Solutions

Solutions to set A2, PHYD38 2025

Problem 2.1 [10p] = Strogatz 3.6.2

Problem 2.2 [10p] = Strogatz 4.3.6

Problem 2.3 [10p] Same matrix, same type of fixed point

Consider a 2-d linear system
dx/dt = A x + c,
where x is a variable vector,c is a constant vector, and A a constant-coefficient 2x2 matrix. Denote the fixed point as x*.

Show that in the new coordinate system, with the origin at the fixed point, the new variable, z= x - x*, obeys a simpler equation
dz/dt = A z
with the same matrix A as original system; therefore analysing the type of the fixed point of the original system, we can disregard the non-zero c.

Solution

The general proof that a change of variable to z =x - x* results in equation z/dt = A z consists of writing the fixed point condition as A x* = c, and substituting into the original equation x = x* + z. The result is
dz/dt = A(z + x*) + c = A z + A x* + c = A z, q.e.d.
(multiplication by a matrix is a linear operation.)

Problem 2.4. [15p] Rotating saddle

Show that any saddle equilibrium point in planar dynamical system is unstable, no matter if and how fast the system in which the coordinates x(t) and y(t) are defined rotates. Assume a quite general form of a saddle below, where the effective potential U(x,y) is due both to gradient force and centrifugal force due to frame rotation. Being a saddle, U has a positive curvature in one direction (call it y) and negative curvature in the perpendicular x direction, such as in U(x,y) = y²/2 - x²/2. This is a general form of saddle point potential, since one can always rotate and rescale axes to yield it, without modifying the problem. Remembering that acceleration is a negative gradient of U(x,y) and that, in a system rotating with angular velocity vector Ω perpendicular to the plane of motion, one needs to also add the Coriolis acceleration:
d²r/dt² = -∇U - 2Ω × dr/dt, where r = (x,y).

Hint: This is a system with dimension 4, but do not panic . Write out the x- and y-components of the equation of motion, substitute trial solutions with time-dependence of the form (x,y) = e^λt(u,v), where u,v do not vary in time.

SOLUTION

First of all, there was a typo in the original text of this problem, time derivative on vector r was missing in Coriolis force term. Some of you noticed it, but rather late. If you noticed, your max points for this problem was 15, if not then 14. The incorrect dynamical system is somewhat interesting on its own, and can be analyzed with the same approach as the correct one; it's instability is conditional, not unconditional (as in the text of the problem and the solution below). The reward for those who noticed the typo is a symbolic 1 pt (out of 55 points for the whole A2, so it's arguably fair).
d²x/dt² = x + 2Ω dy/dt
d²y/dt² = -y - 2Ω dx/dt.

Substitution of (x,y) = e^λt(u,v) and simplification turns this system into
-λ²u = u + 2Ωλ v
-λ²v = -v - 2Ωλ u.
Either writing this as matrix equation and requiring that det(matrix) vanishes, or equivalently eliminating u from equations and canceling v on both sides, we get a bi-quadratic characteristic equation for λ
(1+λ²)(1-λ²) = 4λ²Ω².
or
λ² = -2 Ω² ± (1+4Ω⁴)^1/2.
The lower choice of sign eventually produces complex-conjugate, imaginary λ's (oscillation). But for *any* rotation Ω , the upper sign choice above gives λ² > 0, and thus one real, positive solution for λ. Make sure you see that. That eigenfrequency signifies an exponential instability, because exp(λt) blows up in time.

Problem 2.5 [10p]. Characterize the fixed point (0,0)

dx/dt = - y (1+ ax²)
dy/dt = x (1+ ay²),

for all real parameters a. Is it linearly stable? Is it nonlinearly stable? Is it time-reversible? Does it have an integral of motion like the system in the 2025 midterm?

SOLUTION

The (0,0) fixed point is a linear center. The system is time-reversible (right-hand side odd in y in 1st eq., even in 2nd eq.). But does it make the fixed point a nonlinear center? Yes, since for any a, close to (0,0) trajectories are circles (closed curves, not spirals, which are not symmetric under reflection of 'velocities' y → -y).

The system also has an energy-like integral. Eliminating dt, we get
dx / [-y (1+ ax²)] = dy / [x (1+ ay²)], or
x dx + a x y² dx = -y dy -a y x² dy.
But x y² dx + y x² dy = d(x²y²/2), therefore:
E = x² + y² +a x²y² = const.
on trajectories. Curves around (0,0) have E close to zero, and independent of the value of a are closed. They are the more circular, the closer they are to the center, because a x²y² ≪ x² + y² then. Thus the center is also nonlinearly stable, neither attractive nor repelling.

By the way, try asking the Wolfram Alpha to "plot curve x^2 +y^2 +x^2y^2 = 3" and you will see that for a=1 and E=3 (or 1, or 2, or 100) the trajectories are definitely not circles (though still closed).