to the case of several functions and several constraints. In all cases, the \( \mu \) s are to be regarded as independent functions just like the xs.

There are two parameterizations that we are going to discuss in this appendix: parame-terization of the path by arc length and by time. In the first case, the integral to minimize is

\[ T = { \frac { 1 } { c } } \int n ( \mathbf { x } ( \ell ) ) d \ell , \eqno { ( \mathrm { A 3 } ) } \]

while in the second case

\[ T = \int d t . \eqno ( \mathrm { A 4 } ) \]

As we will see, there are constraints to be appended to these integrals. But before doing that, there is another subtlety that we should discuss first. It is related to the fact that in these integrals the domain of integration is not a priori known and fixed. (For example, if we knew the domain of integration in parameterization with time, then we already knew T.) This is very important, since the standard results we quoted above are usually derived under the assumption that there is a fixed domain of integration and the variations vanish at its boundary.16 So in the rest of this appendix we first address the issue of variable domain in Subsection A 1, and then turn to the length and time parameterizations in Subsections A 2 and A 3, respectively.

# 1. Variation with Variable Domain

In Section II our parametererization was over a fixed domain, namely: 0 \( < s < \) 1. This meant fixed lower and upper limits (0 and 1) in the integral (3) and hence we were eligible to use the Euler-Lagrange equations. However, we may equally well describe the path connecting \( \mathbf { x } _ { 1 } \) and \( \mathbf { x } _ { 2 } \) by a function x(s) on an arbitrary domain \( \sigma _ { 1 } < s < \sigma _ { 2 } \). The travel time is still given by the same integral except for the lower and upper limits of integration:

\[ T = \frac { 1 } { c } \int _ { \sigma _ { 1 } } ^ { \sigma _ { 2 } } n ( { \bf x } ( s ) ) \sqrt { \dot { \bf x } ( s ) \cdot \dot { \bf x } ( s ) } d s . \eqno ( \mathrm { A 5 } ) \]

We wish to find such general parameterizations of the path that minimize T. This appears to be a harder problem in calculus of variations, since: (i) we have to search among a larger class of functions; and (ii) the domain of integration is not fixed as before, but variable.

Our search will be among functions

\[ \mathbf { x } : [ \sigma _ { 1 } , \sigma _ { 2 } ] \to \mathbb { R } ^ { 3 } , \qquad { \mathrm { ~ s u c h ~ t h a t ~ } } \mathbf { x } ( \sigma _ { 1 } ) = \mathbf { x } _ { 1 } { \mathrm { ~ a n d ~ } } \mathbf { x } ( \sigma _ { 2 } ) = \mathbf { x } _ { 2 } . \eqno ( \mathrm { A } 6 ) \]

This is to be contrasted with our previous search scope, \( \mathbf { x } : [ 0 , 1 ] \to \mathbb { R } ^ { 3 } \). Under a variation \( \delta \mathbf { x } \) in the function and \( \delta \sigma _ { 1 , 2 } \) in the domain, the boundary conditions imply

\[ \left( \mathbf { x } + \delta \mathbf { x } \right) \big | _ { \sigma _ { 1 } + \delta \sigma _ { 1 } } = \mathbf { x } _ { 1 } , \qquad \left( \mathbf { x } + \delta \mathbf { x } \right) \big | _ { \sigma _ { 2 } + \delta \sigma _ { 2 } } = \mathbf { x } _ { 2 } , \qquad \qquad \qquad \qquad ( \mathrm { A T } ) \big ( \mathrm { A T } ) \big ( \mathrm { B T } ) \circ \mathrm { A T } . \]

which means

\[ ~ \delta { \bf x } ( \sigma _ { 1 } ) = - \dot { \bf x } ( \sigma _ { 1 } ) \delta \sigma _ { 1 } , \qquad \delta { \bf x } ( \sigma _ { 2 } ) = - \dot { \bf x } ( \sigma _ { 2 } ) \delta \sigma _ { 2 } . \eqno ( \mathrm { A 8 } ) \]

Now the variation of \( \begin{array} { r } { T = \int _ { \sigma _ { 1 } } ^ { \sigma _ { 2 } } L d s } \end{array} \) becomes

\[ \delta T = \int _ { \sigma _ { 1 } } ^ { \sigma _ { 2 } } \left[ { \frac { \partial L } { \partial x _ { i } } } - { \frac { d } { d s } } { \frac { \partial L } { \partial { \dot { x } } _ { i } } } \right] \delta x _ { i } d s + \left[ L - { \dot { x } } _ { i } { \frac { \partial L } { \partial { \dot { x } } _ { i } } } \right] _ { \sigma _ { 2 } } \delta \sigma _ { 2 } - \left[ L - { \dot { x } } _ { i } { \frac { \partial L } { \partial { \dot { x } } _ { i } } } \right] _ { \sigma _ { 1 } } \delta \sigma _ { 1 } , \qquad { \mathrm { ( A 9 ) } } \]

where summation over i is implied. By inspecting (A5) it is clear that L is a homogeneous function of \( \dot { \mathbf { x } } \) and hence the last two terms in \( \delta T \) vanish. Thus we are again left with the standard Euler-Lagrange equation and the result Eq. (5). Again we observe the non-uniqueness in the solutions: If \( \mathbf { x } ( s ) : [ \sigma _ { 1 } , \sigma _ { 2 } ] \to \mathbb { R } ^ { 3 } \) satisfies Eq. (5), then

\[ \mathbf { x } ( \bar { s } ( s ) ) : [ \bar { \sigma } _ { 1 } , \bar { \sigma } _ { 2 } ] \to \mathbb { R } ^ { 3 } \qquad \qquad \qquad \qquad \qquad ( \mathrm { A 1 0 } ) \]

is also a solution, for any monotonic function \( \bar { s } ( s ) \) satisfying \( \bar { s } ( \sigma _ { 1 } ) = \bar { \sigma } _ { 1 } \) and \( \bar { s } ( \sigma _ { 2 } ) = \bar { \sigma } _ { 2 } \).

We could have guessed that the Euler-Lagrange equations remain untouched: A function that minimizes the integral among all functions with variable domains, also does so among the smaller class of functions which share its own domain. So in general, the Euler-Lagrange equations provide a necessary condition for variable-domain problems. The boundary terms that appear in Eq. (A9) may provide additional conditions, although in our case they don’t.

# 2. Parameterization with Length

We are now in a position to perform the variational analysis with length parameterization. So we choose to set the arc length \( \ell \) run from zero to \( \lambda \), the total length of the path, and work with \( { \bf x } ( \ell ) \). This choice corresponds to the constraint \|dx/d\`\| = 1, to handle which we need to employ a Lagrange multiplier function \( \mu ( \ell ) \). Therefore, the quantity to minimize is

\[ \frac { 1 } { c } \int _ { 0 } ^ { \lambda } n ( { \bf x } ( \ell ) ) d \ell + \int _ { 0 } ^ { \lambda } \mu ( \ell ) \left[ \dot { \bf x } ( \ell ) \cdot \dot { \bf x } ( \ell ) - 1 \right] d \ell , \eqno { ( \mathrm { A 1 1 } ) } \]

where dot now means \( d / d \ell \). Again we have a variable domain of integration, since we don’t know the length \( \lambda \) of the desired trajectory before solving for it. Requiring \( \mathbf { x } ( \lambda ) = \mathbf { x } _ { 2 } \) implies

\[ \delta { \bf x } ( \lambda ) = - \dot { \bf x } ( \lambda ) \delta \lambda , \eqno ( \mathrm { A 1 2 } ) \]