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

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Theorem

Let $x: \R \to \R$ and $y: \R \to \R$ 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 $\dot y$ denotes the derivative of $y$ with respect to $t$:

$\dot y = \dfrac {\d y} {\d t}$


Then $J \sqbrk {x, y}$ depends only on the curve in the $x y$-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 function of degree $1$ in $\dot x$ and $\dot y$.


Proof

In the functional:

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

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

In other words, let the curve be described by the vector $\tuple {\map y t,\map x t}$ rather than $\paren {\map y x, x}$.

Then the functional can be rewritten as:

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

The function on the right hand side does not involve $t$ explicitly.

Suppose that 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 all $\lambda > 0$.

Now we will show that the value of such functional depends only on the curve in the $x y$-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.

Let a new parameter $\tau$ be chosen such that $t = \map t \tau$, where $\dfrac {\d t} {\d \tau} > 0$.

Let the interval $\closedint {t_0} {t_1}$ be mapped onto $\closedint {\tau_0} {\tau_1}$.

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

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

$\blacksquare$



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