### Thursday, September 23

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 White-lipped 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"
• If t is not 0, then delta(t)=0
• If g is continuous, then -infty+inftydelta(t)·g(t) dt=g(0).
• The Laplace transform of delta(t) is the constant function 1 (just "plug in" e-st for g(t) in the previous fact) and the Laplace transform of delta(t-t0) (the delta function centered at t0) is e-t0s

We can translate the delta function, just as we translated the Heaviside function. What does delta(t-5) do? The center of interest is then where t-5 is 0, or where t=5. That means, for example, that -inftyinftydelta(t-5)·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 0infinitye-stf(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, 0infinitye-stf(t) dt-->0 That is, lims-->+infinityF(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 0infinity1f(t) dt. Therefore

Asymptotic fact II
As s-->0+, 0infinitye-stf(t) dt-->the net area "under" the bump (as usual, area under the horizontal axis is counted negatively). So lims-->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(t-Pi) 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 right-hand side, delta(t-Pi) 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(t-Pi) is s2Y(s)-sy´(0)-y(0)+Y(s)=e-Pi·s which turns out to be (s2+1)Y(s)=e-Pi·s or Y(s)=e-Pi·s/(s2+1). The predicted vibration of the spring is therefore (!) the inverse Laplace transform of e-Pi·s/(s2+1). I can read that off from the (mythical?) table and the result is y(t)=sin(t-Pi)U(t-Pi): 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(t-Pi)+delta(t-3Pi). Now go through the Laplace transform, using the initial conditions:
s2Y(s)-sy´(0)-y(0)+Y(s)=e-Pi·s+e-3Pi·s becomes
s2Y(s)+Y(s)=e-Pi·s+e-3Pi·s and then
(s2+1)Y(s)=e-Pi·s+e-3Pi·s
so that
Y(s)=[e-Pi·s+e-3Pi·s]/(s2+1).
The inverse Laplace transform tells me that
y(t)=sin(t-Pi)U(t-Pi)+sin(t-3Pi)U(t-3Pi).

It may be useful to understand what this algebraic mess means.

• For example, if 0<t<Pi, y(t)=0: the position of the spring doesn't change from rest before it is hit.
• If Pi<t<3Pi, then y(t)=sin(t-Pi), a sine wave "starting" when time is Pi.
• If 3Pi<t, then y(t)=sin(t-Pi)+sin(t-3Pi). We could think about what this means (actually, it is probably easier to use Maple to graph y(t) but perhaps the effort will eventually help). Well, the sine function is periodic with period 2Pi, so that y(t)=sin(t-Pi)+sin(t-3Pi)=sin(t-Pi)+sin(t-Pi)=2sin(t-Pi). Hey: the function sort of looks the same but its amplitude doubles. The two "hits" reinforced each other, and made the spring deviate from equilibrium more. This is called resonance.
I then asked the class if this was smooth motion: if I were riding the spring, would I feel any jerks (yes, it is easy to make a joke here about feeling jerks). More technically, what I am asking translates to: is y(t) a smooth (=differentiable) function. Certainly, inside the intervals 0<t<Pi and Pi<t<3Pi and 3Pi<t the description we have of y(t) is a rather simple one, and the functions involved have first (and second and third and ...) derivativates. The motion is smooth. At t=Pi, the slope of the tangent line changes from 0 (on the left) to 1 (on the right). At t=3Pi, things are a bit more difficult to see, but the slope changes from 1 (on the left) to 2 on the right: this is not smooth, and the travel of the spring certainly "jerks" there.

And even more resonance
What if the driving force hits every 2Pi after t=Pi? The mathematical model is y´´+Ky´+y=SUMj=0infinitydelta(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)=SUMj=0infinity[e-(2j+1)Pi·s]/[s2+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)=SUMj=0infinitysin(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(t-Pi), and y(t)=SUMj=0infinitysin(t-Pi)U(t-(2j+1)Pi)=sin(t-Pi)SUMj=0infinityU(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.

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=SUMj=0infinitydelta(t-(j+1))with y(0)=0 and y´(0)=0. So
Y(s)=SUMj=0infinity[e-(j+1)Pi·s]/[s2+1] and then
y(t)=SUMj=0infinitysin(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)=SUMj=0infinitysin(t-(j+1)Pi)U(t-(2j+1)Pi)=sin(t)SUMj=0infinity(-1)j+1U(t-(2j+1)Pi).
What is the strange SUMj=0infinity(-1)j+1U(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 half-wave 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.

• -infty+inftydelta(t)·g(t) dt=g(0).
So let me "plug in" U´(t) instead of delta(t) there and look at the result:
-infty+inftyU´(t)·g(t) dt
The computations which are needed to bring Mr. Heaviside's ideas into rough compliance (?) with what we already know are generally l'Hopital's rule and integration by parts. The latter is what I will use here:
-infty+inftyU´(t)·g(t) dt=img src="gifstuff/is4.gif" width=5>-57206U´(t)·g(t) dt since is clearly 0 away from the origin and we don't need those infinite tails in the integral. But take the result and integrate it by parts, where u=g(t) and dv=U´(t) dt. Then du=g´(t) dt and v=U(t). So the result is (u dv=uv-v du):
g(t)U(t)]-57206--57206U(t)·g´(t) dt
The first term, g(t)U(t)]-57206, exactly g(206)U(206)-g(-57)U(-57). But I know all of U's values (hey, either +1 or 0, and it isn't hard to decide which value to use where), so g(206)U(206)-g(-57)U(-57)=g(206).
The second term is -57206U(t)·g´(t) dt=0206g´(t) dt since, again, U(t) is 0 for t<0 and U(t)=1 for t>0. But 0206g´(t) dt is begging for the Fundamental Theorem of Calculus, so it equals g(t)]0206=g(206)-g(0).
Wait: we can get all confused here fairly easily. There is another minus sign wandering around, from the integration by parts: u dv=uv-v du. Let me reassemble things carefully and get the result: g(206)-[g(206)-g(0)]=g(0). That is, whatever the "derivative" of the Heaviside function is, it satisfies the integration prepraty of the delta function. Since the integration property is the key aspect of delta operationally, maybe is is true that U´(t)=delta(t).

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 -1037W´(t) dt=W(t)]-1037=W(37)-W(-10)=1-0=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

1. Take the Laplace transform of everything in sight (specifically, all the equations).
2. Insert information you may have about initial conditions.
3. "Solve" for the unknown Laplace transforms.
4. Find the inverse Laplace transforms.
Of course, as usual, the standard warnings apply: if you are doing this by hand (as in the course, or, rather, in the exams of this course!), it is easy to make mistakes in almost every step. Be careful.

Use the Laplacetransofmr to solve the given system of differential equations.
x´(t)=2y+et
y´(t)=8x-t
with initial conditions x(0)=1 and y(0)=1.

I hope you could think of several ways to do this. But I will take the Laplace transform of the two equations (remembering that x and y are both functions of t):
sX(t)-x(0)=2Y(s)+[1/(s-1)]
sY(s)-y(0)=8X(s)-1/s2
Insert the initial conditions:
sX(s)-1=2Y(s)+[1/(s-1)]
sY(s)-1=8X(s)-1/s2
Now solve for one of the functions. Let's see, maybe since
Y(s)=[8X(s)-1/s2+1]/s I can use this in the first equation to get:
sX(s)-1=2[8X(s)-1/s2+1]/s+[1/(s-1)] so that
X(s)={2[8X(s)-1/s2+1]/s+[1/(s-1)]+1}/s
and then solve for X(s), and get x(t) by taking the inverse Laplace transform. I will finishthis later, if I have time.

The QotD
Suppose we have the ODE y´´-3y´=delta(t-2), along with initial conditions y(0)=0 and y´(0)=1. Find the Laplace transform, Y(s), of the solution to this initial value problem.

 Note This problem satisfies all standards for cruelty-free calculus problems. It uses only small positive integers and uses only well-known 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.

Let's see: a solution should begin like this:
s2Y(s)-sy(0)-y´(0)+3(sY(s)-y(0))=e-3t so, using the initial conditions, we get
s2Y(s)-1+3sY(s)=e-3t and we can solve for Y(s):
Y(s)=[e-3t+1]/[s2+3s].

Maple found this for y(t):
2/3 Heaviside(t - 3) exp(3/2 t - 9/2) sinh(3/2 t - 9/2) + 2/3 exp(3/2 t) sinh(3/2 t)

If you find this answer distressing (I don't!) you can ask Maple to convert it with the instruction convert(name of previous answer,exp) and you will get what follows.

```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=SUMj=0infinitydelta(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.

### Tuesday, September 21

More properties of convolution
ONE
If we take g(t) to be the function which is always 1, then the convolution of f and g is exactly 0tf(tau) dtau. The Laplace transform of this is F(s) multiplied by the Laplace transform of 1, which is 1/s. Therefore,
The Laplace transform of 0tf(tau) dtau is F(s)/s.
You get more s's "downstairs" on the Laplace transform side when you integrate the function on the original side. This is exactly the counterpart of what happens when you differentiate on the original side, except that formula is more complicated due to initial conditions.

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
0tf(tau)g(t-tau) dtau=0tg(tauf(t-tau) 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 t2 and t3 and t4?
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/s2
t2 becomes 2/s3
t3 becomes 6/s4
t4 becomes 24/s5
and the product of the Laplace transforms is 288/s14, whose inverse Laplace transform is (288/13!)t13, 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.

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/s2 and 1/(s2+1). Well, we could use partial fractions (I gave up in class, which was not good). So we would need to solve
1/[s2(s2+1)]=A/s+B/s2+(Cs+D)/(s2+1)
The tops of the fractions (right-hand side fraction combined) should agreed, so
1=As(s2+1)+B(s2+1)+(Cs+D)s2
Now s=0 gives us B=1. Comparing s2 coefficients, we get 0=B+D, so D=-1. The s coefficients give 0=A, and then the s3 coefficients give 0=A+C, so that C=0 also. Therefore we need the inverse Laplace transform of [1/s2]-[1/(s2+1)], and this I can "read off" the table as t-sin(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:

• Almost every reasonable limit should exist.
• Almost every reasonable function should have a derivative.
What follows is my attempt to explain one of the most startling ideas Heaviside created.

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(t-A). Then take the Laplace transform of x´´(t)=F(t) and get, on the left-hand side (with initial conditions built in), s2X(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 s2X(s)=(1/A)[1-e-As] and X(s)=(1/A)1/s2-(1/A)e-As(1/s2. We now take the inverse Laplace transform, and get x(t)=(1/A)(1/2)t2-(1/A)(1/2)(t-A)2U(t-A). For 0A, x(t)=(1/A)(1/2)(2At-A2)=(1/2)(2t-A). 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 Students who were both alert and alive and thinking during class will see that I changed the model for F(t), because the algebra is easier with a box which is A by 1/A (and then A-->0) rather than with one that is 1/A by A (and then A-->infinity). Sorry about that. Limits of the motion and what we expect What happens now as A-->0? The force had Laplace transform (1/A)(1/s)-(1/A)e-As(1/s)= (1/s)(1/A)[1-e-As]. As A-->0, we can try to evaluate the limit since "Almost every reasonable limit should exist." This is the algebraic side of the picture of the particle's motion. Good engineers try first to just "plug in": when A=0 here, the result is 0/0. The next thought should be l'Hopital's rule. The limit of (1/s)[1-e-As]/A as A-->0 is an indeterminate form of appropriate type (0/0). So take the derivative of the top and bottom with respect to A. The result is (1/s)[0-(-s)e-As]/1. As A-->0, this goes to 1. Therefore, the Laplace transform of the limit of the boxes should be 1. Geometrically, the area inside the colored circle is shrinking to the origin at A-->0, and the limiting solution is just two straight half-lines. The impressed force, F(t), becomes more and more instantaneous. The motion becomes more and more like a horizontal line for t<0 and a ray, x(t)=(1/2)t, for t>0.

Dirac delta "function"
Dirac, a Nobel prize-winning 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 -30040delta(t-2) dt: this was 1. The integral -300-200delta(t-2) dt was 0. I tried some further computations. I think one was something like -30040delta(t-2)(t2-5t) 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 t2-5t is close to 4-10=-6. Therefore this integral is just -6.

I wanted to make the Chem E's happy. Here is a two-part problem which I found last year. It is from an essay by Kurt Bryan of the Rose-Hulman 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(1-e-(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(t-20) 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(t-20) 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(1-e-(1/50)t)-5U(t-20)e-(1/50)(t-20). 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(t-20)*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)=(ew-e-w)/2. If you know that, then you can see that Maple's answer is equal to ours.

### Thursday, September 16

Old QotD: two ways
What is the Laplace transform of e-5ttU(t-2)? I use the short table of Laplace transforms which was written last time. Students did this problem in two ways, depending on how they "parsed" the expression for the function. Both methods described below are valid, and the answers are equal.

Left to right
Since e-5ttU(t-2) begins with an exponential, I'll use #5, with a=-5 and f(t)=tU(t-2). 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/s2+2/s (using #0 and #1). Now finish the use of #7: e-2s(1/s2+2/s). So we bump up or back (?) a level to #5, which advises us to change every appearance of s to s-a=s+5. Therefore the final result is e-2[s+5](1/[s+5])2+2/[s+5]).

Right to left
Now e-5ttU(t-2) "begins" (on the write) with a unit jump. So I will start by using #7, with a=2 and g(t)=e-5tt. 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-10e-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/s2+2/s, and then put in s-a=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-10e-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)=0infinitye-stf(t) dt we could try to differentiate F(s). Thus (d/ds)F(s)=(d/ds)0infinitye-stf(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)=0infinity(d/ds)e-stf(t) dt. and (d/ds)e-stf(t)=-te-stf(t) since d/ds only notices appearances of s. Now we have (d/ds)F(s)=0infinity-te-stf(t) dt=-0infinitye-stt·f(t) dt which is minus the Laplace transform of t·f(t).

EntryFunctionLaplace transform
#268 -t·f(t) (d/ds)F(s)
#269 (-1)ntnf(t) (dn/dsn)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(s-3)? Well, I do know that (d/ds)ln(s-3) is 1/(s-3), and that's the Laplace transform of e-3t. Therefore since (d/ds)F(s)=-t·f(t), if F(s)=ln(s-3), then -t·f(t)=e-3t so f(t)=-e-3t/t, and we have

EntryFunctionLaplace transform
#503.2 -e-3t/t ln(s-3)

Computer algebra question (for students!)
Maple 7 was able to verify that the inverse Laplace transform of ln(s-3) 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(s2+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 non-negative 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)=0tf(tau)g(t-tau) 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(4-0)+300e.04(4-2)+200e.04(4-3)
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 0tP(tau)e.04(t-tau)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 t-tau 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 t-tau 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 t2 and t3? This is 0ttau2(t-tau)3dtau. After a huge amount of debate, we decided (please see here or here if this is unfamiliar to you) that (t-tau)3= 1t3-3t2tau1+3t1tau2-1tau3 and therefore
0ttau2(t-tau)3dtau= 0ttau2[1t3-3t2tau1+3t1tau2-1tau3]dtau= 0ttau2t3-3t2tau3+3t1tau4-tau5dtau= (1/3)tau3t3-(3/4)t2tau4+(3/5)t1tau5-(1/6)tau6]tau=0tau=t= (1/3)t6-(3/4)t6+(3/5)t6-(1/6)t6.
Wow, what a computation. Of course, Mr. Klumb (diligent or maniacal?) computed that the answer was more simply (1/60)t6.

QotD, part 2
Suppose A is not equal to B. What is the convolution of eAt and eBt?

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. Pierre-Louis reported that the answer is (1/[B-A])(eAt-eBt). (By the way, what is the convolution of eAt and eAt?)
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.

EntryFunctionLaplace transform
#304 eAt 1/(s-A)
#305 eBt 1/(s-B)
#306 eAt*eBt ???
(1/[B-A])(eAt-eBt) (1/[B-A])([1/{s-A}]-[1/{s-B}])

The ??? conceals an amazing fact. Look: the entry (1/[B-A])([1/{s-A}]-[1/{s-B}]) is obtained by using linearity of the Laplace transform on the direct computation of eAt*eBt. Let us "massage" that entry algebraically:
(1/[B-A])([1/{s-A}]-[1/{s-B}])=(1/[B-A])([{s-B}-{s-A}]/[{s-A}{s-B}])=(1/[B-A])([{A-B]/[{s-A}{s-B}])=(1/[{s-A}{s-B}]).
Therefore the Laplace transform of eAt*eBt is the ordinary product of the Laplace transforms of eAt and eBt. This is an amazing fact, and really worth an early and important entry in thet table.

EntryFunctionLaplace transform
#11 (?) f(t)*g(t) F(s)·G(s)

Again, look at the other example. The convolution of t2 and t3 is (1/60)t6. The Laplace transform of t2 is 2!/s3 and the Laplace transform of t3 is 3!/s4 and the Laplace transform of (1/60)t6 is (1/60)6!/s7. 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)=2t-40tsin(tau)f(t-tau) 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: 0tsin(tau)f(t-tau) dtau=sin(t)*f(t). The general strategy is the following:

1. Take the Laplace transform of the whole equation.
2. Solve for the unknown Laplace transform.
3. Take the inverse Laplace transform of the result to find the unknown function.

Step 1
If F(s) is the Laplace transform of s, then we have F(s)=2/s2-4(1/[s2+1])F(s). The convolution becomes ordinary multiplication in the Laplace transform world.

Step 2
Take F(s)=2/s2-4(1/[s2+1])F(s) and write:
F(s)+4(1/[s2+1])F(s)=2/s2. Then factor:
{1+4(1/[s2+1])}F(s)=2/s2. Combine:
{[s2+1+4]/[s2+1]}F(s)=2/s2. Combine more:
{[s2+5]/[s2+1]}F(s)=2/s2. "Solve":
F(s)=(2/s2){[s2+1]/[s2+5]}

Step 3
We must find the inverse Laplace transform of (2/s2){[s2+1]/[s2+5]}. We can ask Maple (about a tenth of a second) or use partial fractions and then consult the table. We need to write (2s2+2)/(s2[s2+5]) as (A/s)+(B/s2)+([Cs+D]/[s2+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/s2)+([Cs+D]/[s2+5])=[As(s2+5)+B(s2+5)+(Cs+D)s2]/[s2(s2+5)]
We need to make the tops (o.k., the numerators!) equal:
2s2+2=As(s2+5)+B(s2+5)+(Cs+D)s2
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 s3 coefficient on both sides gives 0=A+C, so that C=0 also. Finally, the s2 coefficient gives 2=B+D so that D=2-(2/5)=8/5.

Therefore we need the inverse Laplace transform of [(2/5)/s2]+[{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.

Office hours
I will try to be in my office on Mondays from 3 to 5 and on Wednesdays from 3 to 4. If no one shows up for a while, I may go away. It may be useful to send me e-mail ahead of time telling me you are coming. I am also available by appointment at mutually agreeable times. Talk to me before or after class, or send e-mail.

### Tuesday, September 14

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.

EntryFunctionLaplace transform
#0 c1y1+c2y2
c1 and c2 constants
c1Y1+c2Y2
#1 tn n!/sn+1
#2 eat 1/(s-a)
#3 sin(kt) k/(s2+k2)
#4 cos(kt) s/(s2+k2)
#5 eatf(t) F(s-a)
#6 f(t-a)U(t-a) e-asF(s)
#7 g(t)U(t-a) e-asL(g(t+a))

Example #1
(Credit Mr. Shaw and Ms. Tozour.)
If f(t)=cos(t)U(t-5), 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/(s2+12)}cos(5)-{1/(s2+12)}sin(5)].
Response from Maple which uses Heaviside rather than U:

```>laplace(cos(t)*Heaviside(t-5),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/s3), what is f(t)?
Use #6 (a=5). Then we need f(t) so that F(s)=1/s3, and we use #0 and #1. The answer is:
(1/2!)(t-5)2U(t-5).
Maple told me:

```                      2
(1/2 t  - 5 t + 25/2) Heaviside(t - 5)```

Example #3
Credit who?
If F(s)=s/[(s-1)(s-2)(s-3)], what is f(t)?
Partial fractions. Thus [A/(s-1)]+[B/(s-2)]+[C/(s-3)]=[{A(s-2)(s-3)+B(s-1)(s-3)+C(s-1)(s-2)}/{(s-1)(s-2)(s-3)}] so that (looking at the tops!) A(s-2)(s-3)+B(s-1)(s-3)+C(s-1)(s-2)=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)et+(-2)e2t+(3/2)e3t.
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" t2 at t=1 and turn it off at t=2. So we write f(t)=t2U(t-1)-t2U(t-2). Now we will use linearity (#0) and either #6 or #7. Let's try #7 on t2U(t-1). Here a=1, and g(t)=t2 so that we need the Laplace transform of g(t+1)=(t+1)2=t2+2t+1 whose Laplace transform is (with #0 and #1) 2/s3+2/s2+1/s. We need the other part, the Laplace transform of t2U(t-2) (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=t2+4t+4, or 2/s3+4/s2+4/s. Put all this together, and the answer is
e-s[2/s3+2/s2+1/s]-e-2s[2/s3+4/s2+4/s]
Maple told me:

```>laplace(t^2*Heaviside(t-1)-t^2*Heaviside(t-2),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             s```
Example #5
(Credit Ms. Kadakia and Mr. Shah.)
If the graph of f(t) looks like this, what is F(s)?
This is an obnoxious problem, and is the problem whose discussion led me to make a mess on the board. What a mess! A formula for f(t) between 2 and 3 is t-2, and between 3 and 4 is 4-t. Otherwise the function is 0. We can use this information to write f(t) with various U's. Here it is: f(t)=(t-2)U(t-2)-(t-2)U(t-3)+(4-t)*U(t-3)-(4-t)U(t-4).
If you want to check this, try the Maple instruction plot((t-2)*Heaviside(t-2)-(t-2)*Heaviside(t-3)+(4-t)*Heaviside(t-3)-(4-t)*Heaviside(t-4), t=0..8, thickness=3, scaling=constrained); and the result will look a lot like the graph to the right.
f(t) has four chunks (using linearity) to Laplace transform. Let me do (t-2)U(t-2). According to #6, this is e-2s(1/s). And how about (t-2)U(t-3)? Let's use #7, and get e-3s multiplied by the Laplace transform of ({t+3}-2) which is 1/s2+1/s. Now (4-t)*U(t-3) gives me e3s multiplied by the Laplace transform of 4-(t+3)=1-t, which is 1/s-1/s2. The last piece,(4-t)U(t-4), has Laplace transform -e-4s/s2 using #6. So all together, the Laplace transform is
e-2s(1/s)-e-3s(1/s2+1/s)+ e3s(1/s-1/s2)-e-4s/s2. Wow!
```>laplace((t-2)*Heaviside(t-2)-(t-2)*Heaviside(t-3)+(4-t)*Heaviside(t-3)-(4-t)*Heaviside(t-4),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/[(s-5)2+9], what is f(t)?
This one is complicated. The e-7s tells me that I will have U(t-7) in the answer (a=7 in #6 or #7). What is the inverse Laplace transform of 1/[(s-5)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 s-5? Well, that is handled (see #5) if I multiply by e5t. The inverse Laplace transform of 1/[(s-5)2+9] therefore is e5t(1/3)sin(3t). But #6 tells me that I should substitute t-7 for t in the formula e5t(1/3)sin(3t). Therefore the answer is
e5(t-7)(1/3)sin(3(t-7))U(t-7).
And Maple declares:

```>invlaplace(exp(-7*s)/((s-5)^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

• The techniques requested would all be necessary in this course: no kidding.
• I wrote approximately the following in an e-mail message to a student in the course.  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.]
• A person who is hired for \$50K+benefits (such as a newly graduated student with an engineering degree!) should be able to do elementary computations and discover other facts as needed (in a textbook, conversation, etc.). Such a person should also be able to give simple explanations in writing or orally.

The QotD was to find the Laplace transform of e-5ttU(t-2). Most students answered e-2(s+5){[1/(s+5)2]+[2/(s+5)]}.

### Thursday, September 9

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´=6e3t-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 ert, 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 y1=y and y2=y´. Build a column vector
Y=( y1 y2)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 2-by-2 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 long-term 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´=6e3t-3e-t. At this time please realize that there was a table of Laplace transforms on the board. So the Laplace transform of the right-hand side is easy, using linearity and entries in the table. It is [6/(s-3)]+[-3/(s+1)]. What about the right-hand 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 snY-sn-1y(0)-sn-2y´´(0)-...yn(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 s2Y-sy(0)-y(0). Here we know that y(0)=1 and y´(0)=-1 so that the Laplace transform of y´´ is s2Y-s+1. When n=1, the result states that the Laplace transform of y´ is sY-y(0). Again using y(0)=1, we get sY-1 as the Laplace transform of this y´. Therefore (sigh) the Laplace transform of the left-hand side of the original equation (y´´-4y´) is s2Y-s+1-4(sY-1)=(s2-4s)Y-s+1+4.

Put it all together:
The Laplace transform of
y´´-4y´=6e3t-3e-t, y(0)=1, y´(0)=-1
is
s2Y-s+1-4(sY-1)=[6/(s-3)]+[-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.

s2Y-s+1-4(sY-1)=[6/(s-3)]+[-3/(s+1)] becomes
(s2-4)Y-s+5=[6/(s-3)]+[-3/(s+1)] which gives us
Y=[s/(s2-4)]-[5/(s2-4)] +[6/{(s2-4)(s-3)}]+[-3/{(s2-4)(s+1)}].

[s/(s2-4)] is the Laplace transform of cosh(2t) (an entry in the table!) and ]-[5/(s2-4)] is the Laplace transform of -(5/2)sinh(2t) (an entry in the table slightly modified by linearity).

How about [6/{(s2-4)(s-3)}]? The bottom of this fraction is actually (s-2)(s+2)(s-3), so the partial fractions cell in my brain springs into action: we should be able to write the fraction as the sum of
[A/(s-2)]+[B/(s+2)]+[C/(s-3)]=[{A(s+2)(s-3)+B(s-2)(s-3)+C(s-2)(s+2)}/{(s2-4)(s-3)}].
If we combine the fractions, We know that 6 is supposed to be equal to A(s+2)(s-3)+B(s-2)(s-3)+C(s-2)(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}/(s-2)]+[{3/10}/(s+2)]+[{6/5}/(s-3)] which is {-3/2}e2t+{3/10}e-2t+{6/5}e3t.

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^2-4)*(s-3)),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^2-4)*(s-3)),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)=(et-e-t)/2 and (Euler's formula) sin(t)=(eit-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 eatf(t) is F(s-a).

This is correct because the Laplace transform of eatf(t) is 0infinitye-steatf(t) dt which is the same as 0infinitye-st+atf(t) dt which is the same as 0infinitye-(s-a)tf(t) dt and that is exactly the Laplace transform of f(t) evaluated at s-a, or F(s-a).

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/s14 shifted. So this should be related to the Laplace transform of t13. We needed to fix up the multiplicative constant so that we would get 5. And therefore we multiplied t13 by 5 and divided it by 13! and learned that the Laplace transform of (5/(13!)t13 is 5/s14. Now we needed to shift by -7, and concluded that the Laplace transform of e-t(5/(13!)t13 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(t-3) (the jump is moved to 3) and 4U(t-2)-6U(t-5) (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(t-3)-tU(t-4). The first part, tU(t-3), 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(t-a)f(t-a) is e-asF(s).

Well, the Laplace transform of U(t-a)f(t-a) is (by definition) 0infintye-stU(t-a)f(t-a) dt. But U(t-a) 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(t-a) into 1 in the part with t>a. That is,
0infinityU(t-a)BLAH=0aU(t-a)BLAH+ainfintyU(t-a)BLAH=ainfinityBLAH.
So we have ainfinitye-stf(t-a) dt. I will change variables in this integral. If w=t-a (remember, a is a constant) then dw=dt and t=w+a so that: t=at=infinitye-stf(t-a) dt=w=0w=infinitye-s(w+a)f(w) dw=w=0w=infinitye-swe-saf(w) dw=e-saw=0w=infinitye-swf(w) dw. This is exactly e-asF(s) because w is the variable of integration -- it doesn't matter outside of the integral sign.

What is the Laplace transform of tU(t-3)-tU(t-4)? First look at tU(t-3). I want to make it fit the template of the Second Translation Theorem, which deals with U(t-a)f(t-a). Well, a should be 3. What about f? I want f(t-3) to be t, and therefore, f(t) should be t+3. So the Laplace transform of this first chunk will be e-asF(s) with a=3 and with F(s) equal to the Laplace transform of t+3, which is 1/s2+(3/s). The Laplace transform of tU(t-3) must be e-3s[1/s2+(3/s)]. A similar computation showed that the Laplace transform of tU(t-4) is e-4s[1/s2+(4/s)]. Therefore the Laplace transform of example 1 must be e-3s[1/s2+(3/s)]-e-4s[1/s2+(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.

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(t-a)f(t-a) the function f is be the constant function always equal to 1 and a=Pi/2. Therefore the transform is e-asF(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(t-a)f(t-a)=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-asF(s)=e-[Pi/2]s(s/[s2+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/[s2+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(1-t) for t between 0 and 1, and was 0 elsewhere. The answer is graphed below by Maple.

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!;

1/1245404160```
I 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(t-Pi/2)+Heaviside(t-Pi/2),t,s);

s Pi            s Pi
exp(- ----) s   exp(- ----)
1             2               2
------ - ------------- + -----------
2           2                s
s  + 1      s  + 1```
Finally, 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*(1-t)-t*(1-t)*Heaviside(t-1),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.

HOMEWORK
Please read section 4.3 of the text. Here is another problem to hand in on Tuesday. Write carefully, since I don't yet have a grader!

Special problem #1

You are shown the graph of a function, f(t). It is 0 for t<A, and is also 0 for t>A+B. The function interpolates linearly between (A,0) and (A+B,C). A and B and C are not specific numbers, but letters representing positive real numbers but otherwise not specified) Here is what I would like you to do:

These questions are an effort to get you to think about Laplace transforms a little bit more than looking them up in a table (not that there's anything wrong with that ...). If you have a question, please send me e-mail which I hope to answer.

### Tuesday, September 7

Now I copied Theorem 4.1 (page 193) of the text onto the board. I then checked that the Laplace transform of eat. How did we do this?

0infinitye-steadt= 0infinitye(-s+a)tdt=(Fundamental Theorem of Calculus, but be careful about what variable you are "integrating"!)=(1/[-s+a])e(-s+a)t]0infinity. 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])e0 and e0=1, so that the answer is-(1/[-s+a]) or 1/s-a, 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)|<=CeMt for large t. You could think of this as meaning that the graph of f(t) is caught between +CeMt and -CeMt. If you assume this, then multiplying by e-st when s is greater than M will cause e-stf(t) to decay exponentially (because M-s 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 e-to-the-e-to-the t, for example). There are a bunch of functions which have exponential growth. For example, exponentials (273t). As another example, polynomials. Notice that the phrase exponential growth as used here actually means, "at worst" exponential growth. In fact, t346 does grow more slowly than eMt 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 0inftye-stsin(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])(eit-e-it). Therefore the Laplace transform is (using linearity of Laplace transform) 1/[2i] multiplied by the difference of the Laplace transforms of eit and e-it. The Laplace transform of eit is 1/[s-i] (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/[s-i]}-{1/[s+i]}. Some algebra which we did changes this to (1/[2i])({[s+i]-[s-i]}/{[s-i][s+i]} and this is (1/[2i])([2i]/{s2+1}) which is 1/(s2+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)=0infinitye-stf(t) dt.
Useful Laplace transform asymptotics

1. What happens to F(s) when s-->infinity? Since there is a minus sign in the definition of Laplace transform, the exponential is decreasing. When s gets larger, the exponential gets smaller. In fact, as s-->infinity, e-st-->0. The bump in the integral gets squeezed smaller and smaller, and F(s)-->0 as s-->infinity.
2. What happens to F(s) as s-->0+? Now the e-st factor goes to e0 which is 1, and so F(s) should approach 0infinityf(t) dt: this is the net area underneath f(t) on all of [0,infinity]. So F(s)-->the net area under f(t) as s-->0+.
Let's check this by looking at the Laplace transform of the square wave of height 1 "living on" [1,2]. That Laplace transform is (e-s-e-2s)/s. As s-->infinity, the exponentials on top quickly decrease to 0, and the s on the bottom only squeezes stuff even more. So the first limit above is verified. What happens to(e-s-e-2s)/s as s-->0? The first thing I think of is just "plugging in" s=0. We get (1-1)/0 or 0/0, which doesn't help much. L'Hopital's rule applies here. So:
lims-->0+(e-s-e-2s)/s=[derivative of top]/[derivative of bottom]=lims-->0+(-e-s+2se-2s)/1. But this limit can be evaluated by "plugging in", and its value is (-1+2)/1 which is 1. This is the net area under the square wave.

By the way, if (-1/s)[(e-2s-e-s)/(1-e-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 0infinitye-stf´(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-stdt and v=f(t). Therefore
0infinitye-stf´(t) dt=e-stf(t)]0infinity-0infinity(-s e-stf(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-stf(t)]0infinity. Now s>0, so when t-->infinity, e-stf(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 e0=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/(s2+1). But y(0)=2, so this is sY(s)-2+3Y(s)=s/(s2+1). There are several really neat things to notice.

• The initial condition automatically gets "built into" the Laplace transform version of the ODE.
• The ODE has been "Laplace transformed" to an algebraic equation for Y(s), the Laplace transform of the solution we want.
Now we can solve for Y(s):
sY(s)-2+3Y(s)=s/(s2+1)
(s+3)Y(s)=[s/(s2+1)]+2
Y(s)=[s/{(s+3)(s2+1)}]+[2/(s+3)]
I am running out of different kinds of parentheses!

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)(s2+1)}], seems more difficult to recognize.

We are in the 21st 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.

A student suggested that we use partial fractions on [s/{(s+3)(s2+1)}]. That is, we should write this as [(As+B)/(s2+1)]+[C/(s+3)].

The QotD was to find A and B and C. If we combine the fractions we get (As+B)(s+3)+(s2+1)C on top, and this should equal the top of [s/{(s+3)(s2+1)}]. So we want:
(As+B)(s+3)+(s2+1)C=s
s=-3: ((-3)2+1)C=-3 giving C=-3/10.
The s2 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})/(s2+1)]+[{-3/10}/(s+3)]+[2/(s+3)] which is the same as
[({3/10}s)/(s2+1)]+[({1/10})/(s2+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.

### Thursday, September 2

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´=y2. 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´=y2 are y=1/(C-t). The independent variable in this course will frequently be called t for time, so y is a model of some time-dependent process. How do we know that y=1/(C-t) 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´=y2 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´=y2 and y(0)=1 has the unique solution y=1/(1-t), with domain (-infinity,1). And here's another: y´=y2 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=ert. Then y´´+y=0 becomes magically ert(r2+1)=0. The exponential function is never 0. Therefore the guess solves the ODE exactly when r2+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:

• If y1 is a solution, then y1´´+y1=0.
If y2 is a solution, then y2´´+y2=0.
Add the equations to get y1´´+y1+ y1´´+y1=0. Rearrange and recognize that this is the same as (y1+y2)´´+(y1+y2)=0, so y1+y2 is a solution.
• If y1 is a solution, then y1´´+y1=0.
Multiply the equation by a constant, C, to get C(y1´´+y1)=0. Again, rearrange and recognize that we have (Cy1)´´+(Cy1)=0, so Cy1 is a solution.
This is very pleasant. If we go back momentarily to y´=y2, then look: even though 1/(1-x) and 1/([1/2]-x) are solutions, the sum, {1/(1-x)}+{1/([1/2]-x)} is not a solution, and 7/(1-x) is not a solution. The linearity of the ODE is tremendously convenient. Linearity has an elaborate classical name, the principle of superposition.

Let's use linearity. y´´+y=0 has solutions eix and e-ix, so if C1 and C2 are any constants, then C1eix+C2e-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, yP. What is it?

If yP(x)=C1eit+C2e-it then yP/(x)=iC1eit-iC2e-it. We can get yP(0)=1 by having C1+C2=1. We can get yP(0)=0 by having iC1-iC2=0 which is the same as C1-C2=0. After some massive amount of thought, we managed to solve this system of linear equations: C1=1/2 and C2=1/2. Therefore yP(t)=[eit+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, yV(t). It turns out to be yV(t)=[eit-e-it]/(2i). You should check this!

Why would one want to know the yP and yV 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: 7yP(t)-13yV(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 yP and yV 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:

Euler tells me ...
eit=cos(t)+i sin(t) and cos(t)=[eit+e-it]/2 and sin(t)=[eit-e-it]/(2i)

What happens if we just change a sign in the ODE? If y´´-y=0, the characteristic equation becomes r2-1=0 with roots r=+/-1. Solutions are then C1et+C2e-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:

• yP(t)=(et+e-t)/2 satisfies y´´-y=0 with y(0)=1 and y´(0)=0. This solution is usually called cosh(t), the hyperbolic cosine of t.
• yV(t)=(et+e-t)/2 satisfies y´´-y=0 with y(0)=1 and y´(0)=0. This solution is usually called sinh(t), the hyperbolic sine of t.

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 B2-4AC<0, probably sines and cosines will appear. If B2-4AC>0, the solutions can be written with coshes and sinhes. There will also be some exponential factors (damping or otherwise). What happens if B2-4AC=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!

### Part I of the course: the Laplace transform

Definition Suppose f(t) is defined for all non-negative t: [0,infinity). Then the Laplace transform F(s) of f(t) is 0infinity e-stf(t) dt.

This is a complicated definition.

1. As I remarked in class, the use of the variables t, f(t), s, and F(s) with Laplace transforms is traditional. Almost every text and reference I've seen uses them.
2. The integral is with respect to t, and t is the "dummy variable" in the integral. For example, consider the simpler but strange combination 13x+y dy. I can antidifferentiate this and get xy+(1/2)y2]13 which is 2x+4: quite straightforward, I think. But this is the same as 13x+w dw. The result will again be a function only of x. The y and then the w are dummy variables. Sometimes this can be a bit confusing, and we will need to pay some attention.
3. The integral defining the Laplace transform is an improper integral, and sometimes the behavior of these can be a bit unintuitive. For example, 1infinity(1/x)dx is infinite, but 1infinty(1/x2)dx is finite. I certainly can't tell this by just looking at a rough graph of the two functions. Officially improper integrals are defined as limits. I'll do the first example using limits, quite carefully! But much of the time I won't be so careful and we can sometimes get weird and even wrong results if we are too careless.
Why should we study Laplace transforms?
• The Laplace transform will turn out to be another "trick" which can be used to solve some ODE's. It is a trick that works really well sometimes, and even other times when it doesn't work terrifically, it gives a different way of trying to solve the ODE. I think it is better to know more tricks.
• The Laplace transform will allow us to sove IVP's for ODE's in one computation. We won't need to find homogeneous solutions, look at initial conditions, etc. This is useful.
• Most interestingly, the Laplace transform will allow us to work with rough functions. In calc 1 and much of what follows in the calculus sequece, we look at functions like cos(t17ln(t)). Of course it turns out that such functions may have little relationship to real world data, which can be very rough: things like pounding a piece of metal with a hammer, for example, or abruptly dropping 200 pounds of suger in a 100 gallon tank of water. Both of these situations are probably best modeled by functions with jumps or corners, and those are not easily handled by the delicacy of 250 year old calculus. Laplace transforms will allow us to model them very well.
Example 1 What is the Laplace transform of f(t)=1? We must compute 0infinitye-stdt. I'll work very slowly, with mathematical propriety: 0infinitye-stdt is the limit as A-->infinity of 0Ae-stdt. Now we need an antiderivative with respect to t of e-st. That's -(1/s)e-st. Then we must evaluate -(1/s)e-st]t=0t=A which is -(1/s)e-sA-{-(1/s)}es·0.
I'm trying to go very slowly here, since it our first computation. Now the exponential function's value at 0 is 1. And also the exponential function dies off very quickly for negative real arguments: the limit of ew is 0 as w-->-infinity, and, in fact, it goes so rapidly to 0 that polynomial growth doesn't stop it: w238ew-->0 as w-->-infinity also.
Since A>0, -(1/s)e-sA-{-(1/s)}es·0= -(1/s)e-sA+(1/s)-->1/s as A-->infinity.
Therefore the Laplace transform of 1 is 1/s.

Example 2 What is the Laplace transform of f(t)=t? Now we need to compute 0infinitye-stt dt with a little more care. We need the antiderivative of e-stt 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-stdt so du=dt and v=-(1/s)e-stds.
Now the integral becomes (I am definitely shortening the computation, eliminating all the limit "stuff"):
The pattern:ORIGINAL INTEGRAL=u dv] -v du
0infinitye-stdt=-(t/s)e-st]0infinity-0infinity-(1/s)e-stds.
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)0infinitye-stdt. The integral was computed in example 1 and had value 1/s. So the Laplace transform of t is 1/s2.

Example 3 What is the Laplace transform of f(t)=t2?
Philosophical interruption Much of mathematics is pattern matching, and then attempting to verify the detected patterns. So far we have:
f(t)
F(s)
1
1/s
t
1/s2

which isn't much. I can think of several "patterns" which might extend this table. So let us be a bit more patient, and compute the Laplace transform of t2 directly from the definition: 0infinitye-sttdt. Again, integration by parts works (we're going to do a great deal of integration by parts, so get used to it!):
u=t2 so dv=e-stdt and du=2t dt and v=-(1/s)e-st. The uv] term is -(t2/s)e-st]t=0t=infinity and this vanishes when t=0 surely, and as t-->infinity, since s>0 and exponential decay is faster than polynomial growth the result is again 0. The remaining integral is (keep track of - signs!) -e0infinty-st-(2t/s) dt. We pull out the things which are constant relative to t, and get (2/s)0infinitye-stt dt. But the integral is the Laplace transform of t, which we already know is 1/s2. Therefore we multiply this by 2/s to get the Laplace transform of t<2.
f(t)
F(s)
1
1/s
t
1/s2
t2
2/s3

If we look at the process, we see that if we want to get the Laplace transform of t3 we will need to integrate by parts, and the integration will "spit out" 3/s to multiply by 2/s3. Here is where I hope a good physical scientist will "know" the Laplace transform of nonnegative integer powers of t, and a mathematician will attempt a proof by mathematical induction. In any case:

If n is a nonnegative integer, the Laplace transform of tn is n!/sn+1.

The QotD was: if f(t)=3+7t2-33t4?
Here I wanted people to use the linearity of the Laplace transform.