Global stable and unstable manifolds of quadratic 2D differential system

Question

Show that $x^* = (1, 2)$ is a fixed point of the system

$x_1' = 2 + 3x_1 − 2x_2 − x_1^2 + 2x_1x_2 − x_2^2$

$x_2' = 3 + 4x_1 − 3x_2 − x_1^2 + 2x_1x_2 − x_2^2$

Determine $W^s(x)$ and $W^u(x)$, the global stable and unstable manifolds of the fixed point $x=(1, 2)$ and give a parametric representation of those manifolds.

Answer 1

The parametrization $(x_1,x_2)=(1+u+v,2+u-v)$, yields the differential system $$ u'=u+10v-4v^2,\qquad v'=-v. $$ This proves that $(x_1^*,x_2^*)=(1,2)$, which corresponds to the origin $(0,0)$ of the $(u,v)$-plane, is a fixed point $O$. The linearized system around $O$ is $$ u'=u+10v,\qquad v'=-v, $$ whose eigenvalues are $1$ and $-1$, hence $O$ is a saddle point. The axis $$\color{red}{v=0}$$ is invariant by the original differential system, which reduces there to $u'=u$, $v=0$, hence this axis is the $\color{red}{\text{unstable manifold}}$.

To determine the stable manifold, note that, for every solution $(u,v)$, one has $v(t)=v_0\mathrm e^{-t}$. Solving $u'=u+10v-4v^2$ then yields $$u(t)=-5v_0\mathrm e^{-t}+\tfrac43v_0^2\mathrm e^{-2t}+c\mathrm e^{t}=-5v(t)+\tfrac43v(t)^2+c\mathrm e^t,$$ for some constant $c$. The case $u(t)\to0$ corresponds to $c=0$, thus the $\color{blue}{\text{stable manifold}}$ is defined by the equation $$\color{blue}{u=-5v+\tfrac43v^2}.$$ If need be, these can be translated back in terms of $(x_1,x_2)$ coordinates, using $$ (u,v)=\tfrac12(x_1+x_2-3,x_1-x_2+1). $$

Answer 2

Plug in $(1,2)$ to see that you get $0$, so the point is an equilibrium.

Calculate the jacobian of the right hand side:

$ Df (x_1, x_2)= \left( \begin{array}{ccc} 3-2x_1+2x_2 & -2+2x_1-2x_2 \\ 4-2x_1+2x_2 & -3+2x_1-2x_2 \end{array} \right)$ plug in our equilibrium point:

$ Df (1, 2)= \left( \begin{array}{ccc} 3-2+4 & -2+2-4 \\ 4-2+4& -3+2-4 \end{array} \right)=\left( \begin{array}{ccc} 5 & -4 \\ 6& -5 \end{array} \right) $

Eigenvalues: $(5-\lambda)(-5-\lambda)+24=-(5-\lambda)(5+\lambda)+24=\lambda^2-1=0$

Thus $ \lambda=\pm 1 $ since our first eigenvalue is negative, constitutes to the stable manifold, and the other is positive, and constitutes to the unstable manifolds .

Now our corresponding eigenvectors are $\lambda=1\Rightarrow [1,1]$ and $\lambda=-1 \Rightarrow [1,\frac{2}{3}]$

So our stable subspace:

$E^s=$span$\{$eigenvector, for $\lambda=-1$$\}$=$\operatorname{span}\{[1,\frac{2}{3}]\}=\{(x,y)\in\mathbb{R}^2~|y=\frac{2}{3}x\}$

And unstable subspace:

$E^u=$span$\{$eigenvector, for $\lambda=1$$\}$=$\operatorname{span}\{[1,1]\}=\{(x,y)\in\mathbb{R}^2~|y=x\}$

Then the stable and unstable manifolds $W^s,W^u$ are then tangential to $E^s,E^u$, at $(1,2)$, But I am unsure on how to parameterise them, but they must depend and $t$. And $lim_{t\to\infty}W^s(t)=(1,2)$, and $lim_{t\to-\infty}W^u(t)=(1,2)$.