Show that the vector field $\vec F=(yf(u),xg(u))$ has no potential

Question

I'm trying to prove that the vector field $\vec F=(xf(u),xg(u))$ with $u=xy$ is not conservative. I suppose that there is a function $\phi$ so that $\nabla \phi= \vec F$. So I need to satisfy that: $$\frac {\partial \phi}{\partial x}= yf(u)$$ $$\frac {\partial \phi}{\partial y}= xg(u)$$ Calculating the mixed partial for each I get that: $$\frac {\partial^2 \phi}{\partial x \partial y}= f(u)+y \frac{\partial f(u)}{\partial y}$$ $$\frac {\partial^2 \phi}{\partial y \partial x}= g(u)+x \frac{\partial g(u)}{\partial x}$$ Since mixed partials are equal I have that: $$f(u)+y \frac{\partial f(u)}{\partial y}=g(u)+x \frac{\partial g(u)}{\partial x}$$ and $\frac {\partial f(u)}{\partial y}=\frac {df}{du}\frac {\partial u}{\partial y}= x\frac{df}{du}$, applying this to both sides I end up with: $$f(u)+u \frac{df}{du}=g(u)+u \frac{dg}{du}$$ This is where I'm stuck, I'm not quite sure how to get a contradiction out of this or how I can show that this statement cannot hold. Should I have taken a different approach to this problem?

Answer 1

Your last equation is fine. The function $h:=f-g$ then satisfies $u h'=-h$, and this is satisfied by the functions $h(u)={c\over u}$, $\ c$ a constant. If you insist that the field $\vec F$ should be defined on all of ${\mathbb R}^2$ this implies $c=0$, or $f=g$. On the other hand, when $f(u)=g(u)$ for all $u\in{\mathbb R}$ then the field $\vec F$ satisfies the integrability condition $Q_x=P_y$, hence is conservative, and possesses a potential, contrary to the claim.

If you are satisfied with a field $\vec F$ which defined only in the open first quadrant then you can work with an arbitrary $c\in{\mathbb R}$, and obtain some less obvious solutions.