Analytical Solution to a nonlinear system of equations

Question

Is it possible to analytically solve for x,y,z in the following system, and if so how would one go about it?

$a_{1} (y - 2x + z) - x + 2x^{3} - a_{2}x = 0$

$a_{1} (z - 2y + x) - y + 2y^{3} - a_{2}y = 0$

$a_{1} (x - 2z + y) - z + 2z^{3} - a_{2}z = 0$

I've tried numerous methods such as cyclic permutations of the three equations and substitution but none seem to work, hence any advice would be greatly appreciated.

Answer 1

Notice the symmetry in the equations. Adding them all together, we find that $$-(x+y+z)+2(x^3+y^3+z^3)-a_2(x+y+z) = 0$$ $$2(x^3+y^3+z^3) = (1+a_2)(x+y+z).$$ Now, we utilize the convenient factored form: $$x^3+y^3+z^3 - (x+y+z) = (x+y+z)(x+y+z+1)(x+y+z-1),$$ to write the above equation as $$(x+y+z)(x+y+z+1)(x+y+z-1) = \frac{a_2-1}{2}(x+y+z).$$ Assuming $x+y+z \ne 0$, $$(x+y+z+1)(x+y+z-1) = \frac{a_2-1}{2}.$$ Thus, $$z = \pm \sqrt{\frac{a_2+1}{2}}-x-y.$$ Notice that if we substitute this into the original first equation, we find $$a_1(\sqrt{\frac{a_2+1}{2}}-3x)+2x^3-a_2x = 0.$$ This is a cubic in $x$, which you can solve. Proceed in a similar fashion in the second original equation to find a cubic in terms only of $y$. Finally, solve $z = \pm \sqrt{\frac{a_2+1}{2}}-x-y$ to find $z$.

Note, you will also have to check for solution when $x+y+z = 0$, but I think this will at least give you a good start.

By using $s = x+y+z$ and looking at first part of dxiv's answer, this case is very simple.

Answer 2

Let $\,s = x+y+z\,$ then the first equation can be written as:

$$ \begin{align} 0 &= a_{1} (y - 2x + z) - x + 2x^{3} - a_{2}x \\ &= a_{1} (s - 3x) - x + 2x^{3} - a_{2}x \\ &= 2x^{3} - (3 a_1 + a_2 + 1) x + a_{1} s \end{align} $$

The same equation is satisfied by $\,y\,$ and $\,z\,$. Assuming $\,x,y,z\,$ are distinct, they are the roots of the same depressed cubic, so their sum must be $\,s = 0\,$, and the equations reduce to:

$$ 2t^{3} - (3 a_1 + a_2 + 1) t = 0 \quad\quad \Big|\quad t = x, y, z $$

Therefore the non-trivial distinct solutions of the system are permutations of:

$$ x = \sqrt{\frac{3a_1+a_2+1}{2}} \;,\quad y = - \sqrt{\frac{3a_1+a_2+1}{2}}\;, \quad z = 0 $$

This leaves the case where pairs of $\,x,y,z\,$ are equal to be solved separately.

---

[ EDIT ] The following covers the remaining case where two solutions are equal. Assume WLOG that $\,y=z\,$, then the last equation becomes redundant, and the system reduces to:

$$ \begin{cases} \begin{align} 2x^{3} - (a_2+1) x - 2a_{1} (x-y) &= 0 \\ 2y^{3} - (a_2 + 1)y + a_{1} (x-y) &= 0 \end{align} \end{cases} $$

It can be seen "by inspection" that a solution is $\,x=y\,$ where $\,x\,$ is a root of $\,2x^{3} - (a_2+1) x = 0\,$, which gives the identical triplets $\,x = y = z \in \left\{0, \pm \sqrt{\dfrac{a_2+1}{2}}\right\}\,$ as solutions.

To find all solutions, the first equation gives $\,y\,$ as a cubic in $\,x\,$, then substituting into the second equation gives an equation of degree $\,9\,$ in $\,x\,$ alone. We know from the previous step that $\,2x^{3} - (a_2+1) x\,$ must be a factor, so that leaves a sextic in $\,x\,$, which in the end turns out to be a cubic in $\,x^2\,$ (courtesy WA):

$$ x^2 (a_2 - 2 x^2 + 1)^2 + 6 a_1 x^2 (a_2 - 2 x^2 + 1) - 2 a_1^2 (a_2 - 6 x^2 + 1) - 6 a_1^3 = 0 $$