Sovardipk

Answered

2022-07-01

Show that ${x}^{\ast}=(1,2)$ 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 ${W}^{s}(x)$ 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.

${x}_{1}^{\prime}=2+3{x}_{1}-2{x}_{2}-{x}_{1}^{2}+2{x}_{1}{x}_{2}-{x}_{2}^{2}$

${x}_{2}^{\prime}=3+4{x}_{1}-3{x}_{2}-{x}_{1}^{2}+2{x}_{1}{x}_{2}-{x}_{2}^{2}$

Determine ${W}^{s}(x)$ 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 & Explanation

Ronald Hickman

Expert

2022-07-02Added 18 answers

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

Calculate the jacobian of the right hand side:

$Df({x}_{1},{x}_{2})=\left(\begin{array}{cc}3-2{x}_{1}+2{x}_{2}& -2+2{x}_{1}-2{x}_{2}\\ 4-2{x}_{1}+2{x}_{2}& -3+2{x}_{1}-2{x}_{2}\end{array}\right)$

plug in our equilibrium point:

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

Eigenvalues:

$(5-\lambda )(-5-\lambda )+24=-(5-\lambda )(5+\lambda )+24={\lambda}^{2}-1=0$

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$$$\}$=$\mathrm{span}\{[1,\frac{2}{3}]\}=\{(x,y)\in {\mathbb{R}}^{2}\text{}|y=\frac{2}{3}x\}$

And unstable subspace:

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

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 $li{m}_{t\to \mathrm{\infty}}{W}^{s}(t)=(1,2)$, and $li{m}_{t\to -\mathrm{\infty}}{W}^{u}(t)=(1,2)$.

Calculate the jacobian of the right hand side:

$Df({x}_{1},{x}_{2})=\left(\begin{array}{cc}3-2{x}_{1}+2{x}_{2}& -2+2{x}_{1}-2{x}_{2}\\ 4-2{x}_{1}+2{x}_{2}& -3+2{x}_{1}-2{x}_{2}\end{array}\right)$

plug in our equilibrium point:

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

Eigenvalues:

$(5-\lambda )(-5-\lambda )+24=-(5-\lambda )(5+\lambda )+24={\lambda}^{2}-1=0$

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$$$\}$=$\mathrm{span}\{[1,\frac{2}{3}]\}=\{(x,y)\in {\mathbb{R}}^{2}\text{}|y=\frac{2}{3}x\}$

And unstable subspace:

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

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 $li{m}_{t\to \mathrm{\infty}}{W}^{s}(t)=(1,2)$, and $li{m}_{t\to -\mathrm{\infty}}{W}^{u}(t)=(1,2)$.

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