FRACTAL GEOMETRY, DYNAMICAL SYSTEMS AND CHAOS

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1 FRACTAL GEOMETRY, DYNAMICAL SYSTEMS AND CHAOS MIDTERM SOLUTIONS. Let f : R R be the map on the line generated by the function f(x) = x 3. Find all the fixed points of f and determine the type of their stability. Describe the behavior of all trajectories of the map. Solution. To find the fixed points, we solve the equation f(x) = x, which takes the form x 3 = x, and find three solutions, x =, x = 0, and x =. To determine stability and behaviour of trajectories, we observe that for x >, we have f(x)/x = x >, and so f(x) > x >. Thus the trajectory x n = f n (x 0 ) is increasing for x 0 > : x 0 < x < x <. Every increasing sequence has a limit (perhaps ), and so writing p = lim n x n, we observe that if p <, we have ( ) f(p) = f lim n fn (x 0 ) = lim n fn+ (x 0 ) = p, whence p is fixed. Since there are no fixed points greater than, we must have x n. Similarly, for x <, we find f(x) < x <, and a similar argument shows x n. Finally, for 0 < x <, we see that 0 < f(x) = x x < x <, and so x n converges to a fixed point in [0, ). It follows that x n 0, and the argument for < x < 0 is similar. Thus trajectories which begin in the interval (, ) go to the attracting fixed point 0, while initial conditions with x 0 > or x 0 < lead to trajectories divergies to + or, respectively, and the fixed points and are repelling.. Let f : R R be the map on the line generated by the function f(x) = x. Describe the behavior of the trajectory started at a point x = m where m is a positive integer. (Hint: consider the case when m is even and the case when m is odd). Solution. The map f(x) = x has only one fixed point, x =. Thus for x 0 =, we have x n = f n (x 0 ) = x 0 = for all n N, and the trajectory is fixed.

2 MIDTERM SOLUTIONS For x 0 =, we have x = f(x 0 ) = 0, and x = f(x ) =, and the trajectory is periodic with period. For x 0 >, we have x 0 = a + b, where a N and b = 0 or. A simple induction shows that x a = f a (x 0 ) = b, and then the trajectory displays one of the two behaviours described above. 3. Let Z R be a subset. Prove that dim H Z = dim H f(z) where f(x) = x. Solution. We know that Hausdorff dimension is preserved by bi-lipschitz maps; however, neither f nor its inverse is Lipschitz on R, so we cannot use this fact directly. Observe, however, that since f is continuously differentiable, its restriction to any compact interval [a, b] is Lipschitz. Similarly, f has two branches, x x and x x, each of which is well-defined and continuously differentiable on any compact interval [a, b] (0, ); observe that we must have a > 0 for this to be the case. Now consider the intervals C n = [/n, n], where n N, and observe that C n = (0, ). Similarly, writing C n = [ n, /n], we have C n = (, 0). Let Z n = Z C n, and Z n = Z C n; then we have ( ) ( ) Z = Z n (Z {0}). Because the Hausdorff dimension of a countable union of sets is the supremum of the Hausdorff dimensions of the individual sets, we see that Z n dim H Z = sup max{dim H Z n, dim H Z n}, dim H f(z) = sup max{dim H f(z n ), dim H f(z n)}, using the fact that dim H (Z {0}) = 0. Thus it suffices to prove that dim H f(z n ) = dim H Z n for all n N, and similarly for Z n. But Z n [/n, n], and so f(z n ) [/n, n ], and since both of these intervals are bounded away from 0 and, the desired equality follows from our earlier observation that [ ] [ ] f :, f : n, n [ n,n n ] [,n ] n, n are both Lipschitz on the given intervals. The proof for Z n is similar. 4. Compute the Hausdorff, lower and upper box dimensions of the set A = {(x n, y n ), n =,,... } in the unit square where x n = n and y n = n. Solution #. Because A is countable, we have dim H A = 0.

3 FRACTAL GEOMETRY, DYNAMICAL SYSTEMS AND CHAOS 3 Write z n = (x n, y n ), and observe that d(z n, z n+ ) = (x n x n+ ) + (y n y n+ ) x n x n+ = n(n + ). Now given ε > 0, choose n = n(ε) such that n(n + ) ε < n(n ). It follows that a set of diameter ε can contain at most one of the points z,...,z n, and since each of these points is contained in A, we have N(ε) n. Furthermore if we cover each of these n points with their own ball of diameter ε, then it remains only to cover the points z n+, z n+,..., which all lie in the rectangle R = [ 0, ] [ 0, n + ] n+. Because / n+ /n(n+) ε for sufficiently large n, a ball of radius ε/ centred on the x-axis will contain a rectangle of the form [a, b] [0, / n+ ], where b a > ε/. In particular, b a > /n(n + ), and so we can use n such balls, centred at the points { ( ) k + } 4n(n + ), 0 k = 0,,...,n, to cover R. This shows that N(ε) 3n, and so we have lim n log n log n + log ( + n log N(ε) ) lim lim n log ε n log n + log 3 log n + log ( n Both the outside limits exist and are equal to /, hence dim B A = /. Solution #. Because A is countable, we have dim H A = 0. We now show that the map which takes each point in A to the point directly below it on the x-axis is bi-lipschitz, from which it will follow that A has the same lower and upper box dimension as the set {0,, /, /3,... }; we know from a homework exercise that the box dimension of this set is /. To this end, let g: R R be some function to be specified later, and consider the map f g : R R, (x, y) (x, y + g(x)). This map shifts each vertical line in the plane up or down by a distance g(x) which depends on its x-coordinate. In particular, if we consider the map g(x) = (/) /x, we see that f g (/n, / n ) = (/n, 0). ).

4 4 MIDTERM SOLUTIONS Now if g is Lipschitz, then f g is Lipschitz as well; to see this, suppose that L is the Lipschitz constant for g, and observe that f g (x, y) f g (x, y ) = (x, y + g(x)) (x, y + g(x )) (x, y) (x, y ) + (0, g(x) g(x )) (x, y) (x, y ) + L x x (L + ) (x, y) (x, y ). Furthermore, if g is Lipschitz then g is Lipschitz as well, and so fg = f g is Lipschitz. Thus f g is bi-lipschitz, and preserves box dimension. Observe that this would not have worked if we had moved the sequence onto the y-axis, since then the function giving the distance to shift each line would have been g instead of g, and g is not Lipschitz. 5. Compute the Hausdorff, lower and and upper box dimensions of the Cantor-like set obtained via the following geometric construction in R : Starting with the unit square in R choose two disjoint rectangles with sides λ and µ (where 0 < λ < < µ < ), which are spaced one exactly above the other. At the next step of the construction inside each of these rectangles choose two disjoint rectangles with sides λ and µ, which are spaced one exactly above the other. Continue in the same fashion (see the picture below). µ µ λ λ Solution. Let A n be the union of the basic sets at the n th step of the construction, so that the Cantor-like set in question is A = n A n. Because the basic sets are not similar to each other (the scaling factor depends on the direction since λ µ), we cannot apply Moran s theorem directly. If we begin with the unit square [0, ] [0, ], we see that [ µ n A n, + µn ] [0, ]; the n th step of the construction is contained in a rectangle of width µ n. Since the limiting set A must be contained in every such rectangle, and since µ n 0 as n, we see that A { } [0, ].

5 FRACTAL GEOMETRY, DYNAMICAL SYSTEMS AND CHAOS 5 In particular, A is contained in the vertical line with x-coordinate /, and we see that the same set may be obtained by a Cantor-like construction using intervals on that line, with two ratio coefficients, both equal to λ. Moran s theorem does apply to this construction, and tells us that dim H A = dim B A = dim B A = α, where α solves the equation λ α + λα =. In this case, λ = λ = λ, and so we have α = log log λ.