So far, the cases we've looked at seem to have a number of common features. Each has has a solution with (t) and v(t) constant; there are solutions which immediately tend toward the constant solution (or oscillate around it), and solutions which do some number of loops first (for R = 3, we didn't demonstrate this if the inital angle is 0, a solution needs an initial velocity larger than 86.3 before it will do a loop). How do the possible solutions depend on the value of R? For example, the behavior of solutions for R = 0.1 and R = 0.2 are quite similar, but are dramatically different from R = 0 and R = 3.
One way we can get a better handle on exactly how the behavior of the
solutions depends on R is to examine what happens to solutions near the
constant solution. If we look in the v plane, this solution
corresponds to a single point, which is often referred to as a ``fixed
point''. We explicitly found the fixed point
{(t) = 0, v(t) = 1}
solution for R = 0 in §2 by noticing that
whenever
= 0 and = 0, we must have a constant solution
(and viceversa). Using
solve, we can get maple to tell us the
where the fixed point is for arbitrary values of R. We use convert
to insist that maple represent the result as a radical, instead of using the
RootOf notation.
> R:='R': FixPoint:=convert( solve((v^2cos(theta))/v=0,sin(theta)R*v^2=0, theta,v), radical);
So, we see that there is a fixed point for all values of R (since 1 + R^{2} is always positive, the radical always takes real values). For R > 0, (t) at this fixed solution is negative, so this corresponds to a diving solution. As R increases, the angle of the dive becomes steeper and steeper.
What can we say about behaviour of solutions near this fixed solution?
In order to answer this, notice that we can apply Taylor's theorem to the righthand side of our system of differential equations. That is, think of our differential equation as being in the vector form
First, we give a brief refresher about linear 2×2 systems of
equations. The reader is refered to the texts in the references for more
comprehensive treatment of this topic.
Linear differential equations (with constant coefficients) are the simplest ones to solve and understand. Such equations have the form = AX, where A is an n×n matrix and X is an nvector. We'll be content with n = 2.
First, notice that X = 0 is always a fixed point for this system, and it is the only one. Secondly, if X_{1}(t) and X_{2}(t) are solutions to the system, then so is X_{1}(t) + X_{2}(t). This property is called superposition and will be very useful.
Let's look at the simplest case for the 2×2 system: A = , that is,
=  ax  
=  dy 
A large number of linear systems behave very similarly to the cases above, except the straightline solutions may not lie exactly along the coordinate axes.
Suppose there is a solution of = AX which lies along a straight line, that is, along some vector . Because the system is linear, this means the tangent vectors to this solution must be of the form X for some number . Such a number is called an eigenvalue, and the corresponding vector is the corresponding eigenvector. Note that in the case above, we had eigenvalues a and d with eigenvectors [1, 0] and [0, 1].
To find the eigenvalues, we need to solve
For linear systems, the eigenvalues determine the ultimate fate of solutions
to the ODE. From the above, it should be clear that there are always two
eigenvalues for a 2×2 matrix (although sometimes there might be a
double eigenvalue, when
(TrA)^{2}  4 detA = 0).
Let's call them and .
We've already seen prototypical examples where the eigenvalues are real, nonzero, and distinct. But what if the eigenvalues are complex conjugate (which happens when (TrA)^{2} < 4 detA)? In this case, there is no straight line solution in the reals. Instead, solutions turn around the origin. If the real part of the eigenvalues is positive, solutions spiral away from the origin, and if it is negative, they spiral towards it. If the eigenvalues are purely imaginary, the solutions neither move in nor out; rather, they circle around the origin.
We can summarize this in the table below. We are skipping over dealing with the degenerate cases, when = and when one of the eigenvalues is zero.








Finally, we remark that there is a convenient way to organize this information into a single diagram. Since the eigenvalues of A can be written as
We now return to our nonlinear system, and look at the linearization near the fixed point. As we saw earlier in this section, for every value of R the Phugoid model
> with(linalg): R:='R': J:=jacobian( [(v^2cos(theta))/v, sin(theta)R*v^2], [theta, v]);
Once we know the trace and determinant of the Jacobian matrix at the fixed point, we can use that information to determine how its type (sink, source, saddle, etc.) depends on R. Rather than solve for the eigenvalues directly, we will compute the trace and determinant, which have a simpler form in this case.
First, let's find the fixed point again, issuing the same command we did earlier. Then, we'll substitute that into the result of trace and det to calculate the trace and determinant at the fixed point.
> FixPoint:=convert( solve((v^2cos(theta))/v=0,sin(theta)R*v^2=0, theta,v), radical);
> trfix := simplify(subs(FixPoint, trace(J))); detfix:= simplify(subs(FixPoint, det(J)));
So, we see that for all real values of R, the determinant is positive (meaning saddles are impossible). Furthermore, for R > 0, the trace is negative, so the fixed point is always a sink. (When R = 0, the trace is zero, which means it is possible for this fixed point to be a center.^{3.5}) From the plots we made earlier, we expect that for small values of R, we will have a spiral sink, but it will probably not spiral for sufficiently large values of R. We can check this by solving for the point where Tr(J)^{2} = 4 det(J), which happens when R = 2.
> solve( (trfix)^2 = 4*detfix, R);