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Laplace Transform Question

  1. Feb 17, 2014 #1
    Hello,

    As of recently, I've been working with Laplace transforms and have a question about their relationship to solving differential equations.

    I know the definition of the laplace transform and I know that a function is essentially being transformed from the time domain to complex frequency domain. I also know that they are a useful tool in solving differential equations. They essentially replace solving a differential equation with a simple algebraic equation. However, my question is why?

    I don't completely understand how this occurs and I don't think I have ever been told exactly why this is true. I mean, what if we were to make up some transformation other than the Laplace and apply that to a differential equation. Why couldn't we solve a differential equation with that?

    Is there any good explanation out there for this?
     
  2. jcsd
  3. Feb 18, 2014 #2

    pasmith

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    The advantage of the Laplace transform is that
    [tex]
    \mathcal{L}(f') = \int_0^{\infty} e^{-st} f'(t)\,dt = \left[ e^{-st} f(t) \right]_0^{\infty} + s\int_0^\infty e^{-st} f(t)\,dt = s\mathcal{L}(f) - f(0)
    [/tex]
    which is what converts linear differential equations with constant coefficients into algebraic equations.
     
  4. Feb 18, 2014 #3

    PeroK

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  5. Feb 19, 2014 #4
    This MIT ocw video holds one of the keys to demystifying the Laplace transform.

    http://ocw.mit.edu/courses/mathemat...ure-19-introduction-to-the-laplace-transform/

    One of the key properties of the Laplace transform, shared by the Fourier transform, is the convolution property. Laplace transforms take convolution to multiplication. One thing that's a bit mysterious to me is that, for the Fourier transform, there's a lot of beautiful physical intuition behind this property, whereas, as far as I can tell, for the Laplace transform, it's more algebraic. In the discrete case, convolution turns out to be the same thing as what happens to the coefficients when you multiply polynomials or power series. Laplace transforms are sort of a continuous version of this and convolution of functions of a continuous variable are analogous to polynomial multiplication, so, as I see it, that's why the convolution property still holds.

    Of course, as mentioned by pasmith, in the context of diff eq, the differentiation property is also important (my intuition here is that when you differentiate with respect to e^(st), you get se^(st), but I'm not quite sure how to put my reasoning into words, other than the integration by parts calculation, hence the parantheses).

    There are a lot of interesting things to explore here that might provide more insight than I have at the moment, some of which might take you a bit far afield. The origins of the Laplace transform appear to be in probability. I'm not sure this is what Laplace started with, but in probability, Laplace transforms are essentially moment generating functions of random variables. Convolution comes up in that context when finding the probability density of a sum of independent random variables, which is one reason for the importance of moment generating functions.

    Another interesting thing to look into is the z-transform, which is the discrete version of the Laplace transform.

    Solving diff eq's using Laplace transforms is also sort of analogous to using generating functions (see the free online book, Generatingfunctionology, if you want to see what these are) to solve difference equations (or more generally, recurrence relations). Another precursor of these methods seems to be Heaviside's operational calculus, where he manipulated differential operators algebraically, but that's a story I don't really know very well. It appears that those methods were replaced by Laplace transform methods.

    In electrical engineering, Laplace transforms are kind of nice because you can just put down stuff on the circuit diagram that tells you how to do them, and then you can solve initial-value problems like what happens when you turn on a switch in sort of the same way as you can do problems with a constant voltage source.
     
  6. Feb 21, 2014 #5

    jasonRF

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    As an engineer, the way I think about this (and is along the same lines as what homeomorphic discussed) is that complex exponentials are the eigenfunctions of the differential operator with a constant coefficient:
    [tex]
    \frac{d}{dt} e^{s t} = s e^{s t}
    [/tex]
    which makes it reasonable to think that the space of complex exponentials is appropriate for representing solutions of constant coefficient ODEs. In some sense the Laplace transform (or Fourier transform for that matter) is projecting the function onto this space of complex exponentials.

    For second order differential operators with non-constant coefficients other transforms are useful: Hankel transforms, Mellin transforms, etc., but my hunch is that this is beyond the scope of what you are interested in. In case my hunch is wrong: sometimes these come up naturally when solving boundary value problems (Poisson's equation, heat equation, wave equation, etc.) by separation of variables which yield Sturm-Liouville equations - if the problem allows for a continuum of separation parameters (as opposed to a discrete set) then you "sum up" the solutions using an integral instead of a summation. This typically occurs in problems that are not restricted to a finite volume. In the context of boundary value problems there is also a "recipe" for deriving transform pairs (including Laplace) that involves a contour integral of the appropriate Green's function, but I suspect this is very far beyond the scope of what you care about.
    jason
     
  7. Mar 5, 2014 #6
    Here's an insight to why, as the how has been discussed already:

    A Fourier Transform is a special case of the Laplace transform where if s = σ +/- iω then σ = 0.
    The "σ" term in the Laplace Transform represents a DC offset introduced to a signal. If you expand the exponential e^{st} you get e^{σt}e^{+/-iωt} which in turn is something like e^{σt}(cos(ωt)+sin(ωt)). This is a scaled AC signal (DC offset). That is why this transform is so useful in electronic analysis and signal processing. It represent real occurrences.
     
  8. Mar 11, 2014 #7
    We can, and we do. The Laplace and Fourier transforms are just two examples of Integral transforms.
    Another example of an integral transform is the Hankel Transform

    [itex]H_n \left( f\left(r\right) \right) = \int_0^\infty J_n\left(k r \right) f\left(r\right) rdr[/itex]

    where

    [itex]J_n[/itex] is the n-th order Bessel Function.

    The Hankel transform is useful for solving problems in cylindrical domains.
     
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