Question

A mechanical system with generalised coordinates \[q_{1}\] and \[q_{2}\], has Lagrangian \[L(q_{1}, q_{2}, q_{1}\prime, q_{2}\prime)\]. State Lagrange's equations for the system, and deduce that
\[q _{1}\prime \frac{ \delta L }{ \delta q_{1}\prime } (q_{1}, q_{2}, q_{1}\prime, q_{2}\prime) + q _{2}\prime \frac{ \delta L }{ \delta q_{2}\prime } (q_{1}, q_{2}, q_{1}\prime, q_{2}\prime) - L(q_{1}, q_{2}, q_{1}\prime, q_{2}\prime)\]
is constant during the motion.

Answer 1

Are "Lagrange's equations" the same as the Euler-Lagrange equations?
\[\frac{\partial L}{\partial q_1} = \frac{\text{d}}{\text{d}x}\frac{\partial L}{\partial q_1'}\]
\[\frac{\partial L}{\partial q_2} = \frac{\text{d}}{\text{d}x}\frac{\partial L}{\partial q_2'}\]

I'm assuming that \(q_1\) and \(q_2\) are functions of some variable \(x\) that evolves during "motion".
(\(x\) is probably time).

You want to show that
\[\frac{\text{d}}{\text{d}x}\Big(q_1'\frac{\partial L}{\partial q_1'} + q_2'\frac{\partial L}{\partial q_2'} + L\Big) = 0\]

so

\[\frac{\text{d}}{\text{d}x}\Big(q_1'\frac{\partial L}{\partial q_1'} + q_2'\frac{\partial L}{\partial q_2'} + L\Big) = q_1''\frac{\partial L}{\partial q_1'} +q_1'\frac{\text{d}}{\text{d}x}\frac{\partial L}{\partial q_1'} + q_2''\frac{\partial L}{\partial q_2'} +q_2'\frac{\text{d}}{\text{d}x}\frac{\partial L}{\partial q_2'} - \frac{\partial L}{\partial x}\]

Here it's helpful to write out \(q_1'',q_1'\) etc explicitly and From the Euler-Lagrange equations you can make substitutions.

\[\frac{\partial q_1'}{\partial x}\frac{\partial L}{\partial q_1'} + \frac{\partial q_1}{\partial x}\frac{\partial L}{\partial q_1} + \frac{\partial q_2'}{\partial x}\frac{\partial L}{\partial q_2'} + \frac{\partial q_2}{\partial x}\frac{\partial L}{\partial q_2} - \frac{\partial L}{\partial x} \]

now it should be obvious that the first 4 terms are a chain-rule-like expansion of the last term, hence the whole thing equals zero.