- #1
hauho195
- 1
- 0
Today I've tried to investigate properties of a function [itex]f(x_1,x_2,x_3)[/itex] satisfying [itex]\nabla\times(f\mathbf{x})=0[/itex] in [itex]\mathbb{R}^3[/itex], where some degrees of the differentiability are assumed if needed.
By some basic procedures, I've deduced that: there is a scalar function g such that [itex]f=\frac{1}{x_i}\frac{\partial{g}}{\partial{x_i}}[/itex] for i=1,2,3. Then I guess f is symmetric(i.e., f(x,y,z)=f(x,z,y)=f(y,x,z)=f(y,z,x)=f(z,x,y)=f(z,y,x)). Is it true? If not, is there a counterexample?
By some basic procedures, I've deduced that: there is a scalar function g such that [itex]f=\frac{1}{x_i}\frac{\partial{g}}{\partial{x_i}}[/itex] for i=1,2,3. Then I guess f is symmetric(i.e., f(x,y,z)=f(x,z,y)=f(y,x,z)=f(y,z,x)=f(z,x,y)=f(z,y,x)). Is it true? If not, is there a counterexample?