Proof That Any Vector Field Represented by a Symmetric Jacobian Matrix is Conservative
Any vector field
\[\mathbf{F}\]
represented by a symmetric Jacobian matrix on a region \[B\]
then \[\mathbf{F}\]
is a conservative field.Proof
Take any two points
\[x_0 , \: x_1\]
in \[B\]
and construct a rectangle with edges parallel to the coordinate axis in \[B\]
. The \[K\]
be the set of all paths from \[x_0\]
to \[x_1\]
Any element of
\[K\]
consists of a path which we may represent as by the directions and lengths of the path segment at each vertex on the path eg \[ x_{a_1}e_{a_1} x_{a_2}e_{a_2} ... x_{a_m}e_{a_m}\]
with \[\mathbf{e_{a_l}}\]
being the alth basis vector for the ith axis. Take any two paths\[\gamma_p : x_{p_1}e_{p_1} x_{p_2}e_{p_2} ... x_{p_m}e_{p_m}\]
\[\gamma_q : x_{q_1}e_{q_1} x_{q_2}e_{q_2} ... x_{q_m}e_{q_m}\]
Each path segment occur once in each of the two paths since each is the side of a triangle, opposite sides being the same length, so each path can be transformed into the other by a sequence of interchanging of adjacent vectors. Consider the closed path
\[C\]
consisiting of the segments \[x_{p_s}e_{p_s} x_{p_t}e_{p_t}\]
formed into a reacngle.\[\oint_C \mathbf{F} d \mathbf{x} = \oint_C F_p dx_p + F_q dx_q\]
Green's Theorem tells us
\[\oint_C \mathbf{F} d \mathbf{x} = \int_S ( \frac{\partial F_q}{\partial x_p} - \frac{\partial F_p}{\partial x_q}) dx_p \: x_q =0\]
Since the Jacobian matrix is symmetric, so
\[\frac{\partial F_q}{\partial x_p} = \frac{\partial F_p}{\partial x_q}\]
.This mean a change of path by swapping any two adjacent path segements does not change the integral. Since pthe paths can be changed into each other by a sequence of such operation, each integral gives the same result and the field is conservative.
