Here are some examples of scalars and vectors.

• Geometry and mechanics
  Here vectors are forces and directed line segments in R² (the plane) and R³ (space). These objects may represent forces or fluid flow or heat flow or ... and we add them and multiply them by scalars (in this case, just real numbers). Adding them gives resultants and various components of the vectors have names like flux. We can draw nice simple pictures whose geometry is frequently appealing.
• Vectors in ODE's
  Here the vectors are n-tuples of real or complex numbers. They arise as part of the effort to solve ODE's efficiently. Complex numbers are introduced because diagonalizing matrices allows efficient computation of various things, and you can diagonalize matrices without using complex numbers.
• Laplace transforms If we take the Laplace transform of systems of ODE's, as in section 4.6 of the text, we get collections of equations involving the Laplace transforms of unknown functions with rational functions (quotients of polynomials) as coefficients. For example, if you've done some problems in that section of the book, you would expect several equations similar to what follows: [(s-1)/(s²+4)]X(s)+ {s/(s+2)³]Y(s)=1/(s+7)+2s -7 and then we want to find expressions for one of the unknown functions (X(s) or Y(s)) in terms only of s, and then take the inverse Laplace transform. The initial conditions are all invisibly part of equations like the one I just wrote. The vectors here are X(s) and Y(s), and the scalars are rational functions such as [(s-1)/(s²+4)] and {s/(s+2)³] and 1/(s+7)+2s. You can multiply and divide and add and subtract rational functions. The results will again be rational functions.
• Fourier series We will look at this in the last part of the course. Fourier series look like 3sin(5t)+5cos(7t): sums of constants multiplied by trig functions. The applicable trig functions here will be sine and cosine (not cosecant, thank goodness!). The vectors are functions like sin(5t) and cos(7t), and the scalars in this setting are usually real constants. These series are an extremely useful way to analyze certain partial differential equations, PDE's, and their boundary value problems. These problems occur when trying to understand heat flow, diffusion, vibrations ... lots of stuff. There also turns out to be a complex version of these sums, involving things like (5+2i)e^3i t. Here the scalars are complex numbers, while the vectors are functions like e^3i t.
• Digital signal processing Here the sums are sort of square waves which are the vectors, and the scalars essentially turn out to be (this seems ludicrous when first encountered!) just 0 (off) and 1 (on). Extremely elaborate ideas and algorithms having to due with storage and efficient transmission and transformation of signals are expressed in this language. A chief algorithm is the Fast Fourier Transform (connected with such tools as CAT scans and magnetic resonance), and one way of looking at this algorithm is that it involves writing a matrix as a product in a particularly efficient way. The scalars seem like a ludicrously small collection, but signals are constructed with many, many scalars put together.

HOMEWORK
I would like to give an exam in two weeks, on Monday, October 10. The exam would cover our work on Laplace transforms and some basic linear algebra. Please check the draft version of a formula sheet for the first exam and give me comments.
The exam will cover what we've done on Laplace transforms and two lectures on linear algebra, so you probably should read ahead about linear algebra (see the syllabus & textbook problems.
Further information about the exam will be available soon.

Confession

Thursday, September 22

Laplace transforms

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 rapidly and directly, 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.

1. A simple mechanics problem
2. Some math mumbles
3. Chem engineering problem
4. Several mechanical engineering problems

A simple mechanics problem
I asked people to consider the frictionless eraser. 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. So here we go: I 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(t) 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. What doess the total area represent? Well, 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, as Mr. Wolf suggested. Professor M. Kiessling, a wonderful mathematical physicist, 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 slope is directly proportional to the total net area.

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]/s and X(s)=(1/A)1/s3-(1/A)e-As(1/s3). We now take the inverse Laplace transform. You will need to know the table and the second translation theorem! x(t)=(1/A)(1/2)t2-(1/A)(1/2)(t-A)2U(t-A). For 0<t<A, x(t)=(1/A)(1/2)t2. Since (t-A)2=t2-2At+A2, after some algebra (!!), we see that if t>A, x(t)=(1/A)(1/2)(2At-A2)=(1/2)(2t-A)=t-A/2. It can be checked that x(A)=A/2 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.

Limits of the motion and what we expect What happens now as A-->0? Remember: Almost every reasonable limit should exist! 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)=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 is the limit of these square impulse functions as A-->0⁺. 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 some integrals. They resembled the following:

I computed the integral _-300⁴⁰delta(t-2) dt: this was 1.
The integral _-300^-200delta(t-2) dt was 0. The bump or impulse was centered at 2, and 2 is not in the interval from -300 to -200.
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.

The Laplace transform of delta
In fact, the Laplace transform of delta(t-t₀) is easy. We need to compute ₀^infinitye^-stdelta(t-t₀) dt. But the delta only notices the amplitude of the function multiplying it when delta's argument is 0, so the value of this integral is e^-t₀s.

Another thought about delta
Remember: Almost every reasonable function should have a derivative.
Let's approximate U(t) by another function, f_epsilon(t). Here you should think of epsilon as a very small positive number. For negative t, f_epsilon(t)=0. For positive t>epsilon, f_epsilon(t)=1. In a small interval immediately to the right of 0, f_epsilon(t) increases smoothly from 0 to 1. You are supposed to think that f_epsilon(t) closely approximates U(t). Then f_epsilon´(t) is a bump localized (?) in a small interval to the right of 0. The total integral under the bump is 1, using the Fundamental Theorem of Calculus since _-10³⁷f_epsilon´(t) dt=f_epsilon(t)]_-10³⁷=f_epsilon(37)-f_epsilon(-10)=1-0=1.

If we multiply f_epsilon´(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 f_epsilon´(t)g(t) will be approximately g(0) multiplied by the integral of f_epsilon, which is 1, so the result is approximately g(0). Thus f_epsilon´(t) is approximately delta(t). So as epsilon-->0⁺, f_epsilon-->U and f_epsilon´-->delta. So, "clearly" U´=delta.
I sketched these ideas here because many people in the engineering and applied science communicty (physics, chemistry) find them useful. Maybe you will, also.

A traditional mixing problem
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 expect 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 (it should be easy for you by now!) 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);
                             t                                      t
                  -50 exp(- ----) + 50 + 5 Heaviside(t - 20) exp(- ---- + 2/5)
                             50                                     50

Some pictures Maple made for me

Here's the solution curve for the problem just solved, with an instantaneous increase of 5kg at t=20. The curve jumps up, and then begins to approach 50.	I modified the problem, and dumped in 30 kg of salt at time 60. So you can see the jump above the asymptote, and then the curve trends down towards 50.

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 helps to 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 (!) with a "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 using the second translation theorem 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-2Pi). Now go through the Laplace transform, using the initial conditions:
s²Y(s)-sy´(0)-y(0)+Y(s)=e^-Pi·s+e^-2Pi·s becomes
s²Y(s)+Y(s)=e^-Pi·s+e^-2Pi·s and then
(s²+1)Y(s)=e^-Pi·s+e^-2Pi·s
so that
Y(s)=[e^-Pi·s+e^-2Pi·s]/(s²+1).
The inverse Laplace transform tells me that
y(t)=sin(t-Pi)U(t-Pi)+sin(t-2Pi)U(t-2Pi).

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<2Pi, then y(t)=sin(t-Pi), a sine wave "starting" when time is Pi. We in fact verified, both by pure thought (move the sine wave Pi units) and by using the addition formula for sine, that sin(t-Pi)=-sin(t), so in this interval, y(t)=-sin(t), a positive bump.
• If 2Pi<t, then y(t)=sin(t-Pi)+sin(t-2Pi). Since sine is periodic with period 2Pi, sin(t-2Pi)=sin(t) (again, you can use the addition formula for sine if you wish). But sin(t=Pi) is still -sin(t), so that for these t's, y(t)=-sin(t)+sin(t)=0.
  The two impulses exactly cancel (remember, there is no damping in this model!). That's the story. More next time.

HOMEWORK
The grader, a wonderful human being, talked to me about the homework. She encouraged me to ask some even numbered questions. She also remarked that students should give some reasons for their transitions from the problem statements to the problem solutions. I emphatically agree with that, since this will be needed on exams.
Monday's lecture will be the last on Laplace transforms. Please read the appropriate sections of the book and do more than the minimal number of problems! Please hand in Monday these problems: 4.4: 26, 33 and 4.5: 8, 9.

Find the Laplace transform of tU(t-7)
Use a version of the second translation theorem: the Laplace transform of g(t)U(t-a) is e^-as multiplied by the Laplace transform of g(t+a). So a=7, and g(t)=t, and g(t+a)=t+7 which has Laplace transform 1/s²+7/s. The answer is e^-7s( 1/s²+7/s).

Find the Laplace transform of e^-9tt
We need the first translation theorem: the Laplace transform of e^atf(t) is F(s-a). Here a=-9 and f(t)=t. So F(s)=1/s², and the desired Laplace transform is 1/(s+9)². The + occurs because s-a=s-(-9)=s+9.

Mr. Clark intelligently remembered a fact. So he did the following: the Laplace transform of e^-9t is 1/(s+9), so the Laplace transform of te^-9t is -d/ds[1/(s+9)], and this is 1/(s+9)². I needed to be reminded that there are two minus signs in the result and they cancel. Sigh.

Find the Laplace transform of e^-5ttU(t-2)
This problem is vicious (viscous?). It involves a concatenation of both translation results.
        Today's word concatenation
        To connect or link in a series or chain.
Let's do this two ways.

Find the convolution of e^At and e^Bt
So we must compute ₀^tf(tau)g(t-tau) dtau where f(t)=e^At and g(t)=e^Bt. I'll assume that A is not equal to B. So:
₀^te^Ataue^B(t-tau)dtau and this is the integral of e^Atau+Bt-Btau=e^Bte^(A-B)tau. I separated out the first term because "dtau" it is a constant and can come out of the integral. Then we need to compute more:
e^Bt₀^te^(A-B)taudtau=e^Bt[1/(A-B)]e^(A-B)tau]₀^t=e^Bt[1/(A-B)]{e^(A-B)t-e⁰}. This is a mess. Let me rearrange it a little bit. It is equal to [1/(A-B)]{e^At-e^Bt}.

Think about this result, and let's include a computation from the previous class. Let me rearrange things algebraically a bit.

Function	Its Laplace transform
t2	2/s3
t3	6/s4
(1/60)t6	{6!/60}/s7 Notice that 6!/60 is 12.
eAt	1/(s-A)
eBt	1/(s-B)
[1/(A-B)]{eAt-eBt}.	[1/(A-B)]{1/(s-A)-1/(s-B)}. Notice that, combining fractions, this happens (?) to equal 1/[(s-A)(s-B)]

Convolution and Laplace transform
The Laplace transform of the convolution is the product of the Laplace transforms: the Laplace transform of f(t)_*g(t) is F(s)·G(s).
This is Theorem 4.9 of the text (pp.216-217) and you should at least glance at the proof there, please. The table above contains two examples of this fact.

Theoretical consequences

• Commutativity 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^tf(tau)g(t-tau) dtau=₀^tg(tauf(t-tau) dtau. As Mr. Mostiero noticed, this has the consequence that my Monday and Thursday definitions of convolution agree, which is nice.
  Please 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.
• Associativity 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))

What is the convolution of t and t² and t³ and t⁴?
We can compute t_*t²_*t³t⁴ by computing the Laplace transforms of each function, multiplying them, and then taking the inverse Laplace transform. You should judge if this is "easier" than a direct computation. So:
     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.

What is the convolution of e^3t and sin(5t)?
I'll try the same idea. The Laplace transform of e^3t is 1/(s-3) and the Laplace transform of sin(5t) is 5/(s²+5²). Therefore the Laplace transform of the convolution is the product of these Laplace transforms:
5/[(s-3)(s²+25)]. We can "recognize" this as the Laplace transform of something if we use partial fractions:

       5          A     Bs+C  
-------------- = --- + -------
[(s-3)(s2+25)]   s-3    s2+25

What is the Laplace transform of ₀^tf(tau)dtau?
This is the convolution of f(t) and the function 1 (that is, the function whose value for any t is 1). Therefore the Laplace transform is the product of the Laplace transforms of f(t) and 1: that is, just F(s)/s. This result should be contrasted with what we got for derivative: the Laplace transform of f´(t) is sF(s)-f´(0). The results aren't exactly opposite since the f' has an initial condition built in, but then the integral starts at 0 and is therefore equation to 0 when t=0.

$ and *
Please see here for a discussion of convolution and the "present value of an income stream"

Textbook problem #43 in section 4.4
This problem asks for the solution of an integral equation:
f(t)=1+t-(8/3)₀^t(t-tau)³f(taudtau.

We should do this:

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.

Partial fractions ... World's quickest practical review. This is an algebraic method of transforming quotients of polynomials (rational functions). Students in calculus see it for the first time as a method allowing (in theory!) any rational function to be antidifferentiated in terms of familiar functions. If the quotient is P(t)/Q(t), then there are a sequence of steps: Divide. That is, divide P(t) by Q(t) to get a polynomial plus what's called a proper fraction, something like R(t)/Q(t), where the degree of R is less than the degree of Q. Practicality Very easy, very fast. Now we have R(t)/Q(t). Find the roots of Q(t) in order to factor Q(t). If we're doing this with real coefficients, then the factors will be linear (like t-5 or 3t-17) or "irreducible quadratics" (like t2+78 or 5t2-4t+103 [hey, B2-4AC=16-4·5·103]<0). The irreducible quadratics occur because complex roots happen in conjugate pairs (a+/-bi) for real polynomials. Practicality There are explicit formulas for the roots of polynomials of degree 2, 3, and 4. It has been proved that there are no such algebraic formulas for general polynomials of higher degree. I use the word "general" here because you might get lucky: (t-1)(t-2)(t-22)(t-33)(t+203) certainly "factors". In theory, every polynomial has a complete collection of complex roots (as many roots as the degree of the polynomial, counted with multiplicity -- so (t-55)20 has a multiplicity 20 root at 55). Finding the roots exactly is generally impossible. So one must resort to approximate root finding, and this is possible, but can be lengthy. Now R(t)/Q(t) is R(t)/(some sort of product). If you have ... uhhh ... (t-55)20 then you've got to write SUMj=120Constantj/(t-55)j. If you have irreducible quadratics, then something like (t2+22)10 gives rise to SUMj=110[Another constantj+Yet another constantjt]/(t2+22)j. Etc. Now you add everything together and set it equal to R(t)/Q(t). Solve for all the darn constants. To integrate such things, you may need to do things like "complete the square", etc. This is unappealing but can be done quite readily. Practicality This is solving a big bunch of linear equations. Certainly computationally possible, and can be done fast.

Back to the problem. We need to find the inverse Laplace transform of [s³+2s²]/[s⁴+16]. The poor old prof pooped out (p.o.p.p.o.) here. Well, I can, darn it, factor s⁴+16. First, I solve s⁴=-16. This means s²=+/-4i, and then another square root gives me s=sqrt(2)(+/-1+/-i). What the heck is going on? Well, if I want to do it only with real numbers, I notice that s⁴+16 is never 0 (it is always at least 16 for real s's), so it has no real roots.
I will guess (don't do this yourself!) that s⁴+16=(s²+As+4)(s²+Bs+4. The right-hand side multiplies out to be s⁴+16+other terms. The other terms are
     s terms with coefficients 4(A+B)
     s² terms with coefficients AB+8
     s³ terms with coefficients A+B.
So choose A=-B, then AB+8 will be 0 exactly when A=2sqrt(2) and B=-2sqrt(2). The irreducible factors are
s²+2sqrt(2)s+4 and s²-2sqrt(2)s+4.

YUCK!!!
I do prepare, and I had looked at the answer in the back of the book. The answer there is: (3/8)e^2t+(1/8)e^-2t+(1/2)cos(2t)+(1/4)sin(2t). What the heck! Maple asserts that the inverse Laplace tranform of [s³+2s²]/[s⁴+16] is

> invlaplace((s^3+2*s^2)/(s^4+16),s,t);
         1/2       1/2         1/2                1/2          1/2     1/2          1/2
    1/2 2    sinh(2    t) cos(2    t) + 1/2 cosh(2    t) (sin(2    t) 2    + 2 cos(2    t))

s3+2s2    A     B    Cs+D
------ = --- + ---  -----
s4+16    s-2   s+2   s2+4

If any student sees any errors I made, please tell me. Here is a link to a solution of problem #38 in section 4.4 which I did a year ago. At least I think it did it correctly! It is similar to the problem I attempted here!

HOMEWORK
Please read ahead in the book. The next section (4.5) has the most mind blowing part of Laplace transforms.

Interestingly enough, the Heaviside function is only recently (version 7)
supported by default in MATLAB. But, it works as expected. Looks like they
chose to just leave U(0) undefined, though.

>> x=-5:5

x =

     -5    -4    -3    -2    -1     0     1     2     3     4     5

>> heaviside(x)

ans =

      0     0     0     0     0   NaN     1     1     1     1     1

Trying the function in MATLAB <= 7 or Octave (an open source MATLAB
"clone") results in brokenness.

The Second {Shifting|Translation} Theorem
(Textbook Theorem 4.7, p.208) If the Laplace transform of f(t) is F(s) and a>0, the Laplace transform of f(t-a)U(t-a) is e^-asF(s).

Since this is a 640 course, I should give some justification of this result. So I wrote the following:
The Laplace transform of U(t-a)f(t-a) is (by definition) ₀^infinitye^-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, ₀^infinityU(t-a)BLAH=₀^aU(t-a)BLAH+_a^infintyU(t-a)BLAH=_a^infinityBLAH.
So we have _a^infinitye^-stf(t-a) dt. We can change variables in this integral. If w=t-a (remember, a is a constant) then dw=dt and t=w+a so that:
_t=a^t=infinitye^-stf(t-a) dt=_w=0^w=infinitye^-s(w+a)f(w) dw=_w=0^w=infinitye^-swe^-saf(w) dw=e^-sa_w=0^w=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.

You don't need to think about this verification again. Just learn how to use it.

Examples
What is the Laplace transform of t²U(t-5)?
This is a bit tricky. The "template" of the Second Translation Theorem is f(t-a)U(t-a). I guess a should be 5. But then we need to "recognize" f(t-a)=f(t-5) as t². Here is the trick. t²=([t-5]+5)2. This isn't very profound, but it will work. So ([t-5]+5)2=(t-5)²+2(t-5)+25, and apparently f(t-5)=(t-5)²+2(t-5)+25 so that f(Q)=Q²+2Q+25. Therefore the Laplace transform of f is (2/s³)+2/s²+(25/s). THEREFORE the Laplace transform (according to the Second Translation Theorem) is e^-5s[(2/s³)+2/s²+(25/s)].

What about the inverse Laplace transform of, say, e^-3s/s⁷? This should come from a translated version of Constant·t⁶. The -3 part leads to (t-6)⁵ but we need to adjust for the proper constant. So I think this should be (1/6!)(t-3)⁶U(t-6).

Another version of the Second {Shifting|Translation} Theorem
I find an alternate version of the Second Translation Theorem a bit easier to use. (It's all psychological -- is everything psychological?). Here it is:
The Laplace transform of g(t)U(t-a) is e^-as multiplied by the Laplace transform of g(t+a).

Examples
I think I tried to find the Laplace transform of U(t-Pi)sin(t). This is e^-Pi s multiplied by the Laplace transform of sin(t+Pi). Now sin(t+Pi) is -sin(t), which has Laplace transform -1/(s²+1)

QotD
What is the Laplace transform of the function which is -(t-1)(t-3) for t between 1 and 3, and 0 otherwise?

Weird formula
Here is another formula from the textbook. The Laplace transform of t·f(t) is F´(s). Why is this?

Since F(s)=L(f)(s)=₀^infinitye^-stf(t) dt we could try to differentiate F(s). Thus (d/ds)F(s)=(d/ds)₀^infinitye^-stf(t) dt.

Now push (!) the derivative inside the integral sign:
(d/ds)F(s)=₀^infinity(d/ds)e^-stf(t) dt. and (d/ds)e^-stf(t)=-te^-stf(t) since d/ds only notices appearances of s.

(d/ds)F(s)=₀^infinity-te^-st\f(t) dt=-₀^infinitye^-stt·f(t) dt which is minus the Laplace transform of t·f(t).

Therefore the Laplace transform of -t·f(t) is (d/ds)F(s). The textbook remarks that you can repeat this n times, and the result is:
the Laplace transform of (-1)ⁿtⁿ·f(t) is (dⁿ/dsⁿ)F(s).

I remarked that the interchange of derivative and integral is not always valid, and manipulations of this sort probably helped Heaviside get into trouble with the academic establishment. Look only briefly at the following example which verifies that the interchange is not always true.

A complicated academic example
Here is an intricate example copied from Counterexamples in Analysis by Gelbaum and Olmstead, showing that "differentiation under the integral" is not always valid. Knowing the details of this example is definitely not part of Math 421, certainly, but I include it here so that near maniacs can verify there may be some problems with interchanging differentiation and integration.
The example uses the function f(x,y)=(x^3/y^2)exp(-x^2/y) if y is not 0, and 0 if y=0.

What the heck does this function look like? For fixed x it is continuous in y. For fixed y, it is continuous in x. But it is not continuous in x and y jointly. Look at the curve y=x^2. So f(t,t^2)=(1/t)e^-1, and that is certainly not continuous at t=0. Slices with y=constant reveal a bump which gets higher and moves closer to the origin as y-->0⁺.

The integral with respect to x can be done easily (the function is x³e^-x2 with various constants thrown in!). So _y=0^y=1f(x,y)=xe^-x2, and the derivative of this integral is e^-x2(1-2x²). The value of this at x=0 is 1.

Now the partial derivative of f with respect to x is (for x not equal to 0) e^-x2/y( 3x²/y²-2x⁴/y³ ). The integral of this mess dy from y=0 to y=1 is e^-x2(1-2x²).

For x=0, the partial derivative is 0 (look at the formula!), and the integral is 0.

Notice now that the integral of the partial derivative with respect to x at x=0 is 0, but the derivative of the integral of the function with respect to y at x=0 is 1. These values are not the same!

Convolution
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)=₀^tf(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".

What's the convolution of t² and t³? This is ₀^ttau²(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
₀^ttau²(t-tau)³dtau= ₀^ttau²[1t³-3t²tau¹+3t¹tau²-1tau³]dtau= ₀^ttau²t³-3t²tau³+3t¹tau⁴-tau⁵dtau= (1/3)tau³t³-(3/4)t²tau⁴+(3/5)t¹tau⁵-(1/6)tau⁶]_tau=0^tau=t= (1/3)t⁶-(3/4)t⁶+(3/5)t⁶-(1/6)t⁶.
Wow, what a computation. Of course, the result can be simplified (who could possibly care?) to get (1/60)t⁶.

Mr. O'Sullivan suggested the following web page: http://mathworld.wolfram.com/Convolution.html. It has a discussion of convolution and an animation picture of several convolutions.

I need to give more background about this material, and I will.

HOMEWORK
Please read sections 4.3 and 4.4 and look at the problems from the syllabus for these sections. I think if you practice these problems, life with the Laplace transform will be better, and certainly less confusing.
Please hand in the following problems on Monday, September 19: 4.3.1: 5, 4/3/2: 47, 69, and 4.4: 3.

Diary entry in progress! More to come.

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 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-s^n-1y(0)-s^n-2y´´(0)-...-y^(n-1)(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 them, 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. ME alert n=4: vibrating beam. Please see example #9 on p.211 of the text.

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´, including the stated initial conditions) is s²Y-s+1-4(sY-1)=(s²-4s)Y-s+1+4.

So the Laplace transform of
y´´-4y´=6e^3t-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
(s²-4s)Y-s+5=[6/(s-3)]+[-3/(s+1)] which gives us
Y=[s/(s²-4s)]-[5/(s²-4s)] +[6/{(s²-4s)(s-3)}]+[-3/{(s²-4s)(s+1)}].

In order to find y(t), we should get the inverse Laplace transform of Y(s). Using linearity, we "only" need to get the inverse Laplace transforms of each of the pieces. I think I only analyzed one of the pieces before I gave up: [6/{(s²-4s)(s-3)}].

So we use partial fractions. The bottom of this fraction is actually (s-4)s(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-4)]+[B/(s)]+[C/(s-3)]=[{A(s)(s-3)+B(s-4)(s-3)+C(s-4)(s)}/{(s²-4)(s-3)}].
If we combine the fractions, we know that 6 is supposed to be equal to A(s)(s-3)+B(s-4)(s-3)+C(s-4)(s). The conclusion is easy:
    s=0 gives 6=12B so B=1/2.
    s=-4 gives 6=4A so A=3/2.
    s=3 gives 6=-3C so C=-2.
Therefore we need to find the inverse Laplace transform of [{3/2}/(s-4)]+[{1/2}/(s)]+[-2/(s-3)] which is {3/2}e^4t+{1/2}e^0t+{-2}e^3t.

I just got Maple 10 at home. This is the same version of Maple which is in the Rutgers computer labs. Here's a little bit of the dialogue. # on a line indicates a comment and Maple ignores it.

> with(inttrans);  # loads various useful diff'l equations transforms

[addtable, fourier, fouriercos, fouriersin, hankel, hilbert, invfourier, invhilbert, invlaplace,invmellin, laplace, mellin, savetable]

> invlaplace(6/((s^2-4*s)*(s-3)),s,t);
                                   3/2 exp(4 t) - 2 exp(3 t) + 1/2

> invlaplace(s/(s^2-4*s)-5/(s^2-4*s)+6/((s^2-4*s)*(s-3))-3/((s^2-4*s)*(s+1)),s,t);
                             11
                             -- exp(4 t) + 5/2 - 2 exp(3 t) - 3/5 exp(-t)
                             10

What kinds of functions theory interlude #1
The classical Laplace transform works with functions that have growth no bigger than exponential growth. This means that a function f(t) must satisfy some sort of estimate of the following type:
There are positive constants A and B so that |y(t)|<Ae^Bt for t>C, where C is some additional positive constant.
There's sort of a picture of the situation to the right. The "Stuff happens" doesn't sort of matter. What matters is that for t large enough (t>C), y(t) is "trapped" between Ae^Bt and -Ae^Bt. The two curves occur because there are absolute value signs around the y(t). If y(t) is trapped that way, then in the integral that defines the Laplace transform, ₀^infinitye^-sty(t) dt, when s>B, the integral is less than e^(B-s)t in absolute value, and since this is exponential decrease the integral must converge.

As I remarked, polynomials have exponential growth. Maybe that is "clear" for t², but not for, say, t^216,000. In fact, I bet that for t large enough positive, |t^216,000| is less than even .003e^.000007t. We discussed this, and decided that this could be verified with L'Hopital's rule. Look at the quotient
t^216,000/( .003e^.000007t)
as t-->infinity. Both the top and the bottom go to infinity. Use L'H lots of times. If you use it 216,000 times, the result is
(Some huge constant)/[(Another stupid constant)e^.000007t]
because the polynomial's powers go down to 0, and the bottom is still an exponential with positive exponent, and just "spits out" positive multiplicative constants upon differentiation. But the limit now, after 216,000 uses of L'H is 0, because the fraction is really just
CONSTANT/(exponential growth)
Therefore for large enough t, this is less than, say, 1.

Even something like t^216,000e^30003t has exponential growth. It is bounded eventually by some multiple of e^30004t or even just e^(30003.001)t.

Here is a candidate suggested by a student (who?) for a function which does not have exponential growth:
t^t=e^ln(t)·t.
Indeed, I agree: ln(t) can't be bounded by ANY POSITIVE CONSTANT, so there's no B that will work.

What we're investigating here in Math 421 is the basic, classical Laplace transform. It has been fiddled with in many ways. So actually there are ways that functions with faster growth can be used. And there are ways that, say, difference equations (a discrete analog of differential equations) can be solved with similar tools. But let's learn the classical case well enough so that you will be able to use and understand the variant methods.

Inverse Laplace transforms theory interlude #2
Well, here in the board (I said, motioning to the side blackboard) is a table of Laplace transforms. It turns out that it is also, more or less, sort of, almost, also a table of inverse Laplace transforms. Here in a wonderfully logical 640 subject course what do the words "more or less, sort of, almost ..." mean?

Well, I considered the following example. Certainly our table tells us that the Laplace transform of t² is 2/s². Let's call t² by the name, h(t), and let's further define the function g(t) by the stipulations that g(t) is -1 when t=3 and g(t) is 40 when t=6 and otherwise g(t)=t². I know the darn picture is not drawn to scale, but the poetic truth is there. So, anyway, certainly h and g are distinct functions. But they have the same Laplace transform. Look, the formula for the Laplace transform of y(t) is ₀^infinitye^-sty(t) dt. And the integral doesn't even notice if you change the values of the integrand (the function that's being integrated) at a couple of points. (Look at the darn definition of the integral, if you don't believe it, and take a Riemann sum with very, very narrow width around the jump places.) So there can be two functions whose Laplace transforms are exactly the same. Well, but for all practical purposes if you are in a physical situation, you won't notice that change at a point. And it turns out, more or less, that this is the only kind of problem you'll face.

Lerch's Theorem (I'm sorry but I like the name.)
If two functions have the same Laplace transform, then they differ only on a collection of values that the integral doesn't notice.

Yeah, this can be more precise, but that's all I want to say. But, you may ask, what if we are supposed to find an inverse Laplace transform? For example, the inverse Laplace transform of 2/s³? Well, yeah again, you do have to choose between h(t) and g(t). What everyone chooses is a function that is as continuous as possible. So if the left and right-hand limits agree at a point, that's the value the inverse Laplace transform should have at the point. And the algorithms in symbolic algebra packages are designed to make such a choice. And also, there is an inverse Laplace transform with an integral formula, called Bromwich's integral, which "automatically" gives you the most continuous (there's a phrase!) inverse Laplace transform. All these things can be implemented symbolically and numerically.

You can find lots of references to this stuff on the web.

Our major present goal is to expand our table of Laplace transforms (and, therefore, inverse Laplace transforms). The two shifting or translation theorems are very important in accomplishing this.

The first {shifting|translation} theorem
If F(s) is the Laplace transform of f(t), then the Laplace transform of e^atf(t) is F(s-a).

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

Easy examples
Uhhh, can we find the Laplace transform of e^5tsin(3t)? The table stated that the transform of sin(3t) is 3/(s²+9). But (take a=5) the Laplace transform of e^5tsin(3t) must then be 3/((s-5)²+9).
More ofter we'll need to apply this result backwards. Can we find a function whose Laplace transform is 5/(s+7)¹⁴? We look at the table and see that 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 multiply t¹³ by 5 and divided it by 13! so that the Laplace transform of (5/(13!)t¹³ is 5/s¹⁴. Now we need to shift by -7, and conclude that the Laplace transform of e^-7t(5/(13!)t¹³ is 5/(s+7)¹⁴. Isn't this cool!

The Heaviside function
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. U 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 U(t)+4U(t-3)-2U(t-6) (the graph "starts" at 0 with a jump up at 1, jumps up 4 (to 5) at 3, and then down 2 (to 3) at t=6 - aside from the jumps the graph just is pieced together from horizontal line segments.

Using U
U(t) allows us to write formulas for other piecewise defined functions.

Suppose y(t) is the function which is 0 for t<0, is t² for t in the interval [0,1], is 1 in between 1 and 3, is (t-4)² in the interval [3,4], and is 0 for "later" t. I hope I've drawn a picture of y(t) to the left.
What's a formula for y(t)? I sort of work from left to right in t. First there's nothing. Then we want to "turn on" t² at t=0, so we need t²U(t). Then we need to turn off t² at t=1, and turn on a height of 1. Hey, change the formula to t²U(t)+(-t²+1)U(t-1). Now let's continue to the right, where we need to turn of the height of 1, and turn on the downway parabolic arc. This change is (-1+(t-4)²)U(t-3) and should be added to what we already had. Then, finally, at 4, turn off (t-4)² with -(t-4)²)U(t-4). So a formula for this function is t²U(t)++(-t²+1)U(t-1)+ (-1+(t-4)²)U(t-3) -(t-4)²)U(t-4)

Here is what Maple does. The first instruction defines the function, y(t). This uses the builtin function "Heaviside", what we call U(t). The second instruction plots it. And the plot is shown.

>y:=t->t^2*Heaviside(t)+(-t^2+1)*Heaviside(t-1)+(-1+(t-4)^2)*Heaviside(t-3) -(t-4)^2*Heaviside(t-4):

>plot(y(t), t=-1..6,thickness=3,scaling=constrained);

 

Return of the entrance exam
I retruned the Entrance Exam. I view this exam as a very useful diagnostic, which has shown some correlation with final course grades in Math 421. For me, it is useful because the exam shines a light on two aspects of students. First, it clearly shows student knowledge of representative content: everything asked on the exam will be relevent to parts of the course, and, actually everything asked on the exam is important in sections of the course. Second, given an Entrance Exam like this to a group of somewhat sophisticated educational "consumers" reveals a bit about their psychology: are they willing to work, to review, to ask questions ... students who do these things are more likely to learn, and not just barely survive. As I mentioned in class, this is a fairly advanced undergraduate course, and the person most resposible for teaching you is ... well, you! I hope I will be helpful, but stuff will go by very rapidly. I strongly recommend working with one or more other students in the course on a regular basis, meeting weekly or more often for an hour or two. This is likely to keep you "on task", and working together you will increase the odds of success.

HOMEWORK
You should be reading and doing the problems from sections 4.3 and 4.4, please.

Laplace transform is linear
L(y₁(t)+y₂(t))=₀^infinitye^-st[y₁(t)+y₂(t)] dt=₀^infinitye^-st[y₁(t)]+e^-st[y₂(t)] dt=₀^infinitye^-sty₁(t) dt+₀^infinitye^-sty₂(t) dt=L(y₁(t))+L(y₂(t)).
Similarly, if c is a constant, then L(cy(t))=cL(y(t)).
Linearity is a tremendously useful property, and we will use it constantly.

Then I copied most of Theorem 4.2 (on p. 193 of the textbook). I checked that the Laplace transform of e^at is 1/(s-a) as the theorem stated. Here is what we did.

₀^infinitye^-ste^adt=₀^infinitye^(-s+a)tdt=(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])e⁰ and e⁰=1, so that the answer is-(1/[-s+a]) or 1/s-a, as Theorem 4.1 states.
In many of these computations, there will be plenty of minus signs and much opportunity for error. I don't know how to avoid all errors.

I wanted to verify the theorem's result for the Laplace transform of cos(kt). I remarked that the definition tells me that I need to compute ₀^inftye^-stcos(kt) dt. Maybe this really isn't too difficult to compute directly. I would integrate by parts. But I will need to integrate by parts twice, and the chance for error is large. Let me show you how to do the computation another way.

Last time I reminded people that cos(theta) is (1/2)(e^itheta+e^-itheta). Therefore the Laplace transform of cos(kt) (so kt will be theta) is (using linearity of Laplace transform!) is 1/2 multiplied by the sum of the Laplace transforms of e^ikt and e^-ikt. The Laplace transform of e^ikt is 1/[s-ki] (I'm using the result we just derived, with a changed to ki). The Laplace transform of e^-ikt is 1/[s+ki] (I'm using the result we just derived, now with a changed to -ki). Therefore the Laplace transform of cos(kt) is (1/2)({1/[s-ki]}+{1/[s+ki]}). Some algebra which we did changes this to (1/2)({[s+ki]+[s-ki]}/{[s-ki][s+ki]}) and this is (1/2)([2s]/{s²+k²}) which is s/(s²+k²), the formula I wanted to check.

If you are confused by this, PLEASE try to compute the Laplace transform of sin(kt) using the same method. PLEASE!!!

A Laplace transform of one little triangle
One of great powers of the Laplace transform is its ability to deal with rough functions: functions that may not be differentiable or may not be continuous (or even, as we will see later, may not be functions!). These rough functions are probably more accurate models of many physical situations than the standard functions of calculus. Here is an example. We will consider the piecewise-defined function
     =0 for t<0
y(t)=t for t for t in [0,1]
     =0 for t>1
A graph of this function is shown to the right. It certainly is not continuous at t=1. We will compute its Laplace transform.

So Y(s)=₀¹e^-sty(t) dt. Students immediately observed that we don't "need" the integral where part of the integrand is 0, so we should compute ₀¹e^-stt dt. Hey: this computation demands integration by parts.
Strategic note You will do more computations using integration by parts in this course than ever before during a similar period of your life. (And maybe [likely!] during any similar period for the remainder of your life!)
For integration by parts, I generally write all the details. I have made too many mistakes and I've gotten tired of fouling up. I usually write:

01e-stt dt=
           u dv = uv]   -  v du  

           u=   } { du=
          dv=   } {  v=

Simple Laplace transform asymptotics
It might be useful to have some simple things that can be checked about Laplace transforms. Then we might be able to look for errors, etc. Also, the asymptotic statements I'll make turn out to be what can be observed. Frequently, for example, in chemical kinetics, we can only see what happens for large values of the parameter -- things change too quickly.
Suppose our y(t) is some sort of collection of lumps and bumps, sort of as shown. Well, the Laplace transform formula multiplies this by e^-st, and integrates dt. We will only consider positive s, so the exponential is always decreasing. When s gets larger and larger positive, the exponential decreases more and more rapidly to 0. Yes, it stays at 1 when t=0, but the dropoff gets very rapid indeed. The product of the exponential with y(t) will be some collection {l|b}umps, but the amplitude, the height, will surely -->0. So for the functions we are considering, Y(s)-->0 as s-->infinity.

Now what about s-->0⁺? O.k., the exponential e^-st still decreases, but it decreases more slowly. (Uhhhh ... these heurisitics can be made precise -- did you think this was a math course?). So for more and more t, e^-st gets closer and closer to 1 as s-->0⁺. When we multiply this by y(t), the result for lots of t will be close to y(t), and the total integral will look more and more like the net area (area over the t-axis counted as positive, under the t-axis as negative) of y(t). So I think that as s-->0⁺, Y(s)-->the net area under all of the y(t) function.

Well let's check both of these asymptotic statements with our little triangle function. So Y(s)=(-se^-s-e^-s+1)/s². Certainly as s-->infinity, the e^-s terms decay, and the s² in the bottom takes care of the 1. So Y(s)-->infinity as s-->infinity.
What about as s-->0⁺? Well, heck, you folks are engineering students. I would first try "plugging in" s=0. The result is 0/0. I know what to try next: l'Hopital. So we d/ds the top and bottom of Y(s) separately. The result is (if I do it correctly!) (-e^-s+se^-s+e^-s)/(2s). Now plug in again. Hey, the result is ... uhhhh ... 0/0. You could use l'H again, but in fact, if you clean up the fraction algebraically you'll see that the result is e^-s/2, and as s-->0⁺, this certainly --> 1/2, the area of that little triangle region.

Diary entry in progress! More to come.

The Laplace transform of f´(t)
Here is a major result which will 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^infinitye^-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
₀^infinitye^-stf´(t) dt=e^-stf(t)]₀^infinity-₀^infinity(-s e^-stf(t) 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)]₀^infinity. Now s>0, so when t-->infinity, e^-stf(t)-->0. Technically I am using the fact that f(t) doesn't grow too fast (in fact, eventually it is "killed" by exponentials decreasing fast enough, e^-st for large s), but let's just try to get the mood of this method. 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).

How to solve an initial value problem for an ordinary differential equation
Let us try to solve 1y´(t)+2y(t)=3cos(t) with y(0)=4. (I think some of these numbers were suggested by students. Sigh.) I hope that you could recall several methods from 244 which would allow you to solve the equation. But maybe let us try a slightly different method.
Step 1 Take the Laplace transform of the equation. Then 1y´(t)+2y(t)=3cos(t) becomes -Y(0)+sY(s)+2Y(s)=3s/(s²+1). Here I am using the result we just got previously, and part of the theorem we started with today.
Step 2 Insert the initial condition, and solve for Y(s). So -Y(0)+sY(s)+2Y(s)=3s/(s²+1) becomes -4+sY(s)+2Y(s)=3s/(s²+1) and then we get (s+2)Y(s)=4+[3s/(s²+1)], so that Y(s)=[4/(s+2)]+[(3s)/{(s²+1)(s+2)}].
Step 3 Use any tricks you can to discover what Y(s) is the Laplace transform of (find the "inverse Laplace transform of Y(s)").

The Question of the Day (QotD)
Partial fractions (from the dawn of time: calc 2) states that [(3s)/{(s²+1)(s+2)}]=A/(s+2) +[(Bs+C)/(s²+1)]. Find A and B and C.
Solution Combine the fractions and just look at the tops. Then we get 3s=A(s²+1)+(Bs+C)(s+2), so if s=-2, then -6=5A and A=-6/5. The s² coefficients in the equation must cancel (no s²'s in 3s!) so B=6/5. Considering the constant term gives C=3/5.

Therefore Y(s)=[4/(s+2)]+A/(s+2) +[(Bs+C)/(s²+1)] with A and B and C as above. Look at the results of Theorem 4.2. I predict (linearity!) that y(t)=4e^-2t-(6/5)e^-2t+(6/5)cos(t)+(3/5)sin(t). Wow! You can check that this result does satisfy the given initial condition (not hard) and also satisfies the ODE (more difficult).

This is the whole idea!

HOMEWORK
Read 4.1 and 4.2, and do the appropriate problems in the syllabus. Please hand in on Monday: 4.1:10,34 and 4.2: 13,33

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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. For example, y´=2t. A solution to such an equation is a function which when substituted into the equation makes it "identically" true. For example, in this simple equation y=t²+C (any constant C) solves the equation. An initial condition (IC) is a value, "t₀", for which we require the function to have a certain value, "y₀". For example, we could ask that the solution to this ODE satisfy the IC (3,17). Then we choose C so that 17=3²+C is true: I guess C=8. The combination of ODE and IC is called an initial value problem. We expect, due to simple physical examples, that IVP's will have unique solutions. The independent variable in this course will frequently be called t for time, so the dependent variable y is a model of some process which depends on time.

Here is one version of the major theoretical result of the subject, the
Existence and Uniqueness Theorem for ODE's
Suppose f(t,y) is a differentiable function in both t and y for (t,y)'s in a region in R², and that the point (t₀,y₀) is in the region. Then the ODE y´=f(t,y) has a unique solution going through the point (t₀,y₀).
Another way of writing the IC is y(t₀)=y₀.

This is a wonderful theorem, but its usefulness is definitely limited. You shouldn't read "into" it any more than is already there. So let me show this with some examples.

General disclaimer The examples discussed in class will mostly be very artificial, and chosen so that "hand" computation is practical.

Example 1 y´=e^(t2) and y(0)=0. Then the solution is y(t)=₀^te^(w2)dw. It turns out (and it can be proved!) that this integral can't be simplified or written in terms of any of the standard functions of calculus, using algebraic means of combination, or even using function composition.

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 ... and it even 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.

Example 2 The solutions to y´=ty² are not difficult to get (the instructor of course made some errors). This is a separable ODE. I separated the variables, integrating, cleared up some algebra, and the solutions seemed to be y=2/(C-t²). I verified this in the most direct way possible, by computing y´ and checking that the result was ty².

Let's look at some specific solutions. The solution satisfying the IC (0,100) has C=2/100, so y=2/(.01-t²). The natural domain of this function is (-1/10,1/10). As t-->1/10^-, y-->infinity: the solution explodes. If the IC is (0,2·10⁴) the solution is y=2/(.0001-t²) and it has domain (-1/100,1/100). There is one special case (you need to look critically at the separation of variables process to spot what goes wrong) but if the initial condition is (0,0), the solution is y=0 for all t.
The picture to the right is supposed to an impression of the geometry of the solution curves to this ODE. Notice, please, that the Existence and Uniqueness Theorem is totally correct, but really unhelpful. No inspection of f(t,y)=ty² seems to show anything wrong (!!) with the function, but many of the solutions don't "live" very long, and they live for different amounts of time. 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(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.

Let's use linearity. y´´+y=0 has solutions e^it and e^-it, so if C₁ and C₂ are any constants, then C₁e^ix+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, y_P. What is it?

If y_P(t)=C₁e^it+C₂e^-it then y_P(t)=iC₁e^it-iC₂e^-it. We can get y_P(0)=1 by having C₁+C₂=1. We can get y_P(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 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:

Euler tells me ... (An axiom for Math 421)
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 C₁e^t+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:

• y_P(t)=(e^t+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.
• y_V(t)=(e^t+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.

No, no, no!

Part I of the course: the Laplace transform

This is a complicated definition.
Remarks about the 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, ₁^infinity(1/x)dx is infinite, but ₁^infinty(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.

• 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 sequence, 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 ₀^infinitye^-stdt. I'll work very slowly, with mathematical propriety: ₀^infinitye^-stdt is the limit as A-->infinity of ₀^Ae^-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=0^t=A which is -(1/s)e^-sA-{-(1/s)}e^s·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 e^w is 0 as w-->-infinity, and, in fact, it goes so rapidly to 0 that polynomial growth doesn't stop it: w²³⁸e^w-->0 as w-->-infinity also.
Since A>0, -(1/s)e^-sA-{-(1/s)}e^s·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 ₀^infinitye^-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
₀^infinitye^-stdt=-(t/s)e^-st]₀^infinity-₀^infinity-(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)₀^infinitye^-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: ₀^infinitye^-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=0^t=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!) -e₀^infinty-st-(2t/s) dt. We pull out the things which are constant relative to t, and get (2/s)₀^infinitye^-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.

I observed that the Laplace transform is linear: the sum of Laplace transforms of functions is the Laplace transform of the sum of the functions, and scalar multiplication also works nicely.

The QotD was: if f(t)=5-3t²+9t⁷?
Here I wanted people to use the linearity of the Laplace transform.

HOMEWORK
Please begin to read the chapter in the book about Laplace transforms, and begin the problems. Do the Entrance Exam, and have the result of that ready to hand in next Thursday.