Math, Problem set 3 Solutions September, 13 Problem 1. ẍ = ω x. Solution. Following the general theory of conservative systems with one degree of freedom let us define the kinetic energy T and potential energy U as T = ẋ, x ω ξ dξ = ω x + U and thus the total energy (i.e. the Hamiltonian of the system) as T + U. Note that this system has a unique equilibrium at the origin. We can obtain a detailed plot of the phase curves by considering the graph of U:.1.1.8.6.. 1... 1. This potential has a local minimum at x = and so by the Morse lemma our system has a center singularity at the origin. Moreover the function is bounded below so all phase curves are bounded as well. The phase curves in the (x, ẋ) plane are as follows. 1.... 1. 1.... 1. 1
Problem. ẍ = k x. Solution. We proceed as we did in the previous problem. The potential energy is given by and its graph is x k ξ dξ = k x + U 1... 1....6.8 1. This time we have a local maximum at x = and thus a saddle singularity. Note that any non singular phase curve is unbounded. 1.... 1. 1.... 1. Problem 3. ẍ = x 3 x. Solution. The potential energy is given by and its graph is x ξ 3 ξ dξ = x + x + U..1 1. 1... 1. 1..1
We have a local minimum at the origin which gives rise to a center singularity and two local maximums at ±1 giving rise to saddles; both of these corresponding to the critical value 1. In particular we have heteroclinic curves conecting these two saddles. The phase portrait is as follows: 1. 1.... 1. 1. 1 1 Problem. ẍ = sin x. Solution. The potential energy is given by x sin ξ dξ = cos x + U whose graph is well known. The negative cosine function has local minima at every point of the form kπ, k Z which will correspond to cener singularities. On the other hand we get local maxima at (k + 1)π, k Z. These are saddle singularities with a same critical value. This will create heteroclinic orbits connecting any two adjacent saddles. The phase portrait now follows. 3
Problem. ẍ = sin x + a, a = 1, 1,. Solution. In these problem we are given a perturbation of the equation ẍ = sin x depending on a parameter a. The values of a have been intentionally chosen in such a way to emphasize the arrising bifurcations. In each case the potential energy is given by x sin ξ + a dξ = cos x ax + U. Note that the new term ax will cause the potential U(x) to grow to (or ) as x (x ). Let us consider the case a = 1. The potential function has local maxima and minima at every point where sin x equals 1. The local minima will give rise to center singularities and the local maxima to saddles. Note that for each critical value corresponding to a local maxima there exists a non critical point taking the same valule. This will create homoclinic orbits on each saddle. The phase portrait follows. Now consider the case a = 1. The potential has critical points everytime sin x = 1, however these are not local extrema. In fact, the potential has non positive derivative and is a strictly decreasing function.
In this case the singularities are degenerate and every non singular phase curve is unbounded. Lastly let us consider a =. Clearly this time the potential function has strictly positive derivative and thus our system has no singularities at all. The plot of the potential and the phase curves are as follow. 1 1 1 1
Problem 6. ẋ = x xy, ẏ = y xy. Solution. Yes, the system is Hamiltonian and a Hamiltonian is given by H(x, y) = x y xy. The only singularity of the system is the origin. Since H(, ) = we have a unique critical level set H(x, y) = which clearly corresponds to the union of the lines x =, y = and x y =. The phase portrait is the following. 1 1 1 1 Problem 7. (xy y )dx + (x xy)dy. Solution. Yes, this differential form is exact and it is the differential of the function F (x, y) = x y xy. Note that this function is exactly the Hamiltonian of the system in the previous problem. Since the integral curves are the level curves of F we obtain a similar plot (the difference is that the phase curves on Problem 6 are parametrized curves which have a natural orientation (given by the arrows in the plot) whereas the integral curves of a differential form are just curves embedded in R ). 1 1 1 1 6