Necessary and Sufficient Condition for Integral Parametric Functional to be Independent of Parametric Representation

Theorem
Let $x=\map x t$ and $y=\map y t$ be real functions.

Let $J\sqbrk {x,y}$ be a functional of the form

$\displaystyle J\sqbrk {x,y}=\int_{t_0}^{t_1}\map \Phi {t,x,y,\dot x,\dot y}\rd t$

where $\dot y=\frac{\d y}{\d t}$.

Then $J\sqbrk {x,y}$ depends only on the curve in the xy-plane defined by the parametric equations $x=\map x t$, $y=\map y t$

and not on the choice of the parametric representation of the curve if and only if the integrand $\Phi$ does not involve $t$ explicitly

and is a positive-homogeneous of dregree $1$ in $\dot x$ and $\dot y$.

Proof
Suppose that in the functional


 * $\displaystyle\mathcal J\sqbrk y=\int_{x_0}^{x_1}\map F {x,y,y'}\rd x$

the argument $y$ stands for a curve which is given in a parametric form.

In other words, the curve is described by $ \left ( { y \left ( { t } \right ), x \left ( { t } \right ) } \right ) $ rather than $\paren {\map y x,x}$

Then the functional can be rewritten as

The function on the RHS does not involve $ t $ explicitely.

Suppose, it is positive-homogeneous of degree 1 in $ \dot x \left ( { t } \right ) $ and $ \dot y \left ( { t } \right ) $:

$ \Phi \left ( { x, y, \lambda \dot x, \lambda \dot y } \right ) = \lambda \Phi \left ( { x, y, \dot x, \dot y } \right ) $ for every $ \lambda > 0 $.

Now we will show that the value of such functional depends only on the curve in the xy-plane defined by the parametric equaions $ x = x \left ( { t } \right ) $ and $ y = y \left ( { t } \right ) $,

and not on the functions $ x \left ( { t } \right ) $, $ y \left ( { t } \right ) $ themselves.

Suppose, a new parameter $ \tau $ is chosen such that $ t = t \left ( { \tau } \right ) $, where $ \frac{ d t }{ d \tau } > 0 $, and the interval $ \left [ { t_0 \,. \,. \, t_1 } \right ] $ is mapped onto $ \left [ { \tau_0 \,. \,. \, \tau_1 } \right ] $.

Since $ \Phi $ is positive-homogeneous of degree 1 in $ \dot x $ and $ \dot y $, it follows that