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MAT244--2018F => MAT244--Tests => Final Exam => Topic started by: Victor Ivrii on December 14, 2018, 08:06:54 AM

Title: FE-P6
Post 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^2-16)
\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
Title: Re: FE-P6
Post by: Jingze Wang on December 14, 2018, 09:02:51 AM
part (a)
Find critical points
Let $x'=0, y'=0$
Then $2y(x^2+y^2+4)=0, -2x(x^2+y^2-16)=0$
When $y=0, x=0, x=4, x=-4$
So (0,0) (4,0) (-4,0) are critical points.

Part (b)
$F(x,y)=2y(x^2+y^2+4), G(x,y)=-2x(x^2+y^2-16)$
$F_x=4xy, F_y=2x^2+6y^2+8$
$G_x=-6x^2-2y^2+32, G_y=-4xy$
Then plug in to find J matrix
\begin{equation}    J={ \left[\begin{array}{ccc} 4xy & 2x^2+6y^2+8 \\ -6x^2-2y^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}
$When (4,0)$
\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)
$2x(x^2+y^2-16)dx+2y(x^2+y^2+4)dy=0$
Find this equation is exact, then
$H(x,y)=0.5 x^4+x^2y^2-16x^2+h(y)$
$H_y=2x^2 y +h'(y)$
$h'(y)=2y^3+8y$
$h(y)=0.5 y^4+4y^2$
$H(x,y)=0.5 x^4+x^2y^2-16x^2+0.5 y^4+4y^2=c$

Part (d)
Since it is integrable system
Then center is still center in nonlinear system.
See picture 1.
Title: Re: FE-P6
Post by: Jerry Qinghui Yu on December 14, 2018, 10:49:09 AM
clear picture for part d
Title: Re: FE-P6
Post by: Victor Ivrii on December 15, 2018, 06:46:33 PM
I wonder, nobody sees multiple errors?
Title: Re: FE-P6
Post by: Jingze Wang on December 15, 2018, 07:12:23 PM
Title: Re: FE-P6
Post by: Victor Ivrii on December 16, 2018, 01:44:02 AM
Still ... :(
Title: Re: FE-P6
Post by: Tzu-Ching Yen on December 16, 2018, 10:50:13 AM
$G_y = -4xy$
Title: Re: FE-P6
Post by: Jingze Wang on December 16, 2018, 01:29:15 PM
Thanks, all corrected, finally :)
Title: Re: FE-P6
Post by: Tzu-Ching Yen on December 16, 2018, 02:22:20 PM
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.
Title: Re: FE-P6
Post by: Victor Ivrii on December 16, 2018, 06:50:30 PM
Can anyone write a correct final version of $H(x,y)$?
Title: Re: FE-P6
Post by: Chengyin Ye on December 16, 2018, 07:37:50 PM
dx/dt=2x2y+2y3+8y
dy/dt=-2x3-2xy2+32x
so,dx/dy=2x2y+2y3+8y/-2x3-2xy2+32x
so,(2x3+2xy2-32x)dx+(2x2y+2y3+8y)dy=0
Let M=2x3+2xy2-32x,N=2x2y+2y3+8y
My=4xy,Nx=4xy
My=Nx, so Exact.
There exists a 𝐻(𝑥,𝑦) s.t. 𝐻x(𝑥,𝑦)=M,𝐻y(𝑥,𝑦)=N
𝐻(𝑥,𝑦)=1/2x4+x2y2-16x2+h(y)
𝐻y(𝑥,𝑦)=2x2y+ℎ′(𝑦)
so,ℎ′(𝑦)=2y3+8y
h(y)=1/2y4+4y2+C
Therefore, 𝐻(𝑥,𝑦)=1/2x4+x2y2-16x21/2y4+4y2=C
Title: Re: FE-P6
Post by: Tzu-Ching Yen on December 16, 2018, 07:40:28 PM
Therefore, 𝐻(𝑥,𝑦)=1/2x4+x2y2-16x21/2y4+4y2=C

$+$ missing
Title: Re: FE-P6
Post by: Jingze Wang on December 16, 2018, 08:11:04 PM
Can anyone tell me why Professor said my answer is wrong? Also, to avoid confusion, I use 0.5 instead of 1/2.
Title: Re: FE-P6
Post by: Tzu-Ching Yen on December 16, 2018, 08:15:54 PM
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.
Title: Re: FE-P6
Post by: Victor Ivrii on December 16, 2018, 10:30:25 PM
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:
$$\boxed{H(x,y)=\frac{1}{2}\bigl(x^4 +2x^2y^2 +y^4-32 x^2 +8y^2\bigr).}$$
BTW, making this problem I started from $H(x,y)$ in the form
\begin{align*}
H(x,y)=&\bigl( (x-a)^2 +y^2-b^2\bigr)\bigl( (x+a)^2 +y^2-b^2\bigr)=\\
&\bigl( x^2 +y^2-b^2+a^2 -2ax\bigr)\bigl( x^2 +y^2-b^2+a^2 +2ax\bigr)=\\
&\bigl( x^2 +y^2-b^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 $(0,0)$?
Title: Re: FE-P6
Post by: Jingze Wang on December 17, 2018, 05:44:04 AM
Since system is integrable, so just saddle and center as in linear system :)
Title: Re: FE-P6
Post by: Jingze Wang on December 17, 2018, 05:50:10 AM
I think (-4,0) and (4,0) are maximum and (0,0) is minimum
Title: Re: FE-P6
Post by: Victor Ivrii on December 17, 2018, 07:11:00 AM
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.
Title: Re: FE-P6
Post by: Qinger Zhang on December 17, 2018, 09:08:12 PM
critical points
Title: Re: FE-P6
Post by: Qinger Zhang on December 17, 2018, 09:09:38 PM
I think (0,0) is a saddle. why (4,0) and (-4,0) are not centre?
Title: Re: FE-P6
Post by: Qinger Zhang on December 17, 2018, 09:10:17 PM
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.
Title: FE-P6 Official
Post by: Victor Ivrii on December 17, 2018, 09:18:11 PM
(a)  Since $x^2+y^2+4>0$  all stationary points are at $y=0$, and either $x=0$ or $x^2+y^2-16=0$; thus we have
$A_1=(0,0)$ and $A_{2,3}=(\pm 4,0)$.

(b) Let $f=2y(x^2+y^2+4)$, $g=-2x(x^2+y^2-16)$.

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
$$2x(x^2+y^2-16)\,dx + 2y(x^2+y^2+4)\,dy=0$$
one can check that the form on the left is exact; $H_x= 2(x^3+xy^2-16x)\implies H(x,y)=\frac{1}{2}(x^4 +2x^2y^2 -32 x^2)+h(y)$ 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
$$H(x,y)=\frac{1}{2}\bigl(x^4 +2x^2y^2 +y^4-32 x^2 +8y^2\bigr).$$

Points $(\pm 4,0)$ are minima of $H(x,y)$ and $(0,0)$ is the saddle point. So called symmetric two-well 2D potential.
See 3D plot generated by WolframAlpha