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Let $u=f(x,y)$ and $v=g(x,y)$ be defined in a region "R". Suppose there exists another non constant function $h$, of class $C^{1}$, s.t $h(u,v)=0$ $\forall (x,y) \in$ R. Prove that

$J=\frac{\partial(u,v)}{\partial(x,y)}=0 $, $\forall(x,y)\in$ R.

What I've managed to prove so far is the following. If you derivate $h(u,v)$ with respect to $x$ and $y$ you get a homogeneous system of linear equations. If this system has non trivial solutions, then the problem is solved.

I appreciate any help, Thanks!

Marcus Fermat
  • 301

2 Answers2

1

Derivating $h(u,v)=0$ respect to $x$ and $y$ we have:

\begin{equation} \frac{\partial{h}}{\partial{u}}\frac{\partial{u}}{\partial{x}} + \frac{\partial{h}}{\partial{v}}\frac{\partial{v}}{\partial{x}}=0 \end{equation} and \begin{equation} \frac{\partial{h}}{\partial{u}}\frac{\partial{u}}{\partial{y}} + \frac{\partial{h}}{\partial{v}}\frac{\partial{v}}{\partial{y}}=0 \end{equation}

If we denote $A$ to the system above, then its straightforward to see that

\begin{equation} det(A)= \frac{\partial{(u,v)}}{\partial{(x,y)}} \end{equation}

So , it would be sufficient to prove that $det(A)=0$ $\forall (x,y) \in R$. This will guarantee that the Jacobian is exactly $0$ too. Note that:

  • If A has only the trivial solution, then (in $R$): $$\frac{\partial{h}}{\partial{u}}=\frac{\partial{h}}{\partial{v}} = 0$$

This is not possible because, according to the hypothesis, $h$ is a non constant function. So at least one of these derivatives is not $0$.

$\therefore$ $A$ has a non trivial solution

$\therefore$ $det(A)=0$

\begin{equation} \therefore J=\frac{\partial{(u,v)}}{\partial{(x,y)}}=0 \end{equation}

$\forall (x,y) \in R$

Marcus Fermat
  • 301
0

$$ 0=\frac{\partial h(u(x,y),v(x,y))}{\partial x}=h_u u_x+h_v v_x,\\ 0=\frac{\partial h(u(x,y),v(x,y))}{\partial y}=h_u u_y+h_v v_y, $$ and hence $$ (h_u,h_v)\cdot (u_x,v_x)=(h_u,h_v)\cdot (u_y,u_y)=0, $$ and as $h$ is non-constant, then $(h_u,h_v)$ does not vanish identically, and thus the vectors $$ (u_x,u_y)\quad \text{and}\quad (v_x,v_y), $$ are parallel, and thus $$ \left|\begin{array}{cc} u_x & u_y \\ v_x & v_y \end{array} \right|=0. $$

Yiorgos S. Smyrlis
  • 83,933
  • Thanks Yiorgos, but do you know if there is another possible solution were you use the Implicit Function Theorem. I mean, I understand your solution, but I would like to use the IFT because the hypotheses are met – Marcus Fermat Sep 23 '14 at 21:44
  • Turns out this theorem is not necessary at all!. Thanks! – Marcus Fermat Sep 23 '14 at 23:34