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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 righthand 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....
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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.
 Winter '11
 Kwa
 Differential Equations, Equations

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