GILBERT STRANG: OK.
I'm concentrating now on the
key question of stability.
Do the solutions approach 0 in
the case of linear equations?
Do they approach some constant,
some steady state in the case
of non-linear equations?
So today is the
beginning of non-linear.
I'll start with one equation.
dy dt is some function of y,
not a linear function probably.
And first question, what is a
steady state or critical point?
Easy question.
I'm looking at
special points capital
Y, where the right-hand side
is 0, special points where
the function is 0.
And I'll call those critical
points or steady states.
What's the point?
At a critical point,
here is the solution.
It's a constant.
It's steady.
I'm just checking here that
the equation is satisfied.
The derivative is 0
because it's constant,
and f is 0 because
it's a critical point.
So I have 0 equals 0.
The differential equation
is perfectly good.
So if I start at a critical
point, I stay there.
That's not our central question.
Our key question is, if
I start at other points,
do I approach a critical
point, or do I go away from it?
Is the critical point
stable and attractive,
or is it unstable and repulsive?
So the way to
answer that question
is to look at the
equation when you're
very near the critical point.
Very near the critical point, we
could make the equation linear.
We can linearize the equation,
and that's the whole trick.
And I've spoken before,
and I'll do it again now
for one equation.
But the real message,
the real content
comes with two or
three equations.
That's what we see
in nature very often,
and we want to know,
is the problem stable?
OK.
So what does linearize mean?
Every function is
linear if you look at it
through a microscope.
Maybe I should say if you
blow it up near y equal Y,
every function is linear.
Here is f of y.
Here it's coming
through-- it's a graph
of f of y, whatever it is.
If this we recognize
as the point capital Y,
right, that's where
the function is 0.
And near that point, my function
is almost a straight line.
And the slope of that
tangent is the coefficient,
and everything depends on that.
Everything depends on whether
the slope is going up like
that-- probably that's going to
be unstable-- or coming down.
If it were coming
down, then the slope
would be negative at
the critical point,
and probably that
will be stable.
OK.
So I just have to do
a little calculus.
The whole idea of linearizing
is the central idea of calculus.
That we have curves,
but near a point,
we can pretend-- they are
essentially straight if we
focus in, if we zoom in.
So this is a zooming-in
problem, linearization.
OK.
So if I zoom in the
function at some y.
I'm zooming in around
the point capital Y.
But you remember
the tangent line
stuff is the function
at Y. So little y is
some point close by.
Capital Y is the crossing point.
And this is the y minus Y times
the slope-- that's the slope--
the slope at the critical point
there is all that's-- you see
that the right-hand
side is linear.
And actually, f of Y is 0.
That's the point.
So that I have just a
linear approximation
with that slope and
a simple function.
OK.
So I'll use this approximation.
I'll put that into the
equation, and then I'll
have a linear equation,
which I can easily solve.
Can I do that?
So my plan is, take my
differential equation,
look, focus near
the steady state,
near the critical point
capital Y. Near that point,
this is my good approximation
to f, and I'll just use it.
So I plan to use
that right away.
So now here's the linearized.
So d by dt of y equals f of y.
But I'm going to do
approximately equals this y
minus capital Y times the slope.
So the slope is my
coefficient little a
in my first-order
linear equation.
So I'm going back to chapter
1 for this linearization
for one equation.
But then the next
video is the real thing
by allowing two equations
or even three equations.
So we'll make a
small start on that,
but it's really the next video.
OK.
So that's the equation.
Now, notice that
I could put dy dt
as-- the derivative
of that constant
is 0, so I could
safely put it there.
So what does this tell me?
Let me call that number a.
So I can solve that
equation, and the solution
will be y minus capital
Y. It's just linear.
The derivative is the
thing itself times a.
It's the pure model of steady
growth or steady decay.
y minus Y is, let's
say, some e to the at.
Right?
When I have a coefficient
in the linear equation ay,
I see it in the exponential.
So a less than 0 is stable.
Because a less than
0, that's negative,
and the exponential drops to 0.
And that tells me that
y approaches capital
Y. It goes to the critical
point, to the steady state,
and not away.
Example, example.
Let me just take an
example that you've
seen before, the logistic
equation, where the right side
is, say, 3y minus y squared.
OK.
Not linear.
So I plan to linearize after
I find the critical points.
Critical points, this is 0.
That equals 0 at--
I guess there will
be two critical points because
I have a second-degree equation.
When that is 0, it could be 0
at y equals 0 or at y equals 3.
So two critical points,
and each critical point
has its own linearization, its
slope at that critical point.
So you see, if I
graph f of y here,
this 3y minus y squared has--
there is 3y minus y squared.
There is one critical point, 0.
There is the other
critical point at 3.
Here the slope is
positive-- unstable.
Here the slope is
negative-- stable.
So this is stable, unstable.
And let me just push
through the numbers here.
So the df dy, that's the slope.
So I have to take the
derivative of that.
Notice this is not my
differential equation.
There is my
differential equation.
Here is my linearization
step, my computation
of the derivative, the slope.
So the derivative of
that is 3 minus 2y,
and I've got two
critical points.
At capital Y equal 0, that's 3.
And at capital Y equals 3,
it's 3 minus 6, it's minus 3.
Those are the slopes
we saw on the picture.
Slope up, the
parabola is going up.
Slope down.
So this will
correspond to unstable.
So what does it mean
for this to be unstable?
It means that the solution
Y equals 0, constant 0,
solves the equation, no problem.
If Y stays at 0, it's a
perfectly OK solution.
The derivative is 0.
Everything's 0.
But if I move a
little away from 0,
if I move a little way
from 0, then the 3y minus y
squared, what does it look like?
If I'm moving just
a little away from Y
equals 0, away from
this unstable point,
y squared will be
extremely small.
So it's really 3y.
The y squared will be
small near Y equals 0.
Forget that.
We have exponential
growth, e to the 3t.
We leave the 0 steady
state, and we move on.
Now, eventually
we'll move somewhere
near the other steady state.
At capital Y equals 3, the
slope of this thing is minus 3,
and the negative one
will be the stable point.
So where y minus 3, the
distance to the steady state,
the critical point will grow
like e to the mi-- well,
will decay, sorry, I said grow,
I meant decay-- will decay
like e to the minus 3t because
the minus 3 in the slope
is the minus 3 in the exponent.
OK.
That's not rocket
science, although it's
pretty important for rockets.
Let me just say
what's coming next
and then do it in
the follow-up video.
So what's coming next will be
two equations, dy dt and dz dt.
I have two things.
y and z, they depend
on each other.
So the growth or decay of y
is given by some function f,
and this is given by some
different function g, so
f and g.
Now, when do I
have steady state?
When this is 0.
When they're both 0.
They both have to be 0.
And then dy dt is
0, so y is steady.
dz dt is 0, so z is steady.
So I'm looking for-- I've
got two numbers to look for.
And I've got two
equations, f of y-- oh,
let me call that capital
Y, capital Z-- so those
are numbers now-- equals 0.
So I want to solve-- equals 0,
and g of capital Y, capital Z
equals 0.
Yeah, yeah.
So both right-hand
sides should be 0,
and then I'm in a steady state.
But this is going to be like
more interesting to linearize.
That's really the next video,
is how do you linearize?
What does the
linearized thing look
like when you have two functions
depending on two variables
Y and Z?
You're going to have, we'll
see, [? for ?] slopes-- well,
you'll see it.
So this is what's coming.
And we end up with
a two-by-two matrix
because we have two equations,
two unknowns, and a little more
excitement than the
classical single equation,
like a logistic equation.
OK.
Onward to two.
