Toronto Math Forum
MAT2442018F => MAT244Tests => Final Exam => Topic started by: Victor Ivrii on December 14, 2018, 08:06:54 AM

Typed solutions only. Upload only one picture (a general phase portrait; for general one can use computer generated)
For the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x' = 2y(x^2+y^2+4)\, , \\
&y' = 2x (x^2+y^216)
\end{aligned}\right.
\end{equation*}
(a) Find stationary points.
(b) Linearize the system at stationary points and sketch the phase portrait of this linear system.
(c) Find the equation of the form $H(x,y) = C$, satisfied by the trajectories of the nonlinear system.
(d) Sketch the full phase portrait.
Hint: avoid redundancy: asymptotically (un)stable node, unstable node, stable center

part (a)
Find critical points
Let $x'=0, y'=0$
Then $2y(x^2+y^2+4)=0, 2x(x^2+y^216)=0$
When $y=0, x=0, x=4, x=4$
So (0,0) (4,0) (4,0) are critical points.
Part (b)
$F(x,y)=2y(x^2+y^2+4), G(x,y)=2x(x^2+y^216)$
$F_x=4xy, F_y=2x^2+6y^2+8$
$G_x=6x^22y^2+32, G_y=4xy$
Then plug in to find J matrix
\begin{equation} J={ \left[\begin{array}{ccc} 4xy & 2x^2+6y^2+8 \\ 6x^22y^2+32 & 4xy \end{array} \right ]}, \end{equation}
$When (0,0)$
\begin{equation} J={ \left[\begin{array}{ccc} 0 & 8 \\ 32 & 0 \end{array} \right ]}, \end{equation}
This is a saddle
$When (4,0)$
\begin{equation} J={ \left[\begin{array}{ccc} 0 & 40 \\ 64 & 0 \end{array} \right ]}, \end{equation}
This one is a center
$When (4,0)$
\begin{equation} J={ \left[\begin{array}{ccc} 0 & 40 \\ 64 & 0 \end{array} \right ]}, \end{equation}
This one is also a center
Also, the phase portraits are attached in picture 2
Part (c)
$2x(x^2+y^216)dx+2y(x^2+y^2+4)dy=0$
Find this equation is exact, then
$H(x,y)=0.5 x^4+x^2y^216x^2+h(y)$
$H_y=2x^2 y +h'(y)$
$h'(y)=2y^3+8y$
$h(y)=0.5 y^4+4y^2$
$H(x,y)=0.5 x^4+x^2y^216x^2+0.5 y^4+4y^2=c$
Part (d)
Since it is integrable system
Then center is still center in nonlinear system.
See picture 1.

clear picture for part d

I wonder, nobody sees multiple errors?

Corrected, what about now?

Corrected, what about now?
Still ... :(

$G_y = 4xy$

Thanks, all corrected, finally :)

The term in $H(x, y)$ should be $\frac{1}{2}x^4$ as well. Perhaps an extra constant. Otherwise I couldn't find anything else.

Can anyone write a correct final version of $H(x,y)$?

dx/dt=2x^{2}y+2y^{3}+8y
dy/dt=2x^{3}2xy^{2}+32x
so,dx/dy=2x^{2}y+2y^{3}+8y/2x^{3}2xy^{2}+32x
so,(2x^{3}+2xy^{2}32x)dx+(2x^{2}y+2y^{3}+8y)dy=0
Let M=2x^{3}+2xy^{2}32x,N=2x^{2}y+2y^{3}+8y
M_{y}=4xy,N_{x}=4xy
M_{y}=N_{x}, so Exact.
There exists a 𝐻(𝑥,𝑦) s.t. 𝐻_{x}(𝑥,𝑦)=M,𝐻_{y}(𝑥,𝑦)=N
𝐻(𝑥,𝑦)=1/2x^{4}+x^{2}y^{2}16x^{2}+h(y)
𝐻_{y}(𝑥,𝑦)=2x^{2}y+ℎ′(𝑦)
so,ℎ′(𝑦)=2y^{3}+8y
h(y)=1/2y^{4}+4y^{2}+C
Therefore, 𝐻(𝑥,𝑦)=1/2x^{4}+x^{2}y^{2}16x^{2}1/2y^{4}+4y^{2}=C

Therefore, 𝐻(𝑥,𝑦)=1/2x^{4}+x^{2}y^{2}16x^{2}1/2y^{4}+4y^{2}=C
$+$ missing

Can anyone tell me why Professor said my answer is wrong? Also, to avoid confusion, I use 0.5 instead of 1/2.

Can anyone tell me why Professor said my answer is wrong? Also, to avoid confusion, I use 0.5 instead of 1/2.
No idea tbh. Joyce provided exactly the same answer as your corrected version.

Now it is correct, but before it was not so. Or, may be, because the correct expression was not articulated in the special line, or like this:
$$\boxed{H(x,y)=\frac{1}{2}\bigl(x^4 +2x^2y^2 +y^432 x^2 +8y^2\bigr).}$$
BTW, making this problem I started from $H(x,y)$ in the form
\begin{align*}
H(x,y)=&\bigl( (xa)^2 +y^2b^2\bigr)\bigl( (x+a)^2 +y^2b^2\bigr)=\\
&\bigl( x^2 +y^2b^2+a^2 2ax\bigr)\bigl( x^2 +y^2b^2+a^2 +2ax\bigr)=\\
&\bigl( x^2 +y^2b^2+a^2\bigr)^2 4 a^2x^2
\end{align*}
with $b> a>0$.
Question: For $H(x,y)$ what are points $(\pm 4,0)$ and $(0,0)$?

Since system is integrable, so just saddle and center as in linear system :)

I think (4,0) and (4,0) are maximum and (0,0) is minimum

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.

critical points

I think (0,0) is a saddle. why (4,0) and (4,0) are not centre?

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.

(a) Since $x^2+y^2+4>0$ all stationary points are at $y=0$, and either $x=0$ or $x^2+y^216=0$; thus we have
$A_1=(0,0)$ and $A_{2,3}=(\pm 4,0)$.
(b) Let $f=2y(x^2+y^2+4)$, $g=2x(x^2+y^216)$.
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
$$
2x(x^2+y^216)\,dx + 2y(x^2+y^2+4)\,dy=0
$$
one can check that the form on the left is exact; $H_x= 2(x^3+xy^216x)\implies H(x,y)=\frac{1}{2}(x^4 +2x^2y^2 32 x^2)+h(y)$ 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
$$H(x,y)=\frac{1}{2}\bigl(x^4 +2x^2y^2 +y^432 x^2 +8y^2\bigr).$$
Points $(\pm 4,0)$ are minima of $H(x,y)$ and $(0,0)$ is the saddle point. So called symmetric twowell 2D potential.
See 3D plot generated by WolframAlpha