{
  "index": "1958-A-5",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "5. Show that the integral equation\n\\[\nf(x, y)=1+\\int_{0}^{x} \\int_{0}^{y} f(u, v) d u d v\n\\]\nhas at most one solution continuous for \\( 0 \\leq x \\leq 1,0 \\leq y \\leq 1 \\).",
  "solution": "Solution. Suppose there are two continuous solutions and let \\( g \\) be their difference. Then \\( g \\) is continuous and\n\\[\ng(x, y)=\\int_{0}^{x} \\int_{0}^{v} g(u, v) d u d v\n\\]\n\nSince \\( g \\) is continuous it is bounded on the given square. Let \\( M \\) be a bound. Then\n\\[\n|g(x, y)| \\leq \\int_{0}^{x} \\int_{0}^{\\cdot x}|g(u, v)| d u d v \\leq \\int_{0}^{x} \\int_{0}^{x} M d u d v=M x y\n\\]\nfor \\( 0 \\leq x \\leq 1,0 \\leq y \\leq 1 \\).\nWe now prove that\n\\[\n|g(x, y)| \\leq M \\frac{x^{n}}{n!} \\frac{y^{\\prime \\prime}}{n!}\n\\]\nfor any positive integer \\( n \\). This has been proved for \\( n=1 \\). Assume that it is true for \\( n=k \\); then\n\\[\n\\begin{aligned}\n|g(x, y)| & \\leq \\int_{0}^{x} \\int_{0}^{y}|g(u, v)| d u d v \\\\\n& \\leq \\int_{0}^{x} \\int_{0}^{y} M \\frac{u^{k}}{k!} \\frac{v^{k}}{k!} d u d v=M \\frac{x^{k+1}}{(k+1)!} \\frac{y^{k+1}}{(k+1)!}\n\\end{aligned}\n\\]\n\nThus (1) is established by mathematical induction. But for any fixed \\( x \\) and \\( y \\)\n\\[\n\\lim _{n \\rightarrow \\infty} M \\frac{x^{n}}{n!} \\frac{y^{n}}{n!}=0\n\\]\n\nHence \\( |g(x, y)| \\leq 0 \\), that is \\( g(x, y)=0 \\). Thus there cannot be two different continuous solutions.\n\nRemark. There is a solution to the given integral equation. It is readily found by the power series method to be given by\n\\[\n\\begin{aligned}\nf(x, y) & =1+x y+\\frac{x^{2}}{2!} \\frac{y^{2}}{2!}+\\frac{x^{3}}{3!} \\frac{y^{3}}{3!}+\\cdots+\\frac{x^{n}}{n!} \\frac{y^{n}}{n!}+\\cdots \\\\\n& =J_{0}(\\sqrt{-4 x y})=J_{0}(2 i \\sqrt{x y})\n\\end{aligned}\n\\]\nwhere \\( J_{0} \\) is the Bessel function of the first kind of order zero. See Whittaker and Watson, Modern Analysis, 4th ed., Cambridge University Press, 1940, page 372.\n\nThe problem generalizes immediately to \\( n \\)-dimensions. See Problem 4885, American Mathematical Monthly, vol. 67 (1960), page 87; Solution, vol. 68 (1961), page 73.",
  "vars": [
    "x",
    "y",
    "u",
    "v",
    "f",
    "g",
    "n",
    "k"
  ],
  "params": [
    "M",
    "J_0"
  ],
  "sci_consts": [
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "xcoord",
        "y": "ycoord",
        "u": "uinside",
        "v": "vinside",
        "f": "unknownfn",
        "g": "difference",
        "n": "indexnum",
        "k": "indexk",
        "M": "boundval",
        "J_0": "besselzero"
      },
      "question": "5. Show that the integral equation\n\\[\nunknownfn(xcoord, ycoord)=1+\\int_{0}^{xcoord} \\int_{0}^{ycoord} unknownfn(uinside, vinside) d uinside d vinside\n\\]\nhas at most one solution continuous for \\( 0 \\leq xcoord \\leq 1,0 \\leq ycoord \\leq 1 \\).",
      "solution": "Solution. Suppose there are two continuous solutions and let \\( difference \\) be their difference. Then \\( difference \\) is continuous and\n\\[\ndifference(xcoord, ycoord)=\\int_{0}^{xcoord} \\int_{0}^{vinside} difference(uinside, vinside) d uinside d vinside\n\\]\n\nSince \\( difference \\) is continuous it is bounded on the given square. Let \\( boundval \\) be a bound. Then\n\\[\n|difference(xcoord, ycoord)| \\leq \\int_{0}^{xcoord} \\int_{0}^{\\cdot xcoord}|difference(uinside, vinside)| d uinside d vinside \\leq \\int_{0}^{xcoord} \\int_{0}^{xcoord} boundval d uinside d vinside=boundval xcoord ycoord\n\\]\nfor \\( 0 \\leq xcoord \\leq 1,0 \\leq ycoord \\leq 1 \\).\nWe now prove that\n\\[\n|difference(xcoord, ycoord)| \\leq boundval \\frac{xcoord^{indexnum}}{indexnum!} \\frac{ycoord^{\\prime \\prime}}{indexnum!}\n\\]\nfor any positive integer \\( indexnum \\). This has been proved for \\( indexnum=1 \\). Assume that it is true for \\( indexnum=indexk \\); then\n\\[\n\\begin{aligned}\n|difference(xcoord, ycoord)| & \\leq \\int_{0}^{xcoord} \\int_{0}^{ycoord}|difference(uinside, vinside)| d uinside d vinside \\\\\n& \\leq \\int_{0}^{xcoord} \\int_{0}^{ycoord} boundval \\frac{uinside^{indexk}}{indexk!} \\frac{vinside^{indexk}}{indexk!} d uinside d vinside=boundval \\frac{xcoord^{indexk+1}}{(indexk+1)!} \\frac{ycoord^{indexk+1}}{(indexk+1)!}\n\\end{aligned}\n\\]\n\nThus (1) is established by mathematical induction. But for any fixed \\( xcoord \\) and \\( ycoord \\)\n\\[\n\\lim _{indexnum \\rightarrow \\infty} boundval \\frac{xcoord^{indexnum}}{indexnum!} \\frac{ycoord^{indexnum}}{indexnum!}=0\n\\]\n\nHence \\( |difference(xcoord, ycoord)| \\leq 0 \\), that is \\( difference(xcoord, ycoord)=0 \\). Thus there cannot be two different continuous solutions.\n\nRemark. There is a solution to the given integral equation. It is readily found by the power series method to be given by\n\\[\n\\begin{aligned}\nunknownfn(xcoord, ycoord) & =1+xcoord ycoord+\\frac{xcoord^{2}}{2!} \\frac{ycoord^{2}}{2!}+\\frac{xcoord^{3}}{3!} \\frac{ycoord^{3}}{3!}+\\cdots+\\frac{xcoord^{indexnum}}{indexnum!} \\frac{ycoord^{indexnum}}{indexnum!}+\\cdots \\\\\n& =besselzero(\\sqrt{-4 xcoord ycoord})=besselzero(2 i \\sqrt{xcoord ycoord})\n\\end{aligned}\n\\]\nwhere \\( besselzero \\) is the Bessel function of the first kind of order zero. See Whittaker and Watson, Modern Analysis, 4th ed., Cambridge University Press, 1940, page 372.\n\nThe problem generalizes immediately to \\( indexnum \\)-dimensions. See Problem 4885, American Mathematical Monthly, vol. 67 (1960), page 87; Solution, vol. 68 (1961), page 73."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pineapple",
        "y": "suitcase",
        "u": "lighthouse",
        "v": "waterfall",
        "f": "umbrella",
        "g": "typewriter",
        "n": "alligator",
        "k": "earthquake",
        "M": "harmonica",
        "J_0": "porcupine"
      },
      "question": "5. Show that the integral equation\n\\[\numbrella(pineapple, suitcase)=1+\\int_{0}^{pineapple} \\int_{0}^{suitcase} umbrella(lighthouse, waterfall) d lighthouse d waterfall\n\\]\nhas at most one solution continuous for \\( 0 \\leq pineapple \\leq 1,0 \\leq suitcase \\leq 1 \\).",
      "solution": "Solution. Suppose there are two continuous solutions and let \\( typewriter \\) be their difference. Then \\( typewriter \\) is continuous and\n\\[\ntypewriter(pineapple, suitcase)=\\int_{0}^{pineapple} \\int_{0}^{waterfall} typewriter(lighthouse, waterfall) d lighthouse d waterfall\n\\]\n\nSince \\( typewriter \\) is continuous it is bounded on the given square. Let \\( harmonica \\) be a bound. Then\n\\[\n|typewriter(pineapple, suitcase)| \\leq \\int_{0}^{pineapple} \\int_{0}^{\\cdot pineapple}|typewriter(lighthouse, waterfall)| d lighthouse d waterfall \\leq \\int_{0}^{pineapple} \\int_{0}^{pineapple} harmonica d lighthouse d waterfall=harmonica pineapple suitcase\n\\]\nfor \\( 0 \\leq pineapple \\leq 1,0 \\leq suitcase \\leq 1 \\).\nWe now prove that\n\\[\n|typewriter(pineapple, suitcase)| \\leq harmonica \\frac{pineapple^{alligator}}{alligator!} \\frac{suitcase^{\\prime \\prime}}{alligator!}\n\\]\nfor any positive integer \\( alligator \\). This has been proved for \\( alligator=1 \\). Assume that it is true for \\( alligator=earthquake \\); then\n\\[\n\\begin{aligned}\n|typewriter(pineapple, suitcase)| & \\leq \\int_{0}^{pineapple} \\int_{0}^{suitcase}|typewriter(lighthouse, waterfall)| d lighthouse d waterfall \\\\\n& \\leq \\int_{0}^{pineapple} \\int_{0}^{suitcase} harmonica \\frac{lighthouse^{earthquake}}{earthquake!} \\frac{waterfall^{earthquake}}{earthquake!} d lighthouse d waterfall=harmonica \\frac{pineapple^{earthquake+1}}{(earthquake+1)!} \\frac{suitcase^{earthquake+1}}{(earthquake+1)!}\n\\end{aligned}\n\\]\n\nThus (1) is established by mathematical induction. But for any fixed \\( pineapple \\) and \\( suitcase \\)\n\\[\n\\lim _{alligator \\rightarrow \\infty} harmonica \\frac{pineapple^{alligator}}{alligator!} \\frac{suitcase^{alligator}}{alligator!}=0\n\\]\n\nHence \\( |typewriter(pineapple, suitcase)| \\leq 0 \\), that is \\( typewriter(pineapple, suitcase)=0 \\). Thus there cannot be two different continuous solutions.\n\nRemark. There is a solution to the given integral equation. It is readily found by the power series method to be given by\n\\[\n\\begin{aligned}\numbrella(pineapple, suitcase) & =1+pineapple suitcase+\\frac{pineapple^{2}}{2!} \\frac{suitcase^{2}}{2!}+\\frac{pineapple^{3}}{3!} \\frac{suitcase^{3}}{3!}+\\cdots+\\frac{pineapple^{alligator}}{alligator!} \\frac{suitcase^{alligator}}{alligator!}+\\cdots \\\\\n& =porcupine(\\sqrt{-4 pineapple suitcase})=porcupine(2 i \\sqrt{pineapple suitcase})\n\\end{aligned}\n\\]\nwhere \\( porcupine \\) is the Bessel function of the first kind of order zero. See Whittaker and Watson, Modern Analysis, 4th ed., Cambridge University Press, 1940, page 372.\n\nThe problem generalizes immediately to \\( alligator \\)-dimensions. See Problem 4885, American Mathematical Monthly, vol. 67 (1960), page 87; Solution, vol. 68 (1961), page 73."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "u": "interiorpoint",
        "v": "exteriorpoint",
        "f": "malfunction",
        "g": "summation",
        "n": "fraction",
        "k": "constant",
        "M": "minimumvalue",
        "J_0": "eulerfunction"
      },
      "question": "5. Show that the integral equation\n\\[\nmalfunction(verticalaxis, horizontalaxis)=1+\\int_{0}^{verticalaxis} \\int_{0}^{horizontalaxis} malfunction(interiorpoint, exteriorpoint) d interiorpoint d exteriorpoint\n\\]\nhas at most one solution continuous for \\( 0 \\leq verticalaxis \\leq 1,0 \\leq horizontalaxis \\leq 1 \\).",
      "solution": "Solution. Suppose there are two continuous solutions and let \\( summation \\) be their difference. Then \\( summation \\) is continuous and\n\\[\nsummation(verticalaxis, horizontalaxis)=\\int_{0}^{verticalaxis} \\int_{0}^{exteriorpoint} summation(interiorpoint, exteriorpoint) d interiorpoint d exteriorpoint\n\\]\n\nSince \\( summation \\) is continuous it is bounded on the given square. Let \\( minimumvalue \\) be a bound. Then\n\\[\n|summation(verticalaxis, horizontalaxis)| \\leq \\int_{0}^{verticalaxis} \\int_{0}^{\\cdot verticalaxis}|summation(interiorpoint, exteriorpoint)| d interiorpoint d exteriorpoint \\leq \\int_{0}^{verticalaxis} \\int_{0}^{verticalaxis} minimumvalue d interiorpoint d exteriorpoint=minimumvalue verticalaxis horizontalaxis\n\\]\nfor \\( 0 \\leq verticalaxis \\leq 1,0 \\leq horizontalaxis \\leq 1 \\).\nWe now prove that\n\\[\n|summation(verticalaxis, horizontalaxis)| \\leq minimumvalue \\frac{verticalaxis^{fraction}}{fraction!} \\frac{horizontalaxis^{\\prime \\prime}}{fraction!}\n\\]\nfor any positive integer \\( fraction \\). This has been proved for \\( fraction=1 \\). Assume that it is true for \\( fraction=constant \\); then\n\\[\n\\begin{aligned}\n|summation(verticalaxis, horizontalaxis)| & \\leq \\int_{0}^{verticalaxis} \\int_{0}^{horizontalaxis}|summation(interiorpoint, exteriorpoint)| d interiorpoint d exteriorpoint \\\\\n& \\leq \\int_{0}^{verticalaxis} \\int_{0}^{horizontalaxis} minimumvalue \\frac{interiorpoint^{constant}}{constant!} \\frac{exteriorpoint^{constant}}{constant!} d interiorpoint d exteriorpoint=minimumvalue \\frac{verticalaxis^{constant+1}}{(constant+1)!} \\frac{horizontalaxis^{constant+1}}{(constant+1)!}\n\\end{aligned}\n\\]\n\nThus (1) is established by mathematical induction. But for any fixed \\( verticalaxis \\) and \\( horizontalaxis \\)\n\\[\n\\lim _{fraction \\rightarrow \\infty} minimumvalue \\frac{verticalaxis^{fraction}}{fraction!} \\frac{horizontalaxis^{fraction}}{fraction!}=0\n\\]\n\nHence \\( |summation(verticalaxis, horizontalaxis)| \\leq 0 \\), that is \\( summation(verticalaxis, horizontalaxis)=0 \\). Thus there cannot be two different continuous solutions.\n\nRemark. There is a solution to the given integral equation. It is readily found by the power series method to be given by\n\\[\n\\begin{aligned}\nmalfunction(verticalaxis, horizontalaxis) & =1+verticalaxis horizontalaxis+\\frac{verticalaxis^{2}}{2!} \\frac{horizontalaxis^{2}}{2!}+\\frac{verticalaxis^{3}}{3!} \\frac{horizontalaxis^{3}}{3!}+\\cdots+\\frac{verticalaxis^{fraction}}{fraction!} \\frac{horizontalaxis^{fraction}}{fraction!}+\\cdots \\\\\n& =eulerfunction(\\sqrt{-4 verticalaxis horizontalaxis})=eulerfunction(2 i \\sqrt{verticalaxis horizontalaxis})\n\\end{aligned}\n\\]\nwhere \\( eulerfunction \\) is the Bessel function of the first kind of order zero. See Whittaker and Watson, Modern Analysis, 4th ed., Cambridge University Press, 1940, page 372.\n\nThe problem generalizes immediately to \\( fraction \\)-dimensions. See Problem 4885, American Mathematical Monthly, vol. 67 (1960), page 87; Solution, vol. 68 (1961), page 73."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "u": "mcvbtrpl",
        "v": "skdjflea",
        "f": "lqwertyu",
        "g": "zmxncbvl",
        "n": "asdfghjk",
        "k": "poiulkjh",
        "M": "rtyghjkl",
        "J_0": "cvbnmert"
      },
      "question": "5. Show that the integral equation\n\\[\nlqwertyu(qzxwvtnp, hjgrksla)=1+\\int_{0}^{qzxwvtnp} \\int_{0}^{hjgrksla} lqwertyu(mcvbtrpl, skdjflea) d mcvbtrpl d skdjflea\n\\]\nhas at most one solution continuous for \\( 0 \\leq qzxwvtnp \\leq 1,0 \\leq hjgrksla \\leq 1 \\).",
      "solution": "Solution. Suppose there are two continuous solutions and let \\( zmxncbvl \\) be their difference. Then \\( zmxncbvl \\) is continuous and\n\\[\nzmxncbvl(qzxwvtnp, hjgrksla)=\\int_{0}^{qzxwvtnp} \\int_{0}^{skdjflea} zmxncbvl(mcvbtrpl, skdjflea) d mcvbtrpl d skdjflea\n\\]\n\nSince \\( zmxncbvl \\) is continuous it is bounded on the given square. Let \\( rtyghjkl \\) be a bound. Then\n\\[\n|zmxncbvl(qzxwvtnp, hjgrksla)| \\leq \\int_{0}^{qzxwvtnp} \\int_{0}^{\\cdot qzxwvtnp}|zmxncbvl(mcvbtrpl, skdjflea)| d mcvbtrpl d skdjflea \\leq \\int_{0}^{qzxwvtnp} \\int_{0}^{qzxwvtnp} rtyghjkl d mcvbtrpl d skdjflea=rtyghjkl qzxwvtnp hjgrksla\n\\]\nfor \\( 0 \\leq qzxwvtnp \\leq 1,0 \\leq hjgrksla \\leq 1 \\).\nWe now prove that\n\\[\n|zmxncbvl(qzxwvtnp, hjgrksla)| \\leq rtyghjkl \\frac{qzxwvtnp^{asdfghjk}}{asdfghjk!} \\frac{hjgrksla^{\\prime \\prime}}{asdfghjk!}\n\\]\nfor any positive integer \\( asdfghjk \\). This has been proved for \\( asdfghjk=1 \\). Assume that it is true for \\( asdfghjk=poiulkjh \\); then\n\\[\n\\begin{aligned}\n|zmxncbvl(qzxwvtnp, hjgrksla)| & \\leq \\int_{0}^{qzxwvtnp} \\int_{0}^{hjgrksla}|zmxncbvl(mcvbtrpl, skdjflea)| d mcvbtrpl d skdjflea \\\\\n& \\leq \\int_{0}^{qzxwvtnp} \\int_{0}^{hjgrksla} rtyghjkl \\frac{mcvbtrpl^{poiulkjh}}{poiulkjh!} \\frac{skdjflea^{poiulkjh}}{poiulkjh!} d mcvbtrpl d skdjflea=rtyghjkl \\frac{qzxwvtnp^{poiulkjh+1}}{(poiulkjh+1)!} \\frac{hjgrksla^{poiulkjh+1}}{(poiulkjh+1)!}\n\\end{aligned}\n\\]\n\nThus (1) is established by mathematical induction. But for any fixed \\( qzxwvtnp \\) and \\( hjgrksla \\)\n\\[\n\\lim _{asdfghjk \\rightarrow \\infty} rtyghjkl \\frac{qzxwvtnp^{asdfghjk}}{asdfghjk!} \\frac{hjgrksla^{asdfghjk}}{asdfghjk!}=0\n\\]\n\nHence \\( |zmxncbvl(qzxwvtnp, hjgrksla)| \\leq 0 \\), that is \\( zmxncbvl(qzxwvtnp, hjgrksla)=0 \\). Thus there cannot be two different continuous solutions.\n\nRemark. There is a solution to the given integral equation. It is readily found by the power series method to be given by\n\\[\n\\begin{aligned}\nlqwertyu(qzxwvtnp, hjgrksla) & =1+qzxwvtnp hjgrksla+\\frac{qzxwvtnp^{2}}{2!} \\frac{hjgrksla^{2}}{2!}+\\frac{qzxwvtnp^{3}}{3!} \\frac{hjgrksla^{3}}{3!}+\\cdots+\\frac{qzxwvtnp^{asdfghjk}}{asdfghjk!} \\frac{hjgrksla^{asdfghjk}}{asdfghjk!}+\\cdots \\\\\n& =cvbnmert(\\sqrt{-4 qzxwvtnp hjgrksla})=cvbnmert(2 i \\sqrt{qzxwvtnp hjgrksla})\n\\end{aligned}\n\\]\nwhere \\( cvbnmert \\) is the Bessel function of the first kind of order zero. See Whittaker and Watson, Modern Analysis, 4th ed., Cambridge University Press, 1940, page 372.\n\nThe problem generalizes immediately to \\( asdfghjk \\)-dimensions. See Problem 4885, American Mathematical Monthly, vol. 67 (1960), page 87; Solution, vol. 68 (1961), page 73."
    },
    "kernel_variant": {
      "question": "Let λ be a real constant with \n |λ| < ½.  \nFor the four–variable box  \n Q = [0,1] × [0,1] × [0,1] × [0,1]  \nconsider the integral–differential equation  \n\n f(x,y,z,t)  \n = e^{x y}\\sin (z+t)                                         (0.1)  \n  + ∫_{0}^{x}∫_{0}^{y}∫_{0}^{z}∫_{0}^{t}  \n   (x-u)(y-v)^2 e^{\\,z-w}\\,(t-s)\\;f(u,v,w,s)\\,ds\\,dw\\,dv\\,du  \n\n  + λ ∂/∂x \\biggl[ ∫_{0}^{x}∫_{0}^{y}∫_{0}^{z}∫_{0}^{t}  \n    f(u,v,w,s)\\,ds\\,dw\\,dv\\,du \\biggr],                   (0.2)  \n\ndefined for every (x,y,z,t) ∈ Q.  \n\nProve that there exists at most one C¹–function  \n f : Q → ℝ  \nthat satisfies (0.1)–(0.2) simultaneously on the whole box Q.",
      "solution": "Step 1.  Reduce the problem to a homogeneous equation.  \nAssume f₁ and f₂ are two C¹–solutions of (0.1)–(0.2) and set  \n\n g = f₁ − f₂.                                                (1.1)\n\nSubtracting the two copies of (0.1)–(0.2) gives  \n\n g(x,y,z,t)  \n = ∫_{0}^{x}∫_{0}^{y}∫_{0}^{z}∫_{0}^{t}  \n  (x-u)(y-v)^2 e^{\\,z-w}\\,(t-s)\\;g(u,v,w,s)\\,ds\\,dw\\,dv\\,du   (1.2)  \n\n  + λ ∂/∂x \\biggl[ ∫_{0}^{x}∫_{0}^{y}∫_{0}^{z}∫_{0}^{t}  \n    g(u,v,w,s)\\,ds\\,dw\\,dv\\,du \\biggr].                    \n\nHence g is a continuous (in fact C¹) solution of the homogeneous integral–differential equation (1.2).\n\nStep 2.  Uniform bound for the first (Volterra–type) integral term.  \nPut  \n\n ‖g‖ = sup_{(x,y,z,t)∈Q}|g(x,y,z,t)|.                      (1.3)\n\nInside Q we have 0 ≤ x,y,z,t ≤ 1.  From non–negativity of the factors,\n\n 0 ≤ (x-u) ≤ 1,  0 ≤ (y-v)^2 ≤ 1,  \n 0 ≤ e^{z-w} ≤ e,  0 ≤ (t-s) ≤ 1.\n\nTherefore\n\n | (x-u)(y-v)^2 e^{z-w}(t-s) | ≤ e.                        (1.4)\n\nConsequently,\n\n | ∫ … g(u,v,w,s) |  \n ≤ e‖g‖ ∫_{0}^{x}∫_{0}^{y}∫_{0}^{z}∫_{0}^{t} ds\\,dw\\,dv\\,du  \n = e‖g‖·x·y·z·t  \n ≤ e‖g‖·1·1·1·1  \n = e‖g‖.                                                  (1.5)\n\nA sharper elementary evaluation of the four integrals gives the smaller constant  \n\n A := (1/2)·(1/3)·(e−1)·(1/2) = (e−1)/12 ≈ 0.143,        (1.6)\n\nbut even the rough upper bound e suffices for the following argument.  For definiteness we keep the accurate value A and write\n\n | first integral term | ≤ A‖g‖.                          (1.7)\n\nStep 3.  Bound for the derivative term.  \nDefine  \n\n G(x,y,z,t) = ∫_{0}^{x}∫_{0}^{y}∫_{0}^{z}∫_{0}^{t}  \n    g(u,v,w,s)\\,ds\\,dw\\,dv\\,du.                         (1.8)\n\nBecause g is continuous, G is C¹ and  \n\n ∂G/∂x(x,y,z,t)  \n = ∫_{0}^{y}∫_{0}^{z}∫_{0}^{t} g(x,v,w,s)\\,ds\\,dw\\,dv.   (1.9)\n\nThus,\n\n |∂G/∂x| ≤ ‖g‖·y·z·t ≤ ‖g‖.                              (1.10)\n\nHence the magnitude of the derivative contribution in (1.2) is bounded by\n\n |λ|·|∂G/∂x| ≤ |λ|‖g‖.                                   (1.11)\n\nStep 4.  A global inequality for ‖g‖.  \nCombining (1.7) and (1.11) we obtain, for every (x,y,z,t)∈Q,\n\n |g(x,y,z,t)| ≤ (A+|λ|)‖g‖.                              (1.12)\n\nTaking the supremum over Q yields\n\n ‖g‖ ≤ (A+|λ|)‖g‖.                                       (1.13)\n\nStep 5.  Conclude g≡0 using the smallness of A+|λ|.  \nOur hypothesis |λ| < ½ implies\n\n A+|λ| ≤ 0.143+½ < 1.                                   (1.14)\n\nHence (1.13) forces ‖g‖=0, i.e. g vanishes identically on Q.  \nTherefore f₁≡f₂, and the integral–differential equation (0.1)–(0.2) admits at most one C¹–solution on the four–dimensional box Q.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.506001",
        "was_fixed": false,
        "difficulty_analysis": "1. Additional variables: the problem passes from two and three variables to four variables, expanding the region of integration to a genuine 4-dimensional box.\n2. Weighted kernel: the kernel (x-u)(y-v)²e^{z-w}(t-s) is neither separable nor constant; its anisotropic polynomial-exponential structure forces careful integral estimates.\n3. Integral-differential nature: besides the Volterra–type integral term, a derivative with respect to x is applied to another 4-fold integral of f.  Handling this term requires computing the derivative of a parametrised multiple integral and bounding it systematically.\n4. Parameter dependence: uniqueness must be proved uniformly in a non-trivial parameter λ.  The argument has to isolate a critical constant A and show that uniqueness holds whenever |λ| keeps the sum A+|λ| below 1.\n5. Higher regularity: the solution is sought in C¹ rather than mere continuity, demanding that all manipulations (differentiation under the integral sign, etc.) be justified.\n6. Interacting concepts: the proof blends multidimensional Volterra theory, uniform norm estimates, and an elementary fixed-point (contraction) argument, none of which by itself is sufficient.\n\nAll these layers of complexity make the enhanced kernel variant significantly more demanding—both technically and conceptually—than the original and the simpler 3-variable kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let λ be a real constant with \n |λ| < ½.  \nFor the four–variable box  \n Q = [0,1] × [0,1] × [0,1] × [0,1]  \nconsider the integral–differential equation  \n\n f(x,y,z,t)  \n = e^{x y}\\sin (z+t)                                         (0.1)  \n  + ∫_{0}^{x}∫_{0}^{y}∫_{0}^{z}∫_{0}^{t}  \n   (x-u)(y-v)^2 e^{\\,z-w}\\,(t-s)\\;f(u,v,w,s)\\,ds\\,dw\\,dv\\,du  \n\n  + λ ∂/∂x \\biggl[ ∫_{0}^{x}∫_{0}^{y}∫_{0}^{z}∫_{0}^{t}  \n    f(u,v,w,s)\\,ds\\,dw\\,dv\\,du \\biggr],                   (0.2)  \n\ndefined for every (x,y,z,t) ∈ Q.  \n\nProve that there exists at most one C¹–function  \n f : Q → ℝ  \nthat satisfies (0.1)–(0.2) simultaneously on the whole box Q.",
      "solution": "Step 1.  Reduce the problem to a homogeneous equation.  \nAssume f₁ and f₂ are two C¹–solutions of (0.1)–(0.2) and set  \n\n g = f₁ − f₂.                                                (1.1)\n\nSubtracting the two copies of (0.1)–(0.2) gives  \n\n g(x,y,z,t)  \n = ∫_{0}^{x}∫_{0}^{y}∫_{0}^{z}∫_{0}^{t}  \n  (x-u)(y-v)^2 e^{\\,z-w}\\,(t-s)\\;g(u,v,w,s)\\,ds\\,dw\\,dv\\,du   (1.2)  \n\n  + λ ∂/∂x \\biggl[ ∫_{0}^{x}∫_{0}^{y}∫_{0}^{z}∫_{0}^{t}  \n    g(u,v,w,s)\\,ds\\,dw\\,dv\\,du \\biggr].                    \n\nHence g is a continuous (in fact C¹) solution of the homogeneous integral–differential equation (1.2).\n\nStep 2.  Uniform bound for the first (Volterra–type) integral term.  \nPut  \n\n ‖g‖ = sup_{(x,y,z,t)∈Q}|g(x,y,z,t)|.                      (1.3)\n\nInside Q we have 0 ≤ x,y,z,t ≤ 1.  From non–negativity of the factors,\n\n 0 ≤ (x-u) ≤ 1,  0 ≤ (y-v)^2 ≤ 1,  \n 0 ≤ e^{z-w} ≤ e,  0 ≤ (t-s) ≤ 1.\n\nTherefore\n\n | (x-u)(y-v)^2 e^{z-w}(t-s) | ≤ e.                        (1.4)\n\nConsequently,\n\n | ∫ … g(u,v,w,s) |  \n ≤ e‖g‖ ∫_{0}^{x}∫_{0}^{y}∫_{0}^{z}∫_{0}^{t} ds\\,dw\\,dv\\,du  \n = e‖g‖·x·y·z·t  \n ≤ e‖g‖·1·1·1·1  \n = e‖g‖.                                                  (1.5)\n\nA sharper elementary evaluation of the four integrals gives the smaller constant  \n\n A := (1/2)·(1/3)·(e−1)·(1/2) = (e−1)/12 ≈ 0.143,        (1.6)\n\nbut even the rough upper bound e suffices for the following argument.  For definiteness we keep the accurate value A and write\n\n | first integral term | ≤ A‖g‖.                          (1.7)\n\nStep 3.  Bound for the derivative term.  \nDefine  \n\n G(x,y,z,t) = ∫_{0}^{x}∫_{0}^{y}∫_{0}^{z}∫_{0}^{t}  \n    g(u,v,w,s)\\,ds\\,dw\\,dv\\,du.                         (1.8)\n\nBecause g is continuous, G is C¹ and  \n\n ∂G/∂x(x,y,z,t)  \n = ∫_{0}^{y}∫_{0}^{z}∫_{0}^{t} g(x,v,w,s)\\,ds\\,dw\\,dv.   (1.9)\n\nThus,\n\n |∂G/∂x| ≤ ‖g‖·y·z·t ≤ ‖g‖.                              (1.10)\n\nHence the magnitude of the derivative contribution in (1.2) is bounded by\n\n |λ|·|∂G/∂x| ≤ |λ|‖g‖.                                   (1.11)\n\nStep 4.  A global inequality for ‖g‖.  \nCombining (1.7) and (1.11) we obtain, for every (x,y,z,t)∈Q,\n\n |g(x,y,z,t)| ≤ (A+|λ|)‖g‖.                              (1.12)\n\nTaking the supremum over Q yields\n\n ‖g‖ ≤ (A+|λ|)‖g‖.                                       (1.13)\n\nStep 5.  Conclude g≡0 using the smallness of A+|λ|.  \nOur hypothesis |λ| < ½ implies\n\n A+|λ| ≤ 0.143+½ < 1.                                   (1.14)\n\nHence (1.13) forces ‖g‖=0, i.e. g vanishes identically on Q.  \nTherefore f₁≡f₂, and the integral–differential equation (0.1)–(0.2) admits at most one C¹–solution on the four–dimensional box Q.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.422561",
        "was_fixed": false,
        "difficulty_analysis": "1. Additional variables: the problem passes from two and three variables to four variables, expanding the region of integration to a genuine 4-dimensional box.\n2. Weighted kernel: the kernel (x-u)(y-v)²e^{z-w}(t-s) is neither separable nor constant; its anisotropic polynomial-exponential structure forces careful integral estimates.\n3. Integral-differential nature: besides the Volterra–type integral term, a derivative with respect to x is applied to another 4-fold integral of f.  Handling this term requires computing the derivative of a parametrised multiple integral and bounding it systematically.\n4. Parameter dependence: uniqueness must be proved uniformly in a non-trivial parameter λ.  The argument has to isolate a critical constant A and show that uniqueness holds whenever |λ| keeps the sum A+|λ| below 1.\n5. Higher regularity: the solution is sought in C¹ rather than mere continuity, demanding that all manipulations (differentiation under the integral sign, etc.) be justified.\n6. Interacting concepts: the proof blends multidimensional Volterra theory, uniform norm estimates, and an elementary fixed-point (contraction) argument, none of which by itself is sufficient.\n\nAll these layers of complexity make the enhanced kernel variant significantly more demanding—both technically and conceptually—than the original and the simpler 3-variable kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}