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

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