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Where does the Laplace Transform come from?

  1. Aug 30, 2014 #1
    I took an introduction to ODEs course this past spring semester. It always bothered me where this thing came from. I did a little bit of research and found a video of a professor explaining how it is the continuous analog of an infinite sum. He did a little bit of a derivation using that analogy. That's fine and dandy, but its there a rigorous way to derive this integral expression? What does it mean and what does it tell you about the function you apply it to (i.e., what is the relationship of a function and its Laplace Transform)?
  2. jcsd
  3. Aug 30, 2014 #2

    Simon Bridge

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    Was the video: Mattuck L19: Introduction to the Laplace Transform

    Do you mean the jump from a discrete to a continuous sum?
    That is just the definition of an integral - it is what happens to the discrete sum when the interval between discrete points gets very small. Integration can be thought of as a shortcut method for doing discrete sums - provided the sum converges.

    IRL Laplace came up with the method in bits, by exploration ... not by some logical clear deduction from start to finish. That is why it is so difficult to motivate it.

    Hopefully you can see why expressing part of a DE as a power series is a promising looking idea, and why you'd want to make things powers of e. The whole thought process is about trying to represent some random function as a combination of things we know how to integrate - and can integrate easily.

    You may also gain some insights by comparing the Laplace transform with the Fourier transform.
    Last edited: Aug 31, 2014
  4. Aug 30, 2014 #3
    I believe your link is broken. However, I believe it was the one I referred to because I remember seeing it on the MIT OCW site.

    I don't remember using a change of variables when taking the limit of a Reimann Sum as the length of the partitions approach zero in order to define the definite integral. I'm not sure I understand what you meant by using an integral as a "shortcut" method for a discrete sum. It's not like you can interchange sums and integrals freely, even if they converge. However, I might have misunderstood.

    So is the Laplace Transform something that doesn't have a rigorous derivation? Did he just want to "transform" the discrete sum into an infinite one? And for some reason it just became useful in solving differential equations? What does it mean, though? What relationship does it have with the function it operates on?

    I understand that ##e^x## is a very nice function and how a power series solution is convenient. So far, the only use I see the Laplace Transform has is to make things simpler; to turn a somewhat difficult differential equation into an algebraic equation. Does this mean that the Laplace Transform is just a way of turning the differential equation into an equation without meaning (but not useless) and then revert back using the inverse transform?

    We did not touch on Fourier Transforms in my intro to DE class. Should we have gone over it? I know a tiny scope about the general concept, but that's about it.

    All in all, I know how to work with the Laplace Transform. I just don't like not knowing where it came from. Where is the motivation behind it, other than switching between sums. What is its meaning, if any? I don't want to take it for granted. I don't like learning equations without knowing where they come from (especially knowing how they are derived, proved). Sorry if I seem to skeptical and persistent.
  5. Aug 31, 2014 #4
    I can sympathize--Laplace transforms used to drive me crazy for the very same reasons, but the reason why they don't tell you where it comes from is that it's a bit involved. I'm in favor of more people at least giving the motivation shown in that video. I find the usual approach where they pull it completely from thin air to be in very poor taste.

    There are a lot of different angles from which you can view them, which are kind of deep. If you study Fourier series and transforms, you'll see part of the significance. There are other ways you can look at it, which have to do with generating functions in combinatorics and probability and how they relate to convolutions (convolutions arise, for example when you add up the results of two independent dice rolls in probability and also, in signal processing, when you try to figure out what function a linear, time-invariant system will spit out when you put a given function in). What you've seen in your class is standard, but that doesn't necessarily mean it's a good way for someone like yourself to learn it. Understanding it deeply isn't something that's going to happen over night.
  6. Aug 31, 2014 #5

    Simon Bridge

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    It's one of those situations where you need, and can use, the method well before you have enough maths to understand why it works. But then, quite a bit of life is like that.
  7. Aug 31, 2014 #6

    Simon Bridge

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    Works for me <puzzle>.

    The reiman sums you started with were very simple - but the process holds.

    Just as a sum can approximate an integral, so an integral can approximate a sum.
    The relation is exact in the appropriate limit.
    See also: http://johnmayhk.wordpress.com/2007/09/24/alpm-sum-an-infinite-series-by-definite-integrals/
    ... an example of using an integral to find the sum of an infinite series.

    The relationship of the transform to the function is in the definition - that's it. Like I said - compare with the Fourier transform. It's basically changing domains - but that needn't mean anything.

    The Laplace transform is continuous into continuous - he was looking for a way to understand more difficult DEs.

    An example to start you off...
    If you have y'=f(x) but the f(x) is annoying, then writing ##f(x)=\sum a_nx^n## means that ##y=\sum a_nx^{n+1}/(n+1)## ... and then you can set out to look for the situations where that's an improvement.
    Right away you can see that this depends on the way those ##a_n##'s work out ... which motivates treating it as a function.

    That is pretty much the only use I know for it.
    Most of the maths you've learned - think: long division, multiplication by columns etc - is basically a way to turn a hard calculation into one that is somewhat removed but easier to do.

    However, the Laplace transform is closely related to the Fourier transform's ability to change to a conjugate domain ... i.e. time into frequency, position into momentum, etc. Ultimately whether there is meaning or not depends on the context ... same with any maths.

    Also see:

    There is no special reason you'd meet Fourier transforms ahead of Laplace transforms. I met Fourier first in physics class and ran into Laplace in maths class studying DEs.
  8. Aug 31, 2014 #7
    Hm, I just tried the link today and it works. Last time I got a "doesn't exist" page. Must have been a problem on my end.

    This seems really interesting. I guess, for now, I'll take the Laplace Transform as it is and later in my career I'll probably learn its true nature. I just hope it isn't a sack of magic. I'm still wondering how it works; just because ##e^x## and series solutions are convenient doesn't mean they have to be in the definition. Hopefully, when I encounter Fourier Transforms, I won't have the same problem and I'll learn more about the Laplace Transform, as you mentors have suggested.

    Thanks for your time and for putting up with my stubbornness.
  9. Aug 31, 2014 #8


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    Not magic. It will become very intuitive if you use it much. Since derivatives and integrals of e-St are so simple, that is motivation to represent arbitrary time domain functions in terms of exponentials. The Laplace transform represents how much of each exponential e-St is in the original time function. The real part of S represents a rate of pure exponential growth or decay within the time function. The imaginary part of S represents a constant amplitude frequency within the time function. The combination of the real and imaginary parts of S represent exponential growth or decay of a certain frequency within the time function. The Fourier transform is very similar except that it represents frequencies within the time function using sin and cos rather than using e-i * Im{S} * t
  10. Aug 31, 2014 #9
    I'm not sure I understand this time and frequency business; perhaps I am not that far in my career to know about it. I used the Laplace Transform extensively in my ODE course and I understood its use of simplicity but not what it means or its motivation. Sure exponentials have nice [anti]derivatives and series solutions can be convenient, but why do we use them to solve DEs?

    It feels as if someone said, "OK, I have a power series; let me turn it into an integral. Oh, exponentials are pretty functions; let's put it in there for convenience. Ah, I'm stuck on this differential equation; let me use this transform I made up to help me."
  11. Sep 1, 2014 #10


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    Look at the integral equation for the inverse Laplace transform (see http://en.wikipedia.org/wiki/Inverse_Laplace_transform#Mellin.27s_inverse_formula.

    You can see that it is rebuilding the original function in the time domain by integrating together exponential functions, eSt, where S is a complex number. The Laplace transform tells how much of each exponential the reconstructed time function contains. The exponential function on the complex plane can represent frequencies (determined by Im{S}) which grow or shrink exponentially (determined by Re{S}): eSt = eRe{S}t *(cos(Im{S}t) +i * sin(Im{S}t))

    For me, that is the entire motivation of the Laplace transformation. A nice side benefit is that each frequency component is very easy to take derivatives and integrals of.
  12. Sep 2, 2014 #11
    Ah, I can see how in inverse transform is "integrating away" the ##s## variable to attain the original function ##f(t)##, just as how the regular transform does the opposite. The inverse integral in that link looks pretty intuitive (with some complex constant next to it), but still no derivation. I haven't taken a complex analysis course, but i'm eager for it and hopefully it will bring some more insight.
  13. Sep 3, 2014 #12


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    Exactly. So the inverse Laplace is piecing together frequencies in the time domain to reconstruct the original f(t). The frequencies are in the form of the exponential function. And the Laplace transform of f(t) tells how much of each frequency should be used. Each value of the complex number S represents a frequency.
    Last edited: Sep 3, 2014
  14. Sep 4, 2014 #13
    I think I have a feel for it now that i know a little about its physical use. But, I regress. I suppose I'll leave it until future courses to give me a complete explanation, together with a rigorous derivation and meaning. I'm glad I wrote this topic. It showed me that there are some things that should be left for later. It's hard for me to accept but I have to deal with it. Thanks to you and everyone else!
  15. Sep 10, 2014 #14
    laplace of sin^2 3t, 1-e^t/t
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