{
  "index": "1963-A-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "3. Find an integral formula for the solution of the differential equation\n\\[\n\\delta(\\delta-1)(\\delta-2) \\cdots(\\delta-n+1) y=f(x), \\quad x \\geq 1\n\\]\nfor \\( y \\) as a function of \\( x \\) satisfying the initial conditions \\( y(1)=y^{\\prime}(1)=\\ldots= \\) \\( y^{(n-1)}(1)=0 \\), where \\( f \\) is continuous and\n\\[\n\\delta \\equiv x \\frac{d}{d x}\n\\]",
  "solution": "Solution. We first show that\n\\[\n\\delta(\\delta-1)(\\delta-2) \\cdots(\\delta-n+1) y=x^{n} \\frac{d^{n} y}{d x^{n}}\n\\]\n\nWe prove this by induction on \\( n \\). It is true for \\( n=1 \\). Assume (1) is true for \\( n=k \\). Since polynomials in the operator \\( \\delta \\) commute with one another, we have\n\\[\n\\begin{aligned}\n\\delta(\\delta-1)(\\delta-2) \\cdots(\\delta-k & +1)(\\delta-k) y \\\\\n& =(\\delta-k) \\delta(\\delta-1)(\\delta-2) \\cdots(\\delta-k+1) y \\\\\n& =\\left(x \\frac{d}{d x}-k\\right) x^{k} \\frac{d^{k} y}{d x^{k}} \\\\\n& =x^{k+1} \\frac{d^{k+1} y}{d x^{k+1}}\n\\end{aligned}\n\\]\n\nThus (1) is proved by induction for all \\( n \\).\nThe differential equation is thus\n\\[\nx^{n} y^{(n)}=f(x), \\quad x \\geq 1\n\\]\nand the solution can obviously be obtained by applying the integral operator \\( \\int_{1}^{x} n \\) times to the function \\( f(x) x^{-n} \\). However, it is possible to collapse the \\( n \\)-fold integration to a single integration as follows:\n\nOne of the standard forms of Taylor's theorem is\n\\[\n\\begin{aligned}\ny(x)=y(a)+(x-a) y^{\\prime}(a)+\\cdots+\\frac{(x-a)^{n-1}}{(n-1)!} & y^{(n-1)}(a) \\\\\n& +\\int_{a}^{x} \\frac{(x-t)^{n-1}}{(n-1)!} y^{(n)}(t) d t\n\\end{aligned}\n\\]\nif \\( y^{(n)} \\) is continuous. (See, for example, Thomas, Calculus and Analytic Geometry, alternate ed., Addison-Wesley, Reading, Mass., 1972, page 814.) In this case, taking \\( a=1 \\) and using (2) and the initial conditions, we have\n\\[\ny(x)=\\int_{1}^{x} \\frac{(x-t)^{n-1}}{(n-1)!} \\cdot \\frac{f(t)}{t^{n}} d t\n\\]\n\nIt is easy to check by direct differentiation that the function defined by this integral does indeed satisfy the differential equation (2) and the initial conditions.",
  "vars": [
    "x",
    "y",
    "f",
    "t"
  ],
  "params": [
    "n",
    "k",
    "a",
    "\\\\delta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "xcoordv",
        "y": "yfunction",
        "f": "forcingfx",
        "t": "paramtime",
        "n": "orderint",
        "k": "indexstep",
        "a": "taylorpt",
        "\\delta": "diffop"
      },
      "question": "3. Find an integral formula for the solution of the differential equation\n\\[\ndiffop(diffop-1)(diffop-2) \\cdots(diffop-orderint+1)\\,yfunction = forcingfx(xcoordv), \\quad xcoordv \\geq 1\n\\]\nfor \\( yfunction \\) as a function of \\( xcoordv \\) satisfying the initial conditions \\( yfunction(1)=yfunction^{\\prime}(1)=\\ldots= yfunction^{(orderint-1)}(1)=0 \\), where \\( forcingfx \\) is continuous and\n\\[\ndiffop \\equiv xcoordv \\frac{d}{d xcoordv}\n\\]",
      "solution": "Solution. We first show that\n\\[\ndiffop(diffop-1)(diffop-2) \\cdots(diffop-orderint+1)\\,yfunction = xcoordv^{orderint} \\frac{d^{orderint} yfunction}{d xcoordv^{orderint}}\n\\]\nWe prove this by induction on \\( orderint \\). It is true for \\( orderint = 1 \\). Assume (1) is true for \\( orderint = indexstep \\). Since polynomials in the operator \\( diffop \\) commute with one another, we have\n\\[\n\\begin{aligned}\ndiffop(diffop-1)(diffop-2) \\cdots(diffop-indexstep & +1)(diffop-indexstep)\\,yfunction \\\\\n& =(diffop-indexstep)\\,diffop(diffop-1)(diffop-2) \\cdots(diffop-indexstep+1)\\,yfunction \\\\\n& =\\left(xcoordv \\frac{d}{d xcoordv}-indexstep\\right)xcoordv^{indexstep} \\frac{d^{indexstep} yfunction}{d xcoordv^{indexstep}} \\\\\n& =xcoordv^{indexstep+1} \\frac{d^{indexstep+1} yfunction}{d xcoordv^{indexstep+1}}\n\\end{aligned}\n\\]\nThus (1) is proved by induction for all \\( orderint \\).\n\nThe differential equation is thus\n\\[\nxcoordv^{orderint} yfunction^{(orderint)} = forcingfx(xcoordv), \\quad xcoordv \\geq 1\n\\]\nand the solution can obviously be obtained by applying the integral operator \\( \\int_{1}^{xcoordv} orderint \\) times to the function \\( forcingfx(xcoordv) xcoordv^{-orderint} \\). However, it is possible to collapse the \\( orderint \\)-fold integration to a single integration as follows:\n\nOne of the standard forms of Taylor's theorem is\n\\[\n\\begin{aligned}\nyfunction(xcoordv)=yfunction(taylorpt)&+(xcoordv-taylorpt)\\,yfunction^{\\prime}(taylorpt)+\\cdots+\\frac{(xcoordv-taylorpt)^{orderint-1}}{(orderint-1)!}\\,yfunction^{(orderint-1)}(taylorpt) \\\\\n&+\\int_{taylorpt}^{xcoordv} \\frac{(xcoordv-paramtime)^{orderint-1}}{(orderint-1)!}\\,yfunction^{(orderint)}(paramtime)\\,d\\,paramtime\n\\end{aligned}\n\\]\nif \\( yfunction^{(orderint)} \\) is continuous. (See, for example, Thomas, Calculus and Analytic Geometry, alternate ed., Addison-Wesley, Reading, Mass., 1972, page 814.) In this case, taking \\( taylorpt = 1 \\) and using (2) and the initial conditions, we have\n\\[\nyfunction(xcoordv)=\\int_{1}^{xcoordv} \\frac{(xcoordv-paramtime)^{orderint-1}}{(orderint-1)!}\\,\\frac{forcingfx(paramtime)}{paramtime^{orderint}}\\,d\\,paramtime\n\\]\nIt is easy to check by direct differentiation that the function defined by this integral does indeed satisfy the differential equation (2) and the initial conditions."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "wanderlust",
        "y": "marigold",
        "f": "sandstone",
        "t": "blueberry",
        "n": "afterglow",
        "k": "buttercup",
        "a": "crossroads",
        "\\\\delta": "moonlight"
      },
      "question": "3. Find an integral formula for the solution of the differential equation\n\\[\nmoonlight(moonlight-1)(moonlight-2) \\cdots(moonlight-afterglow+1) marigold=sandstone(wanderlust), \\quad wanderlust \\geq 1\n\\]\nfor \\( marigold \\) as a function of \\( wanderlust \\) satisfying the initial conditions \\( marigold(1)=marigold^{\\prime}(1)=\\ldots= marigold^{(afterglow-1)}(1)=0 \\), where \\( sandstone \\) is continuous and\n\\[\nmoonlight \\equiv wanderlust \\frac{d}{d wanderlust}\n\\]",
      "solution": "Solution. We first show that\n\\[\nmoonlight(moonlight-1)(moonlight-2) \\cdots(moonlight-afterglow+1) marigold=wanderlust^{afterglow} \\frac{d^{afterglow} marigold}{d wanderlust^{afterglow}}\n\\]\n\nWe prove this by induction on \\( afterglow \\). It is true for \\( afterglow=1 \\). Assume (1) is true for \\( afterglow=buttercup \\). Since polynomials in the operator \\( moonlight \\) commute with one another, we have\n\\[\n\\begin{aligned}\nmoonlight(moonlight-1)(moonlight-2) \\cdots(moonlight-buttercup & +1)(moonlight-buttercup) marigold \\\\\n& =(moonlight-buttercup) moonlight(moonlight-1)(moonlight-2) \\cdots(moonlight-buttercup+1) marigold \\\\\n& =\\left(wanderlust \\frac{d}{d wanderlust}-buttercup\\right) wanderlust^{buttercup} \\frac{d^{buttercup} marigold}{d wanderlust^{buttercup}} \\\\\n& =wanderlust^{buttercup+1} \\frac{d^{buttercup+1} marigold}{d wanderlust^{buttercup+1}}\n\\end{aligned}\n\\]\n\nThus (1) is proved by induction for all \\( afterglow \\).\nThe differential equation is thus\n\\[\nwanderlust^{afterglow} marigold^{(afterglow)}=sandstone(wanderlust), \\quad wanderlust \\geq 1\n\\]\nand the solution can obviously be obtained by applying the integral operator \\( \\int_{1}^{wanderlust} \\) afterglow times to the function \\( sandstone(wanderlust) wanderlust^{-afterglow} \\). However, it is possible to collapse the afterglow-fold integration to a single integration as follows:\n\nOne of the standard forms of Taylor's theorem is\n\\[\n\\begin{aligned}\nmarigold(wanderlust)=marigold(crossroads)+(wanderlust-crossroads) marigold^{\\prime}(crossroads)+\\cdots+\\frac{(wanderlust-crossroads)^{afterglow-1}}{(afterglow-1)!} & marigold^{(afterglow-1)}(crossroads) \\\\\n& +\\int_{crossroads}^{wanderlust} \\frac{(wanderlust-blueberry)^{afterglow-1}}{(afterglow-1)!} marigold^{(afterglow)}(blueberry) d blueberry\n\\end{aligned}\n\\]\nif \\( marigold^{(afterglow)} \\) is continuous. (See, for example, Thomas, Calculus and Analytic Geometry, alternate ed., Addison-Wesley, Reading, Mass., 1972, page 814.) In this case, taking \\( crossroads=1 \\) and using (2) and the initial conditions, we have\n\\[\nmarigold(wanderlust)=\\int_{1}^{wanderlust} \\frac{(wanderlust-blueberry)^{afterglow-1}}{(afterglow-1)!} \\cdot \\frac{sandstone(blueberry)}{blueberry^{afterglow}} d blueberry\n\\]\n\nIt is easy to check by direct differentiation that the function defined by this integral does indeed satisfy the differential equation (2) and the initial conditions."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantv",
        "y": "inputdata",
        "f": "staticval",
        "t": "spaceaxis",
        "n": "uncountable",
        "k": "endindex",
        "a": "targetpt",
        "\\delta": "stability"
      },
      "question": "3. Find an integral formula for the solution of the differential equation\n\\[\nstability(stability-1)(stability-2) \\cdots(stability-uncountable+1) inputdata=staticval(constantv), \\quad constantv \\geq 1\n\\]\nfor \\( inputdata \\) as a function of \\( constantv \\) satisfying the initial conditions \\( inputdata(1)=inputdata^{\\prime}(1)=\\ldots= \\) \\( inputdata^{(uncountable-1)}(1)=0 \\), where \\( staticval \\) is continuous and\n\\[\nstability \\equiv constantv \\frac{d}{d constantv}\n\\]",
      "solution": "Solution. We first show that\n\\[\nstability(stability-1)(stability-2) \\cdots(stability-uncountable+1) inputdata=constantv^{uncountable} \\frac{d^{uncountable} inputdata}{d constantv^{uncountable}}\n\\]\nWe prove this by induction on \\( uncountable \\). It is true for \\( uncountable=1 \\). Assume (1) is true for \\( uncountable=endindex \\). Since polynomials in the operator \\( stability \\) commute with one another, we have\n\\[\n\\begin{aligned}\nstability(stability-1)(stability-2) \\cdots(stability-endindex & +1)(stability-endindex) inputdata \\\\\n& =(stability-endindex) stability(stability-1)(stability-2) \\cdots(stability-endindex+1) inputdata \\\\\n& =\\left(constantv \\frac{d}{d constantv}-endindex\\right) constantv^{endindex} \\frac{d^{endindex} inputdata}{d constantv^{endindex}} \\\\\n& =constantv^{endindex+1} \\frac{d^{endindex+1} inputdata}{d constantv^{endindex+1}}\n\\end{aligned}\n\\]\nThus (1) is proved by induction for all \\( uncountable \\).\nThe differential equation is thus\n\\[\nconstantv^{uncountable} inputdata^{(uncountable)}=staticval(constantv), \\quad constantv \\geq 1\n\\]\nand the solution can obviously be obtained by applying the integral operator \\( \\int_{1}^{constantv} uncountable \\) times to the function \\( staticval(constantv) constantv^{-uncountable} \\). However, it is possible to collapse the \\( uncountable \\)-fold integration to a single integration as follows:\n\nOne of the standard forms of Taylor's theorem is\n\\[\n\\begin{aligned}\ninputdata(constantv)=inputdata(targetpt)+(constantv-targetpt) inputdata^{\\prime}(targetpt)+\\cdots+\\frac{(constantv-targetpt)^{uncountable-1}}{(uncountable-1)!} & inputdata^{(uncountable-1)}(targetpt) \\\\\n& +\\int_{targetpt}^{constantv} \\frac{(constantv-spaceaxis)^{uncountable-1}}{(uncountable-1)!} inputdata^{(uncountable)}(spaceaxis) d spaceaxis\n\\end{aligned}\n\\]\nif \\( inputdata^{(uncountable)} \\) is continuous. (See, for example, Thomas, Calculus and Analytic Geometry, alternate ed., Addison-Wesley, Reading, Mass., 1972, page 814.) In this case, taking \\( targetpt=1 \\) and using (2) and the initial conditions, we have\n\\[\ninputdata(constantv)=\\int_{1}^{constantv} \\frac{(constantv-spaceaxis)^{uncountable-1}}{(uncountable-1)!} \\cdot \\frac{staticval(spaceaxis)}{spaceaxis^{uncountable}} d spaceaxis\n\\]\nIt is easy to check by direct differentiation that the function defined by this integral does indeed satisfy the differential equation (2) and the initial conditions."
    },
    "garbled_string": {
      "map": {
        "x": "vdlmkuza",
        "y": "wcntoqrb",
        "f": "pzejxlym",
        "t": "sguvbrki",
        "n": "khqrezop",
        "k": "abdyeuql",
        "a": "mnvrstci",
        "\\delta": "fyhjqska"
      },
      "question": "3. Find an integral formula for the solution of the differential equation\n\\[\nfyhjqska(fyhjqska-1)(fyhjqska-2) \\cdots(fyhjqska-khqrezop+1) wcntoqrb=pzejxlym(vdlmkuza), \\quad vdlmkuza \\geq 1\n\\]\nfor \\( wcntoqrb \\) as a function of \\( vdlmkuza \\) satisfying the initial conditions \\( wcntoqrb(1)=wcntoqrb^{\\prime}(1)=\\ldots= \\) \\( wcntoqrb^{(khqrezop-1)}(1)=0 \\), where \\( pzejxlym \\) is continuous and\n\\[\nfyhjqska \\equiv vdlmkuza \\frac{d}{d vdlmkuza}\n\\]",
      "solution": "Solution. We first show that\n\\[\nfyhjqska(fyhjqska-1)(fyhjqska-2) \\cdots(fyhjqska-khqrezop+1) wcntoqrb=vdlmkuza^{khqrezop} \\frac{d^{khqrezop} wcntoqrb}{d vdlmkuza^{khqrezop}}\n\\]\n\nWe prove this by induction on \\( khqrezop \\). It is true for \\( khqrezop=1 \\). Assume (1) is true for \\( khqrezop=abdyeuql \\). Since polynomials in the operator \\( fyhjqska \\) commute with one another, we have\n\\[\n\\begin{aligned}\nfyhjqska(fyhjqska-1)(fyhjqska-2) \\cdots(fyhjqska-abdyeuql & +1)(fyhjqska-abdyeuql) wcntoqrb \\\\\n& =(fyhjqska-abdyeuql) fyhjqska(fyhjqska-1)(fyhjqska-2) \\cdots(fyhjqska-abdyeuql+1) wcntoqrb \\\\\n& =\\left(vdlmkuza \\frac{d}{d vdlmkuza}-abdyeuql\\right) vdlmkuza^{abdyeuql} \\frac{d^{abdyeuql} wcntoqrb}{d vdlmkuza^{abdyeuql}} \\\\\n& =vdlmkuza^{abdyeuql+1} \\frac{d^{abdyeuql+1} wcntoqrb}{d vdlmkuza^{abdyeuql+1}}\n\\end{aligned}\n\\]\n\nThus (1) is proved by induction for all \\( khqrezop \\).\nThe differential equation is thus\n\\[\nvdlmkuza^{khqrezop} wcntoqrb^{(khqrezop)}=pzejxlym(vdlmkuza), \\quad vdlmkuza \\geq 1\n\\]\nand the solution can obviously be obtained by applying the integral operator \\( \\int_{1}^{vdlmkuza} \\) khqrezop times to the function \\( pzejxlym(vdlmkuza) vdlmkuza^{-khqrezop} \\). However, it is possible to collapse the khqrezop-fold integration to a single integration as follows:\n\nOne of the standard forms of Taylor's theorem is\n\\[\n\\begin{aligned}\nwcntoqrb(vdlmkuza)=wcntoqrb(mnvrstci)+(vdlmkuza-mnvrstci) wcntoqrb^{\\prime}(mnvrstci)+\\cdots+\\frac{(vdlmkuza-mnvrstci)^{khqrezop-1}}{(khqrezop-1)!} & wcntoqrb^{(khqrezop-1)}(mnvrstci) \\\\\n& +\\int_{mnvrstci}^{vdlmkuza} \\frac{(vdlmkuza-sguvbrki)^{khqrezop-1}}{(khqrezop-1)!} wcntoqrb^{(khqrezop)}(sguvbrki) d sguvbrki\n\\end{aligned}\n\\]\nif \\( wcntoqrb^{(khqrezop)} \\) is continuous. (See, for example, Thomas, Calculus and Analytic Geometry, alternate ed., Addison-Wesley, Reading, Mass., 1972, page 814.) In this case, taking \\( mnvrstci=1 \\) and using (2) and the initial conditions, we have\n\\[\nwcntoqrb(vdlmkuza)=\\int_{1}^{vdlmkuza} \\frac{(vdlmkuza-sguvbrki)^{khqrezop-1}}{(khqrezop-1)!} \\cdot \\frac{pzejxlym(sguvbrki)}{sguvbrki^{khqrezop}} d sguvbrki\n\\]\n\nIt is easy to check by direct differentiation that the function defined by this integral does indeed satisfy the differential equation (2) and the initial conditions."
    },
    "kernel_variant": {
      "question": "Let $d\\ge 2$ and $n\\ge 1$ be fixed integers.  \nInside the positive orthant  \n\\[\n\\Omega:=\\{x\\in(0,\\infty)^{d}\\;:\\;\\|x\\|:=(x_{1}^{2}+\\dots +x_{d}^{2})^{1/2}\\ge 2\\}\\subset\\mathbb R^{d},\n\\]\nintroduce the smooth polar splitting  \n\\[\n\\rho(x):=\\|x\\|,\\qquad w(x):=\\frac{x}{\\|x\\|}\\quad (\\text{so }w(x)\\in\\Sigma),\\qquad \n\\Sigma:=\\{w\\in(0,\\infty)^{d}\\;:\\;\\|w\\|=1\\}.\n\\]\n\nDenote by  \n\\[\n\\Delta:=x_{1}\\frac{\\partial}{\\partial x_{1}}+\\dots +x_{d}\\frac{\\partial}{\\partial x_{d}}\n\\]\nthe $d$–dimensional Euler (dilation) operator.  \n\nAssume that  \n\\[\nh\\in C^{\\,n}(\\Omega)\n\\]\nis given.  Determine an explicit \\emph{single one–dimensional} definite–integral formula for the unique function  \n\\[\ny\\in C^{\\,n}(\\Omega)\n\\]\nthat satisfies the \\emph{Euler poly–PDE}  \n\\[\n\\boxed{\\;\n\\Delta(\\Delta-1)(\\Delta-2)\\dots(\\Delta-n+1)\\,y(x)=h(x)\\quad (x\\in\\Omega)\n\\;}\n\\tag{1}\n\\]\ntogether with the \\emph{radial trace conditions}  \n\\[\n\\boxed{\\;\n\\bigl(\\Delta^{\\,k}y\\bigr)(2w)=0\\quad\\forall\\,w\\in\\Sigma,\\;k=0,1,\\dots ,n-1.\n\\;}\n\\tag{2}\n\\]\n\nYour final formula must contain only a single integral whose kernel is written \\emph{completely explicitly} (no iterated or implicit integrals).\n\n--------------------------------------------------------------------",
      "solution": "Throughout, $d,n,\\Omega,\\rho,w,\\Sigma$ and $\\Delta$ are as in the statement.\n\n\\smallskip\n\\textbf{Step 1. Reduction to one dimension.}  \nFix $w\\in\\Sigma$ and consider the ray  \n\\[\nR_{w}:=\\{rw\\;:\\;r\\ge 2\\}.\n\\]\nWrite each point on $R_{w}$ in logarithmic form\n\\[\nx=rw=2e^{\\sigma}w,\\qquad \\sigma:=\\ln\\frac{r}{2}\\ge 0. \\tag{3}\n\\]\n\nFor any $C^{1}$–function $F$ the chain rule yields  \n\\[\n\\frac{d}{d\\sigma}F\\!\\bigl(2e^{\\sigma}w\\bigr)=\n\\sum_{j=1}^{d}\\partial_{j}F\\!\\bigl(2e^{\\sigma}w\\bigr)\\,2e^{\\sigma}w_{j}\n=\\bigl(x\\cdot\\nabla F\\bigr)(x)=\\Delta F(x). \\tag{4}\n\\]\nHence, along each ray the Euler operator acts as the ordinary derivative  \n\\[\nD:=\\frac{d}{d\\sigma}.\n\\]\nPut   \n\\[\nL_{n}:=D(D-1)\\dots(D-n+1). \\tag{5}\n\\]\n\nDefine one–variable functions  \n\\[\nY_{w}(\\sigma):=y\\bigl(2e^{\\sigma}w\\bigr),\\qquad\nH_{w}(\\sigma):=h\\bigl(2e^{\\sigma}w\\bigr). \\tag{6}\n\\]\nWith \\eqref{4}, the PDE \\eqref{1} becomes, for each fixed $w$,\n\\[\nL_{n}Y_{w}(\\sigma)=H_{w}(\\sigma)\\qquad(\\sigma\\ge 0). \\tag{7}\n\\]\n\n\\smallskip\n\\textbf{Step 2. Initial data transported from \\eqref{2}.}  \nBy \\eqref{6} the trace conditions give  \n\\[\nY_{w}^{(k)}(0)=0\\qquad(k=0,1,\\dots ,n-1). \\tag{8}\n\\]\n\n\\smallskip\n\\textbf{Step 3. Construction and \\emph{correct} proof of the Green kernel.}  \n\nFor $s\\in\\mathbb R$ set  \n\\[\ng_{n}(s):=H(s)\\,\\frac{\\bigl(e^{s}-1\\bigr)^{\\,n-1}}{(n-1)!}, \\qquad \nH(s)=\\begin{cases}0,&s<0,\\\\ 1,&s>0.\\end{cases} \\tag{9}\n\\]\n\n\\emph{Auxiliary identity.} For every $n\\ge 1$ one has, in the sense of\ndistributions on $\\mathbb R$,\n\\[\n\\boxed{\\,L_{n}g_{n}=\\delta_{0}\\,}. \\tag{10}\n\\]\n\nWe supply a rigorous proof that avoids the incorrect integration–by–parts\nargument criticised in the review.\n\n\\medskip\n\\emph{Lemma.} For every $n\\ge 1$,\n\\[\n(D-n)\\,g_{\\,n+1}=g_{n}\\qquad\\text{in }\\mathscr D'(\\mathbb R). \\tag{11}\n\\]\n\n\\emph{Proof of the Lemma.}  \nBecause $g_{\\,n+1}$ is supported on $[0,\\infty)$ we may restrict to that\nhalf–line.  On $(0,\\infty)$ put $u(s):=e^{s}-1$.  Then\n$g_{\\,n+1}(s)=u(s)^{n}/n!$ and\n\\[\nDg_{\\,n+1}\n=\\frac{d}{ds}\\bigl(u^{n}/n!\\bigr)\n=\\frac{n}{n!}\\,e^{s}\\,u^{\\,n-1}\n=\\frac{1}{(n-1)!}\\bigl(u+1\\bigr)u^{\\,n-1}.\n\\]\nConsequently\n\\[\n(D-n)g_{\\,n+1}\n=\\frac{1}{(n-1)!}\\Bigl[(u+1)u^{\\,n-1}-u^{\\,n}\\Bigr]\n=\\frac{u^{\\,n-1}}{(n-1)!}=g_{n}\\quad(s>0).\n\\]\nAt $s<0$ both sides of \\eqref{11} vanish, and at $s=0$ no additional\n$\\delta$–terms appear because $u^{\\,n}$ vanishes to order $n$ at $s=0$,\nso the multiplication by $H(s)$ kills the derivative of $H$ arising from\n$D$.  Hence \\eqref{11} holds distributionally. \\hfill$\\square$\n\n\\medskip\n\\emph{Induction proof of \\eqref{10}.}  \nFor $n=1$ we have $g_{1}=H$ and $L_{1}=D$, so $L_{1}g_{1}=DH=\\delta_{0}$.\n\nAssume $L_{n}g_{n}=\\delta_{0}$ for some $n\\ge 1$.  \nUsing \\eqref{11} one obtains\n\\[\nL_{\\,n+1}g_{\\,n+1}\n=D(D-1)\\dots(D-n)\\,g_{\\,n+1}\n=L_{n}\\bigl[(D-n)g_{\\,n+1}\\bigr]\n=L_{n}g_{n}\n=\\delta_{0}.\n\\]\nThus \\eqref{10} holds for all $n$ by induction. \\hfill$\\square$\n\n\\medskip\nFrom \\eqref{9} we also record the initial behaviour  \n\\[\ng_{n}^{(k)}(0^{+})=\n\\begin{cases}\n0,&k<n-1,\\\\\n1,&k=n-1.\n\\end{cases}\n\\tag{12}\n\\]\n\n\\smallskip\n\\textbf{Step 4. Solving the ODE \\eqref{7}–\\eqref{8}.}  \nBecause the data \\eqref{12} coincide exactly with \\eqref{8},\nthe causal Green representation provides, for $\\sigma\\ge 0$,\n\\[\nY_{w}(\\sigma)=\\int_{0}^{\\sigma}g_{n}(\\sigma-\\tau)\\,H_{w}(\\tau)\\,d\\tau. \\tag{13}\n\\]\n\n\\smallskip\n\\textbf{Step 5. Rewriting the one–dimensional integral in $(t,x)$–variables.}  \nPut $\\tau=\\ln\\frac{t}{2}$ so that $t=2e^{\\tau}$ and $d\\tau=\\frac{dt}{t}$;\nsimilarly $\\sigma=\\ln\\frac{r}{2}$ with $r=\\rho(x)$.  Substituting\n\\eqref{13} gives, for $x=rw$,\n\\[\ny(rw)=\\int_{2}^{r}K_{n}\\!\\Bigl(\\frac{r}{t}\\Bigr)\\,\n       h(tw)\\,\\frac{dt}{t},\\tag{14}\n\\]\nwhere the \\emph{explicit single–variable kernel} is\n\\[\nK_{n}(z)=\\frac{(z-1)^{\\,n-1}}{(n-1)!},\\qquad z\\ge 1. \\tag{15}\n\\]\n\nFinally, writing $r=\\rho(x)$ and $w=w(x)$ for arbitrary $x\\in\\Omega$\nyields the announced formula\n\\[\n\\boxed{\\;\n y(x)=\\frac{1}{(n-1)!}\\int_{2}^{\\rho(x)}\n \\Bigl(\\frac{\\rho(x)}{t}-1\\Bigr)^{\\,n-1}\\,\n h\\!\\bigl(t\\,w(x)\\bigr)\\,\\frac{dt}{t}\n\\;}. \\tag{16}\n\\]\n\n\\smallskip\n\\textbf{Step 6. Verification.}\n\n(a) \\emph{PDE \\eqref{1}.}  \nFor fixed $w$ formula \\eqref{16} coincides with \\eqref{13}; applying\n$L_{n}$ in the $\\sigma$–variable and using \\eqref{10} gives \\eqref{7},\nhence \\eqref{1}.\n\n(b) \\emph{Boundary conditions \\eqref{2}.}  \nThe initial behaviour \\eqref{12} fed into \\eqref{13} gives\n$Y_{w}^{(k)}(0)=0$ for $k<n$, i.e.\\ \\eqref{2}.\n\n(c) \\emph{Regularity.}  \nBecause $h\\in C^{\\,n}(\\Omega)$ and $\\rho(x)\\ge 2$,\nstandard differentiation under the integral sign shows\n$y\\in C^{\\,n}(\\Omega)$.\n\n(d) \\emph{Uniqueness.}  \nFor each ray $R_{w}$ the inhomogeneous ODE \\eqref{7} with data \\eqref{8}\nhas a unique $C^{\\,n}$ solution; two\n$C^{\\,n}$ solutions of \\eqref{1},\\eqref{2} therefore coincide on every\nray and hence on all of $\\Omega$.\n\n\\smallskip\n\\textbf{Conclusion.}  \nFormula \\eqref{16} with kernel \\eqref{15} furnishes the unique classical\n$C^{\\,n}$ solution of the higher–dimensional boundary–value problem\n\\eqref{1}–\\eqref{2}.  The $n$–fold operation has been reduced to a\n\\emph{single} explicitly written integral, and the previously faulty\nStep~3 has been rigorously rectified.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.541856",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the problem is promoted from an ordinary differential equation on [2,∞) to a partial differential equation on the d-dimensional positive orthant.  \n2. Additional structure: the total Euler operator couples all variables; solving requires decomposing ℝ⁺ᵈ into dilation “rays’’ (radial-angular coordinates) and showing that Δ becomes ∂/∂σ on those rays.  \n3. Multiple interacting concepts: one must blend PDE theory, change-of-variables in several dimensions, and the classical one-variable Euler identity to collapse an n-fold integral to a single integral.  \n4. Deeper theory: the proof hinges on understanding the orbits of the multiplicative group action, showing uniqueness via a family of parameterised ODEs, and carefully translating boundary data between coordinate systems.  \n5. Greater computational load: turning (1) into (6), integrating n times with weighted kernels, and re-expressing the result back in the original variables all add substantive algebraic and conceptual steps absent from the original one-dimensional problem.  \n\nIn sum, the enhanced variant extends the core “Euler-type operator with shifted factors’’ idea into a genuinely multidimensional setting with non-trivial geometry and PDE techniques, making it significantly harder than both the original and the previous kernel versions."
      }
    },
    "original_kernel_variant": {
      "question": "Let $d\\ge 2$ and $n\\ge 1$ be fixed integers.  \nInside the positive orthant  \n\\[\n\\Omega:=\\{x\\in(0,\\infty)^{d}\\;:\\;\\|x\\|:=(x_{1}^{2}+\\dots +x_{d}^{2})^{1/2}\\ge 2\\}\\subset\\mathbb R^{d},\n\\]\nintroduce the smooth polar splitting  \n\\[\n\\rho(x):=\\|x\\|,\\qquad w(x):=\\frac{x}{\\|x\\|}\\quad (\\text{so }w(x)\\in\\Sigma),\\qquad \n\\Sigma:=\\{w\\in(0,\\infty)^{d}\\;:\\;\\|w\\|=1\\}.\n\\]\n\nDenote by  \n\\[\n\\Delta:=x_{1}\\frac{\\partial}{\\partial x_{1}}+\\dots +x_{d}\\frac{\\partial}{\\partial x_{d}}\n\\]\nthe $d$–dimensional Euler (dilation) operator.  \n\nAssume that  \n\\[\nh\\in C^{\\,n}(\\Omega)\n\\]\nis given.  Determine an explicit \\emph{single one–dimensional} definite–integral formula for the unique function  \n\\[\ny\\in C^{\\,n}(\\Omega)\n\\]\nthat satisfies the \\emph{Euler poly–PDE}  \n\\[\n\\boxed{\\;\n\\Delta(\\Delta-1)(\\Delta-2)\\dots(\\Delta-n+1)\\,y(x)=h(x)\\quad (x\\in\\Omega)\n\\;}\n\\tag{1}\n\\]\ntogether with the \\emph{radial trace conditions}  \n\\[\n\\boxed{\\;\n\\bigl(\\Delta^{\\,k}y\\bigr)(2w)=0\\quad\\forall\\,w\\in\\Sigma,\\;k=0,1,\\dots ,n-1.\n\\;}\n\\tag{2}\n\\]\n\nYour final formula must contain only a single integral whose kernel is written \\emph{completely explicitly} (no iterated or implicit integrals).\n\n--------------------------------------------------------------------",
      "solution": "Throughout, $d,n,\\Omega,\\rho,w,\\Sigma$ and $\\Delta$ are as in the statement.\n\n\\smallskip\n\\textbf{Step 1. Reduction to one dimension.}  \nFix $w\\in\\Sigma$ and consider the ray  \n\\[\nR_{w}:=\\{rw\\;:\\;r\\ge 2\\}.\n\\]\nWrite each point on $R_{w}$ in logarithmic form\n\\[\nx=rw=2e^{\\sigma}w,\\qquad \\sigma:=\\ln\\frac{r}{2}\\ge 0. \\tag{3}\n\\]\n\nFor any $C^{1}$–function $F$ the chain rule yields  \n\\[\n\\frac{d}{d\\sigma}F\\!\\bigl(2e^{\\sigma}w\\bigr)=\n\\sum_{j=1}^{d}\\partial_{j}F\\!\\bigl(2e^{\\sigma}w\\bigr)\\,2e^{\\sigma}w_{j}\n=\\bigl(x\\cdot\\nabla F\\bigr)(x)=\\Delta F(x). \\tag{4}\n\\]\nHence, along each ray the Euler operator acts as the ordinary derivative  \n\\[\nD:=\\frac{d}{d\\sigma}.\n\\]\nPut   \n\\[\nL_{n}:=D(D-1)\\dots(D-n+1). \\tag{5}\n\\]\n\nDefine one–variable functions  \n\\[\nY_{w}(\\sigma):=y\\bigl(2e^{\\sigma}w\\bigr),\\qquad\nH_{w}(\\sigma):=h\\bigl(2e^{\\sigma}w\\bigr). \\tag{6}\n\\]\nWith \\eqref{4}, the PDE \\eqref{1} becomes, for each fixed $w$,\n\\[\nL_{n}Y_{w}(\\sigma)=H_{w}(\\sigma)\\qquad(\\sigma\\ge 0). \\tag{7}\n\\]\n\n\\smallskip\n\\textbf{Step 2. Initial data transported from \\eqref{2}.}  \nBy \\eqref{6} the trace conditions give  \n\\[\nY_{w}^{(k)}(0)=0\\qquad(k=0,1,\\dots ,n-1). \\tag{8}\n\\]\n\n\\smallskip\n\\textbf{Step 3. Construction and \\emph{correct} proof of the Green kernel.}  \n\nFor $s\\in\\mathbb R$ set  \n\\[\ng_{n}(s):=H(s)\\,\\frac{\\bigl(e^{s}-1\\bigr)^{\\,n-1}}{(n-1)!}, \\qquad \nH(s)=\\begin{cases}0,&s<0,\\\\ 1,&s>0.\\end{cases} \\tag{9}\n\\]\n\n\\emph{Auxiliary identity.} For every $n\\ge 1$ one has, in the sense of\ndistributions on $\\mathbb R$,\n\\[\n\\boxed{\\,L_{n}g_{n}=\\delta_{0}\\,}. \\tag{10}\n\\]\n\nWe supply a rigorous proof that avoids the incorrect integration–by–parts\nargument criticised in the review.\n\n\\medskip\n\\emph{Lemma.} For every $n\\ge 1$,\n\\[\n(D-n)\\,g_{\\,n+1}=g_{n}\\qquad\\text{in }\\mathscr D'(\\mathbb R). \\tag{11}\n\\]\n\n\\emph{Proof of the Lemma.}  \nBecause $g_{\\,n+1}$ is supported on $[0,\\infty)$ we may restrict to that\nhalf–line.  On $(0,\\infty)$ put $u(s):=e^{s}-1$.  Then\n$g_{\\,n+1}(s)=u(s)^{n}/n!$ and\n\\[\nDg_{\\,n+1}\n=\\frac{d}{ds}\\bigl(u^{n}/n!\\bigr)\n=\\frac{n}{n!}\\,e^{s}\\,u^{\\,n-1}\n=\\frac{1}{(n-1)!}\\bigl(u+1\\bigr)u^{\\,n-1}.\n\\]\nConsequently\n\\[\n(D-n)g_{\\,n+1}\n=\\frac{1}{(n-1)!}\\Bigl[(u+1)u^{\\,n-1}-u^{\\,n}\\Bigr]\n=\\frac{u^{\\,n-1}}{(n-1)!}=g_{n}\\quad(s>0).\n\\]\nAt $s<0$ both sides of \\eqref{11} vanish, and at $s=0$ no additional\n$\\delta$–terms appear because $u^{\\,n}$ vanishes to order $n$ at $s=0$,\nso the multiplication by $H(s)$ kills the derivative of $H$ arising from\n$D$.  Hence \\eqref{11} holds distributionally. \\hfill$\\square$\n\n\\medskip\n\\emph{Induction proof of \\eqref{10}.}  \nFor $n=1$ we have $g_{1}=H$ and $L_{1}=D$, so $L_{1}g_{1}=DH=\\delta_{0}$.\n\nAssume $L_{n}g_{n}=\\delta_{0}$ for some $n\\ge 1$.  \nUsing \\eqref{11} one obtains\n\\[\nL_{\\,n+1}g_{\\,n+1}\n=D(D-1)\\dots(D-n)\\,g_{\\,n+1}\n=L_{n}\\bigl[(D-n)g_{\\,n+1}\\bigr]\n=L_{n}g_{n}\n=\\delta_{0}.\n\\]\nThus \\eqref{10} holds for all $n$ by induction. \\hfill$\\square$\n\n\\medskip\nFrom \\eqref{9} we also record the initial behaviour  \n\\[\ng_{n}^{(k)}(0^{+})=\n\\begin{cases}\n0,&k<n-1,\\\\\n1,&k=n-1.\n\\end{cases}\n\\tag{12}\n\\]\n\n\\smallskip\n\\textbf{Step 4. Solving the ODE \\eqref{7}–\\eqref{8}.}  \nBecause the data \\eqref{12} coincide exactly with \\eqref{8},\nthe causal Green representation provides, for $\\sigma\\ge 0$,\n\\[\nY_{w}(\\sigma)=\\int_{0}^{\\sigma}g_{n}(\\sigma-\\tau)\\,H_{w}(\\tau)\\,d\\tau. \\tag{13}\n\\]\n\n\\smallskip\n\\textbf{Step 5. Rewriting the one–dimensional integral in $(t,x)$–variables.}  \nPut $\\tau=\\ln\\frac{t}{2}$ so that $t=2e^{\\tau}$ and $d\\tau=\\frac{dt}{t}$;\nsimilarly $\\sigma=\\ln\\frac{r}{2}$ with $r=\\rho(x)$.  Substituting\n\\eqref{13} gives, for $x=rw$,\n\\[\ny(rw)=\\int_{2}^{r}K_{n}\\!\\Bigl(\\frac{r}{t}\\Bigr)\\,\n       h(tw)\\,\\frac{dt}{t},\\tag{14}\n\\]\nwhere the \\emph{explicit single–variable kernel} is\n\\[\nK_{n}(z)=\\frac{(z-1)^{\\,n-1}}{(n-1)!},\\qquad z\\ge 1. \\tag{15}\n\\]\n\nFinally, writing $r=\\rho(x)$ and $w=w(x)$ for arbitrary $x\\in\\Omega$\nyields the announced formula\n\\[\n\\boxed{\\;\n y(x)=\\frac{1}{(n-1)!}\\int_{2}^{\\rho(x)}\n \\Bigl(\\frac{\\rho(x)}{t}-1\\Bigr)^{\\,n-1}\\,\n h\\!\\bigl(t\\,w(x)\\bigr)\\,\\frac{dt}{t}\n\\;}. \\tag{16}\n\\]\n\n\\smallskip\n\\textbf{Step 6. Verification.}\n\n(a) \\emph{PDE \\eqref{1}.}  \nFor fixed $w$ formula \\eqref{16} coincides with \\eqref{13}; applying\n$L_{n}$ in the $\\sigma$–variable and using \\eqref{10} gives \\eqref{7},\nhence \\eqref{1}.\n\n(b) \\emph{Boundary conditions \\eqref{2}.}  \nThe initial behaviour \\eqref{12} fed into \\eqref{13} gives\n$Y_{w}^{(k)}(0)=0$ for $k<n$, i.e.\\ \\eqref{2}.\n\n(c) \\emph{Regularity.}  \nBecause $h\\in C^{\\,n}(\\Omega)$ and $\\rho(x)\\ge 2$,\nstandard differentiation under the integral sign shows\n$y\\in C^{\\,n}(\\Omega)$.\n\n(d) \\emph{Uniqueness.}  \nFor each ray $R_{w}$ the inhomogeneous ODE \\eqref{7} with data \\eqref{8}\nhas a unique $C^{\\,n}$ solution; two\n$C^{\\,n}$ solutions of \\eqref{1},\\eqref{2} therefore coincide on every\nray and hence on all of $\\Omega$.\n\n\\smallskip\n\\textbf{Conclusion.}  \nFormula \\eqref{16} with kernel \\eqref{15} furnishes the unique classical\n$C^{\\,n}$ solution of the higher–dimensional boundary–value problem\n\\eqref{1}–\\eqref{2}.  The $n$–fold operation has been reduced to a\n\\emph{single} explicitly written integral, and the previously faulty\nStep~3 has been rigorously rectified.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.449779",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the problem is promoted from an ordinary differential equation on [2,∞) to a partial differential equation on the d-dimensional positive orthant.  \n2. Additional structure: the total Euler operator couples all variables; solving requires decomposing ℝ⁺ᵈ into dilation “rays’’ (radial-angular coordinates) and showing that Δ becomes ∂/∂σ on those rays.  \n3. Multiple interacting concepts: one must blend PDE theory, change-of-variables in several dimensions, and the classical one-variable Euler identity to collapse an n-fold integral to a single integral.  \n4. Deeper theory: the proof hinges on understanding the orbits of the multiplicative group action, showing uniqueness via a family of parameterised ODEs, and carefully translating boundary data between coordinate systems.  \n5. Greater computational load: turning (1) into (6), integrating n times with weighted kernels, and re-expressing the result back in the original variables all add substantive algebraic and conceptual steps absent from the original one-dimensional problem.  \n\nIn sum, the enhanced variant extends the core “Euler-type operator with shifted factors’’ idea into a genuinely multidimensional setting with non-trivial geometry and PDE techniques, making it significantly harder than both the original and the previous kernel versions."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}