In this video, we're gonna go over
how to convert the system
of second order differential equations
to a system of first order
differential equations.
And we know how to solve those.
So if you've got a system of first order
linear differential equations,
the derivative of some vector
is a matrix times that vector.
We know what the solutions look like.
They're linear combinations of
exponentials of the eigenvalue e^λt
times the eigenvectors.
And then you have some coefficients that
are related to your initial conditions.
Now we also learned how to solve
second order differential equations
where the second derivative
was a matrix times the vector.
And we saw that whenever you had
a positive eigenvalue,
you wind up with cosh's
and sinh's of K (Kappa)t,
where K was the square root
of the eigenvalue
and that whenever you had
negative eigenvalues,
you had terms with cosines
and sines of ω (omega) t,
where ω was the square root of
minus the eigenvalue.
Well what if somebody gives
you a differential equation
of the sort where the second derivative
is a matrix times x
plus another matrix
times the derivative of x?
This happens a fair amount
in the real world.
You know, this might be related
to friction, for example.
This might be a restoring force
from a system of springs,
and this is the friction.
So what happens then?
Well there's a trick to convert
any second order system
into a first order system.
So if you have this
second order system with...
a term that involves x, and the
derivative of x, equals second derivative.
What we do is we define
a new variable; I'm gonna call it y.
Now I apologize that we've been using y
to mean the coordinates of x in a B basis.
This is the different usage
of the letter y.
There are only so many
letters in the alphabet.
So let's let y be the derivative of x,
and then we'll write a differential equation.
What's the derivative of x?
Well, by definition it's y.
And what's the derivative of y?
Well the derivative of y is
the second derivative of x.
And our equation tells us what
the second derivative of x is.
It's Ax + B(dx/dt).
But that's Ax + By.
So the derivative of x is y,
the derivative of y is
Ax + By.
So if you make a big vector
out of x and y,
the derivative of x and y is this
big matrix times x and y.
And we call this big matrix 'M'.
See, if you originally had,
let's say, a 3x3 system,
then this is gonna be a 6x6 matrix.
You're doubling the number of variables
because you have as many y's
as you have x's.
So it's a first order system
in twice as many variables.
And how can you solve that?
Well, you find the eigenvalues
and eigenvectors of this big matrix M,
and you're off to the races
using exactly the same methods
as for a first order system.
Because it is a first order system.
What's more, is you know what
the solutions look like.
The solutions are always exponentials.
So let's do an example
with m equals just 1.
So we had the sec—
the differential equation,
second derivative is -2x + 3(dx/dt).
So then we package x and
the derivative of x into a two vector.
You see we had a second order
system in one variable,
we turn it into a first order system
in two variables:
0, the identity, A, and B.
This is A, this is B.
And then we have to find the
eigenvalues and the eigenvectors
of this M matrix.
Well the eigenvalues are 1 and 2.
You can see the row sums are all 1
and the trace is 3,
so the second eigenvalue has to be 2.
And you can work out that the
eigenvectors are &lt;1, 1&gt; and &lt;1, 2&gt;.
And so our most general solution
using everything we know
about first order systems
is that &lt;x, y&gt; is some constant
times e^t times &lt;1, 1&gt;
plus some other constant e^2t
times &lt;1, 2&gt;.
Now of course, we've never
actually wanted to know about y,
we wanted to know about x.
So we just read off what x is:
(c_1)e^t + (c_2)e^2t.
And of course y—
well the derivative of e^t is e^t,
the derivative of e^2t is 2e^2t.
Or you could read it off from
the second row of this equation.
Okay.
Now this gives a hint about how
things behave in general.
In general, if M is diagonalizable,
then the solutions to our first
order system is that—
This at time t is gonna be some
combination of exponentials.
The exponentials go as e^λt
for each eigenvalue.
These are eigenvalues of the big matrix.
And I'm calling the eigenvector
&lt;b_j, d_j&gt;.
So b_j is the x part of the eigenvalue,
and d_j is the y—sorry, eigenvector.
So in other words, b_j and d_j together
make the eigenvector of M.
And that has eigenvalue λ.
Okay, well let's spell that out.
If M&lt;b_j, d_j&gt; = λ&lt;b_j, d_j&gt;,
then the first equation just says
that d_j = λ(b_j).
So once you know what one of these guys
is, the other comes along for the ride.
It's just λ times it.
And the second equation says that
A(b_j) + B(d_j) = λ(d_j).
And if we substitute this in here,
we can write everything in terms of B,
and there's our equation.
It's not quite an eigenvalue equation,
because it involves two matrices A and B,
and it involves a quadratic equation.
You've got λ^2 over here,
and λB, and an A.
But if you find values of λ
and vectors B that satisfy this equation,
then those are the things that are
gonna go back in this expansion.
And those are the things that are
gonna contribute to x.
Okay. So let's do a 2x2 example
using this machinery.
So on our 2x2 example, we're
gonna say the second derivative
is given by the matrix 3, 3, 3, 3 times x,
minus the identity,
times the derivative of x.
So in other words, A is 3, 3, 3, 3,
B is minus the identity.
And if we wanted, we could
write out what M is.
M is the zero matrix,
the identity matrix,
A, and B.
Now who's up for diagonalizing
a 4x4 matrix?
Well, didn't hear very many voices.
So let's go ahead and do it
using this equation.
We take Ab + λBb = (λ^2)b.
So we're trying to solve this times b
minus λ times b is λ^2 times b.
Okay, in other words, 3, 3, 3, 3 b
is λ^2 + λb.
So in this case, it did turn out
to be an eigenvalue problem,
except the eigenvalue isn't λ,
it's λ^2 + λ.
So since the eigenvalues of
this matrix are 6 and 0,
whenever you have the eigenvalue—
whenever you have &lt;1, 1&gt;,
then this gives you 6b.
And you have to solve λ^2 + λ = 6,
so λ = 2 or -3.
And with these eigenvectors,
this is 0 and you have λ^2 + λ(0).
So λ is either 0 or -1.
So those are our four solutions.
We can put them together,
and our most general solution
is that x is going to be e^2t,
that's the first eigenvalue,
times the b from the first eigenvalue.
e^-3t, -3 was the second eigenvalue,
b from the second.
e^0t, that's a constant,
b from the third.
e^-t, b from the fourth.
And if you try to do everything
in terms of the eigenvalues
and eigenvectors of the original matrix,
so the M matrix,
then you would have found that the
eigenvalues were 2, -3, 0 and -1.
And the eigenvectors were the
first b and twice the first b
because the eigenvalue is 2.
The first b minus 3 times the first b,
because the eigenvalue was -3.
The second b and 0 times that,
the second b and -1 times that.
Okay. Last point
is what about the equation
second derivative = Ax
that we learned how to solve a while ago?
How does all this formalism
work for this particular problem
which we already know how to solve?
Well if B is 0, then our equations just
become Ab = (λ^2)b.
So the vectors we're looking for
are in fact eigenvectors of A.
But they're not eigenvectors
with eigenvalue λ,
they're eigenvectors with eigenvalue λ^2.
In other words, λ is the square root
of the eigenvalue of A.
And it can have either sign.
When the eigenvalue of A is real,
then the square roots are—is positive,
and the square root is real,
and we called it K.
When the eigenvalue was negative,
the square root was imaginary,
and we called it iω.
So for the positive eigenvalues,
you get e^Kt and e^-Kt.
And we package them as cosh's and sinh's.
cosh(Kt) and sinh(Kt).
When the eigenvalues of A were negative,
the square roots that were imaginary—
so we had e^iωt, e^-iωt.
And you know that e^iωt
is cos(ωt) + isin(ωt).
So we repackage that into
sines and cosines of ωt.
So that was a long slog,
but we now see how to take
any second order system
and turn it into a first order system,
and how when you do that
for the familiar ones,
you get the familiar answers.
