{
  "index": "1987-A-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "Let\n\\[\n\\vec{G}(x,y) = \\left( \\frac{-y}{x^2+4y^2}, \\frac{x}{x^2+4y^2},0\n\\right).\n\\]\nProve or disprove that there is a vector-valued function\n\\[\n\\vec{F}(x,y,z) = (M(x,y,z), N(x,y,z), P(x,y,z))\n\\]\nwith the following properties:\n\\begin{enumerate}\n\\item[(i)] $M,N,P$ have continuous partial derivatives for all\n$(x,y,z) \\neq (0,0,0)$;\n\\item[(ii)] $\\mathrm{Curl}\\,\\vec{F} = \\vec{0}$ for all $(x,y,z) \\neq (0,0,0)$;\n\\item[(iii)] $\\vec{F}(x,y,0) = \\vec{G}(x,y)$.\n\\end{enumerate}",
  "solution": "Solution. Let \\( S \\) be a surface not containing \\( (0,0,0) \\) whose boundary \\( \\partial S \\) is the ellipse \\( x^{2}+4 y^{2}-4=z=0 \\) parameterized by \\( (2 \\cos \\theta, \\sin \\theta, 0) \\) for \\( 0 \\leq \\theta \\leq 2 \\pi \\). (For instance, \\( S \\) could be the half of the ellipsoid \\( x^{2}+4 y^{2}+z^{2}=4 \\) with \\( z \\geq 0 \\).) If \\( F \\) exists, then\n\\[\n\\begin{aligned}\n0 & =\\iint_{S}(\\operatorname{Curl} \\vec{F}) \\cdot \\vec{n} d S \\quad(\\text { since Curl } \\vec{F}=\\overrightarrow{0} \\text { on } S) \\\\\n& =\\int_{\\partial S} \\vec{F} \\cdot d \\vec{r} \\quad(\\text { by Stokes' Theorem, e.g. [Ap2, Theorem 12.3]) } \\\\\n& =\\int_{\\partial S} \\vec{G} \\cdot d \\vec{r} \\quad(\\text { since } \\vec{F}=\\vec{G} \\text { on } \\partial S) \\\\\n& =\\int_{0}^{2 \\pi}\\left(\\frac{-\\sin \\theta}{4}, \\frac{2 \\cos \\theta}{4}, 0\\right) \\cdot(-2 \\sin \\theta, \\cos \\theta, 0) d \\theta \\\\\n& =\\int_{0}^{2 \\pi} \\frac{1}{2} d \\theta \\\\\n& =\\pi\n\\end{aligned}\n\\]\na contradiction.",
  "vars": [
    "x",
    "y",
    "z",
    "\\\\theta"
  ],
  "params": [
    "F",
    "G",
    "M",
    "N",
    "P",
    "S",
    "n",
    "r"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "xcoord",
        "y": "ycoord",
        "z": "zcoord",
        "\\theta": "angleth",
        "F": "vectorf",
        "G": "vectorg",
        "M": "funcvalm",
        "N": "funcvaln",
        "P": "funcvalp",
        "S": "surfaces",
        "n": "normalv",
        "r": "pathvec"
      },
      "question": "Let\n\\[\n\\vec{vectorg}(xcoord,ycoord) = \\left( \\frac{-ycoord}{xcoord^2+4ycoord^2}, \\frac{xcoord}{xcoord^2+4ycoord^2},0\n\\right).\n\\]\nProve or disprove that there is a vector-valued function\n\\[\n\\vec{vectorf}(xcoord,ycoord,zcoord) = (funcvalm(xcoord,ycoord,zcoord), funcvaln(xcoord,ycoord,zcoord), funcvalp(xcoord,ycoord,zcoord))\n\\]\nwith the following properties:\n\\begin{enumerate}\n\\item[(i)] $funcvalm,funcvaln,funcvalp$ have continuous partial derivatives for all\n$(xcoord,ycoord,zcoord) \\neq (0,0,0)$;\n\\item[(ii)] $\\mathrm{Curl}\\,\\vec{vectorf} = \\vec{0}$ for all $(xcoord,ycoord,zcoord) \\neq (0,0,0)$;\n\\item[(iii)] $\\vec{vectorf}(xcoord,ycoord,0) = \\vec{vectorg}(xcoord,ycoord)$.\n\\end{enumerate}",
      "solution": "Solution. Let \\( surfaces \\) be a surface not containing \\( (0,0,0) \\) whose boundary \\( \\partial surfaces \\) is the ellipse \\( xcoord^{2}+4 ycoord^{2}-4=zcoord=0 \\) parameterized by \\( (2 \\cos angleth, \\sin angleth, 0) \\) for \\( 0 \\leq angleth \\leq 2 \\pi \\). (For instance, \\( surfaces \\) could be the half of the ellipsoid \\( xcoord^{2}+4 ycoord^{2}+zcoord^{2}=4 \\) with \\( zcoord \\geq 0 \\).) If \\( vectorf \\) exists, then\n\\[\n\\begin{aligned}\n0 & =\\iint_{surfaces}(\\operatorname{Curl} \\vec{vectorf}) \\cdot \\vec{normalv} d surfaces \\quad(\\text { since Curl } \\vec{vectorf}=\\overrightarrow{0} \\text { on } surfaces) \\\\\n& =\\int_{\\partial surfaces} \\vec{vectorf} \\cdot d \\vec{pathvec} \\quad(\\text { by Stokes' Theorem, e.g. [Ap2, Theorem 12.3]) } \\\\\n& =\\int_{\\partial surfaces} \\vec{vectorg} \\cdot d \\vec{pathvec} \\quad(\\text { since } \\vec{vectorf}=\\vec{vectorg} \\text { on } \\partial surfaces) \\\\\n& =\\int_{0}^{2 \\pi}\\left(\\frac{-\\sin angleth}{4}, \\frac{2 \\cos angleth}{4}, 0\\right) \\cdot(-2 \\sin angleth, \\cos angleth, 0) d angleth \\\\\n& =\\int_{0}^{2 \\pi} \\frac{1}{2} d angleth \\\\\n& =\\pi\n\\end{aligned}\n\\]\na contradiction."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lighthouse",
        "y": "crosswind",
        "z": "drumstick",
        "\\theta": "raincloud",
        "F": "kingfisher",
        "G": "snowflake",
        "M": "turnpike",
        "N": "cloudburst",
        "P": "dreamcatch",
        "S": "driftwood",
        "n": "sandpaper",
        "r": "starfruit"
      },
      "question": "Let\n\\[\n\\vec{snowflake}(lighthouse,crosswind) = \\left( \\frac{-crosswind}{lighthouse^{2}+4crosswind^{2}}, \\frac{lighthouse}{lighthouse^{2}+4crosswind^{2}},0\n\\right).\n\\]\nProve or disprove that there is a vector-valued function\n\\[\n\\vec{kingfisher}(lighthouse,crosswind,drumstick) = (turnpike(lighthouse,crosswind,drumstick), cloudburst(lighthouse,crosswind,drumstick), dreamcatch(lighthouse,crosswind,drumstick))\n\\]\nwith the following properties:\n\\begin{enumerate}\n\\item[(i)] $turnpike,cloudburst,dreamcatch$ have continuous partial derivatives for all $(lighthouse,crosswind,drumstick) \\neq (0,0,0)$;\n\\item[(ii)] $\\mathrm{Curl}\\,\\vec{kingfisher} = \\vec{0}$ for all $(lighthouse,crosswind,drumstick) \\neq (0,0,0)$;\n\\item[(iii)] $\\vec{kingfisher}(lighthouse,crosswind,0) = \\vec{snowflake}(lighthouse,crosswind)$.\n\\end{enumerate}",
      "solution": "Solution. Let \\( driftwood \\) be a surface not containing \\( (0,0,0) \\) whose boundary \\( \\partial driftwood \\) is the ellipse \\( lighthouse^{2}+4crosswind^{2}-4=drumstick=0 \\) parameterized by \\( (2 \\cos raincloud, \\sin raincloud, 0) \\) for \\( 0 \\leq raincloud \\leq 2 \\pi \\). (For instance, \\( driftwood \\) could be the half of the ellipsoid \\( lighthouse^{2}+4crosswind^{2}+drumstick^{2}=4 \\) with \\( drumstick \\geq 0 \\).) If \\( kingfisher \\) exists, then\n\\[\n\\begin{aligned}\n0 & =\\iint_{driftwood}(\\operatorname{Curl} \\vec{kingfisher}) \\cdot \\vec{sandpaper} d driftwood \\quad(\\text { since Curl } \\vec{kingfisher}=\\overrightarrow{0} \\text { on } driftwood) \\\\\n& =\\int_{\\partial driftwood} \\vec{kingfisher} \\cdot d \\vec{starfruit} \\quad(\\text { by Stokes' Theorem, e.g. [Ap2, Theorem 12.3]) } \\\\\n& =\\int_{\\partial driftwood} \\vec{snowflake} \\cdot d \\vec{starfruit} \\quad(\\text { since } \\vec{kingfisher}=\\vec{snowflake} \\text { on } \\partial driftwood) \\\\\n& =\\int_{0}^{2 \\pi}\\left(\\frac{-\\sin raincloud}{4}, \\frac{2 \\cos raincloud}{4}, 0\\right) \\cdot(-2 \\sin raincloud, \\cos raincloud, 0) d raincloud \\\\\n& =\\int_{0}^{2 \\pi} \\frac{1}{2} d raincloud \\\\\n& =\\pi\n\\end{aligned}\n\\]\na contradiction."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "z": "surfaceaxis",
        "\\theta": "straightedge",
        "F": "destroyer",
        "G": "unknownfield",
        "M": "combinedfun",
        "N": "singularfun",
        "P": "voidfunct",
        "S": "volumeobj",
        "n": "tangentvec",
        "r": "anchorpoint"
      },
      "question": "Let\n\\[\n\\vec{unknownfield}(verticalaxis,horizontalaxis) = \\left( \\frac{-horizontalaxis}{verticalaxis^{2}+4horizontalaxis^{2}}, \\frac{verticalaxis}{verticalaxis^{2}+4horizontalaxis^{2}},0\n\\right).\n\\]\nProve or disprove that there is a vector-valued function\n\\[\n\\vec{destroyer}(verticalaxis,horizontalaxis,surfaceaxis) = (combinedfun(verticalaxis,horizontalaxis,surfaceaxis), singularfun(verticalaxis,horizontalaxis,surfaceaxis), voidfunct(verticalaxis,horizontalaxis,surfaceaxis))\n\\]\nwith the following properties:\n\\begin{enumerate}\n\\item[(i)] $combinedfun,singularfun,voidfunct$ have continuous partial derivatives for all\n$(verticalaxis,horizontalaxis,surfaceaxis) \\neq (0,0,0)$;\n\\item[(ii)] $\\mathrm{Curl}\\,\\vec{destroyer} = \\vec{0}$ for all $(verticalaxis,horizontalaxis,surfaceaxis) \\neq (0,0,0)$;\n\\item[(iii)] $\\vec{destroyer}(verticalaxis,horizontalaxis,0) = \\vec{unknownfield}(verticalaxis,horizontalaxis)$.\n\\end{enumerate}",
      "solution": "Solution. Let \\( volumeobj \\) be a surface not containing \\( (0,0,0) \\) whose boundary \\( \\partial volumeobj \\) is the ellipse \\( verticalaxis^{2}+4 horizontalaxis^{2}-4=surfaceaxis=0 \\) parameterized by \\( (2 \\cos straightedge, \\sin straightedge, 0) \\) for \\( 0 \\le straightedge \\le 2 \\pi \\). (For instance, \\( volumeobj \\) could be the half of the ellipsoid \\( verticalaxis^{2}+4 horizontalaxis^{2}+surfaceaxis^{2}=4 \\) with \\( surfaceaxis \\ge 0 \\).) If \\( destroyer \\) exists, then\n\\[\n\\begin{aligned}\n0 & =\\iint_{volumeobj}(\\operatorname{Curl} \\vec{destroyer}) \\cdot \\vec{tangentvec} d volumeobj \\quad(\\text { since Curl } \\vec{destroyer}=\\overrightarrow{0} \\text { on } volumeobj) \\\\\n& =\\int_{\\partial volumeobj} \\vec{destroyer} \\cdot d \\vec{anchorpoint} \\quad(\\text { by Stokes' Theorem, e.g. [Ap2, Theorem 12.3]) } \\\\\n& =\\int_{\\partial volumeobj} \\vec{unknownfield} \\cdot d \\vec{anchorpoint} \\quad(\\text { since } \\vec{destroyer}=\\vec{unknownfield} \\text { on } \\partial volumeobj) \\\\\n& =\\int_{0}^{2 \\pi}\\left(\\frac{-\\sin straightedge}{4}, \\frac{2 \\cos straightedge}{4}, 0\\right) \\cdot(-2 \\sin straightedge, \\cos straightedge, 0) d straightedge \\\\\n& =\\int_{0}^{2 \\pi} \\frac{1}{2} d straightedge \\\\\n& =\\pi\n\\end{aligned}\n\\]\na contradiction."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "z": "mndpybxe",
        "\\theta": "fljsgkra",
        "F": "kjdpqraw",
        "G": "czlqevsn",
        "M": "rtfvknaj",
        "N": "wqkpsdml",
        "P": "vnrybega",
        "S": "xeflmpdo",
        "n": "dbchoyva",
        "r": "kpmtosle"
      },
      "question": "Let\n\\[\n\\vec{czlqevsn}(qzxwvtnp,hjgrksla) = \\left( \\frac{-hjgrksla}{qzxwvtnp^2+4hjgrksla^2}, \\frac{qzxwvtnp}{qzxwvtnp^2+4hjgrksla^2},0\n\\right).\n\\]\nProve or disprove that there is a vector-valued function\n\\[\n\\vec{kjdpqraw}(qzxwvtnp,hjgrksla,mndpybxe) = (rtfvknaj(qzxwvtnp,hjgrksla,mndpybxe), wqkpsdml(qzxwvtnp,hjgrksla,mndpybxe), vnrybega(qzxwvtnp,hjgrksla,mndpybxe))\n\\]\nwith the following properties:\n\\begin{enumerate}\n\\item[(i)] $rtfvknaj,wqkpsdml,vnrybega$ have continuous partial derivatives for all\n$(qzxwvtnp,hjgrksla,mndpybxe) \\neq (0,0,0)$;\n\\item[(ii)] $\\mathrm{Curl}\\,\\vec{kjdpqraw} = \\vec{0}$ for all $(qzxwvtnp,hjgrksla,mndpybxe) \\neq (0,0,0)$;\n\\item[(iii)] $\\vec{kjdpqraw}(qzxwvtnp,hjgrksla,0) = \\vec{czlqevsn}(qzxwvtnp,hjgrksla)$.\n\\end{enumerate}",
      "solution": "Solution. Let \\( xeflmpdo \\) be a surface not containing \\( (0,0,0) \\) whose boundary \\( \\partial xeflmpdo \\) is the ellipse \\( qzxwvtnp^{2}+4 hjgrksla^{2}-4=mndpybxe=0 \\) parameterized by \\( (2 \\cos fljsgkra, \\sin fljsgkra, 0) \\) for \\( 0 \\leq fljsgkra \\leq 2 \\pi \\). (For instance, \\( xeflmpdo \\) could be the half of the ellipsoid \\( qzxwvtnp^{2}+4 hjgrksla^{2}+mndpybxe^{2}=4 \\) with \\( mndpybxe \\geq 0 \\).) If \\( kjdpqraw \\) exists, then\n\\[\n\\begin{aligned}\n0 & =\\iint_{xeflmpdo}(\\operatorname{Curl} \\vec{kjdpqraw}) \\cdot \\vec{dbchoyva} d xeflmpdo \\quad(\\text { since Curl } \\vec{kjdpqraw}=\\overrightarrow{0} \\text { on } xeflmpdo) \\\\\n& =\\int_{\\partial xeflmpdo} \\vec{kjdpqraw} \\cdot d \\vec{kpmtosle} \\quad(\\text { by Stokes' Theorem, e.g. [Ap2, Theorem 12.3]) } \\\\\n& =\\int_{\\partial xeflmpdo} \\vec{czlqevsn} \\cdot d \\vec{kpmtosle} \\quad(\\text { since } \\vec{kjdpqraw}=\\vec{czlqevsn} \\text { on } \\partial xeflmpdo) \\\\\n& =\\int_{0}^{2 \\pi}\\left(\\frac{-\\sin fljsgkra}{4}, \\frac{2 \\cos fljsgkra}{4}, 0\\right) \\cdot(-2 \\sin fljsgkra, \\cos fljsgkra, 0) d fljsgkra \\\\\n& =\\int_{0}^{2 \\pi} \\frac{1}{2} d fljsgkra \\\\\n& =\\pi\n\\end{aligned}\n\\]\na contradiction."
    },
    "kernel_variant": {
      "question": "Let the following three C¹–vector fields be prescribed on the punctured coordinate planes of ℝ³:\n\nG₁(x,y) = ( −x²y /(x² + y²)² ,  x y² /(x² + y²)² , 0 )    for (x,y) ≠ (0,0) and z = 0  \nG₂(y,z) = ( 0 , −y²z /(y² + z²)² ,  y z² /(y² + z²)² )    for (y,z) ≠ (0,0) and x = 0  \nG₃(z,x) = ( −z x² /(z² + x²)² , 0 ,  x z² /(z² + x²)² )    for (z,x) ≠ (0,0) and y = 0.\n\nObserve that each field vanishes when one of its two variables is zero, so on every coordinate axis  \n (0,y,0), (x,0,0), (0,0,z)  \nall three prescriptions give the common value (0,0,0); the data are therefore mutually compatible along the pairwise intersections of the planes.\n\nQuestion.  Does there exist a C¹–vector field  \n  F : ℝ³ \\ {(0,0,0)} → ℝ³  (F = (M,N,P))  \nsatisfying simultaneously\n\n(i) curl F ≡ 0 on ℝ³ \\ {(0,0,0)};  \n\n(ii) F coincides with the given traces on the three coordinate planes, i.e.  \n   F(x,y,0)=G₁(x,y), F(0,y,z)=G₂(y,z), F(x,0,z)=G₃(z,x);  \n\n(iii) |F(r)| = O(|r|⁻¹) as |r| → ∞ ?\n\nEither construct such a field or prove that none can exist.",
      "solution": "We show that conditions (i)–(iii) are incompatible; no such vector field F exists.\n\n1.  Topological background  \nThe domain ℝ³\\{0} is simply connected.  Hence any C¹–vector field whose curl is identically zero there is conservative: for every piece-wise smooth closed curve C contained in ℝ³\\{0},\n  ∮_C F·dr = 0.                                              (1)\n\n2.  A closed curve touching all three planes  \nJoin three quarter-circles of radius 1, one in each coordinate plane, head-to-tail:\n\nC₁ : θ∈[0,π/2] → (cos θ, sin θ, 0)   (XY–plane)  \nC₂ : θ∈[0,π/2] → (0, cos θ, sin θ)   (YZ–plane)  \nC₃ : θ∈[0,π/2] → (sin θ, 0, cos θ)   (ZX–plane)\n\nLet C = C₁ ∪ C₂ ∪ C₃.  C is a closed, piece-wise C¹ loop lying in ℝ³\\{0}; it meets each coordinate plane exactly once and avoids the coordinate axes because every component is non–negative and at least one of them is strictly positive along each segment.\n\nBecause of (1),\n  ∮_C F·dr = 0.                                              (2)\n\nOn each quarter-arc the integrand equals the prescribed trace (condition (ii)); we can therefore compute the three contributions explicitly.\n\n3.  Integral over C₁ (XY–plane)  \nParameterisation: r(θ) = (cos θ, sin θ, 0), r′(θ) = (−sin θ, cos θ, 0).\n\nG₁(r(θ)) = ( −cos²θ sin θ , cos θ sin²θ , 0 )          (since (cos²θ+sin²θ)²=1).\n\nDot product:  \nG₁·r′ = (−cos²θ sin θ)(−sin θ) + (cos θ sin²θ)(cos θ)  \n    = 2 cos²θ sin²θ.\n\nHence  \nI₁ := ∫_{C₁} F·dr = ∫₀^{π/2} 2 cos²θ sin²θ dθ  \n   = ∫₀^{π/2} (1/2) sin²(2θ) dθ  \n   = (1/4)∫₀^{π} sin²u du  (u = 2θ)  \n   = (1/4)·(π/2) = π/8.                               (3)\n\n4.  Integral over C₂ (YZ–plane)  \nParameterisation: r(θ) = (0, cos θ, sin θ), r′(θ) = (0, −sin θ, cos θ).\n\nG₂(r(θ)) = ( 0, −cos²θ sin θ , cos θ sin²θ ).\n\nDot product: G₂·r′ = (−cos²θ sin θ)(−sin θ) + (cos θ sin²θ)(cos θ)  \n       = 2 cos²θ sin²θ.\n\nTherefore  \nI₂ := ∫_{C₂} F·dr = π/8.                               (4)\n\n5.  Integral over C₃ (ZX–plane)  \nParameterisation: r(θ) = (sin θ, 0, cos θ), r′(θ) = (cos θ, 0, −sin θ).\n\nG₃(r(θ)) = ( −cos θ sin²θ , 0 , sin θ cos²θ ).\n\nDot product: G₃·r′ = (−cos θ sin²θ)(cos θ) + (sin θ cos²θ)(−sin θ)  \n       = −2 cos²θ sin²θ.\n\nThus  \nI₃ := ∫_{C₃} F·dr = −π/8.                              (5)\n\n6.  Total circulation around C  \nBecause F coincides with the three traces on the corresponding segments,\n\n∮_C F·dr = I₁ + I₂ + I₃  \n     = π/8 + π/8 − π/8 = π/8.                       (6)\n\n7.  Contradiction  \nEquations (2) and (6) are incompatible: (2) asserts the circulation is 0, whereas (6) shows it equals π/8.  This contradiction proves that a C¹ vector field satisfying (i)–(iii) cannot exist.\n\n8.  Remarks  \n•  Compatibility of the data along the coordinate axes (all three traces are identically zero there) removes the elementary “two different values at one point’’ obstruction, forcing one to use a genuine Stokes/circulation argument.  \n•  The decay assumption (iii) is satisfied by each trace (|Gᵢ(r)| ≲ |r|⁻¹) and rules out pathological compensating fields at infinity; it does not alter the contradiction obtained above.  \n•  The loop C spans a surface that avoids the origin, so Stokes’ theorem may legitimately be applied in both directions of the argument.\n\nHence no C¹–vector field F with properties (i)–(iii) exists.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.699466",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimensional constraints: the unknown field must now satisfy three independent boundary conditions—one on each coordinate plane—rather than on a single plane.  \n• Multiple interacting concepts: the argument must keep track of how these separate restrictions interact on a composite spatial path occupying all three planes.  \n• More calculations: three separate (and unequal) rational integrals are required; each demands a change of variables and careful trigonometric/algebraic manipulation.  \n• Deeper theory: understanding why a single path integral must split into pieces but still cancel necessitates a clear grasp of Stokes’ theorem on a piecewise-smooth surface, homology of punctured space, and how line-integral invariants behave under piecewise definitions.  \n• Logical subtlety: none of the three planar data sets alone blocks the existence of F (each can be realised by a standard Aharonov–Bohm potential); the impossibility emerges only when all three are imposed simultaneously, so the solver must design a path that samples every restriction at once.\n\nAll of this represents a significant escalation in technical detail, length of argument, and conceptual depth compared with the original single-plane variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let the following three C¹–vector fields be prescribed on the punctured coordinate planes of ℝ³:\n\nG₁(x,y) = ( −x²y /(x² + y²)² ,  x y² /(x² + y²)² , 0 )    for (x,y) ≠ (0,0) and z = 0  \nG₂(y,z) = ( 0 , −y²z /(y² + z²)² ,  y z² /(y² + z²)² )    for (y,z) ≠ (0,0) and x = 0  \nG₃(z,x) = ( −z x² /(z² + x²)² , 0 ,  x z² /(z² + x²)² )    for (z,x) ≠ (0,0) and y = 0.\n\nObserve that each field vanishes when one of its two variables is zero, so on every coordinate axis  \n (0,y,0), (x,0,0), (0,0,z)  \nall three prescriptions give the common value (0,0,0); the data are therefore mutually compatible along the pairwise intersections of the planes.\n\nQuestion.  Does there exist a C¹–vector field  \n  F : ℝ³ \\ {(0,0,0)} → ℝ³  (F = (M,N,P))  \nsatisfying simultaneously\n\n(i) curl F ≡ 0 on ℝ³ \\ {(0,0,0)};  \n\n(ii) F coincides with the given traces on the three coordinate planes, i.e.  \n   F(x,y,0)=G₁(x,y), F(0,y,z)=G₂(y,z), F(x,0,z)=G₃(z,x);  \n\n(iii) |F(r)| = O(|r|⁻¹) as |r| → ∞ ?\n\nEither construct such a field or prove that none can exist.",
      "solution": "We show that conditions (i)–(iii) are incompatible; no such vector field F exists.\n\n1.  Topological background  \nThe domain ℝ³\\{0} is simply connected.  Hence any C¹–vector field whose curl is identically zero there is conservative: for every piece-wise smooth closed curve C contained in ℝ³\\{0},\n  ∮_C F·dr = 0.                                              (1)\n\n2.  A closed curve touching all three planes  \nJoin three quarter-circles of radius 1, one in each coordinate plane, head-to-tail:\n\nC₁ : θ∈[0,π/2] → (cos θ, sin θ, 0)   (XY–plane)  \nC₂ : θ∈[0,π/2] → (0, cos θ, sin θ)   (YZ–plane)  \nC₃ : θ∈[0,π/2] → (sin θ, 0, cos θ)   (ZX–plane)\n\nLet C = C₁ ∪ C₂ ∪ C₃.  C is a closed, piece-wise C¹ loop lying in ℝ³\\{0}; it meets each coordinate plane exactly once and avoids the coordinate axes because every component is non–negative and at least one of them is strictly positive along each segment.\n\nBecause of (1),\n  ∮_C F·dr = 0.                                              (2)\n\nOn each quarter-arc the integrand equals the prescribed trace (condition (ii)); we can therefore compute the three contributions explicitly.\n\n3.  Integral over C₁ (XY–plane)  \nParameterisation: r(θ) = (cos θ, sin θ, 0), r′(θ) = (−sin θ, cos θ, 0).\n\nG₁(r(θ)) = ( −cos²θ sin θ , cos θ sin²θ , 0 )          (since (cos²θ+sin²θ)²=1).\n\nDot product:  \nG₁·r′ = (−cos²θ sin θ)(−sin θ) + (cos θ sin²θ)(cos θ)  \n    = 2 cos²θ sin²θ.\n\nHence  \nI₁ := ∫_{C₁} F·dr = ∫₀^{π/2} 2 cos²θ sin²θ dθ  \n   = ∫₀^{π/2} (1/2) sin²(2θ) dθ  \n   = (1/4)∫₀^{π} sin²u du  (u = 2θ)  \n   = (1/4)·(π/2) = π/8.                               (3)\n\n4.  Integral over C₂ (YZ–plane)  \nParameterisation: r(θ) = (0, cos θ, sin θ), r′(θ) = (0, −sin θ, cos θ).\n\nG₂(r(θ)) = ( 0, −cos²θ sin θ , cos θ sin²θ ).\n\nDot product: G₂·r′ = (−cos²θ sin θ)(−sin θ) + (cos θ sin²θ)(cos θ)  \n       = 2 cos²θ sin²θ.\n\nTherefore  \nI₂ := ∫_{C₂} F·dr = π/8.                               (4)\n\n5.  Integral over C₃ (ZX–plane)  \nParameterisation: r(θ) = (sin θ, 0, cos θ), r′(θ) = (cos θ, 0, −sin θ).\n\nG₃(r(θ)) = ( −cos θ sin²θ , 0 , sin θ cos²θ ).\n\nDot product: G₃·r′ = (−cos θ sin²θ)(cos θ) + (sin θ cos²θ)(−sin θ)  \n       = −2 cos²θ sin²θ.\n\nThus  \nI₃ := ∫_{C₃} F·dr = −π/8.                              (5)\n\n6.  Total circulation around C  \nBecause F coincides with the three traces on the corresponding segments,\n\n∮_C F·dr = I₁ + I₂ + I₃  \n     = π/8 + π/8 − π/8 = π/8.                       (6)\n\n7.  Contradiction  \nEquations (2) and (6) are incompatible: (2) asserts the circulation is 0, whereas (6) shows it equals π/8.  This contradiction proves that a C¹ vector field satisfying (i)–(iii) cannot exist.\n\n8.  Remarks  \n•  Compatibility of the data along the coordinate axes (all three traces are identically zero there) removes the elementary “two different values at one point’’ obstruction, forcing one to use a genuine Stokes/circulation argument.  \n•  The decay assumption (iii) is satisfied by each trace (|Gᵢ(r)| ≲ |r|⁻¹) and rules out pathological compensating fields at infinity; it does not alter the contradiction obtained above.  \n•  The loop C spans a surface that avoids the origin, so Stokes’ theorem may legitimately be applied in both directions of the argument.\n\nHence no C¹–vector field F with properties (i)–(iii) exists.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.546545",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimensional constraints: the unknown field must now satisfy three independent boundary conditions—one on each coordinate plane—rather than on a single plane.  \n• Multiple interacting concepts: the argument must keep track of how these separate restrictions interact on a composite spatial path occupying all three planes.  \n• More calculations: three separate (and unequal) rational integrals are required; each demands a change of variables and careful trigonometric/algebraic manipulation.  \n• Deeper theory: understanding why a single path integral must split into pieces but still cancel necessitates a clear grasp of Stokes’ theorem on a piecewise-smooth surface, homology of punctured space, and how line-integral invariants behave under piecewise definitions.  \n• Logical subtlety: none of the three planar data sets alone blocks the existence of F (each can be realised by a standard Aharonov–Bohm potential); the impossibility emerges only when all three are imposed simultaneously, so the solver must design a path that samples every restriction at once.\n\nAll of this represents a significant escalation in technical detail, length of argument, and conceptual depth compared with the original single-plane variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}