nonlinear 3d system

Question

I am trying to find all the critical points of my 3D Nonlinear System described as follows:

\begin{array}{ll} \\ \dot{x}_1=\mu-x_1^2+x_3+x_2-2x_1 \quad (1)\\\\ \dot{x}_2=\mu-x_2^2+x_3+x_1-2x_2 \quad (2)\\\\ \dot{x}_3=\mu-x_3^2+x_2+x_1-2x_3 \quad (3)\\ \end{array}

I know that there are at least two critical points such that $(x_1,x_2,x_3)=(\pm\sqrt{\mu},\pm\sqrt{\mu},\pm\sqrt{\mu}).$ Now I want to eliminate my variables from each equation.

So for instance when $\dot{x}_1=0 \implies x_3=x_1^2+2x_1-x_2-\mu$ and substituting this into (2) yields $$\dot{x}_2=x_1^2-x^2_2+3x_1-3x_2$$ Setting this equal to zero gives me the relationship $x_1^2+3x_1=x_2^2+3x_2 \implies x_1=x_2.$

Then putting both these relationships into (3) gives me the quartic equation $$\dot{x}_3=-x^4_1-2x^3_1+(2\mu-3)x^2_1+2\mu x_1+\mu(3-\mu)$$ When $\mu=2$, I found the roots $(-\sqrt{2},-1,\sqrt{2})$, but I feel as if I should of got four distinct roots than a double root ($-1$ is the double root).

I was wondering if I could get some help solving this generally for any $\mu>0$.

Thanks.

Answer 1

The difference of the first two equations factors as $$ 0=x_2^2-x_1^2+3x_2-3x_1=(x_2-x_2)(x_1+x_2+3) $$ so that you get two possibilities, $x_2=x_1$ and additionally $x_2=-3-x_1$.

In the difference between the first and the third equation you get the same variants, $x_3=x_1$ or $x_3=-3-x_1$.

Now insert all 4 combinations in the first of the original equations.

\begin{array}{l|l|ll} x_2=x_1&x_3=x_1&0=\mu-x_1^2&\implies x_1=x_2=x_3=\pm\sqrt\mu\\\hline x_2=x_1&x_3=-3-x_1&0=\mu-x_1^2-3-2x_1&\implies x_1=-1\pm\sqrt{\mu-2}\\\hline x_2=-3-x_1&x_3=-3-x_1&0=\mu-x_1^2-6-4x_1&\implies x_1=-2\pm\sqrt{\mu-2} \end{array}

Answer 2

Solving for $x_1,x_2,x_3$

$$ \left\{ \begin{array}{rcl} \mu-x_1^2-2 x_1+x_2+x_3=0 \\ \mu-x_2^2-2 x_2+x_1+x_3=0 \\ \mu-x_3^2-2 x_3+x_1+x_2=0 \\ \end{array} \right. $$

we have the parameterization

$$ \left[ \begin{array}{ccc} x_1(\mu) & x_2(\mu) & x_3(\mu) \\ -\sqrt{\mu -2}-2 & \sqrt{\mu -2}-1 & \sqrt{\mu -2}-1 \\ -\sqrt{\mu -2}-1 & \frac{1}{2} \left(-\sqrt{4 \mu -4 \sqrt{\mu -2}-7}-3\right) & \frac{1}{2} \left(\sqrt{4 \mu -4 \sqrt{\mu -2}-7}-3\right) \\ -\sqrt{\mu -2}-1 & \frac{1}{2} \left(\sqrt{4 \mu -4 \sqrt{\mu -2}-7}-3\right) & \frac{1}{2} \left(-\sqrt{4 \mu -4 \sqrt{\mu -2}-7}-3\right) \\ \sqrt{\mu -2}-2 & -\sqrt{\mu -2}-1 & -\sqrt{\mu -2}-1 \\ \sqrt{\mu -2}-1 & \frac{1}{2} \left(-\sqrt{4 \mu +4 \sqrt{\mu -2}-7}-3\right) & \frac{1}{2} \left(\sqrt{4 \mu +4 \sqrt{\mu -2}-7}-3\right) \\ \sqrt{\mu -2}-1 & \frac{1}{2} \left(\sqrt{4 \mu +4 \sqrt{\mu -2}-7}-3\right) & \frac{1}{2} \left(-\sqrt{4 \mu +4 \sqrt{\mu -2}-7}-3\right) \\ -\sqrt{\mu } & -\sqrt{\mu } & -\sqrt{\mu } \\ \sqrt{\mu } & \sqrt{\mu } & \sqrt{\mu } \\ \end{array} \right] $$

which parameterize some lines as can be observer in the attached plot

Now we can try to qualify those points according to the linear theory, inserting those values on

$$ J(x(\mu)) = \left( \begin{array}{ccc} -2 x_1(\mu)-2 & 1 & 1 \\ 1 & -2 x_2(\mu)-2 & 1 \\ 1 & 1 & -2 x_3(\mu)-2 \\ \end{array} \right) $$

and analyzing the eigenvalues. This is left as an exercise.