Mikusiński operational calculus

Remarks

Rule 1 follows by induction from $s f = f' + f(0)$, which is the identity $l * f' = f - f(0) l$ divided by $l$. Rule 2 for $k = 1$ follows from rule 1: for $f = e^{\alpha t}$, $s f - 1 = \alpha f$, so $(s - \alpha) f = 1$. The solution found in step 3 is the unique solution, because $\mathcal{M}$ is a field and the embedding of continuous functions is injective. The rules agree with those of the Laplace transform method, with $s$ in the role of the transform variable, but they apply to all continuous functions, without growth conditions.

Example: $y'' + y = 0$, $y(0) = 1$, $y'(0) = 0$. Rule 1 gives $s^2 y - s + y = 0$, so $y = s/(s^2 + 1) = \frac{1}{2}\left(\frac{1}{s - i} + \frac{1}{s + i}\right)$. By rule 2, $y = \frac{1}{2}(e^{it} + e^{-it}) = \cos t$.

Non-example: the operator $s$ itself is not a continuous function, so rule 2 has no analogue for $s$. If $s = g$ for a continuous $g$, then $l * g = l \cdot s = 1$, so $l * g * l = l$ as functions. The left side is a convolution of continuous functions, so it vanishes at $t = 0$, while $l(0) = 1$.