MAT244-2018S > Web Bonus Problems


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Yichen Nie:
The type of critical point is called center, which is neutrally stable.
This occurs when the eigenvalues contain only imaginary numbers.
The trajectories just stay in stable, elliptical orbits as my classmates have shown above.

Victor Ivrii:
there is the big difference between describe (what you see) and explain (why it is so).

Kexin Sun:
The origin is a saddle point,and it is stable.
The ponit (π/2,π/2),(-π/2,π/2),(π/2, -π/2),(-π/2, -π/2) are the centers of closed orbits,respectively.Each of them is orbital stablility.

Victor Ivrii:
See my comment above

Meng Wu:
$\text{My attempt:}$
Let $$\begin{align}\cases{F(x,y)=\sin(x)\cos(y)=0\\G(x,y)=-\cos(x)\sin(y)=0}\end{align}$$
where $k$ can only be $-1,0,1$, since we have $x\in(-4,4)$ and $y\in(-4,4)$, $\pi\cong 3.14159265359$. $\\$
Thus we have the following critical points:$\\$
Case#1: $(0,0);(\pi,0);(-\pi,0);(0,\pi);(0,-\pi);(\pi,\pi);(-\pi,\pi);(-\pi,-\pi);(\pi,-\pi)$. $\\$
Case#2: $({\pi\over2},{\pi\over2});(-{\pi\over2},-{\pi\over2});({\pi\over2},-{\pi\over2});(-{\pi\over2},{\pi\over2})$.
When $(x,y)=$ critical points in Case#1: we get diagonal matrices $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ or $\begin{pmatrix}-1&0\\0&0\end{pmatrix}$, wtih eigenvalues $\lambda=\pm1$, eigenvectors $\xi=\begin{pmatrix}1\\0\end{pmatrix}$ or $\begin{pmatrix}0\\1\end{pmatrix}$. $\\$
Thus, the critical points in Case#1 are all $\text{Saddle Points}$. $\\$
When $(x,y)=$ critical points in Case#2: we get matrices $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ or $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ with eigenvalues $\lambda=\pm i$, eigenvectors $\xi=\begin{pmatrix}1\\i\end{pmatrix}$ or $\begin{pmatrix}1\\-i\end{pmatrix}$. $\\$
Thus, the critical points in Case#2 are all $\text{Center}$. $\\$
Therefore based on the phase portraits for saddle point and center, we can conclude the phase portrait for the given system will be like that.


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