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I'm working on trying to find the phase portrait for a differential equations such that $$ A = \begin{matrix} a_1 & a_2 \\ 0 & a_3 \end{matrix}$$ so that $$x_1'(t) = a_1x_1+a_2x_2 \text { and } x_2'(t) = a_3x_2$$

$$\text{note, that } x_1(0)=x_{10} \text{ and } x_2(0)=x_{20}$$

I was able to solve these solutions fairly easily considering many cases. However when drawing the phase portrait we were told that all we'd need is $$x_1'(t) = a_1x_1+a_2x_2 \text { and } x_2'(t) = a_3x_2$$.

$$\text{it's given that } a_1 \ge a_3 > 0 \text{and that cases } a_2 < 0 \text{ and } a_2 > 0 \text{ will have to be considered.}$$

I'm able to see that when dividing $$ \frac{dx_2}{dx_1} = \frac{a_3x_2}{a_1x_1+a_2x_2}$$ That there is a horizontal asymptote at $$x_2=0$$ and a vertical(ish) asymptote along $$x_2 = \frac{-a_1}{a_2}x_1$$ however I'm not sure how to go about filling in the blanks.

Can this be analysed as before when using critical points and subbing into the first and second derivatives? How does one go about doing this when on a x1/x2 plane? OR am I missing something obvious here?

Any help is much appreciated.

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2 Answers

up vote 1 down vote accepted

One concept that is helpful to draw phase portraits is that of Eigenspaces. Locally, the flow in an eigenspace is invariant, meaning that if a solution curve starts in a point within the eigenspace, it will always stay there.

For your specific case, you can find in principle 2 independent eigenvectors for the matrix A. Such eigenvectors have naturally an eigenvalue associated. So, in the direction of each eigenvector, the flow will be according to the corresponding eigenvalue.

In your case, you have the eigenvalues $\sigma=\lbrace a_1, a_3\rbrace$, and assuming $a_1\neq a_3$, the eigenvectors

$$ v_1=\begin{bmatrix} 1\\0\end{bmatrix}, \; v_2=\begin{bmatrix} \frac{a_2}{a_3-a_1}\\1\end{bmatrix}. $$

So, you only need actually to draw 2 lines to comprehend the entire phase-portrait. One in the $x-$axis direction. The flow in this axis goes towards $\pm\infty$ since $a_1>0$. And the other line according to $v_2$ above. The flow in that line is also "unstable", meaning that grows towards $\pm\infty$ since $a_3>0$. All the other integral curves or flow lines are arranged by the eigenvectors $v_1$ and $v_2$.

The case $a_1=a_3$ is a little bit more subtle. However is not a "typical case". Meaning that if you would choose a $2\times 2$ matrix $A$ at random it is likely that $a_1\neq a_3$. Still, in the cases where 2 independent eigenvectors exist, you can just follow the same thoughts.

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Hints:

  • Find the critical points of the system, that is, where do $x'_1(t)$ and $x'_2(t)$ simultaneously equal zero?
  • Solve for the eigenvalues of the system using the general parameters. How do the signs of the parameters affect the eigenvalues? You were already told to hold two of them positive and mess with the third (did you write those in the order you really wanted them).
  • Those two pieces of information tell you everything you need to know about this particular system.
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This deserves a TU! Nice hints! +1 –  amWhy Sep 18 '13 at 0:09
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