18 Series Solution and Special Functions 18.1 INTRODUCTION Generally the solutions of ordinary differential equations are obtainable in explicit form called a closed form of the solution. However, many differential equations arising in physical problems are linear but have variable coefficients and do not permit a general solution in terms of known functions. For such equations, it is easier to find a solution in the form of an infinite convergent series called power series solution. The series solution of certain differential equations give rise to special functions such as Bessel’s functions, Legendre’s polynomials, Lagurre’s polynomial, Hermite’s polynomial, Chebyshev polynomials. Strum-Liovelle problem based on orthogonality of functions is also included which shows that Bessel’s, Legendre’s and other equations can be determined from a common point of view. 18.2 POWER SERIES SOLUTION OF DIFFERENTIAL EQUATIONS Consider the differential equation 푑2푦 푑푦 푃 푥 + 푃 푥 + 푃 푥 푦 = 0 … (1) 0 푑푥 2 1 푑푥 2 where 푃푖′푠 are polynomials in 푥. If 푃0 푎 ≠ 0, then 푥 = 푎 is called an ordinary point of (1), otherwise a singular point. Ordinary point is also called a regular point of the equation. A singular point 푥 = 푎 of (1) is called regular singular point if, (1) can be put in the form 푑2푦 푄 푥 푑푦 푄 푥 + 1 + 2 푦 = 0 … (2) 푑푥 2 푥−푎 푑푥 푥−푎 2 provided 푄1 푥 and 푄2 푥 both possess derivatives of all orders in the neighborhood of 푎. A singular point which is not regular is called an irregular singular point. Note: The power series method sometimes fails to yield a solution e.g. 푥2푦′′ + 푥푦′ + 푦 = 0 …(3) dividing by 푥2 throughout, 푥2푦′′ + 푥푦′ + 푦 = 0 …(4) 1 1 Here neither of the terms 푃 푥 = and 푃 푥 = is defined at 푥 = 0, so we cannot find a 1 푥 2 푥 2 power series representation for 푃1 푥 or 푃2 푥 that converges in an open interval containing 푥 = 0. Theorem I: If 푥 = 푎 is an ordinary point of the differential equation (1), i.e. 푃0 푎 ≠ 0, then series solution of (1) can be found as: 2 푦 = 푎0 + 푎1(푥 − 푎) + 푎2(푥 − 푎) + ⋯ … (5) 1 푑푦 푑2푦 Calculate the derivatives , from (5), and substitute the values of y and its derivatives in 푑푥 푑푥 2 differential equation (1). The values of the constants 푎2, 푎3, 푎4, … are obtained by equating to zero the coefficients of various powers of 푥. Putting the values of these constants in the solution (5), the desired power series solution of (1) is obtained with 푎0, 푎1 as its arbitrary constants. Theorem II: When 푥 = 푎 is a regular singularity of (1) at least one of the solutions can be expressed as, 푚 2 푦 = (푥 − 푎) [푎0 + 푎1(푥 − 푎) + 푎2(푥 − 푎) + ⋯ ] …(6) Theorem III: The series (5) and (6) are convergent at every point within the circle of convergence at 푎. A solution in series will be valid only if the series is convergent. 풅ퟐ풚 Example 1: Solve in series the equation − 풙풚 = ퟎ. 풅풙ퟐ Solution: Given differential equation is 푑2푦 − 푥푦 = 0 … (1) 푑푥 2 Here 푃0 푥 = 1, so 푃0 0 = 1, i.e. 푥 = 0 is the ordinary point of the differential equation (1). Let the solution of differential equation (1) be 2 3 4 5 푦 = 푎0 + 푎1푥 + 푎2푥 + 푎3푥 + 푎4푥 + 푎5푥 + ⋯ … (2) Differentiating (2) w.r.t. 푥, 푑푦 = 푎 + 2푎 푥 + 3푎 푥2 + 4푎 푥3 + 5푎 푥4 + ⋯ … (3) 푑푥 1 2 3 4 5 Again differentiating w.r.t 푥 푑2푦 = 2푎 + 6푎 푥 + 12푎 푥2 + 20푎 푥3 + ⋯ … (4) 푑푥 2 2 3 4 5 Substitute values of y from (2) and its derivative from (4) in the differential equation (1), we get 2 3 2푎2 + 6푎3푥 + 12푎4푥 + 20푎5푥 + ⋯ 2 3 4 5 −푥 푎0 + 푎1푥 + 푎2푥 + 푎3푥 + 푎4푥 + 푎5푥 + ⋯ = 0 2 3 => 2푎2 + 6푎3 − 푎0 푥 + 12푎4 − 푎1 푥 + 20푎5 − 푎2 푥 + ⋯ = 0 Equating each of the coefficients to zero, we obtain the identities, 2푎2 = 0, 6푎3 − 푎0 = 0, 12푎4 − 푎1 = 0, 20푎5 − 푎2 = 0 1 1 1 which further gives 푎 = 0, 푎 = 푎 , 푎 = 푎 , 푎 = 푎 = 0 2 3 6 0 4 12 1 5 20 2 푎 Generalizing the results, 푎 = 푛−1 … (5) 푛+2 푛+2 (푛+1) 2 Putting 푛 = 4, 5, 6 … in (5), we get 1 1 1 푎 = 푎 = 푎 = 푎 , 6 6 (5) 3 6 6 (5) 0 180 0 1 1 1 푎 = 푎 = 푎 = 푎 , 7 7 (6) 4 12 7 (6) 1 504 1 푎8 = 0. Using the values of the constants in (2), the general solution of differential equation (1) becomes 1 1 1 1 푦 = 푎 1 + 푥3 + 푥6 + ⋯ + 푎 푥 + 푥4 + 푥7 + ⋯ . 0 6 180 1 12 504 Example 2: ASSIGNMENT 18.1 Solve the following differential equations in series 푑2푦 푑푦 1. + 푥 + 푦 = 0. 푑푥 2 푑푥 푑2푦 2. + 푥푦 = 0. 푑푥 2 푑2푦 푑푦 3. 1 − 푥2 − 푥 + 4푦 = 0. 푑푥 2 푑푥 푑2푦 4. + 푦 = 0, given 푦 0 = 0. 푑푥 2 5. 1 − 푥2 푦′′ + 2푦 = 0, 푔푖푣푒푛 푦 0 = 4, 푦′ 0 = 5. ANSWERS 푥 2 푥4 푥 6 푥 3 푥 5 푥 7 1. 푦 = 푎 1 − + − + ⋯ + 푎 푥 − + − + ⋯ 0 2 2.4 2.4.6 1 3 3.5 3.5.7 1 1.4 1∙4.7 2. 푦 = 푎 1 − 푥3 + 푥6 − 푥9 … 0 3! 6! 9! 1.2 2.7 +푎 푥 − 푥4 + 푥7 + ⋯ 1 4! 7! 푥 2 푥 4 3 푥 6 5∙3 푥 8 3. 푦 = 푎 1 − 2푥2 + 푎 푥 1 − − − ∙ − ∙ − ⋯ 0 1 2 8 6 8 8∙6 8 푥 3 푥 5 4. 푦 = 푎 푥 − + − ⋯ 0 3! 5! 5 푥 5 푥 7 5. 푦 = 4 + 5푥 − 4푥2 − 푥3 − − − ⋯ 3 3 7 3 18.3 FROBENIUS METHOD This method is named after a German mathematician F.G. Frobenius (1849 – 1917) who is known for his contributions to the theory of matrices and groups. This method is employed to find the power series solution of the differential equation 푑2푦 푑푦 푃 푥 + 푃 푥 + 푃 푥 푦 = 0 … (1) 0 푑푥 2 1 푑푥 2 when 푥 = 0 is the regular singularity. Working Procedure 푚 2 3 푛 (i) Let 푦 = 푥 (푎0 + 푎1푥 + 푎2푥 + 푎3푥 + ⋯ + 푎푛 푥 + ⋯ ) … (2) be the solution of the differential equation (1), where m is some real or complex number. 푑푦 푑2푦 (ii) Substitute in (1) the values of 푦, , obtained by differentiating (2). 푑푥 푑푥 2 (iii) Find the indicial equation (a quadratic equation) by equating to zero the coefficient of the lowest degree term in x. (iv) Find the values of 푎1, 푎2, 푎3, ⋯ in terms of 푎0 by equating to zero the coefficients of other powers of x. (v) Find the roots 푚1, 푚2 (say) of the indicial equation. The complete solution depends on the nature of roots of the indicial equation. Case I: Roots 풎ퟏ, 풎ퟐ are distinct and do not differ by an integer In this case, the differential equation (1) has two linearly independent solutions of the following forms: 푚1 2 3 푦1 = 푥 (푎0 + 푎1푥 + 푎2푥 + 푎3푥 + ⋯ ) 푚1 2 3 푦2 = 푥 (푏0 + 푏1푥 + 푏2푥 + 푏3푥 + ⋯ ) The complete solution of the differential equation is given by 푦 = 푐1푦1 + 푐2푦2. 풅ퟐ풚 풅풚 Example 3: Solve ퟒ풙 + ퟐ + 풚 = ퟎ 풅풙ퟐ 풅풙 푑2푦 푑푦 Solution: Given 4푥 + 2 + 푦 = 0 … (1) 푑푥 2 푑푥 Here 푥 = 0 is a singular point, let its solution be 푚 푚+1 푚+2 푚+3 푚+4 푦 = 푎0푥 + 푎1푥 + 푎2푥 + 푎3푥 + 푎4푥 + … … (2) 4 From equation (2) 푑푦 = 푚푎 푥푚−1 + 푚 + 1 푎 푥푚 + 푚 + 2 푎 푥푚+1 푑푥 0 1 2 푚+2 + 푚 + 3 푎3푥 + … … … (3) 푑2푦 = 푚 푚 − 1 푎 푥푚−2 + 푚 + 1 푚 푎 푥푚−1 푑푥 2 0 1 푚 + 푚 + 2 푚 + 1 푎2푥 + … … … (4) Putting the above values in equation (1), we get 푚−2 푚−1 푚 4푥 푚 푚 − 1 푎0푥 + 푚 + 1 푚 푎1푥 + 푚 + 2 푚 + 1 푎2푥 + … … 푚−1 푚 푚+1 푚+2 +2 푚푎0푥 + 푚 + 1 푎1푥 + 푚 + 2 푎2푥 + 푚 + 3 푎3푥 + … … 푚 푚+1 푚+2 푚+3 푚+4 + 푎0푥 + 푎1푥 + 푎2푥 + 푎3푥 + 푎4푥 + … … = 0 … (5) Equating the coefficients of 푥푚−1 equal to zero 2 4푚 푚 − 1 푎0 + 2푚푎0 = 0 ⇒ 푎0 4푚 − 4푚 + 2푚 = 0 1 Because 푎 ≠ 0 ⇒ 4푚2 − 2푚 = 0 i.e. 푚 = 0, 0 2 1 ∴ The solution of the indicial equation is 푚 = 0 and 푚 = . 1 2 2 Here, the roots are real, distinct and do not differ by an integer. ∴ Its solution is 푦 = 푐1푦1 + 푐2푦2 … (6) On equating coefficients of 푥푚 , we get 4 푚 + 1 푚푎1 + 2 푚 + 1 푎1 + 푎0 = 0 or 2 푚 + 1 2푚 + 1 푎1 = −푎0 −푎 ⇒ 푎 = 0 … (7) 1 2 푚+1 (2푚+1) Likewise, 4 푚 + 2 푚 + 1 푎2 + 2 푚 + 2 푎2 + 푎1 = 0 푚 + 2 4푚 + 4 + 2 푎2 = −푎1 or 2 푚 + 2 2푚 + 3 푎2 = −푎1 −푎 푎 ⇒ 푎 = 1 = 0 … (8) 2 2 푚+2 (2푚+3) 22 푚+2 푚+1 2푚+1 (2푚+3) and 4 푚 + 3 푚 + 2 푎3 + 2 푚 + 3 푎3 + 푎2 = 0 푚 + 3 4푚 + 8 + 2 푎3 = −푎2 −푎 2 푚 + 3 2푚 + 5 푎 = 0 3 22 푚+2 푚+1 2푚+1 (2푚+3) −푎 ⇒ 푎 = 0 and so on. … (9) 3 23 푚+3 푚+2 푚+1 2푚+1 2푚+3 (2푚+5) 5 Thus, for 푚 = 0, we get 푚 2 푦(푚=0) = 푦1 = 푥 (푎0 + 푎1푥 + 푎2푥 + … … ) 푚=0 1 푥 1 푥 2 1 푥 3 = 푎 1 − + − + … … 0 2 1.1 22 2.1.1.3 23 3.2.1.1.3.5 2 4 6 푥 푥 푥 = 푎 1 − + − + … … = 푎 cos 푥 … (10) 0 2! 4! 6! 0 1 Likewise for 푚 = , we get 2 1 1 푥 1 푥 2 1 푥 3 2 푦 1 = 푦2 = 푎0푥 1 − 3 + 5 3 − 7 5 3 + … … (푚= ) 21 .2 22 .