{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Definition\n\nLet $\\mathcal{M}$ be the [[Mikusiński field of operators|82YAXZMMQ1]].\nIt contains the complex numbers and the continuous functions on $[0, \\infty)$, with the [[convolution on the half-line|FCKX8KFVGN]] as product.\nLet $l$ be the constant function $1$ and $s = 1/l \\in \\mathcal{M}$.\nThe *Mikusiński operational calculus* is the method of solving linear problems in $\\mathcal{M}$ with the following rules.\n\n1. For every $n \\ge 1$ and every $n$ times continuously differentiable $f$ on $[0, \\infty)$, $f^{(n)} = s^n f - s^{n-1} f(0) - s^{n-2} f'(0) - \\dots - f^{(n-1)}(0)$ in $\\mathcal{M}$.\n2. For every complex $\\alpha$, the operator $1/(s - \\alpha)$ is the function $e^{\\alpha t}$, and more generally $1/(s - \\alpha)^{k}$ is the function $t^{k-1} e^{\\alpha t}/(k-1)!$ for $k \\ge 1$.\n3. A [[linear differential equation|0QQRP17Z7S]] with constant coefficients and given initial values is rewritten by rule 1 as a linear equation for $y \\in \\mathcal{M}$. This equation is solved in the field $\\mathcal{M}$, and the solution is identified as a function by rule 2 and partial fractions.\n\n## Remarks\n\nRule 1 follows by induction from $s f = f' + f(0)$, which is the identity $l * f' = f - f(0) l$ divided by $l$.\nRule 2 for $k = 1$ follows from rule 1: for $f = e^{\\alpha t}$, $s f - 1 = \\alpha f$, so $(s - \\alpha) f = 1$.\nThe solution found in step 3 is the unique solution, because $\\mathcal{M}$ is a field and the embedding of continuous functions is injective.\nThe 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.\n\nExample: $y'' + y = 0$, $y(0) = 1$, $y'(0) = 0$.\nRule 1 gives $s^2 y - s + y = 0$, so $y = s/(s^2 + 1) = \\frac{1}{2}\\left(\\frac{1}{s - i} + \\frac{1}{s + i}\\right)$.\nBy rule 2, $y = \\frac{1}{2}(e^{it} + e^{-it}) = \\cos t$.\n\nNon-example: the operator $s$ itself is not a continuous function, so rule 2 has no analogue for $s$.\nIf $s = g$ for a continuous $g$, then $l * g = l \\cdot s = 1$, so $l * g * l = l$ as functions.\nThe left side is a convolution of continuous functions, so it vanishes at $t = 0$, while $l(0) = 1$.\n",
  "commit": "ff8a40749dc16545e3d2fc637f6b0196e81d0bfa",
  "front": {
    "defines": "TD5XK5AHNV",
    "kind": "definition",
    "msc": [
      "44A40"
    ],
    "name": "Mikusiński operational calculus",
    "sources": [
      {
        "url": "https://en.wikipedia.org/wiki/Operational_calculus"
      }
    ],
    "tag": "MTM92ND7HE",
    "uses": [
      "82YAXZMMQ1",
      "FCKX8KFVGN",
      "0QQRP17Z7S"
    ]
  },
  "kind": "definition",
  "link": "https://tamnd.github.io/math2/node/MTM92ND7HE/",
  "links": [
    {
      "field": "defines",
      "target": "TD5XK5AHNV"
    },
    {
      "field": "uses",
      "target": "82YAXZMMQ1"
    },
    {
      "field": "uses",
      "target": "FCKX8KFVGN"
    },
    {
      "field": "uses",
      "target": "0QQRP17Z7S"
    }
  ],
  "markdown": "https://tamnd.github.io/math2/node/MTM92ND7HE.md",
  "msc": [
    "44A40"
  ],
  "name": "Mikusiński operational calculus",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/MTM/MTM92ND7HE.md",
  "status": null,
  "tag": "MTM92ND7HE",
  "uri": "m2://MTM92ND7HE",
  "url": "https://tamnd.github.io/math2/wiki/mikusinski-operational-calculus-definition/",
  "writing": {
    "tag": "MTM92ND7HE",
    "kind": "definition",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      },
      {
        "what": "a source with a locator",
        "have": 0,
        "want": 1
      }
    ]
  }
}