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Unformatted text preview: Lecture 32 Perturbation Methods for ODEs The material for this lecture is to be found in the AMATH 351 Course Notes, Chapter 4, pp. 143149, Eq. (4.21). Lecture 33 The material for this lecture is to be found in the AMATH 351 Course Notes, Chapter 4, pp. 149, starting at Eq. (4.21) and ending at the bottom of p. 152. A major point considering the initial value problem, y ′′ + (1 + ǫ ) 2 y = 0 , y (0) = A, y ′ (0) = 0 , (1) is that the solution of this problem y ( t ) = A cos(1 + ǫ ) t (2) is periodic, with period T = 2 π 1 + ǫ or, equivalently, with frequency ω = 1+ ǫ . The perturbation method outlined in this section attempts to construct a perturbationtype solution to (1) by using the unperturbed solution y ( t ) = A cos t, (3) which has frequency ω = 1. There is a fundamental problem here since the time scales of the two problems do not match. It was Poincar´ e who saw this problem and attempted to correct it by trying to construct perturbation solutions with periods that would match, as best as possible, the solution of the perturbed problem. Lecture 34 We return to the perturbed linear oscillator problem (1) and show how Poincar´ e’s method of matching time scales can be applied. Of course, we are going to pretend that we do not know 1 the exact solution (2). Let us define a new variable as follows, s = ω ( ǫ ) t, (4) where we shall also assume a perturbation expansion for ω ( ǫ ) as follows, ω ( ǫ ) = 1 + ω 1 ǫ + ω 2 ǫ 2 ··· , (5) where the coefficients ω k are to be determined. We shall have to convert the derivatives in t to derivatives in s using the Chain Rule: dy dt = dy ds ds dt = ω dy ds . (6) In a similar way, we find that d 2 y dt 2 = ω 2 d 2 y ds 2 . (7) We shall now substitute these results into the DE in (1): ω 2 d 2 y ds 2 + (1 + ǫ ) 2 y ( s ) = 0 , (8) where we also assume a perturbation expansion for y ( s ) of the form y ( s ) = y ( s ) + ǫy 1 ( s ) ··· . (9) The result is [1 + 2 ω 1 ǫ + ( ω 2 1 + 2 ω 2 ) ǫ 2 ··· ][ y ′′ + ǫy ′′ 1 + ··· ] = [1 + 2 ǫ + ǫ 2 ][ y + ǫy 1 + ··· ] , (10) where the primes now denote differentiation with respect to s . We now collect terms in ǫ k to obtain. The equation in ǫ , which should be the unperturbed DE, is y ′′ + y = 0 , y ( s ) = A, y ′ (0) = 0 . (11) The solution to this IVP is y ( s ) = A cos s. (12) 2 The equation in ǫ 1 is y ′′ 1 + y 1 = 2 ω 1 y ′′ 2 y , y 1 (0) = y ′ 1 (0) = 0 . (13) Substitution of the unperturbed solution into the above DE yields y ′′ 1 + y 1 = 2 A [ ω 1 1] cos s. (14) Note that the righthandside term cos s coincides with one of the components of the homoge neous solution to this equation. As a result, we would have to seek a solution of the form y 1 ( s ) = C 1 s cos s + C 2 s sin s, (15) which is not periodic. In fact, we have once again encountered secular terms....
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This note was uploaded on 09/20/2010 for the course AMATH 351 taught by Professor Sivabalsivaloganathan during the Spring '08 term at Waterloo.
 Spring '08
 SivabalSivaloganathan

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