PlanetPhysics/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 $a_1,a_2,\dots ,a_n$ of $Q(s)$ are simple, then

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

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

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

A special case of the Heaviside formula (1) is Failed to parse (syntax error): {\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\, $s=(\pm 1\pm i)a$,\, we obtain ${ℒ}^{-1}\left\{\frac{s^3}{s^4+4a^4}\right\}=\frac{1}{4}{ℒ}^{-1}\left\{\frac{4s^3}{s^4+4a^4}\right\}=\frac{1}{4}\sum _{\pm }{e}^{(\pm 1\pm i)at}=\frac{e^{at}+e^{-at}}{2}\cdot \frac{e^{iat}+e^{-iat}}{2}=\cosh at\cos at.$\\

Proof of (1). \, Without hurting the generality, we can suppose that $Q(s)$ is monic.\, Therefore $Q(s)=(s-a_1)(s-a_2)\cdots (s-s_n).$ For\, $j=1,2,\dots ,n$,\, 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

$\begin{array}{c}\frac{P(s)}{Q(s)}=\frac{C_1}{s-a_1}+\frac{C_2}{s-a_2}+\dots +\frac{C_n}{s-a_n}\end{array}$

with constants $C_j$.\, According to the linearity and the formula 1 of the parent entry, one gets

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

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

$\begin{array}{c}{C}_j=;\frac{P(a_j)}{Q_j(a_j)}.\end{array}$

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

$\begin{array}{c}{C}_j;=\frac{P(a_j)}{Q^{\prime }(a_j)}.\end{array}$

The values (6) in (4) produce the [[../Formula/|formula]] (1).