By G.C. Layek
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Additional resources for An Introduction to Dynamical Systems and Chaos
Show that /t forms a dynamical group. Is it a commutative group? Solution The solutions of the given system are obtained as below: x_ ¼ dx 1 1 ¼ Àx2 ) ¼ t þ A ) xðtÞ ¼ dt x tþA in any interval of R that does not contain the point x ¼ 0; where A is a constant. If we take starting point xð0Þ ¼ x0 , then A ¼ 1=x0 and so we get xðtÞ ¼ x0 ; 1 þ x0 t t 6¼ À1=x0 : The point x ¼ 0 is not included in this solution. But it is the ﬁxed point of the given system, because x_ ¼ 0 , x ¼ 0: Therefore, /t ð0Þ ¼ 0 for all t 2 R: So the evolution operator of the system is given as /t ðxÞ ¼ 1 þx xt for all x 2 R: The evolution operator /t is not deﬁned for all t 2 R: For example, if t ¼ À1=x; x 6¼ 0; then /t is undeﬁned.
The flow is to the right direction, indicated by the symbol ‘→’, when the velocity x_ [ 0; and to the left direction, indicated by the symbol ‘←’, when x_ \0. We also use solid circle to represent stable ﬁxed point and open circle to represent unstable ﬁxed point. From Fig. 5 we see that the ﬁxed point x ¼ 1 is stable whereas the ﬁxed point x ¼ 0 is unstable. 7 Find the ﬁxed points and analyze the local stability of the following systems (i) x_ ¼ x þ x3 (ii) x_ ¼ x À x3 (iii) x_ ¼ Àx À x3 Solution (i) Here f ð xÞ ¼ x þ x3 .
We 1 can choose e1 ¼ 1, e2 ¼ 1 and e3 ¼ 0, and so we can take one eigenvector 0 1 as @ 1 A. Again, we can choose e1 ¼ 0, e2 ¼ 1 and e3 ¼ 1. Then we obtain another 0 0 1 0 eigenvector @ 1 A. Clearly, these two eigenvectors are linearly independent. Thus, 1 we have two linearly independent eigenvectors corresponding to the repeated eigenvalue −2. Hence, the general solution of the system is given by 0 1 0 1 0 1 1 1 0 x$ ðtÞ ¼ c1 @ 1 Ae4t þ c2 @ 1 AeÀ2t þ c3 @ 1 AeÀ2t 2 0 1 where c1 , c2 and c3 are arbitrary constants.