---
tag: 2C008K3V28
kind: definition
name: Operational calculus
msc: [44A45, 44A40, 44A10]
defines: 8050R3Z6HA
uses: [0V49JEZW06, 07WBTX1MHA, VDMCEPTTXB, 0K090GVHVN, 0JNMJEDD93, 0QQRP17Z7S]
sources:
  - url: https://en.wikipedia.org/wiki/Operational_calculus
---

## Definition

Let $E$ be the space of locally integrable functions $f \colon [0, \infty) \to \C$ of exponential order, that is, with $\abs{f(t)} \le M e^{at}$ for some constants $M, a$ and almost all $t$.
The *operational calculus* based on the [[Laplace transform|0V49JEZW06]] is the correspondence $f \mapsto F = \mathcal{L} f$, $F(s) = \int_0^\infty e^{-st} f(t) \, dt$, used with the following rules.

1. $\mathcal{L}$ is linear and injective up to equality almost everywhere, so $f$ is recovered from $F$ by the [[inverse Laplace transform|07WBTX1MHA]].
2. If $f$ is continuous, $f'$ exists except at finitely many points on each bounded interval, and $f' \in E$, then $\mathcal{L}(f')(s) = s F(s) - f(0)$.
3. The [[convolution on the half-line|VDMCEPTTXB]] becomes a product: $\mathcal{L}(f * g) = \mathcal{L} f \cdot \mathcal{L} g$.

A linear problem for $f$ is solved by transforming it with these rules into an algebraic equation for $F$, solving for $F$, and recovering $f$.

## Remarks

Rule 2 is the [[Laplace transform of a derivative|0K090GVHVN]] and rule 3 is the [[convolution theorem|0JNMJEDD93]].
With them, a [[linear differential equation|0QQRP17Z7S]] with constant coefficients and given initial values becomes a linear equation for $F$.
Its coefficients are polynomials in $s$.
The name operational calculus also covers other frameworks with the same structure.
One is the formal calculus of Heaviside; another is the Mikusiński calculus, where $s$ is an element of a field and no integral transform is used.

Example: $y'' + y = 0$, $y(0) = 0$, $y'(0) = 1$.
Rule 2 twice gives $s^2 Y - 1 + Y = 0$, so $Y(s) = 1/(s^2 + 1)$, which is the transform of $\sin t$.
Check: $\sin'' + \sin = 0$, $\sin 0 = 0$, $\sin' 0 = 1$.

Non-example: $y' = t\, e^{t^2}$ fits the scheme only formally.
Its solutions are $e^{t^2}/2 + C$, which are not of exponential order, so they are outside $E$ and have no Laplace transform.