# How can I show that the action of a SHO is a saddle-point solution if $t_{f} - t_{0} > T/2$?

In this post and in this post, QMechanic claims that a simple harmonic oscillator with Dirichlet boundary conditions has saddle-point solutions if $$t_{f} - t_{0} > T/2$$ where $$T$$ is the characteristic period of the oscillator. I am curious, how can one prove this assertion?

I tried to consider $$t_{0} = 0$$ and $$x(0) = 0$$ for simplicity with Lagrangian $$L = \tfrac{1}{2}m\dot{x}^{2} - \tfrac{1}{2}m\omega^{2}x^{2}$$ where $$T = 2\pi/\omega$$. Going through the usual steps, I found the solution $$x(t) = \frac{x_{f}\sin\omega t}{\sin\omega t_{f}}$$ where $$x_{f} = x(t_{f})$$. Integrating out the Lagrangian gives me the action $$S[x] = \frac{1}{2}m\omega x_{f}^{2} \cot\omega t_{f}.$$ How do I use this to show that $$S[x]$$ is neither a local minimum nor a local maximum if $$t_{f} > T/2$$?

Theorem. Consider a simple harmonic oscillator (SHO) $$S[x]~=~\int_{t_i}^{t_f} \! dt~L, \qquad L~=~\frac{m}{2}\dot{x}^2 - \frac{k}{2}x^2,\tag{1}$$ with characteristic frequency $$\frac{2\pi}{T}~=~\omega~=~\sqrt{\frac{k}{m}},\tag{2}$$ and Dirichlet boundary conditions (BC) $$x(t_i)~=~x_i \quad\text{and}\quad x(t_f)~=~x_f.\tag{3}$$ Then the action $$S$$ has a minimum if $$\Delta t~:=~t_f-t_i~\leq~ \frac{T}{2},\tag{4}$$ (if $$\Delta t=\frac{T}{2}$$ there is a zeromode), and $$S$$ has a saddle point if $$\Delta t > \frac{T}{2}$$.

Sketched proof:

1. The classical solution on the SHO is on the form $$x_{\rm cl}(t)~=~A\cos(\omega t) + B\sin(\omega t).\tag{5}$$

2. An arbitrary virtual path that satisfies the BC (3) is of the form $$x(t)~=~x_{\rm cl}(t)+ y(t),\tag{6}$$ where the fluctuation part is a Fourier series $$y(t)~=~\sum_{n\in\mathbb{N}} a_n\sin\left(n\pi \frac{t-t_i}{\Delta t}\right),\qquad a_n~\in~\mathbb{R}. \tag{7}$$

3. The quadratic action (1) contains no cross-terms $$S[x]~=~S[x_{\rm cl}] + S[y]\tag{8}$$ [because of the EL equation and the BC (3)].

4. The fluctuation part doesn't get any cross-terms contributions: \begin{align} S[y]~=~&\ldots \cr ~=~&\frac{\Delta t}{4}\sum_{n\in\mathbb{N}} a_n^2 \left[m\left(\frac{n\pi}{\Delta t}\right)^2 -k \right]\cr ~=~&\frac{m\pi^2\Delta t}{4}\sum_{n\in\mathbb{N}} a_n^2 \left[\left(\frac{n}{\Delta t}\right)^2 -\left(\frac{2}{T}\right)^2 \right]. \end{align} \tag{9}

5. Case $$\Delta t\leq\frac{T}{2}$$: Then $$S[y]\geq 0$$. If $$S[y]=0$$, then $$y$$ can at most contain an $$n=1$$ mode.

6. Case $$\Delta t<\frac{T}{2}$$: Then $$S[y]=0$$ implies $$y=0$$. It's a minimum.

7. Case $$\Delta t=\frac{T}{2}$$: $$n=1$$ mode is a zeromode.

8. Case $$\Delta t>\frac{T}{2}$$: The $$n=1$$ mode is unbounded from below. It's a saddle point. $$\Box$$