Dynamics, Lectures 5 & 6: Oxford Mathematics 1st Year Student Lecture

Overview

Oxford Mathematics' lecture on linear stability of dynamical systems introduces Taylor expansions to approximate system behavior near equilibrium points. It details how the sign of the derivative of the force function (or second derivative of potential energy) determines stability, leading to oscillatory (stable) or exponential (unstable) solutions. The lecture extends this analysis to multi-variable systems, using matrix methods and eigenvalues to determine stability, and discusses the implications of energy conservation in the absence of damping.

Key takeaways

Linear stability analysis approximates system behavior near equilibrium by linearizing the equations of motion.
The sign of f'(xe) (or v''(xe)) determines stability: negative for stable oscillations, positive for unstable exponential growth.
For multi-variable systems, stability is determined by the eigenvalues of the Jacobian matrix; all eigenvalues must have non-positive real parts for stability.
Conservative systems (like springs without damping) exhibit only pure oscillations or instability, never decaying oscillations.
The three-spring system has two natural frequencies, sqrt(k/m) and sqrt(3k/m), arising from coupled oscillations.

Chapters

0:12 Introduction to Linear Stability Analysis

Focus on one-dimensional dynamical systems of the form m x.. = f(x).
Equilibrium points occur where f(xe) = 0.
Taylor expansion of f(x) around xe is used to approximate behavior near equilibrium.

1:45 Taylor Expansion and Perturbation Analysis

Define psi(t) = x(t) - xe as the perturbation from equilibrium.
Approximate f(xe + psi) ≈ f(xe) + f'(xe) * psi.
Newton's second law becomes m * psi.. ≈ f'(xe) * psi.

8:37 Interpreting the Linearized Equation

The linearized equation m * psi.. = f'(xe) * psi dictates behavior near equilibrium.
If f'(xe) < 0, solutions are sines/cosines (stable oscillation).
If f'(xe) > 0, solutions are exponentials (unstable growth).

11:52 Case 1: Stable Equilibrium (f'(xe) < 0)

f'(xe) < 0 implies v''(xe) > 0 (minimum in potential energy).
Solutions are bounded, e.g., psi(t) = A cos(omega*t + phi).
Omega is defined as sqrt(-f'(xe)/m).

17:02 Case 2: Unstable Equilibrium (f'(xe) > 0)

f'(xe) > 0 implies v''(xe) < 0 (maximum in potential energy).
Solutions involve exponentials, e.g., psi(t) = A*e^(pt) + B*e^(-pt).
p is defined as sqrt(f'(xe)/m).

19:08 Definition of Stability

Stable equilibrium means if the system starts near equilibrium, it stays near.
Unstable equilibrium is characterized by the presence of a positive exponential term, even if a decaying term also exists.

23:30 Case 3: f'(xe) = 0 (Marginal Case)

When f'(xe) = 0, higher-order terms in the Taylor expansion become significant.
This case requires analyzing terms beyond the linear approximation.
Examples like f(x) = -x^3 and f(x) = -x^4 illustrate different behaviors at x=0.

25:14 Example: f(x) = -x^3 vs. f(x) = -x^4

For f(x) = -x^3, f'(0) = 0, f''(0) = 0, f'''(0) = -6. Equilibrium at x=0 is unstable.
For f(x) = -x^4, f'(0) = 0, f''(0) = 0, f'''(0) = 0, f''''(0) = -24. Equilibrium at x=0 is stable.
Energy landscape analogy: x^3 is a saddle point, x^4 is a flat-bottomed valley.

30:04 Complex Example: Mass on a Wire with a Spring

A mass m slides on a horizontal wire, attached by a spring to point (0, d).
The system's position is described by x(t), and the spring has rest length L and constant k.
The goal is to derive Newton's second law and analyze equilibrium stability.

37:13 Deriving the Force Equation

The spring force magnitude depends on the actual length (sqrt(x^2 + d^2)) minus rest length L.
The force vector tau is derived using geometry (cosine and sine of angle theta).
Newton's second law is a vector equation, requiring consideration of normal forces from the wire.

46:59 Simplified Equation of Motion

After algebraic manipulation, the equation of motion for x(t) is derived.
x.. = (k/m) * [ (L / sqrt(d^2 + x^2)) - 1 ] * x.
This equation is analyzed for equilibria and their stability.

48:41 Equilibria and Stability for the Spring System

Equilibria occur when x = 0 or when sqrt(d^2 + x^2) = L.
Stability at x=0 depends on L vs. D: stable if L < D, unstable if L > D.
The second equilibrium (sqrt(d^2 + x^2) = L) exists only if L > D and is stable.

1:00:53 Bifurcation Diagram for the Spring System

A bifurcation diagram plots equilibrium points (XE) against a tuning parameter (e.g., D or L).
Solid lines indicate stable equilibria, dashed lines indicate unstable.
As D decreases below L, the x=0 equilibrium becomes unstable, and new stable equilibria appear.

1:09:10 Stability Analysis without Explicit Derivative Calculation

For a complex system, stability can sometimes be determined by Taylor expanding the perturbation psi around equilibrium.
Example: x.. = x^2 * e^(2x) * cos(x) at xe = pi/2.
Expansion shows psi.. ≈ - (sin(pi/2)) * psi = -psi, indicating stable oscillations.

1:15:00 Introduction to Coupled Oscillators

Extends stability analysis to systems with multiple variables (e.g., two masses, two degrees of freedom).
The core idea is that the dynamics of one variable depend on the others.
The linearized system can be represented by a matrix equation: psi_dot_dot = M * psi.

1:39:11 Linearized Equations and the Jacobian Matrix

The matrix M is the Jacobian matrix of the system's vector field, evaluated at the equilibrium point.
M = [[dF/dx, dF/dy], [dG/dx, dG/dy]] at (xe, ye).
The linearized equation is psi_dot_dot = M * psi, where psi is a vector of perturbations.

1:44:19 Eigenvalue Problem for Coupled Systems

Seeking exponential solutions psi = A * e^(lambda*t).
Plugging into psi_dot_dot = M * psi leads to (M - lambda^2 * I) * A = 0.
This is an eigenvalue problem: lambda^2 must be an eigenvalue of M, and A is the corresponding eigenvector.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Oxford Mathematics.