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Heaviside formula (Definition)

Let $P(s)$ and $Q(s)$ be polynomials with the degree of the former less than the degree of the latter.

  • If all complex zeroes $a_1,\,a_2,\,\ldots,\,a_n$ of $Q(s)$ are simple, then
    $\displaystyle \mathcal{L}^{-1}\left\{\frac{P(s)}{Q(s)}\right\} \;=\; \sum_{j=1}^n\frac{P(a_j)}{Q'(a_j)}e^{a_jt}.$ (1)
  • If the different zeroes $a_1,\,a_2,\,\ldots,\,a_n$ of $Q(s)$ have the multiplicities $m_1,\,m_2,\,\ldots,\,m_n$, respectively, we denote  $F_j(s) := (s\!-\!a_j)^{m_j}P(s)/Q(s)$;  then
    $\displaystyle \mathcal{L}^{-1}\left\{\frac{P(s)}{Q(s)}\right\} \;=\; \sum_{j=1}... ...um_{k=0}^{m_j-1}\frac{F_j^{(k)}(a_j)t^{m_j\!-\!1\!-\!k}}{k!(m_j\!-\!1\!-\!k)!}.$ (2)

A special case of the Heaviside formula (1) is

$\displaystyle \mathcal{L}^{-1}\left\{\frac{Q'(s)}{Q(s)}\right\} \;=\; \sum_{j=1}^ne^{a_jt}.\\ $

Example.  Since the zeros of the binomial $s^4\!+\!4a^4$ are  $s = (\pm1\!\pm\!i)a$,  we obtain

$\displaystyle \mathcal{L}^{-1}\left\{\frac{s^3}{s^4\!+\!4a^4}\right\} \;=\; \fr... ...ac{e^{at}+e^{-at}}{2}\cdot\frac{e^{iat}+e^{-iat}}{2} \;=\; \cosh{at}\,\cos{at}.$

Proof of (1).  Without hurting the generality, we can suppose that $Q(s)$ is monic.  Therefore

$\displaystyle Q(s) \;=\; (s\!-\!a_1)(s\!-\!a_2)\cdots(s\!-\!s_n).$
For  $j = 1,\,2,\;\ldots,\,n$,  denoting

$\displaystyle Q(s) \;:=\; (s\!-\!a_j)Q_j(s),$
one has  $Q_j(a_j) \neq 0$.  We have a partial fraction expansion of the form
$\displaystyle \frac{P(s)}{Q(s)} \;=\; \frac{C_1}{s\!-\!a_1}+\frac{C_2}{s\!-\!a_2}+\ldots+\frac{C_n}{s\!-\!a_n}$ (3)
with constants $C_j$.  According to the linearity and the formula 1 of the parent entry, one gets
$\displaystyle \mathcal{L}^{-1}\left\{\frac{P(s)}{Q(s)}\right\} \;=\; \sum_{j=1}^nC_je^{a_jt}.$ (4)
For determining the constants $C_j$, multiply (3) by $s\!-\!a_j$.  It yields

$\displaystyle \frac{P(s)}{Q_j(s)} = C_j+(s\!-\!a_j)\sum_{\nu \neq j}\frac{C_\nu}{s\!-\!a_\nu}.$
Setting to this identity  $s := a_j$  gives the value
$\displaystyle C_j \;=; \frac{P(a_j)}{Q_j(a_j)}.$ (5)
But since  $Q'(s) = \frac{d}{ds}((s\!-\!a_j)Q_j(s)) = Q_j(s)\!+\!(s\!-\!a_j)Q_j'(s)$,  we see that  $Q'(a_j) = Q_j(a_j)$;  thus the equation (5) may be written
$\displaystyle C_j ;=\; \frac{P(a_j)}{Q'(a_j)}.$ (6)
The values (6) in (4) produce the formula (1).

Bibliography

1
K. V¨AISÄLÄ: Laplace-muunnos.  Handout Nr. 163.    Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1968).



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Physics Classification: 02.30.Uu (Integral transforms)
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