I do not quite remember how to plot a vector function (or maybe I do). Consider the ODE: \begin{equation} x' = \begin{pmatrix}1&1\\-1&1\end{pmatrix}x \end{equation}

I have found the general solution: \begin{equation} x(t) = c_1e^t\begin{pmatrix}-\sin(t)\\ \cos(t)\end{pmatrix} +c_2e^t\begin{pmatrix}\cos(t)\\\sin(t)\end{pmatrix} \end{equation}

I understand that the phase portrait spirals away from the origin as $t\to\infty$. What I do not understand is which direction the spiral starts in. It would make sense to plug in some values for $x$ and see what values of $x'$ we get:

\begin{equation} x=\begin{pmatrix}0\\1\end{pmatrix} \implies x' = \begin{pmatrix}1\\1\end{pmatrix} \end{equation}

Does this mean that at the point $(0,1)$ the slope of the spiral is in the direction $x' = \begin{pmatrix}1\\1\end{pmatrix}$?

Thanks for helping me clarify!


Since your eigenvalues are positive, they spiral outward.

You can figure out the critical point by setting up where the two equations are simultaneously equal to zero, that is:

  • $x+y = 0$
  • $-x+y=0$

This leads to a CP of $x = 0, y = 0$

Here is the phase portrait showing this behavior (notice the direction arrows (green) and the solutions (blue)) and the CP $(x, y) = (0,0)$:

enter image description here

Note: since we have a solution as a function of $t$, you can do a parametric plot of $x(t)$ versus $y(t)$ for a single set of initial conditions, for example, choose $c_1 = c_2 = 1$ and we use WA.

If you repeat this for a bunch of different $c_1, c_2$, you will also see what we have from the phase portrait.

Lastly, you might find these instructive:

  • $\begingroup$ w/o question, deserves a TU $= \large \uparrow$ $\endgroup$ – Namaste Jul 5 '13 at 15:30
  • $\begingroup$ This is nice, I'll bold my question so you know exactly what I am asking. $\endgroup$ – CodeKingPlusPlus Jul 6 '13 at 14:34
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    $\begingroup$ @CodeKingPlusPlus: Apparently, the phase portrait is not clarifying this for you. You need initial conditions in order to make that conclusion. How are you arriving at that determination when you don't know $c_1$ and $c_2$? We evaluate $x(t), x'(t)$ at $t=0$ and then we need $c_1$ and $c_2$, for instance. Repeat this for $t=1$ for $y(t), y'(t)$. What do you get when you do that for the slope? See it? $\endgroup$ – Amzoti Jul 6 '13 at 17:26
  • $\begingroup$ I don't know what you mean by "Repeat this for $t=1$ for $y(t), y'(t)$. I didn't know that the phase portrait was the plot of the family of solutions to the differential equation... hahah Now, I understand what you mean now by needing initial conditions $\endgroup$ – CodeKingPlusPlus Jul 6 '13 at 23:47
  • $\begingroup$ You have an equation for $x(t)$ and $y(t)$, maybe you are calling those $x_1(t)$ and $x_2(t)$. When you choose a $t$, you find the value of the function and then the derivative of each function. The derivative of each function is plotted as the direction field and then we plot the functions on top of those for varying $c_1$ and $c_2$. Regards $\endgroup$ – Amzoti Jul 6 '13 at 23:52

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