1
$\begingroup$

I'm trying to solve the non-linear, two-variable system of equations \begin{align*} y-x+x^5-\frac{y^4x}{2(1+x^2)^2}-\frac{x^3}{1+y^2} &= 0 \\ -x-y+y^5-\frac{x^4y}{2(1+y^2)^2}-\frac{y^3}{1+x^2} &= 0 \end{align*} with $x,y \in \mathbb{R}$. You might notice that the second equation is almost obtained from the first by swapping $x$ and $y$, except for the first term whose signs are opposed. I think the only solution is the origin $x=y=0$, and this is seems to be confirmed numerically on WolframAlpha, but I'm not sure, and I don't know how to prove it. You can also write the system as \begin{align*} y-x+x^5+\frac{1}{4}\partial_x\left(\frac{y^4}{1+x^2}-\frac{x^4}{1+y^2}\right) &= 0 \\ -x-y+y^5-\frac{1}{4}\partial_y\left(\frac{y^4}{1+x^2}-\frac{x^4}{1+y^2}\right) &= 0 \end{align*} Any hints/ideas?

$\endgroup$
2
  • $\begingroup$ The only solution is x=y=0 (desmos.com/calculator/fkxeyiwyks). $\endgroup$
    – Alessio K
    Apr 17 '20 at 15:48
  • $\begingroup$ As I said, this definitely seems true from numerical experiments, but how can I prove it? $\endgroup$
    – smalldog
    Apr 17 '20 at 15:50
1
$\begingroup$

I converted both equations into polynomials and put them into Singular to find all solutions.

LIB "solve.lib";
ring r=0,(x,y),dp;
poly p1=2x9y2+2x9+4x7y2+2x7-4x5+2x4y3+2x4y-4x3y2-6x3+4x2y3+4x2y-xy6-xy4-2xy2-2x+2y3+2y;
poly p2=-x6y-x4y-2x3y4-4x3y2-2x3+2x2y9+4x2y7-4x2y3-2x2y-2xy4-4xy2-2x+2y9+2y7-4y5-6y3-2y;
ideal i=p1,p2;
def T=solve(i,30);

There are $89$ distinct solutions, but $88$ of them involve a complex number.

This shows that the last solution, $x=y=0$, is the only real solution.

$\endgroup$
5
  • $\begingroup$ Thanks for this. Is this library mathematically proven to yield all solutions? $\endgroup$
    – smalldog
    Apr 17 '20 at 16:00
  • $\begingroup$ @chaos Yes. Because it works with Gröbner bases and other exact polynomial root-finding schemes, it is one of the most powerful mathematical libraries around. You should give it a try. $\endgroup$ Apr 17 '20 at 16:06
  • $\begingroup$ Thank you, I was not even aware that exact polynomial root-finding algorithms existed. Is it possible to give exact expressions for all roots, or is it only possible to determine the number of real roots, which is sufficient here? $\endgroup$
    – smalldog
    Apr 17 '20 at 16:09
  • $\begingroup$ @chaos Abel-Ruffini theorem: you cannot determine exact expressions for quintic+ polynomials in the general case. Simply counting roots is sufficient. $\endgroup$ Apr 17 '20 at 16:10
  • $\begingroup$ Yes, that's what I thought. Thanks again! $\endgroup$
    – smalldog
    Apr 17 '20 at 16:13
1
$\begingroup$

According to Maple, the resultant of the numerators of the left sides of the two equations with respect to $y$ is $$ 2\,{x}^{9} \left( 4096\,{x}^{80}+69632\,{x}^{78}+516096\,{x}^{76}+ 2084864\,{x}^{74}+4272128\,{x}^{72}-163840\,{x}^{70}-25825280\,{x}^{68 }-65273856\,{x}^{66}-38729728\,{x}^{64}+154712064\,{x}^{62}+382149632 \,{x}^{60}+144474112\,{x}^{58}-791342976\,{x}^{56}-1359673728\,{x}^{54 }+226609664\,{x}^{52}+3532913024\,{x}^{50}+4419130368\,{x}^{48}- 672814592\,{x}^{46}-8622736384\,{x}^{44}-11440427968\,{x}^{42}- 5943181984\,{x}^{40}+2080361568\,{x}^{38}+6219084784\,{x}^{36}+ 8104484032\,{x}^{34}+13320834677\,{x}^{32}+21495544525\,{x}^{30}+ 25938523882\,{x}^{28}+22524511317\,{x}^{26}+14404728925\,{x}^{24}+ 7384009128\,{x}^{22}+3941633218\,{x}^{20}+2832052128\,{x}^{18}+ 2199851588\,{x}^{16}+1418048480\,{x}^{14}+701995632\,{x}^{12}+ 263885888\,{x}^{10}+74884576\,{x}^{8}+15715584\,{x}^{6}+2310144\,{x}^{ 4}+208896\,{x}^{2}+8192 \right) \left( {x}^{2}+1 \right) ^{6} $$ The only real root of this is $x=0$, which implies $y=0$ from the first equation.

$\endgroup$
3
  • $\begingroup$ Thanks for this. How do you prove that this polynomial of order $80$ has no real roots? $\endgroup$
    – smalldog
    Apr 17 '20 at 16:01
  • $\begingroup$ With Maple's sturm command, which uses Sturm's theorem. $\endgroup$ Apr 17 '20 at 16:06
  • $\begingroup$ Thank you, I was not aware of this theorem. Unfortunately I can only accept one of the two answers, but I will upvote both... I don't know what the policy is here, but I will accept the other simply for providing a little more explanation in the comment, though your answer is equally flawless... $\endgroup$
    – smalldog
    Apr 17 '20 at 16:13

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.