Xjyuk

Consider the system
$$x' = -x^3-xy^2k$$
$$y' = -y^3-x^2ky$$
Where $k$ is a given positive integer.

a.) Find and classify according to stability the equilibrium solutions.

$itHint:$ Let $V(x,y) = x^2 + y^2$

b.) Sketch a phase portrait when $k = 1$

$itHint:$ What are $x'$ and $y'$ when $y=ax$ for some real number $a$?

a.)
Using $V$, we get $fracddtV=2xx'+2yy'$

Plugging in our system , we get:

$$fracddtV=2x(-x^3-xy^2k)+2y(-y^3-x^2ky)$$
$$=-(x^4+y^4)-x^2y^2k-x^2ky^2<0$$
I dropped the $2$ since it doesn't matter to determine stability. We see that our own equilibrium is $(0,0)$ since setting $x'=0$ we get
$$y^2k=-x^2$$
Which only works for $x=y=0$

Therefore our system is asymptotically stable at the origin.

I am having trouble with b.), mostly because the hint is confusing me.

Let $y=ax$, then our system becomes
$$x'=-x^3-a^2x^3=-x^3(1+a^2)$$
$$y'=-a^3x^3-ax^3=-ax^3(1+a^2)$$
I am not sure what to do with this. Using linearization doesn't work since the Jacobian will be the zero vector at the point of interest. I have never had a problem that asks to draw a phase portrait when linearization doesn't work, so I am hoping someone more clever than me can offer some advice.

Although there are many ways to do this, I suspect what the problem is guiding you towards doing is to obtain the flow directly by evaluating over every line that intersects the origin in phase space.

So for a sketch, you would draw the line $y = 0.1 x$, and use the expression you found above for $a = 0.1$ to determine the magnitude and direction of the flow on that line. Then try it for a couple of other different lines, and use common sense to fill in the rest.

Phase portraits - a partial offering

Below are phase portraits for $k=1,2,5$. The red lines indicate the null clines where $doty=0$ and $doty=0$.

$k = 1$

The linear system is

$$beginalign
beginsplit
dotx &= -x^3 - xy^2 = -x left( x^2 + y^2 right) \
doty &= -y^3 - x^2y = -y left( x^2 + y^2 right)
endsplit
endalign$$

$$ dotr = fracx dotx + y dotyr = -r^3 $$

The lone critical point is the origin.

When $y = a x$, $ainmathbbR$, we have
$$beginalign
beginsplit
dotx &= -x^3left( 1 + a^2 right) \
doty &= -a y^3left( 1 + a^2 right)
endsplit
endalign$$

$k = 2$

$$beginalign
beginsplit
dotx &= -x^3 - xy^4 = -x left( x^2 + y^4 right) \
doty &= -y^3 - x^4y = -y left( x^2 + y^2 right)
endsplit
endalign$$

$$ dotr = tfrac18 r^3 left(left(r^2-2right) cos (4 theta )-r^2-6right) $$

The bounding curves for $dotr$ are when $cos 4theta = 1$

$$dotr = -r^3$$

and when $cos 4theta = -1$

$$dotr = -tfrac14 r^3 left(r^2+2right)$$

The bounding curves cross at $r=sqrt2$. At no point is $dotr$ ever positive.