Show that for a gradient system $\bf\dot x= f(x)$, $\frac{\partial f_i}{\partial x_j}-\frac{\partial f_j}{\partial x_i}=0$ for $1 \leq i, j \leq d$

Question

The dynamical system ${\bf \dot x} = {\bf f}({\bf x})$ is called a gradient system if there exists a function $V({\bf x})$ such that $$ {\bf f}({\bf x}) = - \nabla V({\bf x}) $$ Show that if ${\bf \dot x} = {\bf f}({\bf x})$ is a gradient system, then $$ \frac{\partial f_i}{\partial x_j} - \frac{\partial f_j}{\partial x_i} = 0 $$ for $1 \leq i, j \leq d$

So let ${\bf f}({\bf x}) = (f_1({\bf x}), f_2({\bf x}), \dots , f_n({\bf x}))$ and from the definition of a gradient system, we have $$ (f_1({\bf x}), f_2({\bf x}), \dots , f_n({\bf x})) = - \left ( \frac{\partial V({\bf x})}{\partial x_1}, \frac{\partial V({\bf x})}{\partial x_2}, \dots, \frac{\partial V({\bf x})}{\partial x_n} \right ) $$ but then I don't see how to follow this to produce the required equation.

Answer 1

Hint
$V$ has two continuous derivatives, your equation is Schwarz' theorem in disguise. $$\frac{\partial^2}{\partial x_i\partial x_j} V = \frac{\partial^2}{\partial x_j \partial x_i} V$$ now write this in terms of $f$.

Answer 2

$$ \frac{\partial f_i}{\partial x_j}-\frac{\partial f_j}{\partial x_i} = -\left(\frac{\partial}{\partial x_j}\frac{\partial V}{\partial x_i} - \frac{\partial}{\partial x_i} \frac{\partial V}{\partial x_j} \right) = -\left(\frac{\partial^2}{\partial x_j \partial x_i} - \frac{\partial^2}{\partial x_i \partial x_j} \right) V = 0 $$ if $V$ has continuous second partial derivatives.

Answer 3

By definition,

$$\frac{\partial f_{i}}{\partial x_{j}}=-\frac{\partial^{2}V}{\partial x_{j}\partial x_{i}}$$

$$\frac{\partial f_{j}}{\partial x_{i}}=-\frac{\partial^{2}V}{\partial x_{i}\partial x_{j}}$$

So they're equal by Schwarz theorem if $V$ is at least twice differentiable.