0

I do not know if this is an ill-posed question but ... is $\delta(t)e^{-\gamma t}$ equal to $\delta(t)$?

Thanks, biologist

Leptoceratops
  • 1
  • Not every distribution is a function, which is why some mathematicians get really touchy when they hear "delta function". (Myself included. :) ) One possible layman view for distributions is "distributions are functions which only live under integrals", and under the integral these two "functions" are the same, as tabstops answer shows. – Roland Feb 04 '14 at 14:52
  • Thank you. Yes it is supposed to live under an integral as it is a fundamental solution (Green's function). However, in the definition of the fundamental solution there is no integral (the way I see it), e.g. $LF(x) = \delta(x)$. How can the above equation hold then, if there is no integral? Do differential operators act differently on distributions? Thanks. – Leptoceratops Feb 04 '14 at 16:29

2 Answers2

1

If you mean what I think you mean, then: $$\int_{\Bbb R} f(t)\delta(t)\,dt = f(0)$$ and $$\int_{\Bbb R} f(t)\delta(t)e^{-\gamma t}\,dt=f(0)e^{-\gamma\cdot0}=f(0).$$ which works because your "extra" function evaluates to $1$ at the support of $\delta$. So they form identical kernels to integrate against, and that seems like probably what you want to know.

tabstop
  • 1,598
1

Unfortunately, this is a little bit ill-posed. $e^{-\gamma t}$ does not play well with tempered distributions. The delta distribution is a tempered distribution, meaning that it is in the dual space to the Schwartz functions (functions that are smooth and decay exponentially). Examples of such functions are the functions of the form $\exp(-t^{2n})$ where $n$ is a natural number. So you can't be sure that it makes sense to integrate the delta distribution against $\exp(-\gamma t)$. However if you could, then you would be correct.

Cameron Williams
  • 29,432
  • This was in the context of proving that $F(t) = \theta(t)\exp(-\gamma t)$, where $\theta(t)$ is the Heaviside distribution, is a fundamental solution for the linear partial differential operator $L = \frac{\partial}{\partial t} - \gamma$. Are you saying that this distribution equality does not hold for certain functions? – Leptoceratops Feb 05 '14 at 06:46
  • I mean for certain test functions $\varphi(t)$ that can be used in the corresponding inhomogeneous PDE? Thank you. – Leptoceratops Feb 05 '14 at 07:02
  • Oh in this case, that is fine because you are lopping off the "exponentially growing" part of the exponential. If you were not modulating with a Heaviside, you would run into some problems. – Cameron Williams Feb 05 '14 at 15:53
  • @cameronwilliams Inasmuch as the Dirac Delta has support on ${0}$, the functional.$\langle f,\delta\rangle =f(0)$ makes sense for $f(x)=e^{ax}$. – Mark Viola Jun 23 '18 at 17:04