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## O(ǫ): ü1 = −δ1 −

O(ǫ): ü1 = −δ1 − cos 2t to eliminate secular terms we need to choose δ1 = 0 u1(t) = c1 + 1 cos 2t 4 The constant c1 modifies the arbitrarily chosen constant amplitude c1 and can also be chosen arbitrarily. Set c1 = 0. O(ǫ 2 ): � ü2 = −δ2 − c1 + 1 � cos 2t cos 2t 4 = −δ2 − 1 8 − c1 cos 2t − 1 cos 4t 8 to eliminate secular terms we need to choose δ2 = − 1 8 . Thus δ = − 1 8 ǫ2 + O(ǫ 3 ) ii) Case n = 1, i.e. δ0 = 1 O(1): O(ǫ) : using u0 = c1 cost + c2 sin t ü1 + u1 = − (c1 cost + c2 sin t)(δ1 + cos 2t) � = c1 cost −δ1 − 1 � � + c2 sin t −δ1 + 2 1 � 2 1 − c1 2 1 cos 3t − c2 sin 3t 2 cosαcosβ = 1 1 {cos (α + β) + cos (α − β)} sin α cosβ = {sin (α + β) + sin (α − β)} 2 2 To eliminate secular terms one of two cases need to be satisfied O(ǫ 2 ): • δ1 = 1 2 and c1 = 0 u0 = c2 sin t, u1 = c3 cost + c4 sin t + c2 1 sin 3t 16 • δ1 = − 1 2 and c2 = 0 u0 = c1 cost, u1 = c5 cost + c6 sin t + c1 1 cos 3t 16 8

• case δ1 = + 1 2 � � 1 ü2 + u2 = −δ2c2 sin t − c3 cost + c4 sin t + c2 sin 3t (δ1 + cos 2t) 16 � = − sin t δ2c2 + c4δ1 − 1 2 c4 + 1 2 c2 � � 1 − cost c3δ1 + 16 1 2 c3 � � 1 − cos 3t 2 c3 � � 1 − sin 3t 2 c4 + 1 16 c2δ1 � � 1 1 − sin 5t 2 16 c2 � to avoid secular terms need Thus • case δ1 = − 1 2 now we need Thus c3 = 0 δ2 = − 1 32 δ = 1 + 1 1 ǫ − 2 32 ǫ2 + O(ǫ 3 ) � 1 u = c2 sin t + ǫ c4 sin t + c2 16 � sin 3t + . . . � � 1 ü2 + u2 = −δ2c1 cos t − c5 cost + c6 sin t + c1 cos 3t (δ1 + cos 2t) 16 � = − cos t δ2c1 + c5δ1 + 1 2 c5 + 1 2 c1 � � 1 − sin t c6δ1 − 16 1 2 c6 � � 1 − cos 3t 16 c1δ1 + 1 2 c5 � � 1 − sin 3t 2 c6 � � 1 1 − cos 5t 2 16 c1 � c6 = 0 δ2 = − 1 32 δ = 1 − 1 1 ǫ − 2 32 ǫ2 + O(ǫ 3 ) � 1 u = c1 cost + ǫ c5 cost + c1 16 9 � cos 3t + . . .

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