Author Topic: FE-P6  (Read 577 times)

Victor Ivrii

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FE-P6
« 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

Jingze Wang

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Re: FE-P6
« Reply #1 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}
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)
$2x(x^2+y^2-16)dx+2y(x^2+y^2+4)dy=0$
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.
« Last Edit: December 16, 2018, 08:09:20 PM by Jingze Wang »

Jerry Qinghui Yu

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Re: FE-P6
« Reply #2 on: December 14, 2018, 10:49:09 AM »
clear picture for part d

Victor Ivrii

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Re: FE-P6
« Reply #3 on: December 15, 2018, 06:46:33 PM »
I wonder, nobody sees multiple errors?

Jingze Wang

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Re: FE-P6
« Reply #4 on: December 15, 2018, 07:12:23 PM »
Corrected, what about now?
« Last Edit: December 15, 2018, 07:18:33 PM by Jingze Wang »

Victor Ivrii

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Re: FE-P6
« Reply #5 on: December 16, 2018, 01:44:02 AM »

Tzu-Ching Yen

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Re: FE-P6
« Reply #6 on: December 16, 2018, 10:50:13 AM »
$G_y = -4xy$

Jingze Wang

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Re: FE-P6
« Reply #7 on: December 16, 2018, 01:29:15 PM »
Thanks, all corrected, finally :)
« Last Edit: December 16, 2018, 02:53:50 PM by Jingze Wang »

Tzu-Ching Yen

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Re: FE-P6
« Reply #8 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.

Victor Ivrii

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Re: FE-P6
« Reply #9 on: December 16, 2018, 06:50:30 PM »
Can anyone write a correct final version of $H(x,y)$?

Joyce Ye

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Re: FE-P6
« Reply #10 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
« Last Edit: December 16, 2018, 07:42:26 PM by Joyce Ye »

Tzu-Ching Yen

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Re: FE-P6
« Reply #11 on: December 16, 2018, 07:40:28 PM »
Therefore, 𝐻(𝑥,𝑦)=1/2x4+x2y2-16x21/2y4+4y2=C

$+$ missing

Jingze Wang

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Re: FE-P6
« Reply #12 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.

Tzu-Ching Yen

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Re: FE-P6
« Reply #13 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.

Victor Ivrii

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Re: FE-P6
« Reply #14 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 $(0,0)$?
« Last Edit: December 17, 2018, 04:55:53 AM by Victor Ivrii »