Show that the equation $(x^2+y^4)\,f(x,y)+f(x,y)^3=1$ defines a $C ^ 1$ function $f$

Question

Consider the function $f:U\subset\mathbb{R}^2 \rightarrow \mathbb{R}$, defined in some open set $U$ by the equation $$(x^2+y^4)\,f(x,y)+f(x,y)^3=1\ ,$$

and show that $f$ is of $C ^ 1$ class for all $(x,y) \in U$.

I think we can use the theorem of inverse function or implicit function.

Answer 1

HINT.-You must have $f(x,y)>0$ and $[f(x,y)]^3\leq 1$; in fact, let $f(x,y)=a$; it follows $a(x^2+y^4+a^2)=1$ so $x^2+y^4+a^2=\frac 1a$ then $a>0$ and $a^2\leq\frac 1a$

Answer 2

Define $$\begin{array}{l|rcl} F : & \mathbb R^3 & \longrightarrow & \mathbb R\\ & (x,y,z) & \longmapsto & (x^2+y^4)z+z^3-1 \end{array}$$ $F$ is a polynomial function, hence is $\mathcal C^\infty$.

Now pick up $(a,b,c) \in \mathbb R^3$. You have $$\frac{\partial F}{\partial z}(a,b,c)=a^2+b^4+3c^2 >0$$ (for $(a,b,c) \neq (0,0,0)$). Therefore $h \mapsto \frac{\partial F}{\partial z}(a,b,c).h$ is an isomorphism. You can apply the implicit function theorem, which says that there exists an open subset $U_1 \subset \mathbb R^2$ with $(a,b) \in U_1$, an open subset $U_2 \subset \mathbb R$ with $c \in U_1$ such that for $((x,y),z) \in U_1 \times U_2$ you have $$F(x,y,z)=F(a,b,c) \iff z=f_0(x,y)$$ Moreover $f_0$ is $\mathcal C^1$. This proves the local existence of a unique continuous inverse function for all $(a,b,c) \in \mathbb R^3$ with $F(a,b,c)=0$. Now your $f$ is a local continuous inverse. So it is locally equal to the $f_0$ of the implicit function theorem. Hence it is $\mathcal C^1$.

Answer 3

The auxiliary function $$g(t):=t^3+pt-1\qquad(-\infty<t<\infty)$$ with parameter $p\geq0$ has $$g'(t)=3t^2+p\geq0\ .$$ It follows that $g$ is strictly monotonically increasing from $-\infty$ to $\infty$ and has a single real zero $\tau=:\tau_p$. I claim that the function $v:\> p\mapsto \tau_p$ is differentiable. In order to prove this we have to apply the implicit function theorem to the equation $$G(p,t):=t^3+pt-1=0$$ at the points $(p,\tau_p)$. Let a $p\geq0$ be given. Since $${\partial G\over\partial t}\biggr|_{(p,\tau_p)}=3\tau_p^2+p=\left\{\eqalign{&>0\quad(p>0)\cr &=3\quad(p=0)\cr}\right.$$ it follows that $v$ is differentiable in the neighborhood of $p$.

Now to the problem at hand: It follows from the above that the equation $$f^3(x,y)+(x^4+y^4)f(x,y)-1=0$$ defines $$f:\>{\mathbb R}^2\to{\mathbb R},\qquad(x,y)\mapsto v(x^4+y^4)$$ uniquely as a $C^1$-function in the plane.