18.03 Class 34, May 3
Linear Phase Portraits
Vocabulary: (tr,det) plane, critical parabola,
spiral, (proper) node, saddle; center, star, improper node; degeneracy;
Stable: asymptotic, neutral; unstable.
Eigenvalues Rule: this is the moral of today's lecture.
Recall that the characteristic polynomial of a square matrix A is
p_A(lambda) = det(A - lambda I). In the 2x2 case A = [a b ; c d ]
this can be rewritten as
p_A(lambda) = lambda^2 - (tr A) lambda + (det A)
where tr(A) = a + d , deet(A) = ad - bc.
Its roots are the eigenvalues, so
p_A(lambda) = (lambda - lambda1)(lambda - lambda2)
= lambda^2 - (lambda1 + lambda2) lambda + (lambda1 lambda2)
Comparing coefficients,
tr(A) = lambda1 + lambda2 , det(A) = lambda1 lambda2
so the two numbers tr(A) and det(A) , extracted from the four numbers
a, b, c, d, are determined by the eigenvalues. Conversely, they determine
the eigenvalues, as the roots: by the quadratic formula,
lambda1,2 = tr(A)/2 +- sqrt(tr(A)^2/4 - det(A)).
lambda1,2 are not real if det(A) > tr(A)^2/4
are equal if det(A) = tr(A)^2/4
are real and different from each other if det(A) < tr(A)^2/4
The dividing line is the "critical parabola," where det(A) = tr(A)^2/4.
Notice that if the eigenvalues are complex, the real part is tr(A)/2.
If the eigenvalues are real, they have the same sign exactly when their
product is positive, and that sign is positive if their sum is also positive.
Thus:
det
^
. |<-----purely imaginary
. complex roots .
. | .
. Re < 0 | Re > 0 .<----- repeated
. | .
. | .
real < 0 . | . real > 0
.. | ..
.. | ..
------------ ....... ------------> tr
^
real, opposite sign |______ at least one zero e.v.
The corresponding phase portraits exhibit the following behaviors:
stars or det
improper nodes
| ^
.V |<-----centers .
. | .
. stable | unstable .
. spirals | spirals .
. | .
. | .
stable . | . unstable
proper .. | .. proper
nodes .. | .. nodes
------------ ....... ------------> tr
^
saddles |______ degenerate cases
The only important part of this I haven't discussed is the "nodes":
Example: A = [2 -1 ; 0 1 ] .
This matrix is "upper triangular": it's zero below the main diagonal.
In this case the eigenvalues are dead easy to read off: they are precisely
the diagonal entries. This because the eigenvalues are characterized by
the fact that their sum is the trace and their product is the determinant,
and, because of the 0, this is also true of the diagonal entries.
So the eigenvalues here are lambda1 = 2, lambda2 = 1.
lambda = 2: A - 2I : [ 0 -1 ; 0 -1 ] [?;?] = [0;0] : alpha1 = [1;0]
(or any nonzero multiple)
lambda = 1: A - I : [ 1 & -1 ; 0 0 ] [?;?] = [0;0] : alpha2 = [1;1]
(or any nonzero multiple)
so the normal modes are e^{2t}[1;0] and e^t[1;1].
The ray solutions are along the x-axis and along the line with slope 45 deg.
Both increase exponentially in size, but the lambda = 2 eigensolution
blows up like the square of the lambda = 1 solution.
The general solution is a linear combination of these two. Take the sum,
for example; then u(0)= [2;1]. When t << 0, the e^{2t} term is
much smaller than the e^t term, and so the linear combination is
very near the lambda = 1 eigenline, namely the line of slope 45 deg.
The result is a spider like figure, with body along that line and legs
opening out from it. This picture is shown on p. 90 of the Supplementary
Notes.
This is an "unstable proper node," or just an "unstable node" for short.
There are also the special cases that happen along the curves separating
these regions:
. tr = 0 where det > 0 : eigenvalues nonzero and purely imaginary.
The phase portraits are "centers." All trajectories (except the
constant solution at the origin) are ellipses.
. det = 0 : at least one of the eigenvalues is zero.
If alpha is an eigenvector corresponding to this eigenvalue, then
the constant vector valued function u(t) = c alpha is a solution for
any constant c: there is a line (at least) of constant solutions.
Several patterns are possible, and they are illustrated in the
Supplementary Notes.
. det = tr^2/4 , along the critical parabola: repeated real eigenvalues.
The phase portraits are either
stars, in the complete case [ lambda1 0 ; 0 lambda1 ] or
improper nodes , otherwise.
The phase portrait in each one of these borderline cases shows some features
which are not determined purely by the eigenvalues. In addition to these:
when det > tr^2/4, the phase portrait is made up of spirals, but
you can't tell from the eigenvalues alone which
way the spiral is rotating. To discover that in case
A = [ -2 5 ; -2 4 ] for example, let's just evaluate the vector field
at the point [1;0] : it is given by the first column of A, [-2;-2],
and so the vector field points south (and west) at this point, and the
rotation is counterclockwise.
Stability: All linear systems fall into one of the following categories:
Asymptotically stable: all solutions ---> 0 as t ---> infinity
These systems occupy the upper left quadrant, tr < 0 and det > 0,
so the eigenvalues have negative real part.
Neutrally stable: all solutions are periodic
These systems occur only along the ray tr = 0 , det > 0,
so the eigenvalues are nonzero and purely imaginary.
In these linear cases the nonzero trajectories are in fact ellipses.
Unstable: most solutions ---> infinity as t ---> infinity
Saddles and unstable nodes and spirals are examples.
At the end of class I showed an animation of the way the phase portraits
change as you move around a loop in the (tr,det) plane. For each pair
(tr,det), the companion matrix
[ 0 1 ; -det -tr ]
provides an example of a matrix with this trace and determinant. The
corresponding phase portraits are illustrated. This animation can be found at
http://www.awlonline.com/ide/
under Linear Algebra/Linear Classification/Parameter Path Animation Tool.
You might look at other applets in this collection.