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Laplace transform uniqueness

  1. Jul 7, 2012 #1
    Consider one-sided Laplace transform:$$\mathcal{L} \left \{ h(t) \right \}=\int_{0^-}^{\infty}h(t)e^{-st}dt$$

    Q. Is this defined only for the functions of the form f(t)u(t)? If no, then f(t)u(t) and f(t)u(t)+g(t)u(-t-1) are two different functions with the same Laplace transform, and thus, Laplace transform is not unique.

    Now here is my problem. Consider the following differential equation: $$h'(t)+h(t)=\delta(t)$$$$h(0)=0$$
    Taking the Laplace transform, we have: $$sH(s)+H(s)=1$$$$H(s)=\frac{1}{s+1}$$$$h(t)= \mathcal{L}^{-1} \left \{ H(s) \right \}$$

    If the answer to Q is yes, then do I need to show h(t)=h(t)u(t) before taking Laplace transform?

    If the answer to Q is no, then which [itex]\mathcal{L}^{-1} \left \{ H(s) \right \}[/itex] is h(t)? e-t, e-tu(t), e-tu(t+1) or ...?

    Thanks in advance.
  2. jcsd
  3. Jul 8, 2012 #2
    Since this thread is moved from electrical engineering forum, I suppose some clarifications are needed:

    1. u(t) denotes unit step function.

    2. [itex]\delta(t)[/itex] denotes Dirac delta function.
    Last edited: Jul 8, 2012
  4. Jul 8, 2012 #3


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    Functions on the interval [itex][0,\infty)[/itex] have a unique Laplace transform. All of the examples you've given are the same on that interval.

    (Except for f(t)u(t) + g(t)u(-t-1). This would not have the same laplace transform as f(t)u(t). If you had written, say, f(t)u(t) + g(t)u(-t), then that would have the same (one-sided) laplace transform of f(t)u(t)).
  5. Jul 9, 2012 #4
    Thank you Mute.

    Actually I've chosen an improper title for the thread. My question is not merely about the uniqueness of Laplace transform, but how it effects on solving differential equations by using Laplace transform. My problem is to determine which one of e-t, e-tu(t), e-tu(t+1),... is the solution. Although I know that only e-tu(t) satisfies the equation!
    Last edited: Jul 9, 2012
  6. Jul 9, 2012 #5


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    Your solution is defined on the time interval [itex]t \in [0,\infty)[/itex]. My point was that on that time interval, all three of your example "solutions" are exactly the same. On [itex][0,\infty)[/itex] u(t) = 1, u(t+1) = 1,etc. You don't care at all what happens prior to the initial time t = 0. When you do a one-sided Laplace transform of a function f(t), that's the same as the laplace transform of f(t)u(t) or f(t)u(t+1) because u(t) = u(t+1) = 1 on the interval of integration.
  7. Jul 9, 2012 #6
    So, by merely using one-sided Laplace transform, we cannot find what the solution looks like for t<0, and using Laplace transform is not a good idea for finding the complete solution. Is this correct?
  8. Jul 9, 2012 #7


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    You have initial data at t = 0. Typically we don't care about anything which happens prior to that as our function is only defined on t >= 0.

    If you care about the time interval t < 0, the one-side Laplace transform is not an appropriate method to use. You would be better suited to using the Fourier transform, as this is defined for all times. You will in fact find using the Fourier transform that [itex]h(t) = \exp(-t)u(t)[/itex] (i.e., h(t) is zero for t < 0); however, this solution requires that the solution decays to zero faster than exponentially. You don't get to choose the initial data at t = 0. (Well, you can, but you need to do some tricks using the method of images).

    Note that, given your initial data h(0) = 0, you should interpret your solution as being exp(-t) for t > 0 and zero for t <= 0. You can use the step function to make this explicit, if you like in the case of the Laplace transform solution, but it's not a major issue.
  9. Jul 10, 2012 #8
    Thanks a lot.

    In addition, someone suggested:
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