WEBVTT
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GILBERT STRANG: OK.
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Two equations, the question of
stability for two equations,
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stability around
a critical point.
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OK.
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So the idea will
be to linearize,
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to look very near that
critical point, that point.
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But now we're in two dimensions.
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So that's a little more to do.
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So here's the general picture,
and then here is an example.
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So here's the general setup.
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We have an equation
for the changes in y.
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But z is involved.
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And we have an equation for
the rate of change of z.
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But y is involved.
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So they're coupled together.
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It's that coupling
that's going to be new.
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So what's a critical point?
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Critical point is when those
right-hand sides are 0.
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Because then y and
z are both constant.
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So they stay at that point.
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Wherever they are at this
critical point is steady state.
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They stay steady.
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They stay steady.
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They stay at that
constant value.
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So we want that to be 0.
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And we want this to be 0.
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We have two
equations, f equals 0,
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and g equals 0, two equations.
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But we have two
unknowns, y and z.
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So we expect some solutions.
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And each solution has to
be looked at separately.
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Each solution is
a critical point.
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It's like well, you could
think of a golf course,
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with a surface going up.
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So critical points will be
points where the maximum point
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maybe, or the minimum
point, or we'll
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see something called
a saddle point.
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Actually, let's do the example.
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The example is a famous one.
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Predator-prey it's known as.
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Predator like foxes,
prey like rabbits.
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So the foxes eat the rabbits.
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And the question is, what
are the steady states where
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foxes' and rabbits'
constant values could stay?
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So here this is the equation
for what happens to the prey.
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So if the rabbits are left
alone, the prey is the rabbits.
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If they're left alone they
multiply, plenty of grass.
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Go for it.
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But if there are foxes, and
z counts the number of foxes,
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then foxes eat, shall
I say, those rabbits.
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And we lose rabbits.
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So we see that
with a minus sign.
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And the amount of
preying that goes on
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is proportional to
the number of foxes
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times the number of rabbits.
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Because that gives the
number of possible meetings.
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And what about the
foxes, the predator?
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The predator increases.
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So this is from
encountering the rabbits.
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That tends to make the
number of predators increase.
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But if there were no rabbits,
the foxes don't eat grass.
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They're out of luck,
and they decay.
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So I see a minus z there.
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So you see the pattern?
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So starting from 0
and 0, at that point,
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so I've defined critical points.
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So there's my f.
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And that should be 0.
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And there is my g,
and that should be 0.
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And it turns out there are
just two possibilities.
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This is one.
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If y and z are both 0, then
certainly I would get 0's.
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So that's like starting with
a very small number of foxes
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and rabbits.
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Or if y is 1 and z equal 1,
do you see that that would be,
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they would be in
perfect balance?
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If y is 1, and z is 1,
then that's 0 and that's 0.
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So the equation is satisfied.
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We can stay at-- y can stay
at 1, and z can stay at 1.
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It's a steady state.
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And the question is, is that
steady state where rabbits
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are staying--
their population is
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staying at 1, because
they're eating grass, good.
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But they're getting
eaten by the foxes, bad.
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And those two balance, and
give a rate of change of zero.
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And the foxes similarly, the
foxes get a positive push
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from eating rabbits.
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But natural causes
cut them back.
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And they balance
at z equal to 1.
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OK.
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So what I have to
do is linearize.
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And that's the real
point of the lecture.
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That's the real point
for linear-- how
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do linearize for two functions?
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And how do you linearize
for these two functions?
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So let me-- I have to write
the general formula first,
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so you'll see it.
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And then I'll apply it
to those two functions.
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OK.
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So here is the
idea of linearize.
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So I'm linearizing.
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So my first function is whatever
its value is at this point,
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it's like a tangent line.
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But now I've got
two derivatives.
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Because the function
depends on two variables.
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So I have a y minus
capital Y, times now I
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have to do partial derivative.
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So that's the slope in the
y direction, multiplied
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by the movement.
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And then similar term, z minus
capital Z times the movement
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in the z direction.
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And I have to,
because I stopped,
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this is the linear
part of the function.
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I have to put an
approximate symbol.
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Because I've ignored
higher derivatives.
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And of course this is 0.
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So that's why we have
linear in y and linear in z,
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times some numbers, the slopes.
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But we have two more slopes.
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Because we have another
function g of y and z.
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And that again will
just be approximately
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g at the critical point, which
is 0 plus y minus capital Y,
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times dg dy plus z minus
capital Z times dg dz.
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So altogether, the linear
stuff and four numbers,
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the derivatives of f in
the y and z directions,
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the derivatives of g in
the y and the z directions.
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OK.
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Now we had an example going.
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Let me bring that
example down again.
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There was my f.
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There was my g.
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Can easily find those
partial derivatives.
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So let me do it.
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So the partial derivative with
respect to y will be 1 minus z,
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z held constant for
that partial derivative.
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And the partial derivative with
respect to z will be minus y.
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Let me write all
those things down.
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So here is in the example.
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So can I create a
little matrix, df dy?
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It's the nice way, if I've
got four things, 2 by 2 matrix
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is great.
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df, dz; that number,
dg, dy; and dg dz.
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You could say this is the
first derivative matrix.
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It's the matrix of
first derivatives
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and it's always named after
Jacobi who studied these first.
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So it's called the
Jacobian matrix.
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Maybe I'll put his name to
give him credit, Jacobi.
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And the matrix is
the Jacobian matrix.
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And its determinant
is important.
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It's a very important matrix,
important in economics.
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We're doing things--
I'm speaking here
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about predator-prey, little
animals running around.
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But serious stuff
is the economy.
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Is the economy stable?
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If it's a running along
at some steady state
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and we move it a
little bit, does it
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return to that steady
state, or does it
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get totally out of hand?
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So there is the Jacobian matrix.
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And what are those derivatives?
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Remember again, what functions.
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There's my functions.
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So the y derivative
is 1 minus z.
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And the z derivative is minus y.
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The y derivative is z, and
the z derivative is y minus 1.
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Is that all right?
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This is what we need to
know from the functions.
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I've forgotten the functions
in the board that went up.
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Here are their derivatives.
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Now that's my Jacobian matrix.
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That's my matrix of
the-- that matrix
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has these four coefficients,
those four numbers.
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And really, so
the linearization,
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let me call that matrix.
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I should call it J for Jacobian.
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I will call it J for Jacobian.
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OK.
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That's the Jacobian matrix.
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So then my approximate--
what's my linearized equation?
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My linearized equation is the
time derivative of y and z.
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So this left-hand side
is just dy dt, and dz dt.
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So I'm using vector
notations, putting y
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and z together, instead of
separately, no big deal.
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So then I have this matrix
J, this 2 by 2 matrix,
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times you notice here
is a y minus capital Y
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and a z minus capital Z. There
is the linearized problem.
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Linearized because
this is constant,
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and this is linear, single y,
single z, and we have a matrix
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J.
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So I have to find-- so
now my little job is
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find the critical points.
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I've got everything ready.
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I have to find the
critical points.
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Now remember, the critical
points are where f and g are 0.
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And let me remember
what those are.
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So there is the f.
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There's the g.
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One critical point was that one.
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Everything is 0.
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Another critical
point is that one.
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Again, everything is 0.
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So I have two critical
points, two Jacobian matrices,
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one at the first point,
one at the second point.
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So what are those matrices?
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At y and z equals 0, I have
the matrix-- I'll put it here,
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and then I'll copy it.
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If y and z are 0, I have a 1
and a 0 and a 0 and a minus 1.
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I just took y and z to be 0.
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That was the first
critical point.
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The second critical point
gives me the Jacobian
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at that second point.
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The second point
was when z was 1.
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So that's 0 now.
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And y was 1, so that's minus 1.
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z is a 1, y minus 1, y is a 1.
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So that's a 0.
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That's the second Jacobian.
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So we're seeing something
interesting here.
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We're seeing how 2
by 2 matrices will
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work by really nice
examples, 1, 0, minus 1.
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What's that telling me?
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That's telling me
that the rabbits grow.
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Because the rabbits
are the first, the y.
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And the foxes decay.
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And that's what's happening
when the two populations are
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really small.
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When the two populations
are really small,
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multiplying them together
is extremely small.
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So when the two populations
are really small,
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forget the eating.
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There aren't enough
people around, enough
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foxes and rabbits around
to make a decent meal.
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So I just have dy dt equal y.
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Rabbits are growing
from eating grass.
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dz dt is minus z, foxes are
decaying from natural causes.
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So that's what kind of a
stationary point will 0, 0 be?
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Rabbits are growing.
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It's an unstable point.
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We're leaving at 0, 0.
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Rabbits are increasing.
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Now how about the second point?
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The second point was
when they were both 1.
00:14:58.860 --> 00:15:09.090
when they're both 1, then we
got this as the Jacobian matrix.
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Oh, this is the interesting one.
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Can I just stay with
this one to finish?
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So I'm interested
in the-- and I'll
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put these facts on a new board.
00:15:19.670 --> 00:15:26.412
So again, y prime
dy dt is y minus yz.
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z prime is yz, rabbits are
getting eaten, minus z.
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And I'm interested in the
point y equal 1, z equal 1.
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And my matrix of this
is the Jacobian matrix.
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The Jacobian matrix
had the derivatives,
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which were 1 minus z, minus y.
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The y derivative of that is z.
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The z derivative is y minus 1.
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And at this point y and z are 1.
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So that became 0,
minus 1, 1 and 0.
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And what kind of a
problem do I have here?
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So my linearized
equation, so linearized,
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linearized around the point 1,1.
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My equation is y
minus 1 prime, sorry.
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It's the distance to
the critical points.
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the derivative of y minus
1 is, I see here a minus 1.
00:16:52.250 --> 00:16:59.270
I see a minus z minus 1.
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And here a plus y minus 1.
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You've got to understand this
pair of linearized equations.
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If I use some other variable,
the derivative of the first guy
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is minus the second.
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The derivative of the second
guy is plus the first.
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What will happen?
00:17:28.730 --> 00:17:38.050
Initially if I'm a little
bit-- if I have extra foxes,
00:17:38.050 --> 00:17:40.950
the rabbit population will drop.
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The rabbit population--
this will be negative.
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If z, the number of foxes,
is a little higher than 1,
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then the rabbit
population drops.
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And when the rabbit population
drops, z starts dropping.
00:17:56.450 --> 00:18:00.630
As z starts dropping below
1, the rabbit population
00:18:00.630 --> 00:18:01.790
starts increasing.
00:18:01.790 --> 00:18:05.630
I get, what shall we
say, sort of exchange
00:18:05.630 --> 00:18:08.440
of rabbits and
foxes, oscillation
00:18:08.440 --> 00:18:09.845
between rabbits and foxes?
00:18:14.010 --> 00:18:16.560
So this is the--
right in the center
00:18:16.560 --> 00:18:22.640
there, that would be the point
where y is 1 and z equal 1,
00:18:22.640 --> 00:18:23.980
the critical point.
00:18:23.980 --> 00:18:28.250
And if I start out with
some extra rabbits,
00:18:28.250 --> 00:18:33.730
then the number of
rabbits will drop.
00:18:33.730 --> 00:18:35.630
Because foxes are eating them.
00:18:35.630 --> 00:18:37.620
The number of foxes
will increase.
00:18:37.620 --> 00:18:39.050
I'll go up.
00:18:39.050 --> 00:18:46.390
So there I have a little
bit later, I have foxes now,
00:18:46.390 --> 00:18:48.420
but no rabbits to eat.
00:18:48.420 --> 00:18:55.400
So the foxes start
dropping, and what happens?
00:18:55.400 --> 00:18:57.390
I think, yeah.
00:18:57.390 --> 00:19:03.230
The number of rabbit
starts increasing.
00:19:03.230 --> 00:19:05.160
And this is what happens.
00:19:05.160 --> 00:19:07.950
I'll go around and
around in a circle.
00:19:07.950 --> 00:19:16.370
If you remember the pictures of
the paths for 2 by 2 equations,
00:19:16.370 --> 00:19:18.440
there were saddle points.
00:19:18.440 --> 00:19:22.050
That's what this is.
00:19:22.050 --> 00:19:27.240
y equals 0, z equals
0 was a saddle point.
00:19:27.240 --> 00:19:29.775
So no, it's a saddle.
00:19:34.000 --> 00:19:36.960
And what I'm discovering
now for y equal 1
00:19:36.960 --> 00:19:40.970
is an oscillation between
foxes and rabbits.
00:19:40.970 --> 00:19:47.960
So again, I could say
it'll be our center that
00:19:47.960 --> 00:19:53.220
was the very special picture
where it didn't spiral out.
00:19:53.220 --> 00:19:55.010
It didn't spiral in.
00:19:55.010 --> 00:20:00.360
The special numbers,
the eigenvalues
00:20:00.360 --> 00:20:07.290
of that matrix
are-- well, better
00:20:07.290 --> 00:20:09.250
leave eigenvalues
for the future.
00:20:09.250 --> 00:20:13.690
Because they happen to
be i and minus i here.
00:20:13.690 --> 00:20:17.230
It's motion in a circle.
00:20:17.230 --> 00:20:19.710
It comes from this,
the equation here.
00:20:19.710 --> 00:20:28.060
Motion in a circle as in y
double prime plus y equals 0.
00:20:28.060 --> 00:20:29.860
That's motion in a circle.
00:20:29.860 --> 00:20:31.620
And that's what we've got.
00:20:31.620 --> 00:20:33.550
So this is a center.
00:20:33.550 --> 00:20:37.260
Now what about-- do I
call a center stable?
00:20:37.260 --> 00:20:43.140
Not quite, because the rabbits
and foxes don't approach 1.
00:20:43.140 --> 00:20:47.280
They stay on a circle around 1.
00:20:47.280 --> 00:20:50.130
Either I've got extra
rabbits or extra foxes.
00:20:50.130 --> 00:20:56.930
But the total
energy or the total
00:20:56.930 --> 00:20:59.230
stays a constant on that circle.
00:20:59.230 --> 00:21:08.730
And I would call that
neutrally neutral.
00:21:08.730 --> 00:21:11.260
Neutral stability, because
it doesn't blow up.
00:21:11.260 --> 00:21:14.970
I don't leave the area around.
00:21:14.970 --> 00:21:18.420
I stay close to
the critical point.
00:21:18.420 --> 00:21:20.531
But I don't approach it either.
00:21:20.531 --> 00:21:21.030
OK.
00:21:21.030 --> 00:21:28.270
So that's a case where we
could see the stability, based
00:21:28.270 --> 00:21:31.300
on the linearization.
00:21:31.300 --> 00:21:31.920
OK.
00:21:31.920 --> 00:21:34.960
One more example to
come in another lecture.
00:21:34.960 --> 00:21:36.510
Thanks.