Math 4/5027 - Nonlinear Dynamics and Chaos
Solution Set #4 - Spring 1999

Problems are 5 points each. Undergraduate total = 30. Graduate total = 40.

Problem 2. Recall that to show tex2html_wrap_inline94 for any two sets, we must show that whenever tex2html_wrap_inline96 , then tex2html_wrap_inline98 . Recall also that tex2html_wrap_inline100 . Now let's prove that if tex2html_wrap_inline94 , then tex2html_wrap_inline104 . Assume that tex2html_wrap_inline106 which means that there is an tex2html_wrap_inline96 such that y=f(x). But if tex2html_wrap_inline96 , then tex2html_wrap_inline98 (because tex2html_wrap_inline94 ). Thus tex2html_wrap_inline118 . We have shown that for any tex2html_wrap_inline106 , it follows that tex2html_wrap_inline122 . Thus tex2html_wrap_inline104 .

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Problem 4. The transition graph implies that for three intervals A, B, and C, tex2html_wrap_inline130 , tex2html_wrap_inline132 , and tex2html_wrap_inline134 . It is impossible to draw the graph of a continuous functions that satisfies these three conditions. Thus we cannot conclude that period three implies all periods.

*Problem 5. (a) The condition G(C(x)) = C(T(x)) implies by direct substitution (using the branch of T for tex2html_wrap_inline140 ) that

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(b) The family of functions that staisfies tex2html_wrap_inline142 is the family of exponential functions tex2html_wrap_inline144 or tex2html_wrap_inline146 .

(c) If a trial solution of the form tex2html_wrap_inline148 is used, the resulting equation has constant terms in it that cannot be matched with terms of the trial solution.

(d) Using real exponentials as a trial solution may not be worse, but once again we introduce new terms that cannot be matched and eliminated.

(e) Substituting the real-valued function

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into the functional equation give us

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Remember that the goal is to find a,b, and k. Expanding and collecting terms we have

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Notice that we don't introduce any new terms that weren't already in the trial solution (the system is closed). The coefficient of each term must which leaves three independent conditions for a (real and imaginary parts) and b:

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These equations have the solution tex2html_wrap_inline158 and tex2html_wrap_inline160 ; k is undetermined, so far. Thus we have

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Requiring that C be one-to-one on [0,1] with C(0)=0 and C(1)=1 means k=1. So we have tex2html_wrap_inline174 .

*Problem 6. We want to use the conjugacy map to determine tex2html_wrap_inline176 , for any positive integer n, using what we know about the tent map T. The relationship between G and T tells us that tex2html_wrap_inline186 . We first need to find tex2html_wrap_inline188 . Starting with tex2html_wrap_inline190 , it is possible to formally invert this expression and find that

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What about tex2html_wrap_inline192 ? Compositions of T with itself involve combinations of the left branch (T(x) = 2x) and right branch (T(x)=2(1-x)) of T. You can verify that regardless of the order of composition and the value of x, tex2html_wrap_inline204 , where 2M is some even integer (the condition tex2html_wrap_inline208 is not needed). Now we can put it all together. We have

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which is the desired result. We have used tex2html_wrap_inline210 and tex2html_wrap_inline212 .

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Problem 7.13a. The method of nullclines requires sketching equilibrium points, nullclines and the direction field in the phase plane; eigenvalue calculations are not needed. The four equilibrium points are

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The x-nullclines (where x'(t)=0) are x=0 and y = 3-3x. The y-nullclines (where y'(t)=0) are y=0 and tex2html_wrap_inline262 . The nullclines divide the first quadrant into four regions. You must determine the general direction of the direction field in each of these regions. The phase portrait is shown below, indicating that all equilibrium points are unstable except the interior point which is stable. Thus, the two species can coexist in the steady state.