1.7 Evolution of Dynamical State

Lagrange’s equations are ordinary differential equations that the path must
satisfy. They can be used to test if a proposed path is a realizable path of the
system. However, we can also use them to develop a path, starting with initial
conditions.

Assume that the state of a system is given by the tuple $\displaystyle{(t, q, v)}$. If we are
given a prescription for computing the acceleration $\displaystyle{a = A(t, q, v)}$, then

$\displaystyle{D^2 q = A \circ \Gamma[q]}$

and we have as a consequence

$\displaystyle{D^3 q = D( A \circ \Gamma[q]) = D_t A \circ \Gamma[q]}$

and so on.

So the higher-derivative components of the local tuple are given by functions $\displaystyle{D_t A, D_t^2 A, \dots}$. Each of these functions depends on lower-derivative components of the local tuple. All we need to deduce the path from the state is a function that gives the next-higher derivative component of the local description from the state. We use the Lagrange equations to find this function.

— Structure and Interpretation of Classical Mechanics

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Eq. (1.113):

$\displaystyle{ D_t F \circ \Gamma[q] = D(F \circ \Gamma[q]) }$

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Eq. (1.114):

\displaystyle{ \begin{aligned} D_t F (t, q, v, a, ...) &= \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v + \partial_2 F(t, q, v, a, ...) a + ... \\ \end{aligned} }

\displaystyle{ \begin{aligned} D_t F \circ \Gamma[q] (t) &= \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v(t) + \partial_2 F(t, q, v, a, ...) a(t) + ... \\ \end{aligned} }

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The Lagrange equation:

\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial}{\partial \dot q} L (t, q(t), \dot q(t)) \right) - \frac{\partial}{\partial q} L (t, q(t), \dot q(t)) &= 0 \end{aligned}}

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

\displaystyle{ \begin{aligned} \partial_1 L \circ \Gamma[q] &= D ( \partial_2 L \circ \Gamma[q]) \\ \\ &= \partial_0 ( \partial_2 L \circ \Gamma[q]) Dt + \partial_1 ( \partial_2 L \circ \Gamma[q]) Dq + \partial_2 ( \partial_2 L \circ \Gamma[q]) Dv \\ \\ &= \partial_0 \partial_2 L \circ \Gamma[q] + ( \partial_1 \partial_2 L \circ \Gamma[q]) Dq + (\partial_2 \partial_2 L \circ \Gamma[q]) D^2 q \\ \\ \end{aligned}}

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\displaystyle{ \begin{aligned} (\partial_2 \partial_2 L \circ \Gamma[q]) D^2 q &= \partial_1 L \circ \Gamma[q] - \partial_0 \partial_2 L \circ \Gamma[q] - (\partial_1 \partial_2 L \circ \Gamma[q]) Dq \\ \\ D^2 q &= \left[ \partial_2 \partial_2 L \circ \Gamma[q] \right]^{-1} \left\{ \partial_1 L \circ \Gamma[q] - \partial_0 \partial_2 L \circ \Gamma[q] - (\partial_1 \partial_2 L \circ \Gamma[q]) Dq \right\} \\ \\ \end{aligned}}

where $\displaystyle{\left[ \partial_2 \partial_2 L \circ \Gamma \right]}$ is a structure that can be represented by a symmetric square matrix, so we can compute its inverse.

— Me@2022-06-30 11:33:27 AM

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