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Laplace Transform, what does it mean?

  1. Mar 13, 2012 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    We went over the Laplace transformation today in my DE course. We only covered essentially, "how to do it", so:

    [tex]L(f(t)) = \int_{0}^{\infty}f(t)e^{-st}dt[/tex]
    (also, how do you make the fancy f and curly brackets in Latex?)

    Essentially transforming f as a function of t, into a function of s. My book blows over this topic, and simply focuses on it's applications in solving differential equations. My question is, what the heck are we doing? What does it mean to transform the equation? I'm sort of thinking of it like a trig substitution more or less, transforming to something that is easier to work with and still obeys the basic properties of algebra, then transforming it back. What is s-space? I don't get it at all. The text only says that s-space shares a linear proportionality, whatever that may be. I tried looking online, and checking into another book, but I can't seen to find an answer to this question.

    I am starting my homework, and I do have a question on an application problem, which I will post here, as this is "homework help", but I wanted to see if I could figure it out myself provided I can get a better in-depth explanation of what this operation is doing.

    Thanks, QC

    Edit: I am also given to understand that the e^-st portion of the transformation is the "kernel", would that infer that in other similar transformations, they differ only by the kernel used?
     
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  3. Mar 13, 2012 #2

    micromass

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  4. Mar 13, 2012 #3

    micromass

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    Yes. Another big example is the Fourier transform http://en.wikipedia.org/wiki/Fourier_transform

    But you can put in any type of function as the kernel.
     
  5. Mar 13, 2012 #4
    Thanks micromass, I am watching that now.

    I made a typo in my first post. I meant how do you make the script L, not f. The thing that looks like the first character in this image:
    Image1265.gif
     
  6. Mar 13, 2012 #5

    micromass

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    That would likely by \mathcal{L}

    It gives [itex]\mathcal{L}[/itex].

    Be sure to read the entire thread, not only watch the videos. There is some other things you might like (or you might not care about).
     
  7. Mar 13, 2012 #6
    I did read the whole thing. I think the continuous analog of the sum of a power series representation makes more sense to me. I really just wanted a way to think about it, and without going into serious detail (I am positive I will hit this topic again), that explanation will suffice. Thanks, as always.

    So, I tried to apply this technique to a simple DE as follows:

    [tex]y'' + 4y = 0[/tex]
    With initial conditions: y(0)=1 and y'(0)=1

    [tex]\mathcal{L} \{ y'' \} + \mathcal{L} \{ 4y \} = \mathcal{L} \{ 0 \}[/tex]
    [tex](s^{2}Y - sy(0) - y'(0)) + (4Y) = (0)[/tex]
    Where Y is the laplace transform of y, and is a function of s.

    So now I can go ahead and apply my initial conditions to simplify a bit:
    [tex]s^{2}Y - s - 1 + 4Y = 0[/tex]
    [tex]s^{2}Y + 4Y = s + 1[/tex]
    [tex]Y(s^{2} + 4) = s + 1[/tex]
    [tex]Y = \frac{s+1}{s^{2}+4}[/tex]

    Then, expanding a bit so it matches some of the transforms in my table, which is especially clear since I already know in general what the answer will more or less look like by solving the aux. equation.

    [tex]Y = \frac{s}{s^{2}+4} + \frac{1}{s^{2}+4}[/tex]

    Then, to convert back, such that Y becomes y:
    [itex]\frac{s}{s^{2}+4}[/itex] inverse laplaces to [itex]cos(2t)[/itex]

    and similarly:
    [itex]\frac{1}{s^{2}+4} \frac{2}{2}[/itex] inv. laplaces to [itex]\frac{1}{2}sin(2t)[/itex]
    (had to think about that part for a good bit, that's where I was stuck, but multiplying the 2/2 was obvious I can't believe it took so long)

    So now:
    [tex]y = cos(2t) + \frac{1}{2}sin(2t)[/tex]

    Am I doing this properly??
     
  8. Mar 13, 2012 #7

    micromass

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    That looks perfect!!
     
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