Book XXIV contents · ← Ch 1 · master indexChapter 2 · Book XXI Ch 3 →
Principia Orthogona · Book XXIV · Linear and nonlinear · Chapter 2

Superposition: what “linear” buys

Start two motions, add them, and start a third from the sum. If the third is the sum of the first two, the system is linear and a whole theory opens. If not, you are in Strogatz’s other half of the world.

The object. A system x′ = f(x) in the plane, two starting points p and q, and the three motions that start at p, at q and at p + q. The question is whether the third equals the sum of the first two.

drag p (teal) and q (gold) · white = motion from p+q · dashed pink = motion from p plus motion from q
layers

1 · The solutions of a linear system add

Intuition

Pick the spiral sink or the saddle in the plate and drag p and q anywhere. The white curve (the motion that starts at p + q) lies on the dashed pink one (the sum of the two other motions), and the gap on the panel stays at rounding noise. For the pendulum the two separate at once. That difference is what Strogatz means when he says a linear system can be broken into parts, solved separately and recombined, and a nonlinear one cannot (printed pp. 8–9 of Nonlinear Dynamics and Chaos, 2018).

Statement
Proposition 3 (superposition)Let A be a 2×2 matrix. If x1(t) and x2(t) both solve x′ = Ax, then so does c1x1(t) + c2x2(t) for any real c1, c2.
Proof

Differentiate. (c1x1 + c2x2)′ = c1x1′ + c2x2′ = c1Ax1 + c2Ax2 = A(c1x1 + c2x2), the last step because multiplication by A respects sums and scalars. □

Remarks

Three things hold for the set of solutions: it contains the zero solution, it is closed under addition, and it is closed under scaling. A set of vectors with those three properties is the thing Chapter 3 will call a subspace, here a subspace of functions of t. The standard existence-and-uniqueness theorem for linear systems (cited, not proved in this book yet) says that a solution is determined by x(0), so the map from solutions to starting points is a bijection onto the plane; that is the sense in which the solution set is two-dimensional.

2 · The pendulum breaks it

Intuition

Doubling the amplitude of a swinging pendulum does not double the motion. The restoring force grows with the sine of the angle, not with the angle, and sin 2x is not 2 sin x.

Statement
Proposition 4For the pendulum equation θ″ = −sin θ, the solution θ with θ(0) = π/2 and θ′(0) = 0 has the property that 2θ is not a solution.
Proof

Since θ solves the equation, θ″(0) = −sin(π/2) = −1, so (2θ)″(0) = −2. A solution would need (2θ)″(0) = −sin(2θ(0)) = −sin π = 0. The two values differ, so 2θ fails the equation at t = 0. □

Remarks

Proposition 4 refutes scaling by one explicit solution; it does not describe how large the failure is for other starting points. The plate measures that numerically. Its integrator is fourth-order Runge–Kutta, and applied to a linear system one RK4 step is multiplication by a fixed matrix, so the plate’s linear presets are exactly additive and the gap it shows there is arithmetic noise, not a feature of the system. For the pendulum the gap is a property of the system.

Strogatz classifies even the forced harmonic oscillator as nonlinear in his scheme, because he counts time as a third variable. That is his convention, not a theorem, and the pendulum used here is the example he gives himself.

3 · Where this leads

Intuition

The rest of the book goes back and forth. We study linear algebra for its own sake, and at each chapter we ask what it says about a system that is not linear. Near a fixed point a smooth nonlinear system looks linear, and in that neighbourhood the linear picture often decides what happens. Book XXI proves part of that (Hartman–Grobman) and also proves the limits of using “looks the same” as evidence of “is the same.”

Remarks

This is a reading order. It does not claim that the linear and nonlinear halves share a mechanism, only that the second is studied through the first.

4 · Exercises

  1. Show that the solutions of x′ = Ax are closed under scaling by a constant, using the derivative directly.
  2. For x′ = x (one variable), write the solution that starts at p and check Proposition 3 by hand.
  3. Let u and v both solve x′ = Ax + b with b ≠ 0. Show that u − v solves x′ = Ax, and that u + v does not solve x′ = Ax + b. Is the solution set of x′ = Ax + b a subspace? Describe it in terms of one solution and the solutions of x′ = Ax.
  4. For the damped pendulum θ″ = −sin θ − 0.2θ′, show that scaling fails at t = 0 for the same starting point as in Proposition 4.
  5. Run the plate with p and q very small (near the origin) on the pendulum. What happens to the gap, and why?