GILBERT STRANG: OK.
So this is the second
lecture about these pictures,
in the phase plane
that's with axes y and y
prime, for a second order
constant coefficient
linear, good problem.
Good problem.
And you remember that
we study that equation
by looking for special
solutions y equals e to the st.
When we plug that
into the equation,
we get this simple
quadratic equation.
And everything depends on that.
So today this video
is about the case
when the roots are complex.
You remember, so the
roots, complex roots, you
have a real part, plus or
minus an imaginary part.
And this happens when b
squared is smaller than 4ac.
Because you remember,
there's a square root
in the formula for the solution
of a quadratic equation.
There's a square root
of b squared minus 4ac,
the usual formula from school.
And if b squared is smaller,
we have a negative number
under the square root.
And we get complex roots.
So last time the
roots were real.
The pictures in the phase
plane set off to infinity,
or came in to 0, more or less
almost on straight lines.
Now we're going to have
curves and spirals, because
of the complex part.
So here are the three
possibilities now.
We had three last time.
Here are the other three
with complex roots.
So the complex, the
real part, everything
depends on this real
part that the stability
going in, going out,
staying on a circle
depends on that real part.
If the real part is
positive, then we go out.
We have an exponential e
to the a plus i omega t.
And if a is positive that e to
the at would blow up, unstable.
So that's unstable.
Here is a center.
When a is zero, then we just
have e to the i omega t.
That's the nicest example.
I do that one first.
So in that case we're just
going around in a circle
or around in an ellipse.
And finally, the
physical problem
where we have damping,
but not too much damping.
So the roots are still complex.
But they're going in.
Because if a is negative,
e to the at is going to 0.
So that's a stable case.
That's a physical case.
We hope to have a little damping
in our system, and be stable.
This one we could
say neutrally stable.
This one is certainly unstable.
Let me start with that,
the neutrally stable.
Because that's the most famous
equation in second order
equation in mechanics.
It's pure oscillation, a
spring going up and down,
an LC circuit going back
and forth, pure oscillation.
And we see the solutions.
So I've written-- I've taken
this particular equation.
You notice no damping.
There's no y prime term.
OK.
So here is the solution,
famous, famous solution.
And y prime it
will be c1, I guess
the derivative of the
cosine is minus omega times
sine omega t, plus c2.
The derivative of that
is omega cos omega t.
So that's the y and y prime.
So for every t, it's going
to be an easy figure.
Here is y.
Here is y prime.
And that's the phase
plane, phase plane.
So at each time t, I
have a y and a y prime.
And it gives me a point.
So let me put it in there.
As time moves on
that point moves.
And it's the picture
in the phase plane,
the orbit sometimes
you could say,
it's kind of like
a planet or a moon.
So for that, what is
the orbit for that one?
Well it goes around
in an ellipse.
It would be a circle
with-- let me draw it.
This is the case omega equal 1.
In that case, in that
most famous case,
we simply go around a circle.
There's y.
There's y prime.
We have cosine and sine and
cos squared plus sine squared
is 1 squared.
And we're going around a circle
of radius 1, or another circle
depending on the
initial condition.
Here there's a factor
omega, giving an extra push
to y prime.
So if omega was 2, for
example, then we'd have a 2
in y prime from the omega,
which is not in the y.
And that would make y
prime a little larger.
And it would be twice as--
it would go up to twice as--
that's meant to be, meant to
be an ellipse with height 2 up
there, or in general
omega, and 1 there.
So in the y direction
there is no factor omega.
And we just have
cosine and sine.
And that would be a typical
picture in the phase plane.
But if we started with
smaller initial conditions,
we would travel on
another ellipse.
But the point is--
and these are called,
this picture is called a center.
So that's one of the six
possibilities, and in some way,
kind of the most beautiful.
You get ellipsis
in the phase plane.
They close off.
Because the solution just
repeats itself every period.
It's periodic.
y is periodic.
y prime is periodic.
They come around again
and again and again.
No energy is lost, conservation
of energy, perfection.
And I would say neutrally
stable, neutral stability.
The solution doesn't go into 0.
Because there's no damping.
It doesn't go out to infinity.
Because there's constant energy.
And that's the picture
in the phase plane.
OK.
So that's the center.
And now I'll draw one
with a source, or a sink.
I just have to change
the sign on damping
to get source or sink.
So let me do that.
So now I'm going to do
a spiral source or sink.
This is the unstable one,
going out to infinity.
This is the stable
one coming in to 0.
And let me do y double prime,
plus or maybe minus 4y prime,
plus 4y equals 0.
Suppose I take that equation.
Then I have s squared
plus 4s, oh maybe--
maybe 2 is a nicer number.
2 is nicer than 4.
Let me change this
to a 2, and a 2.
And so I have s
squared plus 2s plus 2
or minus 2s plus 2 equals 0.
So those are my-- positive
damping would be with a plus.
So with a plus sign, the roots
are s squared plus 2s plus 2.
The roots are 1, or rather
a minus 1 plus or minus i.
Plus sign, and then the
minus sign, with a minus 2.
Then all the roots have
a plus, plus or minus i.
Everything is depending on
these roots, these exponents,
which are the solutions of
the special characteristic
equation, the simple
quadratic equation.
And you see that depending
on positive damping
or a negative damping, I get
stability or instability.
And let me draw a picture.
I don't if I can try two
pictures in the same thing,
probably not.
That wouldn't be smart.
So what's happening then?
Let's take this one.
So this solution y
is e to the minus t.
That's what's making it
stable coming into 0, times--
and from here we have
c1 cos t and c2 sine t.
That's what we get
from the usual,
as in the case of a center that
carries us around the circle.
So what's happening in this
picture, in this phase plane?
Here's a phase plane
again, y and y prime.
Without the minus
1, we have a center.
We just go around in a circle.
But now because of
the minus 1, which
is the factor e to the
minus t in the solution,
as we go around we come in.
And the word for that
curve is a spiral.
So this would be the center,
going around in a circle.
But now suppose we start here.
Suppose we start at y equal
1, and y prime equal 0,
start there at time 0.
Let time go.
Plot where we go.
Where does this y
and the y prime,
where is the point, y, y prime?
OK.
I'm starting it at-- so I'm
probably taking c1 as 1,
and c2 as 0.
So I'm starting it right there.
And then I'll travel
along, depending on sines.
I would go, I think,
probably this way.
So it will travel
on a-- it comes in
pretty fast, of course.
Because that exponential
is a powerful guy
that e to the minus t.
So this is the solution,
damping out to 0.
That's with the plus
sign, plus damping,
which gives the minus sign
in the s, in the exponent.
And then so that
is a spiral sink.
Sink meaning just as
water in a bathtub
flows in, that's what happens.
Now what happens in a
spiral source that's
what we have with a minus sign.
Now we have a 1.
Now we have an e to the plus t.
Everything is growing.
So instead of decaying, we're
going around but growing-- OK.
I'm off the board, way off
the board with that spiral.
Which is going to
keep going around,
but explode out to infinity.
So those are the
three possibilities
for complex roots, centers,
spiral source, and spiral sink.
For real roots we
had ordinary source,
and ordinary sink, no spiral.
And then the other possibility
was a saddle point,
where almost surely we go out.
But there was one direction
that came into the saddle point.
OK.
Those six pictures
are going to control
the whole problem of stability,
which is our next subject.
Thank you.
