Toronto Math Forum
MAT2442013S => MAT244 MathTests => Final Exam => Topic started by: Victor Ivrii on April 17, 2013, 03:07:40 PM

Consider equation
\begin{equation}
x'' = 2x^3 + 8x.
\end{equation}
(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.

Solutions and phase portrait

Solutions

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
\begin{equation}
\left\{\begin{aligned}
&x'=y,\\
&y'=2x^3+8x.
\end{aligned}\right.
\label{eq1}
\end{equation}
(b) From $\frac{dx}{y}=\frac{dy}{2x^3+8x}$ we conclude $(2x^3+8x)dx=ydy$ and integrating and collecting everything we get
\begin{equation}
H(x,y):=\frac{1}{2}y^2+\frac{1}{2}x^44x^2=C;
\label{eq2}
\end{equation}
(those who got the whole thing with some factor made no error)
(c) (i) Looking for equilibrium points (setting r.h.e. of (\ref{eq1}) 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{eq2}) are empty are closed and cannot escape to infinity as $\frac{1}{2}x^44x^2$ 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/MAT2442011Sforum/index.php?topic=181.0 (http://weyl.math.toronto.edu/MAT2442011Sforum/index.php?topic=181.0)
(http://weyl.math.toronto.edu/MAT2442011Sforum/index.php?action=dlattach;topic=181.0;attach=231;image)