So let's look at the phase
portrait for a stable node. The phase portrait for an
unstable node will be the same, just the directions of
the arrows are reversed. We're looking at the
differential equations X1 dot equals minus 3X1, plus root two X2 and X2 dot equals root two
X1 minus two X2. So the matrix is
minus three root two, root two minus two and we can do an eigen value and eigen
vector analysis of the matrix. We get the first eigen
value is minus four. The first eigen vector is
one minus root two over two. The second eigen value is minus one and the second
eigen vector is one root two. The eigen values
are both negative. So what does the
solution look like? So we can write the solution
in vector form as X1, X2 equals a constant times our vector for the
first eigen value, which would be one minus
root two over two times e to the -4t plus C2 times
the second eigen vector, one root two times
e to the -t. Okay. We want to draw a phase portrait of this general solution right. So the phase portrait includes different values of C1 and C2. So let's draw the diagram. So our x-axis is X1, our y-axis is X2. The origin then is a fixed point. So how do we draw this? Well, the eigen vectors
are the key here. We can consider
initial conditions such that C2 equals zero. So then this whole term
then doesn't contribute and then our solution
follows this first term. Because C1 is a scalar, e to the -4t is a scalar; then the solution will
follow this eigen vector. So what that means is
that there will be a fixed relationship
between X2 and X1. Here X2 is always
going to be equal to minus root two
over two times X1. Okay. Because it's following
the first eigen vector. This is a line,
equation for a line. So on our phase portrait, we can draw this line. This is a line of negative slope. Slope is slightly
larger than minus one. So it's less shallow. So it looks something like this. Okay. So that's the
solution where C2 equals zero and it's decaying
e to the -4t. So that means everything here is headed towards the fixed point. Okay. Then we can draw
the other eigen vector. So if C1 equals zero, so the initial conditions
such that C1 equals zero, the solution follows the
second eigen vector and here X2 is always going to be equal to the square root
of two times X1. That's a line of positive slope, root two, slightly
larger than one. So it looks something like this. Okay. Here the solution
is decaying like e to the -t. So also going
into the fixed point. Okay. So all the solutions then are converging
to the fixed point. These eigen vectors are important because
the solution along this eigen vector converging to the origin much faster
because of e to the -4t, then the solution along
this eigen vector, which is only converging
to the origin as e to the -t. That will give
this phase portrait a rather funky looking picture that you'll be able to see which solution is
decaying faster. I can try and sketch it by hand, but it's much prettier if we let the computer
do that for us. So let's look at
a MATLAB solution for this phase portrait. As you can see, I drew the
lines corresponding to the eigen vectors
and you can see that the solution along the
first eigen vector, this line here, is
proceeding much faster than the solution along
the second eigen vector. Because the first eigen
vector is decaying like e to the -4t and the second eigen vector
is decaying like e to the -t. Okay. So let me review. So in this video, I showed you how to sketch a phase portrait
associated with a node. If it's a stable node, the eigen values are
both negative or the solutions converge
to the origin. If it's an unstable node, the two eigen values will
be both positive and all the solutions will
diverged from the origin, move away from the origin. The key to drawing the
phase-space portrait is to consider separately
initial conditions, where C2 equals zero so that
the solution follows V1, or to consider initial
conditions such that C1 equals zero and the
solution follows V2. I'm Jeff Chasnoff. Thanks for watching, and I'll see you
in the next video.