# Find the equilibria of the following system of ODEs

Find the equilibria of the following system of ODEs

\begin{align*} \dot{x} = ax - y + kx(x^2 + y^2)\\ \dot{y} = x - ay + ky(x^2 + y^2) \end{align*}

where $$a$$ and $$k$$ are constants, $$a > 1$$ and $$a^2 \geq 1$$.

I want to find the equilibria of this system and say something about the behaviour in the phase-plane. Unfortunately, I can't seem to find the equilibria.

What I've tried: I subtracted the second equation from the first one to arrive at \begin{align} ax + ay -y -x+kx(x^2 + y^2) - ky(x^2 + y^2) = 0 \\ \Leftrightarrow a(x + y) - (x+y) + (x - y)k(x^2 + y^2) = 0\\ \Leftrightarrow a - 1 + \dfrac{x - y}{x+y}(x^2 + y^2)k = 0 \end{align} From here I want to find an expression for $$x$$ or $$y$$ but I don't know how to do so.

Question: How should I approach this? If I'd get a hint on how to proceed I think I can solve the problem myself.

Thanks!

• @RodrigodeAzevedo Yes! Jun 8, 2019 at 11:41
• Are you sure about the sign structure of the linear terms? To get a limit cycle I would expect the first term to have a minus sign, $\dot x=-ax-y+...$ Jun 8, 2019 at 11:44

Set $$\dot x=\dot y=0$$. Multiply the first equation with $$x$$, the second with $$y$$ and add to get $$0=a(x^2-y^2)+k(x^2+y^2)^2.$$ Now do everything askew, multiply the first equation with $$y$$, the second with $$x$$ and subtract to get $$0=2axy-(x^2+y^2)$$ Apart from the origin, there are also solutions for $$y=q_\pm x=(a\pm\sqrt{a^2-1})x$$, and from the first equation, $$y^2-x^2=4kax^2y^2\implies q^2-1=4kaq^2x^2,~~ x^2=\frac{q^2-1}{4kaq^2}$$ which only gives real solutions for $$q=q_+=a+\sqrt{a^2-1}>1$$.

Example with $$a=2$$, $$k=3$$ using WolframAlpha

streamplot[{2x-y+3x(x^2+y^2), x-2y+3y(x^2+y^2)}, {x,-1.5,1.5}, {y,-1.5,1.5}]


showing a saddle point at the origin and two additional centers at the other two stationary points. At the boundary of the plot all solutions point outwards.

Solving for the equilibrium poins we have

$$\left\{ \begin{array}{c} a x-y +k x\left(x^2+y^2\right)=0 \\ x-a y+k y\left(x^2+y^2\right)=0 \\ \end{array} \right.$$

giving the equilibrium points. (with the help of Wolfram)

$$\left[ \begin{array}{cc} x & y \\ 0 & 0 \\ -\frac{\sqrt{-\frac{a \left(a^2-1\right) k+\sqrt{a^4 \left(a^2-1\right) k^2}}{a^2 k^2}}}{\sqrt{2}} & \frac{\left(\sqrt{a^4 \left(a^2-1\right) k^2}-a^3 k\right) \sqrt{-\frac{a \left(a^2-1\right) k+\sqrt{a^4 \left(a^2-1\right) k^2}}{a^2 k^2}}}{\sqrt{2} a^2 k} \\ \frac{\sqrt{-\frac{a \left(a^2-1\right) k+\sqrt{a^4 \left(a^2-1\right) k^2}}{a^2 k^2}}}{\sqrt{2}} & \frac{\left(a^3 k-\sqrt{a^4 \left(a^2-1\right) k^2}\right) \sqrt{-\frac{a \left(a^2-1\right) k+\sqrt{a^4 \left(a^2-1\right) k^2}}{a^2 k^2}}}{\sqrt{2} a^2 k} \\ -\frac{\sqrt{\frac{-k a^3+k a+\sqrt{a^4 \left(a^2-1\right) k^2}}{a^2 k^2}}}{\sqrt{2}} & -\frac{\sqrt{\frac{-k a^3+k a+\sqrt{a^4 \left(a^2-1\right) k^2}}{a^2 k^2}} \left(k a^3+\sqrt{a^4 \left(a^2-1\right) k^2}\right)}{\sqrt{2} a^2 k} \\ \frac{\sqrt{\frac{-k a^3+k a+\sqrt{a^4 \left(a^2-1\right) k^2}}{a^2 k^2}}}{\sqrt{2}} & \frac{\sqrt{\frac{-k a^3+k a+\sqrt{a^4 \left(a^2-1\right) k^2}}{a^2 k^2}} \left(k a^3+\sqrt{a^4 \left(a^2-1\right) k^2}\right)}{\sqrt{2} a^2 k} \\ \end{array} \right]$$

for instance, with $$a = 2, k = 3$$ we have

$$\left[ \begin{array}{cc} x & y \\ 0 & 0 \\ -0.19666 & -0.733945 \\ 0.19666 & 0.733945 \\ \end{array} \right]$$

the Jacobian being

$$J = \left( \begin{array}{cc} a+k \left(3 x^2+y^2\right) & 2 k x y-1 \\ k \left(3 x^2+y^2\right)+1 & 2 k x y-a \\ \end{array} \right)$$

and at each equilibrium point

$$\begin{array}{c} \left( \begin{array}{cc} a & -1 \\ 1 & -a \\ \end{array} \right) \\ \left( \begin{array}{cc} \frac{a k+2 \sqrt{a^4 \left(a^2-1\right) k^2}}{a^2 k} & \frac{\sqrt{a^4 \left(a^2-1\right) k^2}}{a^3 k}-1 \\ -a+1+\frac{1}{a}+\frac{2 \sqrt{a^4 \left(a^2-1\right) k^2}}{k a^2} & \frac{\sqrt{a^4 \left(a^2-1\right) k^2}}{a^3 k}-a \\ \end{array} \right) \\ \left( \begin{array}{cc} \frac{a k+2 \sqrt{a^4 \left(a^2-1\right) k^2}}{a^2 k} & \frac{\sqrt{a^4 \left(a^2-1\right) k^2}}{a^3 k}-1 \\ -a+1+\frac{1}{a}+\frac{2 \sqrt{a^4 \left(a^2-1\right) k^2}}{k a^2} & \frac{\sqrt{a^4 \left(a^2-1\right) k^2}}{a^3 k}-a \\ \end{array} \right) \\ \end{array}$$

or assuming $$a = 2, k = 3$$

$$\begin{array}{c} \left( \begin{array}{cc} 2. & -1 \\ 1 & -2. \\ \end{array} \right) \\ \left( \begin{array}{cc} 3.9641 & -0.133975 \\ 1.86603 & 2.9641 \\ \end{array} \right) \\ \left( \begin{array}{cc} 3.9641 & -0.133975 \\ 1.86603 & 2.9641 \\ \end{array} \right) \\ \end{array}$$

with eigenvalues

$$\left( \begin{array}{ccl} -1.73205 & 1.73205 & \text{saddle}\\ 3.4641 & 3.4641 & \text{source}\\ 3.4641 & 3.4641 & \text{source}\\ \end{array} \right)$$

Resuming for $$a = 2, k = 3$$ we have a saddle and two sources.

• @LutzL Tanks for your corrections. I had manipulated some incorrect formulas. Jun 8, 2019 at 16:43