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Consider the system of differential equations:

$$x'=-2x+y-x^3$$ $$y'=-y+x^2$$

a. Determine the fixed points. (Am I correct in thinking that to determine the fixed points, I must set x' and y'=0? I'm not sure if I am doing this right...)

$y'=-y+x^2$

$0=-y+x^2$

$y=x^2$

$x'=-2x+y-x^3$

$0=-2x+y-x^3$

$0=-2x+x^2-x^3$

$0=-x(2-x+x^2)$

so I get $x=0$ and $x=\frac{1±\sqrt{7}i}{2}$

b. Determine the nullclines and the signs of $x'$ and $y'$ in the various regions of the plane.

So I know I have to determine the fixed points to do this, which I did above (not sure if I did that correctly though) and I also need to determine their stability, right?

For stability I have to compute eigenvalues and if the real parts of both are negative, it's stable, if the real parts of one is positive it's unstable, and if one of the real parts is 0, it is L-stable/marginally stable...is this right?

I guess I am not really sure how to compute the eigenvalues in this case, so far I am only really comfortable with computing the eigenvalues when the question is already in matrix form, as opposed to these equations...so I need some guidance...

c. Sketch a rough phase portrait.

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  • $\begingroup$ In (b): "So I know I have to determine the fixed points to do this" Nope. And you should check your resolution of (a)... $\endgroup$ – Did Nov 11 '14 at 23:59
  • $\begingroup$ @Did yea I am not sure about a...so can you give me some direction?? $\endgroup$ – Math Major Nov 12 '14 at 0:01
  • $\begingroup$ How do you solve $x(2−x+x^2)=0$? $\endgroup$ – Did Nov 12 '14 at 0:02
  • $\begingroup$ @Did so I got x=0 (obvious) and then using a=1,b=-1, and c=2 I got $x=1+\frac{\sqrt{-7}}{2}$ and $x=1-\frac{\sqrt{-7}}{2}$ so $x=1+\frac{\sqrt{7}i}{2}$ and $x=1-\frac{\sqrt{7}i}{2}$, right? what am I doing wrong? $\endgroup$ – Math Major Nov 12 '14 at 0:06
  • $\begingroup$ Your two latter "solutions" are not solutions. Recall that one looks for (x,y) in the real plane. $\endgroup$ – Did Nov 12 '14 at 0:12
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$\qquad\qquad\qquad\qquad\qquad$ enter image description here

To prove the convergence to the unique fixed point $(0,0)$, apparent on the phase diagram above, note that $$4(x^2+y^2)'=-2(2x^2-y)^2-(4x-y)^2-5y^2\lt0,$$ for every $(x,y)\ne(0,0)$.

An interesting question about this dynamical system would be to determine an explicit equation for the curve $$x=u(y),$$ also apparent on the phase diagram above, which can be defined by the fact that the reversed dynamics $$\bar x'=2\bar x-\bar y+\bar x^3,\qquad\bar y'=\bar y-\bar x^2,$$ is such that $\bar x(t)\to-\infty$ when $t\to+\infty$ for every $(\bar x(0),\bar y(0))$ such that $\bar x(0)\lt u(\bar y(0))$, and such that $\bar x(t)\to+\infty$ when $t\to+\infty$ for every $(\bar x(0),\bar y(0))$ such that $\bar x(0)\gt u(\bar y(0))$. The function $u$ solves the differential equation $$(z-u^2(z))\cdot u'(z)=u^3(z)+2u(z)-z,$$ with initial condition $u(0)=0$.

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  • $\begingroup$ $\varphi(y)$ is the manifold associated with the eigenspace $(1,1)^T$ and solves the IVP $(y-\varphi^2)\frac{d\varphi}{dy} =\varphi^3 + 2\varphi - y$ with the initial conditions $\varphi(0) = 0$. People are often only interested in local behavior, so one usually produces a truncated Taylor series expansion for $\varphi$. If a closed form expression of $\varphi$ exists, i doubt it's trivial to find. $\endgroup$ – MrSlunk Nov 12 '14 at 0:55

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