5.1, the beginning of (possibly) the last new thing we have to learn.

Chapter 5 is entitled “The Laplace Transform.”

Section 5.1 is entitled “The Definition of the Laplace Transform.”

There are two paragraphs of preface so allow me to summarize those for you: This is another method to solve linear differential equations with constant coefficients. The book offers a picture as a visual for the method so I shall as well.

Anyway, onto our first definition of chapter 5!

“Suppose f(t) is a function defined for 0 < t < ∞. The Laplace transform of f is the function…” (190).

Two things to notice about this definition: The function is of the variable s, and the integral is improper (due to the ∞ symbol, of course). You solve it by taking the limit,

Then the section continues on with a bunch of examples, one more definition, and a theorem. I’ll show two of the examples (and then quote the definition and the theorem) and then post a bunch of websites. Who knows, I might get 5.2 up tonight.

Anyway, the first example (if you don’t remember integration by parts, here’s a handy reminder for you http://mathworld.wolfram.com/IntegrationbyParts.html):

As T gets very large, the growth of an exponential will dominate the growth of T, so

(If you really want to show this with math numbers and such, you can use l’Hôpital’s rule.) The second term actually goes to zero as well (since the exponential is there as well), so the only term we have left is the third one, 1/s². Therefore, the Laplace transform of f(t) = t is

Our second example will be a piecewise continuous function, or a discontinuous function. A function is called piecewise continuous if when it is defined on the interval (0, ∞), “it only has finitely many points of discontinuity over any finite interval and at every point of discontinuity the limit of f exists from both the left and the right” (194).

Now for our example: 

And now for a definition:

“A function f(t) is of exponential order if there are constants C and a such that

And a theorem:

“Suppose f is a piecewise continuous function defined on [0, ∞), which is of exponential order. Then the Laplace transform ℒ(f)(s) exists for large values of s. Specifically, if |f(t)| ≤ Ce^at, then ℒ(f)(s) exists at least for s > a” (196).

You know how I said I might get 5.2 up by tonight as well? No, that’s not going to happen. For the second time during the journey of this blog, Microsoft Word refused to save my work and I had to retype the entire ending bit. If that doesn’t make you angry enough to rage quit for the evening, then I don’t know what will.

Oh. I said I’d post a bunch of websites.

I don’t want to do that either.

That’s what I get for making promises, I suppose.

http://en.wikipedia.org/wiki/Laplace_transform, http://mathworld.wolfram.com/LaplaceTransform.html, and http://tutorial.math.lamar.edu/Classes/DE/LaplaceTransforms.aspx count as a bunch, right? Right.

I’ll see you in 5.2 (eventually).