From this you conclude that these functions are special functions that have been investigated, so that if you need to know some of their properties, you need not start from scratch but can make a search in the literature, notably, in Ref [GR1] in Appendix 1 of AEM, where you will find on p. 1045 in the Index of Notations that Mκ,μz and Wκ,μz are confluent hypergeometric functions , called Whittaker functions (W suggests Whittaker) and which are discussed on p. 505 of Ref. [GR1]. You obtain the coefficients of the power series as given on p. A13 by substituting the usual expressions for y′′, y′ and y, obtaining 2 3 4 5 2a2 + 3 ∙ 2a3 x + 4 ∙ 3a4 x + 5 ∙ 4a5 x + 6 ∙ 5a6 x + 7 ∙ 6a7 x + ⋯ 2 3 4 5 − a1 − 2a2 x − 3a3 x − 4a4 x − 5a5 x − 6a6 x − ⋯ 2 3 4 5 + a0 x + a1 x + a2 x + a3 x + ⋯ = 0. Now equate the sum of the coefficients of each occurring power of x to 0: 0 1 x : a2 = a1, a1 arbitrary 2 1 1 1 x : 2a2 = 3 ∙ 2a3, a3 = a2 = a1 3 6 2 1 1 1 x : a0 3a3 + 12a4 = a0 a1 + 12a4 = 0, a4 = a0 + a1 − − 2 − 12 24 3 1 1 x : a1 4a4 + 20 a5 = a1 4 a0 + a1 + 20a5 − − − 12 24 1 5 1 1 = a0 + a1 + 20a5 = 0 a5 = a0 a1. 3 6 − 60 − 24 With these coefficients depending on both a0 and a1, which remain arbitrary, you can now make up the two series as given on p. A13. A revised edition of [GR1] will appear in 2006 as the Digital Library of Mathematical Functions (http://dlmf.nist.gov). Sec . 5.3 Legendre ’s Equation . Legendre Polynomials Pnx Note well that Legendre’s equation involves the parameter n, so that (1) is actually a whole family of ODEs, with basically different properties for different n. In particular, for integer n = 0,1,2,⋯ one of the series (6) or (7) reduces to a polynomial, and it is remarkable that these “simplest” cases are of particular interest in applications. Chap. 5 Series Solutions of ODEs. Special Functions 43 Problem Set 5.3. Page 180 5. Legendre functions for n = 0. The power series and Frobenius methods were instrumental in establishing large portions of the very extensive theory of special functions (see, for instance, Refs.