### Author Topic: FE-6  (Read 6780 times)

#### Victor Ivrii

• Elder Member
• Posts: 2563
• Karma: 0
##### FE-6
« on: April 17, 2013, 03:07:40 PM »
Consider equation

x'' = -2x^3 + 8x.

(a) Reduce to the first order system in variables $(t,x,y)$ with $y=x'$;

(b)  Find solution in the form $H(x,y)=C$;

(c) Find critical points, linearize the system at these points and classify the linearizations (i.e. specify whether they are nodes, saddles,  etc., indicate stability and, if applicable, the orientation);

(d) Sketch the phase portraits of the linearizations of the system  at every critical point;

(e) Sketch the phase portraits of the nonlinear system near each of the critical points;

(f) Sketch solution on $(x,y)$ plane.

#### Victor Lam

• Jr. Member
• Posts: 13
• Karma: 9
##### Re: FE-6
« Reply #1 on: April 17, 2013, 06:01:07 PM »
Solutions and phase portrait

#### Benny Ho

• Newbie
• Posts: 4
• Karma: 2
##### Re: FE-6
« Reply #2 on: April 17, 2013, 09:02:45 PM »
Solutions

#### Victor Ivrii

• Elder Member
• Posts: 2563
• Karma: 0
##### Re: FE-6
« Reply #3 on: April 18, 2013, 09:55:32 AM »
I finished grading FE6 which gave mean opportunity to be a good cop as with an exception of several borked papers majority did very well

(a) Setting $y'=x$ we arrive to

\left\{\begin{aligned}
&x'=y,\\
&y'=-2x^3+8x.
\end{aligned}\right.
\label{eq-1}

(b) From $\frac{dx}{y}=\frac{dy}{-2x^3+8x}$ we conclude $(-2x^3+8x)dx=ydy$ and integrating and collecting everything we get

H(x,y):=\frac{1}{2}y^2+\frac{1}{2}x^4-4x^2=C;
\label{eq-2}

(those who got the whole thing with some factor made no error)

(c) (i) Looking for equilibrium points (setting r.h.e. of (\ref{eq-1}) to $0$:  $y=-2x^3+8x=0 \implies x=0, \pm 2$, $y=0$ so we have points $(0,0), (-2,0), (2,0)$;
(ii) Linearization at $(0,0)$ leads to the matrix $\begin{pmatrix} 0 & 1\\ 8 &0\end{pmatrix}$ with eigenvalues $\pm \sqrt{8}$ and eigenvectors $\begin{pmatrix}1 \\ \pm \sqrt{8}\end{pmatrix}$ which tells us the slopes of separatrices and which is outgoing (arrows away from $(0,0)$) and which is incoming (arrows towards to $(0,0)$). Those who did not find eigenvectors got 1 pt reduction.
(ii) Linearization at $(\pm2,0)$ leads to the matrix $\begin{pmatrix} 0 & 1\\ -16 &0\end{pmatrix}$ with eigenvalues $\pm 4i$; one does not need to find eigenvectors but needs to observe that $-16<0$ implies clockwise orientation.

(d) (i) Linearized at $(0,0)$ system has a saddle;
(ii) Linearized at $(\pm 2,0)$ systems have centers;

(e) (i) Original system has a saddle at $(0,0)$;
(ii) In the general case center could remain a center o become a stable or unstable spiral point but we have an integrable system which cannot have spiral points; so: centres.

(f) Level lines (\ref{eq-2}) are empty are closed and cannot escape to infinity as $\frac{1}{2}x^4-4x^2$ tends to $+\infty$ as $x\to \pm \infty$; as $C=0$ level line is a separatrice and has a shape of $\Huge{\infty}$, so picture i as below.

One can make an analysis based exclusively on $H(x,y)$ which has minima at $(\pm 2,0)$ and a  saddle at $(0,0)$ (so system has centres and a saddle, respectively) but one should add arrows: if $y>0$ ($y<0$) $x'=y$ so $x$ increases (decreases) and therefore movement is clockwise.

Similar:
http://weyl.math.toronto.edu/MAT244-2011S-forum/index.php?topic=181.0

« Last Edit: April 21, 2013, 08:06:45 AM by Victor Ivrii »