# Direction Field and Trajectories

I am wondering how to draw a direction field and trajectories of a system of linear equations:

$$x'= \left[ \begin{array}{ c c } 4 & -2 \\ 8 & -4 \end{array} \right] x .$$

I remember how to do them for a equation from the first part of ODE where we would have $$y'=2x-y$$ but I do not really understand how to begin with the system in this form. Would it just be:

$$x'=4x_1 -2x_2$$ $$x'=8x_1-4x_2$$ and we would just plot different values of $x_1$ and $x_2$?

In $1-D$, it is called a direction field plot.

In higher dimensions, it is called a phase portrait and these two sets of notes 1 and notes 2 provide a procedure for sketing it by hand.

We look at a handful of items (see the two sets of notes):

• Write the equation as $\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dx}}{\dfrac{dx}{dt}} = 2$

• Sketch the isoclines of horizontal slope (where $y' = \dfrac{dy}{dt} = 0 \implies y = 2x$) - these are typically called nullclines

• Sketch the isoclines of vertical slope (where $x' = \dfrac{dx}{dt} = 0 \implies y = 2x$) - these are typically called nullclines

• Put together the horizontal and vertical arrows and sketch the resultant direction field (note direction and magnitude).

• Trace some sample trajectories.

• You can also find the critical points and evaluate the eigenvalues of the Jacobian at each critical point. For this system, we have eigenvalues $\lambda_{1,2} = 0, 0$, so these are just lines.

A phase portrait shows:

$~~~~~~~~~$

For this system, we can also explicitly solve it and arrive at:

$$x(t) = c_1(4t+1) - 2c_2 t, y(t) = 8c_1 t + c_2(1-4t)$$

You can parametrically plot these solutions and what do you notice? For example, let $x(0) = y(0) = 1$, yielding $x(t) = 2t+1, y(t) = 4t+1$. A parametric plot shows (compare to the phase portrait and add a bunch more of these for different initial conditions):

• Do you mind posting your code for the mathematica plot? Nov 11, 2014 at 13:13
• I just woke up and saw this give me a moment Nov 11, 2014 at 13:17
• So I understand how you got $y=2x$ twice. Then I graphed that line. I do not understand the next step, where you say put together the horizontal and vertical lines and sketch.. Nov 11, 2014 at 13:27
• Yes because I am working with repeated eigenvalues right now :\ Nov 11, 2014 at 13:29
• I think I may understand now. So I took the value $(-1,0)$ and plugged it into my $\frac{dx}{dt}$ and got a negative value so it is moving to the left. If I do the same for my $\frac{dy}{dt}$, I would get a negative value, so it is moving down, so if I combine it, its moving diagonally to the bottom left. If I did the same with a point on the right side of $y=2x$ I would get diagonally moving towards the top right, correct? Nov 11, 2014 at 13:33