John Finn's Comments for Exam 2
Winter 1998
Comment 1
Exam 2 is next Thursday. I already heard Erik say something about having it be
more conceptual than computational, and I give a hearty second to that.
I'm going to give you some hints and comments for the exam. here's the
first.
In class today we worked out
x' = Ax
by saying the answer should be
x = exp(At)*c
and then actually working out what exp(At) is by just eriting down exp(Dt),
where D is the diagonalization of A, and conjugating it by T (or by T-inverse;
I never can remember whether y conjugated by x is x y x-inv or x-inv y x):
exp(At) = T exp(Dt) T-inverse
This works because A = T D T-inverse, and each term of the sum for the
exponential inherits this same conugation.
This conjugation business makes sense: T is the matrix whose columns are the
eigenvectors, call them u and r (so that we can avoid all those *&^*&^
subscripts that the book is so *&^*&6 fond of; Euler would puke). So if we do
T to (10) and (01) we get u and r. If we take u and r as our basis vectors,
and write [a,b] to mean au + br, then u and r are [10] and [01], so what T does
is
T changes a vector in u-r-terms to one in regular terms.
If we write the u-r-vectors with square brackets, then T changes []'s to ()'s.
Now there's only one linear mapping, L, here, but we multiply by A to get it
if we're dealing in () terms, and by D of we're using []'s. So the columns of
D are lambda and mu, since Lu = lambda u and Lr = mu r. Is it A = TDT-inv or
the other way around? Use horse sense to answer that question: A acts on ()'s
and D on []'s, so to use D we first have to translate ()'s to []'s, and that's
what T-inv does. So
T D T-inv x takes an x that's a (),
changes it to its [] form (T-inv x)
mutilplies that by D (D * T-inv x)
and changes the reult back to () form (T * D * T-inv x)
and this is the same as we get by just taking A times x, so
A = T D T-inverse
Solving this for D gives
D = T-inv A T,
which also makes sense. The LHS acts on []'s, hence so must the RHS, and it
takes a [] and changes it to a ()
multiplies by A
converts the resulting () back to a []
The result we got from explicitly figuring out x = exp(At)*c wasn't all that
pretty, and was a combo of fcns that were themselves combos of exp(t) and
exp(-3t), so it's easier just to use those as the basis for our soultion space.
But the point is that you _can_ write down the answer as simply
x = exp(At)*c,
which is exactly the same as in the single-function case. And _that's_ one
thing you want to understand. The other really important thing is the
_picture_, and the role in it of the eigenstuff. This is example 1, page 371,
by the way.
Comment 2
I suggest that if you, like me, try to work out examples and write things
down when reading the book, that you
---------------------------------------
change their notation to something reasonable
--------------------------------------
They put superscripts in ()'s on vectors, and then label the coordiantes of
them bass-ackwards. What you can do is this:
* use an arrow over x, y, etc. instead of boldface.
* indicate n vectors just by writing x, y, ...,z and saying that there are n of
them.
* if a vector is named whatever, then its coordiates are usually called
whatever_1, whatever_2, etc. So if you do want to number your vectors, call
them x_1, x_2, etc, and then their coordinates are
for x_1: x_11, x_12, x_13 etc.
for x_2: x_21, x_22, x_23 etc.
* There's no need to resort to greek letters when you haven't even used more
than a couple of English ones. The book starts using the Greek xi, which is
the hardest Greek letter to write (and to say). I just call that thing b, and
make life _so_ much easier.
* Look at problem 7.4 #2. What they have written there looks more
complicated than quantum mechanics. But if you call the two function
X = (x,y) and R = (r,s), then the Wronskian is just det of
x r
y s
(notice that this wronkian is _different_ from the earlier one, whose rows were
successive derivatives of a function; nice of them, eh?). So
W = xs - yr
and
W' = x's + xs' - (r'y + yr')
Now you have to plug in what x', y', r', s' are, but they're the matrix P times
(x,y) amd (r,s). And you can write P as
P =
a b
c d
if you like, provided you bear in mind that what you;re trying to show is that
you get the Trace of P (the sum f its diagonal elements) times :
W' = TrP * W = (a+d)W.
If you do it this way, you might actually see some meaning in what you do.
Using their horrid notation is not going to mean _anything_, except a bunch of
complicated symbols and a serious loss of ink or pencil lead.
Comment 3
I already mentioned the system x' = Ax where
A =
c -s
s c
where c and s are sin and cos of an angle theta (which I type as "Q").
Actually, this is infinitely many systems, one for each choice of Q.
Anylizing these systems, figuring out hwat the eigen stuff is, and what the
_picture_ is, for different values of Q is an excellent exercise for preparing
yourself for the exam. What do the whiskers look like? What do the solution
curves look like? Where in the picture are the eigenlines? A rotation ain't
got no eigenlines -- obviously. Well...except for a couple of special cases.
What does it mean about the eigenvalues if there are no eigenlines? What are
the special cases? Etc., etc.
Comment 4
The 2nd Math 23 Hour Exam covers material in
* Chapter 3: Second Order Linear Equations
* Chapter 7: Systems of First Order Linear Equations
* Chapter 9, section 9,1: The Phase Plane for Linear Systems
(And we are asking _you_ to read and understand 9.1 on your own).
Am I going to give you an outline here of the material? No; I'm going to
suggest that _you_ write your own outline. But I do have a few suggestions.
There are two aspects to all this material; the computational, and the
theoretical. The subject of variation of parameters illustrates well the
interplay between the two, and I append below some comments about variation of
parameters. Actually, I come up with a theoretical question about variation of
parameters that would make an excellent extra-credit, or alternate exam
problem.
The Theoretical
---------------
One reason we're asking you to read 9.1 on your own is to see just how much
of the theoretical material you've managed to absorb. And the nice thing about
both Second Order Linear Equations (or even higher order ones, if they have
constant coefficients) and Systems of First Order Linear Equations is that much
of the theory is pretty simple, and quite accessible to math 23 students. And
the theory _does_ help you with the practical problem of coming up with
solutions. If you have a system
x' = Ax,
then if you understand the role of A's eigenvalues and their eigenlines, you
can compute them and then
===============================================================
immediately make a good qualitative sketch of the phase space
---------------------------------------------------------------
This gives you right away information about the nature of the solutions. If A
has complex eigenvalues, then you know it has no real eigenlines, and you may
well find that trajectories in the phase space go round and round, indicating
that sine and cosine are involved in the solutions, for example.
So that's one chunk of theory you'll do well to have under your belt, and
exactly what 9.1 talks about; what the phase space looks like, depending on
the nature of the eigenvalues (real distinct, complex, or real repeated).
Another chunk of theory is
===============================================================
some basic facts about linear functions (operators)
between vector spaces.
---------------------------------------------------------------
A _lot_ of what's going on here doesn't depend at all on the fact that we
happen to be looking at differential operators between function spaces. If
L: V --> W
is _any_ linear map between vector spaces V and W, then both the null space and
the range of L are subspaces. If we somehow know the dimension of the null
space of L is n, and we have n vectors in that null space, then they're a
basis of the null space if they're independent. In the context of differential
operators we may choose to call them a "fundamental set of solutions to the
homogeneous equation", but that means nothing more or less than that they're a
basis of the null space of L.
And if L has a 2-dimensional null space, that is, a null space that's a plane
P through the origin, then
(*) LV = w
also has a plane of solutions; in fact, the plane P + v_p, where v_p is any
particular solution to (*). The plane P + v_p is parallel to P, and if the
parametric representation of P is
v = c_1 v_1 + c_1 v_2,
then representation of P + v_p, the plane of solutions to (*) is
v = c_1 v_1 + c_1 v_2 + v_p;
and all of this is true _in general_ if L:V --> W is a linear mapping.
When you start to understand that, even in the context of differential
equations, we're constantly just trying to see whether a few vectors are
independent, you'll realize that for 2 vectors the answer is particularly
simple: v_)1 and v_2 are independent if and only if neither is a multiple of
the other. If the vectors v_1 and v_2 happen to be functions ---and you should
understand that functions _are_ vectors, not only because linear combos of them
are still functions, but also because, like the finite-dimensional vectors
you're more used to thinking of as such, functions are "lists" of values, but
"lists" of a whole continuum of values, since functions are
infinite-dimensional, in fact uncountably-infinite-dimensional vectors, whose
coordinates are every point x in some interval-- that doesn't matter at all.
Functions or not, 2 vectors are independent iff they're not multiples of one
another. Suppose then, that these 2 vectors are e^3x and e^4x. Are they
independent? Sure they are, because e^3x is not some constant times e^4x.
You ain't gotta compute no Wrong Skin to see that!
Do you understand the simple facts of linear maps between vector spaces
that underlie all this differential equation stuff? If not, you're making life
_much_ harder for yourself. And these same facts come up again in our next
subject, Laplace transforms, and again in Fourier transforms, where the dot
product, and understanding that 2 vectors are perpendicular, or orthogonal iff
v dot w = 0, is of central importance.
Do you understand why the EIGENvectors of a matrix are its OWN favorite
vectors? Do you understand that x' = Ax means that we can attach the velocity
vector x' = Ax to each point x in the plane, say, and get a direction field
that shows where x is going (and how fast, if we include the length of x', but
we usually don't do that, since it crowds our picture too much).
Have you played with the Direction Fields program, so that you have a good
intuitive picture of the _dynamic_ significance of the eigenlines of A in terms
of trajectories?
If so, then you'll have a good Big Picture of all this stuff, and you'll be
able to know a lot about the nature of the solutions to a differential
equations, or a system of such equations _before_ you start grinding out these
solutions.
If not -- ouch! All this must seem a bunch of Egyptian hieroglyphics to
memorize.
The Computational
-----------------
Fortunately, much of the computation in chapters 3 and 7 is easy. If we have
an n-th order constant-coefficient homogeneous differential equation, like
Ly = (D-1)(D+1)(D-2)(D-3)(D-4)y = 0,
then we can immediately write down a set of n independent solutions, provided
we can factor L (or, if you prefer, factor its characteristic polynomial).
To solve a non-homogeneous equation
Ly = [D^2 + pD + q]y = g,
if we already have a couple of independent homogeneous solutions, is a just a
matter of finding a _single_ particular solution. And how can we do this? In
many cases, by making EDUCATED GUESSES, and using trial and error. That's
certainly what undetermined coefficients is all about. And to do undetermined
coefficients you don't have to adhere to any strict sequence of steps, as our
text may seem to suggest. Suppose, for instance, we have
Ly = (D-1)(D-2)y = e^3t.
Here we know the basic homogeneous solutions are e^t and e^2t. What kind of
function has derivatives that look like e^3t? Obviously e^3t itself. So plug
y = e^3t into Ly, see what you get, and make the necessary adjustments.
If we have
Ly = (D-1)(D-2)y = e^2t,
then e^2t isn't going to work. There's no problem, though, if we forget that,
for we'll be reminded of it if we try e^2t, and plug it into L, since we'll get
0. OK, what else can we try? You don't need to memorize any list or anything
to know that y = te^2t is worth a try -- so try it!
Later, when we're doing systems, we may come up with a matrix that has 2 as
a double eigenvalue. We have
y = b e^2t
as one solution of our system [Note: Here b is an eigenvector. Notice how
much easier it is to call it plain old "b". instead of using the hardest to
write and say letter of the greek alphabet. Why in the world, if we haven't
used any letters from the English alphabet except x,t,e,and r, do we suddenly
go running off to the Greek alphabet to look for names for things? And why use
subscripts? I could understand resorting to Greek or subscripts if we've used
up _lots_ of ordinary English letters, but otherwise, why not keep things
simple?]
What's a good guess for another solution? Surely y = c te^2t seems a good
place to start. If we try this, though, we find that it doesn't quite work.
But the trial and error will show us that what we need is to include some e^2t
stuff in addition to te^2t; so we try
y = b te^2t + c e^2t,
and this does work, and we can figure out what to make b and e by simply
plugging this in and seeing what adjustments need to be made.
Getting eigenvalues for 2x2 matrices is a matter of solving quadratic
equations, as is finding basic solutions to a 2nd order equation, and that's
stuff we've known since high school.
Where _is_ there any hard computation in chapters 3 and 7? Pretty much only
in variation of parameters, and there it's just a matter of having a couple of
integrals to compute; it's a matter of Math 8 computation, not anything new to
math 23. Still, you probably want to go over the HW problems we assigned on
variation of parameters, and do some nearby problems, just to refresh your
techniques of antidifferentiation skills.
What about determinants? They can certainly be a pain in the butt if
they're dets of 3x3 or 4x4 matrices, but all the ones we deal with here are 2x2
- no problem.
How about diagonalizing a matrix A? All that
A = T D T-inverse
stuff may _look_ complicated, but if you know what's going on it's not hard at
all. Here T has as columns the eigenvectors u and v -- T takes a vector in u,v
notation and changes it back to regular notation (which is easy to see, since
T[1,0] = u and T[0,1] = v, and [1,0] and [0,1] are exactly u and v in u.v
notation. T-inverse does the opposite; it changes a vector in regular
notation to u,v notation. And it's not hard to compute the inverse of a 2x2
matrix. You can check that its right by checking that T-inv u = [1,0] and
T-inv v = [0,1]. And what's D? It's the diagonal matrix that A is if we take
as a basis for R^2 A's OWN (= EIGEN, in German) favorite vectors as a basis.
And since the columns of a matrix just say what it does to the basis vectors,
and since we've _chosen_ u and v because Au = ru and Av = sv, we see that D is
the diagonal matrix with the eigenvalues on the diagonal.
Even raising e to the At isn't a hard computation. For a diagonal matrix
with r,s,... on the diagonal, e^Dt is just the diagonal matrix with e^rt,
e^st,... on the diagonal. And if A = T D T-inverse, then it's easy to see that
e^At = T e^Dt T-inverse.
(But since we haven't asked you to do any of this e^At stuff, you don't need to
worry about it anyhow).
One of the most important things that you _should_ be doing to check your
routine computations of integrals, matrix inverses, etc. is to
USE MAPLE
In real life, is you were a mathematician, physicist, or engineer working with
these diffeq's you'd be using Maple or mathematica to so _all_ your routine
computations. You won't be able to use Maple on the exam, of course, but I
think using it _improves_ your computational skills. In fact, you'll probably
get more out of this course in the long run if you learn to use Maple and
forget every bit of diffeq stuff. A lot of the computations that college math
courses still require you do to are really quite obsolete. It's absolutely
essential that nay working applied mathematician, physicist, engineer -- or for
that matter biologist, chemist, etc., etc. learn to use Maple or Mathematica or
some such tool.
Conclusion
----------
Write _yourself_ an outline of Chapters 3 and 7.
See that you have a good grasp of the simple vector space facts underlying
everything, and good _pictures_ of what's going on. For the 2x2 systems your
pictures should be moving ones, and you can use the program Direction Fields to
check your skills at dynamic pictures of the direction field of a system x' =
Ax.
For the theory, strive to get the Big Picture; to see the forest and not get
lost in the trees.
As for computations, make sure you do them neatly, and in such a way as to
make clear what you're doing. That's how you get partial credit, remember.
_Check_ your computations with Maple , and check that your solutions to a
diffeq make sense by plugging them back into the equation.
DRAW PICTURES of the graphs of your solutions. If a function comes in
pieces, like y = 1/x, for instance, which is defined on the 2 pieces
(-infinity, 0) and (0, infinity), make sure you choose the right piece for your
solution to an initial value problem.
That wraps it up, I think. Follow these suggestions and you'll be much
happier with your performance on the exam.
-----------------------------------------------------------------------------
Variation of Parameters: theoretical and the computational considerations
--------------------------------------------------------------------------
The theoretical content of variation of parameters is that we have a
non-homogeneous equation (2nd order, in this course) whose coefficients need
not be constant, and we assume we have two independent homogeneous solutions.
Thus, we have
(1) Ly = [D^2 + p(t)D + q(t)]y = g(t),
with x(t) and z(t) independent solutions to
Ly = 0.
(I'm calling these solutions x and z to avoid subscripts; I find everything
_much_ clearer when I avoid subscripts, and it's no problem whatever to avoid
them here. This is a good general tip for solving math problems: use your own
notation whenever you think that will make things clearer)
To find a solution y to the non-homogeneous equation, we that y is a linear
combination of x and z, but with coefficients that are functions of t; that
is, with parameters that vary with t; hence the name variation of parameters.
Assuming then that y = ux + vz, we being by writing down y', and then
===========================================================
** we assume that the u' and v' terms in y' add to 0; **
i.e., that
(2) u'x + v'z = 0
-----------------------------------------------------------
Understanding that we make this assumption is crucial to the theory here. Not
only does it considerably simplify y', and hence y'', but (2) comes back into
the argument later, and it's crucial to understand _that_.
We compute y'' from the simplified version of y', and now we plug y into (1),
and
====================================================================
** factor u and v out of the terms that they occur in **
---------------------------------------------------------------------
This observations, also, is crucial to understanding what's going on. For when
we do this, what we get is
(3') Ly = u[Lx] + v[Lz] + u'x' + v'z' = g,
and since x and z are homogeneous solutions, Lx = Lz = 0, so (3') simplifies to
====================================================================
(3) u'x' + v'z' = g
---------------------------------------------------------------------
Now we make the crucial observation that
====================================================================
(2) and (3) make a system of 2 equations in u' and v'
---------------------------------------------------------------------
This is precisely _why_ we made the assumption that led to (2). We solve
this system, and get u' and v', and the rest is straightforward: we "just"
integrate to get u and v, and then plug these values into y = ux + vz to get y.
We can write the solutions to u' and v' in terms of the Wronskian W(x,z),
and can thus write an explicit formula for y = ux + vz, in which u and v are
each an integral of an expression involving W(x,z).
Do I know exactly what this formula is? Nope; don't care. I can easily
enough go through the steps and come up with it myself, and so I am content
with a view of the _forest_, at least for my theoretical understanding of
what's going on, and not concerned with the trees. If I _do_ look at the
details, though, I notice an interesting fact:
====================================================================
The function g(t) is part of the formula for the solution;
the functions p(t) and q(t) are not.
---------------------------------------------------------------------
Why is this so? Doesn't it seem odd that we get the same answer for u and v
regardless of what p and q are?
In fact, if I were going to ask a question on the exam about the
_theoretical_ content of variation of parameters, that would be part of my
question, which would be something like this:
================================================================================
Problem: Derive the solution y we get to
Ly = [D^2 + pD + q]y = g
from the assumptions that
(a) y = ux + vz, where x and z are homogeneous solutions, and that
(b) u'x + v'z = 0.
Write the solution in terms of integrals, and explain why g(t) does appear in
this solution, but p(t) and q(t) do not.
--------------------------------------------------------------------------------
-
I wouldn't, of course, have you come in Thursday evening and find this
problem staring you in the face. Understanding why g does appear but p and q
do not is something that takes time to think about; probably an hour, at
least, of going through the steps that arrive at our solution, and trying to
see why it is that p and q conveniently disappear. What if we don't make
assumption (2)? Everything becomes mush messier, of course, but is it really
this assumptions that makes p and q disappear?
Maybe it doesn't take a full hour of thought, after all. I just came up
with this question, and now already see the answer. Of course, I've been doing
math for 25 years or so, and this is exactly the kind of question i ask myself
over and over when doing it, so it's not surprising that I should be able to
see where p and q vanish more quickly than you.
I'm not going to tell you the answer, though, because it _is_ one that you
_can_ figure out yourself. And I very firmly believe that when you get in the
habit of asking _yourself_ such questions, and finding their answers, doing
mathematics becomes a whole nother ball game for you, and one in which you feel
much more in control.
Maybe I'll let this question be a sort of extra-credit, or alternate, exam
question.
Now -- what about the _computational_ point of view? From this perspective,
of course, it's these integrals involving x and z ---and g-- and the Wronskian
W(x,z), that we actually are going to have to tackle to arrive at a
satisfactory solution. If I were a student facing the exam, I might even take
the time to memorize the integrals, but _only_ if I were fully confident that I
don't _have_ to rely on having memorized them; that I can myself derive them
in a few minutes.
If I actually encountered such a problem in my work, I'd use Maple of
Mathematica to carry out the routine computation of the integrals. And even on
an exam, I might, instead of filling a page computing the integrals, instead
show their derivation and then write down the solution in terms of the
integrals and say:
This _is_ the solution. I choose to leave it in this form, which
is a perfectly explicit mathematical expression of the solution.
That's because I'm lousy at doing computations; if you ask me to add up the
items in a grocery bill by hand, there's no way I'm going to get it right the
first time. In fact, there probably won't even _be_ a first time; I'll
probably just flatly refuse to add a list of some 50 or 60 numbers like 3.45
and 15.98. Hey -- sorry; I'll just take a 0.
But if on an exam I have a variation of parameters problem, and choose
instead of filling the page with computations of the integrals to derive them,
and leave it at that, then certainly I can expect partial credit for the
problem.
And this sort of thing may actually happen to you; you wind up with some
nasty integrals and simply don't have time to compute them. In that case,
================================================================================
maximize your eligibility for partial credit
by making it clear that you know what's going on
--------------------------------------------------------------------------------
If you have a variation of parameters problem, and just write down from memory
the integrals, and then don't have time to work them out, you can't really
expect much partial credit; after all, memorizing the integrals doesn't in
itself show that you have any idea whatever of what's going on. But if you
find yourself without time to compute the integrals, there _are_ things you can
do to maximize your eligibility for partial credit.
-John