Sovardipk

2022-07-01

Show that ${x}^{\ast }=\left(1,2\right)$ is a fixed point of the system
${x}_{1}^{\prime }=2+3{x}_{1}-2{x}_{2}-{x}_{1}^{2}+2{x}_{1}{x}_{2}-{x}_{2}^{2}$
${x}_{2}^{\prime }=3+4{x}_{1}-3{x}_{2}-{x}_{1}^{2}+2{x}_{1}{x}_{2}-{x}_{2}^{2}$
Determine ${W}^{s}\left(x\right)$ and ${W}^{u}\left(x\right)$, the global stable and unstable manifolds of the fixed point $x=\left(1,2\right)$ and give a parametric representation of those manifolds.

Ronald Hickman

Expert

Plug in $\left(1,2\right)$ to see that you get $0$, so the point is an equilibrium.
Calculate the jacobian of the right hand side:
$Df\left({x}_{1},{x}_{2}\right)=\left(\begin{array}{cc}3-2{x}_{1}+2{x}_{2}& -2+2{x}_{1}-2{x}_{2}\\ 4-2{x}_{1}+2{x}_{2}& -3+2{x}_{1}-2{x}_{2}\end{array}\right)$
plug in our equilibrium point:
$Df\left(1,2\right)=\left(\begin{array}{cc}3-2+4& -2+2-4\\ 4-2+4& -3+2-4\end{array}\right)=\left(\begin{array}{cc}5& -4\\ 6& -5\end{array}\right)$
Eigenvalues:
$\left(5-\lambda \right)\left(-5-\lambda \right)+24=-\left(5-\lambda \right)\left(5+\lambda \right)+24={\lambda }^{2}-1=0$
Thus $\lambda =±1$ 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 $\lambda =1⇒\left[1,1\right]$ and $\lambda =-1⇒\left[1,\frac{2}{3}\right]$
So our stable subspace:
${E}^{s}=$ = span$\left\{eigenvector, for\lambda =-1$$\right\}$=
And unstable subspace:
${E}^{u}=$$\left\{$eigenvector, for $\lambda =1$$\right\}$ =
Then the stable and unstable manifolds ${W}^{s},{W}^{u}$ are then tangential to ${E}^{s},{E}^{u}$, at $\left(1,2\right)$, But I am unsure on how to parameterise them, but they must depend and $t$. And $li{m}_{t\to \mathrm{\infty }}{W}^{s}\left(t\right)=\left(1,2\right)$, and $li{m}_{t\to -\mathrm{\infty }}{W}^{u}\left(t\right)=\left(1,2\right)$.

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