Is this class meticulously prepared and impeccably presented?  

/meticulous/ 1. giving great or excessive attention to details. 2. very careful and precise. 
/impeccable/ 1. (of behavior, performance, etc.) faultless, exemplary. 2. [Theol]not liable to sin.  The Whitelipped Peccary 
What happened
I asserted that Mr. Heaviside, one hero of this course, wanted delta(t)
as a model for an instantaneous impulse at t. Therefore
Behavior of the delta "function" 


We can translate the delta function, just as we translated the Heaviside function. What does delta(t5) do? The center of interest is then where t5 is 0, or where t=5. That means, for example, that _{infty}^{infty}delta(t5)·g(t) dt must be g(5) if g is continuous at 5: this delta function is concentrated at 5.
Asymptotics again, and back to delta
I went back and discussed again asymptotics for Laplace transforms. If
f(t) is a function as shown, then the Laplace transform, F(s), is
defined by
_{0}^{infinity}e^{st}f(t) dt. f(t) is
real function, a real bump that we are looking at.
What happens to F(s) as s gets Large
In the integral, t is a positive number (o.k., I am neglecting t=0 but
in the integral there is very little area under t=0). Now
e^{st} for t positive and s LARGE positive will be very
small. Remember, in Mr. Heaviside's philosophy, "Almost every
reasonable limit should exist." Well, here since e^{st}>0
as s>infinity, I bet
Asymptotic fact I
As s>infinity,
_{0}^{infinity}e^{st}f(t) dt>0 That
is, lim_{s>+infinity}F(s)=0.
What happens to F(s) as s gets close to 0
Now look at what happens as t>0^{+}. Here
e^{st}>1, and the integral seems to get closer to
_{0}^{infinity}1f(t) dt. Therefore
Asymptotic fact II
As s>0^{+},
_{0}^{infinity}e^{st}f(t) dt>the
net area "under" the bump (as usual, area under the horizontal axis is
counted negatively). So
lim_{s>0+}F(s)=Net area under
y=f(t) from t=0 to infinity.
Therefore the delta "function" is?
The Laplace transform of delta is just 1. I mean, the Laplace
transform is the function
1 which is 1 for all s. Look: 1 does not
go to 0 as s>infinity. I guess that the delta "function" is
not a standard function. It is something else. Maybe this explains why
academic mathematicians found much of Heaviside's work unappealing,
but the success of his methods in getting correct and useful answers led to the use of his
methods by many many engineers. For much of the time, it is possible
to get the answers with other methods (sometimes awkwardly), but
Laplace transform is fast and seems simple. Here is an example.
Hitting a spring
Let's look at y´´+y=delta(tPi)
with initial conditions y(0)=0
and y´(0)=0. O.k.: to me, what I hope is interesting for the
students in this course is the physical interpretation of
everything. Then we should "solve" it, and then we should consider the
solution from the physical point of view: if it makes sense, this
should help validate the method we used.
Well: y´´+y models a vibrating spring with no damping, a rather ideal situation. The righthand side, delta(tPi) is a "forcing function". In fact, here it correspods to hitting the spring instantaneously (!) force of 1 at time Pi. The initial conditions tell us that the spring is originally in equilibrium. At time Pi, the spring is hit but there is no further force (so far). Probably we should expect that the spring will somehow vibrate in an ideal fashion. Let us try our wonderful Laplace routines. The Laplace transform of y´´+y=delta(tPi) is s^{2}Y(s)sy´(0)y(0)+Y(s)=e^{Pi·s} which turns out to be (s^{2}+1)Y(s)=e^{Pi·s} or Y(s)=e^{Pi·s}/(s^{2}+1). The predicted vibration of the spring is therefore (!) the inverse Laplace transform of e^{Pi·s}/(s^{2}+1). I can read that off from the (mythical?) table and the result is y(t)=sin(tPi)U(tPi): the spring, responding to an instantaneous load of 1 unit, then vibrates sinusiodally.
Hitting a spring again
Keep the initial conditions the same, but change the forcing function.
So here look at
y´´+y=delta(tPi)+delta(t3Pi). Now go through the
Laplace transform, using the initial conditions:
s^{2}Y(s)sy´(0)y(0)+Y(s)=e^{Pi·s}+e^{3Pi·s}
becomes
s^{2}Y(s)+Y(s)=e^{Pi·s}+e^{3Pi·s}
and then
(s^{2}+1)Y(s)=e^{Pi·s}+e^{3Pi·s}
so that
Y(s)=[e^{Pi·s}+e^{3Pi·s}]/(s^{2}+1).
The inverse Laplace transform tells me that
y(t)=sin(tPi)U(tPi)+sin(t3Pi)U(t3Pi).
It may be useful to understand what this algebraic mess means.
And even more resonance
What if the driving force hits every 2Pi after t=Pi? The mathematical
model is
y´´+Ky´+y=SUM_{j=0}^{infinity}delta(t(2j+1))
with,
again, y(0)=0 and y´(0)=0. Let us analyze this as we've already
done. Of course, we could worry about the infinite sum, which
is sort of a limit, but "Almost every reasonable limit should exist."
I won't worry.
Exactly as before, the Laplace transform is
Y(s)=SUM_{j=0}^{infinity}[e^{(2j+1)Pi·s}]/[s^{2}+1].
This Laplace transform already has built into it the initial
conditions and the "rough" inhomogeneity of the forcing term.
The inverse Laplace transform is
y(t)=SUM_{j=0}^{infinity}sin(t(2j+1)Pi)U(t(2j+1)Pi).
Again please remember that sine is periodic with period 2Pi. Therefore
sin(t(2j+1)Pi)=sin(tPi), and
y(t)=SUM_{j=0}^{infinity}sin(tPi)U(t(2j+1)Pi)=sin(tPi)SUM_{j=0}^{infinity}U(t(2j+1)Pi).
What's left in the sum is not a constant function, but it is a
staircase function (the graph looks more or less like 4.3, #62, whose
Laplace transform you were asked for in the last homework assignment
[!]). What about the motion? Is it smooth? Where is it smooth? The
graph is a bit difficult to consider for t large, but if we ask
Maple to "zoom in" near a specific t (I think this is t=3Pi)
you can see a kink in the graph. At 3Pi, the slope changes from 1 to
2. And (3+2j)Pi, motion will be continuous (this spring doesn't fly
apart!) but the graph will not be differentiable: slope changes from j
to j+1.
This is a great vocabulary day:
kink
a. a short backward twist in wire or tubing etc. such as may cause
an obstruction.
b. a tight wave in human or animal hair.
And how about halfway
What if we kick the spring at Pi and at every Pi afterwards? I
asked people to think about about what the solution should be
physically. I remarked that I was surprised by the solution (at least
at the beginning!) but that I hoped engineering students would be able
to "see" the solution. The mathematical model is
y´´+Ky´+y=SUM_{j=0}^{infinity}delta(t(j+1))with
y(0)=0 and y´(0)=0. So
Y(s)=SUM_{j=0}^{infinity}[e^{(j+1)Pi·s}]/[s^{2}+1]
and then
y(t)=SUM_{j=0}^{infinity}sin(t(j+1)Pi)U(t(2j+1)Pi).
Now I need to be a bit careful. While I know that sin(t+/2Pi)=sin(t),
sin(t+/Pi) is sin(t): sine flips when translated horizontally by
half its period. Therefore (watch carefully!)
y(t)=SUM_{j=0}^{infinity}sin(t(j+1)Pi)U(t(2j+1)Pi)=sin(t)SUM_{j=0}^{infinity}(1)^{j+1}U(t(2j+1)Pi).
What is the strange
SUM_{j=0}^{infinity}(1)^{j+1}U(t(2j+1)Pi)
do? In fact, the +/ alternately is 0 and 1. The result (after
multiplying by the sine function) is a graph which the text calls the
halfwave rectification of sine, and this appears in the
homework problems for section 4.3 also, I think. One of our enterprising
students declared that this has something to do with changing AC to DC
or vice versa.
is
And how about limiting the amplitude
Here is a special bonus problem for mechanical
engineering students, and any others who can grasp these concepts. Our
ideal spring y´´+y can have a damping term: y´´+Ky´+y for some real
constant K. This might correspond to the spring vibrating in honey or
10W30 motor oil or ... whatever. Suppose we hit the spring again every
2Pi. I know that the model described above is, indeed, just a
classroom model. It is hard to conceive of Hooke's law applying at the
spring stretches more and more and more (4 light years?). So the
design question: find a value of K so that the spring is limited in
motion (with, say, y(t)<10) for a limited time, say
0<t<200. You must give some supporting evidence that your
candidate for K actually restricts the motion in the required
fashion.
Please note that a math person might actually try to compute the exact
solution, and might then investigate the highest value of y(t),
etc. What could an engineer do? Well, maybe build a spring system
(difficult and perhaps expensive). Or find some other nearly
experimental way to verify that a conjectured value of K works.
Hint After class, I returned to my office, and guessed a value
of K, and got good enough verification within 45 seconds.
Delta as a derivative of U
Let's think this through: Look at U and ask yourself, what
should the derivative of look like, or, better, how should it
behave? Remember, "Almost every reasonable
function should have a derivative." So I know what the graph of
U is. Certainly, U´(t)=0 for t<0 and also for
t>0, since the graph looks locally like a horizontal line for those
t's. What happens at/near t=0? Well, U jumps by 1 in a very
short time, so it should be a rate of change of 1 unit in very very
little time. Uhhh ... what "things" do we know which behave like that?
Well, delta(t) does. So maybe U´(t)=delta(t).
Perhaps my (lack of) logic has not been very clear. Perhaps I have
been too romantic (!) about what a derivative is. Perhaps ... well, I
can identify delta best by its remarkable integral property.
Pedagogical note
In class I tried another approach. I approximated U(t) by
another function, W(t). For large negative t, W(t)=0. For positive t,
W(t)=1. In a small interval immediately to the left of 0, W(t)
increased smoothly from 0 to 1. You are supposed to think that W(t)
closely approximates U(t). Then W´(t) is a bump localized (?)
in that small interval to the left of 0. The total integral under the
bump is 1, using the Fundamental Theorem of Calculus since
_{10}^{37}W´(t) dt=W(t)]_{10}^{37}=W(37)W(10)=10=1.
If we multiply
W´(t) by a continuous function g(t), in the small interval (a
really small interval!) g(t) will hardly vary and will be
almost equal to g(0). So the integral of W´(t)g(t) will be about
1·g(0). Thus W´(t) is about delta(t), so, going
backwards and recognizing again that W(t) is almost U(t),
surely U´=delta.
I these ideas here because many people in the
engineering and applied science communicty (physics, chemistry) find
them useful. Maybe you will, also.
Linear systems
We concluded our study of Laplace transforms by covering section 4.6
in 4.6 seconds. The idea is to apply what we've done to solve some
fairly simple systems by
I tried a problem from the book. I think it was 4.6 #2. The problem is:
Note This problem
satisfies all standards for crueltyfree calculus problems. It
uses only small positive integers and uses only wellknown
techniques, and no ugly computations (more than two steps) are
required. Therefore this problem is approved as safe and appropriate for use in Math 421. 
2/3 Heaviside(t  3) exp(3/2 t  9/2) / 1 \ 1/2 exp(3/2 t  9/2)  1/2  \ exp(3/2 t  9/2)/ / 1 \ + 2/3 exp(3/2 t) 1/2 exp(3/2 t)  1/2  \ exp(3/2 t)/
HOMEWORK
Please complete reading chapter 4 of the text. Even a rapid reading of
the text is probably a good idea. You will get a different point of
view, one which you may find more agreeable, and you will see more
information than I can present in class.
Please hand in these problems on Tuesday:
4.4: 42 (high number
and even, so it may be difficult!) and 4.5: 3, 11, 13
(too many letters) and 4.6: 3, 7, 11.
Special bonus problem
If you hand this in, please do it on a separate piece of paper. Find a
constant K so that if y(0)=0 and y´(0)=0 and
y´´+Ky´+y=SUM_{j=0}^{infinity}delta(t(2j+1))
then for 0<t<200, I know y(t)<10.
Comment
If you choose to do this, you may use
any method you like. You must present some evidence that the K
you suggest is a valid answer, however.
TWO
Since we know that f(t)_{*}g(t) is
F(s)·G(s), and standard multiplication is commutative, we must
have
Convolution is commutative
f(t)_{*}g(t)=g(t)_{*}f(t)
I don't think this is totally obvious. If you want to write it out
using the definition of convolution, we get the more impressive
_{0}^{t}f(tau)g(ttau) dtau=_{0}^{t}g(tauf(ttau) dtau
Therefore notice that the arguments to the "factors" under the
integral must have sum equal to t (the outside variable), and that
there is a minus sign in one, and the integration variable (tau, the
inside variable) in the other. Sometimes this is helpful to me as I
try to compute convolutions.
THREE
Of course, since ordinary multiplication is also associative, so is
convolution (associativity is sometimes very useful, since it
says how products are grouped doesn't matter)
Convolution is associative
(
f(t)_{*}g(t))_{
*}h(t)=
f(t)_{*}
(
g(t)_{
*}h(t))
A silly application of Laplace transform
Here is a possible exam question:
What is the convolution of t and t^{2} and t^{3} and
t^{4}?
First notice that the question is meaningful (huh?). I mean that since
we know that convolution is associative the order we group
products does not matter (there are definitely mathematical structures
[one example is octonions] where what people call
multiplication is not associative).
How difficult is the proposed exam question? I claim it isn't very
difficult. Well, let me take that back. If one the definition of
convolution was used again and again, then the question might take
some time. But if we just recognize the use of Laplace transform:
t becomes 1/s^{2}
t^{2} becomes 2/s^{3}
t^{3} becomes 6/s^{4}
t^{4} becomes
24/s^{5}
and the product of the Laplace transforms is 288/s^{14}, whose
inverse Laplace transform is (288/13!)t^{13}, which is, of
course, the desired convolution. So this is not a very difficult exam
question. Please note that (288/13!) is 1/21,621,600 which I almost
would hate to read on an exam answer  I would not want
students to use time and effort resolving straightforward arithmetic.
But how about ...
In a similar spirit, what is the convolution of t and sin(t)? Now on
the Laplace transform side we need the inverse Laplace transform of
the product of 1/s^{2} and 1/(s^{2}+1). Well, we could
use partial fractions (I gave up in class, which was not good). So we
would need to solve
1/[s^{2}(s^{2}+1)]=A/s+B/s^{2}+(Cs+D)/(s^{2}+1)
The tops of the fractions (righthand side fraction combined) should
agreed, so
1=As(s^{2}+1)+B(s^{2}+1)+(Cs+D)s^{2}
Now s=0 gives us B=1. Comparing s^{2} coefficients, we get
0=B+D, so D=1. The s coefficients give 0=A, and then the
s^{3} coefficients give 0=A+C, so that C=0 also. Therefore we
need the inverse Laplace transform of
[1/s^{2}][1/(s^{2}+1)], and this I can "read off" the
table as tsin(t). I admit that computing the integral for the
convolution is not too difficult (one integration by parts), so
comparing the
efficiency of getting the answer both ways does not reveal an obvious
winner to me.
Heaviside
Oliver
Heaviside wrote:
Should I refuse a good dinner simply because I do not understand the
process of digestion?
Here he referred to his "operational calculus", which was a
nontraditional method of constructing mathematical models. The method
gave useful answers, but in many cases could not be rigorously
justified using techniques then accepted. Generally, I think Heaviside
would have agreed with the following statements:
Simple straight line motion
I tried to slide an eraser along the narrow shelf at the bottom of the
blackboard, asking people to imagine there was no friction. My
mathematical analysis (!) of this problem used F=ma, a wonderful
equation. Here we go: I will use F=ma. I will assume that we are
analyzing the motion of a particle on a line. At time t, the particle
is at x(t). I will also assume (since this is a very simple
model!) that m=1 for all time t. If the particle starts from 0 at time
0 with velocity 0, the initial conditions are x(0)=0 and
x´(0)=0. The force will vary with time. What can one expect of such a
problem?
Then setup
Here's an example of a possible force varying with time, t. Initially, the particle would not move. Then, yes, indeed!, it moves as t increase (once t gets to the region where F(t)>0). Since F(t)>0 there, the particle moves to the right, or (if we consider a graph of t, time, versus, x(t), position) "up". Here is what a graph of x(t0 might look like, qualitatively. What is happening? For early time, before the bump in F(t), the particle doesn't move at all. After the bump in F(t), the particle does move, and the amount of movement, it turns out, depends only on the total area under the bump. The total area determines the slope. The particle moves at uniform speed because there is no new force acting on it. In class, we guessed that the total area under the F(t) graph might be work, but this turns out not to be correct. Work is force·distance, but this is essentially force·time, which in this setup is momentum. (Why? Since x´´(t)=F(t), x´(t)=p(t), the momentum (remember here m=1 always). Thus p´(t)=F(t), and the total area under the F(t) is the change in momentum. I thank Professor M. Kiessling, a wonderful mathematical physicist, who helped (forced?) me to agree to this. In this simple setup, I think that work=energy and momentum gained are proportional. In any case, after the bump, the curve showing positive is linear, with a positive slope. 

The example
The setup Suppose the graph of F(t) is as shown. Let's find the motion of the point. Since x´´(t)=F(t) with x´(0)=0 and x(0)=0, and this is Math 421, we should use the Laplace transform. Well, we can write F(t) in a suitable form: F(t)=(1/A)U(t)(1/A)u(tA). Then take the Laplace transform of x´´(t)=F(t) and get, on the lefthand side (with initial conditions built in), s^{2}X(s)sx´(0)x(0). On the right, we will get (1/A)(e^{0·s}/s)(1/A)(e^{As}/s). Therefore we have s^{2}X(s)=(1/A)[1e^{As}] and X(s)=(1/A)1/s^{2}(1/A)e^{As}(1/s^{2}. We now take the inverse Laplace transform, and get x(t)=(1/A)(1/2)t^{2}(1/A)(1/2)(tA)^{2}U(tA). For 0<t<A, x(t)=(1/A)(1/2)t^{2}. Since (tA)^{2}=t^{2}2At+A^{2}, after some algebra (!!), we see that if t>A, x(t)=(1/A)(1/2)(2AtA^{2})=(1/2)(2tA). It can be checked that x(A)=(1/2)A for either definition, so the motion is continuous (the eraser didn't suddenly and magically jump!) and actually, the function described is differentiable: the tangent lines agree at the "splices" (t=0 and t=A). A graph of the position is shown.  
Pedagogical note
 
Limits of the motion and what we expect 
Dirac delta "function"
Dirac,
a Nobel prizewinning mathematical physicist, worked on quantum
mechanics and relativity. Here is an interesting quote from Dirac: "I
consider that I understand an equation when I can predict the
properties of its solutions, without actually solving it." A contemporary interview with Dirac may
give some idea of his personality.
The Dirac delta function, delta(t), is 0 away from 0. Its total integral is 1. You should think about it as an instantaneous bump, a limit of the boxes above as A>0. I computed the integral _{300}^{40}delta(t2) dt: this was 1. The integral _{300}^{200}delta(t2) dt was 0. I tried some further computations. I think one was something like _{300}^{40}delta(t2)(t^{2}5t) dt. Since the delta function is 0 away from 2, only what multiplies it at (or close to) 2 matters. But close to 2, the function t^{2}5t is close to 410=6. Therefore this integral is just 6.
A traditional mixing problem
I wanted to make the Chem E's happy. Here is a twopart problem which
I found last year. It is from an
essay by Kurt Bryan of the RoseHulman Institute of Technology:
A salt tank contains 100 liters of pure water at time t=0 when water begins flowing into the tank at 2 liters per second. The incoming liquid contains 1/2 kg of salt per liter. The well stirred liquid lows out of the tank at 2 liters per second. Model the situation with a first order ODE and find the amount of the salt in the tank at any time. 
I remarked that I expected students had modeled and solved such problems a number of times in various courses, and certainly in Math 244.
Let m(t) be the kgs of salt in the tank at time t in seconds. We are
given information about how y is changing. Indeed, m(t) is increasing
by (1/2)2 kg/sec (mixture coming in) and decreasing by (2/100)m(t),
the part of the salt in the tank at time t leaving each second. So we
have:
m´(t)=1(1/50)m(t) and m(0)=0 since there is initially no salt in the
tank.
Students seemed to realize that the solution curve should look like
this: a curve starting at the origin, concave
down, increasing, and asymptotic to m=50. In the long term, we expect
about 50 kgs of salt in the tank.
This is easy to solve by a variety of methods, but in 421 we should
use Laplace transforms: so let's look at the Laplace transform of the
equation. We get sM(s)m(0)=(1/s)(1/50)M(s). Use m(0)=0 and solve for
M(s). The result is M(s)=1/[s(s+(1/50))]. This splits by partial
fractions with some guesses to M(s)=50[(1/s)(1/(s+(1/50)))]. It is
easy to find the inverse Laplace transform and write
m(t)=50(1e^{(1/50)t}) which is certainly the expected
solution.
Slightly more interesting ...
Suppose that at time t=20 seconds 5 kgs of salt is instantaneously dropped into the tank. Modify the ODE from the previous part of the problem and solve it. Plot the solution to make sure it is sensible. 
Students should expect a jump in m(t)
at time 20 of 5, but then the solution should continue to go
asymptotically to 50 when t is large. How should the ODE be modified
to reflect the new chunk of salt? Here is one model:
m´(t)=1(1/50)m(t)+5delta(t20) and m(0)=0.
The delta function at t=20 represents the "instantaneous"
change in m(t) at time 20, an immediate impulsive (?) increase in the
amount of salt present.
It is very reasonable to ask if this is a good model. Please keep this is mind!
Back to solving m´(t)=1(1/50)m(t)+5delta(t20) and m(0)=0.
Take the Laplace transform as before, and use the initial condition
as before. The result is now
sM(s)m(0)=(1/s)(1/50)M(s)+5e^{20s} from which we get
M(s)=50[(1/s)(1/(s+(1/50)))]+5[e^{20s}/(s+(1/50))] which has
inverse Laplace transform
m(t)=50(1e^{(1/50)t})5U(t20)e^{(1/50)(t20)}. Of
course I wanted to check my answer, so I used Maple. Here is
the command line and here is Maple's response:
invlaplace(1/(s*(s+(1/50)))+5*exp(20*s)/(s+(1/50)),s,t);
100*exp(1/100*t)*sinh(1/100*t)+5*Heaviside(t20)*exp(1/50*t+2/5)
Well, Maple multiplied 1/50 by 20 to get
2/5. Slightly more interesting is the sinh (hyperbolic sine,
pronounced "cinch") part. Apparently users of Maple are
expected to know that sinh(w)=(e^{w}e^{w})/2. If you
know that, then you can see that Maple's answer is
equal to ours.
More about delta next time.
Left to right
Since e^{5t}tU(t2)
begins with an exponential, I'll
use #5, with a=5 and f(t)=tU(t2). Then we need to find the
Laplace transform of f(t). Here use #7, with a=2 and f(t)=t. We need
L(f(t+2))=L(t+2)=1/s^{2}+2/s
(using #0 and
#1). Now finish the use of #7: e^{2s}(1/s^{2}+2/s). So we
bump up or back (?) a level to #5, which advises us to change
every appearance of s to sa=s+5. Therefore the final result is
e^{2[s+5]}(1/[s+5])^{2}+2/[s+5]).
Right to left
Now e^{5t}tU(t2) "begins" (on the write) with a unit
jump. So I will start by using #7, with a=2 and
g(t)=e^{5t}t. We will need L(g(t+a))=L(g(t+2))=
L(e^{5[t+2]}[t+2]). But
e^{5[t+2]}[t+2]=e^{10}e^{5t}[t+2], and #5
can be used with a=5 here (I am saving e^{10} as a
multiplicative factor at the end). We need the Laplace transform of
t+2, which is 1/s^{2}+2/s, and then put in sa=s(5)=s+5 for
s, which gets us 1/[s+5]^{2}+2/[s+5]. Rolling back to #7, we
must multiply by e^{2s} and also not forget to multiply by
e^{10}. So the result is
e^{10}e^{2s}(1/[s+5]^{2}+2/[s+5]). Thank
goodness that this looks amazingly equal to the previous answer.
Now we progress into section 4.4 with a rather weird fact. Since F(s)=L(f)(s)=_{0}^{infinity}e^{st}f(t) dt we could try to differentiate F(s). Thus (d/ds)F(s)=(d/ds)_{0}^{infinity}e^{st}f(t) dt. Problem 3, especially 3c, was written to encourage you to believe that (d/ds) can be interchanged with the integral sign. In fact, in this case when f(t) is of exponential order, the interchanging is valid. Therefore (d/ds)F(s)=_{0}^{infinity}(d/ds)e^{st}f(t) dt. and (d/ds)e^{st}f(t)=te^{st}f(t) since d/ds only notices appearances of s. Now we have (d/ds)F(s)=_{0}^{infinity}te^{st}f(t) dt=_{0}^{infinity}e^{st}t·f(t) dt which is minus the Laplace transform of t·f(t).
Entry  Function  Laplace transform 

#268  t·f(t)  (d/ds)F(s) 
#269  (1)^{n}t^{n}f(t)  (d^{n}/ds^{n})F(s) 
The silly line #269 is verified by repeating #268 n times. I sincerely hope (and mostly believe) you won't need it very much. But let me show one "use" of #268.
Weird example
What function has Laplace transform equal to ln(s3)? Well, I do know
that (d/ds)ln(s3) is 1/(s3), and that's the Laplace transform of
e^{3t}. Therefore since (d/ds)F(s)=t·f(t), if
F(s)=ln(s3), then t·f(t)=e^{3t} so
f(t)=e^{3t}/t, and we have
Entry  Function  Laplace transform 

#503.2  e^{3t}/t  ln(s3) 
Computer algebra question (for students!)
Maple 7 was able to verify that the inverse Laplace transform
of ln(s3) is what's above, but it was not able to compute the Laplace
transform of the f(t) given. I hope that students will tell me what
Maple 9 and Matlab can do with this computation.
I just tried this on a version of Mathematica. That program
could do neither side of the computation!
Ms. Kadakia reported that
Matlab could not do the computation. Maple 9.5
(available on Eden) could do the inverse transform, but not
the direct Laplace transform. Weird!
QotD, part 1
What is the inverse Laplace transform of ln(s^{2}+5)? I urged
students to try to imitate what was just done.
Convolution, a really neat idea
First I repeated what is in the text:
Suppose that f(t) and g(t) are functions defined for nonnegative
numbers, so their domains include [0,infinity). Then the
convolution of f and g will be another function defined on
[0,infinity) defined by
f_{*}g(t)=_{0}^{t}f(tau)g(ttau) dtau.
Background
The letters t and tau are used everywhere. It is important to
keep them straight. One useful way of doing this is to remember that
the sum of the arguments of f and g is t, one has a minus sign, and
both have tau's.
Google gives me about 12,400 references for "convolution
chemical engineering" and 16,000 for "convolution mechanical engineering".
Convolution and $
Since what I know about real engineering problems could easily be lost
on the head of a pin, yet I would like to offer some insight into the
importance of convolution, I decided to discuss a use of convolution
involving money. We all remember something about compound interest,
with formulas learned and forgotten long ago. What follows may be
correct: if $100 earns 4% interest compounded quarterly for 3 years,
then I think that the amount of money which results is
100·(1+{.04/4})^{4·3}. One of the
things that we learn is more frequent compounding yields more
interest, so compounding 30 times annually will give more money than
compounding quarterly. More detais are here. But what was initially to me almost
unbelievable is that decreasing the interval between compounding
doesn't make much difference in reality after a while. That's because
the limit as n>infinity of (1+{1/n})^{n} is quite close to e
for large n. For example, when n=365 (daily compounding?) this is
about 2.71456. So one could advertise "continuous compounding" and
replace the discrete model of daily compounding by an exponential, and
then compute things more easily and actually not change the answer too
much. Thus I think $100 deposited for 3 years at 4% compounded
continuously would be 100·e^{(.04)3}. I hope this is
clear, and, even more, I hope it is correct. By the way,
$100(1+{.04/365})^{365·3} (daily compounding) is about
$112.7489 while $100(e^{.04·3}) (continous compounding)
is about $112.7496, fairly close. Now something a bit more
complicated.
The present value of an income stream
For simplicity, let me assume that interest is constant at, say,
4%. We're in 2004 (t=4). I might have been given $100 in 2000 (t=0)
and $300 in 2002 (t=2) and $200 in 2003 (t=3). What would be the present
value (value in 2004 including the effect of interest compounded
continuously) of this money? I think it would be
100e^{.04(40)}+300e^{.04(42)}+200e^{.04(43)}
Please notice several things about this sum. First, of course the
principal amount varies (100 and 300 and 200) but also the time
interval changes for each. Of course this is true since the
accumulation of interest occurs over different time intervals. If
there are many payments (an income stream) then maybe the
discrete sum could be approximated fairly accurately by an
integral. Let's see: if time "now" is t, and if the income comes in
from time 0 to t, with an assumed interest rate of 4%, then probably
the present value of the income stream is _{0}^{t}P(tau)e^{.04(ttau)}dtau.
I hope that you can identify this as a convolution of the income
stream with the interest rate function. Of course, this is all a very
very simple model. But models like this are used everywhere in
financial math, an area which has grown tremendously in the last few
years. You can also see easily, I hope, that the delay of ttau
is really important in constructing this model. Of course the model
can easily be made more intricate to take care of interest rates which
vary with time but the ttau has to be there.
In, say, chemical/biochemical engineering we could imagine varying
amounts of an ingredient dropped into a reaction vat at different
times, and each growing or reacting with the vat's content  hey, a
convolution. Etc., etc.
Convolution examples just using the definition
What's the convolution of t^{2} and t^{3}? This is
_{0}^{t}tau^{2}(ttau)^{3}dtau.
After a huge amount of debate, we decided (please see here
or here if
this is unfamiliar to you) that (ttau)^{3}=
1t^{3}3t^{2}tau^{1}+3t^{1}tau^{2}1tau^{3}
and therefore
_{0}^{t}tau^{2}(ttau)^{3}dtau=
_{0}^{t}tau^{2}[1t^{3}3t^{2}tau^{1}+3t^{1}tau^{2}1tau^{3}]dtau=
_{0}^{t}tau^{2}t^{3}3t^{2}tau^{3}+3t^{1}tau^{4}tau^{5}dtau=
(1/3)tau^{3}t^{3}(3/4)t^{2}tau^{4}+(3/5)t^{1}tau^{5}(1/6)tau^{6}]_{tau=0}^{tau=t}=
(1/3)t^{6}(3/4)t^{6}+(3/5)t^{6}(1/6)t^{6}.
Wow, what a computation. Of course, Mr. Klumb (diligent or maniacal?) computed that the answer was more
simply (1/60)t^{6}.
QotD, part 2
Suppose A is not equal to B. What is the convolution of
e^{At} and e^{Bt}?
Doing this exercise is very healthy. One must very conscientiously
distinguish between t and tau and deal with the exponentials
with some care. Even though the manipulations are individually
"trivial", errors certainly can be made. Mr. PierreLouis reported that the answer is
(1/[BA])(e^{At}e^{Bt}). (By the way, what is the
convolution of e^{At} and e^{At}?)
Note
I looked up the spelling of conscientiously. Also, my online
dictionary says it means "diligent and scrupulous".
Let us look at some entries in "the table" now.
Entry  Function  Laplace transform 

#304  e^{At}  1/(sA) 
#305  e^{Bt}  1/(sB) 
#306  e^{At}_{*}e^{Bt}  ??? 
(1/[BA])(e^{At}e^{Bt})  (1/[BA])([1/{sA}][1/{sB}]) 
The ??? conceals an amazing fact. Look: the entry
(1/[BA])([1/{sA}][1/{sB}]) is obtained by using linearity of the
Laplace transform on the direct computation of
e^{At}_{*}e^{Bt}.
Let us "massage" that
entry algebraically:
(1/[BA])([1/{sA}][1/{sB}])=(1/[BA])([{sB}{sA}]/[{sA}{sB}])=(1/[BA])([{AB]/[{sA}{sB}])=(1/[{sA}{sB}]).
Therefore the Laplace transform of
e^{At}_{*}e^{Bt} is the ordinary product of
the Laplace transforms of e^{At} and e^{Bt}. This is
an amazing fact, and really worth an early and important entry in thet
table.
Entry  Function  Laplace transform 

#11 (?)  f(t)_{*}g(t)  F(s)·G(s) 
Again, look at the other example. The convolution of t^{2} and t^{3} is (1/60)t^{6}. The Laplace transform of t^{2} is 2!/s^{3} and the Laplace transform of t^{3} is 3!/s^{4} and the Laplace transform of (1/60)t^{6} is (1/60)6!/s^{7}. Now 3+4=7 so the powers of s agree. Also, 2!·3!=2·6=12 and (1/60)6!=(720/60)=12 so the constants agree. Wonderful.
The general proof of the table entry for Laplace transform of a convolution is in the textbook and depends on change of variable in a double integral (one integral for convolution and one for Laplace transform). I won't do it here.
A textbook problem
I announced that I would do 4.4, #38
(there was applause and gasps, after I requested them). This problem
is an integral equation of a special type called a Volterra
integral equation. Here's the problem:
Find a function f(t) so that
f(t)=2t4_{0}^{t}sin(tau)f(ttau) dtau
I think it would be possible to solve this without Laplace transforms,
but it will be much easier to do it using them. Of course (a textbook
problem!) I hope that you recognize the convolution:
_{0}^{t}sin(tau)f(ttau) dtau=sin(t)_{*}f(t). The general strategy is the
following:
Step 1
If F(s) is the Laplace transform of s, then we have
F(s)=2/s^{2}4(1/[s^{2}+1])F(s). The convolution
becomes ordinary multiplication in the Laplace transform world.
Step 2
Take F(s)=2/s^{2}4(1/[s^{2}+1])F(s) and write:
F(s)+4(1/[s^{2}+1])F(s)=2/s^{2}. Then factor:
{1+4(1/[s^{2}+1])}F(s)=2/s^{2}. Combine:
{[s^{2}+1+4]/[s^{2}+1]}F(s)=2/s^{2}. Combine
more:
{[s^{2}+5]/[s^{2}+1]}F(s)=2/s^{2}. "Solve":
F(s)=(2/s^{2}){[s^{2}+1]/[s^{2}+5]}
Step 3
We must find the inverse Laplace transform of
(2/s^{2}){[s^{2}+1]/[s^{2}+5]}. We can ask
Maple (about a tenth of a second) or use partial
fractions and then consult the table. We need to write
(2s^{2}+2)/(s^{2}[s^{2}+5]) as
(A/s)+(B/s^{2})+([Cs+D]/[s^{2}+5]).
QotD, part 3
If A and B and C and D are known, write a formula for f(t).
Let me try to find the actual values of A and B and C and D. I did the
following algebra wrong in class  I
was certainly hurrying too much.
(A/s)+(B/s^{2})+([Cs+D]/[s^{2}+5])=[As(s^{2}+5)+B(s^{2}+5)+(Cs+D)s^{2}]/[s^{2}(s^{2}+5)]
We need to make the tops (o.k., the numerators!) equal:
2s^{2}+2=As(s^{2}+5)+B(s^{2}+5)+(Cs+D)s^{2}
If s=0, we get 5B=2 so that B=2/5. The s coefficient on both sides
gives 0=5A so that A=0.
The s^{3} coefficient on both sides gives 0=A+C,
so that C=0 also. Finally, the s^{2} coefficient gives 2=B+D so
that D=2(2/5)=8/5.
Therefore we need the inverse Laplace transform of [(2/5)/s^{2}]+[{8/5)/(s^2+5)] and this (the table!) is (2/5)t+(8/5)(1/sqrt(5))sin(sqrt(5)t). A consultation with Maple reads as follows:
>invlaplace((2/s^2)*((s^2+1)/(s^2+5)),s,t); 1/2 1/2 2/5 t + 8/25 5 sin(5 t)so I am fairly happy.
Due to laziness, I had students work on some problems in class. I also managed to make a significant error in an entry in the Laplace transform table, which was stupid. Here is a correct table, I hope.
Entry  Function  Laplace transform 

#0  c_{1}y_{1}+c_{2}y_{2} c_{1} and c_{2} constants  c_{1}Y_{1}+c_{2}Y_{2} 
#1  t^{n}  n!/s^{n+1} 
#2  e^{at}  1/(sa) 
#3  sin(kt)  k/(s^{2}+k^{2}) 
#4  cos(kt)  s/(s^{2}+k^{2}) 
#5  e^{at}f(t)  F(sa) 
#6  f(ta)U(ta)  e^{as}F(s) 
#7  g(t)U(ta)  e^{as}L(g(t+a)) 
Example #1
(Credit Mr. Shaw and Ms. Tozour.)
If f(t)=cos(t)U(t5), what is F(s)?
Use #7 (a=5). Then we will get e^{5s} multiplied by the Laplace
transform of cos(t+5)=cos(t)cos(5)sin(t)sin(5). Now use #0 and #3 and
#4 to get the actual answer:
e^{5s}[{s/(s^{2}+1^{2})}cos(5){1/(s^{2}+1^{2})}sin(5)].
Response from Maple which uses Heaviside rather than U:
>laplace(cos(t)*Heaviside(t5),t,s); /cos(5) s sin(5)\ exp(5 s)     2 2  \ s + 1 s + 1/
Example #2
Credit who?
If F(s)=(e^{5s}/s^{3}), what is f(t)?
Use #6 (a=5). Then we need f(t) so that F(s)=1/s^{3}, and we
use #0 and #1. The answer is:
(1/2!)(t5)^{2}U(t5).
Maple told me:
2 (1/2 t  5 t + 25/2) Heaviside(t  5)
Example #3
Credit who?
If F(s)=s/[(s1)(s2)(s3)], what is f(t)?
Partial fractions. Thus
[A/(s1)]+[B/(s2)]+[C/(s3)]=[{A(s2)(s3)+B(s1)(s3)+C(s1)(s2)}/{(s1)(s2)(s3)}]
so that (looking at the tops!)
A(s2)(s3)+B(s1)(s3)+C(s1)(s2)=s. Now s=1 gives A=1/2 and s=2
gives B=2 and s=3 gives C=3/2. Now use #0 and #2 for the answer:
(1/2)e^{t}+(2)e^{2t}+(3/2)e^{3t}.
Maple told me:
2*exp(2*t)+3/2*exp(3*t)+1/2*exp(t)
Example #4
(Credit Mr. Dow and Mr. Bond.)
If the graph of f(t) looks like this, what is F(s)?
We need to "turn on" t^{2} at t=1 and turn it off at t=2. So
we write f(t)=t^{2}U(t1)t^{2}U(t2). Now we will use
linearity (#0) and either #6 or #7. Let's try #7 on
t^{2}U(t1). Here a=1, and g(t)=t^{2} so that we need
the Laplace transform of g(t+1)=(t+1)^{2}=t^{2}+2t+1
whose Laplace transform is (with #0 and #1)
2/s^{3}+2/s^{2}+1/s. We need the other part, the
Laplace transform of t^{2}U(t2) (there is also a minus sign
joining the two points which we shouldn't forget!). Here a=2, and we
want the Laplace transform of
g(t+1)=(t+2)^{2}=t^{2}+4t+4, or
2/s^{3}+4/s^{2}+4/s. Put all this together, and the
answer is
e^{s}[2/s^{3}+2/s^{2}+1/s]e^{2s}[2/s^{3}+4/s^{2}+4/s]
Maple told me:
>laplace(t^2*Heaviside(t1)t^2*Heaviside(t2),t,s); exp(s) 2 exp(s) 2 exp(s) 4 exp(2 s) 4 exp(2 s) 2 exp(2 s)  +  +        s 2 3 s 2 3 s s s sExample #5
>laplace((t2)*Heaviside(t2)(t2)*Heaviside(t3)+(4t)*Heaviside(t3)(4t)*Heaviside(t4),t,s); exp(2 s) 2 exp(3 s) exp(4 s)    +  2 2 2 s s s
Example #6
Credit who?
If F(s)=e^{7s}/[(s5)^{2}+9], what is f(t)?
This one is complicated. The e^{7s} tells me that I will have
U(t7) in the answer (a=7 in #6 or #7). What is the inverse
Laplace transform
of 1/[(s5)^{2}+9], though? I think it looks sort of like
sine (see #3), in fact, almost like sin(3t). I need to divide by 3 to
compensate for the "extra" k on the top of the last entry in #3. And
how about paying attention to s5? Well, that is handled (see #5) if I
multiply by e^{5t}. The inverse Laplace transform of
1/[(s5)^{2}+9] therefore is e^{5t}(1/3)sin(3t). But
#6 tells me that I should substitute t7 for t in the formula
e^{5t}(1/3)sin(3t). Therefore the answer is
e^{5(t7)}(1/3)sin(3(t7))U(t7).
And Maple declares:
>invlaplace(exp(7*s)/((s5)^2+9),s,t); 1/3 Heaviside(t  7) exp(5 t  35) sin(3 t  21)
How to practice
You could write "things" almost at random, find the Laplace or inverse
Laplace transform "by hand" and use Maple or other software
to check your results.
Return of Entrance Exam
I returned the Entrance Exam. I tried to explain
I do NOT have a course grade selected for you, honestly. I have not prejudged you. But you and all students in the class should know that review of needed topics from calculus is not the object of the course, and that the course builds cumulatively on skills and knowledge acquired in a bunch of courses students should have taken previously. I do not like to surprise students in exams or grades or anything ... the entrance exam, together with my grading, is an attempt to inform [students] honestly about what [they should know at the beginning of the course: about what is expected and will be assumed.] 
The QotD was to find the Laplace transform of e^{5t}tU(t2). Most students answered e^{2(s+5)}{[1/(s+5)^{2}]+[2/(s+5)]}.
It was another glorious day in Math 421, a Thursday, and the Chem. Engineering students have only 19 hours of class today, while Mech. Engineering entertains their students at the usual cocktail party.
I daringly attempted problem #36 in section 4.2. 36 is a large number and even, so the problem must be difficult and there is no answer in the back of the book!
Section 4.2, Problem #36 y´´4y´=6e^{3t}3e^{t}, y(0)=1, y´(0)=1
As I remarked, this is a very simple ODE: second order, linear, constant coefficient. What methods would I expect/hope that students could use on this problem?
Method #1 First solve the "associated homogeneous equation", y´´4y´=0. Plug in e^{rt}, get the characteristic equation, etc. Attempt to get a particular solution using a variety of methods (undetermined coefficients or throwing dice or ...).
Method #2 Let y_{1}=y and y_{2}=y´. Build a
column vector
Y=( y_{1} y_{2})^{t}. Note: the
silly superscript t means transpose (rows go to columns), and I am
using it here (didn't use it in class yet!) because it is harder to
type a column vector in html than a row vector. Sigh. Now if we define
a 2by2 matrix, A, to be this:
(0 1)
(4 0)
then
I hope that the matrix differential equation Y´=AY+S (S=other stuff I
don't want to bother with now) is the same as the original
problem. One can then apply methods of linear algebra (we'll do some
of this later in the course) to solve the matrix DE which in turn will
lead to a solution of the original equation.
Method #3 Use numerical techniques to approximate a solution. For many applications, this is just as good. For some applications such as those requiring longterm asymptotics as a function of some symbolic initial conditions, for example, numerical methods aren't too useful.
Method #4 Use the Laplace transform, which I will write up in a second. I would like to write what I said in response to a student question: why learn another method? Well, in this case almost every method "works". But each method has its own flavor (?) and gives a different perspective on the ODE. Much of the time, none of these methods will apply perfectly or easily. It is the way of thinking which will allow you to find out how to analyze solutions of the ODE. If you know more methods, this is probably better.
So what we do is take the Laplace transform of the entire equation y´´4y´=6e^{3t}3e^{t}. At this time please realize that there was a table of Laplace transforms on the board. So the Laplace transform of the righthand side is easy, using linearity and entries in the table. It is [6/(s3)]+[3/(s+1)]. What about the righthand side? The Laplace transform of the unknown function, y, is usually written, Y. Then what´s the Laplace transform of y´´?
Laplace transforms of derivatives
The Laplace transform of y^{(n)} is
s^{n}Ys^{n1}y(0)s^{n2}y´´(0)...y^{n}(0).
This is proved by repeated integration by parts. I verified the n=1 case last time. This result really shows why the Laplace transform is very well adapted to initial value problems (it was invented for that, darn it!). Most of the time in this course we will use this result with n=1 or n=2 but there are certainly lots of cases in real life with other n's (n=4: vibrating beam).
When n=2, the result states that the Laplace transform of y´´ is s^{2}Ysy(0)y(0). Here we know that y(0)=1 and y´(0)=1 so that the Laplace transform of y´´ is s^{2}Ys+1. When n=1, the result states that the Laplace transform of y´ is sYy(0). Again using y(0)=1, we get sY1 as the Laplace transform of this y´. Therefore (sigh) the Laplace transform of the lefthand side of the original equation (y´´4y´) is s^{2}Ys+14(sY1)=(s^{2}4s)Ys+1+4.
Put it all together:
The Laplace transform of
y´´4y´=6e^{3t}3e^{t}, y(0)=1, y´(0)=1
is
s^{2}Ys+14(sY1)=[6/(s3)]+[3/(s+1)].
(The ODE and the initial conditions are all together in that
one equation!)
Now what? Now we "solve" for Y, the unknown Laplace transform. And we attempt to locate pieces of Y on the Laplace transform side of the table which was on the board. Let's do this.
s^{2}Ys+14(sY1)=[6/(s3)]+[3/(s+1)] becomes
(s^{2}4)Ys+5=[6/(s3)]+[3/(s+1)] which gives us
Y=[s/(s^{2}4)][5/(s^{2}4)]
+[6/{(s^{2}4)(s3)}]+[3/{(s^{2}4)(s+1)}].
[s/(s^{2}4)] is the Laplace transform of cosh(2t) (an entry in the table!) and ][5/(s^{2}4)] is the Laplace transform of (5/2)sinh(2t) (an entry in the table slightly modified by linearity).
How about [6/{(s^{2}4)(s3)}]? The bottom of this fraction is
actually (s2)(s+2)(s3), so the partial fractions cell in my brain
springs into action: we should be able to write the fraction as the
sum of
[A/(s2)]+[B/(s+2)]+[C/(s3)]=[{A(s+2)(s3)+B(s2)(s3)+C(s2)(s+2)}/{(s^{2}4)(s3)}].
If we combine the fractions, We know that 6 is supposed to be equal to
A(s+2)(s3)+B(s2)(s3)+C(s2)(s+2). The conclusion is easy:
s=2 gives 6=A·4·1 so A=3/2.
s=2 gives 6=B·4·5 so B=3/10.
s=3 gives 6=C·1·5 so A=6/5.
Therefore we need to find the inverse Laplace transform of
[{3/2}/(s2)]+[{3/10}/(s+2)]+[{6/5}/(s3)] which is
{3/2}e^{2t}+{3/10}e^{2t}+{6/5}e^{3t}.
By the way, I just checked this with Maple and you may find the results either amusing or irritating or both. One computer at Rutgers has Maple 7, and this is the "dialog":
>invlaplace(6/((s^24)*(s3)),s,t);  3/2 exp(2 t) + 3/10 exp(2 t) + 6/5 exp(3 t)which I guess makes me happy. At home I have a considerably older version of Maple (Maple V Release 5) and here is what I got:
>invlaplace(6/((s^24)*(s3)),s,t); 6/5 exp(3 t)  6/5 cos(sqrt(4) t) + 9/10 sqrt(4) sin(sqrt(4) t)What the heck? Well, there is a relationship between sinh and sine. I know sinh(t)=(e^{t}e^{t})/2 and (Euler's formula) sin(t)=(e^{it}e^{it}/(2i). So I guess i sin(it)=sinh(t). With some effort, I think I could verify that the two answers are the same. "Some effort"!
I think I abandoned problem #36 here. All I needed was a similar computation for the last chunk of Y(s). I hope that in real life (421 is definitely not real!) you will have an electronic friend to do tedious computations. In this class on exams you will need to do things by hand, only with the help of a short table.
The next item on the agenda was expanding the table: that is, showing more Laplace transforms.
The first translation theorem
If F(s) is the Laplace transform of f(t), then the Laplace transform
of e^{at}f(t) is F(sa).
This is correct because the Laplace transform of e^{at}f(t) is _{0}^{infinity}e^{st}e^{at}f(t) dt which is the same as _{0}^{infinity}e^{st+at}f(t) dt which is the same as _{0}^{infinity}e^{(sa)t}f(t) dt and that is exactly the Laplace transform of f(t) evaluated at sa, or F(sa).
Example Actually we'll mostly use this result backwards. I think I asked if students could tell me a function whose Laplace transform is 5/(s+7)^{14}. We looks at the table and saw this seemed to be Constant/s^{14} shifted. So this should be related to the Laplace transform of t^{13}. We needed to fix up the multiplicative constant so that we would get 5. And therefore we multiplied t^{13} by 5 and divided it by 13! and learned that the Laplace transform of (5/(13!)t^{13} is 5/s^{14}. Now we needed to shift by 7, and concluded that the Laplace transform of e^{t}(5/(13!)t^{13} is 5/(s+7)^{14}.
A totally useless discussion followed about the exact value of 13! and whether anyone might care. It is difficult to report whether either the instructor or one student (I will learn the names of the students!) was more silly.
I defined the Heaviside or unit step function, called U(t) in your text (actually with a calligraphic U). Oliver Heaviside was a brilliant English engineer whose life was, overall, rather sad. This is the function which is 0 for t<0 and is 1 for t>=0. There's a jump of 1 at 0, and otherwise the function's graph consists of two half lines. I think I graphed a few examples, like U(t3) (the jump is moved to 3) and 4U(t2)6U(t5) (the graph "starts" at 0, jumps up to 4 at 2, and then down to 2 at t=5  aside from the jumps the graph just is pieced together from horizontal line segments).
But U(t) will also allow us to write convenient formulas for piecewise defined functions. I drew two graphs and asked for descriptions in terms of U(t). I think the examples looked something like what follows.
Example 1
This function is 0 except for a straight line segment from (3,3) to
(4,4). We can write in "algebraically" in terms of U(t) this way:
tU(t3)tU(t4). The first part, tU(t3), starts the diagonal
line at 3. The second part cancels (because of the minus sign)
the same diagonal line at t=4. The graph of the result is what I drew.
Example 2
Maybe this is a little bit more pretty or interesting. This function
begins at (0,0), and then looks like sine until Pi/2. After Pi/2 the
function switches to a horizontal line of height 1. Of course, since
sine has a maximum at (Pi/2,1), this is actually a smooth function
(differentiable). We can write a "formula" for this in terms of
U(t). The formula is sin(t)sin(t)U(t[Pi/2])+U(t[Pi/2]). The first term
creates the graph over the interval [0,Pi/2]. The second term turns
off the sine at Pi/2, and the third term puts in a horizontal line of
height 1 at Pi/2. There was some confusion about this in class and I
believe thinking this way takes practice.
We'll be able to compute the Laplace transforms of such functions easily with
The second translation theorem
If F(s) is the Laplace transform of f(t), then the Laplace transform
of U(ta)f(ta) is e^{as}F(s).
Well, the Laplace transform of U(ta)f(ta) is (by definition)
_{0}^{infinty}e^{st}U(ta)f(ta) dt.
But U(ta) is 0 for t<a and is 1 for t>a. So the integral
doesn't need to start until a and 1 can changes U(ta) into 1 in the
part with t>a. That is,
_{0}^{infinity}U(ta)BLAH=_{0}^{a}U(ta)BLAH+_{a}^{infinty}U(ta)BLAH=_{a}^{infinity}BLAH.
So we have
_{a}^{infinity}e^{st}f(ta) dt. I will
change variables in this integral. If w=ta (remember, a is a
constant) then dw=dt and t=w+a so that:
_{t=a}^{t=infinity}e^{st}f(ta) dt=_{w=0}^{w=infinity}e^{s(w+a)}f(w) dw=_{w=0}^{w=infinity}e^{sw}e^{sa}f(w) dw=e^{sa}_{w=0}^{w=infinity}e^{sw}f(w) dw.
This is exactly e^{as}F(s) because w is the variable of
integration  it doesn't matter outside of the integral sign.
Return to previous example 1
What is the Laplace transform of tU(t3)tU(t4)? First look at
tU(t3). I want to make it fit the template of the Second Translation
Theorem, which deals with U(ta)f(ta). Well, a should be 3. What
about f? I want f(t3) to be t, and therefore, f(t) should be t+3. So
the Laplace transform of this first chunk will be e^{as}F(s)
with a=3 and with F(s) equal to the Laplace transform of t+3, which is
1/s^{2}+(3/s). The Laplace transform of tU(t3) must be
e^{3s}[1/s^{2}+(3/s)]. A similar computation showed
that the Laplace transform of tU(t4) is
e^{4s}[1/s^{2}+(4/s)]. Therefore the Laplace
transform of example 1 must be
e^{3s}[1/s^{2}+(3/s)]e^{4s}[1/s^{2}+(4/s)].
As I mentioned in class, when I prepared this lecture, I did all the
Laplace transforms by hand, but checked the results with
Maple (always wear suspenders with your belt). In this case,
I forgot the  and Maple reminded me.
Return to previous example 2
Here we want the Laplace transform of
sin(t)sin(t)U(t[Pi/2])+U(t[Pi/2]). The transform of the first part is
2/(s2+1) (I read it off the table). The transform of
U(t[Pi/2]) is easy because in
U(ta)f(ta) the function f is be the constant function always equal
to 1 and a=Pi/2. Therefore the transform is
e^{as}F(s)=e^{[Pi/2]s}(1/s).
The interesting part is computing the Laplace transform of sin(t)U(t[Pi/2]) (there is a minus sign which I will not forget when I reassemble the results later). Here U(ta)f(ta)=sin(t)U(t[Pi/2]) so a=Pi/2 and sin(t) is f(t[Pi/2]) so that sin(t+[Pi/2]) is f(t). One way of recognizing f(t) better is to "expand" sin(t+[Pi/2]) using the trig identity sin(A+B)=sin(A)cos(B)+sin(B)cos(A) where A=t and B=Pi/2. Then we learn that sin(t+[Pi/2]) is cos(t). Since f(t)=cos(t), e^{as}F(s)=e^{[Pi/2]s}(s/[s^{2}+1].
I can try to put everything together and conclude that the Laplace transform of the function in example 2 is 2/(s2+1){e^{[Pi/2]s}(s/[s^{2}+1]}+e^{[Pi/2]s}(1/s). Below you can see how Maple confirmed this for me.
The QotD was to write a little parabolic bump followed by 0 as a formula using the Heaviside function. I told people using a graph that the function was t(1t) for t between 0 and 1, and was 0 elsewhere. The answer is graphed below by Maple.
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Here are some more Maple commands and their
results. This loads the package containing various Laplace transform programs. with(inttrans);This command finds the inverse Laplace transform of something which looks a lot like an example I did in class. invlaplace(5/(s+7)^14,s,t); 13 1/1245404160 t exp(7 t)I wanted to be sure that the coefficient was what I thought it should be, so I next checked that the quotient of 5 divided by 13 factorial was the number I thought it should be (what confidence I have!). 5/13!; 1/1245404160I found the Laplace transform of the second example I did for the second Theorem covered today. Notice please that Maple uses Heaviside as the name for what the text calls U, the unit step function. laplace(sin(t)sin(t)*Heaviside(tPi/2)+Heaviside(tPi/2),t,s); s Pi s Pi exp( ) s exp( ) 1 2 2    +  2 2 s s + 1 s + 1Finally, the last Maple command asks that a graph be drawn of the function which is the answer to the QotD. So I checked my answer. plot(t*(1t)t*(1t)*Heaviside(t1),t=0..2,thickness=3,scaling=constrained); The idiotic and disgusting colors are a result of how Maple exports an image to create a gif. I must learn to do this better. 
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I strongly urged students to do homework and study together. I remarked that, after analyzing last spring's statistics, good performance in homework seemed well correlated with a good course grade. So again I recommend practice.
Now I copied Theorem 4.1 (page 193) of the text onto the board. I then checked that the Laplace transform of e^{at}. How did we do this?
_{0}^{infinity}e^{st}e^{a}dt= _{0}^{infinity}e^{(s+a)t}dt=(Fundamental Theorem of Calculus, but be careful about what variable you are "integrating"!)=(1/[s+a])e^{(s+a)t}]_{0}^{infinity}. Now what happens when "t=infinity"? This really means t>infinity. So for large enough s (s bigger than a, for example) the exponential has t multiplied by some constant which is negative. Then the exponential is decaying so as t>infinity, the exponential goes to 0 (more later). When t=0, since we're looking at the bottom of the ], we have (1/[s+a])e^{0} and e^{0}=1, so that the answer is(1/[s+a]) or 1/sa, as Theorem 4.1 states.
I briefly discussed why we might have the phrase "large enough s". The kind of functions that Laplace transform is used for are functions with (at most) exponential growth. A function f(t) has exponential growth if there are constants C>0 and M>0 so that f(t)<=Ce^{Mt} for large t. You could think of this as meaning that the graph of f(t) is caught between +Ce^{Mt} and Ce^{Mt}. If you assume this, then multiplying by e^{st} when s is greater than M will cause e^{st}f(t) to decay exponentially (because Ms will be negative). Therefore the Laplace transform of such functions will be defined if s is large enough. There certainly are functions which will be growing faster than exponentially (hey, try etotheetothe t, for example). There are a bunch of functions which have exponential growth. For example, exponentials (2^{73t}). As another example, polynomials. Notice that the phrase exponential growth as used here actually means, "at worst" exponential growth. In fact, t^{346} does grow more slowly than e^{Mt} for any positive M. The functions we'll consider will all have (at worst!) exponential growth, and I won't say very much more about this.
I then wanted to verify the book's computation of the Laplace transform of sin(t). I remarked that the definition tells me that I need to compute _{0}^{infty}e^{st}sin(t) dt: o.k. this really isn't too difficult: I would integrate by parts. But I will need to integrate by parts twice, and the chance for error is large! Here's another way to verify the Laplace transform of sine.
Last time I reminded people that sin(t) is (1/[2i])(e^{it}e^{it}). Therefore the Laplace transform is (using linearity of Laplace transform) 1/[2i] multiplied by the difference of the Laplace transforms of e^{it} and e^{it}. The Laplace transform of e^{it} is 1/[si] (I'm using the result we just derived, with a changed to i). The Laplace transform of e^{it} is 1/[s+i] (I'm using the result we just derived, with a changed to i). Therefore the Laplace transform of sin(t) is (1/[2i])({1/[si]}{1/[s+i]}. Some algebra which we did changes this to (1/[2i])({[s+i][si]}/{[si][s+i]} and this is (1/[2i])([2i]/{s^{2}+1}) which is 1/(s^{2}+1), the formula I wanted to check.
If you are confused by this, PLEASE find the Laplace transform of cos(t) using the same method. PLEASE!!!
I tried something else, the Laplace transform of a square wave. Here I will refer to a part of my course notes from last semester.
Professor Davidson urged me to go over the following ideas. Suppose
f(t) is some sort of bumpy function, as shown. Of course,
F(s)=_{0}^{infinity}e^{st}f(t) dt.
Useful Laplace transform asymptotics
By the way, if (1/s)[(e^{2s}e^{s})/(1e^{2s})] is the Laplace transform of the sum of all of the square waves (the "heartbeat" model) what should happen as s>0^{+}? I think this limit should be +infinity (that's how much area there is under all the waves). Please check that the limit is infinity. You will need to use l'Hopital's rule again.
Here is a major question which should help to answer "Why are we doing
all this?"
If F(s) is the Laplace transform of f(t), what is the Laplace
transform of f´(t)?
Well, let's start with
_{0}^{infinity}e^{st}f´(t) dt, which
is the Laplace transform of f´(t). Integration by parts can be used,
with u=e^{st} and dv=f´(t) dt, so that
du=s e^{st}dt and v=f(t). Therefore
_{0}^{infinity}e^{st}f´(t) dt=e^{st}f(t)]_{0}^{infinity}_{0}^{infinity}(s e^{st}f(t)&nbps;dt.
There is all sorts of sneaky stuff going on here, and we should be
very careful. Let's see. The boundary term is e^{st}f(t)]_{0}^{infinity}. Now s>0, so when
t>infinity, e^{st}f(t)>0. Technically I am using the
fact that f(t) has (at worst!) exponential growth, but I won't mention
it any more. When t=0, we get f(0) since e^{0}=1. What about
the integral term? Notice that there are two minus signs. They
cancel. What's left is s multiplied by the Laplace transform of f. So
now we know:
L(f´)(s)=f(0)+sL(f)(s).
I was in a hurry to show you what this could be used for, so I tried to solve y´+3y=cos(t) with initial condition y(0)=2. I wanted to take the Laplace transform of the differential equation. The Laplace transform of y is usually written Y. Therefore the Laplace transform of y´+3y=cos(t) is sY(s)y(0)+3Y(s)=s/(s^{2}+1). But y(0)=2, so this is sY(s)2+3Y(s)=s/(s^{2}+1). There are several really neat things to notice.
All we need to do is "recognize" that Y(s) is the Laplace transform of ...... and we will have solve the ODE with initial value. For example, I (and, I hope, we) know that the last "chunk" of Y(s), [2/(s+3)], is the Laplace transform of 2e^{3t}. The first "chunk", [s/{(s+3)(s^{2}+1)}], seems more difficult to recognize.
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We are in the 21^{st} century, I think. Maple's
command invlaplace will get the inverse Laplace transform of
the function in about a twentieth of a second on a computer which
is not very fast. Here are the
commands and the responses: >with(inttrans): # Loads the Laplace "package" >invlaplace(s/((s^2+1)*(s+3)),s,t);  3/10 exp(3 t) + 3/10 cos(t) + 1/10 sin(t) I tend to use Maple because I have a copy at home. Similar capabilities are available on Matlab, Mathematica, etc. 
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The QotD was to find A and B and C. If we combine the fractions
we get (As+B)(s+3)+(s^{2}+1)C on top, and this should equal
the top of [s/{(s+3)(s^{2}+1)}]. So we want:
(As+B)(s+3)+(s^{2}+1)C=s
s=3: ((3)^{2}+1)C=3 giving C=3/10.
The s^{2} coefficients: A+C=0 so A=3/10.
The constant coefficients: 3B+C=0 so B=1/10.
Comment This is certainly not the only way to "solve" the
partial fractions problem. I am lazy and this seems to me like the
least amount of work. Of course, using Maple is even less work!
Now we've written Y(s) as
[({3/10}s+{1/10})/(s^{2}+1)]+[{3/10}/(s+3)]+[2/(s+3)] which
is the same as
[({3/10}s)/(s^{2}+1)]+[({1/10})/(s^{2}+1)]+[{3/10}/(s+3)]+[2/(s+3)]
which we may recognize as the Laplace transform of
(3/10)cos(t)+(1/10)sint(t)+(3/10)e^{3t}+2e^{3t}
and this is the answer to the IVP of the ODE.
Special permissions for math courses can only be gotten by applying on the web.
Due to the delays and bumbling of the instructor, much less was
covered than should have been.
The standard clerical stuff was done. The information discussed, and much more, is available here and here.
I began by reviewing very briefly the idea of initial value problems (IVP's) for ordinary differential equations (ODE's).
An ODE is an equation involving an unknown function of one variable and its derivatives. My initial example was y´=y^{2}. Solutions are found easily since this is a separable ODE.
General disclaimer The examples discussed in class will mostly be very artificial, and chosen so that "hand" computation is practical. Generally, it is impossible to write solutions of a random ODE in terms of familiar functions. Students should know that this is true, even if we use Maple or Matlab or Mathematica or ... It may be very difficult or impractical to approximate solutions numerically. Sigh. The examples and methods that are shown here and in Math 244 really are a collection of tricks which work on many ODE's modeling physical situations. They are not guaranteed to work on all ODE's, or even all ODE's derived from fairly simple physical models. But the tricks are very useful in many examples. Now, back to work.
The solutions to y´=y^{2} are y=1/(Ct). The independent variable in this course will frequently be called t for time, so y is a model of some timedependent process. How do we know that y=1/(Ct) solves the ODE? We can differentiate it and plug everything in. That's the most direct way. Of course, this may be practical only when the darn equation and candidate for a solution are fairly simple. So y´=y^{2} has infinitely many different solutions (all of the different values of C). How does one pick out one solution in practice? One simple way is with the specification of initial values. For example, y´=y^{2} and y(0)=1 has the unique solution y=1/(1t), with domain (infinity,1). And here's another: y´=y^{2} and y(0)=2 has the unique solution y=1/([1/2]t), with domain (infinity,1/2). Notice please that the soolutions are unexpectedly (to me, unexpected, at least!) complicated. Their domains are different. This is no "theoretical" problem. The solutions blow up in the most immediate way (y>infinity) at different values of t. This ODE is nonlinear. We will mostly concentrate on linear ODE's, where it turns out that such problems don't occur.
The most routine example is y´´+y=0, which is the ODE modeling an ideal spring using Hooke's law. We can still learn things from this example! This is a second order ODE (order refers to the highest number of derivatives needed to write the equation). Here the trick is to guess a solution: try y=e^{rt}. Then y´´+y=0 becomes magically e^{rt}(r^{2}+1)=0. The exponential function is never 0. Therefore the guess solves the ODE exactly when r^{2}+1=0 (I think this is called the characteristic equation). The roots of this characteristic equation are +/i.
This is a linear ODE. The particular very very important qualitative consequence is that sums of solutions of solutions of the homogeneous equation are solutions, and so are constant multiples of solutions. Why is this true? Look:
Let's use linearity. y´´+y=0 has solutions e^{ix} and e^{ix}, so if C_{1} and C_{2} are any constants, then C_{1}e^{ix}+C_{2}e^{ix} must be a solution. Let me search for a solution satisfying certain initial conditions. Since y´´+y=0 is a second order equation, the IC's will generally have two parameters. I want a solution satisfying y(0)=1 and y´(0)=0. I will call this an initial position solution, y_{P}. What is it?
If y_{P}(x)=C_{1}e^{it}+C_{2}e^{it} then y_{P}/(x)=iC_{1}e^{it}iC_{2}e^{it}. We can get y_{P}(0)=1 by having C_{1}+C_{2}=1. We can get y_{P}(0)=0 by having iC_{1}iC_{2}=0 which is the same as C_{1}C_{2}=0. After some massive amount of thought, we managed to solve this system of linear equations: C_{1}=1/2 and C_{2}=1/2. Therefore y_{P}(t)=[e^{it}+e^{it}]/2.
Similarly, we can solve the initial value problem y´´+y=0 with y(0)=0 and y´(0)=1, which I'll call the initial velocity solution, y_{V}(t). It turns out to be y_{V}(t)=[e^{it}e^{it}]/(2i). You should check this!
Why would one want to know the y_{P} and y_{V} solutions? Well, they are really neat if you want to "solve" lots of initial value problems for y´´+y=0. The pattern of initial conditions (1 and 0, and 0 and 1) allows us to write a solution for the IVP y(0)=7 and y´(0)=13. Here it is: 7y_{P}(t)13y_{V}(t). This is so easy.
Everything I've written is correct, but of course some important things have not been written! First, writing the solutions as complex exponentials conceals important features of the solutions. For example, since y´´+y=0 models simple harmonic motion, the solutions had better be bounded (they shouldn't grow to infinity in any fashion). I think springs don't do that. And maybe my formulation of y_{P} and y_{V} doesn't entirely display this. But we could compare power series or do other stuff and, in some fashion, remember Euler's formula and its consequences. So here read this, please:
e^{it}=cos(t)+i sin(t) and cos(t)=[e^{it}+e^{it}]/2 and sin(t)=[e^{it}e^{it}]/(2i) 

What happens if we just change a sign in the ODE? If y´´y=0, the characteristic equation becomes r^{2}1=0 with roots r=+/1. Solutions are then C_{1}e^{t}+C_{2}e^{t}. The solutions corresponding to the Position and Velocity initial conditions (which I hope I have convinced you are useful, when combined with linearity, in solving IVP's are as follows:
If we now consider a "general" second order, linear, constant coefficient, homogeneous ODE (by the way, each phrase or word I've just written should make some sense to you and if any do not, you must review material from 244  see the syllabus for suggested reading in our text), Ay´´+By´+Cy=0 then I can tell you what to expect about the solutions. If B^{2}4AC<0, probably sines and cosines will appear. If B^{2}4AC>0, the solutions can be written with coshes and sinhes. There will also be some exponential factors (damping or otherwise). What happens if B^{2}4AC=0? It's a mystery, and you should figure it out.
Part I of the course: Escaping the wolves! 

The snow was coming down thicker and the chill, at first merely uncomfortable, was becoming a serious problem. With nightfall, we could hear the howls of the wolves coming closer. It was time to try for the Duke's castle and safety! I grabbed the child, and jumped on my horse. I called to the others in the party, "Get on your brave steeds, and ride rapidly to the castle ..." 
No, no, no!
This is a complicated definition.
Remarks about the definition
Example 2 What is the Laplace transform of f(t)=t? Now we need
to compute
_{0}^{infinity}e^{st}t dt with a
little more care. We need the antiderivative of e^{st}t with
respect to t. This can be done using integration by parts: if u=t,
then everything else is determined:
u=t and dv=e^{st}dt so du=dt and v=(1/s)e^{st}ds.
Now the integral becomes (I am definitely shortening the computation,
eliminating all the limit "stuff"):
The pattern:ORIGINAL INTEGRAL=u dv] v du
_{0}^{infinity}e^{st}dt=(t/s)e^{st}]_{0}^{infinity}_{0}^{infinity}(1/s)e^{st}ds.
Now (t/s)e^{st} when t=0 is surely 0. And (s>0!) when
t"="infinity (this is shorthand for a limit, really) the value is 0
also, because exponentials with negative argument decrease faster than
polynomials grow. So the boundary term in this integration by
parts is 0 (the stuff arising from ]). As for the other, the
(1/s) can be pulled out of the integral, since it is a dt
integral. And the result (two minus signs make a +) is just
(1/s)_{0}^{infinity}e^{st}dt. The integral
was computed in example 1 and had value 1/s. So the Laplace transform
of t is 1/s^{2}.
Example 3 What is the Laplace transform of f(t)=t^{2}?
Philosophical interruption Much of mathematics is pattern
matching, and then attempting to verify the detected patterns. So far
we have:
If n is a nonnegative integer, the Laplace transform of t^{n} is n!/s^{n+1}.
The QotD was: if f(t)=3+7t^{2}33t^{4}?
Here I wanted people to use the linearity of the Laplace
transform.
Maintained by greenfie@math.rutgers.edu and last modified 9/4/2004.