Existence of a unique function $f:\Bbb R^2\setminus\{(0,0)\}\to\Bbb R, f(x,y)^3=xy\cos(xyf(x,y))-x^2y^2f(x,y), (x,y)\in\Bbb R^2\setminus\{(0,0)\}.$

Question

Does there exist a unique continuous function $f:\Bbb R^2\setminus\{(0,0)\}\to\Bbb R$ such that $$f(x,y)^3=xy\cos(xyf(x,y))-x^2y^2f(x,y),\quad (x,y)\in\Bbb R^2\setminus\{(0,0)\}.$$ Is it of the class $C^1$?

My attempt:

Let $F(x,y,z)=z^3-xy\cos(xyz)+x^2y^2z$ and let $t=xy.$

Then, $F(x,y,1)=1-xy\cos(xy)+x^2y^2=1-t\cos(t)+t^2\ge1-|t|+t^2>0,\forall t\in\Bbb R,$ that is, $\forall (x,y)\in\Bbb R^2$.

Also, $F(x,y,-1)=-1-xy\cos(xy)-x^2y^2=-1-t\cos(t)-t^2\le-1+|t|-t^2<0,\forall t\in\Bbb R,$ that is, $\forall (x,y)\in\Bbb R^2.$

Fix $(x_0,y_0)\in\Bbb R^2$ and define $G:[-1,1]\to\Bbb R,\quad G(z)=F(x_0,y_0,z).$

Since $G$ is continuous and $\lim\limits_{z\to -1^+} G(z)=G(-1)=\ell_1<0$ and $\lim\limits_{z\to 1^-} G(z)=G(1)=\ell_2>0,$ by the mean value theorem,there is $z_0\in(-1,1)$ such that $G(z_0)=F(x_0,y_0,z_0)=0.$

Now, let's look at $G'(z)=\frac{\partial F(x_0,y_0,z)}{\partial z}=3z^2+x^2y^2\sin(xyz)+x^2+y^2=3z^2+x^2y^2(\sin(xyz)+1)\ge 0.$

I think $G'(z)=0\iff z=0$ and $x=0$ or $y=0$.

I would like to apply the implicit function theorem, but, if $x=0$ or $y=0$ and $z=0,$ then I found an example of $F(x_0,y_0,z_0)=0$ and $\frac{\partial F(x_0,y_0,z)}{\partial z}=0$ so I don't know if $f\in C^1(\Bbb R^2\setminus\{(0,0)\})$.

(1) Should, therefore, the domain of the function $f$ be $\Bbb R^2\setminus\{(x,y)\in\Bbb R^2\mid x=0\text{ or } y=0\}?$

In that case, $G'(z)>0\implies G$ is strictly increasing and hence, has a unique root $z_0\in[-1,1]$ and we could define $f(x_0,y_0)=z_0$ for each fixed $(x_0,y_0)\in\Bbb R^2$ and, then, apply the implicit function theorem that guarantees the existence of some open intervals $I,J,K\subseteq\Bbb R$ containing $x_0,y_0,z_0$ respectively and a unique function $g:I\times J\to K$ of the class $C^1$ such that $F(x,y,g(x,y))=0,\forall (x,y)\in I\times J.$

Since $F(x,y,z)=0\iff z=f(x,y),$ we conclude that $f_{\mid I\times J}=g\in C^1$.

As being of the class $C^1$ is a local notion, we conclude $f\in C^1.$

Could anybody verify my answer?

Answer 1

Preliminary remarks:

• You are definitely on the right track. The intermediate value theorem shows that the equation has a unique solution $f(x, y)$ for all $(x, y)$, and the implicit function theorem shows that the solution is differentiable at all points with $xy \ne 0$.

• You also correctly noticed that the implicit function theorem cannot be applied if $x=0$ or $y=0$. However, that does not imply that the solution is not differentiable at those points. It turns out that this is the case, but that must be proven rigorously.

I would proceed as follows. Instead of $$ \tag{1} f(x,y)^3=xy\cos(xyf(x,y))-x^2y^2f(x,y) $$ we consider the equation $$ \tag{2} g(t)^3 = t \cos(t g(t)) - t^2 g(t) $$ for $t \in \Bbb R$. We can show that

• For all $t \in \Bbb R$, $(2)$ has a unique solution $g(t)$.
• The function $g:\Bbb R \to \Bbb R$ is continuous.
• The function $g$ is differentiable on $\Bbb R\setminus \{ 0 \}$, but not at $t=0$.

It then follows that the solution of $(1)$ is given by $f(x, y) =g(xy)$, which is continuous on $\Bbb R^2$ and differentiable on $\Bbb R^2\setminus\{(x,y)\mid x=0\text{ or } y=0\}$.

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In order to solve $(2)$ we consider the function $$ G(t, z) = z^3 -t \cos(tz) + t^2 z \, . $$ If $t=0$ then the unique solution of $G(0, z) = 0$ is given by $z=0$. If $t \ne 0$, then (as you demonstrated) $G(t, -1) < 0 < G(t, 1)$ and $$ \frac{d}{dz}G(t, z) = 3 z^2 + t^2\bigl(\sin(tz) +1 \bigr) > 0 $$ is strictly positive for all $z$, so that $z \mapsto G(t, z)$ is strictly increasing.

In any case, the equation $G(t, z) = 0$ has a unique solution $z$, which we can define as $g(t)$.

Also $\frac{d}{dz}G(t, z) > 0$ for $t \ne 0$, so that the implicit function theorem can be applied, and it follows that $g$ is differentiable on $\Bbb R\setminus \{ 0 \}$.

It remains to investigate the behavior of $g$ in a neighborhood of $t=0$. We know that $-1 < g(t) < 1$. Using the equation $(2)$ we can obtain upper and lower bounds for $g(t)$ for small $t$. The rough idea is that if $g(t)$ is small then $g(t)^3$ is the dominating term.

Concretely: for $|t| < \frac{1}{2\sqrt 2} < \frac{\pi}{4} < 1$ is $$ |g(t)|^3 \le |t| + t^2 \le 2|t| $$ and $$ |g(t)|^3 \ge \frac{|t|}{\sqrt 2} - t^2 \ge \frac{|t|}{2\sqrt 2} \, . $$ We see that there are positive constants $a, b$ such that $$ a |t|^{1/3} \le |g(t)| \le b |t|^{1/3} $$ for sufficiently small $t$. This shows that $g$ is continuous, but not differentiable at $t=0$.