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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.