$\partial_xF+F\partial_yF=0$ implies that $F$=const

Question

Let $F\in \mathit{C}^2(\mathbb{R^2},\mathbb{R})$ and $$\partial_xF+F\partial_yF=0.$$ How do I prove that $F$ is a constant on $\mathbb{R^2}$? I am trying to restrict $F$ to a line passing through the origin and prove that the values of $F$ on the line equal $F(0,0)$, but failed.

Answer 1

Claim: If $F(x_0,y_0)=c$, then $F(x,y)=c$ for all $(x,y)$ satisfying $y=y_0+c(x-x_0)$. Indeed, $$\frac d{dx}F(x,y_0+c(x-x_0))=\partial_xF+c\partial_yF=(c-F)\partial_yF.$$ Since $\partial_yF$ is locally bounded, we therefore have (after restricting to a compact set $K$) $$|\frac d{dx}(c-F)|\leq C|c-F|$$ where we are saying $C$ is $\sup_K|\partial_yF|$. Then by Gronwall’s inequality, we conclude that $F=c$ on the part of this line contained in $K$. Picking a sequence of compact sets that exhaust the plane, we conclude that $F=c$ on the whole line.

Now suppose there are two points $(x_0,y_0)$ and $(x_1,y_1)$ such that $F(x_0,y_0)=c_0$, $F(x_1,y_1)=c_1$. Then by the above claim, $F=c_0$ on a line of slope $c_0$ and $F=c_1$ on a line of slope $c_1$. If $c_0\neq c_1$ these lines would intersect and produce a contradiction, so we must have $c_0=c_1$.

It seems we have only assumed $F\in C^1(\mathbb R^2)$.