Possion’s Lagrange Equation

Structure and Interpretation of Classical Mechanics

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Ex 1.10 Higher-derivative Lagrangians

Derive Lagrange’s equations for Lagrangians that depend on accelerations. In particular, show that the Lagrange equations for Lagrangians of the form \displaystyle{L(t, q, \dot q, \ddot q)} with \displaystyle{\ddot{q}} terms are

\displaystyle{D^2(\partial_3L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q]) + \partial_1 L \circ \Gamma[q] = 0}

In general, these equations, first derived by Poisson, will involve the fourth derivative of \displaystyle{q}. Note that the derivation is completely analogous to the derivation of the Lagrange equations without accelerations; it is just longer. What restrictions must we place on the variations so that the critical path satisfies a differential equation?


Varying the action

\displaystyle{ \begin{aligned}   S[q] (t_1, t_2) &= \int_{t_1}^{t_2} L \circ \Gamma [q] \\   \eta(t_1) &= \eta(t_2) = 0 \\   \end{aligned}}

\displaystyle{ \begin{aligned}   \delta_\eta S[q] (t_1, t_2) &= 0 \\   \end{aligned}}

\displaystyle{ \begin{aligned}   \delta_\eta S[q] (t_1, t_2) &= \int_{t_1}^{t_2} \delta_\eta \left( L \circ \Gamma [q] \right) \\   \end{aligned}}

\displaystyle{ \begin{aligned}     \delta_\eta I [q] &= \eta \\  \delta_\eta g[q] &= D \eta~~~\text{with}~~~g[q] = Dq \\   \end{aligned}}

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Let \displaystyle{h[q] = D^2 q}.

\displaystyle{ \begin{aligned}   \delta_\eta h[q]   &= \lim_{\epsilon \to 0} \frac{h[q+\epsilon \eta] - h[q]}{\epsilon} \\   &= \lim_{\epsilon \to 0} \frac{D^2 (q+\epsilon \eta) - D^2 q}{\epsilon} \\   &= \lim_{\epsilon \to 0} \frac{D^2 q + D^2 \epsilon \eta - D^2 q}{\epsilon} \\   &= \lim_{\epsilon \to 0} \frac{D^2 \epsilon \eta}{\epsilon} \\   &= D^2 \eta \\   \end{aligned}}

\displaystyle{ \begin{aligned}   \Gamma [q] (t) &= (t, q(t), D q(t), D^2 q(t)) \\  \delta_\eta \Gamma [q] (t) &= (0, \eta (t), D \eta (t), D^2 \eta (t)) \\  \end{aligned}}

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Chain rule of functional variation

\displaystyle{ \begin{aligned} &\delta_\eta F[g[q]] \\   &= \delta_\eta (F \circ g)[q] \\   &= \delta_{ \left( \delta_\eta g[q] \right)} F[g] \\ \end{aligned}}

Since variation commutes with integration,

\displaystyle{ \begin{aligned}   \delta_\eta S[q] (t_1, t_2)   &= \delta_\eta \int_{t_1}^{t_2} L \circ \Gamma [q] \\   &= \int_{t_1}^{t_2} \delta_\eta \left( L \circ \Gamma [q] \right) \\   \end{aligned}}

By the chain rule of functional variation:

\displaystyle{ \begin{aligned}   \delta_\eta L \circ \Gamma [q] = \delta_{ \left( \delta_\eta \Gamma[q] \right)} L[\Gamma[q]] \\   \end{aligned}}

If \displaystyle{L} is path-independent,

\displaystyle{ \begin{aligned}   \delta_\eta \left( L \circ \Gamma [q] \right) = \left( DL \circ \Gamma[q] \right) \delta_\eta \Gamma[q] \\   \end{aligned}}

But is \displaystyle{L} path-independent?

The \displaystyle{L \circ \Gamma [.]} is path-dependent. Its input is a path \displaystyle{q}, not just \displaystyle{q(t)}, the value of \displaystyle{q} at the time \displaystyle{t}. However, \displaystyle{L(.)} itself is a path-independent function, because its input is not a path \displaystyle{q}, but a quadruple of values \displaystyle{(t, q(t), Dq(t), D^2 q(t))}.

\displaystyle{ \begin{aligned}   L \circ \Gamma [q] = L(t, q(t), Dq(t), D^2 q(t)) \\   \end{aligned}}

Since \displaystyle{L} is path-independent,

\displaystyle{ \begin{aligned}   \delta_\eta \left( L \circ \Gamma [q] \right)   = \left( DL \circ \Gamma[q] \right) \delta_\eta \Gamma[q] \\   \end{aligned}}

\displaystyle{ \begin{aligned}   &\delta_\eta S[q] (t_1, t_2) \\  &= \int_{t_1}^{t_2} \delta_\eta L \circ \Gamma [q] \\   &= \int_{t_1}^{t_2} \left( D \left( L \circ \Gamma[q] \right) \right) \delta_\eta \Gamma[q]  \\   &= \int_{t_1}^{t_2} \left( D \left( L(t, q, D q, D^2 q) \right) \right) (0, \eta (t), D \eta (t), D^2 \eta (t))  \\   &= \int_{t_1}^{t_2} \left[ \partial_0 L \circ \Gamma[q], \partial_1 L \circ \Gamma[q], \partial_2 L \circ \Gamma[q], \partial_3 L \circ \Gamma[q] \right] (0, \eta (t), D \eta (t), D^2 \eta (t))  \\   &= \int_{t_1}^{t_2} (\partial_1 L \circ \Gamma[q]) \eta + (\partial_2 L \circ \Gamma[q]) D \eta + (\partial_3 L \circ \Gamma[q]) D^2 \eta \\                        &=   \int_{t_1}^{t_2} (\partial_1 L \circ \Gamma[q]) \eta      + \left[ \left. (\partial_2 L \circ \Gamma[q]) \eta \right|_{t_1}^{t_2} - \int_{t_1}^{t_2} D(\partial_2 L \circ \Gamma[q]) \eta \right]     + \int_{t_1}^{t_2} (\partial_3 L \circ \Gamma[q]) D^2 \eta \\                        \end{aligned}}

Since \displaystyle{\eta(t_1) = 0} and \displaystyle{\eta(t_2) = 0},

\displaystyle{ \begin{aligned}   \delta_\eta S[q] (t_1, t_2)   &=   \int_{t_1}^{t_2} (\partial_1 L \circ \Gamma[q]) \eta      - \int_{t_1}^{t_2} D(\partial_2 L \circ \Gamma[q]) \eta      + \int_{t_1}^{t_2} (\partial_3 L \circ \Gamma[q]) D^2 \eta \\                        \end{aligned}}

Here is a trick for integration by parts:

As long as the boundary term \displaystyle{\left. u(t)v(t) \right|_{t_1}^{t_2} = 0},

\displaystyle{\int_{t_1}^{t_2} u(t) dv(t) = - \int_{t_1}^{t_2} v(t) du(t)}

So if \displaystyle{D \eta(t_1) = 0} and \displaystyle{D \eta(t_2) = 0},

\displaystyle{ \begin{aligned}   \delta_\eta S[q] (t_1, t_2)   &= \int_{t_1}^{t_2} (\partial_1 L \circ \Gamma[q]) \eta        - \int_{t_1}^{t_2} D(\partial_2 L \circ \Gamma[q]) \eta        - \int_{t_1}^{t_2} D(\partial_3 L \circ \Gamma[q]) D \eta \\                        \end{aligned}}

Since \displaystyle{\eta(t_1) = 0} and \displaystyle{\eta(t_2) = 0},

\displaystyle{ \begin{aligned}   \delta_\eta S[q] (t_1, t_2)   &= \int_{t_1}^{t_2} (\partial_1 L \circ \Gamma[q]) \eta        - \int_{t_1}^{t_2} D(\partial_2 L \circ \Gamma[q]) \eta        + \int_{t_1}^{t_2} D^2 (\partial_3 L \circ \Gamma[q]) \eta \\                        \end{aligned}}

\displaystyle{ \begin{aligned}   \delta_\eta S[q] (t_1, t_2)   &= \int_{t_1}^{t_2} \left[ (\partial_1 L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q]) + D^2 (\partial_3 L \circ \Gamma[q]) \right] \eta \\                        \end{aligned}}

By the principle of stationary action, \displaystyle{ \delta_\eta S[q] (t_1, t_2) = 0}. So

\displaystyle{ \begin{aligned}   0   &= \int_{t_1}^{t_2} \left[ (\partial_1 L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q]) + D^2 (\partial_3 L \circ \Gamma[q]) \right] \eta \\                        \end{aligned}}

Since this is true for any function \displaystyle{\eta(t)} that satisfies \displaystyle{\eta(t_1) = \eta(t_2) = 0} and \displaystyle{D\eta(t_1) = D\eta(t_2) = 0},

\displaystyle{ \begin{aligned}   (\partial_1 L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q]) + D^2 (\partial_3 L \circ \Gamma[q]) &= 0 \\                        D^2 (\partial_3 L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q]) + \partial_1 L \circ \Gamma[q] &= 0 \\                        \end{aligned}}

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Note:

The notation of the path function \displaystyle{\Gamma} is \displaystyle{\Gamma[q](t)}, not \displaystyle{\Gamma[q(t)]}.

The notation \displaystyle{\Gamma[q](t)} means that \displaystyle{\Gamma} takes a path \displaystyle{q} as input. And then returns a path-independent function \displaystyle{\Gamma[q]}, which takes time \displaystyle{t} as input, returns a value \displaystyle{\Gamma[q](t)}.

The other notation \displaystyle{\Gamma[q(t)]} makes no sense, because \displaystyle{\Gamma[.]} takes a path \displaystyle{q}, not a value \displaystyle{q(t)}, as input.

— Me@2020-11-11 05:37:13 PM

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2020.11.11 Wednesday (c) All rights reserved by ACHK