{
  "index": "1938-B-2",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "9. Find all the solutions of the equation\n\\[\ny y^{\\prime \\prime}-2\\left(y^{\\prime}\\right)^{2}=0\n\\]\nwhich pass through the point \\( x=1, y=1 \\).",
  "solution": "First Solution. \\( 1 / y^{3} \\) is an integrating factor since\n\\[\n\\frac{d}{d x}\\left(\\frac{y^{\\prime}}{y^{2}}\\right)=\\frac{y y^{\\prime \\prime}-2\\left(y^{\\prime}\\right)^{2}}{y^{3}}=0 .\n\\]\n\nTherefore \\( y^{\\prime} / y^{2}=C \\) and \\( -1 / y=C x+D \\) for appropriate constants \\( C \\) and \\( D \\). In order that the solution pass through ( 1,1 ), we require that \\( C+D=-1 \\). Hence\n\\[\ny=\\frac{1}{1+C(1-x)} .\n\\]\n\nConversely, any function of this form satisfies the equation and the initial conditions. If \\( C=0 \\), this is a constant function and its domain may be taken as \\( (-\\infty,+\\infty) \\). If \\( C \\neq 0 \\), the right member of (1) becomes infinite for \\( x=(1+C) / C \\), so the domain of (1) must be restricted to\n\\[\n\\begin{array}{l}\n\\left(-\\infty, \\frac{1+C}{C}\\right) \\text { if } C>0 \\\\\n\\left(\\frac{1+C}{C}, \\infty\\right) \\text { if } C<0\n\\end{array}\n\\]\n\nSecond Solution. Since \\( x \\) does not appear explicitly in the given differential equation, the substitution \\( v=y^{\\prime}, y^{\\prime \\prime}=v d v / d y \\) leads to a firstorder equation\n\\[\nv\\left(y \\frac{d v}{d y}-2 v\\right)=0\n\\]\n\nHence either \\( v=0 \\) and \\( y \\) is constant, or\n\\[\ny \\frac{d v}{d y}-2 v=0\n\\]\n\nIn this case the variables are separable and we obtain\n\\[\nv=C y^{2},\n\\]\nthat is,\n\\[\ny^{\\prime}=C y^{2},\n\\]\nwhich is again separable. We get\n\\[\n-1 / y=C x+D\n\\]\nand the solution proceeds as before.\nNote that the special solution \\( y= \\) constant is subsumed in this general case.\nIf the original equation is solved for the highest derivative\n\\[\ny^{\\prime \\prime}=\\frac{2\\left(y^{\\prime}\\right)^{2}}{y}\n\\]\nit becomes clear that this differential equation is regular in the upper and lower half-planes but may be singular along the \\( x \\)-axis. It is obvious that all solutions (1) are maximal solutions, since they cannot be extended continuously to any larger connected domain. Since none of these solutions passes through any point where the equation might be singular, we are assured that we have found all of the solutions passing through \\( (1,1) \\).",
  "vars": [
    "x",
    "y",
    "v"
  ],
  "params": [
    "C",
    "D"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "v": "velocity",
        "C": "scalefactor",
        "D": "shiftparam"
      },
      "question": "9. Find all the solutions of the equation\n\\[\nordinate \\, ordinate^{\\prime \\prime}-2\\left(ordinate^{\\prime}\\right)^{2}=0\n\\]\nwhich pass through the point \\( abscissa=1, ordinate=1 \\).",
      "solution": "First Solution. \\( 1 / ordinate^{3} \\) is an integrating factor since\n\\[\n\\frac{d}{d abscissa}\\left(\\frac{ordinate^{\\prime}}{ordinate^{2}}\\right)=\\frac{ordinate \\, ordinate^{\\prime \\prime}-2\\left(ordinate^{\\prime}\\right)^{2}}{ordinate^{3}}=0 .\n\\]\n\nTherefore \\( ordinate^{\\prime} / ordinate^{2}=scalefactor \\) and \\( -1 / ordinate=scalefactor \\, abscissa+shiftparam \\) for appropriate constants \\( scalefactor \\) and \\( shiftparam \\). In order that the solution pass through ( 1,1 ), we require that \\( scalefactor+shiftparam=-1 \\). Hence\n\\[\nordinate=\\frac{1}{1+scalefactor(1-abscissa)} .\n\\]\n\nConversely, any function of this form satisfies the equation and the initial conditions. If \\( scalefactor=0 \\), this is a constant function and its domain may be taken as \\( (-\\infty,+\\infty) \\). If \\( scalefactor \\neq 0 \\), the right member of (1) becomes infinite for \\( abscissa=(1+scalefactor) / scalefactor \\), so the domain of (1) must be restricted to\n\\[\n\\begin{array}{l}\n\\left(-\\infty, \\frac{1+scalefactor}{scalefactor}\\right) \\text { if } scalefactor>0 \\\\\n\\left(\\frac{1+scalefactor}{scalefactor}, \\infty\\right) \\text { if } scalefactor<0\n\\end{array}\n\\]\n\nSecond Solution. Since \\( abscissa \\) does not appear explicitly in the given differential equation, the substitution \\( velocity=ordinate^{\\prime},\\; ordinate^{\\prime \\prime}=velocity \\, d velocity / d ordinate \\) leads to a first-order equation\n\\[\nvelocity\\left(ordinate \\frac{d velocity}{d ordinate}-2 \\, velocity\\right)=0\n\\]\n\nHence either \\( velocity=0 \\) and \\( ordinate \\) is constant, or\n\\[\nordinate \\frac{d velocity}{d ordinate}-2 \\, velocity=0\n\\]\n\nIn this case the variables are separable and we obtain\n\\[\nvelocity=scalefactor \\, ordinate^{2},\n\\]\nthat is,\n\\[\nordinate^{\\prime}=scalefactor \\, ordinate^{2},\n\\]\nwhich is again separable. We get\n\\[\n-1 / ordinate=scalefactor \\, abscissa+shiftparam\n\\]\nand the solution proceeds as before.\nNote that the special solution \\( ordinate= \\) constant is subsumed in this general case.\nIf the original equation is solved for the highest derivative\n\\[\nordinate^{\\prime \\prime}=\\frac{2\\left(ordinate^{\\prime}\\right)^{2}}{ordinate}\n\\]\nit becomes clear that this differential equation is regular in the upper and lower half-planes but may be singular along the \\( abscissa \\)-axis. It is obvious that all solutions (1) are maximal solutions, since they cannot be extended continuously to any larger connected domain. Since none of these solutions passes through any point where the equation might be singular, we are assured that we have found all of the solutions passing through \\( (1,1) \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pineapple",
        "y": "screwdriver",
        "v": "tangerine",
        "C": "lawnmower",
        "D": "paperclips"
      },
      "question": "9. Find all the solutions of the equation\n\\[\nscrewdriver screwdriver^{\\prime \\prime}-2\\left(screwdriver^{\\prime}\\right)^{2}=0\n\\]\nwhich pass through the point \\( pineapple=1, screwdriver=1 \\).",
      "solution": "First Solution. \\( 1 / screwdriver^{3} \\) is an integrating factor since\n\\[\n\\frac{d}{d pineapple}\\left(\\frac{screwdriver^{\\prime}}{screwdriver^{2}}\\right)=\\frac{screwdriver screwdriver^{\\prime \\prime}-2\\left(screwdriver^{\\prime}\\right)^{2}}{screwdriver^{3}}=0 .\n\\]\n\nTherefore \\( screwdriver^{\\prime} / screwdriver^{2}=lawnmower \\) and \\( -1 / screwdriver=lawnmower pineapple+paperclips \\) for appropriate constants lawnmower and paperclips. In order that the solution pass through ( 1,1 ), we require that \\( lawnmower+paperclips=-1 \\). Hence\n\\[\nscrewdriver=\\frac{1}{1+lawnmower(1-pineapple)} .\n\\]\n\nConversely, any function of this form satisfies the equation and the initial conditions. If \\( lawnmower=0 \\), this is a constant function and its domain may be taken as \\( (-\\infty,+\\infty) \\). If \\( lawnmower \\neq 0 \\), the right member of (1) becomes infinite for \\( pineapple=(1+lawnmower) / lawnmower \\), so the domain of (1) must be restricted to\n\\[\n\\begin{array}{l}\n\\left(-\\infty, \\frac{1+lawnmower}{lawnmower}\\right) \\text { if } lawnmower>0 \\\\\n\\left(\\frac{1+lawnmower}{lawnmower}, \\infty\\right) \\text { if } lawnmower<0\n\\end{array}\n\\]\n\nSecond Solution. Since \\( pineapple \\) does not appear explicitly in the given differential equation, the substitution \\( tangerine=screwdriver^{\\prime}, screwdriver^{\\prime \\prime}=tangerine d tangerine / d screwdriver \\) leads to a firstorder equation\n\\[\ntangerine\\left(screwdriver \\frac{d tangerine}{d screwdriver}-2 tangerine\\right)=0\n\\]\n\nHence either \\( tangerine=0 \\) and \\( screwdriver \\) is constant, or\n\\[\nscrewdriver \\frac{d tangerine}{d screwdriver}-2 tangerine=0\n\\]\n\nIn this case the variables are separable and we obtain\n\\[\ntangerine=lawnmower screwdriver^{2},\n\\]\nthat is,\n\\[\nscrewdriver^{\\prime}=lawnmower screwdriver^{2},\n\\]\nwhich is again separable. We get\n\\[\n-1 / screwdriver=lawnmower pineapple+paperclips\n\\]\nand the solution proceeds as before.\nNote that the special solution \\( screwdriver= \\) constant is subsumed in this general case.\nIf the original equation is solved for the highest derivative\n\\[\nscrewdriver^{\\prime \\prime}=\\frac{2\\left(screwdriver^{\\prime}\\right)^{2}}{screwdriver}\n\\]\nit becomes clear that this differential equation is regular in the upper and lower half-planes but may be singular along the \\( pineapple \\)-axis. It is obvious that all solutions (1) are maximal solutions, since they cannot be extended continuously to any larger connected domain. Since none of these solutions passes through any point where the equation might be singular, we are assured that we have found all of the solutions passing through \\( (1,1) \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "immutabledatum",
        "y": "independentval",
        "v": "stillnessfactor",
        "C": "varyingentity",
        "D": "mutablefactor"
      },
      "question": "9. Find all the solutions of the equation\n\\[\nindependentval independentval^{\\prime \\prime}-2\\left(independentval^{\\prime}\\right)^{2}=0\n\\]\nwhich pass through the point \\( immutabledatum=1, independentval=1 \\).",
      "solution": "First Solution. \\( 1 / independentval^{3} \\) is an integrating factor since\n\\[\n\\frac{d}{d immutabledatum}\\left(\\frac{independentval^{\\prime}}{independentval^{2}}\\right)=\\frac{independentval independentval^{\\prime \\prime}-2\\left(independentval^{\\prime}\\right)^{2}}{independentval^{3}}=0 .\n\\]\n\nTherefore \\( independentval^{\\prime} / independentval^{2}=varyingentity \\) and \\( -1 / independentval=varyingentity immutabledatum+mutablefactor \\) for appropriate constants \\( varyingentity \\) and \\( mutablefactor \\). In order that the solution pass through ( 1,1 ), we require that \\( varyingentity+mutablefactor=-1 \\). Hence\n\\[\nindependentval=\\frac{1}{1+varyingentity(1-immutabledatum)} .\n\\]\n\nConversely, any function of this form satisfies the equation and the initial conditions. If \\( varyingentity=0 \\), this is a constant function and its domain may be taken as \\( (-\\infty,+\\infty) \\). If \\( varyingentity \\neq 0 \\), the right member of (1) becomes infinite for \\( immutabledatum=(1+varyingentity) / varyingentity \\), so the domain of (1) must be restricted to\n\\[\n\\begin{array}{l}\n\\left(-\\infty, \\frac{1+varyingentity}{varyingentity}\\right) \\text { if } varyingentity>0 \\\\\n\\left(\\frac{1+varyingentity}{varyingentity}, \\infty\\right) \\text { if } varyingentity<0\n\\end{array}\n\\]\n\nSecond Solution. Since \\( immutabledatum \\) does not appear explicitly in the given differential equation, the substitution \\( stillnessfactor=independentval^{\\prime}, independentval^{\\prime \\prime}=stillnessfactor d stillnessfactor / d independentval \\) leads to a firstorder equation\n\\[\nstillnessfactor\\left(independentval \\frac{d stillnessfactor}{d independentval}-2 stillnessfactor\\right)=0\n\\]\n\nHence either \\( stillnessfactor=0 \\) and \\( independentval \\) is constant, or\n\\[\nindependentval \\frac{d stillnessfactor}{d independentval}-2 stillnessfactor=0\n\\]\n\nIn this case the variables are separable and we obtain\n\\[\nstillnessfactor=varyingentity independentval^{2},\n\\]\nthat is,\n\\[\nindependentval^{\\prime}=varyingentity independentval^{2},\n\\]\nwhich is again separable. We get\n\\[\n-1 / independentval=varyingentity immutabledatum+mutablefactor\n\\]\nand the solution proceeds as before.\nNote that the special solution \\( independentval= \\) constant is subsumed in this general case.\nIf the original equation is solved for the highest derivative\n\\[\nindependentval^{\\prime \\prime}=\\frac{2\\left(independentval^{\\prime}\\right)^{2}}{independentval}\n\\]\nit becomes clear that this differential equation is regular in the upper and lower half-planes but may be singular along the \\( immutabledatum \\)-axis. It is obvious that all solutions (1) are maximal solutions, since they cannot be extended continuously to any larger connected domain. Since none of these solutions passes through any point where the equation might be singular, we are assured that we have found all of the solutions passing through \\( (1,1) \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "plomtrqz",
        "v": "snaglyru",
        "C": "frabdexi",
        "D": "dromsple"
      },
      "question": "Find all the solutions of the equation\n\\[\nplomtrqz \\, plomtrqz^{\\prime \\prime}-2\\left(plomtrqz^{\\prime}\\right)^{2}=0\n\\]\nwhich pass through the point \\( qzxwvtnp=1, plomtrqz=1 \\).",
      "solution": "First Solution. \\( 1 / plomtrqz^{3} \\) is an integrating factor since\n\\[\n\\frac{d}{d qzxwvtnp}\\left(\\frac{plomtrqz^{\\prime}}{plomtrqz^{2}}\\right)=\\frac{plomtrqz \\, plomtrqz^{\\prime \\prime}-2\\left(plomtrqz^{\\prime}\\right)^{2}}{plomtrqz^{3}}=0 .\n\\]\n\nTherefore \\( plomtrqz^{\\prime} / plomtrqz^{2}=frabdexi \\) and \\( -1 / plomtrqz=frabdexi \\, qzxwvtnp+dromsple \\) for appropriate constants \\( frabdexi \\) and \\( dromsple \\). In order that the solution pass through ( 1,1 ), we require that \\( frabdexi+dromsple=-1 \\). Hence\n\\[\nplomtrqz=\\frac{1}{1+frabdexi(1-qzxwvtnp)} .\n\\]\n\nConversely, any function of this form satisfies the equation and the initial conditions. If \\( frabdexi=0 \\), this is a constant function and its domain may be taken as \\( (-\\infty,+\\infty) \\). If \\( frabdexi \\neq 0 \\), the right member of (1) becomes infinite for \\( qzxwvtnp=(1+frabdexi) / frabdexi \\), so the domain of (1) must be restricted to\n\\[\n\\begin{array}{l}\n\\left(-\\infty, \\frac{1+frabdexi}{frabdexi}\\right) \\text { if } frabdexi>0 \\\\\n\\left(\\frac{1+frabdexi}{frabdexi}, \\infty\\right) \\text { if } frabdexi<0\n\\end{array}\n\\]\n\nSecond Solution. Since \\( qzxwvtnp \\) does not appear explicitly in the given differential equation, the substitution \\( snaglyru=plomtrqz^{\\prime},\\, plomtrqz^{\\prime \\prime}=snaglyru \\, d snaglyru / d plomtrqz \\) leads to a firstorder equation\n\\[\nsnaglyru\\left(plomtrqz \\frac{d snaglyru}{d plomtrqz}-2 snaglyru\\right)=0\n\\]\n\nHence either \\( snaglyru=0 \\) and \\( plomtrqz \\) is constant, or\n\\[\nplomtrqz \\frac{d snaglyru}{d plomtrqz}-2 snaglyru=0\n\\]\n\nIn this case the variables are separable and we obtain\n\\[\nsnaglyru=frabdexi \\, plomtrqz^{2},\n\\]\nthat is,\n\\[\nplomtrqz^{\\prime}=frabdexi \\, plomtrqz^{2},\n\\]\nwhich is again separable. We get\n\\[\n-1 / plomtrqz=frabdexi \\, qzxwvtnp+dromsple\n\\]\nand the solution proceeds as before.\nNote that the special solution \\( plomtrqz= \\) constant is subsumed in this general case.\nIf the original equation is solved for the highest derivative\n\\[\nplomtrqz^{\\prime \\prime}=\\frac{2\\left(plomtrqz^{\\prime}\\right)^{2}}{plomtrqz}\n\\]\nit becomes clear that this differential equation is regular in the upper and lower half-planes but may be singular along the \\( qzxwvtnp \\)-axis. It is obvious that all solutions (1) are maximal solutions, since they cannot be extended continuously to any larger connected domain. Since none of these solutions passes through any point where the equation might be singular, we are assured that we have found all of the solutions passing through \\( (1,1) \\)."
    },
    "kernel_variant": {
      "question": "Let Ω be a connected open neighbourhood of the origin in ℝ³ and let  \nu : Ω → (0,∞) be a twice–continuously–differentiable function that fulfils the\ncoupled nonlinear system  \n\n(E1) u Δu − 4 |∇u|² = 0,  \n\n(E2) |∇u|² = κ u⁸  for some (a-priori unknown) positive constant κ,  \n\nwhere Δ and ∇ denote, respectively, the three–dimensional Laplacian and gradient.\nAssume the Cauchy data  \n\nu(0,0,0)=2,  ∇u(0,0,0)= (a,0,0) with a ≠ 0.  \n\n(a) Determine explicitly every function u that satisfies (E1)–(E2) together with\nthe above initial data, and express κ in terms of a.  \n\n(b) For each solution, describe the maximal open set Ω_max on which it can be\nextended smoothly.\n\n\n───────────────────────────────────────────────────────────────",
      "solution": "Throughout write x=(x,y,z) and e₁=(1,0,0).\n\nStep 1.  A linearising substitution for (E1).  \nSet  \n\n v := u⁻³.  \n\nA direct computation gives  \n\n ∇v = −3u⁻⁴∇u,  Δv = 12u⁻⁵|∇u|² − 3u⁻⁴Δu.\n\nHence\n\n u⁵Δv = 12|∇u|² − 3uΔu\n    = 12|∇u|² − 3·4|∇u|² = 0,\n\nand because u>0 we conclude\n\n  Δv = 0;  v is harmonic on Ω.           (1)\n\nStep 2.  Consequence of the additional constraint (E2).  \nCondition (E2) together with ∇v = −3u⁻⁴∇u implies  \n\n |∇v|² = 9u⁻⁸|∇u|² = 9κ =: L²,        (2)\n\ni.e. the gradient of v has constant Euclidean length L=3√κ independent of the\npoint x∈Ω.\n\nStep 3.  Classification of harmonic functions with constant gradient length.  \nFor a C² scalar field w the Bochner formula reads\n\n  Δ|∇w|² = 2 |D²w|²,        (3)\n\nwhere D²w is the Hessian of w and |D²w|² the sum of the squares of its entries.\nApplying (3) to the harmonic function v and using (2), the left–hand side of\n(3) vanishes; therefore\n\n  |D²v|² = 0 ⇒ D²v ≡ 0 on Ω.  \n\nHence v is an affine harmonic polynomial of degree 1, i.e. there exist a\nconstant vector p∈ℝ³ and a constant d∈ℝ such that  \n\n  v(x)=p·x+d  for every x∈Ω.        (4)\n\nBecause |p| = |∇v| = L, relation (2) supplies the additional information  \n\n  |p| = 3√κ.           (5)\n\nStep 4.  Imposing the Cauchy data.  \nSince u(0)=2, we have v(0)=1/2³=1/8, so d=1/8.  \nMoreover,\n\n p = ∇v(0) = −3u(0)⁻⁴∇u(0) = −3·2⁻⁴(a,0,0) = −(3a/16) e₁.  (6)\n\nThus\n\n  v(x,y,z)= −(3a/16)x + 1/8.        (7)\n\nTaking norms in (6) and comparing with (5) yields the numerical value of κ:\n\n |p| = 3|a|/16 = 3√κ ⇒ κ = a²/256.    (8)\n\nStep 5.  Recovering u.  \nInverting v=u⁻³ gives\n\n u(x,y,z) = [1/8 − (3a/16)x]^{−1/3}\n     = 2·[1 − (3a/2)x]^{−1/3}.     (9)\n\nStep 6.  Maximal domain of smoothness.  \nThe bracket in (9) must stay positive.  Writing\n\n x₀ := 2/(3a),\n\nwe obtain  \n\n Ω_max = { (x,y,z) ∈ ℝ³  |  x < x₀ } if a>0,  \n Ω_max = { (x,y,z) ∈ ℝ³  |  x > x₀ } if a<0.  (10)\n\nAlong the plane x=x₀ the denominator vanishes and u blows up to +∞, so the\nhalf-space (10) is the maximal connected open set on which u can be extended\nsmoothly.\n\nStep 7.  Verification and uniqueness.  \nA straightforward differentiation of (9) shows that\n\n |∇u|² = (a²/256) u⁸ and uΔu − 4|∇u|² = 0\n\neverywhere on Ω_max, so (E1)–(E2) and the initial data are satisfied.\nConversely, any solution of (E1)–(E2) gives rise, via v=u⁻³, to a harmonic\nfunction v with constant |∇v|; Step 3 forces such a v to be affine, and\nStep 4 then fixes it uniquely.  Hence (9) is the only possible solution.\n\n\n───────────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.339544",
        "was_fixed": false,
        "difficulty_analysis": "1. Dimension jump:  the original one–dimensional ODE is replaced by a\nthree–dimensional system (a nonlinear PDE plus a first–order vector\nconstraint).  \n2. Two coupled equations:  solving (E1) alone still leaves huge freedom;\n(E2) couples all first derivatives and forces an additional potential\nstructure.  \n3. Transformation to harmonic functions in higher dimensions requires\nfacility with vector calculus identities, the maximum principle and\nLiouville’s theorem, none of which appear in the original problem.  \n4. The solution demands identification of an affine harmonic function via\nexploiting both PDE and curl‐condition, an argument that uses\ndivergence–curl tools rather than elementary separation of variables.  \n5. The maximal domain must be deduced from the blow–up set of an affine\ndenominator in ℝ³, not merely from a pole on the real line; the answer\ndepends on the sign of the given constant a.  \n\nAll these features make the enhanced variant substantially harder than both\nthe original and the kernel version: the solver must blend techniques from\nPDE theory, harmonic analysis, vector calculus, and the theory of analytic\ninitial value problems, going well beyond the separable‐ODE tricks that\nsolve the original exercise."
      }
    },
    "original_kernel_variant": {
      "question": "Let Ω be a connected open neighbourhood of the origin in ℝ³ and let  \nu : Ω → (0,∞) be a twice–continuously–differentiable function that fulfils the\ncoupled nonlinear system  \n\n(E1) u Δu − 4 |∇u|² = 0,  \n\n(E2) |∇u|² = κ u⁸  for some (a-priori unknown) positive constant κ,  \n\nwhere Δ and ∇ denote, respectively, the three–dimensional Laplacian and gradient.\nAssume the Cauchy data  \n\nu(0,0,0)=2,  ∇u(0,0,0)= (a,0,0) with a ≠ 0.  \n\n(a) Determine explicitly every function u that satisfies (E1)–(E2) together with\nthe above initial data, and express κ in terms of a.  \n\n(b) For each solution, describe the maximal open set Ω_max on which it can be\nextended smoothly.\n\n\n───────────────────────────────────────────────────────────────",
      "solution": "Throughout write x=(x,y,z) and e₁=(1,0,0).\n\nStep 1.  A linearising substitution for (E1).  \nSet  \n\n v := u⁻³.  \n\nA direct computation gives  \n\n ∇v = −3u⁻⁴∇u,  Δv = 12u⁻⁵|∇u|² − 3u⁻⁴Δu.\n\nHence\n\n u⁵Δv = 12|∇u|² − 3uΔu\n    = 12|∇u|² − 3·4|∇u|² = 0,\n\nand because u>0 we conclude\n\n  Δv = 0;  v is harmonic on Ω.           (1)\n\nStep 2.  Consequence of the additional constraint (E2).  \nCondition (E2) together with ∇v = −3u⁻⁴∇u implies  \n\n |∇v|² = 9u⁻⁸|∇u|² = 9κ =: L²,        (2)\n\ni.e. the gradient of v has constant Euclidean length L=3√κ independent of the\npoint x∈Ω.\n\nStep 3.  Classification of harmonic functions with constant gradient length.  \nFor a C² scalar field w the Bochner formula reads\n\n  Δ|∇w|² = 2 |D²w|²,        (3)\n\nwhere D²w is the Hessian of w and |D²w|² the sum of the squares of its entries.\nApplying (3) to the harmonic function v and using (2), the left–hand side of\n(3) vanishes; therefore\n\n  |D²v|² = 0 ⇒ D²v ≡ 0 on Ω.  \n\nHence v is an affine harmonic polynomial of degree 1, i.e. there exist a\nconstant vector p∈ℝ³ and a constant d∈ℝ such that  \n\n  v(x)=p·x+d  for every x∈Ω.        (4)\n\nBecause |p| = |∇v| = L, relation (2) supplies the additional information  \n\n  |p| = 3√κ.           (5)\n\nStep 4.  Imposing the Cauchy data.  \nSince u(0)=2, we have v(0)=1/2³=1/8, so d=1/8.  \nMoreover,\n\n p = ∇v(0) = −3u(0)⁻⁴∇u(0) = −3·2⁻⁴(a,0,0) = −(3a/16) e₁.  (6)\n\nThus\n\n  v(x,y,z)= −(3a/16)x + 1/8.        (7)\n\nTaking norms in (6) and comparing with (5) yields the numerical value of κ:\n\n |p| = 3|a|/16 = 3√κ ⇒ κ = a²/256.    (8)\n\nStep 5.  Recovering u.  \nInverting v=u⁻³ gives\n\n u(x,y,z) = [1/8 − (3a/16)x]^{−1/3}\n     = 2·[1 − (3a/2)x]^{−1/3}.     (9)\n\nStep 6.  Maximal domain of smoothness.  \nThe bracket in (9) must stay positive.  Writing\n\n x₀ := 2/(3a),\n\nwe obtain  \n\n Ω_max = { (x,y,z) ∈ ℝ³  |  x < x₀ } if a>0,  \n Ω_max = { (x,y,z) ∈ ℝ³  |  x > x₀ } if a<0.  (10)\n\nAlong the plane x=x₀ the denominator vanishes and u blows up to +∞, so the\nhalf-space (10) is the maximal connected open set on which u can be extended\nsmoothly.\n\nStep 7.  Verification and uniqueness.  \nA straightforward differentiation of (9) shows that\n\n |∇u|² = (a²/256) u⁸ and uΔu − 4|∇u|² = 0\n\neverywhere on Ω_max, so (E1)–(E2) and the initial data are satisfied.\nConversely, any solution of (E1)–(E2) gives rise, via v=u⁻³, to a harmonic\nfunction v with constant |∇v|; Step 3 forces such a v to be affine, and\nStep 4 then fixes it uniquely.  Hence (9) is the only possible solution.\n\n\n───────────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.299333",
        "was_fixed": false,
        "difficulty_analysis": "1. Dimension jump:  the original one–dimensional ODE is replaced by a\nthree–dimensional system (a nonlinear PDE plus a first–order vector\nconstraint).  \n2. Two coupled equations:  solving (E1) alone still leaves huge freedom;\n(E2) couples all first derivatives and forces an additional potential\nstructure.  \n3. Transformation to harmonic functions in higher dimensions requires\nfacility with vector calculus identities, the maximum principle and\nLiouville’s theorem, none of which appear in the original problem.  \n4. The solution demands identification of an affine harmonic function via\nexploiting both PDE and curl‐condition, an argument that uses\ndivergence–curl tools rather than elementary separation of variables.  \n5. The maximal domain must be deduced from the blow–up set of an affine\ndenominator in ℝ³, not merely from a pole on the real line; the answer\ndepends on the sign of the given constant a.  \n\nAll these features make the enhanced variant substantially harder than both\nthe original and the kernel version: the solver must blend techniques from\nPDE theory, harmonic analysis, vector calculus, and the theory of analytic\ninitial value problems, going well beyond the separable‐ODE tricks that\nsolve the original exercise."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}