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

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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 $ \paren {\map y t,\map x t} $ rather than $\paren {\map y x,x}$

Then the functional can be rewritten as

\(\displaystyle \mathcal J\sqbrk y\) \(=\) \(\displaystyle \int_{t_0}^{t_1} F\sqbrk {\map x t,\map y t,\frac{\rd\map y t}{\rd \map x t} }\rd \map x t\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_{t_0}^{t_1} F \sqbrk {\map x t,\map y t,\frac{\frac{\d y}{\d t} }{\frac{\d x}{\d t} } } \frac{\d x}{\d t}\rd t\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_{t_0}^{t_1} F\sqbrk {\map x t,\map y t,\frac{\map {\dot y} t} {\map {\dot x} t} } \map {\dot x} t\rd t\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_{t_0}^{t_1}\map \Phi {x,y,\dot x,\dot y}\rd t\) $\quad$ $\quad$

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

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

$\map \Phi {x,y,\lambda\dot x,\lambda\dot y}=\lambda \map \Phi {x,y,\dot x,\dot y}$ 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 equations $x=\map x t$ and $y=\map y t$,

and not on the functions $\map x t$, $\map y t$ themselves.

Suppose, a new parameter $\tau$ is chosen such that $t=\map t \tau$, where $\frac{\d t}{\d\tau}>0$, and the interval $\closedint {t_0} {t_1}$ is mapped onto $\closedint {\tau_0} {tau_1}$.

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

\(\displaystyle \int_{\tau_0}^{\tau_1} \map \Phi {x,y,\frac{\d x}{\d\tau},\frac{\d y}{\d\tau} }\rd\tau\) \(=\) \(\displaystyle \int_{\tau_0}^{\tau_1} \map \Phi {x,y,\dot x\frac{\d t}{\d\tau}, \dot y \frac{\d t}{\d\tau} } \d\tau\) $\quad$ Chain rule for differentiation, $\frac{\d x}{\d\tau}=\frac{\d x}{\d t}\frac{\d t}{\rd\tau}$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_{\tau_0}^{\tau_1}\map \Phi {x,y,\dot x,\dot y} \frac{\d t}{\d\tau} \rd \tau\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_{t_0}^{t_1}\map \Phi {x,y,\dot x,\dot y}\rd t\) $\quad$ $\quad$

$\blacksquare$


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