Mathematical Methods for Physics and Engineering - Matematica.NET

16.3 SERIES SOLUTIONS ABOUT A REGULAR SINGULAR POINT◮Find the power series solutions about z =0of4zy ′′ +2y ′ + y =0.Dividing through by 4z to put the equation into standard form, we obtainy ′′ + 1 2z y′ + 1 y =0, (16.17)4zand on comparing with (16.7) we identify p(z) =1/(2z) andq(z) =1/(4z). Clearly z =0is a singular point of (16.17), but since zp(z) =1/2 andz 2 q(z) =z/4 are finite there, itis a regular singular point. We therefore substitute the Frobenius series y = z ∑ σ ∞n=0 a nz ninto (16.17). Using (16.13) and (16.14), we obtain∞∑(n + σ)(n + σ − 1)a n z n+σ−2 + 1 ∞∑(n + σ)a n z n+σ−1 + 1 ∞∑a n z n+σ =0,2z4zn=0n=0which, on dividing through by z σ−2 ,gives∞∑ [(n + σ)(n + σ − 1) +1(n + σ)+ 1 z] a2 4 n z n =0. (16.18)n=0If we set z = 0 then all terms in the sum with n>0 vanish, and we obtain the indicialequationσ(σ − 1) + 1 σ =0, 2which has roots σ =1/2 andσ = 0. Since these roots do not differ by an integer, weexpect to find two independent solutions to (16.17), in the form of Frobenius series.Demanding that the coefficients of z n vanish separately in (16.18), we obtain therecurrence relation(n + σ)(n + σ − 1)a n + 1 (n + σ)a 2 n + 1 a 4 n−1 =0. (16.19)If we choose the larger root, σ =1/2, of the indicial equation then (16.19) becomes(4n 2 +2n)a n + a n−1 =0 ⇒ a n = −a n−12n(2n +1) .Setting a 0 = 1, we find a n =(−1) n /(2n + 1)!, and so the solution to (16.17) is given byy 1 (z) = √ z∞∑n=0= √ z − (√ z) 33!(−1) n(2n +1)! zn+ (√ z) 55!− ···=sin √ z.To obtain the second solution we set σ = 0 (the smaller root of the indicial equation) in(16.19), which gives(4n 2 − 2n)a n + a n−1 =0 ⇒ a n = − a n−12n(2n − 1) .Setting a 0 =1nowgivesa n =(−1) n /(2n)!, and so the second (independent) solution to(16.17) isy 2 (z) =∞∑n=0n=0(−1) n(2n)! zn =1− (√ z) 2+ (√ 4) 4− ···=cos √ z.2! 4!541