Series Solution of Linear Ordinary Differential Equations

Transcription

1 Series Solution of Linear Ordinary Differential Equations Department of Mathematics IIT Guwahati

2 Aim: To study methods for determining series expansions for solutions to linear ODE with variable coefficients. In particular, we shall obtain the form of the series expansion, a recurrence relation for determining the coefficients, and the interval of convergence of the expansion.

3 Review of power series A series of the form a n (x x 0 ) n = a 0 + a 1 (x x 0 ) + a 2 (x x 0 ) 2 +, (1) is called a power series about the point x 0. Here, x is a variable and a n s are constants. The series (1) converges at x = c if a n(c x 0 ) n converges. That is, the limit of partial sums lim N N a n (c x 0 ) n <. If this limit does not exist, the power series is said to diverge at x = c.

4 Note that a n(x x 0 ) n converges at x = x 0 as a n (x 0 x 0 ) n = a 0. Q. What about convergence for other values of x? Theorem: (Radius of convergence) For each power series of the form (1), there is a number R (0 R ), called the radius of convergence of the power series, such that the series converges absolutely for x x 0 < R and diverge for x x 0 > R. If the series (1) converges for all values of x, then R =. When the series (1) converges only at x 0, then R = 0.

5 Theorem: (Ratio test) If lim a n+1 n a n = L, where 0 L, then the radius of convergence (R) of the power series a n(x x 0 ) n is 1 if 0 < L <, L R = if L = 0, 0 if L =. Remark. If the ratio does not have a limit, then a n+1 a n methods other than the ratio test (e.g. root test) must be used to determine R.

6 Example: Find R for the series ( 2) n (x n+1 3)n. Note that a n = ( 2)n. We have n+1 lim n a n+1 a n = lim n ( 2) n+1 (n + 1) ( 2) n (n + 2) = lim n Thus, R = 1/2. The series converges absolutely for x 3 < 1 2 and diverge for x 3 > 1 2. Next, what happens when x 3 = 1/2? 2(n + 1) (n + 2) = 2 = L. At x = 5/2, the series becomes the harmonic series and hence diverges. When x = 7/2, the series becomes an alternating harmonic series, which converges. Thus, the power series converges for each x (5/2, 7/2]. 1, n+1

7 Given two power series f (x) = a n (x x 0 ) n, g(x) = b n (x x 0 ) n, with nonzero radii of convergence. Then f (x) + g(x) = (a n + b n )(x x 0 ) n has common interval of convergence. The formula for the product is f (x)g(x) = c n (x x 0 ) n, where c n := n a k b n k. (2) k=0 This power series in (2) is called the Cauchy product and will converge for all x in the common interval of convergence for the power series of f and g.

8 Differentiation and integration of power series Theorem: If f (x) = a n(x x 0 ) n has a positive radius of convergence R, then f is differentiable in the interval x x 0 < R and termwise differentiation gives the power series for the derivative: f (x) = na n (x x 0 ) n 1 for x x 0 < R. n=1 Furthermore, termwise integration gives the power series for the integral of f : a n f (x)dx = n + 1 (x x 0) n+1 + C for x x 0 < R.

9 Example: A power series for Since d dx x 2 = 1 x 2 + x 4 x ( 1) n x 2n +. {1/(1 x)} = 1 (1 x) 2, we obtain a power series for 1 (1 x) 2 = 1 + 2x + 3x 2 + 4x nx n 1 +. Since tan 1 x = x 0 termwise to obtain 1 1 dt, integrate the series for 1+t 2 1+x 2 tan 1 x = x 1 3 x x x ( 1)n x 2n n + 1

10 Shifting the summation index The index of a summation in a power series is a dummy index and hence a n (x x 0 ) n = a k (x x 0 ) k = k=0 a i (x x 0 ) i. Shifting the index of summation is particularly important when one has to combine two different power series. Example: x 3 n(n 1)a n x n 2 = n=2 n 2 (n 2)a n x n = i=0 (k + 2)(k + 1)a k+2 x k. k=0 (n 3) 2 (n 5)a n 3 x n. n=3

11 Definition: (Analytic function) A function f is said to be analytic at x 0 if it has a power series representation a n(x x 0 ) n in an neighborhood about x 0, and has a positive radius of convergence. Example: Some analytic functions and their representations: ln x = sin x = e x = x n n!. ( 1) n (2n + 1)! x 2n+1. ( 1) n 1 (x 1) n, x > 0. n n=1

12 Power series solutions to linear ODEs Consider linear ODE of the form: a 2 (x)y (x) + a 1 (x)y (x) + a 0 (x)y(x) = 0, a 2 (x) 0. ( ) Writing in the standard form y (x) + p(x)y (x) + q(x)y(x) = 0, where p(x) := a 1 (x)/a 2 (x) and q(x) := a 0 (x)/a 2 (x). Definition: A point x 0 is called an ordinary point of ( ) if both p(x) = a 1 (x)/a 2 (x) and q(x) = a 0 (x)/a 2 (x) are analytic at x 0. If x 0 is not an ordinary point, it is called a singular point of ( ).

13 Example: Find all the singular point points of Here, xy (x) + x(x 1) 1 y (x) + (sin x)y = 0, x > 0 p(x) = 1 sin x, q(x) = (1 x) x. Note that p(x) is analytic except at x = 1. q(x) is analytic everywhere as it has power series representation q(x) = 1 x 2 3! + x 4 5!. Hence, x = 1 is the only singular point of the given ODE.

14 Power series method about an ordinary point Consider the equation 2y + xy + y = 0. ( ) Let s find a power series solution about x = 0. Seek a power series solution of the form y(x) = a n x n, and then attempt to determine the coefficients a n s. Differentiate termwise to obtain y (x) = na n x n 1, y (x) = n(n 1)a n x n 2. n=1 n=2

15 Substituting these power series in ( ), we find that 2n(n 1)a n x n 2 + na n x n + a n x n = 0. n=2 n=1 By shifting the indices, we rewrite the above equation as 2(k + 2)(k + 1)a k+2 x k + ka k x k + a k x k = 0. k=0 k=1 k=0 Combining the like powers of x in the three summation to obtain 4a 2 + a 0 + [2(k + 2)(k + 1)a k+2 + ka k + a k ]x k = 0. k=1

16 Equating the coefficients of this power series equal to zero yields 4a 2 + a 0 = 0 2(k + 2)(k + 1)a k+2 + (k + 1)a k = 0, k 1. This leads to the recurrence relation Thus, a k+2 = 1 2(k + 2) a k, k 1. a 2 = 1 2 a 0, a 2 3 = a 1 a 4 = a 1 2 = a 0, a 5 = a 3 = a 1

17 With a 0 and a 1 as arbitrary constants, we find that and a 2n+1 = a 2n = ( 1)n 2 2n n! a 0, n 1, ( 1) n 2 n [1 3 5 (2n + 1)] a 1, n 1. From this, we have two linearly independent solutions as y 1 (x) = y 2 (x) = ( 1) n 2 2n n! x 2n, ( 1) n 2 n [1 3 5 (2n + 1)] x 2n+1.

18 Hence the general solution is y(x) = a 0 y 1 (x) + a 1 y 2 (x). Remark. Suppose we are given the value of y(0) and y (0), then a 0 = y(0) and a 1 = y (0). These two coefficients leads to a unique power series solution for the IVP. *** End ***