Friday, November 15, 2013

5.7 - the last end of a chapter you'll ever see D':

Section 5.7 is entitled “Convolutions.”

Let’s review a little bit of what we have learned in the past:


At the end of our summary through 5.6 (i.e. http://differentialequationsjourney.blogspot.com/2013/11/56-last-second-to-last-section-maybe.html) we talked about something called the unit impulse response function. Since the denominator of our Laplace transform is the characteristic polynomial of our second order differential equation, we can rewrite that Laplace transform as

In this case, E(s) is the Laplace transform of the unit impulse response function e(t) for our differential equation. So now we can find a solution to the differential equation by taking the inverse Laplace transform of our rewritten Laplace transform, i.e.

Let’s now look at a proposition:

“Suppose that f and g are functions of exponential order. Suppose that ℒ(f)(s) = F(s) and ℒ(g)(s) = G(s). Then

Note that the choice of the variable of u is merely a choice. The variable of integration could be anything, in theory, since the resulting expression is a function of t at the end of the day.

Now, let’s define something:

“The convolution of two piecewise continuous functions f and g is the function f * g defined by

The means we can rewrite what we said in the proposition as

Now, let’s look at a theorem:

“Suppose f and g are functions of exponential order. Then

Here’s another theorem:
“Suppose f, g, and h are piecewise continuous functions. Then
1. f * g = g * f
2. f * (g + h) = f * g + f * h
3. (f * g) * h = f * (g * h)
4. f * 0 = 0” (235).

(Note: three theorems separate us from the end of this blog.)

Now, we want to think in terms of the general solution to our original initial value problem (which I’ll state once again because why not and also because we’re going to make it a little more general):

Let’s take a look at the theorem that will answer all our questions:

“The solution to the initial value problem [above] can be written as

[where ys(t) is the state-free solution and yi(t) is the input-free solution]. If e(t) is the unit impulse response function for the system, then the state-free response is

and the input-free response is

Let’s now think of properties of the convolution. For one thing, f * 1 is NOT equal to f. This means, unlike a lot of other subjects in mathematics, the function 1 is not the identity for the convolution product. Here’s a theorem for the identity:

“Suppose f is a piecewise continuous function. Then

Allow me to take the time right now to remind you that δ0 is the delta function when p = 0 (http://differentialequationsjourney.blogspot.com/2013/11/56-last-second-to-last-section-maybe.html).

Something nice about this result is that

Finally (and by finally, I mean for the final time in this blog), here’s a theorem:
“Suppose f and g are piecewise differentiable functions. Then

Here are some websites concerning the interesting phenomenon of convolutions:

http://en.wikipedia.org/wiki/Convolution (This will probably be more confusing than helpful but I’m adding it anyway)

http://faculty.atu.edu/mfinan/4243/section76.pdf (This contains more problems for you to work out than examples, but that’s okay, I suppose)


http://www.swarthmore.edu/NatSci/echeeve1/Class/e12/Lectures/ImpulseConvolution/ConvolutionExamples.pdf (This contains quite a few graphs. There were no graphs in the section in the book so this might be more confusing than helpful.)

All right, that’s it. My teacher told me to not worry about chapter 10, so I won’t.
Wow, this is weird. I’m finishing something that has taken up so much time in my semester. This is bittersweet. I like the idea of not having to worry about this blog anymore, but I kind of enjoyed it in a no-one-reads-this-anyway-so-I-don’t-need-to-feel-pressured sort of way.

The journey is over summary-wise but not semester-wise. I still have a couple of weeks before this semester is over so we’ll see what happens.


Goodbye, math blog. (For now.)

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