{
  "index": "1955-A-7",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "7. Consider the function \\( f \\) defined by the differential equation\n\\[\nf^{\\prime \\prime}(x)=\\left(x^{3}+a x\\right) f(x)\n\\]\nand the initial conditions \\( f(0)=1, f^{\\prime}(0)=0 \\). Prove that the roots of \\( f \\) are bounded above but unbounded below.",
  "solution": "First Solution. We show first that the roots of \\( f \\) are bounded above. Let \\( \\alpha \\) be a number such that \\( x^{3}+a x>0 \\) for \\( x \\geq \\alpha \\). We shall prove that no solution of the differential equation\n\\[\ny^{\\prime \\prime}-\\left(x^{3}+a x\\right) y=0\n\\]\nexcept the identically zero solution, has more than one root in [ \\( \\alpha, \\infty \\) ). Suppose \\( g \\) is a non-zero solution of (1) and \\( g\\left(x_{1}\\right)=g\\left(x_{2}\\right)=0 \\) with \\( \\alpha \\leq \\) \\( x_{1}<x_{2} \\). Then \\( g \\) is not zero throughout \\( \\left[x_{1}, x_{2}\\right] \\), so, changing the sign of \\( g \\) if necessary, we assume that \\( g \\) is somewhere positive on \\( \\left[x_{1}, x_{2}\\right. \\) ]. Let \\( g \\) achieve its maximum value on \\( \\left[x_{1}, x_{2}\\right] \\) at \\( x_{3} \\). Then \\( g\\left(x_{3}\\right)>0, g^{\\prime}\\left(x_{3}\\right)=0 \\), and \\( g^{\\prime \\prime}\\left(x_{3}\\right) \\leq 0 \\) by a standard criterion for a maximum. But\n\\[\ng^{\\prime \\prime}\\left(x_{3}\\right)=\\left(x_{3}{ }^{3}+a x_{3}\\right) g\\left(x_{3}\\right)>0\n\\]\nbecause \\( x_{3} \\geq \\alpha \\). This contradiction proves that a non-zero solution of (1), in particular \\( f \\), has at most one root in. \\( [\\alpha, \\infty) \\). Hence the roots of \\( f \\) are bounded above.\n\nWe now show that the roots of \\( f \\) are unbounded below. Let \\( \\beta \\) be a number such that \\( x^{3}+a x<-1 \\) for \\( x \\leq \\beta \\). We shall prove that every solution of (1) has a root on \\( \\left(-\\infty, x_{0}\\right. \\) ] for any choice of \\( x_{0} \\leq \\beta \\). Suppose this is false. Then there is a solution \\( h \\) that has constant sign on \\( \\left(-\\infty, x_{0}\\right] \\). Changing the sign of \\( h \\) if necessary, we can assume that \\( h \\) is positive on this interval. For any choice of \\( x_{1} \\leq x_{0} \\) we have by the extended mean value theorem\n\\[\nh(x)=h\\left(x_{1}\\right)+\\left(x-x_{1}\\right) h^{\\prime}\\left(x_{1}\\right)+\\frac{1}{2}\\left(x-x_{1}\\right)^{2} h^{\\prime \\prime}(\\xi),\n\\]\nwhere \\( \\xi \\) is between \\( x \\) and \\( x_{1} \\). Then if \\( x \\leq x_{0} \\), we have \\( \\xi<x_{0} \\) and \\( h^{\\prime \\prime}(\\xi) \\) \\( =\\left(\\xi^{3}+a \\xi\\right) h(\\xi)<0 \\) and therefore\n\\[\nh(x)<h\\left(x_{1}\\right)+\\left(x-x_{1}\\right) h^{\\prime}\\left(x_{1}\\right) .\n\\]\n\nIf we could choose \\( x_{1} \\) so that \\( h^{\\prime}\\left(x_{1}\\right)>0 \\), this would show that \\( h(x)<0 \\) for large negative \\( x \\), contrary to our assumption. So \\( h^{\\prime}(x) \\leq 0 \\) for all\n\\( x \\leq x_{0} \\). But this implies \\( h(x) \\geq h\\left(x_{0}\\right) \\) for \\( x \\leq x_{0} \\). Then (2) yields\n\\[\nh(x)<h\\left(x_{1}\\right)+\\left(x-x_{1}\\right) h^{\\prime}\\left(x_{1}\\right)-\\frac{1}{2}\\left(x-x_{1}\\right)^{2} h\\left(x_{0}\\right)\n\\]\nsince \\( h^{\\prime \\prime}(\\xi)=\\left(\\xi^{3}+a \\xi\\right) h(\\xi)<-h\\left(x_{0}\\right) \\). But this shows that \\( h(x) \\) is negative for sufficiently large negative values of \\( x \\). This contradiction proves that every solution of (1), in particular \\( f \\). has arbitrarily large negative roots.\n\nSecond Solution. Again we choose \\( \\alpha \\) and \\( \\beta \\) so that \\( x^{3}+a x>0 \\) for \\( x \\geq \\alpha \\), and \\( x^{3}+a x<-1 \\) for \\( x \\leq \\beta \\). We shall apply the Sturm comparison theorem.\n\nOn the interval \\( [\\alpha, \\infty) \\) we compare the differential equations\n\\[\nu^{\\prime \\prime}+0 \\cdot u=0\n\\]\nand\n\\[\ny^{\\prime \\prime}+\\left(-x^{3}-a x\\right) y=0\n\\]\n\nSince \\( 0>-x^{3}-a x \\) on this interval, we conclude that between any two roots of a non-zero solution of (4) appears a root of any solution of (3). But (3) has a constant solution with no roots at all, so we see that no non-zero solution of (4) has two roots in \\( [\\alpha, \\infty) \\). Hence \\( f \\) has at most one root in \\( [\\alpha, \\infty) \\), so its roots are bounded above.\n\nOn the interval ( \\( -\\infty, \\beta \\) ] we compare (4) with\n\\[\nv^{\\prime \\prime}+1 \\cdot v=0\n\\]\n\nSince \\( -x^{3}-a x>1 \\) on this interval, we conclude that between any two roots of any non-zero solution of (5) there is a root of any solution of (4). For any choice of \\( \\gamma, \\sin (x-\\gamma) \\) is a non-zero solution of (5) with roots at \\( \\gamma-\\pi \\) and \\( \\gamma \\). Hence \\( f \\) has a root in the interval \\( (\\gamma-\\pi, \\gamma) \\) for any \\( \\gamma \\leq \\beta \\). Thus the roots of \\( f \\) are unbounded below.\n\nA proof of the Sturm comparison theorem is given on page 451.",
  "vars": [
    "x",
    "f",
    "y",
    "g",
    "h",
    "u",
    "v",
    "x_0",
    "x_1",
    "x_2",
    "x_3",
    "\\\\xi"
  ],
  "params": [
    "a",
    "\\\\alpha",
    "\\\\beta",
    "\\\\gamma"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "f": "mainfunc",
        "y": "genericf",
        "g": "altfuncg",
        "h": "altfunch",
        "u": "altfuncu",
        "v": "altfuncv",
        "x_0": "pointzer",
        "x_1": "pointone",
        "x_2": "pointtwo",
        "x_3": "pointthr",
        "\\xi": "interxi",
        "a": "coeffpar",
        "\\alpha": "boundalp",
        "\\beta": "boundbet",
        "\\gamma": "boundgam"
      },
      "question": "7. Consider the function \\( mainfunc \\) defined by the differential equation\n\\[\nmainfunc^{\\prime \\prime}(abscissa)=\\left(abscissa^{3}+coeffpar\\,abscissa\\right) mainfunc(abscissa)\n\\]\nand the initial conditions \\( mainfunc(0)=1, mainfunc^{\\prime}(0)=0 \\). Prove that the roots of \\( mainfunc \\) are bounded above but unbounded below.",
      "solution": "First Solution. We show first that the roots of \\( mainfunc \\) are bounded above. Let \\( boundalp \\) be a number such that \\( abscissa^{3}+coeffpar\\,abscissa>0 \\) for \\( abscissa \\geq boundalp \\). We shall prove that no solution of the differential equation\n\\[\ngenericf^{\\prime \\prime}-\\left(abscissa^{3}+coeffpar\\,abscissa\\right) genericf=0\n\\]\nexcept the identically zero solution, has more than one root in \\( [ boundalp, \\infty ) \\). Suppose \\( altfuncg \\) is a non-zero solution of (1) and \\( altfuncg(pointone)=altfuncg(pointtwo)=0 \\) with \\( boundalp \\leq pointone<pointtwo \\). Then \\( altfuncg \\) is not zero throughout \\( [pointone,pointtwo] \\), so, changing the sign of \\( altfuncg \\) if necessary, we assume that \\( altfuncg \\) is somewhere positive on \\( [pointone,pointtwo] \\). Let \\( altfuncg \\) achieve its maximum value on \\( [pointone,pointtwo] \\) at \\( pointthr \\). Then \\( altfuncg(pointthr)>0,\\;altfuncg^{\\prime}(pointthr)=0, \\) and \\( altfuncg^{\\prime \\prime}(pointthr)\\le 0 \\) by a standard criterion for a maximum. But\n\\[\naltfuncg^{\\prime \\prime}(pointthr)=\\left(pointthr^{3}+coeffpar\\,pointthr\\right) altfuncg(pointthr)>0\n\\]\nbecause \\( pointthr \\geq boundalp \\). This contradiction proves that a non-zero solution of (1), in particular \\( mainfunc \\), has at most one root in \\( [boundalp,\\infty) \\). Hence the roots of \\( mainfunc \\) are bounded above.\n\nWe now show that the roots of \\( mainfunc \\) are unbounded below. Let \\( boundbet \\) be a number such that \\( abscissa^{3}+coeffpar\\,abscissa<-1 \\) for \\( abscissa \\leq boundbet \\). We shall prove that every solution of (1) has a root on \\( (-\\infty,pointzer] \\) for any choice of \\( pointzer \\leq boundbet \\). Suppose this is false. Then there is a solution \\( altfunch \\) that has constant sign on \\( (-\\infty,pointzer] \\). Changing the sign of \\( altfunch \\) if necessary, we can assume that \\( altfunch \\) is positive on this interval. For any choice of \\( pointone \\le pointzer \\) we have by the extended mean value theorem\n\\[\naltfunch(abscissa)=altfunch(pointone)+(abscissa-pointone) altfunch^{\\prime}(pointone)+\\frac{1}{2}(abscissa-pointone)^{2} altfunch^{\\prime \\prime}(interxi),\n\\]\nwhere \\( interxi \\) is between \\( abscissa \\) and \\( pointone \\). Then if \\( abscissa \\le pointzer \\), we have \\( interxi<pointzer \\) and \\( altfunch^{\\prime \\prime}(interxi)=\\left(interxi^{3}+coeffpar\\,interxi\\right) altfunch(interxi)<0 \\), and therefore\n\\[\naltfunch(abscissa)<altfunch(pointone)+(abscissa-pointone) altfunch^{\\prime}(pointone).\n\\]\nIf we could choose \\( pointone \\) so that \\( altfunch^{\\prime}(pointone)>0 \\), this would show that \\( altfunch(abscissa)<0 \\) for large negative \\( abscissa \\), contrary to our assumption. So \\( altfunch^{\\prime}(abscissa)\\le 0 \\) for all \\( abscissa \\le pointzer \\). But this implies \\( altfunch(abscissa)\\ge altfunch(pointzer) \\) for \\( abscissa \\le pointzer \\). Then (2) yields\n\\[\naltfunch(abscissa)<altfunch(pointone)+(abscissa-pointone) altfunch^{\\prime}(pointone)-\\frac{1}{2}(abscissa-pointone)^{2} altfunch(pointzer)\n\\]\nsince \\( altfunch^{\\prime \\prime}(interxi)=\\left(interxi^{3}+coeffpar\\,interxi\\right) altfunch(interxi)<-altfunch(pointzer) \\). But this shows that \\( altfunch(abscissa) \\) is negative for sufficiently large negative values of \\( abscissa \\). This contradiction proves that every solution of (1), in particular \\( mainfunc \\), has arbitrarily large negative roots.\n\nSecond Solution. Again we choose \\( boundalp \\) and \\( boundbet \\) so that \\( abscissa^{3}+coeffpar\\,abscissa>0 \\) for \\( abscissa \\ge boundalp \\), and \\( abscissa^{3}+coeffpar\\,abscissa<-1 \\) for \\( abscissa \\le boundbet \\). We shall apply the Sturm comparison theorem.\n\nOn the interval \\( [boundalp,\\infty) \\) we compare the differential equations\n\\[\naltfuncu^{\\prime \\prime}+0\\cdot altfuncu=0\n\\]\nand\n\\[\ngenericf^{\\prime \\prime}+\\left(-abscissa^{3}-coeffpar\\,abscissa\\right) genericf=0\n\\]\nSince \\( 0>-abscissa^{3}-coeffpar\\,abscissa \\) on this interval, we conclude that between any two roots of a non-zero solution of (4) appears a root of any solution of (3). But (3) has a constant solution with no roots at all, so we see that no non-zero solution of (4) has two roots in \\( [boundalp,\\infty) \\). Hence \\( mainfunc \\) has at most one root in \\( [boundalp,\\infty) \\), so its roots are bounded above.\n\nOn the interval \\( (-\\infty,boundbet] \\) we compare (4) with\n\\[\naltfuncv^{\\prime \\prime}+1\\cdot altfuncv=0\n\\]\nSince \\( -abscissa^{3}-coeffpar\\,abscissa>1 \\) on this interval, we conclude that between any two roots of any non-zero solution of (5) there is a root of any solution of (4). For any choice of \\( boundgam,\\; \\sin(abscissa-boundgam) \\) is a non-zero solution of (5) with roots at \\( boundgam-\\pi \\) and \\( boundgam \\). Hence \\( mainfunc \\) has a root in the interval \\( (boundgam-\\pi,\\,boundgam) \\) for any \\( boundgam\\le boundbet \\). Thus the roots of \\( mainfunc \\) are unbounded below.\n\nA proof of the Sturm comparison theorem is given on page 451."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "marigold",
        "f": "tortoise",
        "y": "shoelace",
        "g": "porcupine",
        "h": "raincloud",
        "u": "floorlamp",
        "v": "gingerale",
        "x_0": "lemonade",
        "x_1": "toothpaste",
        "x_2": "basketball",
        "x_3": "chandelier",
        "\\xi": "butterfly",
        "a": "hedgehog",
        "\\alpha": "pineapple",
        "\\beta": "snowflake",
        "\\gamma": "grapefruit"
      },
      "question": "7. Consider the function \\( tortoise \\) defined by the differential equation\n\\[\ntortoise^{\\prime \\prime}(marigold)=\\left(marigold^{3}+hedgehog marigold\\right) tortoise(marigold)\n\\]\nand the initial conditions \\( tortoise(0)=1, tortoise^{\\prime}(0)=0 \\). Prove that the roots of \\( tortoise \\) are bounded above but unbounded below.",
      "solution": "First Solution. We show first that the roots of \\( tortoise \\) are bounded above. Let \\( pineapple \\) be a number such that \\( marigold^{3}+hedgehog marigold>0 \\) for \\( marigold \\geq pineapple \\). We shall prove that no solution of the differential equation\n\\[\nshoelace^{\\prime \\prime}-\\left(marigold^{3}+hedgehog marigold\\right) shoelace=0\n\\]\nexcept the identically zero solution, has more than one root in [ \\( pineapple, \\infty \\) ). Suppose \\( porcupine \\) is a non-zero solution of (1) and \\( porcupine\\left(toothpaste\\right)=porcupine\\left(basketball\\right)=0 \\) with \\( pineapple \\leq \\) \\( toothpaste<basketball \\). Then \\( porcupine \\) is not zero throughout \\( \\left[toothpaste, basketball\\right] \\), so, changing the sign of \\( porcupine \\) if necessary, we assume that \\( porcupine \\) is somewhere positive on \\( \\left[toothpaste, basketball\\right. \\) ]. Let \\( porcupine \\) achieve its maximum value on \\( \\left[toothpaste, basketball\\right] \\) at \\( chandelier \\). Then \\( porcupine\\left(chandelier\\right)>0, porcupine^{\\prime}\\left(chandelier\\right)=0 \\), and \\( porcupine^{\\prime \\prime}\\left(chandelier\\right) \\leq 0 \\) by a standard criterion for a maximum. But\n\\[\nporcupine^{\\prime \\prime}\\left(chandelier\\right)=\\left(chandelier{ }^{3}+hedgehog chandelier\\right) porcupine\\left(chandelier\\right)>0\n\\]\nbecause \\( chandelier \\geq pineapple \\). This contradiction proves that a non-zero solution of (1), in particular \\( tortoise \\), has at most one root in. \\( [pineapple, \\infty) \\). Hence the roots of \\( tortoise \\) are bounded above.\n\nWe now show that the roots of \\( tortoise \\) are unbounded below. Let \\( snowflake \\) be a number such that \\( marigold^{3}+hedgehog marigold<-1 \\) for \\( marigold \\leq snowflake \\). We shall prove that every solution of (1) has a root on \\( \\left(-\\infty, lemonade\\right. \\) ] for any choice of \\( lemonade \\leq snowflake \\). Suppose this is false. Then there is a solution \\( raincloud \\) that has constant sign on \\( \\left(-\\infty, lemonade\\right] \\). Changing the sign of \\( raincloud \\) if necessary, we can assume that \\( raincloud \\) is positive on this interval. For any choice of \\( toothpaste \\leq lemonade \\) we have by the extended mean value theorem\n\\[\nraincloud(marigold)=raincloud\\left(toothpaste\\right)+\\left(marigold-toothpaste\\right) raincloud^{\\prime}\\left(toothpaste\\right)+\\frac{1}{2}\\left(marigold-toothpaste\\right)^{2} raincloud^{\\prime \\prime}(butterfly),\n\\]\nwhere \\( butterfly \\) is between \\( marigold \\) and \\( toothpaste \\). Then if \\( marigold \\leq lemonade \\), we have \\( butterfly<lemonade \\) and \\( raincloud^{\\prime \\prime}(butterfly) \\)\n\\( =\\left(butterfly^{3}+hedgehog butterfly\\right) raincloud(butterfly)<0 \\) and therefore\n\\[\nraincloud(marigold)<raincloud\\left(toothpaste\\right)+\\left(marigold-toothpaste\\right) raincloud^{\\prime}\\left(toothpaste\\right) .\n\\]\n\nIf we could choose \\( toothpaste \\) so that \\( raincloud^{\\prime}\\left(toothpaste\\right)>0 \\), this would show that \\( raincloud(marigold)<0 \\) for large negative \\( marigold \\), contrary to our assumption. So \\( raincloud^{\\prime}(marigold) \\leq 0 \\) for all\n\\( marigold \\leq lemonade \\). But this implies \\( raincloud(marigold) \\geq raincloud\\left(lemonade\\right) \\) for \\( marigold \\leq lemonade \\). Then (2) yields\n\\[\nraincloud(marigold)<raincloud\\left(toothpaste\\right)+\\left(marigold-toothpaste\\right) raincloud^{\\prime}\\left(toothpaste\\right)-\\frac{1}{2}\\left(marigold-toothpaste\\right)^{2} raincloud\\left(lemonade\\right)\n\\]\nsince \\( raincloud^{\\prime \\prime}(butterfly)=\\left(butterfly^{3}+hedgehog butterfly\\right) raincloud(butterfly)<-raincloud\\left(lemonade\\right) \\). But this shows that \\( raincloud(marigold) \\) is negative for sufficiently large negative values of \\( marigold \\). This contradiction proves that every solution of (1), in particular \\( tortoise \\). has arbitrarily large negative roots.\n\nSecond Solution. Again we choose \\( pineapple \\) and \\( snowflake \\) so that \\( marigold^{3}+hedgehog marigold>0 \\) for \\( marigold \\geq pineapple \\), and \\( marigold^{3}+hedgehog marigold<-1 \\) for \\( marigold \\leq snowflake \\). We shall apply the Sturm comparison theorem.\n\nOn the interval \\( [pineapple, \\infty) \\) we compare the differential equations\n\\[\nfloorlamp^{\\prime \\prime}+0 \\cdot floorlamp=0\n\\]\nand\n\\[\nshoelace^{\\prime \\prime}+\\left(-marigold^{3}-hedgehog marigold\\right) shoelace=0\n\\]\n\nSince \\( 0>-marigold^{3}-hedgehog marigold \\) on this interval, we conclude that between any two roots of a non-zero solution of (4) appears a root of any solution of (3). But (3) has a constant solution with no roots at all, so we see that no non-zero solution of (4) has two roots in \\( [pineapple, \\infty) \\). Hence \\( tortoise \\) has at most one root in \\( [pineapple, \\infty) \\), so its roots are bounded above.\n\nOn the interval ( \\( -\\infty, snowflake \\) ] we compare (4) with\n\\[\ngingerale^{\\prime \\prime}+1 \\cdot gingerale=0\n\\]\n\nSince \\( -marigold^{3}-hedgehog marigold>1 \\) on this interval, we conclude that between any two roots of any non-zero solution of (5) there is a root of any solution of (4). For any choice of \\( grapefruit, \\sin (marigold-grapefruit) \\) is a non-zero solution of (5) with roots at \\( grapefruit-\\pi \\) and \\( grapefruit \\). Hence \\( tortoise \\) has a root in the interval \\( (grapefruit-\\pi, grapefruit) \\) for any \\( grapefruit \\leq snowflake \\). Thus the roots of \\( tortoise \\) are unbounded below.\n\nA proof of the Sturm comparison theorem is given on page 451."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "stationary",
        "f": "constantvalue",
        "y": "unchanging",
        "g": "stagnant",
        "h": "staticfunc",
        "u": "steadyvar",
        "v": "fixedval",
        "x_0": "stationaryzero",
        "x_1": "stationaryone",
        "x_2": "stationarytwo",
        "x_3": "stationarythree",
        "\\xi": "knownvalue",
        "a": "variableconst",
        "\\alpha": "lastletter",
        "\\beta": "firstletter",
        "\\gamma": "middlelet"
      },
      "question": "7. Consider the function \\( constantvalue \\) defined by the differential equation\n\\[\nconstantvalue^{\\prime \\prime}(stationary)=\\left(stationary^{3}+variableconst\\ stationary\\right) constantvalue(stationary)\n\\]\nand the initial conditions \\( constantvalue(0)=1, constantvalue^{\\prime}(0)=0 \\). Prove that the roots of \\( constantvalue \\) are bounded above but unbounded below.",
      "solution": "First Solution. We show first that the roots of \\( constantvalue \\) are bounded above. Let \\( lastletter \\) be a number such that \\( stationary^{3}+variableconst\\ stationary>0 \\) for \\( stationary \\geq lastletter \\). We shall prove that no solution of the differential equation\n\\[\nunchanging^{\\prime \\prime}-\\left(stationary^{3}+variableconst\\ stationary\\right) unchanging=0\n\\]\nexcept the identically zero solution, has more than one root in [ \\( lastletter, \\infty \\) ). Suppose \\( stagnant \\) is a non-zero solution of (1) and \\( stagnant\\left(stationaryone\\right)=stagnant\\left(stationarytwo\\right)=0 \\) with \\( lastletter \\leq  stationaryone<stationarytwo \\). Then \\( stagnant \\) is not zero throughout \\( \\left[stationaryone, stationarytwo\\right] \\), so, changing the sign of \\( stagnant \\) if necessary, we assume that \\( stagnant \\) is somewhere positive on \\( \\left[stationaryone, stationarytwo\\right. \\) ]. Let \\( stagnant \\) achieve its maximum value on \\( \\left[stationaryone, stationarytwo\\right] \\) at \\( stationarythree \\). Then \\( stagnant\\left(stationarythree\\right)>0, stagnant^{\\prime}\\left(stationarythree\\right)=0 \\), and \\( stagnant^{\\prime \\prime}\\left(stationarythree\\right) \\leq 0 \\) by a standard criterion for a maximum. But\n\\[\nstagnant^{\\prime \\prime}\\left(stationarythree\\right)=\\left(stationarythree{ }^{3}+variableconst\\ stationarythree\\right) stagnant\\left(stationarythree\\right)>0\n\\]\nbecause \\( stationarythree \\geq lastletter \\). This contradiction proves that a non-zero solution of (1), in particular \\( constantvalue \\), has at most one root in. \\( [lastletter, \\infty) \\). Hence the roots of \\( constantvalue \\) are bounded above.\n\nWe now show that the roots of \\( constantvalue \\) are unbounded below. Let \\( firstletter \\) be a number such that \\( stationary^{3}+variableconst\\ stationary<-1 \\) for \\( stationary \\leq firstletter \\). We shall prove that every solution of (1) has a root on \\( \\left(-\\infty, stationaryzero\\right. \\) ] for any choice of \\( stationaryzero \\leq firstletter \\). Suppose this is false. Then there is a solution \\( staticfunc \\) that has constant sign on \\( \\left(-\\infty, stationaryzero\\right] \\). Changing the sign of \\( staticfunc \\) if necessary, we can assume that \\( staticfunc \\) is positive on this interval. For any choice of \\( stationaryone \\leq stationaryzero \\) we have by the extended mean value theorem\n\\[\nstaticfunc(stationary)=staticfunc\\left(stationaryone\\right)+\\left(stationary-stationaryone\\right) staticfunc^{\\prime}\\left(stationaryone\\right)+\\frac{1}{2}\\left(stationary-stationaryone\\right)^{2} staticfunc^{\\prime \\prime}(knownvalue),\n\\]\nwhere \\( knownvalue \\) is between \\( stationary \\) and \\( stationaryone \\). Then if \\( stationary \\leq stationaryzero \\), we have \\( knownvalue<stationaryzero \\) and \\( staticfunc^{\\prime \\prime}(knownvalue)  =\\left(knownvalue^{3}+variableconst\\ knownvalue\\right) staticfunc(knownvalue)<0 \\) and therefore\n\\[\nstaticfunc(stationary)<staticfunc\\left(stationaryone\\right)+\\left(stationary-stationaryone\\right) staticfunc^{\\prime}\\left(stationaryone\\right) .\n\\]\n\nIf we could choose \\( stationaryone \\) so that \\( staticfunc^{\\prime}\\left(stationaryone\\right)>0 \\), this would show that \\( staticfunc(stationary)<0 \\) for large negative \\( stationary \\), contrary to our assumption. So \\( staticfunc^{\\prime}(stationary) \\leq 0 \\) for all\n\\( stationary \\leq stationaryzero \\). But this implies \\( staticfunc(stationary) \\geq staticfunc\\left(stationaryzero\\right) \\) for \\( stationary \\leq stationaryzero \\). Then (2) yields\n\\[\nstaticfunc(stationary)<staticfunc\\left(stationaryone\\right)+\\left(stationary-stationaryone\\right) staticfunc^{\\prime}\\left(stationaryone\\right)-\\frac{1}{2}\\left(stationary-stationaryone\\right)^{2} staticfunc\\left(stationaryzero\\right)\n\\]\nsince \\( staticfunc^{\\prime \\prime}(knownvalue)=\\left(knownvalue^{3}+variableconst\\ knownvalue\\right) staticfunc(knownvalue)<-staticfunc\\left(stationaryzero\\right) \\). But this shows that \\( staticfunc(stationary) \\) is negative for sufficiently large negative values of \\( stationary \\). This contradiction proves that every solution of (1), in particular \\( constantvalue \\). has arbitrarily large negative roots.\n\nSecond Solution. Again we choose \\( lastletter \\) and \\( firstletter \\) so that \\( stationary^{3}+variableconst\\ stationary>0 \\) for \\( stationary \\geq lastletter \\), and \\( stationary^{3}+variableconst\\ stationary<-1 \\) for \\( stationary \\leq firstletter \\). We shall apply the Sturm comparison theorem.\n\nOn the interval \\( [lastletter, \\infty) \\) we compare the differential equations\n\\[\nsteadyvar^{\\prime \\prime}+0 \\cdot steadyvar=0\n\\]\nand\n\\[\nunchanging^{\\prime \\prime}+\\left(-stationary^{3}-variableconst\\ stationary\\right) unchanging=0\n\\]\n\nSince \\( 0>-stationary^{3}-variableconst\\ stationary \\) on this interval, we conclude that between any two roots of a non-zero solution of (4) appears a root of any solution of (3). But (3) has a constant solution with no roots at all, so we see that no non-zero solution of (4) has two roots in \\( [lastletter, \\infty) \\). Hence \\( constantvalue \\) has at most one root in \\( [lastletter, \\infty) \\), so its roots are bounded above.\n\nOn the interval ( \\( -\\infty, firstletter \\) ] we compare (4) with\n\\[\nfixedval^{\\prime \\prime}+1 \\cdot fixedval=0\n\\]\n\nSince \\( -stationary^{3}-variableconst\\ stationary>1 \\) on this interval, we conclude that between any two roots of any non-zero solution of (5) there is a root of any solution of (4). For any choice of \\( middlelet, \\sin (stationary-middlelet) \\) is a non-zero solution of (5) with roots at \\( middlelet-\\pi \\) and \\( middlelet \\). Hence \\( constantvalue \\) has a root in the interval \\( (middlelet-\\pi, middlelet) \\) for any \\( middlelet \\leq firstletter \\). Thus the roots of \\( constantvalue \\) are unbounded below.\n\nA proof of the Sturm comparison theorem is given on page 451."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "f": "hjgrksla",
        "y": "mkdlfrqa",
        "g": "sdhlmvne",
        "h": "flrmqzpt",
        "u": "fznxkqwp",
        "v": "jrdwpqsn",
        "x_0": "pzrtkmla",
        "x_1": "cmvzgktr",
        "x_2": "tfqknpsj",
        "x_3": "lqhpdwre",
        "\\xi": "rskdmbfj",
        "a": "kwjqsbzd",
        "\\alpha": "jtqrlmzn",
        "\\beta": "nxphsrqd",
        "\\gamma": "dflgzrcw"
      },
      "question": "7. Consider the function \\( hjgrksla \\) defined by the differential equation\n\\[\nhjgrksla^{\\prime \\prime}(qzxwvtnp)=\\left(qzxwvtnp^{3}+kwjqsbzd \\, qzxwvtnp\\right) hjgrksla(qzxwvtnp)\n\\]\nand the initial conditions \\( hjgrksla(0)=1, hjgrksla^{\\prime}(0)=0 \\). Prove that the roots of \\( hjgrksla \\) are bounded above but unbounded below.",
      "solution": "First Solution. We show first that the roots of \\( hjgrksla \\) are bounded above. Let \\( jtqrlmzn \\) be a number such that \\( qzxwvtnp^{3}+kwjqsbzd \\, qzxwvtnp>0 \\) for \\( qzxwvtnp \\geq jtqrlmzn \\). We shall prove that no solution of the differential equation\n\\[\nmkdlfrqa^{\\prime \\prime}-\\left(qzxwvtnp^{3}+kwjqsbzd \\, qzxwvtnp\\right) mkdlfrqa=0\n\\]\nexcept the identically zero solution, has more than one root in \\( [jtqrlmzn, \\infty) \\). Suppose \\( sdhlmvne \\) is a non-zero solution of (1) and \\( sdhlmvne\\!\\left(cmvzgktr\\right)=sdhlmvne\\!\\left(tfqknpsj\\right)=0 \\) with \\( jtqrlmzn \\leq cmvzgktr<tfqknpsj \\). Then \\( sdhlmvne \\) is not zero throughout \\( [cmvzgktr, tfqknpsj] \\), so, changing the sign of \\( sdhlmvne \\) if necessary, we assume that \\( sdhlmvne \\) is somewhere positive on \\( [cmvzgktr, tfqknpsj] \\). Let \\( sdhlmvne \\) achieve its maximum value on \\( [cmvzgktr, tfqknpsj] \\) at \\( lqhpdwre \\). Then \\( sdhlmvne\\!\\left(lqhpdwre\\right)>0, sdhlmvne^{\\prime}\\!\\left(lqhpdwre\\right)=0 \\), and \\( sdhlmvne^{\\prime \\prime}\\!\\left(lqhpdwre\\right) \\leq 0 \\) by a standard criterion for a maximum. But\n\\[\nsdhlmvne^{\\prime \\prime}\\!\\left(lqhpdwre\\right)=\\left(lqhpdwre^{3}+kwjqsbzd \\, lqhpdwre\\right) sdhlmvne\\!\\left(lqhpdwre\\right)>0\n\\]\nbecause \\( lqhpdwre \\geq jtqrlmzn \\). This contradiction proves that a non-zero solution of (1), in particular \\( hjgrksla \\), has at most one root in \\( [jtqrlmzn, \\infty) \\). Hence the roots of \\( hjgrksla \\) are bounded above.\n\nWe now show that the roots of \\( hjgrksla \\) are unbounded below. Let \\( nxphsrqd \\) be a number such that \\( qzxwvtnp^{3}+kwjqsbzd \\, qzxwvtnp<-1 \\) for \\( qzxwvtnp \\leq nxphsrqd \\). We shall prove that every solution of (1) has a root on \\( \\left(-\\infty, pzrtkmla\\right] \\) for any choice of \\( pzrtkmla \\leq nxphsrqd \\). Suppose this is false. Then there is a solution \\( flrmqzpt \\) that has constant sign on \\( \\left(-\\infty, pzrtkmla\\right] \\). Changing the sign of \\( flrmqzpt \\) if necessary, we can assume that \\( flrmqzpt \\) is positive on this interval. For any choice of \\( cmvzgktr \\leq pzrtkmla \\) we have by the extended mean value theorem\n\\[\nflrmqzpt(qzxwvtnp)=flrmqzpt\\!\\left(cmvzgktr\\right)+\\left(qzxwvtnp-cmvzgktr\\right) flrmqzpt^{\\prime}\\!\\left(cmvzgktr\\right)+\\frac{1}{2}\\left(qzxwvtnp-cmvzgktr\\right)^{2} flrmqzpt^{\\prime \\prime}\\!\\left(rskdmbfj\\right),\n\\]\nwhere \\( rskdmbfj \\) is between \\( qzxwvtnp \\) and \\( cmvzgktr \\). Then if \\( qzxwvtnp \\leq pzrtkmla \\), we have \\( rskdmbfj<pzrtkmla \\) and \\( flrmqzpt^{\\prime \\prime}\\!\\left(rskdmbfj\\right)=\\left(rskdmbfj^{3}+kwjqsbzd \\, rskdmbfj\\right) flrmqzpt\\!\\left(rskdmbfj\\right)<0 \\) and therefore\n\\[\nflrmqzpt(qzxwvtnp)<flrmqzpt\\!\\left(cmvzgktr\\right)+\\left(qzxwvtnp-cmvzgktr\\right) flrmqzpt^{\\prime}\\!\\left(cmvzgktr\\right).\n\\]\n\nIf we could choose \\( cmvzgktr \\) so that \\( flrmqzpt^{\\prime}\\!\\left(cmvzgktr\\right)>0 \\), this would show that \\( flrmqzpt(qzxwvtnp)<0 \\) for large negative \\( qzxwvtnp \\), contrary to our assumption. So \\( flrmqzpt^{\\prime}(qzxwvtnp) \\leq 0 \\) for all \\( qzxwvtnp \\leq pzrtkmla \\). But this implies \\( flrmqzpt(qzxwvtnp) \\geq flrmqzpt\\!\\left(pzrtkmla\\right) \\) for \\( qzxwvtnp \\leq pzrtkmla \\). Then (2) yields\n\\[\nflrmqzpt(qzxwvtnp)<flrmqzpt\\!\\left(cmvzgktr\\right)+\\left(qzxwvtnp-cmvzgktr\\right) flrmqzpt^{\\prime}\\!\\left(cmvzgktr\\right)-\\frac{1}{2}\\left(qzxwvtnp-cmvzgktr\\right)^{2} flrmqzpt\\!\\left(pzrtkmla\\right),\n\\]\nsince \\( flrmqzpt^{\\prime \\prime}\\!\\left(rskdmbfj\\right)=\\left(rskdmbfj^{3}+kwjqsbzd \\, rskdmbfj\\right) flrmqzpt\\!\\left(rskdmbfj\\right)<-flrmqzpt\\!\\left(pzrtkmla\\right) \\). But this shows that \\( flrmqzpt(qzxwvtnp) \\) is negative for sufficiently large negative values of \\( qzxwvtnp \\). This contradiction proves that every solution of (1), in particular \\( hjgrksla \\), has arbitrarily large negative roots.\n\nSecond Solution. Again we choose \\( jtqrlmzn \\) and \\( nxphsrqd \\) so that \\( qzxwvtnp^{3}+kwjqsbzd \\, qzxwvtnp>0 \\) for \\( qzxwvtnp \\geq jtqrlmzn \\), and \\( qzxwvtnp^{3}+kwjqsbzd \\, qzxwvtnp<-1 \\) for \\( qzxwvtnp \\leq nxphsrqd \\). We shall apply the Sturm comparison theorem.\n\nOn the interval \\( [jtqrlmzn, \\infty) \\) we compare the differential equations\n\\[\nfznxkqwp^{\\prime \\prime}+0 \\cdot fznxkqwp=0\n\\]\nand\n\\[\nmkdlfrqa^{\\prime \\prime}+\\left(-qzxwvtnp^{3}-kwjqsbzd \\, qzxwvtnp\\right) mkdlfrqa=0\n\\]\n\nSince \\( 0>-qzxwvtnp^{3}-kwjqsbzd \\, qzxwvtnp \\) on this interval, we conclude that between any two roots of a non-zero solution of (4) appears a root of any solution of (3). But (3) has a constant solution with no roots at all, so we see that no non-zero solution of (4) has two roots in \\( [jtqrlmzn, \\infty) \\). Hence \\( hjgrksla \\) has at most one root in \\( [jtqrlmzn, \\infty) \\), so its roots are bounded above.\n\nOn the interval \\( (-\\infty, nxphsrqd] \\) we compare (4) with\n\\[\njrdwpqsn^{\\prime \\prime}+1 \\cdot jrdwpqsn=0\n\\]\n\nSince \\( -qzxwvtnp^{3}-kwjqsbzd \\, qzxwvtnp>1 \\) on this interval, we conclude that between any two roots of any non-zero solution of (5) there is a root of any solution of (4). For any choice of \\( dflgzrcw, \\sin (qzxwvtnp-dflgzrcw) \\) is a non-zero solution of (5) with roots at \\( dflgzrcw-\\pi \\) and \\( dflgzrcw \\). Hence \\( hjgrksla \\) has a root in the interval \\( (dflgzrcw-\\pi, dflgzrcw) \\) for any \\( dflgzrcw \\leq nxphsrqd \\). Thus the roots of \\( hjgrksla \\) are unbounded below.\n\nA proof of the Sturm comparison theorem is given on page 451."
    },
    "kernel_variant": {
      "question": "Let b be a real constant and let the function f be defined by\n\\[\n  f''(x)=\\bigl(x^{5}+b\\,x\\bigr)\\,f(x), \\qquad f(0)=1,\\;f'(0)=0.\n\\]\nProve that the zeros of f are bounded above, while to the left they are unbounded: there exists M>0 such that f(x)=0 has **no** solution for x\\ge M, but for every T<0 the equation f(x)=0 possesses at least one root with x\\le T.",
      "solution": "Let f be the unique solution of\n\n  f''(x) = (x^5 + b x)\\,f(x),   f(0)=1,\n  f'(0)=0.\n\nWe split into two parts.\n\n1. Zeros are bounded above.\n\nSince x^5 + b x → +∞ as x → +∞, there is α>0 so that q(x)=x^5+bx>0 for all x ≥ α.  Consider any nontrivial solution y of\n\n  y'' = q(x)\\,y.   (⇒ y'' - q(x)y = 0.)\n\nSuppose y had two zeros x₁<x₂ with x₁,x₂ ≥ α.  Then on [x₁,x₂], y is not identically zero, so without loss of generality y>0 at some interior point.  Let x₃∈(x₁,x₂) be where y attains its maximum.  Then y(x₃)>0, y'(x₃)=0 and y''(x₃)≤0.  But the equation gives\n\n  y''(x₃) = q(x₃)\\,y(x₃) > 0,\n\na contradiction.  Hence y has at most one real zero ≥α.  In particular f has at most one zero in [α,∞), so there is M≥α such that f(x)≠0 for all x ≥ M.  Thus the roots of f are bounded above.\n\n2. Zeros are unbounded below.\n\nSince x^5+bx → −∞ as x→−∞, choose β<0 so that x^5+bx < −1 for all x ≤ β.  Then on (−∞,β] we have\n\n  P(x) = −(x^5+bx) > 1,\n\nand f satisfies\n\n  f'' + P(x)\\,f = 0.\n\nCompare this with the constant‐coefficient equation v'' + 1⋅v = 0, whose nontrivial solution v(x)=sin x has zeros exactly at kπ (k∈ℤ).  By Sturm’s Comparison Theorem, whenever a solution of v''+1⋅v=0 has two consecutive zeros, any solution of f''+P(x)f=0 must have at least one zero between them, provided P(x) ≥1 on the interval of comparison.\n\nNow choose an integer N so that\n\n  (N+1)π ≤ β.\n\nThen for every integer j ≤ N the closed interval [jπ,(j+1)π] lies entirely in (−∞,β], and sin x has consecutive zeros at x=jπ and x=(j+1)π.  By Sturm comparison, f has at least one zero in each open interval (jπ,(j+1)π).  Since there are infinitely many integers j≤N (extending arbitrarily negative), f has infinitely many real zeros tending to −∞.\n\nConclusion.  We have exhibited M>0 so that f(x)≠0 for x≥M, yet shown that for every T<0 there is a root of f in (−∞,T].  Equivalently, the real zeros of f are bounded above but unbounded below, as required.",
      "_meta": {
        "core_steps": [
          "Choose α so that the coefficient q(x)=x³+ax is positive for all x≥α",
          "Prove any non-zero solution has at most one zero on [α,∞) (extremum test or Sturm comparison)",
          "Choose β so that q(x) is uniformly negative on (−∞,β]",
          "Compare with an easier ODE (or use Taylor) to force a zero in every interval far to the left, proving zeros are unbounded below"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "The cubic power 3 in the leading term of q(x)=x³+ax; any odd exponent (or any function tending to ±∞ with opposite signs at ±∞) works.",
            "original": "3"
          },
          "slot2": {
            "description": "Linear-term coefficient in q(x); any real parameter keeps the reasoning.",
            "original": "a"
          },
          "slot3": {
            "description": "Negative constant chosen so q(x)<constant<0 on (−∞,β]; only the fact that it is <0 matters.",
            "original": "-1"
          },
          "slot4": {
            "description": "Zero potential used for the comparison ODE u''+0·u=0; any constant ≤ min q(x) on [α,∞) would do.",
            "original": "0"
          },
          "slot5": {
            "description": "Positive potential in the comparison ODE v''+1·v=0; any positive constant smaller in magnitude than max(−q) on (−∞,β] works.",
            "original": "1"
          },
          "slot6": {
            "description": "Root-spacing π of sin used in Sturm comparison; changes with the constant in slot5 (π/√slot5).",
            "original": "π"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}