MATH.13:5.3 - A symmetry of motion and what a numerical step preserves

Consider the planar model qdot=v, vdot=-q. Applying the same planar rotation R to q and v preserves these equations. Thus a trajectory starting at (q0,v0) produces a rotated trajectory starting at (R*q0,R*v0).

For q0=(1,0), v0=(0,1), a quarter-turn changes the initial data. The transformed trajectory solves another initial-value problem. Uniqueness for the original data therefore does not imply that its trajectory is fixed by every rotation.

In this model, J=x*vy-y*vx is angular momentum for unit mass. Direct differentiation gives

Jdot = x*ay-y*ax = x*(-y)-y*(-x)=0.

This establishes conservation along its trajectories. A radial law vdot=-k(t)*q gives the same cancellation. The proof uses the acceleration relation; the rotational description alone is not the argument.

Now compute with explicit Euler:

q_next=q+h*v; v_next=v-h*q.

The update is equivariant under the same rotations because R distributes over the linear combinations. Yet substitution gives

J_next=(1+h^2)*J.

Starting with J=1, h=0.1 and taking one hundred steps yields about 2.7048. Rotation equivariance of the scheme has survived while conservation of J has failed. If the question uses angular momentum, the computational construction must be changed or its error made acceptable for that use.

For this model, the update v_next=v-h*q; q_next=q+h*v_next preserves J by substitution. Its preservation of this quantity is one property of that scheme; questions about phase or other errors retain their own numerical analysis.