Heaviside formula

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 (http://planetmath.org/Zero) $a_1,a_2,\mathrm{\dots },a_n$ of $Q(s)$ are simple, then

  ${ℒ}^{-1}\left\{{\displaystyle \frac{P(s)}{Q(s)}}\right\}={\displaystyle \sum _{j=1}^n}{\displaystyle \frac{P(a_j)}{Q^{\prime }(a_j)}}{e}^{a_{j}t}.$ (1)

• If the different zeroes $a_1,a_2,\mathrm{\dots },a_n$ of $Q(s)$ have the multiplicities $m_1,m_2,\mathrm{\dots },m_n$, respectively, we denote  $F_j(s):={(s-a_j)}^{m_j}P(s)/Q(s)$;  then

  ${ℒ}^{-1}\left\{{\displaystyle \frac{P(s)}{Q(s)}}\right\}={\displaystyle \sum _{j=1}^n}{e}^{a_{j}t}{\displaystyle \sum _{k=0}^{m_j-1}}{\displaystyle \frac{F_j^{(k)}(a_j)t^{m_j-1-k}}{k!(m_j-1-k)!}}.$ (2)

A special case of the Heaviside formula (1) is

${ℒ}^{-1}\left\{{\displaystyle \frac{Q^{\prime }(s)}{Q(s)}}\right\}={\displaystyle \sum _{j=1}^n}{e}^{a_{j}t}.$ (3)

Example.  Since the zeros of the binomial $s^4+4a^4$ are  $s=(\pm 1\pm i)a$,  we can calculate by (3) as follows:

$${ℒ}^{-1}\left\{\frac{s^3}{s^4+4a^4}\right\}=\frac{1}{4}{ℒ}^{-1}\left\{\frac{4s^3}{s^4+4a^4}\right\}=\frac{1}{4}\sum _{\pm }{e}^{(\pm 1\pm i)at}=\frac{e^{at}+e^{-at}}{2}\cdot \frac{e^{iat}+e^{-iat}}{2}=\cosh at\cos at$$

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

$$Q(s)=(s-a_1)(s-a_2)\mathrm{\cdots }(s-s_n).$$

For  $j=1,2,\mathrm{\dots },n$,  denoting

$$Q(s):=(s-a_j)Q_j(s),$$

one has  $Q_j(a_j)\ne 0$.  We have a partial fraction expansion of the form

${\displaystyle \frac{P(s)}{Q(s)}}={\displaystyle \frac{C_1}{s-a_1}}+{\displaystyle \frac{C_2}{s-a_2}}+\mathrm{\dots }+{\displaystyle \frac{C_n}{s-a_n}}$ (4)

with constants $C_j$.  According to the linearity and the formula 1 of the parent entry (http://planetmath.org/LaplaceTransform), one gets

${ℒ}^{-1}\left\{{\displaystyle \frac{P(s)}{Q(s)}}\right\}={\displaystyle \sum _{j=1}^n}{C}_j{e}^{a_{j}t}.$ (5)

For determining the constants $C_j$, multiply (3) by $s-a_j$.  It yields

$$\frac{P(s)}{Q_j(s)}=C_j+(s-a_j)\sum _{\nu \ne j}\frac{C_{\nu }}{s-a_{\nu }}.$$

Setting to this identity  $s:=a_j$  gives the value

$C_j={\displaystyle \frac{P(a_j)}{Q_j(a_j)}}.$ (6)

But since  $Q^{\prime }(s)=\frac{d}{ds}((s-a_j)Q_j(s))=Q_j(s)+(s-a_j)Q_j^{\prime }(s)$,  we see that  $Q^{\prime }(a_j)=Q_j(a_j)$;  thus the equation (5) may be written

$C_j={\displaystyle \frac{P(a_j)}{Q^{\prime }(a_j)}}.$ (7)

The values (6) in (4) produce the formula (1).

Title	Heaviside formula
Canonical name	HeavisideFormula
Date of creation	2014-03-19 9:14:46
Last modified on	2014-03-19 9:14:46
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	11
Author	pahio (2872)
Entry type	Topic
Classification	msc 44A10
Synonym	Heaviside expansion formula
Synonym	inverse Laplace transform of rational function
Related topic	HyperbolicFunctions
Related topic	ComplexSineAndCosine