Central Field is Field of Functional

Theorem
Let $ \mathbf y $ be an N-dimensional vector.

Let $ J $ be a functional, such that:


 * $ \displaystyle J \left [ { \mathbf y } \right ] = \int_a^b F \left ( { x, \mathbf y, \mathbf y' } \right ) \mathrm d x $

Let the following be a central field:


 * $ \displaystyle \mathbf y' \left ( { x } \right ) = \boldsymbol \psi \left ( { x, \mathbf y } \right ) $

Then this central field is a field of functional $ J $.

Proof
Suppose:


 * $ \displaystyle g \left ( { x, \mathbf y } \right ) = \int_c^{ \left ( { x, \mathbf y } \right ) } F \left ( { x, \hat{ \mathbf y }, \hat{ \mathbf y }' } \right ) \mathrm d x $

where $ \hat{ \mathbf y } $ is an extremal of $ J $ connecting points $ c $ and $ \left ( { x, \mathbf y } \right ) $.

Define a field of directions in $ R $ by the following:


 * $ \mathbf p \left ( { x, \mathbf y, \mathbf y' } \right ) = g_{ \mathbf y } \left ( { x, \mathbf y } \right ) $

where $ \mathbf p $ stands for momentum.

Note, that it does not depend on the path defined by $ \hat{ \mathbf y } $, only on its endpoint at $ \left ( { x, \mathbf y } \right ) $.

By definition, $ g \left ( { x, \mathbf y } \right ) $ is a geodetic distance $ S $.

Hence, $ g_\mathbf y $ does not explicitly depend on $ \mathbf y' $.

Then:


 * $ g_{ \mathbf y } \left ( { x, \mathbf y } \right ) = \mathbf p \left ( { x, \mathbf y, \mathbf z } \right ) $

where $ \mathbf z = \mathbf z \left ( { x, \mathbf y } \right ) $ denotes slope of the curve joining $ c $ and $ \left ( { x, \mathbf y } \right ) $ at the point $ \left ( { x, \mathbf y } \right ) $.

Furthermore, since $ g $ is a geodetic distance, it satisfies Hamilton-Jacobi equation:


 * $ \displaystyle \frac{ \partial g \left ( { x, \mathbf y } \right ) }{ \partial x } + H \left ( { x, \mathbf y, g_{ \mathbf y } \left ( { x, \mathbf y } \right )  } \right ) = 0 $

which together with the previous relation results in a system of equations:


 * $ \begin{cases}

\displaystyle \frac{ \partial g \left ( { x, \mathbf y } \right ) }{ \partial x } + H \left ( { x, \mathbf y, \mathbf p \left ( { x, \mathbf y, \mathbf z } \right )  } \right ) = 0 \\ \displaystyle \mathbf z \left ( { x, \mathbf y } \right ) = \mathbf y' \left ( { x } \right ) \end{cases} $

Differentiation $ \mathbf y $ yields:


 * $ \begin{cases}

\displaystyle \displaystyle \frac{ \partial \mathbf p }{ \partial x} = - \frac{ \partial H }{ \partial \mathbf y } \\ \displaystyle \mathbf z \left ( { x, \mathbf y } \right ) = \mathbf y' \left ( { x } \right ) \end{cases} $

These are Euler's equations in canonical coordinates with a constraint on derivative.

Hence, the previously defined field of directions coincides with the field of functional.