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Laplace transform

  1. Dec 10, 2007 #1
    Laplace transform....

    I cant help but keep asserting how me textbook of differential equations just keeps stating new ideas without any background on WHY and HOW they were created. I've just started Laplace transform and the first thing i see is a blue DEFINITION box that just states the tranform equation asking us to memorise it. Can you please explain WHY is that specific function so important and HOW did they figure it out?
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  3. Dec 10, 2007 #2


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    The Laplace transiorm is important in differential equations because the Laplace transform of a derivative is a simple algebraic function of the
    Laplace transform of the function itself.
  4. Dec 10, 2007 #3


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    Its a tool. Learn how to use it.

    What do you know about the history of a hammer and nails? Yet you use them without thought.

    If you want background do a search on Wiki, there is some history there.
  5. Dec 11, 2007 #4
    already searched wiki. everything there is just stated. All I'm asking is: why is the laplace transform defined the way its defined now? how did laplace figure it out in its form?
  6. Dec 11, 2007 #5


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    the same way all such techniques are developed: try this and it didn't work, try that and it didn't work, keep going until you find something that works!

    Laplace recognised what I said above:
    "the Laplace transform of a derivative is a simple algebraic function of the
    Laplace transform of the function itself."
  7. Dec 11, 2007 #6

    Ben Niehoff

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    I have to say I rather strongly disagree with this sentiment. We must understand the principles and the motivations behind such things if we ever hope to see the broader implications.
  8. Dec 11, 2007 #7


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    There is a difference between the HISTORY of a method and understanding how to use it.
    He is asking about the HISTORY, while that is nice it certainly is not critical, and may even be a waste of his time if he pursues that rather then understanding the transform and how to use it.
  9. Dec 11, 2007 #8


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    okay, OJ, i'll try to give you an answer where other folks here declined to. this is the pedagogy that electrical engineers have with the Laplace Transform, it may not be congruent to the actual history.

    it starts out with Fourier series. using a sum of sinusoids to represent a periodic function:

    [tex] x(t+T) = x(t) \quad \forall t [/tex]

    Fourier thought that sines and cosines having the same period T might be able to add up to the same x(t):

    [tex] x(t) = \frac{a_0}{2} + \sum_{n=1}^{+\infty} a_n \cos( 2 \pi n t / T ) - b_n \sin( 2 \pi n t / T ) [/tex]


    [tex] x(t) = \frac{a_0}{2} + \sum_{n=1}^{+\infty} g_n \cos( 2 \pi n t / T + \phi_n ) [/tex]


    [tex] x(t) = \sum_{n=-\infty}^{+\infty} c_n e^{ i 2 \pi n t / T } [/tex]

    if you fiddle around with trig identities and complex numbers, you can see that the three expressions are the same (with some mapping between an, bn and gn or cn). The latter expression is the most convenient; it's pretty easy to solve for cn :

    [tex] c_n = \frac{1}{T} \int_{t_0}^{t_0+T} x(t) e^{ -i 2 \pi n t / T } dt \quad \forall t_0 [/tex]

    note that cn is not a function of t and indicates the strength of the frequency component at frequency of n/T. now if you plug that cn back into the x(t) expression (and change the letter for the dummy variable of integration, you get:

    [tex] x(t) = \sum_{n=-\infty}^{+\infty} \frac{1}{T} \int_{t_0}^{t_0+T} x(u) e^{ -i 2 \pi n u / T } du \quad e^{ i 2 \pi n t / T } [/tex]

    this is a self-contained expression for a periodic x(t) of any finite period T, so, if you want to represent a function that is not periodic, you can let T get very large (like a year), and if you're careful with your Riemann sums, you can show that they become a Riemann integral. first define the arbitrary t0 = -T/2,

    [tex] x(t) = \sum_{n=-\infty}^{+\infty} \int_{-T/2}^{+T/2} x(u) e^{ -i 2 \pi n u / T } du \quad e^{ i 2 \pi n t / T } \frac{1}{T} [/tex]

    as [itex] T \to \infty [/itex], the 1/T becomes a differential amount of frequency and the summation becomes this integral:

    [tex] x(t) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} x(u) e^{ -i 2 \pi f u } du \quad e^{ i 2 \pi f t } df [/tex]


    [tex] x(t) = \int_{-\infty}^{+\infty} X(f) e^{ i 2 \pi f t } df [/tex]


    [tex] X(f) = \int_{-\infty}^{+\infty} x(t) e^{ -i 2 \pi f t } dt [/tex]

    so, instead of representing x(t) as a sum of discrete frequencies of n/T that are getting closer and closer to each other as T goes to infinity, we are representing it as a continuoum of frequencies that are infinitesimally close to eacy other. that is the Fourier Integral.

    there are some functions that the Fourier Integral does not do to well with (doesn't converge nicely), like the Heaviside unit step function, so we generalize the purely imaginary frequency term in the exponent [itex]i 2 \pi f [/itex] by adding a little real part to it to make the integral converge.

    [tex] \int_{-\infty}^{+\infty} x(t) e^{ -(\sigma + i 2 \pi f) t } dt [/tex]

    if you define this complex value as [itex]s = \sigma + i 2 \pi f[/itex] or [itex]s = \sigma + i \omega[/itex], the above integral is the so-called two-sided Laplace Transform and it converges for some functions that the Fourier Transform does not.

    [tex] X(s) = \int_{-\infty}^{+\infty} x(t) e^{ -s t } dt [/tex]

    That's the motivation behind the definition of the Laplace Transform.
  10. Dec 12, 2007 #9
    There is an enourmous amount of book just about that topic. How to get to laplace transform is actually very complex, but if you search online I'm sure you can find a book about it.
  11. Dec 12, 2007 #10


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    There's a problem with that: Laplace developed the "Laplace transform" method in 1785, Fourier developed Fourier series in 1822. Fourier series was hardly the "motivation behind the definition of the Laplace Transform"!

    Wikipedia has a good article on the Laplace transform- including the history and motivation:
    Last edited by a moderator: Dec 12, 2007
  12. Dec 12, 2007 #11
    My advice would be working out the relation between Fourier and Laplace (one-sided, two sided) transforms. And the other integral kernels with zillion examples to get into the Laplace's mind (You start speeding up after a couple of hundreds! ) . The common analogy is the integral transforms...

    I have to admit, if you start from the mid-point (in the sense of using the tool without knowing) it takes so much time to get onto the right track because, these are presented to us like an item in the menu but they require a lot of insight to interpret. For me it is a pedological bug in the system. Check also the conformal mapping etc.

    Hmm, HallsofIvy was faster about the wiki page
  13. Dec 12, 2007 #12
    im sorry if im being annoying. But reading that history section in wikipedia, i can tell you what's presented there doesnt really suffice. Could you link me to somewhere a bit more detialed and elaborate?
  14. Dec 12, 2007 #13
  15. Dec 12, 2007 #14
    O.J. -

    Which text book are you using, and in your course will you be studying special functions next? (like Bessel functions or Legendre polynomials..)
  16. Dec 12, 2007 #15

    Chris Hillman

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    PCM (and I don't mean "Pulse-Code Modulation"!)

    When it comes out, this should be just what you want:
    I know there is a chapter on Fourier transform so I assume there will be one on Laplace transform.

    From my perspective, the formal properties of both Laplace and Fourier transform are what really make these so "functionals" (functions of functions) so important. (Have you seen the Chauvenet Prize volumes published by the MAA?) The point of the formal properties are that Laplace transform is perfectly suited for solving initial value problems involving functions of time, while Fourier transform is perfectly suited for solving linear PDEs, often involving spatial differentiations, and these can be combined. The result is that we can reduce a PDE to an algebraic equation which can be solved by elementary high school algebra, and the transforms can then be inverted to give the desired solution. I have given elsewhere specific examples of this lovely trick!

    Again trying to stress the importance of the formal properties: we have certain group actions on function spaces and our formal properties basically say that we have an equivariant mapping from the space under one action to the space under the other action. This is one way of understanding how a differentiation turns into a scalar multiplication. There are also powerful connections here with the notion of eigenthings, representation theory, and invariant theory.

    Oh yeah, Wikipedia: don't believe what you read there: this website has a nice appearance but unfortunately it is too unstable and unreliable to be useful as an information resource on anything more important than shoot-em-up-game or pop culture trivia.
    Last edited: Dec 12, 2007
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