= Double pendulum equations of motion
Write $\Delta=\theta_1-\theta_2$. Applying the <Euler-Lagrange equations> to the <double pendulum Lagrangian>, then dividing by $m\ell^2$, gives
$$
\begin{aligned}
2\ddot\theta_1+\cos\Delta\,\ddot\theta_2+\sin\Delta\,\dot\theta_2^2+\frac{2g}{\ell}\sin\theta_1&=0,\\
\cos\Delta\,\ddot\theta_1+\ddot\theta_2-\sin\Delta\,\dot\theta_1^2+\frac{g}{\ell}\sin\theta_2&=0.
\end{aligned}
$$
For example, the upper <generalized momentum> is $\partial L/\partial\dot\theta_1=m\ell^2(2\dot\theta_1+\cos\Delta\,\dot\theta_2)$. Its <time> derivative contains $-m\ell^2\sin\Delta(\dot\theta_1-\dot\theta_2)\dot\theta_2$. Subtracting $\partial L/\partial\theta_1=-m\ell^2\sin\Delta\,\dot\theta_1\dot\theta_2-2mg\ell\sin\theta_1$ cancels the mixed product and leaves the positive $\sin\Delta\,\dot\theta_2^2$ term in the first equation. This cancellation is a useful check on the sign convention.
The <matrix> multiplying the two angular <accelerations> is
$$
\begin{pmatrix}2&\cos\Delta\\\cos\Delta&1\end{pmatrix},\qquad
\det=2-\cos^2\Delta\geq1.
$$
It is invertible for every configuration, so specified <angles> and <angular velocities> determine both angular <accelerations>. For the same reason $T>0$ whenever the <angular velocities> are not both zero. The <mass> cancels from the motion equations, while the timescale is set by $\sqrt{\ell/g}$.
As an exact constructed release, put $\theta_1(0)=\alpha$, $\theta_2(0)=0$, and both <angular velocities> zero. Solving the two equations at release yields
$$
\ddot\theta_1(0)=-\frac{2g}{\ell}\frac{\sin\alpha}{2-\cos^2\alpha},\qquad
\ddot\theta_2(0)=\frac{2g}{\ell}\frac{\sin\alpha\cos\alpha}{2-\cos^2\alpha}.
$$
For $0<\alpha<\pi/2$, the upper rod initially moves back toward vertical, while the lower rod initially moves away from it. The lower pivot's <acceleration> makes this possible even though the lower rod starts vertical. The <small-angle approximation> gives $\ddot\theta_1(0)\simeq-2g\alpha/\ell$ and $\ddot\theta_2(0)\simeq2g\alpha/\ell$.
The conserved <mechanical energy> is $E=T+V$, because the <Lagrangian> has no explicit <time> dependence. Directly, $dE/dt$ is $m\ell^2$ times the first equation multiplied by $\dot\theta_1$ plus the second multiplied by $\dot\theta_2$, so the equations imply $dE/dt=0$. This also supplies a check when computing a <trajectory>.
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