Solving for one of variables locally when the Implicit Function Theorem does not apply

Question

I'm having trouble deciding whether certain functions can be locally solved. I have some examples:

• Can $xye^{xz} - z\log y =0$ be locally solved in $(0,1,0)$ for x? y? z?

In this case, I used Implicit Function Theorem to answer affirmatively that $x$ can be solved locally in function of $y$ and $z$. As $d/dy$ and $d/dz$ are equal zero, I can't use the theorem in those cases. I tried proving that $x$ has a local maximum at $(0,1,0)$ and could prove the first derivatives of $x$ with respect to $y$ and $z$ are zero, but I couldn't go further.

• If $f(x,y) = \frac{\sin(xy^2)}{xy}$ if $x\neq y$ and $f(x,y)=0$ otherwise, can $f(x,y) = 0$ be locally solved for $y=y(x)$ around $(0,0)$?

I couldn't apply implicit function theorem again and I have a clue that we can solve this locally for $y$, using $sin$ power series.

Is there a general procedure for problems like that? Any help is appreciated.

Answer 1

When implicit function theorem does not apply, the approach is usually ad hoc, based on the details of the function. Two things to try:

1. Change the variables. Some substitutions, like $u=x^3$, can reduce the function to a form in which the implicit function theorem applies.
2. Restrict $f$ to some line parallel to the coordinate of interest. If the zero set has multiple points, one can't solve for that coordinate.

First example

Restriction to lines works here:

• The line $x=0=z$ is in the zero set, which shows you can't solve for $y$
• The line $x=0$, $y=1$ is in the zero set, which shows you can't solve for $z$

Second example

The above tricks are not needed: the structure of particular function helps. Indeed, $f(x,y)$ turns into zero precisely when $\sin(xy^2)=0$. And since we are in a neighborhood of $(0,0)$, that means $xy^2=0$, which is equivalent to $x=0$ or $y=0$. Now the local structure of the set $f=0$ is clear.