finding a function $f$ such that $S=f^{-1}(\{0\})$

Question

I'm having some trouble with the following exercise:

Consider the function $\phi:\mathbb R^2\to \mathbb R^3$ given by: $$\phi(u,v)=(u-v,u+v,2(u^2+v^2))$$ Prove that $S=\phi(\mathbb R^2)$ is a regular surface, and find a function $f:\mathbb R^3\to \mathbb R$ such that $S=f^{-1}(\{0\})$ and $0$ is a regular value of $f$, i.e. $\nabla f(x,y,z)\neq (0,0,0)$ when $f(x,y,z) = 0$.

I was able to prove easily that $S$ is a regular surface, but I'm having some trouble finding such function $f$.

Is there any general procedure/technique to solve this kind of problem, or at least some intuition behind how to find such solutions?

Answer 1

Let $(x,y,z)\in S$, then there exists $(u,v)\in\mathbb{R}^2$ such that $(x,y,z)=\phi(u,v)$. Then $$ x = u-v, \quad y=u+v, \quad z=2(u^2+v^2). $$ Solve for $u$ and $v$ in the first two equations and substitute the obtained values in the third to obtain $$ z = x^2 + y^2. $$ So if we let $T=\{(x,y,z)\in\mathbb{R}^3 \; | \; z-x^2-y^2=0\}$, we proved that $S\subseteq T$. Try to show that $S=T$ (its not difficult at all). Then if we define $$ f:\mathbb{R}^3\to\mathbb{R}, \quad (x,y,z)\mapsto z-x^2-y^2, $$ you obtain that $S=f^{-1}(\{0\})$, and all you need is to verify that $0$ is indeed a regular value of $f$.

Answer 2

In this particular case, it helps to write the third coordinate of $\phi(u,v)$ in terms of the first and second. So, because we have $$ x = u-v, \ y = u+v, \ z = 2(u^2+v^2), $$ we can solve the $x$ and $y$ equations to get $u$ and $v$ as functions of $x$ and $y$, then substitute for $z$. We find $$ u = \frac12(x+y), \ v = \frac12(y-x) $$ which gives $z = x^2+y^2$. So $S$ is precisely the paraboloid given by $z = x^2+y^2$.

Now you'll find that the map $f : \mathbb{R}^3 \to \mathbb{R}$ given by $f(x,y,z) = z - (x^2+y^2)$ does the trick. This is because $\nabla f(x,y,z) = (-2x,-2y,1)$, so the gradient is never zero. In general, if you can get one coordinate as an explicit function of the others, then that function will always work for your problem.

Answer 3

Take $x=u-v,$ $y=u+v$ and $z=2(u^2+v^2)$. Then it satisfies the equation $$x^2+y^2=z,$$ which is a paraboloid, so the surface $S$ is given by this equation. Now define $f:\mathbb{R}^3\to\mathbb{R},$ given by $f(x,y,z)=x^2+y^2-z$. Then $S=f^{-1}(\{0\}).$ Its straightforward to prove that is a regular value because the gradient of the function is $(2x,2y,1)$, which never vanishes.