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Laplace transform of Dirac's delta distribution (Definition)

A Dirac $\delta$ symbol can be interpreted as a linear functional, i.e. a linear mapping from a function space, consisting e.g. of certain real functions, to $\mathbb{R}$ (or $\mathbb{C}$), having the property

$\displaystyle \delta[f] \;=\; f(0).$
One may think this as the inner product

$\displaystyle \langle f,\,\delta\rangle \;=\; \int_0^\infty\!f(t)\delta(t)\,dt$
of a function $f$ and another “function” $\delta$, when the well-known formula

$\displaystyle \int_0^\infty\!f(t)\delta(t)\,dt \;=\; f(0)$
is true.  Applying this to  $f(t) := e^{-st}$,  one gets

$\displaystyle \int_0^\infty\!e^{-st}\delta(t)\,dt \;=\; e^{-0},$
i.e. the Laplace transform
$\displaystyle \mathcal{L}\{\delta(t)\} \;=\; 1.$ (0.1)
By the delay theorem, this result may be generalised to

$\displaystyle \mathcal{L}\{\delta(t\!-\!a))\} \;=\; e^{-as}.$

When introducing a so-called “Dirac delta function”, for example

\begin{align*}\eta_\varepsilon(t) \;:=\; \begin{cases} \frac{1}{\varepsilon} \qu... ...arepsilon,\ 0 \quad \mbox{for} \qquad t > \varepsilon, \end{cases}\end{align*}    

as an “approximation” of Dirac delta, we obtain the Laplace transform

$\displaystyle \mathcal{L}\{\eta_\varepsilon(t)\} \;=\; \int_0^\infty\!e^{-st}\e... ...^\varepsilon\!e^{-st}\,dt \;=\; \frac{1\!-\!e^{-\varepsilon s}}{\varepsilon s}.$
As the Taylor expansion shows, we then have

$\displaystyle \lim_{\varepsilon\to0+}\mathcal{L}\{\eta_\varepsilon(t)\} \;=\; 1,$
according to ref.(2).

Laplace transform of Dirac delta

The Dirac delta, $\delta$, can be correctly defined as a linear functional, i.e. a linear mapping from a function space, consisting e.g. of certain real functions, to $\mathbb{R}$ (or $\mathbb{C}$), having the property

$\displaystyle \delta[f] \;=\; f(0).$
One may think of this as an inner product

$\displaystyle \langle f,\,\delta\rangle \;=\; \int_0^\infty\!f(t)\delta(t)\,dt$
of a function $f$ and another “function” $\delta$, when the well-known formula

$\displaystyle \int_0^\infty\!f(t)\delta(t)\,dt \;=\; f(0)$
holds.  By applying this to   $f(t) := e^{-st}$,  one gets

$\displaystyle \int_0^\infty\!e^{-st}\delta(t)\,dt \;=\; e^{-0},$
i.e. the Laplace transform
$\displaystyle \mathcal{L}\{\delta(t)\} \;=\; 1.$ (0.2)

By the delay theorem, this result may be generalised to:

$\displaystyle \mathcal{L}\{\delta(t\!-\!a)\} \;=\; e^{-as}.$

Bibliography

1
Schwartz, L. (1950–1951), ThÃ�orie des distributions, vols. 1–2, Hermann: Paris.
2
W. Rudin, Functional Analysis, McGraw-Hill Book Company, 1973.
3
L. Hörmander, The Analysis of Linear Partial Differential Operators I, (Distribution theory and Fourier Analysis), 2nd ed, Springer-Verlag, 1990.



"Laplace transform of Dirac's delta distribution" is owned by bci1.

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See Also: Laplace transform, table of Laplace transforms

Also defines:  Dirac's delta Laplace transform
Keywords:  Laplace transforms, Dirac delta

Cross-references: theorem, Laplace transform, inner product, function

This is version 27 of Laplace transform of Dirac's delta distribution, born on 2010-03-05, modified 2010-10-06.
Object id is 846, canonical name is LaplaceTransformOfDiracsDelta.
Accessed 1191 times total.

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Physics Classification: 00. (GENERAL)
Pending Errata and Addenda
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server problem by bloftin on 2010-05-07 14:59:22
For some reason this entry created a latex process that never finished and took up almost 100% of cpu for a long time.  I have a crude fix in place for similar events now.  If there is anything out of the ordinary with regards to the latex in this article let me know.

Ben
[ reply | up ]

Testing some escape charachters for html category with a generator has an injective cogenerator" now escape ” with "