## Problem #16

For the given system of equations,

(a) Determine the eigenvalues in terms of .

(b) Find the critical value or values of where the qualitiative nature of the phase portrait for the system changes.

(c) Draw a phase portrait for a value of slightly below, and for another value slightly above, each critical value.

## Part (a)

Using pen and paper, or symbolic manipulation technology, we find
that the eigenvalues of the coefficient matrix in terms of
are

## Part (b)

From the equation of the eigenvalues, we see that if < 0, then the eigenvalues are complex, while if > 0,
then the eigenvalues are real and distinct. Thus, in part (c) below,
we'll look at phase
portraits (generated by dirfield2.m) for
= -0.1 and = 0.1.

## Part (c)

If < 0, then the eigenvalues are complex, so we
can expect that the phase portrait will be a spiral. Here is the phase
portrait for = - 0.1. It is a spiral, but not as tightly
curved as most.

If > 0, then the eigenvalues are real and distinct,
so the origin is a node. In the graph below, the lines through the
eigenvectors are drawn in black so that we can more clearly see that the
origin is a node.

The figure below shows phase portraits for 9 values of
. In this figure, we can see how as moves from -2 through 0 to +2, the phase portraits change from having the
origin clearly a spiral point to an improper node and then to a node. For
the values of near 0, the phase portraits do not display "classic" behavior: The spirals and nodes are more obscure than for larger magnitudes of .

**Note:** The figure above was generated in MATLAB using the module
Ch07Sec06Prob16.m, which was adapted from the
code that generates solution curves for the 'dirfield2d' module.