{
  "index": "2019-B-4",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $\\mathcal{F}$ be the set of functions $f(x,y)$ that are twice continuously differentiable for $x \\geq 1$, $y \\geq 1$ and that satisfy the following two equations (where subscripts denote partial derivatives):\n\\begin{gather*}\nxf_x + yf_y = xy \\ln(xy), \\\\\nx^2 f_{xx} + y^2 f_{yy} = xy.\n\\end{gather*}\nFor each $f \\in \\mathcal{F}$, let\n\\[\nm(f) = \\min_{s \\geq 1} \\left(f(s+1,s+1) - f(s+1,s) - f(s,s+1) + f(s,s) \\right).\n\\]\nDetermine $m(f)$, and show that it is independent of the choice of $f$.",
  "solution": "\\noindent\n\\textbf{Solution 1.}\nWe compute that $m(f) = 2 \\ln 2 - \\frac{1}{2}$.\nLabel the given differential equations by (1) and (2). If we write, e.g., $x\\frac{\\partial}{\\partial x}(1)$ for the result of differentiating (1) by $x$ and multiplying the resulting equation by $x$, then the combination\n$x\\frac{\\partial}{\\partial x}(1)+y\\frac{\\partial}{\\partial y}(1)-(1)-(2)$ gives the equation\n$2xyf_{xy} = xy\\ln(xy)+xy$, whence $f_{xy} = \\frac{1}{2} (\\ln(x)+\\ln(y)+1)$.\n\nNow we observe that\n\\begin{align*}\n\\lefteqn{f(s+1,s+1)-f(s+1,s)-f(s,s+1)+f(s,s)} \\\\\n &= \\int_s^{s+1} \\int_s^{s+1} f_{xy}\\,dy\\,dx \\\\\n&= \\frac{1}{2} \\int_s^{s+1} \\int_s^{s+1} (\\ln(x)+\\ln(y)+1)\\,dy\\,dx \\\\\n&= \\frac{1}{2} + \\int_s^{s+1} \\ln(x)\\,dx.\n\\end{align*}\n\nSince $\\ln(x)$ is increasing, $\\int_s^{s+1} \\ln(x)\\,dx$ is an increasing function of $s$, and so it is minimized over $s \\in [1,\\infty)$ when $s=1$. We conclude that\n\\[\nm(f) = \\frac{1}{2} + \\int_1^2 \\ln(x)\\,dx = 2 \\ln 2-\\frac{1}{2}\n\\]\nindependent of $f$.\n\n\\noindent\n\\textbf{Remark.}\nThe phrasing of the question suggests that solvers were not expected to prove that $\\mathcal{F}$ is nonempty,\neven though this is necessary to make the definition of $m(f)$ logically meaningful. Existence will be explicitly established in the next solution.\n\n\\noindent\n\\textbf{Solution 2.}\nWe first verify that \n\\[\nf(x,y) = \\frac{1}{2}(xy\\ln(xy)-xy)\n\\]\nis an element of $\\mathcal{F}$, by computing that\n\\begin{gather*}\nxf_x = yf_y = \\frac{1}{2} xy \\ln(xy) \\\\\nx^2 f_{xx} = y^2 f_{yy} = xy.\n\\end{gather*}\n(See the following remark for motivation for this guess.)\n\nWe next show that the only elements of $\\mathcal{F}$ are $f+a \\ln(x/y) + b$ where $a,b$ are constants.\nSuppose that $f+g$ is a second element of $\\mathcal{F}$. As in the first solution, we deduce that $g_{xy} = 0$; this implies that\n$g(x,y) = u(x) + v(y)$ for some twice continuously differentiable functions $u$ and $v$. We also have\n$xg_x + yg_y = 0$, which now asserts that $xg_x = - yg_y$ is equal to some constant $a$. This yields that\n$g = a \\ln(x/y) + b$ as desired.\n\nWe next observe that \n\\[\ng(s+1,s+1)-g(s+1,s)-g(s,s+1)+g(s,s) = 0,\n\\]\nso $m(f) = m(f+g)$. It thus remains to compute $m(f)$. To do this, we verify that\n\\[\nf(s+1,s+1)-f(s+1,s)-f(s,s+1)+f(s,s)\n\\]\nis nondecreasing in $s$ by computing its derivative to be $\\ln (s+1) - \\ln(s)$\n(either directly or using the integral representation from the first solution).\nWe thus minimize by taking $s=1$ as in the first solution.\n\n\\noindent\n\\textbf{Remark.}\nOne way to make a correct guess for $f$ is to notice that the given equations are both symmetric in $x$ and $y$\nand posit that $f$ should also be symmetric. Any symmetric function of $x$ and $y$ can be written in terms of the variables $u = x+y$ and $v = xy$, so in principle we could translate the equations into those variables and solve. However, before trying this, we observe that $xy$ appears explicitly in the equations, so it is reasonable to make a first guess of the form $f(x,y) = h(xy)$. \nFor such a choice, we have\n\\[\nx f_x + y f_y = 2xy h' = xy \\ln(xy)\n\\]\nwhich forces us to set $h(t)  = \\frac{1}{2}(t \\ln(t) - t)$.",
  "vars": [
    "x",
    "y",
    "s",
    "t",
    "f",
    "g",
    "h",
    "u",
    "v",
    "m",
    "f_x",
    "f_y",
    "f_xx",
    "f_yy",
    "f_xy",
    "g_x",
    "g_y",
    "g_xy"
  ],
  "params": [
    "a",
    "b"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "s": "slider",
        "t": "parameter",
        "f": "funcmain",
        "g": "funcauxi",
        "h": "helperfn",
        "u": "auxvaru",
        "v": "auxvarv",
        "m": "minimum",
        "f_x": "partialx",
        "f_y": "partialy",
        "f_xx": "secondx",
        "f_yy": "secondy",
        "f_xy": "mixedxy",
        "g_x": "auxpartialx",
        "g_y": "auxpartialy",
        "g_xy": "auxmixedxy",
        "a": "constanta",
        "b": "constantb"
      },
      "question": "Let $\\mathcal{F}$ be the set of functions $\\funcmain(\\abscissa,\\ordinate)$ that are twice continuously differentiable for $\\abscissa \\geq 1$, $\\ordinate \\geq 1$ and that satisfy the following two equations (where subscripts denote partial derivatives):\n\\begin{gather*}\n\\abscissa\\partialx + \\ordinate\\partialy = \\abscissa \\ordinate \\ln(\\abscissa\\ordinate), \\\\\n\\abscissa^2 \\secondx + \\ordinate^2 \\secondy = \\abscissa \\ordinate.\n\\end{gather*}\nFor each $\\funcmain \\in \\mathcal{F}$, let\n\\[\n\\minimum(\\funcmain) = \\min_{\\slider \\geq 1} \\left(\\funcmain(\\slider+1,\\slider+1) - \\funcmain(\\slider+1,\\slider) - \\funcmain(\\slider,\\slider+1) + \\funcmain(\\slider,\\slider) \\right).\n\\]\nDetermine $\\minimum(\\funcmain)$, and show that it is independent of the choice of $\\funcmain$.",
      "solution": "\\noindent\n\\textbf{Solution 1.}\nWe compute that $\\minimum(\\funcmain) = 2 \\ln 2 - \\frac{1}{2}$.\nLabel the given differential equations by (1) and (2). If we write, e.g., $\\abscissa\\frac{\\partial}{\\partial \\abscissa}(1)$ for the result of differentiating (1) by $\\abscissa$ and multiplying the resulting equation by $\\abscissa$, then the combination\n$\\abscissa\\frac{\\partial}{\\partial \\abscissa}(1)+\\ordinate\\frac{\\partial}{\\partial \\ordinate}(1)-(1)-(2)$ gives the equation\n$2\\abscissa\\ordinate\\mixedxy = \\abscissa\\ordinate\\ln(\\abscissa\\ordinate)+\\abscissa\\ordinate$, whence $\\mixedxy = \\frac{1}{2} (\\ln(\\abscissa)+\\ln(\\ordinate)+1)$.\n\nNow we observe that\n\\begin{align*}\n\\lefteqn{\\funcmain(\\slider+1,\\slider+1)-\\funcmain(\\slider+1,\\slider)-\\funcmain(\\slider,\\slider+1)+\\funcmain(\\slider,\\slider)} \\\\\n &= \\int_{\\slider}^{\\slider+1} \\int_{\\slider}^{\\slider+1} \\mixedxy\\,d\\ordinate\\,d\\abscissa \\\\\n&= \\frac{1}{2} \\int_{\\slider}^{\\slider+1} \\int_{\\slider}^{\\slider+1} (\\ln(\\abscissa)+\\ln(\\ordinate)+1)\\,d\\ordinate\\,d\\abscissa \\\\\n&= \\frac{1}{2} + \\int_{\\slider}^{\\slider+1} \\ln(\\abscissa)\\,d\\abscissa.\n\\end{align*}\n\nSince $\\ln(\\abscissa)$ is increasing, $\\int_{\\slider}^{\\slider+1} \\ln(\\abscissa)\\,d\\abscissa$ is an increasing function of $\\slider$, and so it is minimized over $\\slider \\in [1,\\infty)$ when $\\slider=1$. We conclude that\n\\[\n\\minimum(\\funcmain) = \\frac{1}{2} + \\int_1^2 \\ln(\\abscissa)\\,d\\abscissa = 2 \\ln 2-\\frac{1}{2}\n\\]\nindependent of $\\funcmain$.\n\n\\noindent\n\\textbf{Remark.}\nThe phrasing of the question suggests that solvers were not expected to prove that $\\mathcal{F}$ is nonempty,\neven though this is necessary to make the definition of $\\minimum(\\funcmain)$ logically meaningful. Existence will be explicitly established in the next solution.\n\n\\noindent\n\\textbf{Solution 2.}\nWe first verify that \n\\[\n\\funcmain(\\abscissa,\\ordinate) = \\frac{1}{2}(\\abscissa\\ordinate\\ln(\\abscissa\\ordinate)-\\abscissa\\ordinate)\n\\]\nis an element of $\\mathcal{F}$, by computing that\n\\begin{gather*}\n\\abscissa\\partialx = \\ordinate\\partialy = \\frac{1}{2} \\abscissa\\ordinate \\ln(\\abscissa\\ordinate) \\\\\n\\abscissa^2 \\secondx = \\ordinate^2 \\secondy = \\abscissa\\ordinate.\n\\end{gather*}\n(See the following remark for motivation for this guess.)\n\nWe next show that the only elements of $\\mathcal{F}$ are $\\funcmain+\\constanta \\ln(\\abscissa/\\ordinate) + \\constantb$ where $\\constanta,\\constantb$ are constants.\nSuppose that $\\funcmain+\\funcauxi$ is a second element of $\\mathcal{F}$. As in the first solution, we deduce that $\\auxmixedxy = 0$; this implies that\n$\\funcauxi(\\abscissa,\\ordinate) = \\auxvaru(\\abscissa) + \\auxvarv(\\ordinate)$ for some twice continuously differentiable functions $\\auxvaru$ and $\\auxvarv$. We also have\n$\\abscissa\\auxpartialx + \\ordinate\\auxpartialy = 0$, which now asserts that $\\abscissa\\auxpartialx = - \\ordinate\\auxpartialy$ is equal to some constant $\\constanta$. This yields that\n$\\funcauxi = \\constanta \\ln(\\abscissa/\\ordinate) + \\constantb$ as desired.\n\nWe next observe that \n\\[\n\\funcauxi(\\slider+1,\\slider+1)-\\funcauxi(\\slider+1,\\slider)-\\funcauxi(\\slider,\\slider+1)+\\funcauxi(\\slider,\\slider) = 0,\n\\]\nso $\\minimum(\\funcmain) = \\minimum(\\funcmain+\\funcauxi)$. It thus remains to compute $\\minimum(\\funcmain)$. To do this, we verify that\n\\[\n\\funcmain(\\slider+1,\\slider+1)-\\funcmain(\\slider+1,\\slider)-\\funcmain(\\slider,\\slider+1)+\\funcmain(\\slider,\\slider)\n\\]\nis nondecreasing in $\\slider$ by computing its derivative to be $\\ln (\\slider+1) - \\ln(\\slider)$\n(either directly or using the integral representation from the first solution).\nWe thus minimize by taking $\\slider=1$ as in the first solution.\n\n\\noindent\n\\textbf{Remark.}\nOne way to make a correct guess for $\\funcmain$ is to notice that the given equations are both symmetric in $\\abscissa$ and $\\ordinate$\nand posit that $\\funcmain$ should also be symmetric. Any symmetric function of $\\abscissa$ and $\\ordinate$ can be written in terms of the variables $\\auxvaru = \\abscissa+\\ordinate$ and $\\auxvarv = \\abscissa\\ordinate$, so in principle we could translate the equations into those variables and solve. However, before trying this, we observe that $\\abscissa\\ordinate$ appears explicitly in the equations, so it is reasonable to make a first guess of the form $\\funcmain(\\abscissa,\\ordinate) = \\helperfn(\\abscissa\\ordinate)$. \nFor such a choice, we have\n\\[\n\\abscissa \\partialx + \\ordinate \\partialy = 2\\abscissa\\ordinate \\helperfn' = \\abscissa\\ordinate \\ln(\\abscissa\\ordinate)\n\\]\nwhich forces us to set $\\helperfn(\\parameter)  = \\frac{1}{2}(\\parameter \\ln(\\parameter) - \\parameter)$. "
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pineapple",
        "y": "strawberry",
        "s": "blueberry",
        "t": "raspberry",
        "f": "marshmallow",
        "g": "butterscotch",
        "h": "bubblegum",
        "u": "cheesecake",
        "v": "gingerbread",
        "m": "chocolate",
        "f_x": "lollipop",
        "f_y": "tangerine",
        "f_xx": "watermelon",
        "f_yy": "cantaloupe",
        "f_xy": "persimmon",
        "g_x": "pomegranate",
        "g_y": "passionfruit",
        "g_xy": "nectarine",
        "a": "blackberry",
        "b": "boysenberry"
      },
      "question": "Let $\\mathcal{F}$ be the set of functions $marshmallow(pineapple,strawberry)$ that are twice continuously differentiable for $pineapple \\geq 1$, $strawberry \\geq 1$ and that satisfy the following two equations (where subscripts denote partial derivatives):\n\\begin{gather*}\npineapple\\,lollipop + strawberry\\,tangerine = pineapple\\,strawberry \\ln(pineapple strawberry), \\\\\npineapple^2\\,watermelon + strawberry^2\\,cantaloupe = pineapple\\,strawberry.\n\\end{gather*}\nFor each $marshmallow \\in \\mathcal{F}$, let\n\\[\nchocolate(marshmallow) = \\min_{blueberry \\geq 1} \\left(marshmallow(blueberry+1,blueberry+1) - marshmallow(blueberry+1,blueberry) - marshmallow(blueberry,blueberry+1) + marshmallow(blueberry,blueberry) \\right).\n\\]\nDetermine $chocolate(marshmallow)$, and show that it is independent of the choice of $marshmallow$.",
      "solution": "\\noindent\n\\textbf{Solution 1.}\nWe compute that $chocolate(marshmallow) = 2 \\ln 2 - \\frac{1}{2}$.\nLabel the given differential equations by (1) and (2). If we write, e.g., $pineapple\\frac{\\partial}{\\partial pineapple}(1)$ for the result of differentiating (1) by $pineapple$ and multiplying the resulting equation by $pineapple$, then the combination\n$pineapple\\frac{\\partial}{\\partial pineapple}(1)+strawberry\\frac{\\partial}{\\partial strawberry}(1)-(1)-(2)$ gives the equation\n$2 pineapple strawberry persimmon = pineapple strawberry\\ln(pineapple strawberry)+pineapple strawberry$, whence $persimmon = \\frac{1}{2} (\\ln(pineapple)+\\ln(strawberry)+1)$.\n\nNow we observe that\n\\begin{align*}\n\\lefteqn{marshmallow(blueberry+1,blueberry+1)-marshmallow(blueberry+1,blueberry)-marshmallow(blueberry,blueberry+1)+marshmallow(blueberry,blueberry)} \\\\\n &= \\int_{blueberry}^{blueberry+1} \\int_{blueberry}^{blueberry+1} persimmon\\,d strawberry\\,d pineapple \\\\\n&= \\frac{1}{2} \\int_{blueberry}^{blueberry+1} \\int_{blueberry}^{blueberry+1} (\\ln(pineapple)+\\ln(strawberry)+1)\\,d strawberry\\,d pineapple \\\\\n&= \\frac{1}{2} + \\int_{blueberry}^{blueberry+1} \\ln(pineapple)\\,d pineapple.\n\\end{align*}\n\nSince $\\ln(pineapple)$ is increasing, $\\int_{blueberry}^{blueberry+1} \\ln(pineapple)\\,d pineapple$ is an increasing function of $blueberry$, and so it is minimized over $blueberry \\in [1,\\infty)$ when $blueberry=1$. We conclude that\n\\[\nchocolate(marshmallow) = \\frac{1}{2} + \\int_1^2 \\ln(pineapple)\\,d pineapple = 2 \\ln 2-\\frac{1}{2}\n\\]\nindependent of $marshmallow$.\n\n\\noindent\n\\textbf{Remark.}\nThe phrasing of the question suggests that solvers were not expected to prove that $\\mathcal{F}$ is nonempty,\neven though this is necessary to make the definition of $chocolate(marshmallow)$ logically meaningful. Existence will be explicitly established in the next solution.\n\n\\noindent\n\\textbf{Solution 2.}\nWe first verify that \n\\[\nmarshmallow(pineapple,strawberry) = \\frac{1}{2}(pineapple strawberry\\ln(pineapple strawberry)-pineapple strawberry)\n\\]\nis an element of $\\mathcal{F}$, by computing that\n\\begin{gather*}\npineapple lollipop = strawberry tangerine = \\frac{1}{2} pineapple strawberry \\ln(pineapple strawberry) \\\\\npineapple^2 watermelon = strawberry^2 cantaloupe = pineapple strawberry.\n\\end{gather*}\n(See the following remark for motivation for this guess.)\n\nWe next show that the only elements of $\\mathcal{F}$ are $marshmallow+blackberry \\ln(pineapple/strawberry) + boysenberry$ where $blackberry,boysenberry$ are constants.\nSuppose that $marshmallow+butterscotch$ is a second element of $\\mathcal{F}$. As in the first solution, we deduce that $nectarine = 0$; this implies that\n$butterscotch(pineapple,strawberry) = cheesecake(pineapple) + gingerbread(strawberry)$ for some twice continuously differentiable functions $cheesecake$ and $gingerbread$. We also have\n$pineapple pomegranate + strawberry passionfruit = 0$, which now asserts that $pineapple pomegranate = - strawberry passionfruit$ is equal to some constant $blackberry$. This yields that\n$butterscotch = blackberry \\ln(pineapple/strawberry) + boysenberry$ as desired.\n\nWe next observe that \n\\[\nbutterscotch(blueberry+1,blueberry+1)-butterscotch(blueberry+1,blueberry)-butterscotch(blueberry,blueberry+1)+butterscotch(blueberry,blueberry) = 0,\n\\]\nso $chocolate(marshmallow) = chocolate(marshmallow+butterscotch)$. It thus remains to compute $chocolate(marshmallow)$. To do this, we verify that\n\\[\nmarshmallow(blueberry+1,blueberry+1)-marshmallow(blueberry+1,blueberry)-marshmallow(blueberry,blueberry+1)+marshmallow(blueberry,blueberry)\n\\]\nis nondecreasing in $blueberry$ by computing its derivative to be $\\ln (blueberry+1) - \\ln(blueberry)$\n(either directly or using the integral representation from the first solution).\nWe thus minimize by taking $blueberry=1$ as in the first solution.\n\n\\noindent\n\\textbf{Remark.}\nOne way to make a correct guess for $marshmallow$ is to notice that the given equations are both symmetric in $pineapple$ and $strawberry$\nand posit that $marshmallow$ should also be symmetric. Any symmetric function of $pineapple$ and $strawberry$ can be written in terms of the variables $cheesecake = pineapple+strawberry$ and $gingerbread = pineapple strawberry$, so in principle we could translate the equations into those variables and solve. However, before trying this, we observe that $pineapple strawberry$ appears explicitly in the equations, so it is reasonable to make a first guess of the form $marshmallow(pineapple,strawberry) = bubblegum(pineapple strawberry)$. \nFor such a choice, we have\n\\[\npineapple lollipop + strawberry tangerine = 2 pineapple strawberry bubblegum' = pineapple strawberry \\ln(pineapple strawberry)\n\\]\nwhich forces us to set $bubblegum(t)  = \\frac{1}{2}(t \\ln(t) - t)$.",
      "confidence": "0.14"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "s": "termination",
        "t": "spacestate",
        "f": "constantvalue",
        "g": "fixedscalar",
        "h": "nonvarying",
        "u": "difference",
        "v": "quotient",
        "m": "greatestvalue",
        "f_x": "constantvaluederivx",
        "f_y": "constantvaluederivy",
        "f_xx": "constantvaluederivxx",
        "f_yy": "constantvaluederivyy",
        "f_xy": "constantvaluederivxy",
        "g_x": "fixedscalarderivx",
        "g_y": "fixedscalarderivy",
        "g_xy": "fixedscalarderivxy",
        "a": "variable",
        "b": "nonconstant"
      },
      "question": "Let $\\mathcal{F}$ be the set of functions $\\constantvalue(\\verticalaxis,\\horizontalaxis)$ that are twice continuously differentiable for $\\verticalaxis \\geq 1$, $\\horizontalaxis \\geq 1$ and that satisfy the following two equations (where subscripts denote partial derivatives):\n\\begin{gather*}\n\\verticalaxis\\constantvaluederivx + \\horizontalaxis\\constantvaluederivy = \\verticalaxis\\horizontalaxis \\ln(\\verticalaxis\\horizontalaxis), \\\\\n\\verticalaxis^{2}\\constantvaluederivxx + \\horizontalaxis^{2}\\constantvaluederivyy = \\verticalaxis\\horizontalaxis.\n\\end{gather*}\nFor each $\\constantvalue \\in \\mathcal{F}$, let\n\\[\n\\greatestvalue(\\constantvalue) = \\min_{\\termination \\geq 1} \\left(\\constantvalue(\\termination+1,\\termination+1) - \\constantvalue(\\termination+1,\\termination) - \\constantvalue(\\termination,\\termination+1) + \\constantvalue(\\termination,\\termination) \\right).\n\\]\nDetermine $\\greatestvalue(\\constantvalue)$, and show that it is independent of the choice of $\\constantvalue$.",
      "solution": "\\noindent\n\\textbf{Solution 1.}\nWe compute that $\\greatestvalue(\\constantvalue) = 2 \\ln 2 - \\frac{1}{2}$.  \nLabel the given differential equations by (1) and (2). If we write, e.g., $\\verticalaxis \\frac{\\partial}{\\partial \\verticalaxis}(1)$ for the result of differentiating (1) by $\\verticalaxis$ and multiplying the resulting equation by $\\verticalaxis$, then the combination\n$\\verticalaxis \\frac{\\partial}{\\partial \\verticalaxis}(1)+\\horizontalaxis \\frac{\\partial}{\\partial \\horizontalaxis}(1)-(1)-(2)$ gives the equation\n$2\\verticalaxis\\horizontalaxis\\constantvaluederivxy = \\verticalaxis\\horizontalaxis\\ln(\\verticalaxis\\horizontalaxis)+\\verticalaxis\\horizontalaxis$, whence $\\constantvaluederivxy = \\frac{1}{2} (\\ln(\\verticalaxis)+\\ln(\\horizontalaxis)+1)$.\n\nNow we observe that\n\\begin{align*}\n\\lefteqn{\\constantvalue(\\termination+1,\\termination+1)-\\constantvalue(\\termination+1,\\termination)-\\constantvalue(\\termination,\\termination+1)+\\constantvalue(\\termination,\\termination)} \\\\\n &= \\int_{\\termination}^{\\termination+1} \\int_{\\termination}^{\\termination+1} \\constantvaluederivxy\\,d\\horizontalaxis\\,d\\verticalaxis \\\\\n&= \\frac{1}{2} \\int_{\\termination}^{\\termination+1} \\int_{\\termination}^{\\termination+1} (\\ln(\\verticalaxis)+\\ln(\\horizontalaxis)+1)\\,d\\horizontalaxis\\,d\\verticalaxis \\\\\n&= \\frac{1}{2} + \\int_{\\termination}^{\\termination+1} \\ln(\\verticalaxis)\\,d\\verticalaxis.\n\\end{align*}\n\nSince $\\ln(\\verticalaxis)$ is increasing, $\\int_{\\termination}^{\\termination+1} \\ln(\\verticalaxis)\\,d\\verticalaxis$ is an increasing function of $\\termination$, and so it is minimized over $\\termination \\in [1,\\infty)$ when $\\termination=1$. We conclude that\n\\[\n\\greatestvalue(\\constantvalue) = \\frac{1}{2} + \\int_1^2 \\ln(\\verticalaxis)\\,d\\verticalaxis = 2 \\ln 2-\\frac{1}{2}\n\\]\nindependent of $\\constantvalue$.\n\n\\noindent\n\\textbf{Remark.}\nThe phrasing of the question suggests that solvers were not expected to prove that $\\mathcal{F}$ is nonempty,\neven though this is necessary to make the definition of $\\greatestvalue(\\constantvalue)$ logically meaningful. Existence will be explicitly established in the next solution.\n\n\\noindent\n\\textbf{Solution 2.}\nWe first verify that \n\\[\n\\constantvalue(\\verticalaxis,\\horizontalaxis) = \\frac{1}{2}(\\verticalaxis\\horizontalaxis\\ln(\\verticalaxis\\horizontalaxis)-\\verticalaxis\\horizontalaxis)\n\\]\nis an element of $\\mathcal{F}$, by computing that\n\\begin{gather*}\n\\verticalaxis\\constantvaluederivx = \\horizontalaxis\\constantvaluederivy = \\frac{1}{2} \\verticalaxis\\horizontalaxis \\ln(\\verticalaxis\\horizontalaxis) \\\\\n\\verticalaxis^{2}\\constantvaluederivxx = \\horizontalaxis^{2}\\constantvaluederivyy = \\verticalaxis\\horizontalaxis.\n\\end{gather*}\n(See the following remark for motivation for this guess.)\n\nWe next show that the only elements of $\\mathcal{F}$ are $\\constantvalue+\\variable \\ln(\\verticalaxis/\\horizontalaxis) + \\nonconstant$ where $\\variable,\\nonconstant$ are constants.\nSuppose that $\\constantvalue+\\fixedscalar$ is a second element of $\\mathcal{F}$. As in the first solution, we deduce that $\\fixedscalarderivxy = 0$; this implies that\n$\\fixedscalar(\\verticalaxis,\\horizontalaxis) = \\difference(\\verticalaxis) + \\quotient(\\horizontalaxis)$ for some twice continuously differentiable functions $\\difference$ and $\\quotient$. We also have\n$\\verticalaxis\\fixedscalarderivx + \\horizontalaxis\\fixedscalarderivy = 0$, which now asserts that $\\verticalaxis\\fixedscalarderivx = - \\horizontalaxis\\fixedscalarderivy$ is equal to some constant $\\variable$. This yields that\n$\\fixedscalar = \\variable \\ln(\\verticalaxis/\\horizontalaxis) + \\nonconstant$ as desired.\n\nWe next observe that \n\\[\n\\fixedscalar(\\termination+1,\\termination+1)-\\fixedscalar(\\termination+1,\\termination)-\\fixedscalar(\\termination,\\termination+1)+\\fixedscalar(\\termination,\\termination) = 0,\n\\]\nso $\\greatestvalue(\\constantvalue) = \\greatestvalue(\\constantvalue+\\fixedscalar)$. It thus remains to compute $\\greatestvalue(\\constantvalue)$. To do this, we verify that\n\\[\n\\constantvalue(\\termination+1,\\termination+1)-\\constantvalue(\\termination+1,\\termination)-\\constantvalue(\\termination,\\termination+1)+\\constantvalue(\\termination,\\termination)\n\\]\nis nondecreasing in $\\termination$ by computing its derivative to be $\\ln (\\termination+1) - \\ln(\\termination)$\n(either directly or using the integral representation from the first solution).\nWe thus minimize by taking $\\termination=1$ as in the first solution.\n\n\\noindent\n\\textbf{Remark.}\nOne way to make a correct guess for $\\constantvalue$ is to notice that the given equations are both symmetric in $\\verticalaxis$ and $\\horizontalaxis$\nand posit that $\\constantvalue$ should also be symmetric. Any symmetric function of $\\verticalaxis$ and $\\horizontalaxis$ can be written in terms of the variables $\\difference = \\verticalaxis+\\horizontalaxis$ and $\\quotient = \\verticalaxis\\horizontalaxis$, so in principle we could translate the equations into those variables and solve. However, before trying this, we observe that $\\verticalaxis\\horizontalaxis$ appears explicitly in the equations, so it is reasonable to make a first guess of the form $\\constantvalue(\\verticalaxis,\\horizontalaxis) = \\nonvarying(\\verticalaxis\\horizontalaxis)$. \nFor such a choice, we have\n\\[\n\\verticalaxis \\constantvaluederivx + \\horizontalaxis \\constantvaluederivy = 2\\verticalaxis\\horizontalaxis \\nonvarying' = \\verticalaxis\\horizontalaxis \\ln(\\verticalaxis\\horizontalaxis)\n\\]\nwhich forces us to set $\\nonvarying(\\spacestate)  = \\frac{1}{2}(\\spacestate \\ln(\\spacestate) - \\spacestate)$. "
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "s": "mpqldvre",
        "t": "blfaxnku",
        "f": "vczmsefh",
        "g": "lqtrkpaw",
        "h": "zxyrumad",
        "u": "rnpvgdls",
        "v": "ktcjwfob",
        "m": "sgqvleno",
        "f_x": "uvpshmca",
        "f_y": "efacijrb",
        "f_xx": "qkphzntu",
        "f_yy": "gmcratev",
        "f_xy": "birsowke",
        "g_x": "lxzueyod",
        "g_y": "naqpmhsf",
        "g_xy": "vzctxqre",
        "a": "jwksyehm",
        "b": "dmqrapon"
      },
      "question": "Let $\\mathcal{F}$ be the set of functions $vczmsefh(qzxwvtnp,hjgrksla)$ that are twice continuously differentiable for $qzxwvtnp \\geq 1$, $hjgrksla \\geq 1$ and that satisfy the following two equations (where subscripts denote partial derivatives):\n\\begin{gather*}\nqzxwvtnp\\,uvpshmca + hjgrksla\\,efacijrb = qzxwvtnp hjgrksla \\ln(qzxwvtnp hjgrksla), \\\\\nqzxwvtnp^2 qkphzntu + hjgrksla^2 gmcratev = qzxwvtnp hjgrksla.\n\\end{gather*}\nFor each $vczmsefh \\in \\mathcal{F}$, let\n\\[\nsgqvleno(vczmsefh) = \\min_{mpqldvre \\geq 1} \\left(vczmsefh(mpqldvre+1,mpqldvre+1) - vczmsefh(mpqldvre+1,mpqldvre) - vczmsefh(mpqldvre,mpqldvre+1) + vczmsefh(mpqldvre,mpqldvre) \\right).\n\\]\nDetermine $sgqvleno(vczmsefh)$, and show that it is independent of the choice of $vczmsefh$.",
      "solution": "\\noindent\n\\textbf{Solution 1.}\nWe compute that $sgqvleno(vczmsefh) = 2 \\ln 2 - \\tfrac12$.\nLabel the given differential equations by (1) and (2). If we write, e.g., $qzxwvtnp\\tfrac{\\partial}{\\partial qzxwvtnp}(1)$ for the result of differentiating (1) by $qzxwvtnp$ and multiplying the resulting equation by $qzxwvtnp$, then the combination\n$qzxwvtnp\\tfrac{\\partial}{\\partial qzxwvtnp}(1)+hjgrksla\\tfrac{\\partial}{\\partial hjgrksla}(1)-(1)-(2)$ gives the equation\n$2qzxwvtnp hjgrksla\\,birsowke = qzxwvtnp hjgrksla\\ln(qzxwvtnp hjgrksla)+qzxwvtnp hjgrksla$, whence $birsowke = \\tfrac12 (\\ln(qzxwvtnp)+\\ln(hjgrksla)+1)$.\n\nNow we observe that\n\\begin{align*}\n\\lefteqn{vczmsefh(mpqldvre+1,mpqldvre+1)-vczmsefh(mpqldvre+1,mpqldvre)-vczmsefh(mpqldvre,mpqldvre+1)+vczmsefh(mpqldvre,mpqldvre)} \\\\ & = \\int_{mpqldvre}^{mpqldvre+1} \\int_{mpqldvre}^{mpqldvre+1} birsowke\\,dhjgrksla\\,dqzxwvtnp \\\\ &= \\tfrac12 \\int_{mpqldvre}^{mpqldvre+1} \\int_{mpqldvre}^{mpqldvre+1} (\\ln(qzxwvtnp)+\\ln(hjgrksla)+1)\\,dhjgrksla\\,dqzxwvtnp \\\\ &= \\tfrac12 + \\int_{mpqldvre}^{mpqldvre+1} \\ln(qzxwvtnp)\\,dqzxwvtnp.\n\\end{align*}\n\nSince $\\ln(qzxwvtnp)$ is increasing, $\\int_{mpqldvre}^{mpqldvre+1} \\ln(qzxwvtnp)\\,dqzxwvtnp$ is an increasing function of $mpqldvre$, and so it is minimized over $mpqldvre \\in [1,\\infty)$ when $mpqldvre=1$. We conclude that\n\\[\nsgqvleno(vczmsefh) = \\tfrac12 + \\int_1^2 \\ln(qzxwvtnp)\\,dqzxwvtnp = 2 \\ln 2-\\tfrac12\n\\]\nindependent of $vczmsefh$.\n\n\\noindent\n\\textbf{Remark.}\nThe phrasing of the question suggests that solvers were not expected to prove that $\\mathcal{F}$ is nonempty,\neven though this is necessary to make the definition of $sgqvleno$ logically meaningful. Existence will be explicitly established in the next solution.\n\n\\noindent\n\\textbf{Solution 2.}\nWe first verify that \n\\[\nvczmsefh(qzxwvtnp,hjgrksla) = \\tfrac12(qzxwvtnp hjgrksla\\,\\ln(qzxwvtnp hjgrksla)-qzxwvtnp hjgrksla)\n\\]\nis an element of $\\mathcal{F}$, by computing that\n\\begin{gather*}\nqzxwvtnp\\,uvpshmca = hjgrksla\\,efacijrb = \\tfrac12 qzxwvtnp hjgrksla \\ln(qzxwvtnp hjgrksla) \\\\\nqzxwvtnp^2 qkphzntu = hjgrksla^2 gmcratev = qzxwvtnp hjgrksla.\n\\end{gather*}\n(See the following remark for motivation for this guess.)\n\nWe next show that the only elements of $\\mathcal{F}$ are $vczmsefh+jwksyehm \\ln(qzxwvtnp/hjgrksla) + dmqrapon$ where $jwksyehm,dmqrapon$ are constants.\nSuppose that $vczmsefh+lqtrkpaw$ is a second element of $\\mathcal{F}$. As in the first solution, we deduce that $vzctxqre = 0$; this implies that\n$lqtrkpaw(qzxwvtnp,hjgrksla) = rnpvgdls(qzxwvtnp) + ktcjwfob(hjgrksla)$ for some twice continuously differentiable functions $rnpvgdls$ and $ktcjwfob$. We also have\n$qzxwvtnp\\,lxzueyod + hjgrksla\\,naqpmhsf = 0$, which now asserts that $qzxwvtnp\\,lxzueyod = - hjgrksla\\,naqpmhsf$ is equal to some constant $jwksyehm$. This yields that\n$lqtrkpaw = jwksyehm \\ln(qzxwvtnp/hjgrksla) + dmqrapon$ as desired.\n\nWe next observe that \n\\[\nlqtrkpaw(mpqldvre+1,mpqldvre+1)-lqtrkpaw(mpqldvre+1,mpqldvre)-lqtrkpaw(mpqldvre,mpqldvre+1)+lqtrkpaw(mpqldvre,mpqldvre) = 0,\n\\]\nso $sgqvleno(vczmsefh) = sgqvleno(vczmsefh+lqtrkpaw)$. It thus remains to compute $sgqvleno(vczmsefh)$. To do this, we verify that\n\\[\nvczmsefh(mpqldvre+1,mpqldvre+1)-vczmsefh(mpqldvre+1,mpqldvre)-vczmsefh(mpqldvre,mpqldvre+1)+vczmsefh(mpqldvre,mpqldvre)\n\\]\nis nondecreasing in $mpqldvre$ by computing its derivative to be $\\ln (mpqldvre+1) - \\ln(mpqldvre)$\n(either directly or using the integral representation from the first solution).\nWe thus minimize by taking $mpqldvre=1$ as in the first solution.\n\n\\noindent\n\\textbf{Remark.}\nOne way to make a correct guess for $vczmsefh$ is to notice that the given equations are both symmetric in $qzxwvtnp$ and $hjgrksla$\nand posit that $vczmsefh$ should also be symmetric. Any symmetric function of $qzxwvtnp$ and $hjgrksla$ can be written in terms of the variables $rnpvgdls = qzxwvtnp+hjgrksla$ and $ktcjwfob = qzxwvtnp hjgrksla$, so in principle we could translate the equations into those variables and solve. However, before trying this, we observe that $qzxwvtnp hjgrksla$ appears explicitly in the equations, so it is reasonable to make a first guess of the form $vczmsefh(qzxwvtnp,hjgrksla) = zxyrumad(qzxwvtnp hjgrksla)$. \nFor such a choice, we have\n\\[\nqzxwvtnp vczmsefh_{qzxwvtnp} + hjgrksla vczmsefh_{hjgrksla} = 2qzxwvtnp hjgrksla\\,zxyrumad' = qzxwvtnp hjgrksla \\ln(qzxwvtnp hjgrksla)\n\\]\nwhich forces us to set $zxyrumad(blfaxnku)  = \\tfrac12(blfaxnku \\ln(blfaxnku) - blfaxnku)$.}",
      "confidence": "0.16"
    },
    "kernel_variant": {
      "question": "Let ℋ denote the class of real–valued functions  \n\n  h :  [2,∞)³  → ℝ,  h ∈ C³,  \n\nthat satisfy the three (coupled) partial–differential relations  \n\n(1) x h_x + y h_y + z h_z = 3 xyz ln(xyz),  \n(2) x²h_{xx}+y²h_{yy}+z²h_{zz} = 3 xyz,  \n(3) h_{xyz} = ln x + ln y + ln z + 2.  \n\nFor h∈ℋ define the “discrete three–dimensional curvature’’  \n\n  Δ_h(s)=∑_{ε_x ,ε_y ,ε_z∈{0,1}} (–1)^{ε_x+ε_y+ε_z}\\;\n                   h(s+ε_x , s+ε_y , s+ε_z),  s ≥ 2,\n\nand put  \n\n  N(h)=min_{s ≥ 2} Δ_h(s).\n\n(a)  Show that ℋ is non–empty and determine all its members.  \n(b)  Prove that for every h∈ℋ the function Δ_h is strictly increasing on [2,∞).  \n(c)  Deduce that N(h) is the same for every element of ℋ and compute its common value explicitly.",
      "solution": "Notation.  Write w = xyz and ω = ln w = ln x + ln y + ln z throughout.\n\n————————————————————————————————  \nStep 0.  One explicit element of ℋ.  \nSet  \n\n  h₀(x,y,z) = w(ω – 1) = xyz (ln(xyz) – 1).\n\nDirect differentiation gives  \n\n  h₀_x = yz ω, h₀_{xx} = yz/x, h₀_{xyz} = ω + 2,\n\nso h₀ satisfies (1)–(3); hence ℋ ≠ ∅.\n\n————————————————————————————————  \nStep 1.  The full six–parameter family of solutions of (1)–(3).  \n\nPut δ := h – h₀ ∈ C³([2,∞)³).  Then δ obeys the homogeneous system\n\n  x δ_x + y δ_y + z δ_z = 0,     (4)  \n  x²δ_{xx}+y²δ_{yy}+z²δ_{zz} = 0,  (5)  \n  δ_{xyz} = 0.            (6)\n\n1 a.  Reduction to two variables.  \nThe characteristic system of (4) is\n\n  dx/x = dy/y = dz/z,\n\nso δ is constant along every ray (λx,λy,λz).  Equivalently\n\n  δ(x,y,z) = Φ(u,v), u = ln(y/x), v = ln(z/x), Φ ∈ C³(ℝ²). (7)\n\n1 b.  Translation of (5) and (6).  \nA routine calculation yields\n\n  x²δ_{xx}+y²δ_{yy}+z²δ_{zz} = 2(Φ_{uu}+Φ_{uv}+Φ_{vv}), (8)\n\nhence (5) becomes the planar elliptic equation  \n\n  Φ_{uu}+Φ_{uv}+Φ_{vv}=0.             (9)\n\nFurther,\n\n  δ_{xyz}= –(Φ_{uuv}+Φ_{uvv})/(xyz).       (10)\n\nCondition (6) is therefore  \n\n  Φ_{uuv}+Φ_{uvv}=0.             (11)\n\nCombining (9) and (11) and differentiating twice gives Φ_{vvv}=0 and Φ_{uuu}=0, i.e. Φ is at most quadratic separately in u and v.\n\n1 c.  An explicit quadratic ansatz and its consequences.  \nWrite  \n\n  Φ(u,v) = A(u)v² + B(u)v + C(u).        (12)\n\nInsert (12) into (9) and match the coefficients of v², v and 1:\n\n  A''(u) = 0,                   (13a)  \n  B''(u) + 2A'(u) = 0,              (13b)  \n  C''(u) + B'(u) + 2A(u) = 0.          (13c)\n\nSolve the linear ODEs:\n\n  A(u) = αu + β,  \n  B(u) = –αu² + γu + δ,  \n  C(u) = –½(γ+2β)u² + εu + ζ,        (14)\n\nwith arbitrary real constants  \n\n  α,β,γ,δ,ε,ζ ∈ ℝ.\n\nNo further restriction arises from (11), so (14) is the most general C³ solution.\n\n1 d.  Back to x, y, z.  \nLet u = ln(y/x), v = ln(z/x); then\n\n  δ(x,y,z) =\n   (αu + β)v² + (–αu² + γu + δ)v\n   + (–½(γ+2β)u² + εu + ζ).    (15)\n\nHence every h ∈ ℋ can be written uniquely as  \n\n  h(x,y,z)= xyz(ln(xyz)–1) + δ(x,y,z),     (16)\n\nwhere δ is given by (15).  Conversely, (15)–(16) always satisfy (4)–(6), so (16) indeed parametrises the whole class ℋ.  Thus ℋ is a six–parameter family.\n\n————————————————————————————————  \nStep 2.  The universal mixed third derivative.  \nBecause every δ satisfies (6), we have for every h ∈ ℋ\n\n  h_{xyz}=h₀_{xyz}=ω+2.            (17)\n\n————————————————————————————————  \nStep 3.  The alternating–sum identity.  \nFor any C³-function on ℝ³ and any s ≥ 2\n\n  Δ_h(s)=∭_{[\\,s,s+1]^3} h_{xyz}(x,y,z)\\,dx\\,dy\\,dz  (18)\n\n(three successive one–dimensional integrations by parts).  With (17)\n\n  Δ_h(s)=∭_{[s,s+1]^3}(ω+2)\\,dx\\,dy\\,dz.     (19)\n\nBy symmetry the three logarithmic integrals coincide; put  \n\n  L(s):=∫_{s}^{s+1} ln t dt.          (20)\n\nThen\n\n  Δ_h(s)=3L(s)+2.              (21)\n\n————————————————————————————————  \nStep 4.  Monotonicity of Δ_h.  \nSince ln t is strictly increasing,\n\n  L′(s)=ln(s+1)–ln s > 0  (s ≥ 2),     (22)\n\nso L and therefore Δ_h are strictly increasing on [2,∞).  Hence  \n\n  N(h)=Δ_h(2).              (23)\n\n————————————————————————————————  \nStep 5.  Evaluation of the minimum.  \nA short computation gives  \n\n  L(2)=3 ln 3 – 2 ln 2 – 1.        (24)\n\nInsert (24) into (21) and (23):\n\n  N(h)=3L(2)+2 = 9 ln 3 – 6 ln 2 – 1.   (25)\n\nThe right–hand side is independent of the six parameters α,β,γ,δ,ε,ζ and therefore of the choice of h ∈ ℋ.\n\n————————————————————————————————  \nAnswer.  \n(a)  ℋ consists of the six–parameter family (16) with δ given by (15); in particular ℋ is non-empty.  \n(b)  For every h∈ℋ the map s ↦ Δ_h(s) is strictly increasing on [2,∞).  \n(c)  All members of ℋ share the common value  \n\n  N(h)=9 ln 3 – 6 ln 2 – 1.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.863966",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the problem moves from 2 to 3 variables, turning a planar argument\n  into a spatial one and replacing a 2×2 finite difference by an 8-term “cube’’ difference.  \n\n• Higher-order analysis: one now needs the third mixed partial h_{xyz}, not merely a\n  mixed second derivative; the derivation demands two successive non-trivial\n  manipulations of the PDE system.\n\n• Coupled PDE system: the solver must recognise that the two given equations are tightly\n  linked and that a delicate combination (followed by another differentiation) is required\n  to isolate h_{xyz}.  Naïve pattern-matching fails.\n\n• Functional rigidity: showing that h_{xyz} is unique for all solutions requires analysing\n  the kernel of the linear operator defined by (I)–(II), a step absent in the original task.\n\n• Integration in higher dimension: translating the 8-term discrete curvature into a\n  triple integral, evaluating it with the universal h_{xyz}, and proving monotonicity of the\n  resulting expression all add conceptual and computational layers.\n\nAltogether these added dimensions, derivatives, and algebraic interdependencies force\nthe use of multivariable calculus, homogeneous PDE techniques, and careful symmetry\narguments, substantially raising the technical bar relative to the original problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let ℋ denote the class of real–valued functions  \n\n  h :  [2,∞)³  → ℝ,  h ∈ C³,  \n\nthat satisfy the three (coupled) partial–differential relations  \n\n(1) x h_x + y h_y + z h_z = 3 xyz ln(xyz),  \n(2) x²h_{xx}+y²h_{yy}+z²h_{zz} = 3 xyz,  \n(3) h_{xyz} = ln x + ln y + ln z + 2.  \n\nFor h∈ℋ define the “discrete three–dimensional curvature’’  \n\n  Δ_h(s)=∑_{ε_x ,ε_y ,ε_z∈{0,1}} (–1)^{ε_x+ε_y+ε_z}\\;\n                   h(s+ε_x , s+ε_y , s+ε_z),  s ≥ 2,\n\nand put  \n\n  N(h)=min_{s ≥ 2} Δ_h(s).\n\n(a)  Show that ℋ is non–empty and determine all its members.  \n(b)  Prove that for every h∈ℋ the function Δ_h is strictly increasing on [2,∞).  \n(c)  Deduce that N(h) is the same for every element of ℋ and compute its common value explicitly.",
      "solution": "Notation.  Write w = xyz and ω = ln w = ln x + ln y + ln z throughout.\n\n————————————————————————————————  \nStep 0.  One explicit element of ℋ.  \nSet  \n\n  h₀(x,y,z) = w(ω – 1) = xyz (ln(xyz) – 1).\n\nDirect differentiation gives  \n\n  h₀_x = yz ω, h₀_{xx} = yz/x, h₀_{xyz} = ω + 2,\n\nso h₀ satisfies (1)–(3); hence ℋ ≠ ∅.\n\n————————————————————————————————  \nStep 1.  The full six–parameter family of solutions of (1)–(3).  \n\nPut δ := h – h₀ ∈ C³([2,∞)³).  Then δ obeys the homogeneous system\n\n  x δ_x + y δ_y + z δ_z = 0,     (4)  \n  x²δ_{xx}+y²δ_{yy}+z²δ_{zz} = 0,  (5)  \n  δ_{xyz} = 0.            (6)\n\n1 a.  Reduction to two variables.  \nThe characteristic system of (4) is\n\n  dx/x = dy/y = dz/z,\n\nso δ is constant along every ray (λx,λy,λz).  Equivalently\n\n  δ(x,y,z) = Φ(u,v), u = ln(y/x), v = ln(z/x), Φ ∈ C³(ℝ²). (7)\n\n1 b.  Translation of (5) and (6).  \nA routine calculation yields\n\n  x²δ_{xx}+y²δ_{yy}+z²δ_{zz} = 2(Φ_{uu}+Φ_{uv}+Φ_{vv}), (8)\n\nhence (5) becomes the planar elliptic equation  \n\n  Φ_{uu}+Φ_{uv}+Φ_{vv}=0.             (9)\n\nFurther,\n\n  δ_{xyz}= –(Φ_{uuv}+Φ_{uvv})/(xyz).       (10)\n\nCondition (6) is therefore  \n\n  Φ_{uuv}+Φ_{uvv}=0.             (11)\n\nCombining (9) and (11) and differentiating twice gives Φ_{vvv}=0 and Φ_{uuu}=0, i.e. Φ is at most quadratic separately in u and v.\n\n1 c.  An explicit quadratic ansatz and its consequences.  \nWrite  \n\n  Φ(u,v) = A(u)v² + B(u)v + C(u).        (12)\n\nInsert (12) into (9) and match the coefficients of v², v and 1:\n\n  A''(u) = 0,                   (13a)  \n  B''(u) + 2A'(u) = 0,              (13b)  \n  C''(u) + B'(u) + 2A(u) = 0.          (13c)\n\nSolve the linear ODEs:\n\n  A(u) = αu + β,  \n  B(u) = –αu² + γu + δ,  \n  C(u) = –½(γ+2β)u² + εu + ζ,        (14)\n\nwith arbitrary real constants  \n\n  α,β,γ,δ,ε,ζ ∈ ℝ.\n\nNo further restriction arises from (11), so (14) is the most general C³ solution.\n\n1 d.  Back to x, y, z.  \nLet u = ln(y/x), v = ln(z/x); then\n\n  δ(x,y,z) =\n   (αu + β)v² + (–αu² + γu + δ)v\n   + (–½(γ+2β)u² + εu + ζ).    (15)\n\nHence every h ∈ ℋ can be written uniquely as  \n\n  h(x,y,z)= xyz(ln(xyz)–1) + δ(x,y,z),     (16)\n\nwhere δ is given by (15).  Conversely, (15)–(16) always satisfy (4)–(6), so (16) indeed parametrises the whole class ℋ.  Thus ℋ is a six–parameter family.\n\n————————————————————————————————  \nStep 2.  The universal mixed third derivative.  \nBecause every δ satisfies (6), we have for every h ∈ ℋ\n\n  h_{xyz}=h₀_{xyz}=ω+2.            (17)\n\n————————————————————————————————  \nStep 3.  The alternating–sum identity.  \nFor any C³-function on ℝ³ and any s ≥ 2\n\n  Δ_h(s)=∭_{[\\,s,s+1]^3} h_{xyz}(x,y,z)\\,dx\\,dy\\,dz  (18)\n\n(three successive one–dimensional integrations by parts).  With (17)\n\n  Δ_h(s)=∭_{[s,s+1]^3}(ω+2)\\,dx\\,dy\\,dz.     (19)\n\nBy symmetry the three logarithmic integrals coincide; put  \n\n  L(s):=∫_{s}^{s+1} ln t dt.          (20)\n\nThen\n\n  Δ_h(s)=3L(s)+2.              (21)\n\n————————————————————————————————  \nStep 4.  Monotonicity of Δ_h.  \nSince ln t is strictly increasing,\n\n  L′(s)=ln(s+1)–ln s > 0  (s ≥ 2),     (22)\n\nso L and therefore Δ_h are strictly increasing on [2,∞).  Hence  \n\n  N(h)=Δ_h(2).              (23)\n\n————————————————————————————————  \nStep 5.  Evaluation of the minimum.  \nA short computation gives  \n\n  L(2)=3 ln 3 – 2 ln 2 – 1.        (24)\n\nInsert (24) into (21) and (23):\n\n  N(h)=3L(2)+2 = 9 ln 3 – 6 ln 2 – 1.   (25)\n\nThe right–hand side is independent of the six parameters α,β,γ,δ,ε,ζ and therefore of the choice of h ∈ ℋ.\n\n————————————————————————————————  \nAnswer.  \n(a)  ℋ consists of the six–parameter family (16) with δ given by (15); in particular ℋ is non-empty.  \n(b)  For every h∈ℋ the map s ↦ Δ_h(s) is strictly increasing on [2,∞).  \n(c)  All members of ℋ share the common value  \n\n  N(h)=9 ln 3 – 6 ln 2 – 1.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.656478",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the problem moves from 2 to 3 variables, turning a planar argument\n  into a spatial one and replacing a 2×2 finite difference by an 8-term “cube’’ difference.  \n\n• Higher-order analysis: one now needs the third mixed partial h_{xyz}, not merely a\n  mixed second derivative; the derivation demands two successive non-trivial\n  manipulations of the PDE system.\n\n• Coupled PDE system: the solver must recognise that the two given equations are tightly\n  linked and that a delicate combination (followed by another differentiation) is required\n  to isolate h_{xyz}.  Naïve pattern-matching fails.\n\n• Functional rigidity: showing that h_{xyz} is unique for all solutions requires analysing\n  the kernel of the linear operator defined by (I)–(II), a step absent in the original task.\n\n• Integration in higher dimension: translating the 8-term discrete curvature into a\n  triple integral, evaluating it with the universal h_{xyz}, and proving monotonicity of the\n  resulting expression all add conceptual and computational layers.\n\nAltogether these added dimensions, derivatives, and algebraic interdependencies force\nthe use of multivariable calculus, homogeneous PDE techniques, and careful symmetry\narguments, substantially raising the technical bar relative to the original problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}