Question concerning implicit function theorem

Question

I am studying the implicit function theorem. I have a question about the condition $F_y(x_0,y_0)\neq 0.$

More precisely, let $w=F(x,y)$ be a $C^1$ function on an open rectangle $R=(a,b) \times (c,d)$. Consider $F(x,y)=C,$ where $C \in \mathbb{R}$ is fixed, and $F(x_0,y_0)=C$ for some $(x_0,y_0) \in R.$

By implicit function theorem, if $F_y(x_0,y_0)\neq 0,$ $\exists$ a $C^1$ function $y=y(x)$ near $(x_0,y_0)$ satisfying $F(x,y(x))=C.$

If $\exists$ a differential function $y=y(x)$ near $(x_0,y_0)$ satisfying $F(x,y(x))=C,$ then, by chain rule, $F_x(x_0,y_0)+F_y(x_0,y_0)y'(x_0)=0,$ which implies that the case $F_x(x_0,y_0)\neq 0$ and $F_y(x_0,y_0)=0$ is impossible.

My question: Is it possible that $F_x(x_0,y_0)= F_y(x_0,y_0)=0$?

I would be grateful if you give any comment for my question. Thanks in advance.

Answer 1

Sure. Let's focus on the case $(x_0,y_0)=(0,0)$ (up to translation of the plane this is general). There can be several related phenomena producing $F_x=F_y=0$:

1) There could be several branches of $F(x,y)=0$ near $(0,0)$. For example $F(x,y)=x^2-y^2$ has $F_x=F_y=0$ but of course there is $y(x)=x$ such that $F(x,y(x))=0$. In fact for $y(x)=-x$ we also have $F(x,y(x))=0$, which is the second branch (similarly for $F(x,y)=xy$, with the second branch $x=0$ vertical). Note that the extra branches may be not visible over real numbers, for example $F(x,y)=(x-y)(x^2+y^2)$ which over reals is equivalent to $y=x$, but over complex numbers has other solutions $y=ix$ and $y=-ix$.

2) The multiplicity of the single branch could be more than 1. For example $F(x,y)=y^2$ with $y(x)=0$ or $F(x,y)=(x-y)^2$ with $y(x)=x$ (or $F(x,y)=y^5$ etc.). This is like multiple branches, but with two or more "different" branches becoming equal.