2
$\begingroup$

Prove with implicit function theorem that $xy^2+y^3z^4+z^5x^6=1$ has a $C^1$ solution with a form of $(x,g(x,z),z)$ in an open neighborhood about the point $x_0=(0,1,-1)$.

What I have gotten so far:

Let $f(x,y,z)=xy^2+y^3z^4+z^5x^6-1$.

Now $f(0,1,-1)=0\cdot1^2+1^3\cdot(-1)^4+(-1)^5\cdot0^6-1=0$.

Also $D_yf(x,y,z)=2xy+3z^4y^2$ and $D_yf(0,1,-1)=2\cdot0\cdot1+3\cdot(-1)^4\cdot1^2=3\ne0$.

Now I can conclude with implicit function theorem that the function has a $C^1$ solution in an open neighborhood about the point $x_0=(0,1,-1)$ but how do I know it takes the form $(x,g(x,z),z)$?

$\endgroup$
7
  • $\begingroup$ Hint: consider when the condition does not hold. $\endgroup$
    – Trebor
    Dec 29, 2018 at 9:19
  • 3
    $\begingroup$ IFT specifically correlates the non-vanishing of the partial derivatie with respect to $y$ with the ability to "implicitly" solve $y$ locally as a $C^1$-function of the other coordinates. If you want to "solve" $z$ in terms of $x$ and $y$ instead of $y$, then you need to check that $D_zf$ does not vanish. $\endgroup$ Dec 29, 2018 at 9:30
  • $\begingroup$ @JyrkiLahtonen Hi! Firstly, thank you for answering. However, I'm afraid I don't quite understand. How do I know I have to solve z in terms of x and y? Would the same statement hold if I choose to solve x in terms of y and z? $\endgroup$
    – Joe
    Dec 29, 2018 at 10:24
  • 1
    $\begingroup$ If you want to solve $x$, then you need to check that $D_x f$ does not vanish at the point of interest. $\endgroup$ Dec 29, 2018 at 10:27
  • 2
    $\begingroup$ In a sense IFT is about linearizing the solution set (locally and approximately). So, when teaching, I emphasize the analogy with linear equations (and systems of such). From $$ax+by+cz=d$$ you can solve $x$ when $a\neq0$, $y$ when $b\neq0$ and $z$ when $c\neq0$. And, guess what, those $a,b,c$ are exactly the partial derivatives of the function here! The same with several constraints, Jacobians and the relevant minors of matrices of the coefficients of a linear system of equations. $\endgroup$ Dec 29, 2018 at 10:31

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.