MAT244-2014F > FE



Victor Ivrii:
For the system of ODEs
&x'_t =4x^2y-2x^2-4xy+2y, \\
&y'_t =-4xy^2+2y^2+4xy-2x

(a)  linearize the system at $\ x_0 = 1\, , \, y_0 = 1\ $ and sketch the phase portrait of this linear system,

(b) find the equation of the form $\ H(x,y) = C\ $ satisfied by the trajectories of the nonlinear system,

(c) describe the type of the critical point $\ x_0 = 12 \, , \, y_0 = 10\ $ of the nonlinear system.


(a) Let $f= 4x^2y -2x^2-4xy +2y$, $g= -4xy^2 + 2y^2+4xy -2x$; then $f_x(1,1)=0$, $f_y(1,1)= 2$,  $g_x(1,1)=-2$, $f_y(1,1)= 0$ and the linearized system is
&X'_t =2Y, \\
&Y'_t = -2X
with phase portrait consisting of clock-wise circles.

(b) Rewriting system as $fdx-gdy=0$ we get
 (4xy^2-2y^2 -4xy+2x)\, dx+  (4x^2 y-2x^2 -4xy+2y)\,dy =0
which is exact; then
 H_x= 4xy^2-2y^2 -4xy+2x , \quad H_y=  4x^2 y-2x^2 -4xy+2y)$$
and the first equation implies that
 H= 2x^2 y^2 -2xy^2 -2x^2y +x ^2 + \phi(y)$$
and the second equation implies that $\phi'=2y$ and $y=y^2$ and then
 H=2x^2 y^2 -2xy^2 -2x^2y +x ^2 + y^2 = x^2(y-1)^2 + y^2(x-1)^2.
(c) Since linearized system has a center and original system has a solution $H(x,y)=C$ the type of the stationary point is a center.

In fact the system has also critical point  $(0,0)$ of the type center, and critical point  $(\frac{1}{2},\frac{1}{2})$ of the type saddle


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