MAT244--2018F > Final Exam


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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
&x'  = 2y(x^2+y^2+4)\, , \\
&y'  = -2x (x^2+y^2-16)

(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)
Find this equation is exact, then
$H(x,y)=0.5 x^4+x^2y^2-16x^2+h(y)$
$H_y=2x^2 y +h'(y)$
$h(y)=0.5 y^4+4y^2$
$H(x,y)=0.5 x^4+x^2y^2-16x^2+0.5 y^4+4y^2=c$

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?


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