WEBVTT
Kind: captions
Language: en
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In this lecture, we deal with fixed points
and linerazation. So, consider the system
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x dot = f of xy, y dot = g of xy. And we suppose
that x*, y* is a fixed point, so f of x* y*
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= 0 and gs of x* and y = 0. So let u = x - x*
or v = y -y*, be small disturbances from the
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fixed point, now we need to work out, if the
disturbances grow or decay. So, we now derive
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differential equations for u and v.
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So, let us first focus on the u equation,
so u dot = x dot and that is as x* is a constant,
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this is = f of x* + u and y* + v and this
is by the simple substitution and this expands
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to f of x* y* + u times df dx + v times df
dy + terms that are order u square v square
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and uv. And this comes by employing a Taylor
series expansion and this is = u df dx + v
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df dy + terms which are order u square v square
uv, as f of x* y* = 0. Now note that df dx
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and df dy are to be evaluated at the fixed
point x*t y*.
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So in a similar way v dot = u dg dx + v dg
dy + terms which are order u square v square
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and uv. Note that order u square v square
uv, denotes quadratic terms in u and v and
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since u and v are small disturbances. The
quadratic terms are in fact very small. So,
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the disturbance u v evolves according to u
dot v dot = df dx, df dy, dg dx, dg dy times
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uv + quadratic terms.
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So, the matrix A = df dx, df dy, dg dx, dg
dy evaluated at x* y* is called the Jacobian
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matrix at the fixed points x*, y*. So, the
nonlinear system is x dot = f of xy, y dot
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= g of xy, where x*, y* is a fixed point and
the associated linearized system is u dot
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v dot = df dx, df dy, dg dx, dg dy evaluated
at x* y* times u v.
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Now let us consider the impact of small nonlinear
terms. So, the question we have is the following:
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Is it really safe to neglect the quadratic
terms in the original nonlinear system? So,
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another way to ask the question is the following;
does the linearized system give a qualitatively
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correct picture of the phase portrait near
the fixed point x* y*? The short answer is
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yes, but we have to be very careful.
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So if the linearized system predicts a saddle,
a node or a spiral when the fixed point really
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is a saddle, node or a spiral for the original
nonlinear system, the border line cases that
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is centers, degenerates nodes, stars or non
isolated fixed points have to be treated much
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more carefully.
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So, let us consider an example, find all the
fixed points of the system x dot = - x + x
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cube and y dot = -dy and use the technique
of linearization to classify them. Additionally,
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check the conclusions by deriving the phase
portrait for the full nonlinear system. We
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know that the fixed points occur where x dot
and y dot are equal to zero. And hence x = 0
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or x = plus minus 1 and y = 0 are the fixed
points.
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So, we have three fixed points 0 0 1 0 -1
0. Now the Jacobian matrix at a general point
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xy is A = dx dot dx, dx dot dy, dy dot dx,
dy dot dy = -1 + 3x square 0 0 and 2. Now
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we evaluate A at the fixed points at 0 0,
A = -1 0 0 -2. And so 0 0 is a stable node.
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At plus minus 1 0 A = 2 0 0 -2 and so 1 0
and -1 0 are both saddle points. As we have
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stable nodes and saddle points the fixed points
for the nonlinear system are in fact predicted
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correctly.
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Now let us check the conclusions by deriving
the phase portrait for the original nonlinear
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system. Note that the x and y equations are
uncoupled, so we have two independent first
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order systems at right angles to each other.
In the y direction, all the trajectories decay
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exponentially to y = 0. In the x direction,
the trajectories are attracted to x=0 and
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repelled from x=plus minus 1. The vertical
lines x = 0 and x = plus minus 1 or in variant
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because x dot =0 on them.
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So, any trajectory that starts on these lines
will stay on them forever. Similarly, y = 0
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is an invariant horizontal line. Finally note
that the phase portrait would be symmetric
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in both the x and the y axis as the equations
are invariant under the transformations x
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to -x and y to -y. So, we now put this together
to arrive at the phase portrait. So that is
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one fixed point, that is the second fixed
point, and that is the third fixed point
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and so we go ahead and fill out the rest of
phase portrait for this nonlinear system.
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Note that 0 0 is a stable node plus minus
1 0 are saddles and this is exactly as expected
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from the linearization.
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We now offer some comments on hyperbolic fixed
points, topological equivalence and structural
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stability. If the real part of the lambda
is not equal to zero for both Eigen values,
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then the fixed points are called hyperbolic.
The stability of hyperbolic fixed points is
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unaffected by small nonlinear terms. Non-hyperbolic
fixed points are the fragile ones. So here
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is a simple example of hyperbolicity from
our study of vector fields on the line.
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Recall x dot = f of x and that the stability
of a fixed point is accurately predicted by
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the linearization as long as f prime of x*
is not equal to zero. Now this is the same
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as saying that the real part of the lambda
is not equal to zero. So, the fixed points
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of an nth order system is hyperbolic, if all
the Eigen values of the linearization lie
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off the imaginary axis ie real part of lambda
i is not equal to zero for i = 1 to n.
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The Hartman Grobmen theorem states the following;
the local phase portrait near a hyperbolic
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fixed point is topologically equivalent to
the face portrait of the linearization. In
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particular, the stability type of the fixed
points is captured by the linearization. Topologically
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equivalent essentially means that there a
homeomorphism which is a continuous deformation
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with a continuous inverse that maps one local
phase portrait on to the other.
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Such that trajectories map on to trajectories
and the sense of time meaning the direction
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of the arrows is actually preserved. Here
is another way of thinking about it, two phase
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portraits are topologically equivalent. If
one is simply a distorted version of the other,
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hyperbolic fixed points also highlight the
notion of the structural stability. A phase
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portrait is structurally stable if its topology
cannot be changed by an arbitrarily small
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perturbation to the vector field.
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For example, the phase portrait of a saddle
point is structurally stable but a small amount
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of damping can actually convert a center into
a spiral.
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Now in this lecture, we dealt with a very
important topic called linearization. So you
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can start with a two dimensional flow of the
form x dot = f of xy and y dot = gs of xy,
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where this nonlinear system as a equilibrium
point denoted as x* and y*, so then what one
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can do is the following; We introduced small
disturbances u and v around the equilibrium
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point. We introduced this into the original
nonlinear system and we take a Taylor series
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expansion around the equilibrium.
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In the Taylor series expansion, we only retain
the linear terms. The quadratic and whole
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higher order terms are discarded. So, the
resulting equation would be linear and so
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essentially we have a linearized equation
associated with the original nonlinear system
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around the equilibrium x* y*. So fundamental
question that you really want to know as the
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following; to what extent does the linearized
version give a qualitatively correct picture
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of the phase portrait around the equilibrium.
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So to some extent how much can we trust anything
that we get out of this linearized and the
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answer is that the if the linearized version
predicts a saddle, a node or a spiral. Then
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the fixed point really is a saddle, a node
or a spiral for the original nonlinear system.
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So, in the sense this technique of linearization
can be very, very powerful to get qualitative
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aspect about the phase portrait of the original
nonlinear system. And we can trust this as
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long as we have a saddle, a node or a spiral.