6.1 Preliminaries

Series Solution This chapter lays emphasis on the classical theory of series solution for ODE by using power series expansion around ordinary and singular points. We mainly focus on Legendre, Hermite and Bessel differential equations. We also deduce various interesting properties of Legendre and Hermite polynomials as well as Bessel functions. 6.1 Preliminaries A power series in powers of (t − t0 ) is an infinite series of the form ∞ X k=0 ck (t − t0 )k = c0 + c1 (t − t0 ) + c2 (t − t0 )2 + · · · (6.1.1) where t is a variable and c0 , c1 , c2 , · · · are constants. Recall that a series ∞ X ck (t − t0 )k is convergent at a point t if the limit of the partial sums sn (t) = k=0 n X k=0 ck (t − t0 )k exists. This limit f (t) is denoted as the sum of the series at the point t. A series which does not converge is said to be a divergent series. Example 6.1.1 Consider the geometric power series ∞ X k=0 tk = 1 + t + t 2 + · · · The partial sums sn (t) = 1 + t + · · · + tn satisfy the relation tsn (t) = t + t2 + · · · + tn + tn+1 and hence sn (t) = 1 − tn+1 . 1−t This gives f (t) = lim sn (t) = 1 for |t| < 1. 1−t Definition 6.1.1 A power series at a point t if the series ∞ X k=0 ∞ X k=0 ck (t − t0 )k is said to converge absolutely |ck (t − t0 )k | converges. One can show that if the series converges absolutely, then it also converges. However, the converse is not true. We have the following tests for checking the convergence or divergence of a series of real numbers. (i) Comparision test (a) Let a series ∞ X ak of real numbers be given and let there exists a k=0 convergent series ∞ X bk of nonnegative real numbers such that k=0 |ak | ≤ bk , Then the orginal series ∞ X k≥1 ak converges. k=0 (b) Let a series ∞ X k=0 divergent series ak of real numbers be given and let there exists a ∞ X dk of nonnegative real numbers such that k=0 |ak | ≥ dk , Then the orginal series ∞ X k≥1 ak diverges. k=0 (ii) Ratio test Let the series ∞ X k=0 (a) The series (b) The series ak ( with an 6= 0) be such that lim ∞ X k=0 ∞ X k→∞ ak converges absolutely if L < 1. ak diverges if L > 1. k=0 (c) No conclusion if L = 1. ak+1 = L. ak 6.1. PRELIMINARIES (iii) Root test ∞ X If the series ak is such that lim k→∞ k=0 (a) The series (b) The series ∞ X k=0 ∞ X  1 (|ak |) k  = L. ak converges absolutely of L < 1. ak diverges if L > 1. k=0 (c) No conclusion if L = 1. Theorem 6.1.1 If the power series given by Eq. (6.1.1) converges at a point t = t1 , then it converges absolutely for every t for which |t − t0 | < |t1 − t0 |. Proof : Since the series given by Eq. (6.1.1) converges for t = t1 , it follows that the partial sums sn (t1 ) converge and hence sn (t1 ) is Cauchy. This implies that sn+1 (t1 ) − sn (t1 ) → 0 as n → ∞ This, in turn, implies that ck (t1 − t0 )k −→ 0 as k → ∞. Hence the elements ck (t1 − t0 )k of the series given by Eq. (6.1.1) are bounded at t = t1 . That is, |ck (t1 − t0 )k | ≤ M for all k ≥ 1 (6.1.2) Eq. (6.1.2) implies that |ck (t − t0 )k | = ck (t1 − t0 )k ≤ M If |t−t0 | < |t1 −t0 |, the series test the series ∞ X k=0 t − t0 t1 − t 0 ∞ X t − t0 t1 − t 0  k t − t0 t1 − t 0 k (6.1.3) k converges and hence by comparision k=0 ck (t − t0 )k converges absolutely. Definition 6.1.2 If the series ∞ X k=0 ck (t − t0 )k converges absolutely for |t − t0 | < r and diverges for |t − t0 | > r, then r is called the radius of convergence. ∞ k X t The power series converges for |t| < 1 and diverges for |t| > 1. At t = 1, k k=1 it diverges and t = −1, it converges. Thus, the radius of convergence of this series is 1. The radius of convergence of the power series given by Eq. (6.1.1) may be determined from the coefficients of the series as follows. Theorem 6.1.2 (Radius of convergence) 1 (i) Suppose that the sequence {|ck | k } converges . Let L denotes its limit. 1 Then, if L 6= 0, the radius of convergence r of the power series is . L (ii) If L = 0, then r = ∞ and hence the series Eq. (6.1.1) converges for all t. 1 where l = l 1 sup{|ck | k }. If this sequence is not bounded, then r = 0 and hence the series is convergent only for t = t0 . 1 (iii) If {|ck | k } does not converge, but it is bounded, then r = Let ∞ X k=0 ck (t − t0 )k be a power series with non zero radius of convergence r. Then the sum of the series is a function f (t) of t and we write f (t) = ∞ X k=0 ck (t − t0 )k (6.1.4) One can easily show the following. (i) The function f (t) in Eq. (6.1.4) is continuous at t = t0 . (ii) The same function f (t) can not be represented by two different power series with the same centre. That is, if f (t) = ∞ X k=0 ck (t − t0 )k = ∞ X k=0 dk (t − t0 )k in a disk: |t − t0 | < r, then ck = dk for all k ≥ 0. We can carry out the standard operations on power series with ease - addition and subtraction, multiplication, term by term differentiation and integration. (i) Addition and Subtraction P∞ P∞ Two power series k=0 ck (t − t0 )k and k=0 dk (t − t0 )k can be added and subtracted in the common radius of convergence. If f (t) = g(t) = ∞ X k=0 ∞ X k=0 Then f (t) ± g(t) = ∞ X k=0 ck (t − t0 )k in |t − t0 | < r1 and dk (t − t0 )k in |t − t0 | < r2 (ck ± dk )(t − t0 )k in |t − t0 | < r = min(r1 , r2 ). 6.1. PRELIMINARIES (ii) Multiplication If f (t) = g(t) = ∞ X k=0 ∞ X k=0 ck (t − t0 )k in |t − t0 | < r1 and dk (t − t0 )k in |t − t0 | < r2 Then h(t) = f (t)g(t) defined within the radius of convergence of each series and ∞ X h(t) = ck (t − t0 )k k=0 where ck = k X cm dk−m . That is m=0 h(t) = c0 d0 + (c0 d1 + c1 d0 ) t + (c0 d2 + c1 d1 + c2 d0 ) t2 + · · · The series converges absolutely within the radius of convergence of each series. (iii) Differentiation Let f (t) = ∞ X ck (t − t0 )k in |t − t0 | < r ∞ X ck k(t − t0 )k−1 in |t − t0 | < r ∞ X ck (t − t0 )k in |t − t0 | < r k=0 Then df = f˙ = dt k=1 (iv) Integration Let f (t) = k=0 Then Z f (t)dt = ∞ X ck (t − t0 )k+1 in |t − t0 | < r k+1 k=0 Definition 6.1.3 We shall say that a function f (t) is analytic at t = t0 if it ∞ X can be expanded as a sum of a power series of the form ck (t − t0 )k with k=0 a radius of convergence r. It is clear that if f (t) is analytic at t0 , then ck = f (k) (t0 ) , k = 0, 1, 2, · · · ( f k (t0 ) denotes the k th derivative of f at t0 ). k! 180 Example 6.1.2 We have the binomial expansion for (1 − t)−k for a fixed positive number k to give us (1 − t)−k = 1 + kt + k(k + 1) 2 k(k + 1)...(k + r − 1) r t + t + · · · for |t| < 1 2! r! We denote by ur = k(k + 1)...(k + r − 1) r! Then (1 − t)−k = ∞ X ur t r , |t| < 1 r=0 As a power series can be differentiated term by term in its interval of convergence, we have ∞ X k(1 − t)−k−1 = rur tr−1 , |t| < 1 r=1 This gives ∞ k(1 − t)−k X 1 = rur tr−1 , (1 − t) r=1 Again, using the power series expansion for ∞ X k t r r=0 ! ∞ X ur t r=0 r ! |t| < 1 1 = 1 + t + t2 + · · · , we get 1−t = ∞ X rur tr−1 r=1 Using the product formula for LHS, we get k ∞ X r=0 tr r X ul = ∞ X (r + 1)ur+1 tr r=0 l=0 Uniqueness of power series representation gives (r + 1)ur+1 = k r X ul (6.1.5a) l=0 where ur = k(k + 1).....(k + r − 1) , k is a fixed integer r! (6.1.5b) We now make an attempt to define the concept of uniform convergence. To define this concept, we consider the following series whose terms are functions ∞ {fk (t)}k=0 ∞ X k=0 fk (t) = f0 (t) + f1 (t) + f2 (t) + · · · Note that for fk (t) = ck (t − t0 )k , we get the power series. (6.1.6) 6.2. LINEAR SYSTEM WITH ANALYTIC COEFFICIENTS Definition 6.1.4 We shall say that the series given by Eq. (6.1.6) with sum f (t) in an interval I ⊂ < is uniformly convergent if for every  > 0, we can find N = N (t), not depending on t, such that |f (t) − sn (t)| <  for all n ≥ N () where sn (t) = f0 (t) + f1 (t) + · · · + fn (t). Theorem 6.1.3 A power series ∞ X k=0 ck (t − t0 )k with nonzero radius of conver- gennt is uniformly convergent in every closed interval |t − t0 | ≤ p of radius p < r. Proof : For |t − t0 | ≤ p and any positive integers n and l we have cn+1 (t − t0 )n+1 + · · · + cn+l (t − t0 )n+l ≤ |cn+1 | pn+1 + · · · + |cn+l | pn+l The series ∞ X k=0 (6.1.7) ck (t − t0 )k converges absolutely if |t − t0 | ≤ p < r ( by Theorem 6.1.1) and hence by Cauchy convergence, given  > 0, ∃ N () such that |cn+1 | pn+1 + · · · + |cn+p | pn+l <  for n ≥ N (), l = 1, 2, .. Eq. (6.1.7) gives cn+1 (t − t0 )n+1 + · · · + cn+p (t − t0 )n+l ≤ |cn+1 | pn+1 + · · · + |cn+p | pn+l <  for n ≥ N (), |t − t0 | ≤ p < r This implies that, given  > 0, ∃ N () such that |f (t) − sn (t)| <  for n ≥ N () n X and all t where sn (t) = ck (t − t0 )k . This proves the uniform convergence of k=0 the power series inside the interval |t − t0 | ≤ p < r. Example 6.1.3 The geometric series 1 + t + t2 + · · · is uniformly convergent in the interval |t| ≤ p < 1. It is not uniformly convergent in the whole interval |t| < 1.