{
  "index": "1960-B-7",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "7. Let \\( g(t) \\) and \\( h(t) \\) be real, continuous functions for \\( t \\geq 0 \\). Show that any function \\( v(t) \\) satisfying the differential inequality\n\\[\n\\frac{d v}{d t}+g(t) v \\geq h(t), \\quad v(0)=c\n\\]\nsatisfies the further inequality \\( v(t) \\geq u(t) \\) where\n\\[\n\\frac{d u}{d t}+g(t) u=h(t), \\quad u(0)=c\n\\]\n\nFrom this, conclude that for sufficiently small \\( t>0 \\), the solution of\n\\[\n\\frac{d v}{d t}+g(t) v=v^{2}, \\quad v(0)=c_{1}\n\\]\nmay be written\n\\[\nv=\\max _{w}\\left[c_{1} e^{-\\int_{0}^{t}[g(s)-2 w(s)] d s}-\\int_{0}^{t} e^{-\\int_{0}^{t}\\left[g\\left(s_{1}\\right)-2 w\\left(s_{1}\\right)\\right] d s_{1}} w^{2}(s) d s\\right]\n\\]\nwhere the maximization is over all continuous functions \\( \\boldsymbol{w}(t) \\) defined over some \\( t \\)-interval \\( \\left[0, t_{0}\\right] \\).",
  "solution": "Solution. Let\n\\[\nG(t)=\\int_{0}^{t} g(s) d s .\n\\]\n\nThen \\( G^{\\prime}=g \\). Put \\( x=u e^{G}, y=v e^{G} \\). Then\n\\[\n\\begin{array}{l}\n\\frac{d x}{d t}=\\left(\\frac{d u}{d t}+g u\\right) e^{G}=h e^{G} \\\\\n\\frac{d y}{d t}=\\left(\\frac{d v}{d t}+g v\\right) e^{G} \\geq h e^{G}\n\\end{array}\n\\]\nsince \\( e^{G}>0 \\) for all \\( t \\). Thus \\( (d / d t)(y-x) \\geq 0 \\). Now \\( x(0)=y(0)=c \\), and we conclude \\( x(t) \\leq y(t) \\) and therefore \\( u(t) \\leq v(t) \\) for all \\( t \\geq 0 \\).\n\nFor the second part of the problem, we suppose that \\( v \\) satisfies\n\\[\n\\frac{d v}{d t}+g v=v^{2} \\text { for } t \\in\\left[0, t_{0}\\right], \\quad v(0)=c_{1}\n\\]\nand we let \\( w \\) be any continuous function on \\( \\left[0, t_{0}\\right] \\). Then\n\\[\n\\begin{array}{c}\n(v-w)^{2} \\geq 0 \\quad \\text { for all } t \\\\\n\\frac{d v}{d t}+(g-2 w) v=v^{2}-2 w v \\geq-w^{2}\n\\end{array}\n\\]\n\nIf \\( W(t)=\\int_{0}^{1} w(s) d s \\) and \\( y=v e^{G 2 w} \\) then\n\\[\n\\begin{array}{c}\n\\frac{d y}{d t}=\\left(\\frac{d v}{d t}+(g-2 w) v\\right) e^{G-2 w} \\geq-w^{2} e^{G} 2 w \\\\\ny(t) \\geq c_{1}-\\int_{0}^{\\prime} w^{2}(s) \\exp [G(s)-2 W(s)] d s \\\\\nv(t) \\geq c_{1} \\exp [-G(t)+2 W(t)] \\\\\n-\\int_{0}^{\\prime} \\exp [G(s)-G(t)-2(W(s)-W(t))] w^{2}(s) d s\n\\end{array}\n\\]\n\nSo\n\\[\n\\begin{aligned}\nv(t) \\geq \\max _{w}\\{ & \\left\\{c_{1} \\exp [-G(t)+2 W(t)]\\right. \\\\\n& \\left.-\\int_{0}^{\\prime \\prime} \\exp [G(s)-G(t)-2(W(s)-W(t))] w^{2}(s) d s\\right\\}\n\\end{aligned}\n\\]\n\nNow if we take \\( w=v \\), then all of the preceding inequalities become equalities so we conclude\n\\[\n\\begin{aligned}\nv(t)=\\max _{n^{\\prime}}\\{ & c_{1} \\exp [-G(t)+2 W(t)] \\\\\n& \\left.-\\int_{0}^{t} \\exp [G(s)-G(t)-2(W(s)-W(t))] w^{2}(s) d s\\right\\}\n\\end{aligned}\n\\]\nwhich is equivalent to the form given.\nThe solution of the equation\n\\[\n\\frac{d v}{d t}+g v=v^{2}\n\\]\nsubject to the initial condition \\( v(0)=c_{1} \\), may be unbounded on a finite interval, say \\( v \\rightarrow \\infty \\) as \\( t \\mid t_{1} \\). Then the preceding analysis is valid only for \\( t<t_{1} \\). If we pick any \\( t_{0} \\) with \\( 0<t_{0}<t_{1} \\), then it is valid for [0, \\( \\left.t_{0}\\right] \\).",
  "vars": [
    "t",
    "s",
    "s_1",
    "v",
    "u",
    "w",
    "x",
    "y",
    "G",
    "W"
  ],
  "params": [
    "g",
    "h",
    "c",
    "c_1",
    "t_0",
    "t_1"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "t": "timevar",
        "s": "sample",
        "s_1": "sampleone",
        "v": "statefn",
        "u": "auxfunc",
        "w": "trialfn",
        "x": "transvar",
        "y": "transyvar",
        "G": "accumg",
        "W": "accumw",
        "g": "coeffg",
        "h": "inhomfn",
        "c": "initcst",
        "c_1": "initone",
        "t_0": "boundzero",
        "t_1": "boundone"
      },
      "question": "Let \\( coeffg(timevar) \\) and \\( inhomfn(timevar) \\) be real, continuous functions for \\( timevar \\geq 0 \\). Show that any function \\( statefn(timevar) \\) satisfying the differential inequality\n\\[\n\\frac{d statefn}{d timevar}+coeffg(timevar) statefn \\geq inhomfn(timevar), \\quad statefn(0)=initcst\n\\]\nsatisfies the further inequality \\( statefn(timevar) \\geq auxfunc(timevar) \\) where\n\\[\n\\frac{d auxfunc}{d timevar}+coeffg(timevar) auxfunc=inhomfn(timevar), \\quad auxfunc(0)=initcst\n\\]\n\nFrom this, conclude that for sufficiently small \\( timevar>0 \\), the solution of\n\\[\n\\frac{d statefn}{d timevar}+coeffg(timevar) statefn=statefn^{2}, \\quad statefn(0)=initone\n\\]\nmay be written\n\\[\nstatefn=\\max _{trialfn}\\left[initone e^{-\\int_{0}^{timevar}[coeffg(sample)-2 trialfn(sample)] d sample}-\\int_{0}^{timevar} e^{-\\int_{0}^{timevar}\\left[coeffg\\left(sampleone\\right)-2 trialfn\\left(sampleone\\right)\\right] d sampleone} trialfn^{2}(sample) d sample\\right]\n\\]\nwhere the maximization is over all continuous functions \\( \\boldsymbol{trialfn}(timevar) \\) defined over some \\( timevar \\)-interval \\( \\left[0, boundzero\\right] \\).",
      "solution": "Solution. Let\n\\[\naccumg(timevar)=\\int_{0}^{timevar} coeffg(sample) d sample .\n\\]\n\nThen \\( accumg^{\\prime}=coeffg \\). Put \\( transvar=auxfunc e^{accumg}, \\; transyvar=statefn e^{accumg} \\). Then\n\\[\n\\begin{array}{l}\n\\frac{d transvar}{d timevar}=\\left(\\frac{d auxfunc}{d timevar}+coeffg auxfunc\\right) e^{accumg}=inhomfn e^{accumg} \\\\\n\\frac{d transyvar}{d timevar}=\\left(\\frac{d statefn}{d timevar}+coeffg statefn\\right) e^{accumg} \\geq inhomfn e^{accumg}\n\\end{array}\n\\]\nsince \\( e^{accumg}>0 \\) for all \\( timevar \\). Thus \\( (d / d timevar)(transyvar-transvar) \\geq 0 \\). Now \\( transvar(0)=transyvar(0)=initcst \\), and we conclude \\( transvar(timevar) \\leq transyvar(timevar) \\) and therefore \\( auxfunc(timevar) \\leq statefn(timevar) \\) for all \\( timevar \\geq 0 \\).\n\nFor the second part of the problem, we suppose that \\( statefn \\) satisfies\n\\[\n\\frac{d statefn}{d timevar}+coeffg statefn=statefn^{2} \\text { for } timevar \\in\\left[0, boundzero\\right], \\quad statefn(0)=initone\n\\]\nand we let \\( trialfn \\) be any continuous function on \\( \\left[0, boundzero\\right] \\). Then\n\\[\n\\begin{array}{c}\n(statefn-trialfn)^{2} \\geq 0 \\quad \\text { for all } timevar \\\\\n\\frac{d statefn}{d timevar}+(coeffg-2 trialfn) statefn=statefn^{2}-2 trialfn statefn \\geq-trialfn^{2}\n\\end{array}\n\\]\n\nIf \\( accumw(timevar)=\\int_{0}^{1} trialfn(sample) d sample \\) and \\( transyvar=statefn e^{accumg-2 accumw} \\) then\n\\[\n\\begin{array}{c}\n\\frac{d transyvar}{d timevar}=\\left(\\frac{d statefn}{d timevar}+(coeffg-2 trialfn) statefn\\right) e^{accumg-2 accumw} \\geq-trialfn^{2} e^{accumg-2 accumw} \\\\\ntransyvar(timevar) \\geq initone-\\int_{0}^{\\prime} trialfn^{2}(sample) \\exp [accumg(sample)-2 accumw(sample)] d sample \\\\\nstatefn(timevar) \\geq initone \\exp [-accumg(timevar)+2 accumw(timevar)] \\\\\n-\\int_{0}^{\\prime} \\exp [accumg(sample)-accumg(timevar)-2(accumw(sample)-accumw(timevar))] trialfn^{2}(sample) d sample\n\\end{array}\n\\]\n\nSo\n\\[\n\\begin{aligned}\nstatefn(timevar) \\geq \\max _{trialfn}\\{ & \\left\\{initone \\exp [-accumg(timevar)+2 accumw(timevar)]\\right. \\\\\n& \\left.-\\int_{0}^{\\prime \\prime} \\exp [accumg(sample)-accumg(timevar)-2(accumw(sample)-accumw(timevar))] trialfn^{2}(sample) d sample\\right\\}\n\\end{aligned}\n\\]\n\nNow if we take \\( trialfn=statefn \\), then all of the preceding inequalities become equalities so we conclude\n\\[\n\\begin{aligned}\nstatefn(timevar)=\\max _{n^{\\prime}}\\{ & initone \\exp [-accumg(timevar)+2 accumw(timevar)] \\\\\n& \\left.-\\int_{0}^{timevar} \\exp [accumg(sample)-accumg(timevar)-2(accumw(sample)-accumw(timevar))] trialfn^{2}(sample) d sample\\right\\}\n\\end{aligned}\n\\]\nwhich is equivalent to the form given.\n\nThe solution of the equation\n\\[\n\\frac{d statefn}{d timevar}+coeffg statefn=statefn^{2}\n\\]\nsubject to the initial condition \\( statefn(0)=initone \\), may be unbounded on a finite interval, say \\( statefn \\rightarrow \\infty \\) as \\( timevar \\mid boundone \\). Then the preceding analysis is valid only for \\( timevar<boundone \\). If we pick any \\( boundzero \\) with \\( 0<boundzero<boundone \\), then it is valid for [0, \\( \\left.boundzero\\right] \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "t": "lantern",
        "s": "meadowbrook",
        "s_1": "cottonseed",
        "v": "orchardbird",
        "u": "riverstone",
        "w": "pinecone",
        "x": "ironbridge",
        "y": "silvercloud",
        "G": "foxglove",
        "W": "hummingway",
        "g": "mapleroot",
        "h": "windwhisper",
        "c": "rainshadow",
        "c_1": "stairwell",
        "t_0": "moondance",
        "t_1": "amberlight"
      },
      "question": "7. Let \\( mapleroot(lantern) \\) and \\( windwhisper(lantern) \\) be real, continuous functions for \\( lantern \\geq 0 \\). Show that any function \\( orchardbird(lantern) \\) satisfying the differential inequality\n\\[\\frac{d orchardbird}{d lantern}+mapleroot(lantern)\\,orchardbird(lantern) \\geq windwhisper(lantern), \\quad orchardbird(0)=rainshadow\\]\nsatisfies the further inequality \\( orchardbird(lantern) \\geq riverstone(lantern) \\) where\n\\[\\frac{d riverstone}{d lantern}+mapleroot(lantern)\\,riverstone(lantern)=windwhisper(lantern), \\quad riverstone(0)=rainshadow\\]\n\nFrom this, conclude that for sufficiently small \\( lantern>0 \\), the solution of\n\\[\\frac{d orchardbird}{d lantern}+mapleroot(lantern)\\,orchardbird(lantern)=orchardbird^{2}, \\quad orchardbird(0)=stairwell\\]\nmay be written\n\\[orchardbird=\\max _{pinecone}\\left[stairwell\\,e^{-\\int_{0}^{lantern}[mapleroot(meadowbrook)-2\\,pinecone(meadowbrook)]\\,d meadowbrook}-\\int_{0}^{lantern}e^{-\\int_{0}^{lantern}\\left[mapleroot\\left(cottonseed\\right)-2\\,pinecone\\left(cottonseed\\right)\\right]d cottonseed}\\,pinecone^{2}(meadowbrook)\\,d meadowbrook\\right]\\]\nwhere the maximization is over all continuous functions \\( \\boldsymbol{pinecone}(lantern) \\) defined over some \\( lantern \\)-interval \\( \\left[0, moondance\\right] \\).",
      "solution": "Solution. Let\n\\[foxglove(lantern)=\\int_{0}^{lantern} mapleroot(meadowbrook)\\,d meadowbrook .\\]\n\nThen \\( foxglove^{\\prime}=mapleroot \\). Put \\( ironbridge=riverstone e^{foxglove},\\; silvercloud=orchardbird e^{foxglove} \\). Then\n\\[\\begin{array}{l}\n\\frac{d ironbridge}{d lantern}=\\left(\\frac{d riverstone}{d lantern}+mapleroot\\,riverstone\\right)e^{foxglove}=windwhisper e^{foxglove}\\\\\n\\frac{d silvercloud}{d lantern}=\\left(\\frac{d orchardbird}{d lantern}+mapleroot\\,orchardbird\\right)e^{foxglove}\\geq windwhisper e^{foxglove}\n\\end{array}\\]\nsince \\( e^{foxglove}>0 \\) for all \\( lantern \\). Thus \\( (d/ d lantern)(silvercloud-ironbridge)\\geq0 \\). Now \\( ironbridge(0)=silvercloud(0)=rainshadow \\), and we conclude \\( ironbridge(lantern)\\leq silvercloud(lantern) \\) and therefore \\( riverstone(lantern)\\leq orchardbird(lantern) \\) for all \\( lantern\\geq0 \\).\n\nFor the second part of the problem, we suppose that \\( orchardbird \\) satisfies\n\\[\\frac{d orchardbird}{d lantern}+mapleroot\\,orchardbird=orchardbird^{2}\\text{ for } lantern\\in[0,moondance],\\quad orchardbird(0)=stairwell\\]\nand we let \\( pinecone \\) be any continuous function on \\([0,moondance]\\). Then\n\\[\\begin{array}{c}\n(orchardbird-pinecone)^{2}\\geq0\\quad\\text{ for all } lantern\\\\\n\\frac{d orchardbird}{d lantern}+(mapleroot-2 pinecone)\\,orchardbird=orchardbird^{2}-2 pinecone\\,orchardbird\\geq-pinecone^{2}\n\\end{array}\\]\n\nIf \\( hummingway(lantern)=\\int_{0}^{1} pinecone(meadowbrook)\\,d meadowbrook \\) and \\( silvercloud=orchardbird\\,e^{foxglove 2 pinecone} \\) then\n\\[\\begin{array}{c}\n\\frac{d silvercloud}{d lantern}=\\left(\\frac{d orchardbird}{d lantern}+(mapleroot-2 pinecone)\\,orchardbird\\right)e^{foxglove-2 pinecone}\\geq-pinecone^{2}e^{foxglove}2 pinecone\\\\\nsilvercloud(lantern)\\geq stairwell-\\int_{0}^{\\prime} pinecone^{2}(meadowbrook)\\exp[foxglove(meadowbrook)-2 hummingway(meadowbrook)]\\,d meadowbrook\\\\\norchardbird(lantern)\\geq stairwell\\exp[-foxglove(lantern)+2 hummingway(lantern)]\\\\\n-\\int_{0}^{\\prime} \\exp[foxglove(meadowbrook)-foxglove(lantern)-2(hummingway(meadowbrook)-hummingway(lantern))] pinecone^{2}(meadowbrook)\\,d meadowbrook\n\\end{array}\\]\n\nSo\n\\[\\begin{aligned}\norchardbird(lantern)\\geq \\max_{pinecone}\\{&\\,stairwell\\exp[-foxglove(lantern)+2 hummingway(lantern)]\\\\\n&-\\int_{0}^{\\prime\\prime}\\exp[foxglove(meadowbrook)-foxglove(lantern)-2(hummingway(meadowbrook)-hummingway(lantern))] pinecone^{2}(meadowbrook)\\,d meadowbrook\\}\n\\end{aligned}\\]\n\nNow if we take \\( pinecone=orchardbird \\), then all of the preceding inequalities become equalities so we conclude\n\\[\\begin{aligned}\norchardbird(lantern)=\\max_{n^{\\prime}}\\{&\\,stairwell\\exp[-foxglove(lantern)+2 hummingway(lantern)]\\\\\n&-\\int_{0}^{lantern}\\exp[foxglove(meadowbrook)-foxglove(lantern)-2(hummingway(meadowbrook)-hummingway(lantern))] pinecone^{2}(meadowbrook)\\,d meadowbrook\\}\n\\end{aligned}\\]\nwhich is equivalent to the form given.\n\nThe solution of the equation\n\\[\\frac{d orchardbird}{d lantern}+mapleroot\\,orchardbird=orchardbird^{2}\\]\nsubject to the initial condition \\( orchardbird(0)=stairwell \\), may be unbounded on a finite interval, say \\( orchardbird\\to\\infty \\) as \\( lantern\\mid amberlight \\). Then the preceding analysis is valid only for \\( lantern<amberlight \\). If we pick any \\( moondance \\) with \\( 0<moondance<amberlight \\), then it is valid for [0, \\( \\left.moondance\\right] \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "t": "spacedist",
        "s": "stillness",
        "s_1": "stillnessone",
        "v": "reststate",
        "u": "chaosfunc",
        "w": "steadfun",
        "x": "depthval",
        "y": "sidevalue",
        "G": "disintegral",
        "W": "emptiness",
        "g": "levityfun",
        "h": "unrestfun",
        "c": "variable",
        "c_1": "volatile",
        "t_0": "endlesses",
        "t_1": "eternity"
      },
      "question": "7. Let \\( levityfun(spacedist) \\) and \\( unrestfun(spacedist) \\) be real, continuous functions for \\( spacedist \\geq 0 \\). Show that any function \\( reststate(spacedist) \\) satisfying the differential inequality\n\\[\n\\frac{d reststate}{d spacedist}+levityfun(spacedist) reststate \\geq unrestfun(spacedist), \\quad reststate(0)=variable\n\\]\nsatisfies the further inequality \\( reststate(spacedist) \\geq chaosfunc(spacedist) \\) where\n\\[\n\\frac{d chaosfunc}{d spacedist}+levityfun(spacedist) chaosfunc=unrestfun(spacedist), \\quad chaosfunc(0)=variable\n\\]\n\nFrom this, conclude that for sufficiently small \\( spacedist>0 \\), the solution of\n\\[\n\\frac{d reststate}{d spacedist}+levityfun(spacedist) reststate=reststate^{2}, \\quad reststate(0)=volatile\n\\]\nmay be written\n\\[\nreststate=\\max _{steadfun}\\left[volatile e^{-\\int_{0}^{spacedist}[levityfun(stillness)-2 steadfun(stillness)] d stillness}-\\int_{0}^{spacedist} e^{-\\int_{0}^{spacedist}\\left[levityfun\\left(stillnessone\\right)-2 steadfun\\left(stillnessone\\right)\\right] d stillnessone} steadfun^{2}(stillness) d stillness\\right]\n\\]\nwhere the maximization is over all continuous functions \\( \\boldsymbol{steadfun}(spacedist) \\) defined over some \\( spacedist \\)-interval \\( \\left[0, endlesses\\right] \\).",
      "solution": "Solution. Let\n\\[\ndisintegral(spacedist)=\\int_{0}^{spacedist} levityfun(stillness) d stillness .\n\\]\n\nThen \\( disintegral^{\\prime}=levityfun \\). Put \\( depthval=chaosfunc e^{disintegral}, sidevalue=reststate e^{disintegral} \\). Then\n\\[\n\\begin{array}{l}\n\\frac{d depthval}{d spacedist}=\\left(\\frac{d chaosfunc}{d spacedist}+levityfun chaosfunc\\right) e^{disintegral}=unrestfun e^{disintegral} \\\\\n\\frac{d sidevalue}{d spacedist}=\\left(\\frac{d reststate}{d spacedist}+levityfun reststate\\right) e^{disintegral} \\geq unrestfun e^{disintegral}\n\\end{array}\n\\]\nsince \\( e^{disintegral}>0 \\) for all \\( spacedist \\). Thus \\( (d / d spacedist)(sidevalue-depthval) \\geq 0 \\). Now \\( depthval(0)=sidevalue(0)=variable \\), and we conclude \\( depthval(spacedist) \\leq sidevalue(spacedist) \\) and therefore \\( chaosfunc(spacedist) \\leq reststate(spacedist) \\) for all \\( spacedist \\geq 0 \\).\n\nFor the second part of the problem, we suppose that \\( reststate \\) satisfies\n\\[\n\\frac{d reststate}{d spacedist}+levityfun reststate=reststate^{2} \\text { for } spacedist \\in\\left[0, endlesses\\right], \\quad reststate(0)=volatile\n\\]\nand we let \\( steadfun \\) be any continuous function on \\( \\left[0, endlesses\\right] \\). Then\n\\[\n\\begin{array}{c}\n(reststate-steadfun)^{2} \\geq 0 \\quad \\text { for all } spacedist \\\\\n\\frac{d reststate}{d spacedist}+(levityfun-2 steadfun) reststate=reststate^{2}-2 steadfun reststate \\geq-steadfun^{2}\n\\end{array}\n\\]\n\nIf \\( emptiness(spacedist)=\\int_{0}^{1} steadfun(stillness) d stillness \\) and \\( sidevalue=reststate e^{disintegral 2 steadfun} \\) then\n\\[\n\\begin{array}{c}\n\\frac{d sidevalue}{d spacedist}=\\left(\\frac{d reststate}{d spacedist}+(levityfun-2 steadfun) reststate\\right) e^{disintegral-2 steadfun} \\geq-steadfun^{2} e^{disintegral} 2 steadfun \\\\\nsidevalue(spacedist) \\geq volatile-\\int_{0}^{\\prime} steadfun^{2}(stillness) \\exp [disintegral(stillness)-2 emptiness(stillness)] d stillness \\\\\nreststate(spacedist) \\geq volatile \\exp [-disintegral(spacedist)+2 emptiness(spacedist)] \\\\\n-\\int_{0}^{\\prime} \\exp [disintegral(stillness)-disintegral(spacedist)-2(emptiness(stillness)-emptiness(spacedist))] steadfun^{2}(stillness) d stillness\n\\end{array}\n\\]\n\nSo\n\\[\n\\begin{aligned}\nreststate(spacedist) \\geq \\max _{steadfun}\\{ & \\left\\{volatile \\exp [-disintegral(spacedist)+2 emptiness(spacedist)]\\right. \\\\\n& \\left.-\\int_{0}^{\\prime \\prime} \\exp [disintegral(stillness)-disintegral(spacedist)-2(emptiness(stillness)-emptiness(spacedist))] steadfun^{2}(stillness) d stillness\\right\\}\n\\end{aligned}\n\\]\n\nNow if we take \\( steadfun=reststate \\), then all of the preceding inequalities become equalities so we conclude\n\\[\n\\begin{aligned}\nreststate(spacedist)=\\max _{n^{\\prime}}\\{ & volatile \\exp [-disintegral(spacedist)+2 emptiness(spacedist)] \\\\\n& \\left.-\\int_{0}^{spacedist} \\exp [disintegral(stillness)-disintegral(spacedist)-2(emptiness(stillness)-emptiness(spacedist))] steadfun^{2}(stillness) d stillness\\right\\}\n\\end{aligned}\n\\]\nwhich is equivalent to the form given.\nThe solution of the equation\n\\[\n\\frac{d reststate}{d spacedist}+levityfun reststate=reststate^{2}\n\\]\nsubject to the initial condition \\( reststate(0)=volatile \\), may be unbounded on a finite interval, say \\( reststate \\rightarrow \\infty \\) as \\( spacedist \\mid eternity \\). Then the preceding analysis is valid only for \\( spacedist<eternity \\). If we pick any \\( endlesses \\) with \\( 0<endlesses<eternity \\), then it is valid for [0, \\( \\left.endlesses\\right] \\)."
    },
    "garbled_string": {
      "map": {
        "t": "akpsejrm",
        "s": "fnduqzlo",
        "s_1": "htrivcpm",
        "v": "plxmoqns",
        "u": "ykgrtabe",
        "w": "dzlhvsep",
        "x": "jbqnarud",
        "y": "kigwesom",
        "G": "bvcayzth",
        "W": "mljxprwa",
        "g": "szepidra",
        "h": "qlonfuvk",
        "c": "wesdjnol",
        "c_1": "ytmkzabe",
        "t_0": "kxbrtahu",
        "t_1": "nsgpdewi"
      },
      "question": "7. Let \\( szepidra(akpsejrm) \\) and \\( qlonfuvk(akpsejrm) \\) be real, continuous functions for \\( akpsejrm \\geq 0 \\). Show that any function \\( plxmoqns(akpsejrm) \\) satisfying the differential inequality\n\\[\n\\frac{d plxmoqns}{d akpsejrm}+szepidra(akpsejrm) plxmoqns \\geq qlonfuvk(akpsejrm), \\quad plxmoqns(0)=wesdjnol\n\\]\nsatisfies the further inequality \\( plxmoqns(akpsejrm) \\geq ykgrtabe(akpsejrm) \\) where\n\\[\n\\frac{d ykgrtabe}{d akpsejrm}+szepidra(akpsejrm) ykgrtabe=qlonfuvk(akpsejrm), \\quad ykgrtabe(0)=wesdjnol\n\\]\n\nFrom this, conclude that for sufficiently small \\( akpsejrm>0 \\), the solution of\n\\[\n\\frac{d plxmoqns}{d akpsejrm}+szepidra(akpsejrm) plxmoqns=plxmoqns^{2}, \\quad plxmoqns(0)=ytmkzabe\n\\]\nmay be written\n\\[\nplxmoqns=\\max _{dzlhvsep}\\left[ytmkzabe e^{-\\int_{0}^{akpsejrm}[szepidra(fnduqzlo)-2 dzlhvsep(fnduqzlo)] d fnduqzlo}-\\int_{0}^{akpsejrm} e^{-\\int_{0}^{akpsejrm}\\left[szepidra\\left(htrivcpm\\right)-2 dzlhvsep\\left(htrivcpm\\right)\\right] d htrivcpm} dzlhvsep^{2}(fnduqzlo) d fnduqzlo\\right]\n\\]\nwhere the maximization is over all continuous functions \\( \\boldsymbol{dzlhvsep}(akpsejrm) \\) defined over some \\( akpsejrm \\)-interval \\( \\left[0, kxbrtahu\\right] \\).",
      "solution": "Solution. Let\n\\[\nbvcayzth(akpsejrm)=\\int_{0}^{akpsejrm} szepidra(fnduqzlo) d fnduqzlo .\n\\]\n\nThen \\( bvcayzth^{\\prime}=szepidra \\). Put \\( jbqnarud=ykgrtabe e^{bvcayzth}, kigwesom=plxmoqns e^{bvcayzth} \\). Then\n\\[\n\\begin{array}{l}\n\\frac{d jbqnarud}{d akpsejrm}=\\left(\\frac{d ykgrtabe}{d akpsejrm}+szepidra ykgrtabe\\right) e^{bvcayzth}=qlonfuvk e^{bvcayzth} \\\\\n\\frac{d kigwesom}{d akpsejrm}=\\left(\\frac{d plxmoqns}{d akpsejrm}+szepidra plxmoqns\\right) e^{bvcayzth} \\geq qlonfuvk e^{bvcayzth}\n\\end{array}\n\\]\nsince \\( e^{bvcayzth}>0 \\) for all \\( akpsejrm \\). Thus \\( (d / d akpsejrm)(kigwesom-jbqnarud) \\geq 0 \\). Now \\( jbqnarud(0)=kigwesom(0)=wesdjnol \\), and we conclude \\( jbqnarud(akpsejrm) \\leq kigwesom(akpsejrm) \\) and therefore \\( ykgrtabe(akpsejrm) \\leq plxmoqns(akpsejrm) \\) for all \\( akpsejrm \\geq 0 \\).\n\nFor the second part of the problem, we suppose that \\( plxmoqns \\) satisfies\n\\[\n\\frac{d plxmoqns}{d akpsejrm}+szepidra plxmoqns=plxmoqns^{2} \\text { for } akpsejrm \\in\\left[0, kxbrtahu\\right], \\quad plxmoqns(0)=ytmkzabe\n\\]\nand we let \\( dzlhvsep \\) be any continuous function on \\( \\left[0, kxbrtahu\\right] \\). Then\n\\[\n\\begin{array}{c}\n(plxmoqns-dzlhvsep)^{2} \\geq 0 \\quad \\text { for all } akpsejrm \\\\\n\\frac{d plxmoqns}{d akpsejrm}+(szepidra-2 dzlhvsep) plxmoqns=plxmoqns^{2}-2 dzlhvsep plxmoqns \\geq-dzlhvsep^{2}\n\\end{array}\n\\]\n\nIf \\( mljxprwa(akpsejrm)=\\int_{0}^{1} dzlhvsep(fnduqzlo) d fnduqzlo \\) and \\( kigwesom=plxmoqns e^{bvcayzth 2 dzlhvsep} \\) then\n\\[\n\\begin{array}{c}\n\\frac{d kigwesom}{d akpsejrm}=\\left(\\frac{d plxmoqns}{d akpsejrm}+(szepidra-2 dzlhvsep) plxmoqns\\right) e^{bvcayzth-2 dzlhvsep} \\geq-dzlhvsep^{2} e^{bvcayzth} 2 dzlhvsep \\\\\nkigwesom(akpsejrm) \\geq ytmkzabe-\\int_{0}^{\\prime} dzlhvsep^{2}(fnduqzlo) \\exp [bvcayzth(fnduqzlo)-2 mljxprwa(fnduqzlo)] d fnduqzlo \\\\\nplxmoqns(akpsejrm) \\geq ytmkzabe \\exp [-bvcayzth(akpsejrm)+2 mljxprwa(akpsejrm)] \\\\\n-\\int_{0}^{\\prime} \\exp [bvcayzth(fnduqzlo)-bvcayzth(akpsejrm)-2( mljxprwa(fnduqzlo)- mljxprwa(akpsejrm))] dzlhvsep^{2}(fnduqzlo) d fnduqzlo\n\\end{array}\n\\]\n\nSo\n\\[\n\\begin{aligned}\nplxmoqns(akpsejrm) \\geq \\max _{dzlhvsep}\\{ & \\left\\{ytmkzabe \\exp [-bvcayzth(akpsejrm)+2 mljxprwa(akpsejrm)]\\right. \\\\\n& \\left.-\\int_{0}^{\\prime \\prime} \\exp [bvcayzth(fnduqzlo)-bvcayzth(akpsejrm)-2( mljxprwa(fnduqzlo)- mljxprwa(akpsejrm))] dzlhvsep^{2}(fnduqzlo) d fnduqzlo\\right\\}\n\\end{aligned}\n\\]\n\nNow if we take \\( dzlhvsep=plxmoqns \\), then all of the preceding inequalities become equalities so we conclude\n\\[\n\\begin{aligned}\nplxmoqns(akpsejrm)=\\max _{n^{\\prime}}\\{ & ytmkzabe \\exp [-bvcayzth(akpsejrm)+2 mljxprwa(akpsejrm)] \\\\\n& \\left.-\\int_{0}^{akpsejrm} \\exp [bvcayzth(fnduqzlo)-bvcayzth(akpsejrm)-2( mljxprwa(fnduqzlo)- mljxprwa(akpsejrm))] dzlhvsep^{2}(fnduqzlo) d fnduqzlo\\right\\}\n\\end{aligned}\n\\]\nwhich is equivalent to the form given.\nThe solution of the equation\n\\[\n\\frac{d plxmoqns}{d akpsejrm}+szepidra plxmoqns=plxmoqns^{2}\n\\]\nsubject to the initial condition \\( plxmoqns(0)=ytmkzabe \\), may be unbounded on a finite interval, say \\( plxmoqns \\rightarrow \\infty \\) as \\( akpsejrm \\mid nsgpdewi \\). Then the preceding analysis is valid only for \\( akpsejrm<nsgpdewi \\). If we pick any \\( kxbrtahu \\) with \\( 0<kxbrtahu<nsgpdewi \\), then it is valid for [0, \\( \\left.kxbrtahu\\right] \\)."
    },
    "kernel_variant": {
      "question": "Let n∈ℕ and a∈ℝ.  Denote by 𝑆ⁿ the space of real symmetric n×n matrices, partially ordered by the Loewner cone  \nA ≽ B  ⇔  A−B is positive–semidefinite.  All matrix entries are Lebesgue–measurable and every integral below is a Lebesgue integral.\n\n1.  (Matrix–valued linear comparison.)  \n    Let  \n       G:[a,∞)→ℝ^{n×n},  H:[a,∞)→𝑆ⁿ  \n    be locally integrable, and let V:[a,∞)→𝑆ⁿ be absolutely continuous and satisfy, for a.e. t≥a,  \n       V′(t)+G(t)ᵀV(t)+V(t)G(t) ≽ H(t),  V(a)=S∈𝑆ⁿ.  \n    Denote by U the unique absolutely continuous solution of the matrix ODE  \n       U′(t)+G(t)ᵀU(t)+U(t)G(t)=H(t), U(a)=S.  \n    Prove the comparison principle  \n       V(t) ≽ U(t) for every t≥a.\n\n2.  (Variational representation for a matrix Riccati equation.)  \n    Fix δ>0 and let  \n       Q:[a,a+δ)→𝑆ⁿ  \n    be locally integrable.  Suppose V:[a,a+δ)→𝑆ⁿ is absolutely continuous and solves the matrix Riccati equation  \n       V′(t)+G(t)ᵀV(t)+V(t)G(t)=V(t)²+Q(t), V(a)=S₀∈𝑆ⁿ.  (★)  \n    For any W∈L²_loc([a,a+δ);𝑆ⁿ) define Φ_W:[a,a+δ)→ℝ^{n×n} by  \n       Φ_W′(t)=(G(t)−W(t))Φ_W(t), Φ_W(a)=I_n.  (†)\n\n    Show that for every t∈[a,a+δ)  \n       V(t)= sup_{W∈L²_loc} Ξ(t,W),                  (‡)  \n    where  \n       Ξ(t,W)=Φ_W(t)^{−ᵀ} S₀ Φ_W(t)^{−1}  \n               −∫_a^{t} Φ_W(t)^{−ᵀ} Φ_W(s)ᵀ[W(s)²−Q(s)]Φ_W(s) Φ_W(t)^{−1} ds.  \n\n    (The supremum is taken with respect to the Loewner order on 𝑆ⁿ; it is actually a maximum and is attained for the choice W=V.)\n\n    You may assume δ is chosen so that V is finite on [a,a+δ).",
      "solution": "Part 1: Linear comparison in the Loewner order  \n---------------------------------------------------------------\nStep 1.  Fundamental matrix.  \nSolve the auxiliary linear system  \n   Ψ′(t)=G(t)Ψ(t), Ψ(a)=I_n.  \nLocal integrability of G guarantees an absolutely continuous, point-wise invertible Ψ with Ψ,Ψ^{−1} locally absolutely continuous.\n\nStep 2.  Conjugation.  \nPut X(t):=Ψ(t)ᵀ V(t) Ψ(t).  Because V and Ψ are absolutely continuous, so is X.  Differentiating,\n\n   X′(t)=Ψ(t)ᵀ[ V′(t)+G(t)ᵀV(t)+V(t)G(t) ]Ψ(t).\n\nUsing the hypothesis V′+GᵀV+VG ≽ H,\n\n   X′(t) ≽ Ψ(t)ᵀ H(t) Ψ(t).               (1)\n\nStep 3.  Comparison with the linear solution.  \nDefine X_U(t):=Ψ(t)ᵀ U(t) Ψ(t); since U satisfies the equality,\n\n   X_U′(t)=Ψ(t)ᵀ H(t) Ψ(t).               (2)\n\nSet Y(t):=X(t)−X_U(t).  From (1)–(2) we have Y′(t) ≽ 0 a.e. and Y(a)=0, hence Y(t) ≽ 0 for all t≥a.  Multiplying by Ψ(t)^{−ᵀ} (left) and Ψ(t)^{−1} (right) preserves the Loewner order, giving V(t) ≽ U(t).\n\nPart 2: Variational representation for the Riccati solution  \n----------------------------------------------------------------\nThroughout let W∈L²_loc([a,a+δ);𝑆ⁿ) be arbitrary.\n\nStep 1.  A quadratic matrix inequality.  \nFor symmetric matrices Z,W the square (Z−W)² is positive-semidefinite; expanding,\n\n   (Z−W)² = Z² − ZW − WZ + W² ≽ 0  \n   ⇒ Z² ≽ ZW + WZ − W².               (3)\n\nTaking Z=V(t) (symmetric) in (3) we obtain\n\n   V(t)² ≽ V(t)W(t) + W(t)V(t) − W(t)².        (4)\n\nStep 2.  Inserting (4) into the Riccati equation.  \nEquation (★) and (4) give, for a.e. t,\n\n   V′ + GᵀV + VG \n     = V² + Q\n     ≽ V W + W V − W² + Q.             (5)\n\nRe-grouping the mixed terms,\n\n   V′ + (G−W)ᵀV + V(G−W) ≽ Q − W².        (6)\n\nNotice that (G−W)ᵀV+V(G−W) is symmetric because V and W are.\n\nStep 3.  Conjugation by the flow Φ_W.  \nLet Φ_W solve (†) and set  \n\n   X_W(t):=Φ_W(t)ᵀ V(t) Φ_W(t).           (7)\n\nBecause Φ_W′=(G−W)Φ_W and Φ_W is invertible, differentiating (7) yields\n\n   X_W′(t)=Φ_W(t)ᵀ\n            [ V′ + (G−W)ᵀV + V(G−W) ](t)\n            Φ_W(t)\n          ≽ Φ_W(t)ᵀ[ Q(t) − W(t)² ] Φ_W(t).   (8)\n\nStep 4.  Integration.  \nIntegrate (8) from a to t (Φ_W(a)=I_n):\n\n   X_W(t) ≽ S₀ − ∫_a^{t} Φ_W(s)ᵀ[ W(s)² − Q(s) ] Φ_W(s) ds. (9)\n\nStep 5.  Returning to V.  \nPremultiplying (9) by Φ_W(t)^{−ᵀ} and post-multiplying by Φ_W(t)^{−1} gives\n\n   V(t) ≽ Φ_W(t)^{−ᵀ}S₀Φ_W(t)^{−1}\n          −∫_a^{t} Φ_W(t)^{−ᵀ} Φ_W(s)ᵀ[ W(s)²−Q(s) ]\n                     Φ_W(s) Φ_W(t)^{−1} ds\n        = Ξ(t,W).                 (10)\n\nHence V(t) dominates every Ξ(t,W); therefore\n\n   V(t) ≽ sup_{W∈L²_loc} Ξ(t,W).          (11)\n\nStep 6.  Equality for the maximising choice W=V.  \nChoose W=V.  In (3) we then have equality, and consequently every “≽” in (5)–(10) becomes “=”.  Thus Ξ(t,V)=V(t), so V(t) realises the supremum.  Therefore the supremum in (‡) is a maximum attained at W=V and (11) holds with equality:\n\n   V(t)= sup_{W∈L²_loc} Ξ(t,W)=Ξ(t,V).  \n\nStep 7.  Well-posedness remarks.  \n• For each W the flow Φ_W exists and is invertible because G−W is locally integrable.  \n• The integrand in Ξ(t,W) is locally square-integrable, hence the integral is finite on [a,t).  \n• Although (𝑆ⁿ,≽) is not a lattice when n≥2, the supremum in (‡) exists since it is achieved at W=V.\n\nThis completes the proof of (‡).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.525673",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimensional structure: The scalar functions of the original problem are replaced by symmetric n×n matrices, forcing the use of the non-trivial Loewner partial order and properties of matrix positive–semidefiniteness.  \n• Non-commutativity: Conjugation by a fundamental matrix and matrix products introduce non-commuting factors, greatly complicating differentiation and order preservation.  \n• Riccati dynamics: The non-linear part is now matrix-quadratic (V²), the natural multivariate analogue of the scalar v², and requires careful handling via the inequality (Z−W)² ≽ 0.  \n• Control–theoretic flavour: The representation involves an auxiliary matrix control W and a matrix differential equation (†) for Φ_W, mirroring advanced linear–quadratic optimal–control theory.  \n• Cone-valued analysis: All arguments must respect the Loewner cone; standard scalar inequalities do not directly apply and have to be re-established for matrices.  \n• Technical prerequisites: Knowledge of fundamental matrices, matrix ODEs, invariance of the cone under similarity transforms, and integration of operator-valued functions is required.  \nHence the enhanced variant is substantially more intricate and cannot be solved by a straightforward lift of the original scalar techniques."
      }
    },
    "original_kernel_variant": {
      "question": "Let n∈ℕ and a∈ℝ.  Denote by 𝑆ⁿ the space of real symmetric n×n matrices, partially ordered by the Loewner cone  \nA ≽ B  ⇔  A−B is positive–semidefinite.  All matrix entries are Lebesgue–measurable and every integral below is a Lebesgue integral.\n\n1.  (Matrix–valued linear comparison.)  \n    Let  \n       G:[a,∞)→ℝ^{n×n},  H:[a,∞)→𝑆ⁿ  \n    be locally integrable, and let V:[a,∞)→𝑆ⁿ be absolutely continuous and satisfy, for a.e. t≥a,  \n       V′(t)+G(t)ᵀV(t)+V(t)G(t) ≽ H(t),  V(a)=S∈𝑆ⁿ.  \n    Denote by U the unique absolutely continuous solution of the matrix ODE  \n       U′(t)+G(t)ᵀU(t)+U(t)G(t)=H(t), U(a)=S.  \n    Prove the comparison principle  \n       V(t) ≽ U(t) for every t≥a.\n\n2.  (Variational representation for a matrix Riccati equation.)  \n    Fix δ>0 and let  \n       Q:[a,a+δ)→𝑆ⁿ  \n    be locally integrable.  Suppose V:[a,a+δ)→𝑆ⁿ is absolutely continuous and solves the matrix Riccati equation  \n       V′(t)+G(t)ᵀV(t)+V(t)G(t)=V(t)²+Q(t), V(a)=S₀∈𝑆ⁿ.  (★)  \n    For any W∈L²_loc([a,a+δ);𝑆ⁿ) define Φ_W:[a,a+δ)→ℝ^{n×n} by  \n       Φ_W′(t)=(G(t)−W(t))Φ_W(t), Φ_W(a)=I_n.  (†)\n\n    Show that for every t∈[a,a+δ)  \n       V(t)= sup_{W∈L²_loc} Ξ(t,W),                  (‡)  \n    where  \n       Ξ(t,W)=Φ_W(t)^{−ᵀ} S₀ Φ_W(t)^{−1}  \n               −∫_a^{t} Φ_W(t)^{−ᵀ} Φ_W(s)ᵀ[W(s)²−Q(s)]Φ_W(s) Φ_W(t)^{−1} ds.  \n\n    (The supremum is taken with respect to the Loewner order on 𝑆ⁿ; it is actually a maximum and is attained for the choice W=V.)\n\n    You may assume δ is chosen so that V is finite on [a,a+δ).",
      "solution": "Part 1: Linear comparison in the Loewner order  \n---------------------------------------------------------------\nStep 1.  Fundamental matrix.  \nSolve the auxiliary linear system  \n   Ψ′(t)=G(t)Ψ(t), Ψ(a)=I_n.  \nLocal integrability of G guarantees an absolutely continuous, point-wise invertible Ψ with Ψ,Ψ^{−1} locally absolutely continuous.\n\nStep 2.  Conjugation.  \nPut X(t):=Ψ(t)ᵀ V(t) Ψ(t).  Because V and Ψ are absolutely continuous, so is X.  Differentiating,\n\n   X′(t)=Ψ(t)ᵀ[ V′(t)+G(t)ᵀV(t)+V(t)G(t) ]Ψ(t).\n\nUsing the hypothesis V′+GᵀV+VG ≽ H,\n\n   X′(t) ≽ Ψ(t)ᵀ H(t) Ψ(t).               (1)\n\nStep 3.  Comparison with the linear solution.  \nDefine X_U(t):=Ψ(t)ᵀ U(t) Ψ(t); since U satisfies the equality,\n\n   X_U′(t)=Ψ(t)ᵀ H(t) Ψ(t).               (2)\n\nSet Y(t):=X(t)−X_U(t).  From (1)–(2) we have Y′(t) ≽ 0 a.e. and Y(a)=0, hence Y(t) ≽ 0 for all t≥a.  Multiplying by Ψ(t)^{−ᵀ} (left) and Ψ(t)^{−1} (right) preserves the Loewner order, giving V(t) ≽ U(t).\n\nPart 2: Variational representation for the Riccati solution  \n----------------------------------------------------------------\nThroughout let W∈L²_loc([a,a+δ);𝑆ⁿ) be arbitrary.\n\nStep 1.  A quadratic matrix inequality.  \nFor symmetric matrices Z,W the square (Z−W)² is positive-semidefinite; expanding,\n\n   (Z−W)² = Z² − ZW − WZ + W² ≽ 0  \n   ⇒ Z² ≽ ZW + WZ − W².               (3)\n\nTaking Z=V(t) (symmetric) in (3) we obtain\n\n   V(t)² ≽ V(t)W(t) + W(t)V(t) − W(t)².        (4)\n\nStep 2.  Inserting (4) into the Riccati equation.  \nEquation (★) and (4) give, for a.e. t,\n\n   V′ + GᵀV + VG \n     = V² + Q\n     ≽ V W + W V − W² + Q.             (5)\n\nRe-grouping the mixed terms,\n\n   V′ + (G−W)ᵀV + V(G−W) ≽ Q − W².        (6)\n\nNotice that (G−W)ᵀV+V(G−W) is symmetric because V and W are.\n\nStep 3.  Conjugation by the flow Φ_W.  \nLet Φ_W solve (†) and set  \n\n   X_W(t):=Φ_W(t)ᵀ V(t) Φ_W(t).           (7)\n\nBecause Φ_W′=(G−W)Φ_W and Φ_W is invertible, differentiating (7) yields\n\n   X_W′(t)=Φ_W(t)ᵀ\n            [ V′ + (G−W)ᵀV + V(G−W) ](t)\n            Φ_W(t)\n          ≽ Φ_W(t)ᵀ[ Q(t) − W(t)² ] Φ_W(t).   (8)\n\nStep 4.  Integration.  \nIntegrate (8) from a to t (Φ_W(a)=I_n):\n\n   X_W(t) ≽ S₀ − ∫_a^{t} Φ_W(s)ᵀ[ W(s)² − Q(s) ] Φ_W(s) ds. (9)\n\nStep 5.  Returning to V.  \nPremultiplying (9) by Φ_W(t)^{−ᵀ} and post-multiplying by Φ_W(t)^{−1} gives\n\n   V(t) ≽ Φ_W(t)^{−ᵀ}S₀Φ_W(t)^{−1}\n          −∫_a^{t} Φ_W(t)^{−ᵀ} Φ_W(s)ᵀ[ W(s)²−Q(s) ]\n                     Φ_W(s) Φ_W(t)^{−1} ds\n        = Ξ(t,W).                 (10)\n\nHence V(t) dominates every Ξ(t,W); therefore\n\n   V(t) ≽ sup_{W∈L²_loc} Ξ(t,W).          (11)\n\nStep 6.  Equality for the maximising choice W=V.  \nChoose W=V.  In (3) we then have equality, and consequently every “≽” in (5)–(10) becomes “=”.  Thus Ξ(t,V)=V(t), so V(t) realises the supremum.  Therefore the supremum in (‡) is a maximum attained at W=V and (11) holds with equality:\n\n   V(t)= sup_{W∈L²_loc} Ξ(t,W)=Ξ(t,V).  \n\nStep 7.  Well-posedness remarks.  \n• For each W the flow Φ_W exists and is invertible because G−W is locally integrable.  \n• The integrand in Ξ(t,W) is locally square-integrable, hence the integral is finite on [a,t).  \n• Although (𝑆ⁿ,≽) is not a lattice when n≥2, the supremum in (‡) exists since it is achieved at W=V.\n\nThis completes the proof of (‡).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.439488",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimensional structure: The scalar functions of the original problem are replaced by symmetric n×n matrices, forcing the use of the non-trivial Loewner partial order and properties of matrix positive–semidefiniteness.  \n• Non-commutativity: Conjugation by a fundamental matrix and matrix products introduce non-commuting factors, greatly complicating differentiation and order preservation.  \n• Riccati dynamics: The non-linear part is now matrix-quadratic (V²), the natural multivariate analogue of the scalar v², and requires careful handling via the inequality (Z−W)² ≽ 0.  \n• Control–theoretic flavour: The representation involves an auxiliary matrix control W and a matrix differential equation (†) for Φ_W, mirroring advanced linear–quadratic optimal–control theory.  \n• Cone-valued analysis: All arguments must respect the Loewner cone; standard scalar inequalities do not directly apply and have to be re-established for matrices.  \n• Technical prerequisites: Knowledge of fundamental matrices, matrix ODEs, invariance of the cone under similarity transforms, and integration of operator-valued functions is required.  \nHence the enhanced variant is substantially more intricate and cannot be solved by a straightforward lift of the original scalar techniques."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}