GILBERT STRANG: This is a
topic I think is interesting.
I like this one.
It's about stability or
instability of a steady state.
So let me show you the
differential equation.
It could be linear, but
might be non linear.
dy dt is f of y.
I'm going to-- I keep it that
right hand side not depending
on t, so just a function of y.
And when do I have
a steady state?
There's a steady state
when the derivative is 0.
So if the derivative is
0 when f of y equals 0,
let me call those special
y's by a capital letter.
So capital Y is a
number, a starting value,
where the right hand side
of the equation is 0.
And if the right hand
side of the equation is 0,
the left side of the equation
is 0, and dy dt is 0,
and we don't go anywhere.
So the solution-- if
f or y is equal to 0,
then we have y stays at y.
It's a constant for all
time, and my question
is, if we start near capital
Y, do we approach capital Y
as time goes on?
It's, in that case, I
would say, stable-- or does
the solution when we start
near y go far away from Y?
From capital Y?
Leave the steady state?
In that case, I would call
the steady state unstable.
So stable or unstable,
and it's very important
to know which it is.
And let me just
do some examples,
and you'll see the whole point.
So here is first starting
with a linear equation.
So what is capital
Y in this case?
Well, this is f of y here.
So if I set that to
0, the steady state
is capital Y-- capital--
equals 0 in this case.
That is 0.
So if I start at 0, I stay at 0.
Here is a second example, the
logistic equation, where I've
taken the coefficients to be 1.
What are the steady states
for the logistic equation?
Again, I set the
right hand side to 0.
I find two possible
steady states--
capital Y equals 0 or 1.
That right hand side
is 0 for both of those,
so in both cases, those are
both constant solutions,
steady states.
If the solution starts
at 0, it stays there
because the derivative is 0.
Has no reason to move.
And finally, now I let y
minus y cubed equals 0.
I solve y equals y cubed,
and I find three solutions,
three steady states.
Y could be 0 again.
It could be 1 again,
or it could be minus 1.
y equals y cubed.
Then y can be any of
those three groups,
and of course,
these are examples.
The actual problem
could have sines,
and cosines, and
exponentials, but these
are three clear cases, and
of course, the linear case
is always the good guide.
So in the linear case, when
does a solution stay near 0?
If I start small, when do I
go to 0, and when do I leave?
So I'm ready for
the answer here.
So, stable or not.
In this example, y equals 0.
That's stable if-- well,
do you see what's coming?
The solution is e to the at
if I start-- or constant times
e to the at.
When does that go to 0?
When does it approach
the steady state?
I need a to be negative.
That's going to be
the key to everything.
That number a
should be negative.
Now, over here, we
don't have an a.
The key point will
be to see what
is that that should be
negative in these examples.
And can I tell you the answer?
So the thing to look at,
negative or positive, stable
or unstable, is the derivative.
Look at the derivative of
that right hand side at y
equals y at the steady state.
And if the derivative df dy
is negative, then stable.
That was correct in
this linear case.
The derivative of
ay was just a, we
know that we get stable
when a is negative
because the solution
has an e to the at.
a is negative.
We go to 0.
What about examples
two and three?
So with those two examples
you'll see the whole idea.
So look at the second
example, y minus y squared.
f is y minus y squared.
We look at its derivative.
Its derivative is 1 minus 2y.
The derivative of-- so
I'm looking at 1 minus 2y.
That's df dy.
So what's the story on that?
If y is 0, then that derivative
is 1 plus 1 unstable.
So y equals 0 is now unstable,
and the other possibility,
y equals 1, I think, will be
stable, because when y is 1,
1 minus 2y-- that derivative
that we check-- 1 minus 2y
comes out minus 1
now-- negative--
and that's the
test for stability.
So capital Y equals
1-- you remember
how those S curves went up and
approached the horizontal line,
the steady state
capital Y equals 1?
So OK with two different steady
states there-- one unstable
and one stable.
And now here we have
three steady states,
and in other examples,
we could have many,
or they might be hard
to find, but here we
can see exactly
what's happening.
Now, I look at the
derivative df dy.
It's the derivative
of y minus y cubed.
So that's 1 minus 3y squared.
So again, y equals
0 is bad news.
y equals 0 I get 1--
positive number, unstable.
So y equals 0, unstable.
Whereas y equals 1
or minus 1-- those
are the other two
steady states--
then 1 minus 3y squared.
y squared will be
1 in those cases.
So I have 1 minus 3 minus 2.
[INAUDIBLE] it's negative.
So those are stable.
Do you see how easy the test is?
Compute the derivative df
dy at the steady state,
and just see is it stable
or is it not stable,
and that gives the-- see
whether is it negative
or is it positive.
That decides stable or unstable.
Now, I just want
to show why briefly
and then show you an example
by throwing the book,
and this would be an
example in three dimensions
that we will get to when we're
doing a system of equations.
So for something flying
in three dimensions,
we'll need three differential
equations, and all
this discussion, which is coming
to the end of first order-- one
first order equation.
So this stability is
one of the nice topics.
Now, what's the
reasoning behind it?
Behind this test?
Here's our test.
If df dy is 0, that's our
test, and why is that our test?
Can I explain it here?
I want to look at the difference
between y and the steady state.
And my question is, if that goes
to 0, I have something stable.
If that blows up, if
y goes further away
from this steady
state, it's unstable.
So dy dt is f of y.
d capital Y dt-- well,
that's actually 0.
Capital Y is that
constant steady state,
and at the same
time, f of Y is 0.
So I've just put a 0 on the left
side and a 0 on the right side,
remembering that capital
Y solves the equation
with no movement at all.
It's just steady.
Now I have f of y
minus f at capital Y.
I'm going to use calculus.
The difference between
the function at a point
and the function at a nearby
point is approximately--
and the mean value
theorem tells me
that it really is-- is
approximately the derivative df
dy times y minus 1.
That's the whole
point of calculus
actually-- to be able to
estimate the difference
between f at two points.
This is delta f, if you
like, and this is delta Y.
And delta f divided by delta
Y is approximately df dy,
and approximately means
more and more approximately,
closer and closer, as
these points-- as little y
and capital Y come close.
So in other words, what
I have is approximately
the linear equation-- the linear
equation where the test is this
is my a.
Here is the a.
Well, it's only
approximate because this
isn't a truly linear equation.
We are allowing more
terms, but calculus
says it's better and
better when you're close,
and so our question
is, do we get closer
or do we not get close?
And the answer is that when
that a is negative, then
it's just like the
linear equation.
The exponential of at goes to 0.
y minus capital Y
goes to 0-- stable--
when this thing is
positive or maybe even 0.
0 is kind of a marginal case.
I don't know whether I
shoot off or go to 0,
so I'm only going to say if
that is negative, it's stable,
and if it's positive, then
my e to the at blows up.
And that e to the at
is y minus capital Y.
It gets bigger and
bigger-- unstable.
So that's the reasoning
behind the beautiful, simple,
easy to apply test, which is,
if the derivative is negative,
then stable.
That's good, and now
I'm ready to show you
the example of a tumbling box.
That is, I'm more or less ready.
I'm going to take
a copy of the book,
and I'm going to
throw it in the air.
Well, I have put
on a rubber band
to hold it together,
because the book is rather
precious to be throwing around.
And let me say here I
learned about this experiment
from Professor Alar Toomre and
just today I've asked him would
he like to do the
experiment on a video?
If yes, then he
will do it properly.
If no, I will do more with it
when we get to three equations,
because we're in
three dimensions,
but let me just
show you the point.
So the point is I'm going
to throw this book up.
Does it wobble all
over the place--
unstable-- or does it
turn nicely on its axis?
I'm going to throw it up
on the narrow axis here,
the thin axis,
half an inch or so.
To me, that's stable.
I'm not as good as Professor
Toomre at catching it,
but with a rubber
band on it I caught.
Now, that's one
axis, but I'm in 3D.
Here's another way to
throw it-- is this way.
Now, you can try this
on somebody else's book.
I'm going to throw
it now this way.
I'm going to start it this
way, and the question is,
does it turn steadily
this way or not?
No.
Absolutely not.
It went all over the place.
Shall I do that one again?
You see how it tumbles?
Not so easy to catch.
So that is unstable, and then
there is a third direction.
Let's see.
I've done the very narrow
one, the middle one.
Probably the third
direction is this one,
and if I do it-- I'm
going to leave well enough
alone-- it will come out stable.
So two directions-- stable, one
unstable-- for a tumbling box,
and the website and the
book have lots more details,
and we'll do more.
One more thing I want to add
to this board for these three
examples-- can I do that?
One more thing.
I want a picture that
shows-- so here's
a line off to plus infinity,
and this way to minus infinity.
And if I took this example,
0 in example one, say,
for dy dt equals minus
y a negative is stable.
It is stable.
So the solutions here
are approaching 0.
This is approaching--
so my first example
is dy dt equals minus y.
I draw a line of y's.
There's y equals 0, and the
solution, wherever it starts,
approaches 0.
Now I'm ready to do
the logistic equation.
dy dt equals y minus y squared.
Now I have y equals
0 is unstable now.
y equals 0 is now unstable.
y doesn't approach 0 anymore.
It goes away from 0.
And what does it go to?
The other steady state, if
you remember, was y equals 1.
1 minus 1 is 0.
The derivative is 0.
That's a steady state.
Let me put it in here-- 1-- and
that was a stable steady state.
So that arrow is correct,
and it goes to 1,
and it's also going
to 1 from above.
So there is the stability line.
Let me call this the
stability line of y's that
shows in the simplest
possible picture
what direction what direction
the solution moves, which
is the same as showing
me the sine of dy dt.
The sine of dy dt is positive,
and this y minus y squared
is positive for y
between 0 and 1.
Between 0 and 1, 1/2 would
be a 1/2 minus 1/4 positive.
So it increases,
but it approaches 1.
And now finally can I add in--
can I create the stability line
for y minus y cubed?
This is still correct.
y equals 1 is still
a stable point.
0 is an unstable
point, but now I
have 0, 1, or that other
possibility, minus 1.
So let me put that into
the picture-- minus 1.
That is a stable one.
So for the y minus
y cubed example,
I'm stable, unstable--
you see the arrow is
going away and going
into 1 and minus 1,
and then they go
in from both sides.
Isn't that a simple
picture to put together
what we discovered
from the derivatives
of this thing, the 1 minus 3y
squared at those three points?
At this point, the
derivative was negative.
We go to it.
At this point, the derivative,
one minus 3y squared,
was positive.
We leave it.
At this point, this
is another stable one.
The derivative df dy
is negative there,
and the solution
approaches y equals 1.
We didn't have any
formula for the solution.
That's the nice thing.
We're getting this
essential information
by just taking the derivative
of that simple function
and looking to see is
it negative or positive
and getting that picture
without a formula.
So tumbling book,
stability, and instability,
and more to do in
higher dimensions.
Thank you.
