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Messages - Jingze Wang

Pages: [1] 2
1
Final Exam / Re: FE-P6
« on: December 17, 2018, 05:50:10 AM »
I think (-4,0) and (4,0) are maximum and (0,0) is minimum

2
Final Exam / Re: FE-P6
« on: December 17, 2018, 05:44:04 AM »
Since system is integrable, so just saddle and center as in linear system :)

3
Final Exam / Re: FE-P6
« 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.

4
Final Exam / Re: FE-P6
« on: December 16, 2018, 01:29:15 PM »
Thanks, all corrected, finally :)

5
Final Exam / Re: FE-P6
« on: December 15, 2018, 07:12:23 PM »
Corrected, what about now?

6
Final Exam / Re: FE-P5
« on: December 14, 2018, 10:42:58 AM »
This is the computer generated global phase portrait.

We already know that Wolfram Alpha provides rather crappy pictures here. V.I.

7
Final Exam / Re: FE-P1
« on: December 14, 2018, 10:18:57 AM »
Sorry I have not finished my typed solution at that time, so you just saw part of my solution, but we get the same answer finally :)

8
Final Exam / Re: FE-P1
« on: December 14, 2018, 09:53:16 AM »
Let $M=2x\sin(y)+1, N=4x^2\cos(y)+3x\cot(y)+5\sin(2y)$
$M_y=2x\cos(y), N_x=8x\cos(y)+3\cot(y)$
Check and find this is not exact
Then try to find integrating factor
$N_x-M_y=8x\cos(y)+3\cot(y)-2x\cos(y)=6x\cos(y)+3\cot(y)$
By observation, $\frac{N_x-M_y}{M}=3\cot(y)$
Therefore, the integrating factor is $\sin^{3}(y)$
$M'=2x\sin^4(y)+\sin^3(y)$
$\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+h(y)$
$\psi_y=4x^2\sin^3(y)\cos(y) + x \sin^2(y)\cos(y)+h'(y)$
$h'(y)=10\sin^4(y) \cos(y)$
$h(y)=2\sin^5 (y)$
$\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+2\sin^5 (y)$

9
Final Exam / Re: FE-P6
« 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.

10
Quiz-7 / Re: Q7 TUT 0401
« on: November 30, 2018, 08:33:46 PM »
Here is computer generated picture

11
Quiz-7 / Re: Q7 TUT 0201
« on: November 30, 2018, 08:31:23 PM »
I think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted

12
Quiz-7 / Re: Q7 TUT 0301
« on: November 30, 2018, 05:27:47 PM »
c) For the first critical point(s), the eigenvalues are $\pm i$. This means the phaseportrait is centre (CW).
Hi Michael, I think you should mention that phase portrait is counterclockwise for the first matrix since -1<0 rather than imaginary eigenvalues

13
Quiz-7 / Re: Q7 TUT 5102
« on: November 30, 2018, 05:17:58 PM »
This is computer generated picture

14
Quiz-7 / Re: Q7 TUT 0101
« on: November 30, 2018, 04:35:45 PM »
This is computer generated picture :)

15
Quiz-7 / Re: Q7 TUT 0501
« on: November 30, 2018, 04:29:02 PM »
To find critical points, let $x'=0, y'=0$
Thus, $-2x-y-x(x^2+y^2)=0, x-y+y(x^2+y^2)=0$
$ (0,0)$ is one crtical point
Also, $(0.330757,1.09242), (0.330757,-1.09242)$ are also critical points

Let $f(x)=-2x-y-x(x^2+y^2), g(x)=x-y+y(x^2+y^2)$
$f_x=-2-3x^2-y^2, f_y=-1-2xy$
Also, $ g_x=1+2xy, g_y=-1+x^2+3y^2$

$J=\begin{bmatrix}
-2-3x^2-y^2&-1-2xy\\
1+2xy&-1+x^2+3y^2\\
\end{bmatrix}$

Plug in the critical points to find eigenvalues of each linear system
For $(0,0)$,

We get $\begin{bmatrix}
-2&--1\\
1&-1\\
\end{bmatrix}$,
$r=\frac{-3\pm\sqrt{-3}i}{2}$
Therefore, it is stable at $(0,0)$

Similarly, for others critical points,
Plug in and eigenvalues are $-3.5092, 2.6771$
Thus, it is a saddle point therefore unstable




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