stability of $ \dot x = (x-1)(y-2), ~~~\dot y=(x-3)(y-2)$

Question

Question:

I want to determine the point of equilibrium and the stability (asymptotically stable, stable, or intable) $$ \dot x = (x-1)(y-2), ~~~\dot y=(x-3)(y-2)$$ Attempted solution:

So it has to be $$ 0=(x_0-1)(y_0-2), ~~~0=(x_0-3)(y_0-2) $$ Only case to consider is
$$y_0=2 \Rightarrow x_0=\alpha, \alpha \in \mathbb{R}$$ since $x_0=1$ and $x_0 =3$ is included.So the equilibrium solutions are $$(x_1(t),x_2(t)=(\alpha, 2), \alpha \in \mathbb{R}$$ It is $$J_v(\vec x) = \begin {bmatrix} y-2 & x-1 \\ y-2 & x-3 \end{bmatrix}$$ So $$J_v(\alpha, 2) = \begin {bmatrix} 0 & \alpha-1 \\ 0 & \alpha-3 \end{bmatrix}$$ with eigenvalues $\lambda_1=0, \lambda_2=\alpha-3$. Since $$Ker \begin {bmatrix} y-2 & x-1 \\ y-2 & x-3 \end{bmatrix}= \begin {bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$ the geometric multiplicity g and algebraic multiplicity a are equal if $\alpha \neq 3$. $$g(\lambda_1)=1=a(\lambda_1)$$ Futhermore the system is instable for $\alpha >3$ and stable for $\alpha < 3$ and for $\alpha = 0$ it's either asymptotically stable or stable or intable .

Could somebody please tell me, if this is correct?

Answer 1

It is the first time that I am studying such system. Let me tell you what I understand from this system. The solution $(x,y)$ can be plotted in the $xyt$-space. As we easily see, $y=2$ plane is the equlibrium plane, i.e., no solution curve with initial condititions $(x_{0},y_{0},t_{0})$ will intersect this plane in any time as we can show that the solutions exist and they are unique. If we denote the sign of the derivative of the solution in arrows, we can draw the following.

Show[{ParametricPlot[{{t,2},{1,t},{3,t}},{t,-1,5},PlotStyle->{{Red,Thick},{Blue,Dashed},{Blue,Dashed}},PlotRange->{{-1,5},{-1,5}},FrameTicks->{{1,3},2},None,None},Frame->True],ListPlot[{{1,2},{3,2}},PlotStyle->PointSize[Large]],Graphics[{Text["(\[DownArrow],\[DownArrow])",{0,4}],Text["(\[UpArrow],\[DownArrow])",{2,4}],Text["(\[UpArrow],\[UpArrow])",{4,4}],Text["(\[DownArrow],\[DownArrow])",{4,0}],Text["(\[DownArrow],\[UpArrow])",{2,0}],Text["(\[UpArrow],\[UpArrow])",{0,0}]}]},ImageSize->200]

So, what we may infer from this figure is the following.

1. If $x_{0}<1$ and $y_{0}>2$, then $(x(t),y(t))\to(-\infty,2)$.
2. If $1<x_{0}<3$ and $y_{0}>2$, then $(x(t),y(t))\to(1,2)$.
3. If $x_{0}>3$ and $y_{0}>2$, then $(x(t),y(t))\to(\infty,\infty)$.
4. If $x_{0}<1$ and $y_{0}<2$, then $(x(t),y(t))\to(1,2)$.
5. If $1<x_{0}<3$ and $y_{0}<2$, then $(x(t),y(t))\to(1,2)$.
6. If $x_{0}>3$ and $y_{0}<2$, then the $(x(t),y(t))\to(3,\infty)$.
7. If $-\infty<x_{0}<\infty$ and $y_{0}=2$, then $(x(t),y(t))\to(x_{0},2)$.
8. If $x_{0}=1$ and $y_{0}>2$, then $(x(t),y(t))\to(1,2)$.
9. If $x_{0}=1$ and $y_{0}<2$, then $(x(t),y(t))\to(1,2)$.
10. If $x_{0}=3$ and $y_{0}>2$, then $(x(t),y(t))\to(\infty,y_{0})$.
11. If $x_{0}=3$ and $y_{0}<2$, then $(x(t),y(t))\to(-\infty,y_{0})$.

Note that in Cases 7--11 above, we can explicitly solve the system.

I don't see any atractor points here, i.e., there is no point $P=(a,b)$ such that any initial condition $X_{0}=(x_{0},y_{0})$ close to it (i.e., $\|X_{0}-P\|<\delta$) implies $X=X(x_{0},y_{0},t)\to{}P$ as $t\to\infty$. I think the system is always unstable.