FE-P6

[Victor Ivrii]

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]

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.

[Jerry Qinghui Yu]

clear picture for part d

[Victor Ivrii]

I wonder, nobody sees multiple errors?

[Jingze Wang]

Corrected, what about now?

[Victor Ivrii]

Corrected, what about now?

Still ...

[Tzu-Ching Yen]

$G_y = -4xy$

[Jingze Wang]

Thanks, all corrected, finally

[Tzu-Ching Yen]

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]

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

[Chengyin Ye]

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
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)=1/2x⁴+x²y²-16x²+h(y)
𝐻_y(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

[Tzu-Ching Yen]

Therefore, 𝐻(𝑥,𝑦)=1/2x⁴+x²y²-16x²1/2y⁴+4y²=C

$+$ missing

[Jingze Wang]

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]

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]

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 $(0,0)$?