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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!

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  • $\begingroup$ @RodrigodeAzevedo Yes! $\endgroup$
    – Titus
    Jun 8, 2019 at 11:41
  • $\begingroup$ 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, $\dot x=-ax-y+...$ $\endgroup$ Jun 8, 2019 at 11:44

2 Answers 2

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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}]

enter image description here

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.

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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.

enter image description here

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  • $\begingroup$ @LutzL Tanks for your corrections. I had manipulated some incorrect formulas. $\endgroup$
    – Cesareo
    Jun 8, 2019 at 16:43

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