Consider the system of equations $$ x^2+2y^2+u^2+v=6,\quad 2x^3+4y^2+u+v^2=9. $$ Around $(1,-1,-1,2)$ express $(u,v)$ in terms of $(x,y)$. Is there a maximal neighborhood where this expression is defined?
- Here, $\mathbf{F} : \Bbb{R}^4\to \Bbb{R}^2$ with $\mathbf{F}(x,y,u,v)=(x^2+2y^2+u^2+v-6,\ 2x^3+4y^2+u+v^2-9).$
- Clearly, $\mathbf{F} $ is $C^1$ for the whole domain $\Bbb{R}^4.$
- $\mathbf{F}(1,-1,-1,2)=0 .$
- $D\mathbf{F}(x,y,u,v)=\left. \begin{bmatrix} 2x & 2y & 2u & 1\\ 6x^2& 8y & 1 & 2v \end{bmatrix}\right|_{(1,-1,-1,2)}=\begin{bmatrix} 2&-2&-2&1\\6&-8&1&4 \end{bmatrix}$. Since, $\begin{bmatrix} -2&1\\1&4 \end{bmatrix}$ is invertible so all the hypothesis of implicit theorem are satisfied.
Hence, there exists $V$ and $W$ neighborhoods of $(1,-1)$ and $(-1,2)$ and $\mathbf{G}:V\to W$ such that $\mathbf{G}(1,-1)=0$ and $\mathbf{F}(x,y,\mathbf{G}(x,y))=0$, forall $(x,y)\in V.$ So, using this theorem I can assure that $(u,v)$ can be expressed as a function of $(x,y)$, but I am unable to do it that "how to find it"?