# Toronto Math Forum

## MAT244-2013S => MAT244 Math--Tests => Term Test 2 => Topic started by: Jason Hamilton on March 27, 2013, 10:02:33 PM

Title: TT2 Question 2
Post by: Jason Hamilton on March 27, 2013, 10:02:33 PM

Consider the second order equation
\begin{equation*}
x''=x^4-5x^2+4
\end{equation*}

(a) Reduce to the first order system in variables $(x, y, t)$  with $y = x'$, i.e.
\begin{equation*}
\left\{ \begin{array}{ll}
x'=\ldots\\
y'=\ldots\\
\end{array}\right.
\end{equation*}

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

(c) Find critical points and linearize system in these points.

(d)  Classify the linearizations at the critical points (i.e. specify  whether they are nodes, saddles, etc., indicate stability and, if applicable,  orientation) and sketch their phase portraits.

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

(f) Sketch the solutions on $(x,y)$ plane.

Title: Re: TT2 Question 2
Post by: Jeong Yeon Yook on March 27, 2013, 10:03:43 PM
q2 part e) and bonus
Title: Re: TT2 Question 2
Post by: Jeong Yeon Yook on March 27, 2013, 10:04:35 PM
q2 part a) b) and c)
Title: Re: TT2 Question 2
Post by: Jeong Yeon Yook on March 27, 2013, 10:11:33 PM
#2 part d)
Title: Re: TT2 Question 2
Post by: Jeong Yeon Yook on March 27, 2013, 10:17:35 PM
part d) continued
(2, 0) is a saddle and unstable.
Title: Re: TT2 Question 2
Post by: Jeong Yeon Yook on March 27, 2013, 10:23:47 PM
#2 d) continued. Last file

The files are in order except for part e) and bonus
Title: Re: TT2 Question 2
Post by: Victor Lam on March 28, 2013, 12:59:14 AM
Solutions & sketches, etc
Title: Re: TT2 Question 2
Post by: Victor Ivrii on March 28, 2013, 03:41:14 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
\label{A}
\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 $(-1,0)$

Title: Re: TT2 Question 2
Post by: Sabrina (Man) Luo 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 $(-1,0)$
Title: Re: TT2 Question 2
Post by: Victor Ivrii on March 28, 2013, 09:13:03 AM
Sabrina, 3D picture is useful but you need just slightly increase the span for $x$ from $[-2,2]$ (which is insufficient as $x=\pm 2$ are critical points) to $[-2.3,2.3]$ f.e. (going too large would be bad as fast growth of $V(x)$ there would dwarf the important domain)

But I meant something more individual: look at the computer picture (download it first) and just trace two separatrices)