 Solution

Set $l=m=g=1$. At $({\theta }_1,{\theta }_2)=(0,0)$ the spring has $d=r$, so its potential energy has zero gradient: ${\partial }_j{V}_s=k(d-r){\partial }_{j}d=0$. The gravitational potential energy has ${\partial }_j{V}_g=\sin {\theta }_j=0$ there. Zero velocities therefore satisfy the Euler-Lagrange equations, proving this is an equilibrium.

For fixed $r>0$, let $\epsilon =\max (|{\theta }_1|,|{\theta }_2|)$ and $\delta ={\theta }_2-{\theta }_1$. The Taylor series giveThus, near the downward equilibrium, the horizontal separation is positive andIn particular, $V_s=k{\delta }^2/2+O({\epsilon }^4)$ and $V_g=-2+({\theta }_1^2+{\theta }_2^2)/2+O({\epsilon }^4)$. Dropping the additive constant and terms beyond quadratic order gives the harmonic approximationFor physical spring constant $k\ge 0$, the quadratic potential energy is a positive-definite quadratic form. The downward position is therefore a stable equilibrium in Lagrangian mechanics, and its nearby motion admits small oscillations.