For the Laplace transform, there is a rule to handle convolutions: $$\mathcal{L}\{u*v\}=\mathcal{L}\{u\}\cdot\mathcal{L}\{v\}.$$ In Fourier transform, there is a similar formular and furthermore, there is a formular to invert this convolution and multiplication theorem. Is there something similar for the Laplace transform, like $$\mathcal{L}\{u\cdot v\} = f(U * V)$$ for any or certain classes of functions $U, V$ in general or under certain conditions?

  • $\begingroup$ Are you looking for $$(u*v)(t)=\int_0^t u(\tau)v(t-\tau)\,d\tau$$? $\endgroup$ – Denis28 Feb 2 '18 at 19:58
  • $\begingroup$ No, I would like to have something like $$\mathcal{L}\{u\cdot v\}=\;?\;\cdot (U*V)$$ $\endgroup$ – Kutsubato Feb 2 '18 at 20:00
  • $\begingroup$ I am not really sure there exists such identity, but I’ll leave others prove this $\endgroup$ – Denis28 Feb 2 '18 at 20:20

Unfortunately, after doing some research, it seems that there is no general, simply form to solve my problem. Thanks for all contributions.


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