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This Dynamical systems

In Chaotic Dynamical Systems.pdf - Foxit Reader ora Find @ An Marquee 123 Loupe ABC Weights AutoScroll Magnifier Read Navigation Status Word Panels Bar Count Assistant View Setting Review pdf Free PPT to PDF UKC Converter WEICO wa TCO pom Pi, then plot ( C PI). If the orbit of 0 under Qc, is attracted to a 2-cycle, and 92, then plot (C2.91) and (C2,92). If there is no pattern to the orbit of O. then use a histogram to determine approximately the interval (or intervals) RE CHAPTER 6 BIFURCATIONS 67 Exercises 1. Each of the following functions undergoes a bifurcation of fixed points at the given parameter value. In each case, use algebraic or graphical methods to identify this bifurcation as either a saddle-node or period-doubling bifur- cation, or neither of these. In each case, sketch the phase portrait for typical parameter values below, at, and above the bifurcation value. a. F(I) = 1 + 2 + 1, 1 = 0 b. F. (2) = 1 ++X, = -1 c. G (2) = pi + , u = -1 d. G (2) = + , =1 e. S(x) = sinz, y = 1 f. Sy(z) = sinr, = -1 g. F.(2) = 1 +c, c = 2/3V3 h. Ex(z) = (e- 1), X = -1 i. Ex(+) = X(e" - 1), X = 1 j. H(2) = 1 + cz, c=0 k. F.(2) = + + cz? + ', c=0 over dis- 25. The next four exercises apply to the family Q(x)=x+e. 2. Verify the formulas for the fixed points pt and the 2-cycle 4 the text. E OUT 240.91% · GOTH given in - 5:21 PM Links A 2 Links 1/27/2020

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Answer 1

b. F_$\lambda$(x)=x+x²+$\lambda$, $\lambda$ =-1

to find the fixed points of F ,solve F(x)=x for x.to get the solutions x=sqrt(-$\lambda$) and x= -sqrt (-$\lambda$).in order to determine the behavior of these fixed points,we need to evaluate the equation F'(x)=2x+1 at sqrt(-$\lambda$)and -sqrt (-$\lambda$).for $\lambda$ between -1 and 0, it is easy to see that F^'(-sqrt(-$\lambda$)) falls between -1 and 1, which implies sqrt (-$\lambda$) is an attracting fixed point.on the other hand sqrt(-$\lambda$) is repelling fixed point for all $\lambda$ less than 0 ,since F^'(sqrt(-$\lambda$)) is greater than 1. at $\lambda$ = -1 ,the attracting fixed point -sqrt(-(-1))= -1 becomes neutral ,with F^' (-1)=2(-1)+ 1= -1.this implies that F(x) x+x²+$\lambda$.

undergoes a period - doubling bifurcation at $\lambda$ =1

c. G_$\mu$(x) =$\mu$x +x^{3 ,}$\mu$=-1

first set G(x) =x and solve for x.the solutions are x=0 , x=sqrt(1 - $\mu$ ) and x= - sqrt (1-$\mu$) . for $\mu$ is greater than or equal to 1 , there is only one reveal value fixed point at the origin , and there are three for $\mu$ 1. From the equation

G^'(x) =$\mu$ +3x² , we see that G^'(0)=$\mu$, which implies that the fixed point at the origin is attracting for $\mu$ between -1 and 1 .evaluating G^'(x) =at the other two fixed points ,sqrt(1 -$\mu$) and - sqrt (1-$\mu$), yields the same response of 3 -2$\mu$ . This implies that both fixed points are attracting for $\mu$ between 1 and 2 . so , when $\mu$ = -1 , we have G^'(0) = -1 and the derivative at the other two points is greater than one ,indicating a period doubling bifurcation at the origin with the other two fixed points repelling.

e . S$\mu$(x) =$\mu$ sin x ,$\mu$ =1

S(x) has a fixed point at 0 for all values of $\mu$ . for $\mu$ between -1 and 1, this is the only fixed point of the function.

S^' (x) =$\mu$cos x indicates that at the origin we have S^'(x) =$\mu$. this implies that for $\mu$ between -1 and 1 , the origin is repelling.The origin becomes neutral at $\mu$ =1 ,and is repelling for values of  $\mu$ greater than 1. at the bifurcation point,the function gives rise to two other fixed points ,one less than 0 and one greater than 0. this can easily seen by graphing the function .This can easily seen by graphing the function .These two fixed points will be attracting for $\mu$ 1.

i .E$\lambda$(x) = $\lambda$ (e^x-1),$\lambda$ = 1

E(x) has a fixed point at the origin for all values of $\lambda$ .Since E^'(x) =$\lambda$ e^x, E^'(0) =$\lambda$.Therefore , when the absolute value of $\lambda$ is less than one , the origin is attracting . At the bifurcation point of $\lambda$ =1, the fixed point is neutral.and when the absolute value of $\lambda$ is greater than 1,0 is repelling. Also , for values of $\lambda$ greater than one , another fixed point appears.

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