MATH.10:5.2 - Compare histories between fixed endpoints

Consider a one-dimensional free particle with mass m>0, elapsed time T>0, and fixed positions q(0)=0, q(T)=L. The model’s action is

J[q] = integral from 0 to T of (m/2)*(qdot(t))^2 dt.

Take the straight history q0(t)=L*t/T. For any continuously differentiable eta with eta(0)=eta(T)=0, construct q_h(t)=q0(t)+h*eta(t). Expanding the square gives

DeltaJ(h) = (m*h*L/T)*integral eta_dot(t) dt + (m*h^2/2)*integral eta_dot(t)^2 dt.

The first integral is eta(T)-eta(0)=0. The remaining term is nonnegative, and it vanishes for a changed history only when its derivative change is identically zero; the fixed endpoints then force the change itself to vanish. Every continuously differentiable history with these endpoints can be expressed as q0 plus such a change. The straight history therefore uniquely minimizes this functional over that class.

The construction exposes the reason: the fixed endpoints remove the cross term, while every remaining velocity deviation adds a nonnegative contribution.

Now include a restoring potential. In dimensionless variables, use the harmonic-oscillator action J[q]=integral from 0 to 2*pi of (qdot^2-q^2)/2 dt, with q(0)=q(2*pi)=0. At the history q=0, every admissible family q_h=h*eta has zero first variation. Yet the admissible directions eta=sin(t/2) and eta=sin(3*t/2) give respectively DeltaJ=-3*pi*h^2/8 and DeltaJ=5*pi*h^2/8: integration uses integral eta^2 dt=pi and integral eta_dot^2 dt=n^2*pi/4 for these directions with n=1 and n=3. Arbitrarily small changes of both lower and higher action are available. The stationary history is therefore a saddle, not a minimum.

Changing an endpoint condition also requires rebuilding the allowed histories before reusing the free-particle argument. Its vanishing cross term depended on both endpoint values.

This example uses a supplied classical model. Constructing an appropriate action for an unfamiliar physical system is a further physical and mathematical modeling task.