I would like to solve the following system: $$ \begin{cases} x =\sin(u)\\ y=\sin(2u) \end{cases} $$ with${ \;0\leq u \leq\pi}$ (and not until $2\pi$!). This is illutrated with the following picture:

enter image description here

When I solve the equation we can obtain: $$ y^2=4x^2(1-x^2) $$

However this gives a famous butterfly shape: enter image description here

How do I obtain an equation as above without restricting $x$ to be positive?

  • $\begingroup$ Maybe you could try to write $\sin{2u}$ in terms of $\sin{u}$? $\endgroup$ – Matti P. Nov 8 at 12:12
  • $\begingroup$ $u=\frac {\pi} 2+\arctan (-\frac y {2x^{2}})$. You get this using the fact that $\sin (2u)=2 \sin (u)\cos (u)$. $\endgroup$ – Kabo Murphy Nov 8 at 12:20
  • $\begingroup$ Yes, thank you. I did this and it gave me the solution $y^2=4x^2(1-x^2)$. But I do not know how to obtain an equation describing a single part without having to restrict the range of $x$. $\endgroup$ – Ulysse Nov 8 at 12:22
  • $\begingroup$ What do you mean by "solve the system" ??? $\endgroup$ – Yves Daoust Nov 8 at 13:04

This is virtually impossible.

You can look for a solution of the form


where the function $f$ ensures that the product remains negative in the negative $x$, i.e. $$f(x)<0\iff x\in(-1,0).$$

This is obtained with $f(x)=\text{sgn}(x(1+x))$, leading to


The RHS has the merit of being defined everywhere and to be continuous. But there is a but: it also cancels at $x=-1$, giving an extra isolated point, and this is irreparable, because this zero is independent of the function $f$.

You can also work additively and design a function which is zero in $[0,1]$ and sufficiently negative elsewhere. For instance

$$y^2=4x^2(1-x^2)+1-|x|-|x-1|.$$ The plot shows the RHS (in magenta the original function, in green the negative term, in blue the result).

enter image description here

If the absolute value function is allowed, then this fulfills the request. But it is so ugly !

You can also add a term that restricts the domain to the non-negatives and is neutral in the positives, such as $\sqrt x-\sqrt{|x|}$, but IMO this is even worse.


Long story short: This does the trick... $$ y^2=4|x|x(1-x^2) $$

  • $\begingroup$ It doesn't. For instance $(-1,0)$ does fulfill this implicit equation. $\endgroup$ – Yves Daoust Nov 8 at 13:08
  • 1
    $\begingroup$ To the upvoter: before upvoting, check the answers. $\endgroup$ – Yves Daoust Nov 8 at 13:08

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.