Author Topic: FE-P6  (Read 879 times)

Jingze Wang

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Re: FE-P6
« Reply #15 on: December 17, 2018, 05:44:04 AM »
Since system is integrable, so just saddle and center as in linear system :)

Jingze Wang

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Re: FE-P6
« Reply #16 on: December 17, 2018, 05:50:10 AM »
I think (-4,0) and (4,0) are maximum and (0,0) is minimum
« Last Edit: December 17, 2018, 12:42:29 PM by Jingze Wang »

Victor Ivrii

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Re: FE-P6
« Reply #17 on: December 17, 2018, 07:11:00 AM »
There cannot be spiral in the integrable case.

However my question is different and Calculus II related: function $H(x,y)$ has the same critical points as the system. What types they are: maximum, minimum or saddle for each of three points found.

Qinger Zhang

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Re: FE-P6
« Reply #18 on: December 17, 2018, 09:08:12 PM »
critical points

Qinger Zhang

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Re: FE-P6
« Reply #19 on: December 17, 2018, 09:09:38 PM »
I think (0,0) is a saddle. why (4,0) and (-4,0) are not centre?
« Last Edit: December 17, 2018, 09:15:05 PM by Qinger Zhang »

Qinger Zhang

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Re: FE-P6
« Reply #20 on: December 17, 2018, 09:10:17 PM »
Question: For $H(x,y)$ what are points $(\pm 4,0)$ and $(0,0)$?
From the grape, I don't think any of them is max or min.
« Last Edit: December 17, 2018, 09:16:16 PM by Qinger Zhang »

Victor Ivrii

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FE-P6 Official
« Reply #21 on: December 17, 2018, 09:18:11 PM »
(a)  Since $x^2+y^2+4>0$  all stationary points are at $y=0$, and either $x=0$ or $x^2+y^2-16=0$; thus we have
$A_1=(0,0)$ and $A_{2,3}=(\pm 4,0)$.

(b) Let $f=2y(x^2+y^2+4)$, $g=-2x(x^2+y^2-16)$.

At $A_1$ we have $f_x=0$, $f_y=8$, $g_x=32$, $g_y=0$ and linearization is
$$\begin{pmatrix}X\\ Y\end{pmatrix}'=\begin{pmatrix} 0 &8\\32 &0\end{pmatrix}\begin{pmatrix}X\\Y\end{pmatrix}$$
 with eigenvalues $\pm 16$ and eigenvectors $\begin{pmatrix}1 \\\pm2\end{pmatrix}$ so  it is a saddle:

At points $A_{2,3}$ we have $f_x=0$, $f_y=16$, $g_x=-16$, $g_y=0$ with eigenvalues $16i$ and $-16i$; in virtue of (c) it is a center, and since the bottom left element is negative it is clockwise.


Rewriting the system as
$$
2x(x^2+y^2-16)\,dx + 2y(x^2+y^2+4)\,dy=0
$$
one can check that the form on the left is exact; $H_x= 2(x^3+xy^2-16x)\implies H(x,y)=\frac{1}{2}(x^4 +2x^2y^2 -32 x^2)+h(y)$ and plugging to $H_y=2x^2y +h'(y)= 2yx^2+2y^3 +8y\implies h'(y)=2y^3 +8y\implies h(y)=\frac{1}{2}y^4+4y^2$ (we take a constant equal $0$). Then
$$H(x,y)=\frac{1}{2}\bigl(x^4 +2x^2y^2 +y^4-32 x^2 +8y^2\bigr).$$

Points $(\pm 4,0)$ are minima of $H(x,y)$ and $(0,0)$ is the saddle point. So called symmetric two-well 2D potential.
See 3D plot generated by WolframAlpha
« Last Edit: December 17, 2018, 09:19:47 PM by Victor Ivrii »