Central Field is Field of Functional

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $J$ be a functional, such that:

$\displaystyle J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$

Let the following be a central field:

$\map {\mathbf y'} x = \map {\boldsymbol \psi} {x, \mathbf y}$


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


Proof

Suppose:

$\displaystyle \map g {x, \mathbf y} = \int_c^{\paren {x, \mathbf y} } \map F {x, \hat {\mathbf y}, \hat {\mathbf y}'} \rd x$

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



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

$\map {\mathbf p} {x, \mathbf y, \mathbf y'} = \map {g_{\mathbf y} } {x, \mathbf y}$

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 $\tuple {x, \mathbf y}$.

By definition, $ \map g {x, \mathbf y}$ is a geodetic distance $S$.

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

Then:

$\map {g_{\mathbf y} } {x, \mathbf y} = \map {\mathbf p} {x, \mathbf y, \mathbf z}$

where $\mathbf z = \map {\mathbf z} {x, \mathbf y}$ denotes slope of the curve joining $c$ and $\tuple {x, \mathbf y}$ at the point $\tuple {x, \mathbf y}$.

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

$\displaystyle \frac {\partial \map g {x, \mathbf y} } {\partial x} + \map H {x, \mathbf y, \map {g_{\mathbf y} } {x, \mathbf y} } = 0$

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

$\begin{cases} \dfrac {\partial \map g {x, \mathbf y} } {\partial x} + \map H {x, \mathbf y, \map {\mathbf p} {x, \mathbf y, \mathbf z} } = 0 \\ \map {\mathbf z} {x, \mathbf y} = \map{\mathbf y'} x \end{cases}$

Differentiation with respect to $\mathbf y$ yields:

$\begin{cases} \dfrac {\partial \mathbf p} {\partial x} = -\dfrac {\partial H} {\partial \mathbf y} \\ \map {\mathbf z} {x,\mathbf y} = \map{\mathbf y'} x \end{cases}$

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.

$\blacksquare$


Sources