-- and lift-off on
differential equations. So, this section is
about how to solve a system of first order,
first derivative, constant coefficient linear equations. And if we do it right, it turns
directly into linear algebra. The key idea is the solutions
to constant coefficient linear equations
are exponentials. So if you look for
an exponential, then all you have to find is
what's up there in the exponent and what multiplies
the exponential and that's the linear algebra. So -- and the result --
one thing we will fine -- it's completely parallel
to powers of a matrix. So the last lecture was
about how would you compute A to the K or A to the 100? How do you compute high
powers of a matrix? Now it's not powers anymore,
but it's exponentials. That's the natural thing
for differential equation. Okay. But can I begin with an example? And I'll just go
through the mechanics. How would I solve
the differential -- two differential equations? So I'm going to make it -- I'll have a two by two
matrix and the coefficients are minus one two, one minus
two and I'd better give you some initial condition. So suppose it starts u at
times zero -- this is u1, u2 -- let it -- let it -- suppose everything is
in u1 at times zero. So -- at -- at the
start, it's all in u1. But what happens as time
goes on, du2/dt will -- will be positive,
because of that u1 term, so flow will move into the u2
component and it will go out of the u1 component. So we'll just follow
that movement as time goes forward by looking at the
eigenvalues and eigenvectors of that matrix. That's a first job. Before you do anything
else, find the -- find the matrix and its
eigenvalues and eigenvectors. So let me do that. Okay. So here's our matrix. Maybe you can tell
me right away what -- what are the eigenvalues
and -- eigenvalues anyway. And then we can check. But can you spot any of the
eigenvalues of that matrix? We're looking for
two eigenvalues. Do you see -- I mean, if I just wrote
that matrix down, what -- what do you notice about it? It's singular, right. That -- that's a
singular matrix. That tells me right away
that one of the eigenvalues is lambda equals zero. I can -- that's a
singular matrix, the second column is minus
two times the first column, the determinant is zero,
it's -- it's singular, so zero is an eigenvalue and
the other eigenvalue will be -- from the trace. I look at the
trace, the sum down the diagonal is minus three. That has to agree with
the sum of the eigenvalue, so that second eigenvalue
better be minus three. I could, of course -- I could compute -- why
don't I over here -- compute the determinant
of A minus lambda I, the determinant of this minus
one minus lambda two one minus two minus lambda matrix. But we know what's coming. When I do that multiplication,
I get a lambda squared. I get a two lambda and a one
lambda, that's a three lambda. And then -- now I'm going
to get the determinant, which is two minus
two which is zero. So there's my characteristic
polynomial, this determinant. And of course I factor that into
lambda times lambda plus three and I get the two eigenvalues
that we saw coming. What else do I need? The eigenvectors. So before I even think about the
differential equation or what -- how to solve it, let me
find the eigenvectors for this matrix. Okay. So take lambda equals zero -- so that -- that's
the first eigenvalue. Lambda one equals zero
and the second eigenvalue will be lambda two
equals minus three. By the way, I -- I already know something
important about this. The eigenvalues are
telling me something. You'll see how it comes
out, but let me point to -- these numbers are
-- this eigenvalue, a negative eigenvalue,
is going to disappear. There's going to be an e to the
minus three t in the answer. That e to the minus
three t as times goes on is going to be very, very small. The other part of the
answer will involve an e to the zero t. But e to the zero t is
one and that's a constant. So I'm expecting that this
solution'll have two parts, an e to the zero t part and an
e to the minus three t part, and that -- and as time goes
on, the second part'll disappear and the first part
will be a steady It won't move. state. It will be -- at the end of
-- as t approaches infinity, this part disappears
and this is the -- the e to the zero t
part is what I get. And I'm very interested in these
steady states, so that's -- I get a steady state
when I have a zero eigenvalue. Okay. What about those eigenvectors? So what's the eigenvector that
goes with eigenvalue zero? Okay. The matrix is singular as
it is, the eigenvector is -- is the guy in the null space,
so what vector is in the null space of that matrix? Let's see. I guess I probably give the
free variable the value one and I realize that if I want to
get zero I need a two up here. Okay? So Ax1 is zero x1. A x1 is zero x1. Fine. Okay. What about the other eigenvalue? Lambda two is minus three. Okay. How do I get the other
eigenvalue, then? For the moment --
can I mentally do it? I subtract minus three
along the diagonal, which means I add three -- can I -- I'll just do it with an
erase -- erase for the moment. So I'm going to add
three to the diagonal. So this minus one will
become a two and -- I'll make it in big
loopy letters -- and when I add three to this
guy, the minus two becomes -- well, I can't make one
very loopy, but how's that? Okay. Now that's A minus three I -- A plus three I, sorry. That's A plus three I. It's supposed to
be singular, right? I-- if things --
if I did it right, this matrix should be
singular and the x2, the eigenvector should
be in its null space. Okay. What do I get for the
null space of this? Maybe minus one one,
or one minus one. Doesn't matter. Those are both perfectly good. Right? Because that's in the
null space of this. Now I'll -- because A times
that vector is three times that vector. Ax2 is minus three x2. Good. Okay. Can I get A again so
we see that correctly? That was a minus one and
that was a minus two. Good. Okay. That -- that's the first job. eigenvalues and eigenvectors. And already the
eigenvalues are telling me the most important
information about the answer. But now, what is the answer? The answer is -- the
solution will be U of T -- okay. Now, wh- now I use those
eigenvalues and eigenvectors. The solution is some --
there are two eigenvalues. So I -- it -- so there're going
to be two special solutions here. Two pure exponential solutions. The first one is going to
be either the lambda one tx1 and the -- so that solves the
equation, and so does this one. They both are solutions to
the differential equation. That's the general solution. The general solution
is a combination of that pure
exponential solution and that pure
exponential solution. Can I just see that those
guys do solve the equation? So let me just check -- check
on this one, for example. I -- I want to check that
the -- my equation -- let's Check. remember, the
equation -- du/dt is Au. I plug in e to the
lambda one t x1 and let's just see that
the equation's okay. I believe this is a
solution to that equation. So just plug it in. On the left-hand side, I
take the time derivative -- so the left-hand side will be
lambda one, e to the lambda one t x1, right? The time derivative -- this is
the term that depends on time, it's just ordinary exponential,
its derivative brings down a lambda one. On the other side of the
equation it's A times this thing. A times either the lambda one t
x one, and does that check out? Do we have equality there? Yes, because either the lambda
one t appears on both sides and the other one is Ax1
equal lambda one x1 -- check. Do you -- so, the -- we've come
to the first point to remember. These pure solutions. Those pure solutions are the
-- those pure exponentials are the differential
equations analogue of -- last time we had pure powers. Last time -- so -- so last time, the
analog was lambda -- lambda one to the K-th power
x1, some amount of that, plus some amount of lambda
two to the K-th power x2. That was our formula
from last time. I put it up just to -- so
your eye compares those two formulas. Powers of lambda in the -- in the difference equation
-- that -- this was in the -- this was for the equation
uk plus one equals A uk. That was for the finite
step -- stepping by one. And we got powers,
now this is the one we're interested in,
the exponentials. So -- so that's --
that's the solution -- what are c1 and c2? Then we're through. What are c1 and c2? Well, of course we know
these actual things. Let me just -- let
me come back to this. c1 is -- we haven't figured out
yet, but e to the lambda one t, the lambda one is zero so that's
just a one times x1 which is two one. So it's c1 times this one that's
not moving times the vector, the eigenvector two
one and c2 times -- what's e to the lambda two t? Lambda two is minus three. So this is the term
that has the minus three t and its eigenvector
is this one minus one. So this vector
solves the equation and any multiple of it. This vector solves the equation
if it's got that factor e to the minus three t. We've got the answer
except for c1 and c2. So -- so everything I've done
is immediate as soon as you know the eigenvalues
and eigenvectors. So how do we get c1 and c2? That has to come from
the initial condition. So now I -- now I use -- u
of zero is given as one zero. So this is the initial condition
that will find c1 and c2. So let me do that on
the board underneath. At t equals zero, then -- I get c1 times this guy plus
c2 and now I'm at times zero. So that's a one and
this is a one minus one and that's supposed to agree
with u of zero one zero. Okay. That should be two equations. That should give me c1 and
c2 and then I'm through. So what are c1 and c2? Let's see. I guess we could
actually spot them by eye or we could solve two
equations in two unknowns. Let's see. If these were both ones
-- so I'm just adding -- then I would get three zero. So what's the -- what's
the solution, then? If -- if c1 and c2 are both
ones, I get three zero, so I want, like,
one third of that, because I want to get one zero. So I think it's c1 equals
a third, c2 equals a third. So finally I have the answer. Let me keep it in the
-- in this board here. Finally the answer is one third
of this plus one third of this. Do you see what -- what's
actually happening with this flow? This flow started out at --
the solution started out at one zero. Started at one zero. Then as time went on,
people moved, essentially. Some fraction of
this one moved here. And -- and in the limit, there's
-- there's the limit, as -- right? As t goes to infinity,
as t gets very large, this disappears and this
is the steady state. So the steady state is -- so the steady state -- u -- we could call it u at
infinity is one third of two and one. It's -- it's two
thirds of one third. So that's the -- we really -- I mean, you're
getting, like, total, insight into the
behavior of the solution, what the differential
equation does. Of course, we don't -- wouldn't
always have a steady state. Sometimes we would
approach zero. Sometimes we would blow up. Can we straighten
out those cases? The eigenvalue should tell us. So when do we get -- so -- so let me ask first,
when do we get stability? That's u of t going to zero. When would the solution
go to zero no matter what the initial condition is? Negative eigenvalues, right. Negative eigenvalues. But now I have to -- I have to ask you
for one more step. Suppose the eigenvalues
are complex numbers? Because we know they could be. Then we want stability --
this -- this -- we want -- we need all these e to the
lambda t-s all going to zero and somehow that asks us
to have lambda negative. But suppose lambda
is a complex number? Then what's the test? What -- if lambda's a
complex number like, oh, suppose lambda is negative
plus an imaginary part? Say lambda is minus
three plus six i? What -- what happens then? Can we just, like,
do a -- a case here? If -- if this lambda is
minus three plus six it, how big is that number? Does this -- does this imaginary
part play a -- play a -- play a role here or not? Or how big is -- what's the absolute value
of that -- of that quantity? It's just e to the
minus three t, right? Because this other part, this --
the -- the magnitude -- the -- this -- e to the six it -- what
-- that has absolute value one. Right? That's just this cosine of
six t plus i, sine of six t. And the absolute
value squared will be cos squared plus sine
squared will be one. This is -- this complex number
is running around the unit circle. This com- this -- the -- it's
the real part that matters. This is what I'm trying to do. Real part of lambda
has to be negative. If lambda's a complex
number, it's the real part, the minus three, that
either makes us go to zero or doesn't, or let
-- or blows up. The imaginary part won't
-- will just, like, oscillate between
the two components. Okay. So that's stability. And what about -- what about a steady state? When would we have,
a steady state, always in the same direction? So let me -- I'll take this part away -- when -- so that was, like,
checking that it's -- that it's the real part
that we care about. Now, we have a
steady state when -- when lambda one is zero and the
other eigenvalues have what? So I'm looking -- like,
that example was, like, perfect for a steady state. We have a zero eigenvalue
and the other eigenvalues, we want those to disappear. So the other eigenvalues
have real part negative. And we blow up, for sure -- we blow up if any real
part of lambda is positive. So if I -- if I reverse the
sign of A -- of the matrix A -- suppose instead of the matrix
I had, the A that I had, I changed it -- I changed all its sines. What would that do to the
eigenvalues and eigenvectors? If I -- if -- if I know the
eigenvalues and eigenvectors of A, tell me about minus A. What happens to the eigenvalues? For minus A, they'll
all change sine. So I'll have blow up. This -- instead of the
e to the minus three t, if I change that to minus --
if I change the sines in that matrix, I would change
the trace to plus three, I would have an e to the plus
three t and I would have blow up. Of course the zero eigenvalue
would stay at zero, but the other guy
is taking off in -- if I reversed all the sines. Okay. So this is -- this is
the stability picture. And for a two by two
matrix, we can actually pin down even more
closely what that means. Can I -- let -- can I do that? Let me do that -- I want to -- for a two by two matrix, I
can tell whether the real part of the eigenvalues is
negative, I -- well, let me -- let me tell you what I
have in mind for that. So two by two stability -- let me -- this is
just a little comment. Two by two stability. So my matrix, now,
is just a b c d and I'm looking for the real
parts of both eigenvalues to be negative. Okay. What -- how can I tell
from looking at the matrix, without computing
its eigenvalues, whether the two
eigenvalues are negative, or at least their real
parts are negative? What would that tell
me about the trace? So -- so the trace -- that's this a plus d -- what can you tell me about
the trace in the case of a two by two stable matrix? That means the eigenvalues
have -- are negative, or at least the real parts of
those eigenvalues are negative -- then, when I take the -- when
I look at the matrix and find its trace, what -- what
do I know about that? It's negative, right. This is the sum of
the -- this equals -- this equals lambda one plus
lambda two, so it's negative. The two eigenvalues, by
the way, will have -- if they're complex -- will have
plus six i and minus six i. The complex parts will -- will
be conjugates of each other and disappear when we add and
the trace will be negative. Okay, the trace
has to be negative. Is that enough -- is a negative trace enough
to make the matrix stable? Shouldn't be enough, right? Can I -- can you make -- what's
a matrix that has a negative trace but still it's not stable? So it -- it has a blow -- it
still has a blow-up factor and a -- and a --
and a decaying one. So what would be a -- so
just -- just to see -- maybe I just put that here. This -- now I'm looking for an
example where the trace could be negative but still blow up. Of course -- yeah,
let's just take one. Oh, look, let me -- let me make
it minus two zero zero one. Okay. There's a case where that
matrix has negative trace -- I know its
eigenvalues of course. They're minus two and
one and it blows up. It's got -- it's got a
plus one eigenvalue here, so there would be an e to
the plus t in the solution and it'll blow up if it has
any second component at all. I need another condition. And it's a condition
on the determinant. And what's that condition? If I know that the
two eigenvalues -- suppose I know they're
negative, both negative. What does that tell me
about the determinant? Let me ask again. If I know both the
eigenvalues are negative, then I know the
trace is negative but the determinant is
positive, because it's the product of the
two eigenvalues. So this determinant is
lambda one times lambda two. This is -- this is lambda
one times lambda two and if they're both negative,
the product is positive. So positive determinant,
negative trace. I can easily track down the --
this condition also for the -- if -- if there's an imaginary
part hanging around. Okay. So that's a -- like a
small but quite useful, because two by two is
always important -- comment on stability. Okay. So let's just look
at the picture again. Okay. The main part of my
lecture, the one thing you want to be able to,
like, just do automatically is this step of
solving the system. Find the eigenvalues,
find the eigenvectors, find the coefficients. And notice -- what's the matrix
-- in this linear system here, I can't help -- if I -- if I
have equations like that -- that's my equations
column at a time -- what's the matrix
form of that equation? So -- so this -- this
system of equations is -- is some matrix multiplying
c1, c2 to give u of zero. One zero. What's the matrix? Well, it's obviously
two one, one minus one, but we have a name, or
at least a letter -- actually a name for that matrix. Wh- what matrix
are we s- are we -- are we dealing with here in
this step of finding the c-s? Its letter is S -- it's the eigenvector matrix. Of course. These are the
eigenvectors, there in the columns of our matrix. So this is S c
equals u of zero -- is the -- that step where you
find the actual coefficients, you find out how much of
each pure exponential is in the solution. By getting it right at the
start, then it stays right forever. I -- you're seeing this
picture that each -- each pure exponential goes on
its own way once you start it from u of zero. So you start it by
figuring out how much each one is present in u of
zero and then off they go. Okay. So -- so that's a system of
two equations in two unknowns coupled -- the matrix sort of couples
u1 and u2 and the eigenvalues and eigenvectors uncouple
it, diagonalize it. Actually -- okay, now can I -- can I think in terms
of S and lambda? So I want to write
the solution down, again in terms of S and lambda. Okay. I'll do that on this far board. Okay. So I'm coming back to -- I'm coming back to our
equation du/dt equals Au. Now this matrix A couples them. The whole point of
eigenvectors is to uncouple. So one way to see that is
introduce set u equal A -- not A. It's S, the eigenvector
matrix that uncouples it. So I'm -- I'm taking this
equation as I'm given, coupled with -- with A has --
is probably full of non-zeroes, but I'm -- by uncoupling it,
I mean I'm diagonalizing it. If I can get a
diagonal matrix, I'm -- I'm in. Okay. So I plug that in. This is A S v. And this is S dv/dt. S is a constant. It's -- this it the
eigenvector matrix. This is the eigenvector matrix. Okay. Now I'm going to bring
S inverse over here. And what have I got? That combination S inverse A S
is lambda, the diagonal matrix. That's -- that's the
point, that in -- if I'm using the
eigenvectors as my basis, then my system of
equations is just diagonal. I -- each -- there's
no coupling anymore -- dv1/dt is lambda one v1. So let's just write that
down. dv1/ dt is lambda one v1 and so on for all
n of the equations. It's a system of equations
but they're not connected, so they're easy to solve
and why don't I just write down the solution? v -- well, v is now some
e to the lambda one t -- well, okay -- I guess my idea here now is to
use, the natural notation -- v of T is e to the
lambda tv of zero. And u of t will be Se to the
lambda t S inverse, u of zero. This is the -- this is the,
formula I'm headed for. This -- this matrix, S e
to the lambda t S inverse, that's my exponential. That's my e to the A t, is this
S e to the lambda t S inverse. So my -- my job really now is
to explain what's going on with this matrix up in
the exponential. What does that mean? What does it mean to
say e to a matrix? This ought to be easier because
this is e to a diagonal matrix, but still it's a matrix. What do we mean by e to the A t? Because really e to the
A t is my answer here. My -- my answer to
this equation is -- this u of t, this is my -- this
is my e to the A t u of zero. So it's -- my job is
really now to say what's -- what does that mean? What's the exponential
of a matrix and why is that formula correct? Okay. So I'll put that on
the board underneath. What's the exponential
of a matrix? Let me come back here. So I'm talking about
matrix exponentials. e to the At. Okay. How are we going to define
the exponential of a -- of something? The trick -- the idea is --
the thing to go back to is exponential -- there's a
power series for exponentials. That's how you -- you -- the
good way to define e to the x is the power series one plus
x plus one half x squared, one six x cubed and we'll
do it now when the -- when we have a matrix. So the one becomes the
identity, the x is At, the x squared is At squared
and I divide by two. The cube, the x cube
is At cubed over six, and what's the
general term in here? At to the n-th
power divided by -- and it goes on. But what do I divide by? So, you see the pattern
-- here I divided by one, here I divided by one by two by
six, those are the factorials. It's n factorial. That was, like, the one
beautiful Taylor series. The one beautiful Taylor series
-- well, there are two -- there are two beautiful
Taylor series in this world. The Taylor series
for e to the x is the s with x to the
n-th over n factorial. And all I'm doing is doing
the same thing for matrixes. The other beautiful
Taylor series is just the sum of x to the
n-th not divided by n factorial. Can you -- do you know
what function that one is? So if I take --
this is the series, all these sums are going
from zero to infinity. What's -- what
function have I got -- this is like a side comment -- this is one plus x plus x
squared plus x cubed plus x to the fourth not divided
by anything, what's -- what's that function? One plus x plus x squared plus
x cubed plus x fourth forever is one over one minus x. It's the geometric series, the
nicest power series of all. So, actually, of course, there
would be a similar thing here. If -- if I wanted, I minus
A t inverse would be -- now I've got matrixes. I've got matrixes everywhere,
but it's just like that series with -- and just like this
one without the divisions. It's I plus At plus At squared
plus At cubed and forever. So that's actually a --
a reasonable way to find the inverse of a matrix. If I chop it off -- well, it's reasonable
if t is small. If t is a small number, then -- then t squared is
extremely small, t cubed is even smaller,
so approximately that inverse is I plus At. I can keep more terms if I like. Do you see what I'm doing? I'm just saying we can do the
same thing for matrixes that we do for ordinary functions
and the good thing about the exponential
series -- so in a way, this series is
better than this one. Why? Because this one
always converges. I'm dividing by these
bigger and bigger numbers, so whatever matrix A and however
large t is, that series -- these terms go to zero. The series adds up to a finite
sum, e to the At is a -- is -- is completely defined. Whereas this second
guy could fail, right? If At is big -- somehow if At has an
eigenvalue larger than one, then when I square it it'll
have that eigenvalue squared, when I cube it the
eigenvalue will be cubed -- that series will blow up
unless the eigenvalues of At are smaller than one. So when the eigenvalues of
At are smaller than one -- so I'd better put that in. The -- all eigenvalues
of At below one -- then that series converges
and gives me the inverse. Okay. So this is the guy I'm chiefly
interested in, and I wanted to connect it to -- oh, okay. What's -- how do I -- how do
I get -- this is my, like, main thing now to do -- how do I get e to the At -- how do I see that e to the
At is the same as this? In other words, I can find e to
the At by finding S and lambda, because now e to the lambda t is easy. Lambda's a diagonal matrix
and we can write down either the lambda t -- and will
right -- in a minute. But how -- do you see what -- do you see that
we're hoping for a -- we're hoping that we can
compute either the A T from S and lambda -- and I have to look at that
definition and say, okay, how do -- how do I get an S and
the lambda to come out of that? Okay, can -- do you see how I -- I want to connect that to
that, from that definition. So let me erase this --
the geometric series, which isn't part of the
differential equations case and get the S and the
lambda into this picture. Oh, okay. Here we go. So identity is fine. Now -- all right, you --
you -- you'll see how I'm -- how I'm -- how I going to
get A replaced by S and S is in lambda's? Well I use the fundamental
formula of this whole chapter. A is S lambda S inverse
and then times t. That's At. Okay. What's A squared t? I can -- I've got
the divide by two, I've got the t squared
and I've got an A squared. All right, I -- so I've got
a -- there's A -- there's A. Now square it. So what happens
when I square it? We've seen that before. When I square it, I get S
lambda squared S inverse, right? When I square that thing,
the -- there's an S and a -- an S cancels out an S inverse. I'm left with the S
on the left, the S inverse on the right and
lambda squared in the middle. And so on. The next one'll be S
lambda cubed, S inverse -- times t cubed over
three factorial. And now -- what do I do now? I want to pull an S
out from everything. I want an S out of
the whole thing. Well, look, I'd better
write the identity how? I -- I want to be able to pull
an S out and an S inverse out from the other side, so I just
write the identity as S times S inverse. So I have an S there and
an S inverse from this side and what have I
got in the middle? If I pull out an S
and an S inverse, what have I got in the middle? I've got the
identity, a lambda t, a lambda squared t
squared over two -- I've got e to the lambda t. That's what's in the middle. That's my formula
for e to the At. Oh, now I have to ask you. Does this formula always work? This formula always works -- well, except it's
an infinite series. But what do I mean
by always work? And this one doesn't
always work and I just have to remind you
of what assumption is built into this
formula that's not built into the original. The assumption that A
can be diagonalized. You'll remember that
there are some small -- sm- some subset of
matrixes that don't have n independent
eigenvectors, so we don't have an S inverse
for those matrixes and the whole
diagonalization breaks down. We could still
make it triangular. I'll tell you about that. But diagonal we can't do for
those particular degenerate matrixes that don't have enough
independent eigenvectors. But otherwise, this is golden. Okay. So that's the formula --
that's the matrix exponential. Now it just remains for me to
say what is e to the lambda t? Can I just do that? Let me just put that
in the corner here. What is the exponential
of a diagonal matrix? Remember lambda is diagonal,
lambda one down to lambda n. What's the exponential
of that diagonal matrix? Because our whole point is
that this ought to be simple. Our whole point is that to take
the exponential of a diagonal matrix ought to be
completely decoupled -- it ought to be diagonal,
in other words, and it is. It's just e to the lambda
one t, e to the lambda two t, e to the lambda n t, all zeroes. So -- so if we have a diagonal
matrix and I plug it into this exponential formula,
everything's diagonal and the diagonal terms are
just the ordinary scaler exponentials e to
the lambda one t. Okay, so that's -- that's -- in a sense, I'm doing here, on
this board, with -- with, like, formulas what I did on the -- in the first half of the
lecture with specific matrix A and specific eigenvalues
and eigenvectors. The -- so let me show
you the formulas again. But the -- so you -- I guess -- what should
you know from this lecture? You should know what this
matrix exponential is and, like, when does it go to zero? Tell me again, now,
the answer to that. When does e to
the At approach -- get smaller and
smaller as t increases? Well, the S and the S
inverse aren't moving. It's this one that has to
get smaller and smaller and that one has this
simple diagonal form. And it goes to zero provided
every one of these lambdas -- I -- I need to have each one
of these guys go to zero, so I need every real part of
every eigenvalue negative. Right? If the real part is
negative, that's -- that takes the exponential
-- that forces -- the exponential goes to zero. Okay, so that -- that's
really the difference. If I can just draw the -- here's
a picture of the -- of the -- this is the complex plain. Here's the real axis and
here's the imaginary axis. And where do the
eigenvalues have to be for stability in
differential equations? They have to be over here,
in the left half plain. So the left half plain is this
plain, real part of lambda, less than zero. Where do the ma- where
do the eigenvalues have to be for powers of the
matrix to go to zero? Powers of the matrix go to zero
if the eigenvalues are in here. So this is the stability
region for powers -- this is the region -- absolute
value of lambda, less than one. That's the stability for -- that
tells us that the powers of A go to zero, this tells us
that the exponential of A goes to zero. Okay. One final example. Let me just write down how
to deal with a final example. Let's see. So my final example will be a
single equation, u''+bu'+Ku=0. One equation, second order. How do I -- and maybe I should have used -- I'll use -- I prefer
to use y here, because that's what you see
in differential equations. And I want u to be a vector. So how do I change one second
order equation into a two by two first order system? Just the way I
did for Fibonacci. I'll let u be y prime and y. What I'm going to do is I'm
going to add an extra equation, y prime equals y prime. So I take this -- so by -- so using this as
the vector unknown, now my equation is u prime. My first order
system is u prime, that'll be y double prime y
prime, the derivative of u, okay, now the differential
equation is telling me that y double prime is m- so
I'm just looking for -- what's this matrix? y prime y. I'm looking for the matrix A. What's the matrix in case I have
a single first order equation and I want to make it
into a two by two system? Okay, simple. The first row of the matrix
is given by the equation. So y''-by'-ky -- no problem. And what's the second
row on the matrix? Then we're done. The second row of the
matrix I want just to be the trivial equation
y prime equals y prime, so I need a one
and a zero there. So matrixes like these,
the gen- the general case, if I had a five by five -- if
I had a fifth order equation and I wanted a five
by five matrix, I would see the coefficients of
the equation up there and then my four trivial equations
would put ones here. This is the kind of matrix
that goes from a fifth order to a five by five first order. So the -- and the eigenvalues
will come out in a natural way connected to the differential equation. Okay, that's
differential equations. The -- a parallel lecture
compared to powers of a matrix we can now do exponentials. Thanks.
