Heaviside formula

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

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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