= Solution
Take a <variation> $x+\varepsilon\xi$ with $\xi(a)=\xi(b)=0$. Expanding the <action> gives
$$
S[x+\varepsilon\xi]
=S[x]+\varepsilon\int_a^b\bigl(\dot x\dot\xi-V'(x)\xi\bigr)\,dt
+\frac{\varepsilon^2}{2}\int_a^b\bigl(\dot\xi^2-V''(x)\xi^2\bigr)\,dt
+O(\varepsilon^3).
$$
An <integration by parts> and the fixed endpoint conditions give the <first variation>
$$
\delta S=-\int_a^b\bigl(\ddot x+V'(x)\bigr)\xi\,dt.
$$
The <fundamental lemma of the calculus of variations> therefore yields the <Euler-Lagrange equation>
$$
\boxed{\ddot x+V'(x)=0},
$$
and the <second variation> is
$$
\boxed{\delta^2S=\frac12\int_a^b\bigl(\dot\xi^2-V''(x)\xi^2\bigr)\,dt}.
$$
Linearizing the equation of motion about $x$ gives
$$
0=\ddot x+V'(x)+\varepsilon\bigl(\ddot u+V''(x)u\bigr)+O(\varepsilon^2),
$$
so the <Jacobi equation> is
$$
\ddot u+V''(x)u=0.
$$
When $u$ has no zero on $[a,b]$,
$$
\dot\xi^2-V''(x)\xi^2
=\left(\dot\xi-\frac{\dot u}{u}\xi\right)^2
+\frac d{dt}\left(\frac{\dot u}{u}\xi^2\right).
$$
The total derivative integrates to zero because $\xi(a)=\xi(b)=0$, hence
$$
\boxed{\delta^2S=\frac12\int_a^b
\left(\dot\xi-\frac{\dot u}{u}\xi\right)^2dt\geq0}.
$$
For the <simple harmonic oscillator>, the Jacobi equation is $\ddot u+\omega^2u=0$. Put $t_0=(a+b)/2$ and choose
$$
u(t)=\cos\bigl(\omega(t-t_0)\bigr).
$$
If $b-a<\pi/\omega$, then $|\omega(t-t_0)|<\pi/2$ throughout $[a,b]$, so $u$ is positive there. The preceding square identity proves that the classical path is a local minimum of the action whenever the elapsed time is less than half an <oscillation period>.
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