{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe differentiate the explicit formula for $\\phi$ under the integral sign.\n\n1. Write $F = (F_1, F_2, F_3)$ and $\\partial_i = \\partial / \\partial x_i$. By the definition of the [[curl|0GDNHS9J2X]], $\\nabla \\times F = 0$ means $\\partial_2 F_3 = \\partial_3 F_2$, $\\partial_3 F_1 = \\partial_1 F_3$ and $\\partial_1 F_2 = \\partial_2 F_1$. So $\\partial_j F_i = \\partial_i F_j$ for all $i, j$.\n2. Fix $p \\in U$. For $x \\in U$ and $t \\in [0, 1]$, the point $p + t (x - p)$ lies in $U$, since $U$ is [[convex|0K06X9QXDX]]. So $\\phi(x) = \\int_0^1 F(p + t (x - p)) \\cdot (x - p) \\, dt$ is defined, with a continuous integrand.\n3. Each $F_i$ is $C^1$, so it is differentiable by [[continuous partial derivatives imply differentiability|0KXSK8TEEB]]. Its derivative in a direction $v$ is $\\nabla F_i \\cdot v$, by the definition of the [[gradient|0NS2QYVC4Y]].\n4. Fix $x \\in U$ and $j \\in \\set{1, 2, 3}$. Choose $r \u003e 0$ such that $x + h e_j \\in U$ for $\\abs{h} \u003c 2 r$, where $e_j$ is the $j$-th basis vector.\n5. For $t \\in [0, 1]$ and $\\abs{h} \u003c 2 r$, let $y_h = x + h e_j$ and $q(t, h) = p + t (y_h - p)$. By step 2, $q(t, h) \\in U$.\n6. Let $g(t, h) = F(q(t, h)) \\cdot (y_h - p) = \\sum_i F_i(q(t, h)) \\, (y_{h, i} - p_i)$. Then $\\phi(x + h e_j) = \\int_0^1 g(t, h) \\, dt$, and $g$ is continuous on $[0, 1] \\times [-r, r]$.\n7. For fixed $t$, the curve $h \\mapsto q(t, h)$ has derivative $t e_j$. By the [[chain rule along a curve|0A8TJ0VHC7]] and step 3, $\\frac{\\partial}{\\partial h} F_i(q(t, h)) = t \\, \\partial_j F_i(q(t, h))$.\n8. Also $\\frac{\\partial}{\\partial h} (y_{h, i} - p_i)$ is $1$ for $i = j$ and $0$ otherwise. By the [[product rule|0V9A43H7WV]], $\\frac{\\partial g}{\\partial h}(t, h) = t \\sum_i \\partial_j F_i(q(t, h)) \\, (y_{h, i} - p_i) + F_j(q(t, h))$.\n9. This [[partial derivative|05AXG3QRKR]] is continuous on $[0, 1] \\times [-r, r]$, since the $\\partial_j F_i$ are continuous. By [[differentiation under the integral sign|0AR06W1E1B]] at $h = 0$, $\\partial_j \\phi(x) = \\int_0^1 \\frac{\\partial g}{\\partial h}(t, 0) \\, dt$.\n10. Let $c(t) = p + t (x - p)$, so $c(t) = q(t, 0)$. By step 1, $\\frac{\\partial g}{\\partial h}(t, 0) = t \\sum_i \\partial_i F_j(c(t)) \\, (x_i - p_i) + F_j(c(t))$.\n11. The curve $c$ has derivative $x - p$. By the chain rule along a curve and step 3, $\\frac{d}{dt} F_j(c(t)) = \\sum_i \\partial_i F_j(c(t)) \\, (x_i - p_i)$ for $t \\in (0, 1)$.\n12. By the product rule and steps 10 and 11, $\\frac{\\partial g}{\\partial h}(t, 0) = \\frac{d}{dt} \\big( t \\, F_j(c(t)) \\big)$ for $t \\in (0, 1)$. The function $t \\mapsto t F_j(c(t))$ is continuous on $[0, 1]$.\n13. By the [[fundamental theorem of calculus|0NHGY0N0JB]], $\\partial_j \\phi(x) = 1 \\cdot F_j(c(1)) - 0 \\cdot F_j(c(0)) = F_j(x)$.\n14. So the partial derivatives of $\\phi$ are $F_1, F_2, F_3$, which are continuous. Hence $\\phi$ is of [[class $C^1$|0TCBZMEYGN]], and $\\nabla \\phi = F$ on $U$ by the coordinate form of the gradient.\n15. Therefore $F$ is [[conservative|05R9NF2J49]] on $U$, with the potential $\\phi$ of the statement.\n\n## Idea\n\nThe candidate potential is the work of $F$ along the straight segment from $p$ to $x$. Differentiating in $x_j$ under the integral produces terms $\\partial_j F_i$. The curl-free condition turns them into $\\partial_i F_j$, and then the integrand becomes an exact $t$-derivative of $t F_j$, which integrates to $F_j(x)$.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "26B12"
    ],
    "name": "Proof that a curl-free vector field on a convex set is conservative",
    "proves": "0B22P010D6",
    "sources": [
      {
        "locator": "Lemma 3.11",
        "url": "https://www.maths.tcd.ie/~zaitsev/topics-ca/wedhorn-Complex-Analysis.pdf"
      }
    ],
    "tag": "242AXRRT28",
    "uses": [
      "0GDNHS9J2X",
      "05R9NF2J49",
      "0K06X9QXDX",
      "0AR06W1E1B",
      "0A8TJ0VHC7",
      "0KXSK8TEEB",
      "0NS2QYVC4Y",
      "0V9A43H7WV",
      "0NHGY0N0JB",
      "0TCBZMEYGN",
      "05AXG3QRKR"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/242AXRRT28/",
  "links": [
    {
      "field": "proves",
      "target": "0B22P010D6"
    },
    {
      "field": "uses",
      "target": "0GDNHS9J2X"
    },
    {
      "field": "uses",
      "target": "05R9NF2J49"
    },
    {
      "field": "uses",
      "target": "0K06X9QXDX"
    },
    {
      "field": "uses",
      "target": "0AR06W1E1B"
    },
    {
      "field": "uses",
      "target": "0A8TJ0VHC7"
    },
    {
      "field": "uses",
      "target": "0KXSK8TEEB"
    },
    {
      "field": "uses",
      "target": "0NS2QYVC4Y"
    },
    {
      "field": "uses",
      "target": "0V9A43H7WV"
    },
    {
      "field": "uses",
      "target": "0NHGY0N0JB"
    },
    {
      "field": "uses",
      "target": "0TCBZMEYGN"
    },
    {
      "field": "uses",
      "target": "05AXG3QRKR"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/242AXRRT28.md",
  "msc": [
    "26B12"
  ],
  "name": "Proof that a curl-free vector field on a convex set is conservative",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/242/242AXRRT28.md",
  "status": null,
  "tag": "242AXRRT28",
  "uri": "m2://242AXRRT28",
  "url": "https://math2.tamnd.com/wiki/proof-that-a-curl-free-vector-field-on-a-convex-set-is-conservative/",
  "writing": {
    "tag": "242AXRRT28",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}