Central Field is Field of Functional

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

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

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 $\paren{x,\mathbf y}$ at the point $\paren{x,\mathbf y}$.

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:

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

Differentiation with respect to $\mathbf y$ yields:

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

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