Procedure to draw a phase portrait by hand ?

sid9221

So I want to be able to draw the phase portrait for linear systems such as:

x'=x-2y
y'=3x-4y

I am completely confused, but this is what I have come up with so far:

Step 1: Write out the system in the form of a matrix.
Step 2: Find the eigenvalues and eigenvectors for the matrix.
Step 3: Using the eigenvectors draw the eigenlines.
Step 4: Using the the eigenvalues label the direction of the eigenlines[(+) = away, (-)= towards]
Step 5: Using the eigenvalues determine the type of the system. Eg: node, star, spiral etc.
Step 6: Fill in a few trajectories.

My issues appear at steps 5 and 6, I can't figure out how to draw the trajectories. Which eigenline should they be based around ?
Also with complex eigenvalues or when there is 1 eigenvector what do I have to do ?

Any help or link to a webpage would be greatly appreciated.

wisvuze

since you have a linear system, as with all linear systems, you want to put it in a "cannonical form", and draw your phase portrait as a linear shift from the cannonical form phase portrait ( with the change of basis operator as your shift )
the cannonical forms are exactly the situations where you have 2 different eigenvalues, 1 eigenvector... et c if you haven't learnt about these, I will elaborate

sid9221

No idea what that means...

HallsofIvy

Okay for this problem that is
[tex]\begin{bmatrix}x' \\ y'\end{bmatrix}= \begin{bmatrix}1 & -2 \\ 3 & -4\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}[/tex]

[quote]Step 2: Find the eigenvalues and eigenvectors for the matrix.[quote]
The eigenvalue equation for this matrix is
[tex]\left|\begin{array}{cc}1-\lambda & -2 \\ 3 & -4-\lambda\end{array}\right|= \lambda^2+ 3\lambda+ 2= (\lambda+ 1)(\lambda+ 2)= 0[/tex]
so the eigenvalues are -1 and -2.

If <x y> is an eigenvector corresponding to eigenvalue -1, then
[tex]\begin{bmatrix}1 & -2 \\ 3 & -4\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}x- 2y \\ 3x- 4y\end{bmatrix}= \begin{bmatrix}-x \\ -y\end{bmatrix}[/tex]
which gives the two equations x- 2y= -x and 3x- 4y= -y both of which reduce to y=x. An eigenvector corresponding to eigenvalue -1 is <1, 1>.

If <x y> is an eigenvector corresponding to eigenvalue -2, then
[tex]\begin{bmatrix}1 & -2 \\ 3 & -4\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}x- 2y \\ 3x- 4y\end{bmatrix}= \begin{bmatrix}-2x \\ -2y\end{bmatrix}[/tex]
which gives the two equations x- 2y= -2x and 3x- 4y= -2y both of which reduce to 2y=3x. An eigenvector corresponding to eigenvalue -2 is <2, 3>.

They are, of course, the lines y= x and 2y= 3x. They intersect at the origin which is the "equilibrium point" for this system.

An eigenvalue essentially gives the "flow" on that eigenvalue. Here they are both negative so flow along each is toward the origin. I prefer to mark little arrows on each line showing the direction of flow. If both eigenvalues had been positive, we would mark them with arrows outward, away from the origin. If one eigenvalue is positive and one is negative, mark each one accordingly- the line corresponding to positive eigenvalue outward, the line corresponding to negative eigenvalue inward.

Since all flow is toward the origin, this is a focus. If the flow were all outward, it would be a star. If one inward, the other outward, it would be a node. You will have something like a spiral (generally "circular" motion) only if the imaginary part of the eigenvalues is non-zero.

The crucial point here is the "existance and uniqueness" theorem. Except where the right side of the system is 0 (at the origin, the equilibrium point) there must be only a single trajectory through any point other than the origin. Also solutions are "continuous" in the sense that trajectories through near by points must be similar. Here, that means that trajectories must all move toward the origin. Flow close to the line with eigenvalue -2 must be twice as fast as that near the line with eigenvalue -1. If you wanted to be very accurate, you could draw more lines close to the line with eigenvalue -2 to indicate that but probably your teacher will not require it. Essentially you just have lines through the origin with arrows pointing toward the origin. If both eigenvalues were positive, you would have the same thing but with arrows pointing away from the origin.

The "interesting" case is when one eigenvalue is positive and the other negative. In that case, trajectories close to the line with negative eigenvalue must be toward the origin and trajectories close to the line with positive eigenvalue must be away form the origin. Also, they cannot pass through the origin yet must be continuous lines- the only way that can happen is if the curve close to the origin so that a line toward the origin turns into a line away from the origin. Again, trajectories cannot cross except at the origin.

rsaad

@HallsofIvy: Thank you so much! That was such an elaborate reply! helped me a lot!