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### Messages - Sabrina (Man) Luo

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1
##### Ch 9 / Final Review
« on: April 16, 2013, 02:29:29 AM »
As you said that if it is node, then the system is not integrable. However, what will be the way to know more about globally phase portrait?

Another one:We studied energy equation for saddle and centre in non-linear system, when it is integrable, so can we think that if it is integrable, then by looking for energy equation, the centre won't become a spiral?

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##### Quiz 5 / Re: Night Sections
« on: April 04, 2013, 08:46:34 AM »
Benny, nice try, but ...

Devangi â€“ good, Frankâ€”even better. But phase portrait is still missing (and bonus question too)

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##### Quiz 5 / Re: Night Sections
« on: April 04, 2013, 08:20:40 AM »
Benny, nice try, but ...

Devangi â€“ good, Frankâ€”even better. But phase portrait is still missing (and bonus question too)

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##### Quiz 5 / Day Section's Quiz Problem 2
« on: April 03, 2013, 09:12:44 AM »
(2) Find an equation of the form H(x,y)=c satisfied by solutions to
\begin{equation*}
\left\{\begin{aligned}
&dx/dt=2x^2y-3x^2-4y,\\
&dy/dt=-2xy^2+6xy
\end{aligned}
\right.\end{equation*}

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##### Quiz 5 / Day Section's Quiz - Problem 1
« on: April 03, 2013, 09:11:21 AM »
(1) For the system
\begin{equation*}
\left\{\begin{aligned}
&dx/dt=y+x(1-x^2-y^2),\\
&dy/dt=-x+y(1-x^2-y^2)
\end{aligned}
\right.\end{equation*}
determine all critical points, linearize around each critical point, and determine what conclusion can be made about the nonlinear system at each critical point based on the linearization. Draw a phase portrait for the nonlinear system.

6
##### Term Test 2 / Re: TT2 Question 2
« on: March 28, 2013, 08:33:10 AM »
This is an integrable system: denoting $y=x'$ we get
\begin{equation*}
H(x,y) := \frac{1}{2}y^2 \underbrace{-\frac{1}{5}x^5+\frac{5}{3}x^3 -4x}_{V(x)}=E
\end{equation*}
wehre $V(x)=-\int f(x)\,dx$, $f(x)=x^4-5x^2+4$ is the r.h.e.  $V(x)$ is a potential.

Since $V(x)$ has non-degenerate maxima at $x=-1$ and $x=2$ and minima at $x=-2$ and $x=1$ and $\frac{1}{2}y^2$ has non-degenerate minimum at $y=0$ we have 4 non-degenerate critical points of $H(x,y)$: namely, $(-1,0)$ and $(2,0)$ are saddle points and $(-2,0)$ and $(1,0)$ are minima.

For dynamics two former are saddle points and two latter re centers (recall that integrable systems cannot have spiral or nodal points and centers are detectable!)

What is missing for everyone? Plot of $V(x)$ which shows that $V(-1)>V(-2)>V(2)>V(1)$ (one needs just calculates them) and therefore picture must be like on Jason' computer generated and not Yeong' drawing since separatrix passing through saddle $(-1,0)$ is "higher" and therefore envelops saddle $(2,0)$ and cannot pass through it as Yeong drew. And therefore separatrix passing through $(2,0)$ stops short and goes back to the right from $(-1,0)$ (not passing through  it). If in some examples we studied before separatrices passing through different saddles are the same it is not the general rule.

To analyze Jason's picture:

a) Look at the centers
b) Find separatrix passing through $(2,0)$
c) Find separatrix passing through $(-1,0)$

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##### Quiz 4 / Re: Quiz 4--Problem (night sections)
« on: March 21, 2013, 11:01:47 PM »
I am talking about the second picture (classification of part b), the written solution of the third equilibrium part. There is a calculating error. I just corrected it.

8
##### Quiz 4 / Re: Quiz 4--Problem (night sections)
« on: March 21, 2013, 01:23:04 PM »
I am talking about the second picture (classification of part b), the written solution of the third equilibrium part. There is a calculating error. I just corrected it.

9
##### Quiz 4 / Re: Quiz 4--Problem (night sections)
« on: March 21, 2013, 06:37:10 AM »
Some classifications for part b
For the Thrid Matrix A: Det(A) should be r^2+r+3=0, and r=-1/2Â±i sqrt 11/2

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