18.03 Class 39, May 17
Dangers of linearization; limit cycles; chaos; next steps in mathematics
[1] The method we sketched on Monday works well "generically," i.e.
almost all the time.
Things get tricky if the trace and determinant of the Jacobian matrix
lie on one of the dividing curves in the (tr,det) plane.
If
tr = 0
and
det > 0 , the linearization gives centers, ellipses,
periodic trajectories. But the actual trajectories of the nonlinear
system
may not be periodic. They may be densely packed spirals. These spirals
will
rotate in the same direction as the predicted ellipse, but they may be
either stable or unstable. It's as if the nonlinearity jostles the
linear
phase portrait off onto one of the regions bounded by the
tr = 0
line.
Similarly, if
det = (tr/2)^2 > 0 , the linear phase plane is a star or
a defective node. The actual phase plane near the equilibrium may be
more
like a spiral, or more like a node.
None of this is too bad, but there is one case where the actual phase
portrait
may not resemble the linear one at all. This is when the Jacobian matrix
is
degenerate, i.e. has determinant zero, i.e. has at least one eigenvalue
which
is zero.
For example,
x'
y'
=
=
4xy
x^2 - y^2
x' = 0
if
either
x = 0
or
y = 0 .
y' = 0
So the only critical point is at
(0,0).
if x = +- y .
J(x,y) = [ f_x , f_y ; g_x , g_y ] = [ 4y , 4x ; 2x ; -2y ]
so
J(0,0) = [ 0 0 ; 0 0 ]
What is the phase portrait of this linear system? It just says
u' = 0
so every point is a constant solution. The eigenvalues are both zero,
every vector is an eigenvector, it is a degenerate system.
On the other hand, I showed a Matlab plot of the phase portrait: it has