Consider one-sided Laplace transform:$$\mathcal{L} \left \{ h(t) \right \}=\int_{0^-}^{\infty}h(t)e^{-st}dt$$(adsbygoogle = window.adsbygoogle || []).push({});

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^{-t}u(t), e^{-t}u(t+1) or ...?

Thanks in advance.

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# Laplace transform uniqueness

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