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00:00:00,500 --> 00:00:01,420
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?

167
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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.

175
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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.

179
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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.

182
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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,

185
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times you notice here
is a y minus capital Y

186
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and a z minus capital Z. There
is the linearized problem.

187
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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.

190
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So I have to find-- so
now my little job is

191
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find the critical points.

192
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I've got everything ready.

193
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I have to find the
critical points.

194
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Now remember, the critical
points are where f and g are 0.

195
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And let me remember
what those are.

196
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So there is the f.

197
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There's the g.

198
00:12:38,216 --> 00:12:41,480
One critical point was that one.

199
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Everything is 0.

200
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Another critical
point is that one.

201
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Again, everything is 0.

202
00:12:47,120 --> 00:12:51,380
So I have two critical
points, two Jacobian matrices,

203
00:12:51,380 --> 00:12:55,960
one at the first point,
one at the second point.

204
00:12:55,960 --> 00:12:57,930
So what are those matrices?

205
00:12:57,930 --> 00:13:02,830
At y and z equals 0, I have
the matrix-- I'll put it here,

206
00:13:02,830 --> 00:13:04,680
and then I'll copy it.

207
00:13:04,680 --> 00:13:10,510
If y and z are 0, I have a 1
and a 0 and a 0 and a minus 1.

208
00:13:13,340 --> 00:13:16,070
I just took y and z to be 0.

209
00:13:16,070 --> 00:13:17,960
That was the first
critical point.

210
00:13:17,960 --> 00:13:21,150
The second critical point
gives me the Jacobian

211
00:13:21,150 --> 00:13:23,260
at that second point.

212
00:13:23,260 --> 00:13:26,020
The second point
was when z was 1.

213
00:13:26,020 --> 00:13:27,930
So that's 0 now.

214
00:13:27,930 --> 00:13:30,890
And y was 1, so that's minus 1.

215
00:13:30,890 --> 00:13:34,520
z is a 1, y minus 1, y is a 1.

216
00:13:34,520 --> 00:13:36,030
So that's a 0.

217
00:13:36,030 --> 00:13:38,440
That's the second Jacobian.

218
00:13:38,440 --> 00:13:41,070
So we're seeing something
interesting here.

219
00:13:41,070 --> 00:13:43,480
We're seeing how 2
by 2 matrices will

220
00:13:43,480 --> 00:13:49,120
work by really nice
examples, 1, 0, minus 1.

221
00:13:49,120 --> 00:13:51,090
What's that telling me?

222
00:13:51,090 --> 00:13:55,000
That's telling me
that the rabbits grow.

223
00:13:55,000 --> 00:13:58,380
Because the rabbits
are the first, the y.

224
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And the foxes decay.

225
00:14:01,850 --> 00:14:05,280
And that's what's happening
when the two populations are

226
00:14:05,280 --> 00:14:06,650
really small.

227
00:14:06,650 --> 00:14:11,900
When the two populations
are really small,

228
00:14:11,900 --> 00:14:15,070
multiplying them together
is extremely small.

229
00:14:15,070 --> 00:14:17,510
So when the two populations
are really small,

230
00:14:17,510 --> 00:14:19,150
forget the eating.

231
00:14:19,150 --> 00:14:21,090
There aren't enough
people around, enough

232
00:14:21,090 --> 00:14:25,120
foxes and rabbits around
to make a decent meal.

233
00:14:25,120 --> 00:14:27,810
So I just have dy dt equal y.

234
00:14:27,810 --> 00:14:30,370
Rabbits are growing
from eating grass.

235
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dz dt is minus z, foxes are
decaying from natural causes.

236
00:14:37,920 --> 00:14:44,760
So that's what kind of a
stationary point will 0, 0 be?

237
00:14:44,760 --> 00:14:46,590
Rabbits are growing.

238
00:14:46,590 --> 00:14:48,520
It's an unstable point.

239
00:14:48,520 --> 00:14:50,400
We're leaving at 0, 0.

240
00:14:50,400 --> 00:14:52,290
Rabbits are increasing.

241
00:14:52,290 --> 00:14:54,890
Now how about the second point?

242
00:14:54,890 --> 00:14:58,860
The second point was
when they were both 1.

243
00:14:58,860 --> 00:15:09,090
when they're both 1, then we
got this as the Jacobian matrix.

244
00:15:09,090 --> 00:15:10,900
Oh, this is the interesting one.

245
00:15:10,900 --> 00:15:13,430
Can I just stay with
this one to finish?

246
00:15:13,430 --> 00:15:17,140
So I'm interested
in the-- and I'll

247
00:15:17,140 --> 00:15:19,670
put these facts on a new board.

248
00:15:19,670 --> 00:15:26,412
So again, y prime
dy dt is y minus yz.

249
00:15:26,412 --> 00:15:36,060
z prime is yz, rabbits are
getting eaten, minus z.

250
00:15:36,060 --> 00:15:41,250
And I'm interested in the
point y equal 1, z equal 1.

251
00:15:41,250 --> 00:15:46,930
And my matrix of this
is the Jacobian matrix.

252
00:15:46,930 --> 00:15:50,690
The Jacobian matrix
had the derivatives,

253
00:15:50,690 --> 00:15:54,960
which were 1 minus z, minus y.

254
00:15:54,960 --> 00:15:57,590
The y derivative of that is z.

255
00:15:57,590 --> 00:16:01,050
The z derivative is y minus 1.

256
00:16:01,050 --> 00:16:05,120
And at this point y and z are 1.

257
00:16:05,120 --> 00:16:12,250
So that became 0,
minus 1, 1 and 0.

258
00:16:12,250 --> 00:16:15,510
And what kind of a
problem do I have here?

259
00:16:15,510 --> 00:16:19,460
So my linearized
equation, so linearized,

260
00:16:19,460 --> 00:16:25,930
linearized around the point 1,1.

261
00:16:25,930 --> 00:16:35,840
My equation is y
minus 1 prime, sorry.

262
00:16:35,840 --> 00:16:44,430
It's the distance to
the critical points.

263
00:16:47,310 --> 00:16:52,250
the derivative of y minus
1 is, I see here a minus 1.

264
00:16:52,250 --> 00:16:59,270
I see a minus z minus 1.

265
00:16:59,270 --> 00:17:03,580
And here a plus y minus 1.

266
00:17:07,579 --> 00:17:12,710
You've got to understand this
pair of linearized equations.

267
00:17:12,710 --> 00:17:18,420
If I use some other variable,
the derivative of the first guy

268
00:17:18,420 --> 00:17:20,310
is minus the second.

269
00:17:20,310 --> 00:17:22,945
The derivative of the second
guy is plus the first.

270
00:17:25,660 --> 00:17:28,730
What will happen?

271
00:17:28,730 --> 00:17:38,050
Initially if I'm a little
bit-- if I have extra foxes,

272
00:17:38,050 --> 00:17:40,950
the rabbit population will drop.

273
00:17:40,950 --> 00:17:43,700
The rabbit population--
this will be negative.

274
00:17:43,700 --> 00:17:48,490
If z, the number of foxes,
is a little higher than 1,

275
00:17:48,490 --> 00:17:51,550
then the rabbit
population drops.

276
00:17:51,550 --> 00:17:56,450
And when the rabbit population
drops, z starts dropping.

277
00:17:56,450 --> 00:18:00,630
As z starts dropping below
1, the rabbit population

278
00:18:00,630 --> 00:18:01,790
starts increasing.

279
00:18:01,790 --> 00:18:05,630
I get, what shall we
say, sort of exchange

280
00:18:05,630 --> 00:18:08,440
of rabbits and
foxes, oscillation

281
00:18:08,440 --> 00:18:09,845
between rabbits and foxes?

282
00:18:14,010 --> 00:18:16,560
So this is the--
right in the center

283
00:18:16,560 --> 00:18:22,640
there, that would be the point
where y is 1 and z equal 1,

284
00:18:22,640 --> 00:18:23,980
the critical point.

285
00:18:23,980 --> 00:18:28,250
And if I start out with
some extra rabbits,

286
00:18:28,250 --> 00:18:33,730
then the number of
rabbits will drop.

287
00:18:33,730 --> 00:18:35,630
Because foxes are eating them.

288
00:18:35,630 --> 00:18:37,620
The number of foxes
will increase.

289
00:18:37,620 --> 00:18:39,050
I'll go up.

290
00:18:39,050 --> 00:18:46,390
So there I have a little
bit later, I have foxes now,

291
00:18:46,390 --> 00:18:48,420
but no rabbits to eat.

292
00:18:48,420 --> 00:18:55,400
So the foxes start
dropping, and what happens?

293
00:18:55,400 --> 00:18:57,390
I think, yeah.

294
00:18:57,390 --> 00:19:03,230
The number of rabbit
starts increasing.

295
00:19:03,230 --> 00:19:05,160
And this is what happens.

296
00:19:05,160 --> 00:19:07,950
I'll go around and
around in a circle.

297
00:19:07,950 --> 00:19:16,370
If you remember the pictures of
the paths for 2 by 2 equations,

298
00:19:16,370 --> 00:19:18,440
there were saddle points.

299
00:19:18,440 --> 00:19:22,050
That's what this is.

300
00:19:22,050 --> 00:19:27,240
y equals 0, z equals
0 was a saddle point.

301
00:19:27,240 --> 00:19:29,775
So no, it's a saddle.

302
00:19:34,000 --> 00:19:36,960
And what I'm discovering
now for y equal 1

303
00:19:36,960 --> 00:19:40,970
is an oscillation between
foxes and rabbits.

304
00:19:40,970 --> 00:19:47,960
So again, I could say
it'll be our center that

305
00:19:47,960 --> 00:19:53,220
was the very special picture
where it didn't spiral out.

306
00:19:53,220 --> 00:19:55,010
It didn't spiral in.

307
00:19:55,010 --> 00:20:00,360
The special numbers,
the eigenvalues

308
00:20:00,360 --> 00:20:07,290
of that matrix
are-- well, better

309
00:20:07,290 --> 00:20:09,250
leave eigenvalues
for the future.

310
00:20:09,250 --> 00:20:13,690
Because they happen to
be i and minus i here.

311
00:20:13,690 --> 00:20:17,230
It's motion in a circle.

312
00:20:17,230 --> 00:20:19,710
It comes from this,
the equation here.

313
00:20:19,710 --> 00:20:28,060
Motion in a circle as in y
double prime plus y equals 0.

314
00:20:28,060 --> 00:20:29,860
That's motion in a circle.

315
00:20:29,860 --> 00:20:31,620
And that's what we've got.

316
00:20:31,620 --> 00:20:33,550
So this is a center.

317
00:20:33,550 --> 00:20:37,260
Now what about-- do I
call a center stable?

318
00:20:37,260 --> 00:20:43,140
Not quite, because the rabbits
and foxes don't approach 1.

319
00:20:43,140 --> 00:20:47,280
They stay on a circle around 1.

320
00:20:47,280 --> 00:20:50,130
Either I've got extra
rabbits or extra foxes.

321
00:20:50,130 --> 00:20:56,930
But the total
energy or the total

322
00:20:56,930 --> 00:20:59,230
stays a constant on that circle.

323
00:20:59,230 --> 00:21:08,730
And I would call that
neutrally neutral.

324
00:21:08,730 --> 00:21:11,260
Neutral stability, because
it doesn't blow up.

325
00:21:11,260 --> 00:21:14,970
I don't leave the area around.

326
00:21:14,970 --> 00:21:18,420
I stay close to
the critical point.

327
00:21:18,420 --> 00:21:20,531
But I don't approach it either.

328
00:21:20,531 --> 00:21:21,030
OK.

329
00:21:21,030 --> 00:21:28,270
So that's a case where we
could see the stability, based

330
00:21:28,270 --> 00:21:31,300
on the linearization.

331
00:21:31,300 --> 00:21:31,920
OK.

332
00:21:31,920 --> 00:21:34,960
One more example to
come in another lecture.

333
00:21:34,960 --> 00:21:36,510
Thanks.