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GILBERT STRANG: This is a
topic I think is interesting.

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I like this one.

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It's about stability or
instability of a steady state.

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So let me show you the
differential equation.

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It could be linear, but
might be non linear.

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dy dt is f of y.

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I'm going to-- I keep it that
right hand side not depending

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on t, so just a function of y.

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And when do I have
a steady state?

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There's a steady state
when the derivative is 0.

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So if the derivative is
0 when f of y equals 0,

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let me call those special
y's by a capital letter.

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So capital Y is a
number, a starting value,

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where the right hand side
of the equation is 0.

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And if the right hand
side of the equation is 0,

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the left side of the equation
is 0, and dy dt is 0,

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and we don't go anywhere.

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So the solution-- if
f or y is equal to 0,

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then we have y stays at y.

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It's a constant for all
time, and my question

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is, if we start near capital
Y, do we approach capital Y

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as time goes on?

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It's, in that case, I
would say, stable-- or does

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the solution when we start
near y go far away from Y?

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From capital Y?

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Leave the steady state?

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In that case, I would call
the steady state unstable.

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So stable or unstable,
and it's very important

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to know which it is.

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And let me just
do some examples,

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and you'll see the whole point.

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So here is first starting
with a linear equation.

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So what is capital
Y in this case?

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Well, this is f of y here.

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So if I set that to
0, the steady state

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is capital Y-- capital--
equals 0 in this case.

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That is 0.

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So if I start at 0, I stay at 0.

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Here is a second example, the
logistic equation, where I've

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taken the coefficients to be 1.

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What are the steady states
for the logistic equation?

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Again, I set the
right hand side to 0.

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I find two possible
steady states--

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capital Y equals 0 or 1.

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That right hand side
is 0 for both of those,

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so in both cases, those are
both constant solutions,

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steady states.

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If the solution starts
at 0, it stays there

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because the derivative is 0.

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Has no reason to move.

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And finally, now I let y
minus y cubed equals 0.

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I solve y equals y cubed,
and I find three solutions,

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three steady states.

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Y could be 0 again.

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It could be 1 again,
or it could be minus 1.

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y equals y cubed.

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Then y can be any of
those three groups,

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and of course,
these are examples.

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The actual problem
could have sines,

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and cosines, and
exponentials, but these

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are three clear cases, and
of course, the linear case

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is always the good guide.

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So in the linear case, when
does a solution stay near 0?

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If I start small, when do I
go to 0, and when do I leave?

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So I'm ready for
the answer here.

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So, stable or not.

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In this example, y equals 0.

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That's stable if-- well,
do you see what's coming?

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The solution is e to the at
if I start-- or constant times

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e to the at.

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When does that go to 0?

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When does it approach
the steady state?

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I need a to be negative.

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That's going to be
the key to everything.

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That number a
should be negative.

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Now, over here, we
don't have an a.

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The key point will
be to see what

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is that that should be
negative in these examples.

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And can I tell you the answer?

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So the thing to look at,
negative or positive, stable

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or unstable, is the derivative.

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Look at the derivative of
that right hand side at y

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equals y at the steady state.

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And if the derivative df dy
is negative, then stable.

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That was correct in
this linear case.

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The derivative of
ay was just a, we

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know that we get stable
when a is negative

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because the solution
has an e to the at.

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a is negative.

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We go to 0.

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What about examples
two and three?

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So with those two examples
you'll see the whole idea.

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So look at the second
example, y minus y squared.

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f is y minus y squared.

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We look at its derivative.

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Its derivative is 1 minus 2y.

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The derivative of-- so
I'm looking at 1 minus 2y.

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That's df dy.

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So what's the story on that?

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If y is 0, then that derivative
is 1 plus 1 unstable.

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So y equals 0 is now unstable,
and the other possibility,

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y equals 1, I think, will be
stable, because when y is 1,

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1 minus 2y-- that derivative
that we check-- 1 minus 2y

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comes out minus 1
now-- negative--

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and that's the
test for stability.

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So capital Y equals
1-- you remember

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how those S curves went up and
approached the horizontal line,

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the steady state
capital Y equals 1?

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So OK with two different steady
states there-- one unstable

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and one stable.

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And now here we have
three steady states,

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and in other examples,
we could have many,

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or they might be hard
to find, but here we

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can see exactly
what's happening.

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Now, I look at the
derivative df dy.

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It's the derivative
of y minus y cubed.

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So that's 1 minus 3y squared.

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So again, y equals
0 is bad news.

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y equals 0 I get 1--
positive number, unstable.

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So y equals 0, unstable.

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Whereas y equals 1
or minus 1-- those

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are the other two
steady states--

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then 1 minus 3y squared.

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y squared will be
1 in those cases.

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So I have 1 minus 3 minus 2.
[INAUDIBLE] it's negative.

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So those are stable.

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Do you see how easy the test is?

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Compute the derivative df
dy at the steady state,

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and just see is it stable
or is it not stable,

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and that gives the-- see
whether is it negative

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or is it positive.

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That decides stable or unstable.

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Now, I just want
to show why briefly

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and then show you an example
by throwing the book,

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and this would be an
example in three dimensions

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that we will get to when we're
doing a system of equations.

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So for something flying
in three dimensions,

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we'll need three differential
equations, and all

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this discussion, which is coming
to the end of first order-- one

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first order equation.

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So this stability is
one of the nice topics.

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Now, what's the
reasoning behind it?

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Behind this test?

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Here's our test.

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If df dy is 0, that's our
test, and why is that our test?

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Can I explain it here?

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I want to look at the difference
between y and the steady state.

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And my question is, if that goes
to 0, I have something stable.

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If that blows up, if
y goes further away

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from this steady
state, it's unstable.

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So dy dt is f of y.

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d capital Y dt-- well,
that's actually 0.

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Capital Y is that
constant steady state,

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and at the same
time, f of Y is 0.

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So I've just put a 0 on the left
side and a 0 on the right side,

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remembering that capital
Y solves the equation

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with no movement at all.

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It's just steady.

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Now I have f of y
minus f at capital Y.

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I'm going to use calculus.

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The difference between
the function at a point

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and the function at a nearby
point is approximately--

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and the mean value
theorem tells me

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that it really is-- is
approximately the derivative df

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dy times y minus 1.

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That's the whole
point of calculus

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actually-- to be able to
estimate the difference

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between f at two points.

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This is delta f, if you
like, and this is delta Y.

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And delta f divided by delta
Y is approximately df dy,

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and approximately means
more and more approximately,

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closer and closer, as
these points-- as little y

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and capital Y come close.

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So in other words, what
I have is approximately

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the linear equation-- the linear
equation where the test is this

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is my a.

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Here is the a.

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Well, it's only
approximate because this

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isn't a truly linear equation.

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We are allowing more
terms, but calculus

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says it's better and
better when you're close,

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and so our question
is, do we get closer

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or do we not get close?

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And the answer is that when
that a is negative, then

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it's just like the
linear equation.

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The exponential of at goes to 0.

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y minus capital Y
goes to 0-- stable--

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when this thing is
positive or maybe even 0.

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0 is kind of a marginal case.

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I don't know whether I
shoot off or go to 0,

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so I'm only going to say if
that is negative, it's stable,

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and if it's positive, then
my e to the at blows up.

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And that e to the at
is y minus capital Y.

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It gets bigger and
bigger-- unstable.

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So that's the reasoning
behind the beautiful, simple,

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easy to apply test, which is,
if the derivative is negative,

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then stable.

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That's good, and now
I'm ready to show you

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the example of a tumbling box.

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That is, I'm more or less ready.

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I'm going to take
a copy of the book,

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and I'm going to
throw it in the air.

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Well, I have put
on a rubber band

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to hold it together,
because the book is rather

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precious to be throwing around.

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And let me say here I
learned about this experiment

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from Professor Alar Toomre and
just today I've asked him would

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he like to do the
experiment on a video?

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If yes, then he
will do it properly.

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If no, I will do more with it
when we get to three equations,

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because we're in
three dimensions,

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but let me just
show you the point.

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So the point is I'm going
to throw this book up.

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Does it wobble all
over the place--

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unstable-- or does it
turn nicely on its axis?

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I'm going to throw it up
on the narrow axis here,

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the thin axis,
half an inch or so.

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To me, that's stable.

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I'm not as good as Professor
Toomre at catching it,

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but with a rubber
band on it I caught.

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Now, that's one
axis, but I'm in 3D.

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Here's another way to
throw it-- is this way.

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Now, you can try this
on somebody else's book.

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I'm going to throw
it now this way.

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I'm going to start it this
way, and the question is,

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does it turn steadily
this way or not?

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No.

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Absolutely not.

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It went all over the place.

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Shall I do that one again?

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You see how it tumbles?

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Not so easy to catch.

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So that is unstable, and then
there is a third direction.

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Let's see.

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I've done the very narrow
one, the middle one.

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Probably the third
direction is this one,

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and if I do it-- I'm
going to leave well enough

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alone-- it will come out stable.

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So two directions-- stable, one
unstable-- for a tumbling box,

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and the website and the
book have lots more details,

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and we'll do more.

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One more thing I want to add
to this board for these three

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examples-- can I do that?

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One more thing.

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I want a picture that
shows-- so here's

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a line off to plus infinity,
and this way to minus infinity.

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And if I took this example,
0 in example one, say,

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for dy dt equals minus
y a negative is stable.

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It is stable.

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So the solutions here
are approaching 0.

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This is approaching--
so my first example

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is dy dt equals minus y.

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I draw a line of y's.

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There's y equals 0, and the
solution, wherever it starts,

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approaches 0.

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00:17:33,820 --> 00:17:45,090
Now I'm ready to do
the logistic equation.

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dy dt equals y minus y squared.

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Now I have y equals
0 is unstable now.

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y equals 0 is now unstable.

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y doesn't approach 0 anymore.

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It goes away from 0.

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And what does it go to?

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The other steady state, if
you remember, was y equals 1.

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1 minus 1 is 0.

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The derivative is 0.

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That's a steady state.

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Let me put it in here-- 1-- and
that was a stable steady state.

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So that arrow is correct,
and it goes to 1,

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and it's also going
to 1 from above.

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So there is the stability line.

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Let me call this the
stability line of y's that

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shows in the simplest
possible picture

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what direction what direction
the solution moves, which

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is the same as showing
me the sine of dy dt.

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The sine of dy dt is positive,
and this y minus y squared

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is positive for y
between 0 and 1.

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Between 0 and 1, 1/2 would
be a 1/2 minus 1/4 positive.

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So it increases,
but it approaches 1.

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And now finally can I add in--
can I create the stability line

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for y minus y cubed?

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This is still correct.

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00:19:30,920 --> 00:19:33,472
y equals 1 is still
a stable point.

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00:19:33,472 --> 00:19:36,050
0 is an unstable
point, but now I

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00:19:36,050 --> 00:19:42,950
have 0, 1, or that other
possibility, minus 1.

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00:19:42,950 --> 00:19:46,280
So let me put that into
the picture-- minus 1.

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00:19:46,280 --> 00:19:49,400
That is a stable one.

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00:19:49,400 --> 00:19:53,050
So for the y minus
y cubed example,

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I'm stable, unstable--
you see the arrow is

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00:19:56,120 --> 00:20:01,030
going away and going
into 1 and minus 1,

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00:20:01,030 --> 00:20:04,585
and then they go
in from both sides.

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00:20:07,215 --> 00:20:11,160
Isn't that a simple
picture to put together

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00:20:11,160 --> 00:20:14,130
what we discovered
from the derivatives

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00:20:14,130 --> 00:20:18,970
of this thing, the 1 minus 3y
squared at those three points?

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00:20:18,970 --> 00:20:21,190
At this point, the
derivative was negative.

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00:20:21,190 --> 00:20:22,270
We go to it.

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00:20:22,270 --> 00:20:25,190
At this point, the derivative,
one minus 3y squared,

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was positive.

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00:20:26,180 --> 00:20:27,310
We leave it.

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00:20:27,310 --> 00:20:31,160
At this point, this
is another stable one.

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00:20:31,160 --> 00:20:35,410
The derivative df dy
is negative there,

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00:20:35,410 --> 00:20:39,150
and the solution
approaches y equals 1.

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00:20:39,150 --> 00:20:41,940
We didn't have any
formula for the solution.

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00:20:41,940 --> 00:20:43,410
That's the nice thing.

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00:20:43,410 --> 00:20:46,620
We're getting this
essential information

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00:20:46,620 --> 00:20:50,570
by just taking the derivative
of that simple function

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00:20:50,570 --> 00:20:54,020
and looking to see is
it negative or positive

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00:20:54,020 --> 00:20:59,220
and getting that picture
without a formula.

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00:20:59,220 --> 00:21:03,950
So tumbling book,
stability, and instability,

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00:21:03,950 --> 00:21:08,580
and more to do in
higher dimensions.

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00:21:08,580 --> 00:21:10,370
Thank you.