WEBVTT

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GILBERT STRANG: OK.

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I'm concentrating now on the
key question of stability.

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Do the solutions approach 0 in
the case of linear equations?

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Do they approach some constant,
some steady state in the case

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of non-linear equations?

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So today is the
beginning of non-linear.

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I'll start with one equation.

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dy dt is some function of y,
not a linear function probably.

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And first question, what is a
steady state or critical point?

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Easy question.

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I'm looking at
special points capital

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Y, where the right-hand side
is 0, special points where

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the function is 0.

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And I'll call those critical
points or steady states.

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What's the point?

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At a critical point,
here is the solution.

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It's a constant.

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It's steady.

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I'm just checking here that
the equation is satisfied.

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The derivative is 0
because it's constant,

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and f is 0 because
it's a critical point.

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So I have 0 equals 0.

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The differential equation
is perfectly good.

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So if I start at a critical
point, I stay there.

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That's not our central question.

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Our key question is, if
I start at other points,

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do I approach a critical
point, or do I go away from it?

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Is the critical point
stable and attractive,

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or is it unstable and repulsive?

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So the way to
answer that question

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is to look at the
equation when you're

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very near the critical point.

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Very near the critical point, we
could make the equation linear.

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We can linearize the equation,
and that's the whole trick.

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And I've spoken before,
and I'll do it again now

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for one equation.

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But the real message,
the real content

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comes with two or
three equations.

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That's what we see
in nature very often,

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and we want to know,
is the problem stable?

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OK.

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So what does linearize mean?

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Every function is
linear if you look at it

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through a microscope.

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Maybe I should say if you
blow it up near y equal Y,

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every function is linear.

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Here is f of y.

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Here it's coming
through-- it's a graph

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of f of y, whatever it is.

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If this we recognize
as the point capital Y,

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right, that's where
the function is 0.

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And near that point, my function
is almost a straight line.

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And the slope of that
tangent is the coefficient,

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and everything depends on that.

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Everything depends on whether
the slope is going up like

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that-- probably that's going to
be unstable-- or coming down.

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If it were coming
down, then the slope

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would be negative at
the critical point,

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and probably that
will be stable.

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OK.

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So I just have to do
a little calculus.

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The whole idea of linearizing
is the central idea of calculus.

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That we have curves,
but near a point,

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we can pretend-- they are
essentially straight if we

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focus in, if we zoom in.

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So this is a zooming-in
problem, linearization.

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OK.

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So if I zoom in the
function at some y.

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I'm zooming in around
the point capital Y.

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But you remember
the tangent line

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stuff is the function
at Y. So little y is

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some point close by.

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Capital Y is the crossing point.

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And this is the y minus Y times
the slope-- that's the slope--

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the slope at the critical point
there is all that's-- you see

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that the right-hand
side is linear.

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And actually, f of Y is 0.

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That's the point.

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So that I have just a
linear approximation

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with that slope and
a simple function.

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OK.

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So I'll use this approximation.

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I'll put that into the
equation, and then I'll

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have a linear equation,
which I can easily solve.

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Can I do that?

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So my plan is, take my
differential equation,

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look, focus near
the steady state,

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near the critical point
capital Y. Near that point,

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this is my good approximation
to f, and I'll just use it.

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So I plan to use
that right away.

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So now here's the linearized.

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So d by dt of y equals f of y.

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But I'm going to do
approximately equals this y

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minus capital Y times the slope.

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So the slope is my
coefficient little a

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in my first-order
linear equation.

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So I'm going back to chapter
1 for this linearization

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for one equation.

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But then the next
video is the real thing

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by allowing two equations
or even three equations.

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So we'll make a
small start on that,

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but it's really the next video.

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OK.

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So that's the equation.

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Now, notice that
I could put dy dt

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as-- the derivative
of that constant

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is 0, so I could
safely put it there.

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So what does this tell me?

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Let me call that number a.

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So I can solve that
equation, and the solution

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will be y minus capital
Y. It's just linear.

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The derivative is the
thing itself times a.

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It's the pure model of steady
growth or steady decay.

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y minus Y is, let's
say, some e to the at.

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Right?

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When I have a coefficient
in the linear equation ay,

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I see it in the exponential.

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So a less than 0 is stable.

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Because a less than
0, that's negative,

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and the exponential drops to 0.

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And that tells me that
y approaches capital

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Y. It goes to the critical
point, to the steady state,

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and not away.

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Example, example.

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Let me just take an
example that you've

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seen before, the logistic
equation, where the right side

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is, say, 3y minus y squared.

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OK.

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Not linear.

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So I plan to linearize after
I find the critical points.

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Critical points, this is 0.

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That equals 0 at--
I guess there will

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be two critical points because
I have a second-degree equation.

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When that is 0, it could be 0
at y equals 0 or at y equals 3.

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So two critical points,
and each critical point

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has its own linearization, its
slope at that critical point.

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So you see, if I
graph f of y here,

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this 3y minus y squared has--
there is 3y minus y squared.

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There is one critical point, 0.

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There is the other
critical point at 3.

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Here the slope is
positive-- unstable.

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Here the slope is
negative-- stable.

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So this is stable, unstable.

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And let me just push
through the numbers here.

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So the df dy, that's the slope.

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So I have to take the
derivative of that.

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Notice this is not my
differential equation.

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There is my
differential equation.

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Here is my linearization
step, my computation

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of the derivative, the slope.

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So the derivative of
that is 3 minus 2y,

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and I've got two
critical points.

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At capital Y equal 0, that's 3.

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And at capital Y equals 3,
it's 3 minus 6, it's minus 3.

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Those are the slopes
we saw on the picture.

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Slope up, the
parabola is going up.

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Slope down.

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So this will
correspond to unstable.

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So what does it mean
for this to be unstable?

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It means that the solution
Y equals 0, constant 0,

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solves the equation, no problem.

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If Y stays at 0, it's a
perfectly OK solution.

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The derivative is 0.

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Everything's 0.

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But if I move a
little away from 0,

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if I move a little way
from 0, then the 3y minus y

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squared, what does it look like?

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If I'm moving just
a little away from Y

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equals 0, away from
this unstable point,

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y squared will be
extremely small.

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So it's really 3y.

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The y squared will be
small near Y equals 0.

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Forget that.

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We have exponential
growth, e to the 3t.

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We leave the 0 steady
state, and we move on.

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Now, eventually
we'll move somewhere

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near the other steady state.

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At capital Y equals 3, the
slope of this thing is minus 3,

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and the negative one
will be the stable point.

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So where y minus 3, the
distance to the steady state,

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the critical point will grow
like e to the mi-- well,

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will decay, sorry, I said grow,
I meant decay-- will decay

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like e to the minus 3t because
the minus 3 in the slope

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is the minus 3 in the exponent.

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OK.

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That's not rocket
science, although it's

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pretty important for rockets.

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Let me just say
what's coming next

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and then do it in
the follow-up video.

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So what's coming next will be
two equations, dy dt and dz dt.

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I have two things.

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y and z, they depend
on each other.

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So the growth or decay of y
is given by some function f,

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and this is given by some
different function g, so

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f and g.

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Now, when do I
have steady state?

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When this is 0.

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When they're both 0.

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They both have to be 0.

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And then dy dt is
0, so y is steady.

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dz dt is 0, so z is steady.

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So I'm looking for-- I've
got two numbers to look for.

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And I've got two
equations, f of y-- oh,

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let me call that capital
Y, capital Z-- so those

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are numbers now-- equals 0.

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So I want to solve-- equals 0,
and g of capital Y, capital Z

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equals 0.

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Yeah, yeah.

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So both right-hand
sides should be 0,

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and then I'm in a steady state.

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But this is going to be like
more interesting to linearize.

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That's really the next video,
is how do you linearize?

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What does the
linearized thing look

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like when you have two functions
depending on two variables

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Y and Z?

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You're going to have, we'll
see, [? for ?] slopes-- well,

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you'll see it.

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So this is what's coming.

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And we end up with
a two-by-two matrix

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because we have two equations,
two unknowns, and a little more

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excitement than the
classical single equation,

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like a logistic equation.

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OK.

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Onward to two.