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Hello i was wondering if anyone could tell me if the phase diagram i have plotted is correct for the system below

$\frac{dx}{dt}=-2x-y+2$ and $\frac{dy}{dt}=xy$

I calculated the equilibrium points to be $(\bar x,\bar y)=(0,2)$ and $(\bar x,\bar y)=(1,0)$

For the first equilibrium point i calculated that the eigenvalues to be $\lambda=-1\pm 2i$ which implies that this is a stable spiral?

For the second equilibrium point the eigenvalues are $\lambda_1=-2,\lambda_2=-1$ which implies that this equilibrium point is a stable node?

Now for the maple diagram my code was

     DEplot(sys, [x(t), y(t)], t = 10 .. -10, x = -3 .. 3, y = -3 .. 3, [[x(0) = 0, y(0) = 2]], [[x(0) = 0, y(0) = 1]], stepsize = .1, linecolor = blue, thickness = 2, arrows = medium);

where 

           sys := {diff(x(t), t) = -2*x(t)-y(t)+2, diff(y(t), t) = x(t)*y(t)}

This gave the following plot The diagram

if this is incorrect can anyone tell me where i'm going wrong? it doesnt look right to me but i cant think of any other way plotting it using maple.

1 Answers 1

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The Jacobian is given by

$$J[x, y] =\begin{bmatrix} \dfrac{\partial x'}{\partial x} & \dfrac{\partial x'}{\partial y} \\ \dfrac{\partial y'}{\partial x} & \dfrac{\partial y'}{\partial y} \end{bmatrix}=\begin{bmatrix} -2 & - 1 \\ y & x \end{bmatrix}$$

When we evaluate this at the first critical (equilibrium) point, $(x, y) = (2, 0)$, we have

$$J[x, y] = \begin{bmatrix} -2 & - 1 \\ 0 & 2 \end{bmatrix} \implies \lambda_{1,2} = -1 \pm i \implies \mbox{Stable Spiral}.$$

When we evaluate this at the second critical point, $(x, y) = (0,1)$, we have

$$J[x, y] = \begin{bmatrix} -2 & - 1 \\ 0 & 1 \end{bmatrix} \implies \lambda_{1,2} = -2, 1 \implies \mbox{Unstable Saddle}.$$

As noted in comments, a saddle is always an unbstable equilibrium.

In this second matrix, we find the roots of characteristic polynomial using $|A-\lambda I| = 0 \implies \lambda ^2+\lambda -2 = (\lambda -1) (\lambda +2) = 0 \implies \lambda_1 = -2, \lambda_2 = 1$.

The phase portrait is correct. Here is another variant

enter image description here

Update

In the first matrix, the characteristic polynomial is given by:

$$ = \lambda ^2+2 \lambda +2 = (\lambda + (1 - i)) (\lambda + (1 + i)) \implies \lambda_{1,2} = -1 \pm i$$

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    Just a very small correction: saddle itself is always unstable equilibrium :) – 2017-01-22
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    @moo thanks i made an error in my working i know understand how you calculated that characteristic poly – 2017-01-22
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    @moo in your solution it should be (0,2) not (2,0)? for the first equilibrium point – 2017-01-22
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    @Moo, no, no, everything is ok :) I'm just saying that you don't have to write "unstable saddle": saddle itself is inherently unstable equilibrium because the definition requires having eigenvalues with both positive and negative real values. Thus it is always unstable. – 2017-01-22
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    @Evgeny: Got it - thanks! – 2017-01-22
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    @moo is the reason why there is only one blue line on my diagram because the other equilibrium point is a saddle? & it should be 1-2i not 1-i? – 2017-01-22
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    Added update. If you look at the phase portrait, it is giving you a bunch of solution curves for any initial condition to help you see the qualitative behavior of your system. Your diagram only highlights a single I.C. to draw the solid curve. If you were to add a bunch more, you'd see what happens to the phase portrait. For example, replot your solution using an IC in each quadrant and on either side of the equilibrium. – 2017-01-22
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    ok thank you that makes sense, for the first equilibrium point can you explain why it is a saddle? i thought if it was complex with a negative real part it is a spiral – 2017-01-22
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    Sorry, just a typo, correcting. – 2017-01-22