 Solution

Write $u=\pi x^2/l_0^2$ and assume $V_0>0$. The derivative of the potential is $V^{\prime }(x)=-2\pi V_{0}x\sin u/l_0^2$. Its equilibrium positions are zero and $x=\pm l_0\sqrt{n}$ for positive integers $n$. At a nonzero equilibrium,Thus the stable equilibria and their small-oscillation normal mode frequencies areEven positive $n$ give unstable maxima. At zero the quadratic term vanishes but $V(x)=V_0-V_0{\pi }^2{x}^4/(2l_0^4)+\cdots$, so the equilibrium is still unstable; a zero linearized frequency does not prove stability. If the unspecified sign of $V_0$ were chosen negative, even-$n$ positions would be stable instead, together with a degenerate quartic minimum at zero whose frequency tends to zero with oscillation amplitude.