{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Example\n\nLet $b \u003e 0$ and $0 \u003c \\alpha \u003c 1$.\nThe [[Riemann–Liouville integral|JTKCZ44NN3]] of order $\\alpha$ of a continuous function $\\varphi$ on $[0, b]$ is\n$$(I^\\alpha \\varphi)(x) = \\frac{1}{\\Gamma(\\alpha)} \\int_0^x (x - t)^{\\alpha - 1} \\varphi(t) \\, dt,$$\nwhere $\\Gamma$ is the [[Gamma function|0H5SYBSED9]].\nThe *Riemann–Liouville fractional integral equation* of order $\\alpha$ is $I^\\alpha \\varphi = f$ for a given $f$ on $[0, b]$.\nIt is a [[Volterra equation of the first kind|H6ZGEDM4VV]] of [[convolution type|SNFYTC1TTA]], with the weakly singular kernel $k(u) = u^{\\alpha - 1} / \\Gamma(\\alpha)$.\n\nCheck on power functions: let $\\beta \u003e -1$.\nThe substitution $t = xu$ gives\n$$I^\\alpha (t^\\beta)(x) = \\frac{x^{\\alpha + \\beta}}{\\Gamma(\\alpha)} \\int_0^1 (1 - u)^{\\alpha - 1} u^\\beta \\, du = \\frac{x^{\\alpha + \\beta} B(\\alpha, \\beta + 1)}{\\Gamma(\\alpha)} = \\frac{\\Gamma(\\beta + 1)}{\\Gamma(\\alpha + \\beta + 1)} \\, x^{\\alpha + \\beta},$$\nusing the [[Beta function|QHNWVSC6AQ]].\nSo for $\\mu \u003e \\alpha - 1$, the equation $I^\\alpha \\varphi = x^\\mu$ is solved by\n$$\\varphi(t) = \\frac{\\Gamma(\\mu + 1)}{\\Gamma(\\mu + 1 - \\alpha)} \\, t^{\\mu - \\alpha}.$$\nFor instance, $\\alpha = \\tfrac12$ and $\\mu = 1$ give $I^{1/2} \\varphi = x$ with $\\varphi(t) = \\frac{\\Gamma(2)}{\\Gamma(3/2)} t^{1/2} = \\frac{2}{\\sqrt{\\pi}} \\sqrt{t}$.\n\n## Remarks\n\nWhen a continuous solution exists, it is $\\varphi = \\frac{d}{dx} I^{1 - \\alpha} f$, the Riemann–Liouville fractional derivative of order $\\alpha$ of $f$.\nThis follows from the semigroup law $I^{1 - \\alpha} I^\\alpha = I^1$, and it is the [[Abel inversion formula|TDMXFY06A2]] in other notation.\nFor $\\alpha = \\tfrac12$, the equation is [[Abel's integral equation|G4GWBYTW7B]] divided by $\\Gamma(\\tfrac12) = \\sqrt{\\pi}$.\n\n## Explanations\n\n### L1 human\n\nIntegrating $n$ times can be written as one integral with the kernel $(x - t)^{n - 1}/(n - 1)!$.\nReplacing $n$ by a fraction $\\alpha$ gives the fractional integral $I^\\alpha$.\nSolving $I^\\alpha \\varphi = f$ means taking a derivative of order $\\alpha$ of $f$.\nThe power functions show how this works: integrating to order $\\alpha$ raises the exponent by $\\alpha$, and solving lowers it by $\\alpha$.\n",
  "commit": "ff8a40749dc16545e3d2fc637f6b0196e81d0bfa",
  "front": {
    "instance_of": [
      "H6ZGEDM4VV",
      "SNFYTC1TTA"
    ],
    "kind": "example",
    "msc": [
      "45D05",
      "26A33"
    ],
    "name": "Riemann–Liouville fractional integral equation",
    "sources": [
      {
        "url": "https://en.wikipedia.org/wiki/Riemann%E2%80%93Liouville_integral"
      }
    ],
    "tag": "81JHJAVRV4",
    "uses": [
      "JTKCZ44NN3",
      "0H5SYBSED9",
      "QHNWVSC6AQ"
    ]
  },
  "kind": "example",
  "link": "https://tamnd.github.io/math2/node/81JHJAVRV4/",
  "links": [
    {
      "field": "instance_of",
      "target": "H6ZGEDM4VV"
    },
    {
      "field": "instance_of",
      "target": "SNFYTC1TTA"
    },
    {
      "field": "uses",
      "target": "JTKCZ44NN3"
    },
    {
      "field": "uses",
      "target": "0H5SYBSED9"
    },
    {
      "field": "uses",
      "target": "QHNWVSC6AQ"
    }
  ],
  "markdown": "https://tamnd.github.io/math2/node/81JHJAVRV4.md",
  "msc": [
    "45D05",
    "26A33"
  ],
  "name": "Riemann–Liouville fractional integral equation",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/81J/81JHJAVRV4.md",
  "status": null,
  "tag": "81JHJAVRV4",
  "uri": "m2://81JHJAVRV4",
  "url": "https://tamnd.github.io/math2/wiki/riemann-liouville-fractional-integral-equation/",
  "writing": {
    "tag": "81JHJAVRV4",
    "kind": "example",
    "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
      }
    ]
  }
}