A particle of unit mass moves in a smooth one-dimensional potential $V(x)$. Its path $x(t)$ is such that the action integral

$$S[x]=\int_{a}^{b} L(x, \dot{x}) d t$$

has a stationary value, where $a$ and $b>a$ are constants, a dot denotes differentiation with respect to time $t$

$$L(x, \dot{x})=\frac{1}{2} \dot{x}^{2}-V(x)$$

is the Lagrangian function and the initial and final positions $x(a)$ and $x(b)$ are fixed.

By considering $S[x+\epsilon \xi]$ for suitably restricted functions $\xi(t)$, derive the differential equation governing the motion of the particle and obtain an integral expression for the second variation $\delta^{2} S$.

If $x(t)$ is a solution of the equation of motion and $x(t)+\epsilon u(t)+O\left(\epsilon^{2}\right)$ is also a solution of the equation of motion in the limit $\epsilon \rightarrow 0$, show that $u(t)$ satisfies the equation

$$\ddot{u}+V^{\prime \prime}(x) u=0$$

If $u(t)$ satisfies this equation and is non-vanishing for $a \leqslant t \leqslant b$, show that

$$\delta^{2} S=\frac{1}{2} \int_{a}^{b}\left(\dot{\xi}-\frac{\dot{u} \xi}{u}\right)^{2} d t$$

Consider the simple harmonic oscillator, for which

$$V(x)=\frac{1}{2} \omega^{2} x^{2}$$

where $2 \pi / \omega$ is the oscillation period. Show that the solution of the equation of motion is a local minimum of the action integral, provided that the time difference $b-a$ is less than half an oscillation period.