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

Theorem
A necessary and sufficient condition for the functional

$\displaystyle\int_{t_0}^{t_1}\Phi\left( t,~x,~y,~\dot{x},~\dot{y} \right)\mathrm{d}{t}$

to depend only on the curve in the xy-plane defined by the parametric equations $x=x(t)$, $y=y(t)$ and not on the choice of the parametric representation pf the curve, is that the integrand $\Phi$ should not involve $t$ explicitely and should be a positive-homogeneous functions of dregree $1$ in $\dot{x}$ and $\dot{y}$.

Proof
Suppose that in the functional

$\int_{x_0}^{x_1}F(x, y, y')\mathrm{d}{x}$

we wish to regard the argument $y$ as a curve which is given in parametric form.

Then the functional can be rewritten as

$\displaystyle\int_{t_0}^{t_1}F\left[x(t),~y(t),~\frac{\dot{y}(t)}{\dot{x}(t)}\right]\dot{x}(t)\mathrm{d}{t}=\int_{t_0}^{t_1}\Phi\left(x,~y,~\dot{x},~\dot{y}\mathrm{d}{t}\right)$

The function on the RHS does not involve $t$ explicitely, and is positive-homogeneous of degree 1 in $\dot{x}(t)$ and $\dot{y}(t)$, which means that

$\Phi\left(x,~,y~,~\lambda\dot{x},~\lambda\dot{y}\right)=\lambda\Phi\left(x,~y,~\dot{x},~\dot{y} \right)$

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 $x=x(t)$ and $y=y(t)$, and not on the functions $x(t)$, $y(t)$ themselves.

In other words, if we go from $t$ to some new parameter $\tau$ by setting $t=t(\tau)$, where $\frac{d t}{d\tau}>0$ and the interval [t_0, t_1] goes into $[\tau_0, \tau_1]$, then

$\int_{\tau_0}^{\tau_1}\Phi\left(x,y,\frac{dx}{d\tau},\frac{dy}{d\tau} \right)\mathrm{d}{\tau}= \int_{t_0}^{t_1}\Phi(x,y,\dot{x},\dot{y})\mathrm{d}{t}$

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

$\int_{\tau_0}^{\tau_1}\Phi\left(x,y,\frac{dx}{d\tau},\frac{dy}{d\tau}\right)\mathrm{d}{\tau}=\int_{\tau_0}^{\tau_1}\Phi\left(x,y,dot{x}\frac{dt}{d\tau},\dot{y}\frac{dt}{d\tau}\right)\mathrm{d}{\tau}=\int_{\tau_0}^{\tau_1}\Phi\left(x,y,\dot{x},\dot{y}\right)\frac{dt}{d\tau}\mathrm{d}{\tau}=\int_{t_0}^{t_1}\Phi\left(x,y,\dot{x},\dot{y}\right)\mathrm{d}{t}$