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Note: 1 lecture, part of §6.2 in [EP], parts of §7.5 and §7.6 in [BD]
Let us take a moment to talk about constant coefficient linear homogeneous systems in the plane. Much intuition can be obtained by studying this simple case. Suppose we have a matrix and the system
The system is autonomous (compare this section to § 1.6) and so we can draw a vector field (see end of § 3.1). We will be able to visually tell what the vector field looks like and how the solutions behave, once we find the eigenvalues and eigenvectors of the matrix . For this section, we assume that has two eigenvalues and two corresponding eigenvectors.
Case 1. Suppose that the eigenvalues of are real and positive. We find two corresponding eigenvectors and plot them in the plane. For example, take the matrix . The eigenvalues are 1 and 2 and corresponding eigenvectors are and . See Figure 3.3.
Now suppose that and are on the line determined by an eigenvector for an eigenvalue . That is, for some scalar . Then
The derivative is a multiple of and hence points along the line determined by . As , the derivative points in the direction of when is positive and in the opposite direction when is negative. Let us draw the lines determined by the eigenvectors, and let us draw arrows on the lines to indicate the directions. See Figure 3.4.
We fill in the rest of the arrows for the vector field and we also draw a few solutions. See Figure 3.5. Notice that the picture looks like a source with arrows coming out from the origin. Hence we call this type of picture a source or sometimes an unstable node.
Case 2. Suppose both eigenvalues were negative. For example, take the negation of the matrix in case 1, . The eigenvalues are and and corresponding eigenvectors are the same, and . The calculation and the picture are almost the same. The only difference is that the eigenvalues are negative and hence all arrows are reversed. We get the picture in Figure 3.6. We call this kind of picture a sink or sometimes a stable node.
Case 3. Suppose one eigenvalue is positive and one is negative. For example the matrix . The eigenvalues are 1 and and corresponding eigenvectors are and . We reverse the arrows on one line (corresponding to the negative eigenvalue) and we obtain the picture in Figure 3.7. We call this picture a saddle point.
For the next three cases we will assume the eigenvalues are complex. In this case the eigenvectors are also complex and we cannot just plot them in the plane.
Case 4. Suppose the eigenvalues are purely imaginary. That is, suppose the eigenvalues are . For example, let . The eigenvalues turn out to be and eigenvectors are and . Consider the eigenvalue and its eigenvector . The real and imaginary parts of are
Case 5. Now suppose the complex eigenvalues have a positive real part. That is, suppose the eigenvalues are for some . For example, let . The eigenvalues turn out to be and eigenvectors are and . We take and its eigenvector and find the real and imaginary of are
Case 6. Finally suppose the complex eigenvalues have a negative real part. That is, suppose the eigenvalues are for some . For example, let . The eigenvalues turn out to be and eigenvectors are and . We take and its eigenvector and find the real and imaginary of are
We summarize the behavior of linear homogeneous two dimensional systems in Table 3.1.
|real and both positive||source / unstable node|
|real and both negative||sink / stable node|
|real and opposite signs||saddle|
|purely imaginary||center point / ellipses|
|complex with positive real part||spiral source|
|complex with negative real part||spiral sink|
Exercise 3.5.1: Take the equation , with , , for the mass-spring system. a) Convert this to a system of first order equations. b) Classify for what do you get which behavior. c) Can you explain from physical intuition why you do not get all the different kinds of behavior here?