5.2we - 5.2: SERIES SOLUTIONS NEAR AN ORDINARY POINT KIAM...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5.2: SERIES SOLUTIONS NEAR AN ORDINARY POINT KIAM HEONG KWA Consider the differential equation (1) P ( x ) d 2 y dx 2 + Q ( x ) dy dx + R ( x ) y = 0 , where P ( x ), Q ( x ), and R ( x ) are general analytic functions 1 . A point x at which both Q ( x ) /P ( x ) and R ( x ) /P ( x ) are analytic is called an ordinary point . In particular, P ( x ) 6 = 0 for an ordinary point x . By continuity, P ( x ) is never zero in an interval I about the ordinary point x . Hence within such an interval I , (1) is equivalent to the equation (2) d 2 y dx 2 + p ( x ) dy dx + q ( x ) y = 0 , where p ( x ) = Q ( x ) /P ( x ) and q ( x ) = R ( x ) /P ( x ). Therefore by the results in section 3.2, there exists a unique solution of (2), and hence of (1), on I that satisfies the initial conditions y ( x ) = y and y ( x ) = y for arbitrarily prescribed values of y and y . On the other hand, a point x at which either p ( x ) or q ( x ) is not analytic is called a singular point . This occurs, for instance, if P ( x ) = 0; in this case, if either Q ( x ) 6 = 0 or R ( x ) 6 = 0, then either p ( x ) or q ( x ) becomes unbounded as x approaches x . Hence the results in section 3.2 does not apply in this case. We will say more about computing solutions near a singular point in later sections. Remark 1. If P ( x ) , Q ( x ) , and R ( x ) in (1) are polynomials, then x is an ordinary point if and only if P ( x ) 6 = 0 ; otherwise, x is a singular point. Date : February 6, 2011. 1 Recall that for a function f ( x ) to be (real) analytic at a point x ∈ R , it means that f ( x ) can be locally expressed as a power series near x in the sense that f ( x ) = ∞ X n =0 a n ( x- x ) n , where the right-hand side is a convergent power series in the interval ( x- ρ,x + ρ ) for some ρ > 0. 1 2 KIAM HEONG KWA To calculate the solution of (1) near an ordinary point x , the first assumption we make is that the solution y ( x ) can be expressed as a power series about x , i.e., y ( x ) = ∞ X k =0 a k ( x- x ) k (3) = a + a 1 ( x- x ) + a 2 ( x- x ) 2 + ··· + a k ( x- x ) k + ··· , and that this series converges for | x- x | < ρ for some ρ > 2 . Recall that a power series can be differentiated term by term within its interval of convergence, so that y ( x ) = ∞ X k =1 ka k ( x- x ) k- 1 (4) = a 1 + 2 a 2 ( x- x ) + 3 a 3 ( x- x ) 2 + ··· + ka k ( x- x ) k- 1 + ··· and y 00 ( x ) = ∞ X k =2 ( k- 1) ka k ( x- x ) k- 2 (5) = 1 · 2 a 2 + 2 · 3 a 3 ( x- x ) + ··· + ( k- 1) ka k ( x- x ) k- 2 + ··· . By substituting (3)-(5) into (1), the problem reduces to determining the coefficients a k ’s, k = 0 , 1 , ··· . Remark 2. An interested reader may refer to [1, Ch. 4] for a rather readable exposition to power series solutions of ordinary differential equations....
View Full Document

This note was uploaded on 11/11/2011 for the course MATH 255.01 taught by Professor Kwa during the Winter '11 term at Ohio State.

Page1 / 22

5.2we - 5.2: SERIES SOLUTIONS NEAR AN ORDINARY POINT KIAM...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online