{
"type": "Article",
"authors": [
{
"type": "Person",
"familyNames": [
"Rassias"
],
"givenNames": [
"Michael",
"Th."
]
}
],
"identifiers": [],
"references": [
{
"type": "Article",
"id": "bib-bib1",
"authors": [],
"title": "\nJ. Chinis, Siegel zeros and Sarnak’s conjecture,\npreprint, arXiv:2105.14653 (2021)\n",
"url": "https://arxiv.org/abs/2105.14653"
},
{
"type": "Article",
"id": "bib-bib2",
"authors": [],
"title": "\nS. Chowla, The Riemann hypothesis and Hilbert’s tenth problem.\nMathematics and its Applications 4, Gordon and Breach Science Publishers, New York (1965)\n"
},
{
"type": "Article",
"id": "bib-bib3",
"authors": [],
"title": "\nP. D. T. A. Elliott, On the correlation of multiplicative functions.\nNotas Soc. Mat. Chile 11, 1–11 (1992)\n"
},
{
"type": "Article",
"id": "bib-bib4",
"authors": [],
"title": "\nG. H. Hardy and J. E. Littlewood, Some problems of ‘Partitio numerorum’; III: On the expression of a number as a sum of primes.\nActa Math. 44, 1–70 (1923)\n"
},
{
"type": "Article",
"id": "bib-bib5",
"authors": [],
"title": "\nD. R. Heath-Brown, Prime twins and Siegel zeros.\nProc. London Math. Soc. (3) 47, 193–224 (1983)\n"
},
{
"type": "Article",
"id": "bib-bib6",
"authors": [],
"title": "\nK. Matomäki, J. Merikoski, Siegel zeros, twin primes, Goldbach’s conjecture, and primes in short intervals,\npreprint, arXiv:2112.11412 (2021)\n",
"url": "https://arxiv.org/abs/2112.11412"
},
{
"type": "Article",
"id": "bib-bib7",
"authors": [],
"title": "\nK. Matomäki and M. Radziwiłł, Multiplicative functions in short intervals.\nAnn. of Math. (2) 183, 1015–1056 (2016)\n"
},
{
"type": "Article",
"id": "bib-bib8",
"authors": [],
"title": "\nK. Matomäki, M. Radziwiłł and T. Tao, An averaged form of Chowla’s conjecture.\nAlgebra Number Theory 9, 2167–2196 (2015)\n"
},
{
"type": "Article",
"id": "bib-bib9",
"authors": [],
"title": "\nK. Matomäki, M. Radziwiłł and T. Tao, Fourier uniformity of bounded multiplicative functions in short intervals on average.\nInvent. Math. 220, 1–58 (2020)\n"
},
{
"type": "Article",
"id": "bib-bib10",
"authors": [],
"title": "\nP. Sarnak, Mobius randomness and dynamics.\nNot. S. Afr. Math. Soc. 43, 89–97 (2012)\n"
},
{
"type": "Article",
"id": "bib-bib11",
"authors": [],
"title": "\nW. Sawin and M. Shusterman, On the Chowla and twin primes conjectures over 𝔽q[T].\nAnn. of Math. (2) 196, 457–506 (2022)\n"
},
{
"type": "Article",
"id": "bib-bib12",
"authors": [],
"title": "\nT. Tao, The logarithmically averaged Chowla and Elliott conjectures for two-point correlations.\nForum Math. Pi 4, Paper No. e8 (2016)\n"
},
{
"type": "Article",
"id": "bib-bib13",
"authors": [],
"title": "\nT. Tao, Equivalence of the logarithmically averaged Chowla and Sarnak conjectures.\nIn Number Theory—Diophantine Problems, Uniform Distribution and Applications, Springer, Cham, 391–421 (2017)\n"
},
{
"type": "Article",
"id": "bib-bib14",
"authors": [],
"title": "\nT. Tao and J. Teräväinen, Odd order cases of the logarithmically averaged Chowla conjecture.\nJ. Théor. Nombres Bordeaux 30, 997–1015 (2018)\n"
},
{
"type": "Article",
"id": "bib-bib15",
"authors": [],
"title": "\nT. Tao, J. Teräväinen, The Hardy–Littlewood–Chowla conjecture in the presence of a Siegel zero,\npreprint, arXiv:2109.06291 (2021)\n",
"url": "https://arxiv.org/abs/2109.06291"
},
{
"type": "Article",
"id": "bib-bib16",
"authors": [],
"title": "\nT. Tao and J. Teräväinen, The structure of logarithmically averaged correlations of multiplicative functions, with applications to the Chowla and Elliott conjectures.\nDuke Math. J. 168, 1977–2027 (2019)\n"
}
],
"title": "Solved and unsolved problems",
"meta": {},
"content": [
{
"type": "Paragraph",
"id": "p1",
"content": [
"The present column is devoted to Number Theory."
]
},
{
"type": "Heading",
"id": "S1",
"depth": 1,
"content": [
"I Six new problems – solutions solicited"
]
},
{
"type": "Paragraph",
"id": "S1.p1",
"content": [
"Solutions will appear in a subsequent issue."
]
},
{
"type": "Heading",
"id": "S1.SSx1",
"depth": 2,
"content": [
"269"
]
},
{
"type": "Paragraph",
"content": [
"Consider two positive integers ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "n≥1",
"meta": {
"altText": "n\\geq 1"
}
},
" and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "a≥2",
"meta": {
"altText": "a\\geq 2"
}
},
" such that"
]
},
{
"type": "MathBlock",
"id": "S1.Ex1",
"mathLanguage": "mathml",
"text": "a2n+an+1",
"meta": {
"altText": "a^{2n}+a^{n}+1"
}
},
{
"type": "Paragraph",
"content": [
"is a prime.\nProve that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "n",
"meta": {
"altText": "n"
}
},
" is a power of ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "3",
"meta": {
"altText": "3"
}
},
"."
]
},
{
"type": "Paragraph",
"id": "S1.SSx1.p2",
"content": [
{
"type": "Emphasis",
"content": [
"Dorin Andrica and George Cătălin Ţurcaş (Babeş-Bolyai University, Cluj-Napoca, Romania)"
]
}
]
},
{
"type": "Heading",
"id": "S1.SSx2",
"depth": 2,
"content": [
"270"
]
},
{
"type": "Paragraph",
"content": [
"The Collatz map is defined as follows:"
]
},
{
"type": "MathBlock",
"id": "S1.Ex2",
"mathLanguage": "mathml",
"text": "Col(n):={n/2ifnis even,3n+1ifnis odd.",
"meta": {
"altText": "\\operatorname{Col}(n):=\\begin{cases}n/2&\\textrm{if}\\ n\\ \\textrm{is even},\\\\\n3n+1&\\textrm{if}\\ n\\ \\textrm{is odd}.\\end{cases}"
}
},
{
"type": "Paragraph",
"content": [
"Let"
]
},
{
"type": "MathBlock",
"id": "S1.Ex3",
"mathLanguage": "mathml",
"text": "0:\\operatorname{Col}^{m}(n)\\geq x).\" display=\"block\">tm,x≔min(n>0:Colm(n)≥x).",
"meta": {
"altText": "t_{m,x}≔\\min(n>0:\\operatorname{Col}^{m}(n)\\geq x)."
}
},
{
"type": "Paragraph",
"content": [
"That is, ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "tm,x",
"meta": {
"altText": "t_{m,x}"
}
},
" is the smallest integer such that, if we apply the Collatz map ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "m",
"meta": {
"altText": "m"
}
},
" times, the result is larger than ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x",
"meta": {
"altText": "x"
}
},
"."
]
},
{
"type": "Paragraph",
"id": "S1.SSx2.p2",
"content": [
"(a) Find ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "t3,1000",
"meta": {
"altText": "t_{3,1000}"
}
},
" and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "t4,1000",
"meta": {
"altText": "t_{4,1000}"
}
},
"."
]
},
{
"type": "Paragraph",
"content": [
"(b) Show that, for ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x",
"meta": {
"altText": "x"
}
},
" large enough (larger than (say) ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "1000",
"meta": {
"altText": "1000"
}
},
"), we have"
]
},
{
"type": "MathBlock",
"id": "S1.Ex4",
"mathLanguage": "mathml",
"text": "t4,x≡3mod4ort4,x≡6mod8.",
"meta": {
"altText": "t_{4,x}\\equiv 3\\bmod 4\\quad\\textrm{or}\\quad t_{4,x}\\equiv 6\\bmod 8."
}
},
{
"type": "Paragraph",
"id": "S1.SSx2.p4",
"content": [
"(c) In general, for ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "m",
"meta": {
"altText": "m"
}
},
" odd and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x",
"meta": {
"altText": "x"
}
},
" large enough, there exists a constant ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "Xm,x",
"meta": {
"altText": "X_{m,x}"
}
},
" such that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "tm,x",
"meta": {
"altText": "t_{m,x}"
}
},
" is the smallest ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "X_{m,x}\" display=\"inline\">n>Xm,x",
"meta": {
"altText": "n>X_{m,x}"
}
},
" such that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "n≡cmmodMm",
"meta": {
"altText": "n\\equiv c_{m}\\bmod M_{m}"
}
},
".\nFind ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "Mm",
"meta": {
"altText": "M_{m}"
}
},
" and relate ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "cm",
"meta": {
"altText": "c_{m}"
}
},
" to ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "cm−1",
"meta": {
"altText": "c_{m-1}"
}
},
".\n"
]
},
{
"type": "Paragraph",
"id": "S1.SSx2.p5",
"content": [
{
"type": "Emphasis",
"content": [
"Christopher Lutsko (Department of Mathematics, Rutgers University, Piscataway, USA)"
]
}
]
},
{
"type": "Heading",
"id": "S1.SSx3",
"depth": 2,
"content": [
"271"
]
},
{
"type": "Paragraph",
"id": "S1.SSx3.p1",
"content": [
"The light-bulb problem: Alice and Bob are in jail for trying to divide by ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "0",
"meta": {
"altText": "0"
}
},
".\nThe jailer proposes the following game to decide their freedom: Alice will be shown an ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "n×n",
"meta": {
"altText": "n\\times n"
}
},
" grid of light bulbs.\nThe jailer will point to a light bulb of his choice and Alice will decide whether it should be on or off.\nThen the jailer will point to another bulb of his choice and Alice will decide on/off.\nThis continues until the very last bulb, when the jailer will decide whether this bulb is on or off.\nSo the jailer controls the order of the selection, and the state of the final bulb.\nAlice is now removed from the room, and Bob is brought in.\nBob’s goal is to choose ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "n",
"meta": {
"altText": "n"
}
},
" bulbs such that his selection includes the final bulb (the one determined by the jailer)."
]
},
{
"type": "Paragraph",
"id": "S1.SSx3.p2",
"content": [
"Is there a strategy that Alice and Bob can use to guarantee success?\nWhat if Bob does not know the orientation in which Alice saw the board (i.e., what if Bob does not know which are the rows and which are the columns)?"
]
},
{
"type": "Paragraph",
"id": "S1.SSx3.p3",
"content": [
{
"type": "Emphasis",
"content": [
"Christopher Lutsko (Department of Mathematics, Rutgers University, Piscataway, USA)"
]
}
]
},
{
"type": "Heading",
"id": "S1.SSx4",
"depth": 2,
"content": [
"272"
]
},
{
"type": "Paragraph",
"id": "S1.SSx4.p1",
"content": [
"Let ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "p",
"meta": {
"altText": "p"
}
},
" and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "q",
"meta": {
"altText": "q"
}
},
" be coprime integers greater than or equal to ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "2",
"meta": {
"altText": "2"
}
},
".\nLet ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "invq(p)",
"meta": {
"altText": "\\operatorname{inv}_{q}(p)"
}
},
" and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "invp(q)",
"meta": {
"altText": "\\operatorname{inv}_{p}(q)"
}
},
" denote the modular inverse of ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "pmodq",
"meta": {
"altText": "p\\bmod q"
}
},
" and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "qmodp",
"meta": {
"altText": "q\\bmod p"
}
},
", respectively.\nThat is, ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "invq(p)p≡1modq",
"meta": {
"altText": "\\operatorname{inv}_{q}(p)p\\equiv 1\\bmod q"
}
},
" and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "invp(q)q≡1modp",
"meta": {
"altText": "\\operatorname{inv}_{p}(q)q\\equiv 1\\bmod p"
}
},
"."
]
},
{
"type": "Paragraph",
"content": [
"(a) Show that"
]
},
{
"type": "MathBlock",
"id": "S1.Ex5",
"mathLanguage": "mathml",
"text": "\\frac{q}{2}.\" display=\"block\">invp(q)≤p2if and only ifinvq(p)>q2.",
"meta": {
"altText": "\\operatorname{inv}_{p}(q)\\leq\\frac{p}{2}\\quad\\textrm{if and only if}\\quad\\operatorname{inv}_{q}(p)>\\frac{q}{2}."
}
},
{
"type": "Paragraph",
"content": [
"(b) Show by providing an example that, if ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "1≤u<v",
"meta": {
"altText": "1\\leq uα≔u/v",
"meta": {
"altText": "\\alpha≔u/v"
}
},
", then the statement"
]
},
{
"type": "MathBlock",
"id": "S1.E1",
"mathLanguage": "mathml",
"text": "(1-\\alpha)q\" display=\"block\">invp(q)≤αp if and only if invq(p)>(1−α)q",
"meta": {
"altText": "\\operatorname{inv}_{p}(q)\\leq\\alpha p\\quad\\textrm{ if and only if }\\quad\\operatorname{inv}_{q}(p)>(1-\\alpha)q"
}
},
{
"type": "Paragraph",
"content": [
"is not necessarily true."
]
},
{
"type": "Paragraph",
"id": "S1.SSx4.p4",
"content": [
"(c) What additional assumption should ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "p",
"meta": {
"altText": "p"
}
},
" and/or ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "q",
"meta": {
"altText": "q"
}
},
" satisfy so that the equivalence (",
{
"type": "Cite",
"target": "S1-E1",
"content": [
"1"
]
},
") holds?\n"
]
},
{
"type": "Paragraph",
"id": "S1.SSx4.p5",
"content": [
{
"type": "Emphasis",
"content": [
"Athanasios Sourmelidis (Institut für Analysis und Zahlentheorie, Technische Universität Graz, Austria)"
]
}
]
},
{
"type": "Heading",
"id": "S1.SSx5",
"depth": 2,
"content": [
"273"
]
},
{
"type": "Paragraph",
"content": [
"Let ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "cn(k)",
"meta": {
"altText": "c_{n}(k)"
}
},
" denote the Ramanujan sum defined as the sum of ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "k",
"meta": {
"altText": "k"
}
},
"th powers of the primitive ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "n",
"meta": {
"altText": "n"
}
},
"th roots of unity.\nShow that, for any integer ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "m≥1",
"meta": {
"altText": "m\\geq 1"
}
},
","
]
},
{
"type": "MathBlock",
"id": "S1.Ex6",
"mathLanguage": "mathml",
"text": "∑[n,k]=mcn(k)=φ(m),",
"meta": {
"altText": "\\sum_{[n,k]=m}c_{n}(k)=\\varphi(m),"
}
},
{
"type": "Paragraph",
"content": [
"where the sum is over all ordered pairs ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "(n,k)",
"meta": {
"altText": "(n,k)"
}
},
" of positive integers ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "n,k",
"meta": {
"altText": "n,k"
}
},
" such that their lcm is ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "m",
"meta": {
"altText": "m"
}
},
", and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "φ",
"meta": {
"altText": "\\varphi"
}
},
" is Euler’s totient function.\n"
]
},
{
"type": "Paragraph",
"id": "S1.SSx5.p2",
"content": [
{
"type": "Emphasis",
"content": [
"László Tóth (University of Pécs, Hungary)"
]
}
]
},
{
"type": "Heading",
"id": "S1.SSx6",
"depth": 2,
"content": [
"274"
]
},
{
"type": "Paragraph",
"content": [
"Show that, for every integer ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "n≥1",
"meta": {
"altText": "n\\geq 1"
}
},
", we have the polynomial identity"
]
},
{
"type": "MathBlock",
"id": "S1.Ex7",
"mathLanguage": "mathml",
"text": "∏k=1(k,n)=1n(x(k−1,n)−1)=∏d∣nΦd(x)φ(n)/φ(d),",
"meta": {
"altText": "\\prod_{\\begin{subarray}{c}k=1\\\\\n(k,n)=1\\end{subarray}}^{n}(x^{(k-1,n)}-1)=\\prod_{d\\mid n}\\Phi_{d}(x)^{\\varphi(n)/\\varphi(d)},"
}
},
{
"type": "Paragraph",
"content": [
"where ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "Φd(x)",
"meta": {
"altText": "\\Phi_{d}(x)"
}
},
" are the cyclotomic polynomials and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "φ",
"meta": {
"altText": "\\varphi"
}
},
" denotes Euler’s totient function."
]
},
{
"type": "Paragraph",
"id": "S1.SSx6.p2",
"content": [
{
"type": "Emphasis",
"content": [
"László Tóth (Department of Mathematics, University of Pécs, Hungary)"
]
}
]
},
{
"type": "Heading",
"id": "S2",
"depth": 1,
"content": [
"II Open problem"
]
},
{
"type": "Heading",
"id": "S2.SSx1",
"depth": 2,
"content": [
"275*. Chowla’s conjecture and its relatives"
]
},
{
"type": "Paragraph",
"id": "S2.SSx1.p1",
"content": [
{
"type": "Emphasis",
"content": [
"by Terence Tao (UCLA, Department of Mathematics, Los Angeles, USA)"
]
}
]
},
{
"type": "Paragraph",
"content": [
"Let ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "λ:ℕ→{−1,+1}",
"meta": {
"altText": "\\lambda\\colon\\mathbb{N}\\to\\{-1,+1\\}"
}
},
" denote the Liouville function.\nIn [",
{
"type": "Cite",
"target": "bib-bib2",
"content": [
"2"
]
},
"], Chowla conjectured that"
]
},
{
"type": "MathBlock",
"id": "S2.E2",
"mathLanguage": "mathml",
"text": "∑n≤xλ(n+h1)⋯λ(n+hk)=o(x)",
"meta": {
"altText": "\\sum_{n\\leq x}\\lambda(n+h_{1})\\cdots\\lambda(n+h_{k})=o(x)"
}
},
{
"type": "Paragraph",
"content": [
"as ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x→∞",
"meta": {
"altText": "x\\to\\infty"
}
},
", for any distinct natural numbers ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "h1,…,hk",
"meta": {
"altText": "h_{1},\\ldots,h_{k}"
}
},
" (in fact, Chowla made the more general conjecture that"
]
},
{
"type": "MathBlock",
"id": "S2.Ex8",
"mathLanguage": "mathml",
"text": "∑n≤xλ(P(n))=o(x)",
"meta": {
"altText": "\\sum_{n\\leq x}\\lambda(P(n))=o(x)"
}
},
{
"type": "Paragraph",
"content": [
"whenever ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "P",
"meta": {
"altText": "P"
}
},
" is a square-free polynomial mapping from ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "ℕ",
"meta": {
"altText": "\\mathbb{N}"
}
},
" to ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "ℕ",
"meta": {
"altText": "\\mathbb{N}"
}
},
").\nChowla’s conjecture was extended to other bounded multiplicative functions by Elliott [",
{
"type": "Cite",
"target": "bib-bib3",
"content": [
"3"
]
},
"] (see also a technical correction to the conjecture in [",
{
"type": "Cite",
"target": "bib-bib8",
"content": [
"8"
]
},
"])."
]
},
{
"type": "Paragraph",
"content": [
"One can view (",
{
"type": "Cite",
"target": "S2-E2",
"content": [
"2"
]
},
") as a less difficult cousin of the notorious ",
{
"type": "Emphasis",
"content": [
"Hardy–Littlewood prime tuples conjecture"
]
},
" [",
{
"type": "Cite",
"target": "bib-bib4",
"content": [
"4"
]
},
"], which conjectures an asymptotic of the form"
]
},
{
"type": "MathBlock",
"id": "S2.E3",
"mathLanguage": "mathml",
"text": "∑n≤xΛ(n+h1)⋯Λ(n+hk)=𝔖x+o(x)",
"meta": {
"altText": "\\sum_{n\\leq x}\\Lambda(n+h_{1})\\cdots\\Lambda(n+h_{k})=\\mathfrak{S}x+o(x)"
}
},
{
"type": "Paragraph",
"content": [
"where the ",
{
"type": "Emphasis",
"content": [
"singular series"
]
},
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "𝔖",
"meta": {
"altText": "\\mathfrak{S}"
}
},
" is an explicit product over primes of factors involving the numbers ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "h1,…,hk",
"meta": {
"altText": "h_{1},\\ldots,h_{k}"
}
},
".\nFor ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "k=1",
"meta": {
"altText": "k=1"
}
},
", both conjectures follow readily from the prime number theorem, but they remain open for higher ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "k",
"meta": {
"altText": "k"
}
},
".\nHowever, the analogue of (",
{
"type": "Cite",
"target": "S2-E2",
"content": [
"2"
]
},
") (and (",
{
"type": "Cite",
"target": "S2-E3",
"content": [
"3"
]
},
") for ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "k≤2",
"meta": {
"altText": "k\\leq 2"
}
},
") were recently established in certain function fields [",
{
"type": "Cite",
"target": "bib-bib11",
"content": [
"11"
]
},
"], and are also known to hold in the presence of a Siegel zero [",
{
"type": "Cite",
"target": "bib-bib1",
"content": [
"1"
]
},
", ",
{
"type": "Cite",
"target": "bib-bib5",
"content": [
"5"
]
},
", ",
{
"type": "Cite",
"target": "bib-bib15",
"content": [
"15"
]
},
", ",
{
"type": "Cite",
"target": "bib-bib6",
"content": [
"6"
]
},
"].\nThe conjecture (",
{
"type": "Cite",
"target": "S2-E2",
"content": [
"2"
]
},
") is also known if one performs enough averaging in the ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "h1,…,hk",
"meta": {
"altText": "h_{1},\\dots,h_{k}"
}
},
" variables [",
{
"type": "Cite",
"target": "bib-bib8",
"content": [
"8"
]
},
"]."
]
},
{
"type": "Paragraph",
"content": [
"The logarithmically averaged version"
]
},
{
"type": "MathBlock",
"id": "S2.E4",
"mathLanguage": "mathml",
"text": "∑n≤xλ(n+h1)⋯λ(n+hk)n=o(logx)",
"meta": {
"altText": "\\sum_{n\\leq x}\\frac{\\lambda(n+h_{1})\\cdots\\lambda(n+h_{k})}{n}=o(\\log x)"
}
},
{
"type": "Paragraph",
"content": [
"turns out to be more tractable, as it can be analyzed by the “entropy decrement method” [",
{
"type": "Cite",
"target": "bib-bib12",
"content": [
"12"
]
},
"], which has been successfully used to establish the conjecture for ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "k=2",
"meta": {
"altText": "k=2"
}
},
" (see [",
{
"type": "Cite",
"target": "bib-bib12",
"content": [
"12"
]
},
"], building upon the breakthrough work [",
{
"type": "Cite",
"target": "bib-bib7",
"content": [
"7"
]
},
"]) and for odd ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "k",
"meta": {
"altText": "k"
}
},
" (see [",
{
"type": "Cite",
"target": "bib-bib14",
"content": [
"14"
]
},
", ",
{
"type": "Cite",
"target": "bib-bib16",
"content": [
"16"
]
},
"]).\nThe conjecture (",
{
"type": "Cite",
"target": "S2-E4",
"content": [
"4"
]
},
") for arbitrary ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "k",
"meta": {
"altText": "k"
}
},
" is also known to be equivalent [",
{
"type": "Cite",
"target": "bib-bib13",
"content": [
"13"
]
},
"] to the (logarithmically averaged) Sarnak conjecture [",
{
"type": "Cite",
"target": "bib-bib10",
"content": [
"10"
]
},
"], which asserts that"
]
},
{
"type": "MathBlock",
"id": "S2.Ex9",
"mathLanguage": "mathml",
"text": "∑n≤xλ(n)F(Tnx0)n=o(logx)",
"meta": {
"altText": "\\sum_{n\\leq x}\\frac{\\lambda(n)F(T^{n}x_{0})}{n}=o(\\log x)"
}
},
{
"type": "Paragraph",
"content": [
"whenever ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "T:X→X",
"meta": {
"altText": "T\\colon X\\to X"
}
},
" is a compact dynamical system of zero entropy, ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x0",
"meta": {
"altText": "x_{0}"
}
},
" is a point in ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "X",
"meta": {
"altText": "X"
}
},
", and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "F:X→ℂ",
"meta": {
"altText": "F\\colon X\\to\\mathbb{C}"
}
},
" is continuous.\nMany special cases of this conjecture are known, unfortunately too many to list here."
]
},
{
"type": "Paragraph",
"content": [
"The conjecture (",
{
"type": "Cite",
"target": "S2-E4",
"content": [
"4"
]
},
") would also follow from a ",
{
"type": "Emphasis",
"content": [
"higher-order local Fourier uniformity conjecture"
]
},
" [",
{
"type": "Cite",
"target": "bib-bib12",
"content": [
"12"
]
},
"], which is somewhat complicated to state in full generality here; however, the first unsolved special case of this conjecture asserts that"
]
},
{
"type": "MathBlock",
"id": "S2.Ex10",
"mathLanguage": "mathml",
"text": "∫X2Xsupα∈ℝ/ℤ|∑x≤n≤x+Hλ(n)e(αn)|dx=o(XH)asX→∞",
"meta": {
"altText": "\\int_{X}^{2X}\\sup_{\\alpha\\in\\mathbb{R}/\\mathbb{Z}}\\Bigl|\\sum_{x\\leq n\\leq x+H}\\lambda(n)e(\\alpha n)\\Bigr|\\,dx=o(XH)\\quad\\textrm{as}\\ X\\to\\infty"
}
},
{
"type": "Paragraph",
"content": [
"whenever ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "1≤H=H(X)≤X",
"meta": {
"altText": "1\\leq H=H(X)\\leq X"
}
},
" is such that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "H(X)→∞",
"meta": {
"altText": "H(X)\\to\\infty"
}
},
" as ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "X→∞",
"meta": {
"altText": "X\\to\\infty"
}
},
", where ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "e(θ)≔e2πiθ",
"meta": {
"altText": "e(\\theta)≔e^{2\\pi i\\theta}"
}
},
".\nThis is currently only established in the regime ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "H≥Xε",
"meta": {
"altText": "H\\geq X^{\\varepsilon}"
}
},
" for a fixed ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "0\" display=\"inline\">ε>0",
"meta": {
"altText": "\\varepsilon>0"
}
},
" (see [",
{
"type": "Cite",
"target": "bib-bib9",
"content": [
"9"
]
},
"])."
]
},
{
"type": "Heading",
"id": "S3",
"depth": 1,
"content": [
"III Solutions"
]
},
{
"type": "Heading",
"id": "S3.SSx1",
"depth": 2,
"content": [
"260"
]
},
{
"type": "Paragraph",
"id": "S3.SSx1.p1",
"content": [
"Let ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "f:[0,∞)→ℝ",
"meta": {
"altText": "f\\colon[0,\\infty)\\to\\mathbb{R}"
}
},
" be a ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "C1",
"meta": {
"altText": "C^{1}"
}
},
"-differentiable and convex function with ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "f(0)=0",
"meta": {
"altText": "f(0)=0"
}
},
"."
]
},
{
"type": "Paragraph",
"content": [
"(i) Prove that, for every ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x∈[0,∞)",
"meta": {
"altText": "x\\in[0,\\infty)"
}
},
", the following inequality holds:\n"
]
},
{
"type": "MathBlock",
"id": "S3.Ex11",
"mathLanguage": "mathml",
"text": "∫0xf(t)𝑑t≤x22f′(x).",
"meta": {
"altText": "\\int_{0}^{x}f(t)\\,dt\\leq\\frac{x^{2}}{2}f^{\\prime}(x)."
}
},
{
"type": "Paragraph",
"id": "S3.SSx1.p3",
"content": [
"(ii) Determine all functions ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "f",
"meta": {
"altText": "f"
}
},
" for which we have equality.\n"
]
},
{
"type": "Paragraph",
"id": "S3.SSx1.p4",
"content": [
{
"type": "Emphasis",
"content": [
"Dorin Andrica (“Babeş-Bolyai” University, Cluj-Napoca, Romania) and\nMihai Piticari (“Dragoş Vodă” National College, Câmpulung Moldovenesc, Romania)"
]
}
]
},
{
"type": "Heading",
"id": "S3.SSx1.SSSx1",
"depth": 3,
"content": [
"Solution by the proposers"
]
},
{
"type": "Paragraph",
"content": [
"(i) We have"
]
},
{
"type": "MathBlock",
"id": "S3.Ex12",
"mathLanguage": "mathml",
"text": "∫0xf(t)𝑑t=∫0x(f(t)−f(0))𝑑t.",
"meta": {
"altText": "\\int_{0}^{x}f(t)\\,dt=\\int_{0}^{x}\\bigl(f(t)-f(0)\\bigr)\\,dt."
}
},
{
"type": "Paragraph",
"content": [
"By the Mean Value Theorem,"
]
},
{
"type": "MathBlock",
"id": "S3.Ex13",
"mathLanguage": "mathml",
"text": "f(t)−f(0)=tf′(ct)",
"meta": {
"altText": "f(t)-f(0)=tf^{\\prime}(c_{t})"
}
},
{
"type": "Paragraph",
"content": [
"for some ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "ct∈(0,t)⊂[0,x]",
"meta": {
"altText": "c_{t}\\in(0,t)\\subset[0,x]"
}
},
".\nSince ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "f",
"meta": {
"altText": "f"
}
},
" is convex, it follows that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "f′",
"meta": {
"altText": "f^{\\prime}"
}
},
" is increasing; hence ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "f′(ct)≤f′(x)",
"meta": {
"altText": "f^{\\prime}(c_{t})\\leq f^{\\prime}(x)"
}
},
".\nTherefore, ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "f(t)=tf′(ct)≤tf′(x)",
"meta": {
"altText": "f(t)=tf^{\\prime}(c_{t})\\leq tf^{\\prime}(x)"
}
},
", and we obtain\n"
]
},
{
"type": "MathBlock",
"id": "S3.Ex14",
"mathLanguage": "mathml",
"text": "∫0xf(t)𝑑t≤∫0xtf′(x)𝑑t=f′(x)∫0xt𝑑t=x22f′(x).",
"meta": {
"altText": "\\int_{0}^{x}f(t)\\,dt\\leq\\int_{0}^{x}tf^{\\prime}(x)\\,dt=f^{\\prime}(x)\\int_{0}^{x}t\\,dt=\\frac{x^{2}}{2}f^{\\prime}(x)."
}
},
{
"type": "Paragraph",
"content": [
"(ii) We have to find all the solutions to the equation"
]
},
{
"type": "MathBlock",
"id": "S3.Ex15",
"mathLanguage": "mathml",
"text": "∫0xf(t)dt=x22f′(x).",
"meta": {
"altText": "\\mathop{\\smash[b]{\\int_{0}^{x}}}f(t)\\,dt=\\frac{x^{2}}{2}f^{\\prime}(x)."
}
},
{
"type": "Paragraph",
"content": [
"Denoting"
]
},
{
"type": "MathBlock",
"id": "S3.Ex16",
"mathLanguage": "mathml",
"text": "g(x)=∫0xf(t)dt,x∈[0,∞),",
"meta": {
"altText": "g(x)=\\mathop{\\smash[t]{\\int_{0}^{x}}}f(t)\\,dt,\\quad x\\in[0,\\infty),"
}
},
{
"type": "Paragraph",
"content": [
"the above equation is equivalent to the second-order differential equation"
]
},
{
"type": "MathBlock",
"id": "S3.Ex17",
"mathLanguage": "mathml",
"text": "g(x)=x22g′′(x),x∈[0,∞).",
"meta": {
"altText": "g(x)=\\smash[t]{\\frac{x^{2}}{2}}g^{\\prime\\prime}(x),\\quad x\\in[0,\\infty)."
}
},
{
"type": "Paragraph",
"content": [
"Note that if ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "g",
"meta": {
"altText": "g"
}
},
" is a solution, then"
]
},
{
"type": "MathBlock",
"id": "S3.Ex18",
"mathLanguage": "mathml",
"text": "g(x)+xg′(x)=x22g′′(x)+xg′(x),",
"meta": {
"altText": "g(x)+xg^{\\prime}(x)=\\smash[b]{\\frac{x^{2}}{2}}g^{\\prime\\prime}(x)+xg^{\\prime}(x),"
}
},
{
"type": "Paragraph",
"content": [
"whence"
]
},
{
"type": "MathBlock",
"id": "S3.Ex19",
"mathLanguage": "mathml",
"text": "(xg(x))′=(x22g′(x))′.",
"meta": {
"altText": "(xg(x))^{\\prime}=\\smash{\\Bigl(\\frac{x^{2}}{2}g^{\\prime}(x)\\Bigr)^{\\prime}}."
}
},
{
"type": "Paragraph",
"content": [
"It follows that"
]
},
{
"type": "MathBlock",
"id": "S3.Ex20",
"mathLanguage": "mathml",
"text": "(xg(x)−x22g′(x))′=0,",
"meta": {
"altText": "\\smash{\\Bigl(xg(x)-\\frac{x^{2}}{2}g^{\\prime}(x)\\Bigr)^{\\prime}}=0,"
}
},
{
"type": "Paragraph",
"content": [
"and so"
]
},
{
"type": "MathBlock",
"id": "S3.Ex21",
"mathLanguage": "mathml",
"text": "x22g′(x)−xg(x)=C1,x∈[0,∞),",
"meta": {
"altText": "\\smash[t]{\\frac{x^{2}}{2}}g^{\\prime}(x)-xg(x)=C_{1},\\quad x\\in[0,\\infty),"
}
},
{
"type": "Paragraph",
"content": [
"where ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "C1",
"meta": {
"altText": "C_{1}"
}
},
" is an arbitrary constant.\nThis last equation is equivalent to"
]
},
{
"type": "MathBlock",
"id": "S3.Ex22",
"mathLanguage": "mathml",
"text": "x2g′(x)−2xg(x)x4=2C1x4,x∈(0,∞),",
"meta": {
"altText": "\\smash[b]{\\frac{x^{2}g^{\\prime}(x)-2xg(x)}{x^{4}}=\\frac{2C_{1}}{x^{4}}},\\quad x\\in(0,\\infty),"
}
},
{
"type": "Paragraph",
"content": [
"or"
]
},
{
"type": "MathBlock",
"id": "S3.Ex23",
"mathLanguage": "mathml",
"text": "(g(x)x2)′=2C1x4,x∈(0,∞).",
"meta": {
"altText": "\\smash{\\Bigl(\\frac{g(x)}{x^{2}}\\Bigr)^{\\prime}=\\frac{2C_{1}}{x^{4}}},\\quad\nx\\in(0,\\infty)."
}
},
{
"type": "Paragraph",
"content": [
"Consequently,\n"
]
},
{
"type": "MathBlock",
"id": "S3.Ex24",
"mathLanguage": "mathml",
"text": "g(x)x2=−2C13x3+C2,x∈(0,∞),",
"meta": {
"altText": "\\smash[b]{\\frac{g(x)}{x^{2}}=\\frac{-2C_{1}}{3x^{3}}}+C_{2},\\quad x\\in(0,\\infty),"
}
},
{
"type": "Paragraph",
"content": [
"and we get"
]
},
{
"type": "MathBlock",
"id": "S3.Ex25",
"mathLanguage": "mathml",
"text": "g(x)=−2C13x+C2x2,x∈(0,∞).",
"meta": {
"altText": "g(x)=\\smash[t]{\\frac{-2C_{1}}{3x}}+C_{2}x^{2},\\quad x\\in(0,\\infty)."
}
},
{
"type": "Paragraph",
"content": [
"On the other hand, ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "g",
"meta": {
"altText": "g"
}
},
" is continuous and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "g(0)=0",
"meta": {
"altText": "g(0)=0"
}
},
".\nThis implies that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "C1=0",
"meta": {
"altText": "C_{1}=0"
}
},
" and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "g(x)=C2x2",
"meta": {
"altText": "g(x)=C_{2}x^{2}"
}
},
".\nWe conclude that the sought-for functions are necessarily of the form ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "f(x)=g′(x)=2C2x",
"meta": {
"altText": "f(x)=g^{\\prime}(x)=2C_{2}x"
}
},
", i.e., ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "f(x)=Ax",
"meta": {
"altText": "f(x)=Ax"
}
},
", where ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "A",
"meta": {
"altText": "A"
}
},
" is an arbitrary real constant."
]
},
{
"type": "Heading",
"id": "S3.SSx2",
"depth": 2,
"content": [
"261"
]
},
{
"type": "Paragraph",
"content": [
"Let ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y(x)",
"meta": {
"altText": "y(x)"
}
},
" be the unknown function of the following fractional-order derivative Cauchy problem:"
]
},
{
"type": "MathBlock",
"id": "S3.E5",
"mathLanguage": "mathml",
"text": "{Dαy=f(x,y),0<α<1,y(0)=y*.",
"meta": {
"altText": "\\left\\{\\begin{aligned} D^{\\alpha}y&=f(x,y),\\quad 0<\\alpha<1,\\\\\ny(0)&=y^{*}.\\end{aligned}\\right."
}
},
{
"type": "Paragraph",
"content": [
"Find the solution of this problem by solving an equivalent first-order ordinary Cauchy problem, with a solution independent of the kernel of the fractional operator.\n"
]
},
{
"type": "Paragraph",
"id": "S3.SSx2.p2",
"content": [
{
"type": "Emphasis",
"content": [
"Carlo Cattani (Engineering School, DEIM, University “La Tuscia”, Viterbo, Italy)"
]
}
]
},
{
"type": "Heading",
"id": "S3.SSx2.SSSx1",
"depth": 3,
"content": [
"Solution by the proposer"
]
},
{
"type": "Paragraph",
"id": "S3.SSx2.SSSx1.p1",
"content": [
"Before we give the solution of (",
{
"type": "Cite",
"target": "S3-E5",
"content": [
"5"
]
},
"), let us make some preliminary remarks about the most popular definitions of fractional derivatives."
]
},
{
"type": "Paragraph",
"content": [
"The Riemann–Liouville integral of fractional order ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "ν≥0",
"meta": {
"altText": "\\nu\\geq 0"
}
},
" of a function ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "f(x)",
"meta": {
"altText": "f(x)"
}
},
" is defined as"
]
},
{
"type": "MathBlock",
"id": "S3.Ex26",
"mathLanguage": "mathml",
"text": "0,\\\\[1.5pt]\nf(t),&\\nu=0.\\end{cases}\" display=\"block\">(Jνf)(t)={1Γ(ν)∫0t(t−τ)ν−1f(τ)𝑑τ,ν>0,f(t),ν=0.",
"meta": {
"altText": "(J^{\\nu}f)(t)=\\begin{cases}\\displaystyle\\frac{1}{\\Gamma(\\nu)}\\int_{0}^{t}(t-\\tau)^{\\nu-1}{f(\\tau)}\\,d\\tau,&\\nu>0,\\\\[1.5pt]\nf(t),&\\nu=0.\\end{cases}"
}
},
{
"type": "Paragraph",
"content": [
"The corresponding Riemann–Liouville fractional derivative of order ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "0\" display=\"inline\">α>0",
"meta": {
"altText": "\\alpha>0"
}
},
" is defined as"
]
},
{
"type": "MathBlock",
"id": "S3.E6",
"mathLanguage": "mathml",
"text": "DRLαf(t)=dndtnJn−αf(t),n∈ℕ,n−1<α≤n.",
"meta": {
"altText": "D^{\\alpha}_{\\mathrm{RL}}f(t)=\\frac{d^{n}}{dt^{n}}J^{n-\\alpha}f(t),\\quad n\\in\\mathbb{N},\\,n-1<\\alpha\\leq n."
}
},
{
"type": "Paragraph",
"content": [
"The main problem with this derivative is that it assigns a nonzero value to a constant function.\nTo avoid this issue, people often use the so-called order-",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "α",
"meta": {
"altText": "\\alpha"
}
},
" Caputo fractional derivative, defined as"
]
},
{
"type": "MathBlock",
"id": "S3.Ex27",
"mathLanguage": "mathml",
"text": "0,\\\\[-3.0pt]\n&0\\leq n-1<\\alphaDCαf(x)={dnf(x)dxn,0<α∈ℕ,1Γ(n−α)∫0xf(n)(τ)(x−τ)α−n+1𝑑τ,t>0,0≤n−1<α<n,",
"meta": {
"altText": "D^{\\alpha}_{\\mathrm{C}}f(x)=\\begin{cases}\\dfrac{d^{n}f(x)}{dx^{n}},&0<\\alpha\\in\\mathbb{N},\\\\[3.0pt]\n\\displaystyle\\frac{1}{\\Gamma(n-\\alpha)}\\int_{0}^{x}\\frac{f^{(n)}(\\tau)}{(x-\\tau)^{\\alpha-n+1}}\\,d\\tau,&t>0,\\\\[-3.0pt]\n&0\\leq n-1<\\alphan",
"meta": {
"altText": "n"
}
},
" is an integer, ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "0\" display=\"inline\">x>0",
"meta": {
"altText": "x>0"
}
},
", and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "f∈𝒞n",
"meta": {
"altText": "f\\in\\mathcal{C}^{n}"
}
},
"."
]
},
{
"type": "Paragraph",
"content": [
"Riemann–Liouville (RL) and Caputo (C) derivatives are the most popular and have been used in many applications; nevertheless, they both have some drawbacks.\nFor this reason, many authors have introduced some more flexible fractional operators.\nThe most general fractional derivative with a given kernel ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "K(x,α)",
"meta": {
"altText": "K(x,\\alpha)"
}
},
" is defined as"
]
},
{
"type": "MathBlock",
"id": "S3.E7",
"mathLanguage": "mathml",
"text": "0,\\\\[-3.0pt]\n&0\\leq n-1<\\alphaDαf(x)={dnf(x)dxn,0<α∈ℕ,∫0xf(n)(τ)K(x−τ,α)𝑑τ,t>0,0≤n−1<α<n.",
"meta": {
"altText": "D^{\\alpha}f(x)=\\begin{cases}\\dfrac{d^{n}f(x)}{dx^{n}},&0<\\alpha\\in\\mathbb{N},\\\\[6.45831pt]\n\\displaystyle\\int_{0}^{x}f^{(n)}(\\tau)K(x-\\tau,\\alpha)\\,d\\tau,&t>0,\\\\[-3.0pt]\n&0\\leq n-1<\\alphalimα→0K(x−τ,α)=1andlimα→1K(x−τ,α)=δ(x−τ).",
"meta": {
"altText": "\\lim_{\\alpha\\to 0}K(x-\\tau,\\alpha)=1\\quad\\textrm{and}\\quad\\lim_{\\alpha\\to 1}K(x-\\tau,\\alpha)=\\delta(x-\\tau)."
}
},
{
"type": "Paragraph",
"content": [
"Moreover, to ensure that one is dealing with a non-singular kernel, one requires that"
]
},
{
"type": "MathBlock",
"id": "S3.Ex29",
"mathLanguage": "mathml",
"text": "limx→τK(x−τ,α)≠0∀α.",
"meta": {
"altText": "\\lim_{x\\to\\tau}K(x-\\tau,\\alpha)\\neq 0\\quad\\forall\\alpha."
}
},
{
"type": "Paragraph",
"id": "S3.SSx2.SSSx1.p5",
"content": [
"Although several definitions of fractional derivatives are available, they all depend on the proposed kernel, thus implying a subjective and a priori unjustified choice of the fractional operator each time one studies a fractional differential problem.\nThis issue can be avoided by using the following simple definition, which is based on an intuitive interpolation."
]
},
{
"type": "Paragraph",
"content": [
"Limiting ourselves to the case ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "n=1",
"meta": {
"altText": "n=1"
}
},
", the general structure of the Caputo-type fractional derivative"
]
},
{
"type": "MathBlock",
"id": "S3.Ex30",
"mathLanguage": "mathml",
"text": "DCαf(x)=∫0xf′(τ)K(x−τ,α)𝑑τ,0<α<1,",
"meta": {
"altText": "D^{\\alpha}_{\\mathrm{C}}f(x)=\\int_{0}^{x}f^{\\prime}(\\tau)K(x-\\tau,\\alpha)\\,d\\tau,\\quad 0<\\alpha<1,"
}
},
{
"type": "Paragraph",
"content": [
"is based on the kernel ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "K(x−τ,α)",
"meta": {
"altText": "K(x-\\tau,\\alpha)"
}
},
", which is a positive function that decays at infinity (to ensure convergence), while according to (",
{
"type": "Cite",
"target": "S3-E6",
"content": [
"6"
]
},
"), the general structure of the Riemann–Liouville first-order derivative is"
]
},
{
"type": "MathBlock",
"id": "S3.Ex31",
"mathLanguage": "mathml",
"text": "DRLαf(x)=1Γ(1−α)ddx∫0xf(τ)(x−τ)−α𝑑τ,0<α≤1.",
"meta": {
"altText": "D^{\\alpha}_{\\mathrm{RL}}f(x)=\\frac{1}{\\Gamma(1-\\alpha)}\\frac{d}{dx}\\int_{0}^{x}f(\\tau)(x-\\tau)^{-\\alpha}\\,d\\tau,\\quad 0<\\alpha\\leq 1."
}
},
{
"type": "Paragraph",
"content": [
"Usually, to find the solution of (",
{
"type": "Cite",
"target": "S3-E5",
"content": [
"5"
]
},
"), we should first choose the kernel of the fractional operator and then solve the fractional problem by using a suitable numerical method, which roughly consists in constructing and solving an equivalent algebraic/differential (of integer order) problem.\nIn any case, the solution will depend not only on the independent variable ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x",
"meta": {
"altText": "x"
}
},
" and the initial condition ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y*",
"meta": {
"altText": "y^{*}"
}
},
", but also on the kernel and on the fractional-order parameter:"
]
},
{
"type": "MathBlock",
"id": "S3.Ex32",
"mathLanguage": "mathml",
"text": "y(x,y*,K(x−τ),α).",
"meta": {
"altText": "y(x,y^{*},K(x-\\tau),\\alpha)."
}
},
{
"type": "Paragraph",
"content": [
"The dependence on the fractional parameter ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "α",
"meta": {
"altText": "\\alpha"
}
},
" is essential in solving fractional-order problems.\nHowever, the dependence on the kernel leads to an inessential “struggle” about the best choice of the kernel and about its physical/mathematical meaning – an obviously subjective and non-unique choice.\nBecause of this lack of uniqueness, fractional calculus is missing a strong mathematical motivation.\nOn the other hand, there exist many useful mathematical tools, important for the solution of differential problems, that require making choices, such as wavelets, orthogonal polynomials, integral transforms, and many more.\nTherefore, one can either ignore the uniqueness problem, or try to defend a specific choice by using some reasoning that may still be not sufficient to convince the mathematics community.\nIn the following, we give a solution which is both independent of the choice of the kernel and can be analytically obtained by reduction to an equivalent ordinary differential problem having the same solution as (",
{
"type": "Cite",
"target": "S3-E5",
"content": [
"5"
]
},
")."
]
},
{
"type": "Paragraph",
"content": [
"We search the solution by assuming that the fractional derivative is obtained by linear interpolation between a function and its first-order derivative; consequently, we do not need the integral definition (",
{
"type": "Cite",
"target": "S3-E7",
"content": [
"7"
]
},
") and the accompanying choice of kernel.\nThis is an acceptable assumption, based on the original simple idea in the fundamentals of fractional calculus that the fractional parameter describes a family of interpolation curves.\nThus, we set"
]
},
{
"type": "MathBlock",
"id": "S3.Ex33",
"mathLanguage": "mathml",
"text": "Dαy=(1−α)(y−y*)+αdydx",
"meta": {
"altText": "D^{\\alpha}y=(1-\\alpha)(y-y^{*})+\\alpha\\frac{dy}{dx}"
}
},
{
"type": "Paragraph",
"content": [
"so that the initial value problem (",
{
"type": "Cite",
"target": "S3-E5",
"content": [
"5"
]
},
") becomes"
]
},
{
"type": "MathBlock",
"id": "S3.Ex34",
"mathLanguage": "mathml",
"text": "αdydx+(1−α)y=(1−α)y*+f(x,y).",
"meta": {
"altText": "\\alpha\\frac{dy}{dx}+(1-\\alpha)y=(1-\\alpha)y^{*}+f(x,y)."
}
},
{
"type": "Paragraph",
"content": [
"Now, for ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "0<α<1",
"meta": {
"altText": "0<\\alpha<1"
}
},
", we easily obtain the following ordinary differential problem that is equivalent to (",
{
"type": "Cite",
"target": "S3-E5",
"content": [
"5"
]
},
"):"
]
},
{
"type": "MathBlock",
"id": "S3.E8",
"mathLanguage": "mathml",
"text": "{dydx−(1−1α)y=−(1−1α)y*+1αf(x,y),y(0)=y0.",
"meta": {
"altText": "\\left\\{\\begin{aligned} \\frac{dy}{dx}-\\Bigl(1-\\frac{1}{\\alpha}\\Bigr)y&=-\\Bigl(1-\\frac{1}{\\alpha}\\Bigr)y^{*}+\\frac{1}{\\alpha}f(x,y),\\\\\ny(0)&=y_{0}.\\end{aligned}\\right."
}
},
{
"type": "Paragraph",
"id": "S3.SSx2.SSSx1.p10",
"content": [
"For the moment, we can suppose that the initial conditions of (",
{
"type": "Cite",
"target": "S3-E5",
"content": [
"5"
]
},
") and (",
{
"type": "Cite",
"target": "S3-E8",
"content": [
"8"
]
},
") do not necessarily coincide, i.e., ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y*≠y0",
"meta": {
"altText": "y^{*}\\neq y_{0}"
}
},
".\nHowever, some further assumptions can be made when the function ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "f(x,y)",
"meta": {
"altText": "f(x,y)"
}
},
" is given explicitly.\nIn particular, to achieve a perfect equivalence between (",
{
"type": "Cite",
"target": "S3-E5",
"content": [
"5"
]
},
") and (",
{
"type": "Cite",
"target": "S3-E8",
"content": [
"8"
]
},
"), we can set ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y*=(1−α)τ+αy0∀τ∈ℝ",
"meta": {
"altText": "y^{*}=(1-\\alpha)\\tau+\\alpha y_{0}\\,\\forall\\tau\\in\\mathbb{R}"
}
},
", and then let ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "α→1",
"meta": {
"altText": "\\alpha\\to 1"
}
},
"."
]
},
{
"type": "Heading",
"id": "S3.SSx3",
"depth": 2,
"content": [
"262"
]
},
{
"type": "Paragraph",
"content": [
"Let ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y(x)",
"meta": {
"altText": "y(x)"
}
},
" be the unknown function of the following Bernoulli fractional-order Cauchy problem:"
]
},
{
"type": "MathBlock",
"id": "S3.E9",
"mathLanguage": "mathml",
"text": "{Dαy=g(x)yβ,0<α<1,β≠0,1,y(0)=y*,",
"meta": {
"altText": "\\left\\{\\begin{aligned} D^{\\alpha}y&=g(x)y^{\\beta},\\quad 0<\\alpha<1,\\,\\beta\\neq\n0,1,\\\\\ny(0)&=y^{*},\\end{aligned}\\right."
}
},
{
"type": "Paragraph",
"content": [
"where ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "g(x)",
"meta": {
"altText": "g(x)"
}
},
" is a continuous function in the interval ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "I=[0,∞)",
"meta": {
"altText": "I=[0,\\infty)"
}
},
"."
]
},
{
"type": "Paragraph",
"id": "S3.SSx3.p2",
"content": [
"Find the solution of this problem by solving an equivalent first-order ordinary Cauchy problem, with a solution independent of the kernel of the fractional operator."
]
},
{
"type": "Paragraph",
"id": "S3.SSx3.p3",
"content": [
{
"type": "Emphasis",
"content": [
"Carlo Cattani (Engineering School, DEIM, University “La Tuscia”, Viterbo, Italy)"
]
}
]
},
{
"type": "Heading",
"id": "S3.SSx3.SSSx1",
"depth": 3,
"content": [
"Solution by the proposer"
]
},
{
"type": "Paragraph",
"id": "S3.SSx3.SSSx1.p1",
"content": [
"We search the solution by simply assuming that the fractional derivative is a linear interpolation between a function and its first-order derivative so that we do not need the usual integral definition of the fractional operator which requires us to choose the underlying kernel."
]
},
{
"type": "Paragraph",
"content": [
"Thus, we set"
]
},
{
"type": "MathBlock",
"id": "S3.E10",
"mathLanguage": "mathml",
"text": "Dαy=(1−α)(y−y*)+αdydx",
"meta": {
"altText": "D^{\\alpha}y=(1-\\alpha)(y-y^{*})+\\alpha\\smash{\\frac{dy}{dx}}"
}
},
{
"type": "Paragraph",
"content": [
"so that (",
{
"type": "Cite",
"target": "S3-E9",
"content": [
"9"
]
},
") becomes"
]
},
{
"type": "MathBlock",
"id": "S3.E11",
"mathLanguage": "mathml",
"text": "αdydx+(1−α)y=(1−α)y*+g(x)yβ.",
"meta": {
"altText": "\\alpha\\frac{dy}{dx}+(1-\\alpha)y=(1-\\alpha)y^{*}+g(x)y^{\\beta}."
}
},
{
"type": "Paragraph",
"content": [
"Then taking ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "0<α<1",
"meta": {
"altText": "0<\\alpha<1"
}
},
" and using equations (",
{
"type": "Cite",
"target": "S3-E10",
"content": [
"10"
]
},
") and (",
{
"type": "Cite",
"target": "S3-E11",
"content": [
"11"
]
},
"), we easily get the following ordinary differential problem equivalent to (",
{
"type": "Cite",
"target": "S3-E9",
"content": [
"9"
]
},
"):"
]
},
{
"type": "MathBlock",
"id": "S3.E12",
"mathLanguage": "mathml",
"text": "{dydx−(1−1α)y=−(1−1α)y*+1αg(x)yβ,β≠{0,1},y(0)=y0.",
"meta": {
"altText": "\\left\\{\\begin{aligned} \\frac{dy}{dx}-\\Bigl(1-\\frac{1}{\\alpha}\\Bigr)y&=-\\Bigl(1-\\frac{1}{\\alpha}\\Bigr)y^{*}+\\frac{1}{\\alpha}g(x)y^{\\beta},\\\\[-3.0pt]\n&\\hskip 40.00006pt\\beta\\neq\\{0,1\\},\\\\[-3.0pt]\ny(0)&=y_{0}.\\end{aligned}\\right."
}
},
{
"type": "Paragraph",
"content": [
"For the moment, we search a general solution of (",
{
"type": "Cite",
"target": "S3-E9",
"content": [
"9"
]
},
") by assuming that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y*≠y0",
"meta": {
"altText": "y^{*}\\neq y_{0}"
}
},
".\nSolving separately the two equations"
]
},
{
"type": "MathBlock",
"id": "S3.Ex35X",
"mathLanguage": "mathml",
"text": "dydx−(1−1α)y=−(1−1α)y*,",
"meta": {
"altText": "\\displaystyle\\frac{dy}{dx}-\\Bigl(1-\\frac{1}{\\alpha}\\Bigr)y=-\\Bigl(1-\\frac{1}{\\alpha}\\Bigr)y^{*},"
}
},
{
"type": "MathBlock",
"id": "S3.Ex35Xa",
"mathLanguage": "mathml",
"text": "dydx−(1−1α)y=1αg(x)yβ,β≠{0,1},",
"meta": {
"altText": "\\displaystyle\\frac{dy}{dx}-\\Bigl(1-\\frac{1}{\\alpha}\\Bigr)y=\\frac{1}{\\alpha}g(x)y^{\\beta},\\quad\\beta\\neq\\{0,1\\},"
}
},
{
"type": "Paragraph",
"content": [
"we obtain the respective solutions"
]
},
{
"type": "MathBlock",
"id": "S3.Ex36",
"mathLanguage": "mathml",
"text": "y(x)=y*+k1e(1−1α)x,",
"meta": {
"altText": "\\displaystyle y(x)=y^{*}+k_{1}e^{(1-\\frac{1}{\\alpha})x},"
}
},
{
"type": "MathBlock",
"id": "S3.Ex37",
"mathLanguage": "mathml",
"text": "y(x)1−β=e(1−β)(1−1α)x[k2+1−βα∫0xg(ξ)e1−βαξ𝑑ξ],β≠{0,1}.",
"meta": {
"altText": "\\displaystyle\\begin{split}y(x)^{1-\\beta}&=e^{(1-\\beta)(1-\\frac{1}{\\alpha})x}\\biggl[k_{2}+\\frac{1-\\beta}{\\alpha}\\int_{0}^{x}g(\\xi)e^{\\frac{1-\\beta}{\\alpha}\\xi}\\,d\\xi\\biggr],\\\\[-3.0pt]\n&\\hskip 40.00006pt\\beta\\neq\\{0,1\\}.\\end{split}"
}
},
{
"type": "Paragraph",
"content": [
"Consequently, the solution of problem (",
{
"type": "Cite",
"target": "S3-E12",
"content": [
"12"
]
},
"), which is also the solution of problem (",
{
"type": "Cite",
"target": "S3-E9",
"content": [
"9"
]
},
"), takes the forms listed below."
]
},
{
"type": "Paragraph",
"content": [
"1. Let ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y0≠y*",
"meta": {
"altText": "y_{0}\\neq y^{*}"
}
},
" and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "β≠{0,1}",
"meta": {
"altText": "\\beta\\neq\\{0,1\\}"
}
},
".\nThen"
]
},
{
"type": "MathBlock",
"id": "S3.Ex38",
"mathLanguage": "mathml",
"text": "y(x)=y*+12(y0−y*)e(1−1α)x+e(1−β)(1−1α)x[12(y0−y*)+1−βα∫0xg(ξ)e1−βαξ𝑑ξ].",
"meta": {
"altText": "\\begin{split}y(x)&=y^{*}+\\frac{1}{2}(y_{0}-y^{*})e^{(1-\\frac{1}{\\alpha})x}\\\\[-3.0pt]\n&\\qquad+e^{(1-\\beta)(1-\\frac{1}{\\alpha})x}\\biggl[\\frac{1}{2}(y_{0}-y^{*})+\\frac{1-\\beta}{\\alpha}\\int_{0}^{x}g(\\xi)e^{\\frac{1-\\beta}{\\alpha}\\xi}\\,d\\xi\\biggr].\\end{split}"
}
},
{
"type": "Paragraph",
"content": [
"2. Let ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y0=y*",
"meta": {
"altText": "y_{0}=y^{*}"
}
},
" and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "β≠{0,1}",
"meta": {
"altText": "\\beta\\neq\\{0,1\\}"
}
},
".\nThen"
]
},
{
"type": "MathBlock",
"id": "S3.Ex39",
"mathLanguage": "mathml",
"text": "y(x)=y*+e(1−β)(1−1α)x[1−βα∫0xg(ξ)e1−βαξ𝑑ξ].",
"meta": {
"altText": "y(x)=y^{*}+e^{(1-\\beta)(1-\\frac{1}{\\alpha})x}\\biggl[\\frac{1-\\beta}{\\alpha}\\int_{0}^{x}g(\\xi)e^{\\frac{1-\\beta}{\\alpha}\\xi}\\,d\\xi\\biggr]."
}
},
{
"type": "Paragraph",
"content": [
"3. Let ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y0=0",
"meta": {
"altText": "y_{0}=0"
}
},
", ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y*≠0",
"meta": {
"altText": "y^{*}\\neq 0"
}
},
" and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "β≠{0,1}",
"meta": {
"altText": "\\beta\\neq\\{0,1\\}"
}
},
".\nIn this case, note that, in order to solve the given fractional-order Cauchy problem (",
{
"type": "Cite",
"target": "S3-E9",
"content": [
"9"
]
},
") via an equivalent ordinary differential problem, we can simply set ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y0=0",
"meta": {
"altText": "y_{0}=0"
}
},
" in (",
{
"type": "Cite",
"target": "S3-E12",
"content": [
"12"
]
},
") and thus obtain the solution"
]
},
{
"type": "MathBlock",
"id": "S3.Ex40",
"mathLanguage": "mathml",
"text": "y(x)=y*−12y*e(1−1α)x+e(1−β)(1−1α)x[−12y*+1−βα∫0xg(ξ)e1−βαξ𝑑ξ].",
"meta": {
"altText": "\\begin{split}y(x)&=y^{*}-\\frac{1}{2}y^{*}e^{(1-\\frac{1}{\\alpha})x}\\\\[-3.0pt]\n&\\qquad+e^{(1-\\beta)(1-\\frac{1}{\\alpha})x}\\biggl[-\\frac{1}{2}y^{*}+\\frac{1-\\beta}{\\alpha}\\int_{0}^{x}g(\\xi)e^{\\frac{1-\\beta}{\\alpha}\\xi}\\,d\\xi\\biggr].\\end{split}"
}
},
{
"type": "Paragraph",
"content": [
"4. Let ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y0=y*=0",
"meta": {
"altText": "y_{0}=y^{*}=0"
}
},
" and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "β≠{0,1}",
"meta": {
"altText": "\\beta\\neq\\{0,1\\}"
}
},
".\nThen"
]
},
{
"type": "MathBlock",
"id": "S3.Ex41",
"mathLanguage": "mathml",
"text": "y(x)=1−βαe(1−β)(1−1α)x∫0xg(ξ)e1−βαξ𝑑ξ.",
"meta": {
"altText": "y(x)=\\frac{1-\\beta}{\\alpha}e^{(1-\\beta)(1-\\frac{1}{\\alpha})x}\\int_{0}^{x}g(\\xi)e^{\\frac{1-\\beta}{\\alpha}\\xi}\\,d\\xi."
}
},
{
"type": "Heading",
"id": "S3.SSx4",
"depth": 2,
"content": [
"263"
]
},
{
"type": "Paragraph",
"content": [
"Let ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "g",
"meta": {
"altText": "g"
}
},
" be a real-valued ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "C2",
"meta": {
"altText": "C^{2}"
}
},
"-function defined on ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "(0,∞)",
"meta": {
"altText": "(0,\\infty)"
}
},
", strictly increasing, such that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "1\" display=\"inline\">g(x)>1",
"meta": {
"altText": "g(x)>1"
}
},
" for all ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x∈(0,∞)",
"meta": {
"altText": "x\\in(0,\\infty)"
}
},
" and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "g(2)<4",
"meta": {
"altText": "g(2)<4"
}
},
".\nConsider the boundary value problem"
]
},
{
"type": "MathBlock",
"id": "S3.Ex42",
"mathLanguage": "mathml",
"text": "y′′=−g(x)y,y(0)=1,y′(0)=0.",
"meta": {
"altText": "y^{\\prime\\prime}=-g(x)y,\\quad y(0)=1,\\quad y^{\\prime}(0)=0."
}
},
{
"type": "Paragraph",
"content": [
"Prove that the solution ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y",
"meta": {
"altText": "y"
}
},
" has exactly one zero in ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "(0,π/2)",
"meta": {
"altText": "(0,\\pi/2)"
}
},
", i.e., there exists a unique point ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x0∈(0,π/2)",
"meta": {
"altText": "x_{0}\\in(0,\\pi/2)"
}
},
" such that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y(x0)=0",
"meta": {
"altText": "y(x_{0})=0"
}
},
", and give a positive lower bound for ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x0",
"meta": {
"altText": "x_{0}"
}
},
"."
]
},
{
"type": "Paragraph",
"id": "S3.SSx4.p2",
"content": [
{
"type": "Emphasis",
"content": [
"Luz Roncal (BCAM – Basque Center for Applied Mathematics, Bilbao, Spain, Ikerbasque Basque Foundation for Science, Bilbao, Spain and Universidad del País Vasco/Euskal Herriko Unibertsitatea, Bilbao, Spain)"
]
}
]
},
{
"type": "Heading",
"id": "S3.SSx4.SSSx1",
"depth": 3,
"content": [
"Solution by the proposer"
]
},
{
"type": "Paragraph",
"content": [
"First suppose that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "0\" display=\"inline\">y(x)>0",
"meta": {
"altText": "y(x)>0"
}
},
" for ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x∈(0,π/2)",
"meta": {
"altText": "x\\in(0,\\pi/2)"
}
},
".\nThe function ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "z(x)=cosx",
"meta": {
"altText": "z(x)=\\cos x"
}
},
" is the solution to the auxiliary initial value problem"
]
},
{
"type": "MathBlock",
"id": "S3.Ex43",
"mathLanguage": "mathml",
"text": "z′′=−z,z(0)=1,z′(0)=0.",
"meta": {
"altText": "z^{\\prime\\prime}=-z,\\quad z(0)=1,\\quad z^{\\prime}(0)=0."
}
},
{
"type": "Paragraph",
"content": [
"Therefore,"
]
},
{
"type": "MathBlock",
"id": "S3.Ex44",
"mathLanguage": "mathml",
"text": "y′′z−yz′′=(1−g(x))zy.",
"meta": {
"altText": "y^{\\prime\\prime}z-yz^{\\prime\\prime}=(1-g(x))zy."
}
},
{
"type": "Paragraph",
"content": [
"Integrating this equality over the interval ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "(0,π/2)",
"meta": {
"altText": "(0,\\pi/2)"
}
},
", we obtain"
]
},
{
"type": "MathBlock",
"id": "S3.Ex45",
"mathLanguage": "mathml",
"text": "∫0π/2(1−g(x))zy𝑑x=∫0π/2(y′′z−yz′′)𝑑x=(y′z−yz′)|0π/2=y(π/2),",
"meta": {
"altText": "\\begin{split}\\int_{0}^{\\pi/2}(1-g(x))zy\\,dx&=\\int_{0}^{\\pi/2}(y^{\\prime\\prime}z-yz^{\\prime\\prime})\\,dx\\\\\n&=(y^{\\prime}z-yz^{\\prime})|_{0}^{\\pi/2}=y(\\pi/2),\\end{split}"
}
},
{
"type": "Paragraph",
"content": [
"and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y(π/2)≥0",
"meta": {
"altText": "y(\\pi/2)\\geq 0"
}
},
" by the continuity of ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y",
"meta": {
"altText": "y"
}
},
".\nBut"
]
},
{
"type": "MathBlock",
"id": "S3.Ex46",
"mathLanguage": "mathml",
"text": "(1−g(x))zy<0in(0,π/2),",
"meta": {
"altText": "(1-g(x))zy<0\\quad\\textrm{in}\\ (0,\\pi/2),"
}
},
{
"type": "Paragraph",
"content": [
"so we reached a contradiction.\nThus, ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y",
"meta": {
"altText": "y"
}
},
" has at least one zero in ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "(0,π/2)",
"meta": {
"altText": "(0,\\pi/2)"
}
},
"."
]
},
{
"type": "Paragraph",
"content": [
"Next, observe that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "π/2<2",
"meta": {
"altText": "\\pi/2<2"
}
},
" and by assumption ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "g(2)<4",
"meta": {
"altText": "g(2)<4"
}
},
".\nConsider the function ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "w(x)=cos(2x)",
"meta": {
"altText": "w(x)=\\cos(2x)"
}
},
", which is a solution to the second auxiliary initial value problem"
]
},
{
"type": "MathBlock",
"id": "S3.Ex47",
"mathLanguage": "mathml",
"text": "w′′=−4w,w(0)=1,w′(0)=0",
"meta": {
"altText": "w^{\\prime\\prime}=-4w,\\quad w(0)=1,\\quad w^{\\prime}(0)=0"
}
},
{
"type": "Paragraph",
"content": [
"and satisfies ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "0\" display=\"inline\">w(x)>0",
"meta": {
"altText": "w(x)>0"
}
},
" for ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x∈(0,π/4)",
"meta": {
"altText": "x\\in(0,\\pi/4)"
}
},
".\nSuppose that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y",
"meta": {
"altText": "y"
}
},
" has (at least) one zero in ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "(0,π/4)",
"meta": {
"altText": "(0,\\pi/4)"
}
},
".\nDenote the smallest such zero by ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x1",
"meta": {
"altText": "x_{1}"
}
},
".\nThen ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y(x)",
"meta": {
"altText": "y(x)"
}
},
" is positive for ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x∈(0,x1)",
"meta": {
"altText": "x\\in(0,x_{1})"
}
},
" (recall that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y(0)=1",
"meta": {
"altText": "y(0)=1"
}
},
") and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y(x1)=0",
"meta": {
"altText": "y(x_{1})=0"
}
},
"; hence ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y′(x1)≤0",
"meta": {
"altText": "y^{\\prime}(x_{1})\\leq 0"
}
},
".\nAn argument analogous to the one above shows that"
]
},
{
"type": "MathBlock",
"id": "S3.Ex48",
"mathLanguage": "mathml",
"text": "∫0x1(g(x)−4)wy𝑑x=∫0π/2(w′′y−wy′′)𝑑x=(w′y−wy′)|0x1=−w(x1)y′(x1).",
"meta": {
"altText": "\\begin{split}\\int_{0}^{x_{1}}(g(x)-4)wy\\,dx&=\\int_{0}^{\\pi/2}(w^{\\prime\\prime}y-wy^{\\prime\\prime})\\,dx\\\\\n&=(w^{\\prime}y-wy^{\\prime})|_{0}^{x_{1}}=-w(x_{1})y^{\\prime}(x_{1}).\\end{split}"
}
},
{
"type": "Paragraph",
"content": [
"Note that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "−w(x1)y′(x1)≥0",
"meta": {
"altText": "-w(x_{1})y^{\\prime}(x_{1})\\geq 0"
}
},
", but the integrand ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "(g(x)−4)wy<0",
"meta": {
"altText": "(g(x)-4)wy<0"
}
},
", so again we reached a contradiction.\nThus, ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y",
"meta": {
"altText": "y"
}
},
" has no zero in ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "(0,π/4)",
"meta": {
"altText": "(0,\\pi/4)"
}
},
", so ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "π/4",
"meta": {
"altText": "\\pi/4"
}
},
" is a positive lower bound for the zeros of ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y",
"meta": {
"altText": "y"
}
},
"."
]
},
{
"type": "Paragraph",
"content": [
"Finally, suppose that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y",
"meta": {
"altText": "y"
}
},
" has more than one zero in ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "(π/4,π/2)",
"meta": {
"altText": "(\\pi/4,\\pi/2)"
}
},
", namely, there exist at least two points ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x2,x3∈(π/4,π/2)",
"meta": {
"altText": "x_{2},x_{3}\\in(\\pi/4,\\pi/2)"
}
},
" such that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x2<x3",
"meta": {
"altText": "x_{2}y(x2)=y(x3)=0",
"meta": {
"altText": "y(x_{2})=y(x_{3})=0"
}
},
".\nTake the function"
]
},
{
"type": "MathBlock",
"id": "S3.Ex49",
"mathLanguage": "mathml",
"text": "v(x)=acos(2x+b),",
"meta": {
"altText": "v(x)=a\\cos(2x+b),"
}
},
{
"type": "Paragraph",
"content": [
"where ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "a,b",
"meta": {
"altText": "a,b"
}
},
" are chosen in such a way that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "v(x2)=0",
"meta": {
"altText": "v(x_{2})=0"
}
},
" and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "v(x)",
"meta": {
"altText": "v(x)"
}
},
" is negative for ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x∈(x2,x3)",
"meta": {
"altText": "x\\in(x_{2},x_{3})"
}
},
"; see Figure ",
{
"type": "Cite",
"target": "S3-F1",
"content": [
"1"
]
},
"."
]
},
{
"type": "Figure",
"id": "S3-F1",
"caption": [
{
"type": "Paragraph",
"content": [
"Assuming ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y",
"meta": {
"altText": "y"
}
},
" has more than one zero in ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "(π/4,π/2)",
"meta": {
"altText": "(\\pi/4,\\pi/2)"
}
}
]
}
],
"content": [
{
"type": "ImageObject",
"contentUrl": "mag-127-vy.svg",
"mediaType": "image/svg+xml",
"meta": {
"inline": false
}
}
]
},
{
"type": "Paragraph",
"content": [
"We have ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "v′′=−4v",
"meta": {
"altText": "v^{\\prime\\prime}=-4v"
}
},
", so ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y′′v−yv′′=(4−g(x))yv",
"meta": {
"altText": "y^{\\prime\\prime}v-yv^{\\prime\\prime}=(4-g(x))yv"
}
},
", which is positive on the interval ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "(x2,x3)",
"meta": {
"altText": "(x_{2},x_{3})"
}
},
".\nTherefore,"
]
},
{
"type": "MathBlock",
"id": "S3.Ex50",
"mathLanguage": "mathml",
"text": "0.\" display=\"block\">∫x2x3(y′′v−yv′′)𝑑x>0.",
"meta": {
"altText": "\\int_{x_{2}}^{x_{3}}(y^{\\prime\\prime}v-yv^{\\prime\\prime})\\,dx>0."
}
},
{
"type": "Paragraph",
"content": [
"On the other hand,"
]
},
{
"type": "MathBlock",
"id": "S3.Ex51",
"mathLanguage": "mathml",
"text": "∫x2x3(y′′v−yv′′)𝑑x=(y′v−yv′)|x2x3=y′(x3)v(x3),",
"meta": {
"altText": "\\int_{x_{2}}^{x_{3}}(y^{\\prime\\prime}v-yv^{\\prime\\prime})\\,dx=(y^{\\prime}v-yv^{\\prime})|_{x_{2}}^{x_{3}}=y^{\\prime}(x_{3})v(x_{3}),"
}
},
{
"type": "Paragraph",
"content": [
"and the right-hand side is negative since ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y(x3)=0",
"meta": {
"altText": "y(x_{3})=0"
}
},
" and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "y(x)<0",
"meta": {
"altText": "y(x)<0"
}
},
" for ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "x∈(x2,x3)",
"meta": {
"altText": "x\\in(x_{2},x_{3})"
}
},
"."
]
},
{
"type": "Heading",
"id": "S3.SSx5",
"depth": 2,
"content": [
"264"
]
},
{
"type": "Paragraph",
"id": "S3.SSx5.p1",
"content": [
"We propose an interesting stochastic-source scattering problem in acoustics.\nThe stochastic nature for such problems forces us to deal with stochastic partial differential equations (SPDEs), rather than partial differential equations (PDEs) which hold for the corresponding deterministic counterparts.\nIn particular, the results of our proposed model will be applied to establish existence and uniqueness for the stochastic solution of a finite element approximation of the stochastic-source Helmholtz equation."
]
},
{
"type": "Paragraph",
"content": [
"Consider the following approximation problem of a stochastic-source Helmholtz equation:"
]
},
{
"type": "MathBlock",
"id": "S3.E13",
"mathLanguage": "mathml",
"text": "Δu+k2u=finD,",
"meta": {
"altText": "\\displaystyle\\Delta u+{k^{2}}u=f\\quad\\textrm{in}\\ D,"
}
},
{
"type": "MathBlock",
"id": "S3.Ex52",
"mathLanguage": "mathml",
"text": "u=0,x∈∂D,",
"meta": {
"altText": "\\displaystyle u=0,\\quad x\\in\\partial D,"
}
},
{
"type": "Paragraph",
"content": [
"where ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "f=∑afaHa",
"meta": {
"altText": "f=\\sum_{a}f_{a}H_{a}"
}
},
" is a generalized stochastic source.\nFor the stochastic problem (",
{
"type": "Cite",
"target": "S3-E13",
"content": [
"13"
]
},
"), we use the equations"
]
},
{
"type": "MathBlock",
"id": "S3.Ex53",
"mathLanguage": "mathml",
"text": "u=∑auaHa,f=∑afaHa,",
"meta": {
"altText": "u=\\sum_{a}u_{a}H_{a},\\quad f=\\sum_{a}f_{a}H_{a},"
}
},
{
"type": "Paragraph",
"content": [
"and we get the collection of deterministic problems"
]
},
{
"type": "MathBlock",
"id": "S3.E14",
"mathLanguage": "mathml",
"text": "Δua+k2ua=fainD,",
"meta": {
"altText": "\\displaystyle\\Delta u_{a}+k^{2}u_{a}=f_{a}\\quad\\textrm{in}\\ D,"
}
},
{
"type": "MathBlock",
"id": "S3.Ex54",
"mathLanguage": "mathml",
"text": "ua=0,x∈∂D.",
"meta": {
"altText": "\\displaystyle u_{a}=0,\\quad x\\in\\partial D."
}
},
{
"type": "Paragraph",
"content": [
"Assume that ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "ua∈H01(D)",
"meta": {
"altText": "u_{a}\\in H_{0}^{1}(D)"
}
},
" solves problem (",
{
"type": "Cite",
"target": "S3-E14",
"content": [
"14"
]
},
").\nThen prove that, for all ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "v∈H01(D)",
"meta": {
"altText": "v\\in H_{0}^{1}(D)"
}
},
", the solution ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "ua∈H01(D)",
"meta": {
"altText": "u_{a}\\in H_{0}^{1}(D)"
}
},
" satisfies"
]
},
{
"type": "MathBlock",
"id": "S3.Ex55",
"mathLanguage": "mathml",
"text": "−∫D∇ua⋅∇vdx+∫Dk2uav𝑑x=∫Dfav𝑑x.",
"meta": {
"altText": "-\\int_{D}\\nabla u_{a}\\cdot\\nabla v\\,dx+\\int_{D}k^{2}u_{a}v\\,dx=\\int_{D}f_{a}v\\,dx."
}
},
{
"type": "Paragraph",
"content": [
{
"type": "Emphasis",
"content": [
"George Kanakoudis, Konstantinos G. Lallas and Vassilios Sevroglou (Department of Statistics and Insurance Science, University of Piraeus, Greece)"
]
}
]
},
{
"type": "Heading",
"id": "S3.SSx5.SSSx1",
"depth": 3,
"content": [
"Solution by the proposers"
]
},
{
"type": "Paragraph",
"id": "S3.SSx5.SSSx1.p1",
"content": [
"We decompose our problem into a hierarchy of deterministic evolution (BVPs), and we give their corresponding variational formulations."
]
},
{
"type": "Paragraph",
"content": [
"For ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "|a|=0",
"meta": {
"altText": "\\lvert a\\rvert=0"
}
},
", we get"
]
},
{
"type": "MathBlock",
"id": "S3.E15",
"mathLanguage": "mathml",
"text": "{Δu0+k2u0=f0inD,u0=0on∂D.",
"meta": {
"altText": "\\left\\{\\begin{aligned} \\Delta u_{0}+k^{2}u_{0}&=f_{0}&&\\quad\\textrm{in}\\ D,\\\\\nu_{0}&=0&&\\quad\\textrm{on}\\ \\partial D.\\end{aligned}\\right."
}
},
{
"type": "Paragraph",
"content": [
"The estimation of a solution of problem (",
{
"type": "Cite",
"target": "S3-E15",
"content": [
"15"
]
},
") is"
]
},
{
"type": "MathBlock",
"id": "S3.Ex56",
"mathLanguage": "mathml",
"text": "∥u0∥H1(D)≤c0∥f0∥L2(D).",
"meta": {
"altText": "\\lVert u_{0}\\rVert_{H^{1}(D)}\\leq c_{0}\\lVert f_{0}\\rVert_{L^{2}(D)}."
}
},
{
"type": "Paragraph",
"content": [
"For ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "|a|=1",
"meta": {
"altText": "\\lvert a\\rvert=1"
}
},
", we get"
]
},
{
"type": "MathBlock",
"id": "S3.E16",
"mathLanguage": "mathml",
"text": "Δu1+k2u1=f1inD,",
"meta": {
"altText": "\\displaystyle\\Delta u_{1}+k^{2}u_{1}=f_{1}\\quad\\textrm{in}\\ D,"
}
},
{
"type": "MathBlock",
"id": "S3.Ex57",
"mathLanguage": "mathml",
"text": "u1=0on∂D.",
"meta": {
"altText": "\\displaystyle u_{1}=0\\quad\\textrm{on}\\ \\partial D."
}
},
{
"type": "Paragraph",
"content": [
"We take an arbitrary ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "v∈H01(D)",
"meta": {
"altText": "v\\in H_{0}^{1}(D)"
}
},
" and multiply equation (",
{
"type": "Cite",
"target": "S3-E16",
"content": [
"16"
]
},
") by ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "v",
"meta": {
"altText": "v"
}
},
".\nThen we get\n"
]
},
{
"type": "MathBlock",
"id": "S3.E17",
"mathLanguage": "mathml",
"text": "(Δu1)v+k2u1v=f1v",
"meta": {
"altText": "(\\Delta u_{1})v+k^{2}u_{1}v=f_{1}v"
}
},
{
"type": "Paragraph",
"content": [
"and integrate over ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "D",
"meta": {
"altText": "D"
}
},
".\nEvery term is integrable since ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "ua∈H01(D)",
"meta": {
"altText": "u_{a}\\in H_{0}^{1}(D)"
}
},
", and hence we have ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "Δu1∈H01(D)",
"meta": {
"altText": "\\Delta u_{1}\\in H_{0}^{1}(D)"
}
},
" and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "v∈H01(D)",
"meta": {
"altText": "v\\in H_{0}^{1}(D)"
}
},
", so"
]
},
{
"type": "MathBlock",
"id": "S3.Ex58",
"mathLanguage": "mathml",
"text": "(Δu1)v∈H01(D),k2∈L∞(D),u1∈H01(D),v∈H01(D).",
"meta": {
"altText": "(\\Delta u_{1})v\\in H_{0}^{1}(D),\\quad k^{2}\\in L^{\\infty}(D),\\quad u_{1}\\in H_{0}^{1}(D),\\quad v\\in H_{0}^{1}(D)."
}
},
{
"type": "Paragraph",
"content": [
"Therefore, ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "k2u1v∈H01(D)",
"meta": {
"altText": "k^{2}u_{1}v\\in H_{0}^{1}(D)"
}
},
" and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "f1∈L2(D)",
"meta": {
"altText": "f_{1}\\in L^{2}(D)"
}
},
", so ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "f1v∈H01(D)",
"meta": {
"altText": "f_{1}v\\in H_{0}^{1}(D)"
}
},
".\nWe obtain"
]
},
{
"type": "MathBlock",
"id": "S3.Ex59",
"mathLanguage": "mathml",
"text": "∫D(Δu1)v𝑑x+∫Dk2u1v𝑑x=∫Df1v𝑑x.",
"meta": {
"altText": "\\int_{D}(\\Delta u_{1})v\\,dx+\\int_{D}k^{2}u_{1}v\\,dx=\\int_{D}f_{1}v\\,dx."
}
},
{
"type": "Paragraph",
"content": [
"We use Green’s formula according to which"
]
},
{
"type": "MathBlock",
"id": "S3.Ex60",
"mathLanguage": "mathml",
"text": "∫D(Δu1)v𝑑x=−∫D∇u1⋅∇vdx+∫∂Dγ1(u1)γ0(v)𝑑Γ",
"meta": {
"altText": "\\int_{D}(\\Delta u_{1})v\\,dx=-\\int_{D}\\nabla u_{1}\\cdot\\nabla v\\,dx+\\int_{\\partial D}\\gamma_{1}(u_{1})\\gamma_{0}(v)\\,d\\Gamma"
}
},
{
"type": "Paragraph",
"content": [
"since ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "v∈H01(D)",
"meta": {
"altText": "v\\in H_{0}^{1}(D)"
}
},
" is equivalent to ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "γ0(v)=0",
"meta": {
"altText": "\\gamma_{0}(v)=0"
}
},
"."
]
},
{
"type": "Paragraph",
"content": [
"Let ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "H01(D)",
"meta": {
"altText": "H_{0}^{1}(D)"
}
},
" be a stochastic Hilbert space.\nIf we now assume the bilinear form on ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "H01(D)×H01(D)",
"meta": {
"altText": "H_{0}^{1}(D)\\times H_{0}^{1}(D)"
}
},
","
]
},
{
"type": "MathBlock",
"id": "S3.Ex61",
"mathLanguage": "mathml",
"text": "a(u1,v)=∫D(−∇u1⋅∇v+k2u1v)𝑑x,",
"meta": {
"altText": "a(u_{1},v)=\\int_{D}(-\\nabla u_{1}\\cdot\\nabla v+k^{2}u_{1}v)\\,dx,"
}
},
{
"type": "Paragraph",
"content": [
"and the linear functional on ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "H01(D)",
"meta": {
"altText": "H_{0}^{1}(D)"
}
},
","
]
},
{
"type": "MathBlock",
"id": "S3.Ex62",
"mathLanguage": "mathml",
"text": "ℓ(v)=∫Df1v𝑑x,",
"meta": {
"altText": "\\ell(v)=\\int_{D}f_{1}v\\,dx,"
}
},
{
"type": "Paragraph",
"content": [
"then the variational formulation of problem (",
{
"type": "Cite",
"target": "S3-E17",
"content": [
"17"
]
},
") is"
]
},
{
"type": "MathBlock",
"id": "S3.E18",
"mathLanguage": "mathml",
"text": "a(u1,v)=ℓ(v)∀v∈H01(D).",
"meta": {
"altText": "a(u_{1},v)=\\ell(v)\\quad\\forall v\\in H_{0}^{1}(D)."
}
},
{
"type": "Paragraph",
"content": [
"The estimation of a solution of problem (",
{
"type": "Cite",
"target": "S3-E18",
"content": [
"18"
]
},
") is"
]
},
{
"type": "MathBlock",
"id": "S3.Ex63",
"mathLanguage": "mathml",
"text": "∥u1∥H1(D)≤c1∥f1∥L2(D).",
"meta": {
"altText": "\\lVert u_{1}\\rVert_{H^{1}(D)}\\leq c_{1}\\lVert f_{1}\\rVert_{L^{2}(D)}."
}
},
{
"type": "Paragraph",
"content": [
"For ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "|a|=n",
"meta": {
"altText": "\\lvert a\\rvert=n"
}
},
", we get"
]
},
{
"type": "MathBlock",
"id": "S3.E19",
"mathLanguage": "mathml",
"text": "{Δun+k2un=fninD,un=0on∂D.",
"meta": {
"altText": "\\left\\{\\begin{aligned} \\Delta u_{n}+k^{2}u_{n}&=f_{n}&&\\quad\\textrm{in}\\ D,\\\\\nu_{n}&=0&&\\quad\\textrm{on}\\ \\partial D.\\end{aligned}\\right."
}
},
{
"type": "Paragraph",
"content": [
"The estimation of a solution of problem (",
{
"type": "Cite",
"target": "S3-E19",
"content": [
"19"
]
},
") is"
]
},
{
"type": "MathBlock",
"id": "S3.Ex64",
"mathLanguage": "mathml",
"text": "∥un∥H1(D)≤cn∥fn∥L2(D).",
"meta": {
"altText": "\\lVert u_{n}\\rVert_{H^{1}(D)}\\leq c_{n}\\lVert f_{n}\\rVert_{L^{2}(D)}."
}
},
{
"type": "Paragraph",
"content": [
"Via the above variational formulations and taking into account ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "u=∑auaHa",
"meta": {
"altText": "u=\\sum_{a}u_{a}H_{a}"
}
},
", we can prove that the solution ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "u",
"meta": {
"altText": "u"
}
},
" of the stochastic boundary value problem (",
{
"type": "Cite",
"target": "S3-E13",
"content": [
"13"
]
},
") satisfies the following inequality:"
]
},
{
"type": "MathBlock",
"id": "S3.Ex65",
"mathLanguage": "mathml",
"text": "∥u∥H1(D)≤c0∥f0∥L2(D)+c1∥f1∥L2(D)+⋯+cn∥fn∥L2(D),",
"meta": {
"altText": "\\lVert u\\rVert_{H^{1}(D)}\\leq c_{0}\\lVert f_{0}\\rVert_{L^{2}(D)}+c_{1}\\lVert f_{1}\\rVert_{L^{2}(D)}+\\cdots+c_{n}\\lVert f_{n}\\rVert_{L^{2}(D)},\\\\\n"
}
},
{
"type": "Paragraph",
"content": [
"where ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "ci",
"meta": {
"altText": "c_{i}"
}
},
", ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "i=0,1,…,n",
"meta": {
"altText": "i=0,1,\\ldots,n"
}
},
", are considered to be in agreement with appropriate built-in weights.\nThe solution ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "u",
"meta": {
"altText": "u"
}
},
" belongs to the space ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "{H1(D),Ω,F,μ}",
"meta": {
"altText": "\\{H^{1}(D),\\Omega,F,\\mu\\}"
}
},
" which is a stochastic Hilbert space with ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "μ",
"meta": {
"altText": "\\mu"
}
},
" the probability measure defined by ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "Ha(ω)",
"meta": {
"altText": "H_{a}(\\omega)"
}
},
"."
]
},
{
"type": "Heading",
"id": "S3.SSx6",
"depth": 2,
"content": [
"265"
]
},
{
"type": "Paragraph",
"content": [
"For a Newtonian incompressible fluid, the Navier–Stokes momentum equation, in vector form, reads [",
{
"type": "Cite",
"target": "biba-bib4",
"content": [
"4"
]
},
"]"
]
},
{
"type": "MathBlock",
"id": "S3.E20X",
"mathLanguage": "mathml",
"text": "ρ(∂𝐮∂t+𝐮⋅∇𝐮)=−∇p+μ∇2𝐮+𝐅,",
"meta": {
"altText": "\\displaystyle\\rho\\Bigl(\\frac{\\partial\\mathbf{u}}{\\partial t}+\\mathbf{u}\\cdot\\nabla\\mathbf{u}\\Bigr)=-\\nabla p+\\mu\\nabla^{2}\\mathbf{u}+\\mathbf{F},"
}
},
{
"type": "MathBlock",
"id": "S3.E20Xa",
"mathLanguage": "mathml",
"text": "𝐮=𝐮(𝐱,t),𝐮:𝐑n×(0,∞)→𝐑n.",
"meta": {
"altText": "\\displaystyle\\mathbf{u}=\\mathbf{u}(\\mathbf{x},t),\\quad\\mathbf{u}\\colon\\mathbf{R}^{n}\\times(0,\\infty)\\to\\mathbf{R}^{n}."
}
},
{
"type": "Paragraph",
"content": [
"Here, ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "ρ",
"meta": {
"altText": "\\rho"
}
},
" is the fluid density, ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "𝐮",
"meta": {
"altText": "\\mathbf{u}"
}
},
" is the velocity vector field, ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "p",
"meta": {
"altText": "p"
}
},
" is the pressure, ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "μ",
"meta": {
"altText": "\\mu"
}
},
" is the viscosity, and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "𝐅",
"meta": {
"altText": "\\mathbf{F}"
}
},
" is an external force field."
]
},
{
"type": "Paragraph",
"content": [
"(i) Assuming that both the pressure drop ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "∇p",
"meta": {
"altText": "\\nabla p"
}
},
" and the external field ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "𝐅",
"meta": {
"altText": "\\mathbf{F}"
}
},
" are negligible, it is easy to show that equation (",
{
"type": "Cite",
"target": "S3-E20",
"content": [
"20"
]
},
") reduces to"
]
},
{
"type": "MathBlock",
"id": "S3.E21",
"mathLanguage": "mathml",
"text": "∂𝐮∂t+𝐮⋅∇𝐮=ν∇2𝐮,",
"meta": {
"altText": "\\frac{\\partial\\mathbf{u}}{\\partial t}+\\mathbf{u}\\cdot\\nabla\\mathbf{u}=\\nu\\nabla^{2}\\mathbf{u},"
}
},
{
"type": "Paragraph",
"content": [
"and finally to equation (",
{
"type": "Cite",
"target": "S3-E22",
"content": [
"22"
]
},
"), where ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "ν=μρ",
"meta": {
"altText": "\\nu=\\frac{\\mu}{\\rho}"
}
},
" is the so-called kinematic viscosity [",
{
"type": "Cite",
"target": "biba-bib5",
"content": [
"5"
]
},
"]."
]
},
{
"type": "Paragraph",
"content": [
"(ii) Regarding the one-dimensional viscous Burgers equation\n"
]
},
{
"type": "MathBlock",
"id": "S3.E22",
"mathLanguage": "mathml",
"text": "∂u∂t+u∂u∂x=ν∂2u∂x2,u=u(x,t),",
"meta": {
"altText": "\\frac{\\partial u}{\\partial t}+u\\frac{\\partial u}{\\partial x}=\\nu\\frac{\\partial^{2}u}{\\partial x^{2}},\\quad u=u(x,t),"
}
},
{
"type": "Paragraph",
"content": [
"prove that an analytical solution can be obtained by means of the Tanh Method [",
{
"type": "Cite",
"target": "biba-bib2",
"content": [
"2"
]
},
", ",
{
"type": "Cite",
"target": "biba-bib3",
"content": [
"3"
]
},
", ",
{
"type": "Cite",
"target": "biba-bib5",
"content": [
"5"
]
},
"] as"
]
},
{
"type": "MathBlock",
"id": "S3.Ex66",
"mathLanguage": "mathml",
"text": "0.\" display=\"block\">u(x,t)=λ[1−tanh(λ2ν(x−λt))],λ>0.",
"meta": {
"altText": "u(x,t)=\\lambda\\Bigl[1-\\operatorname{tanh}\\Bigl(\\frac{\\lambda}{2\\nu}(x-\\lambda t)\\Bigr)\\Bigr],\\quad\\lambda>0."
}
},
{
"type": "Paragraph",
"content": [
{
"type": "Emphasis",
"content": [
"M. A. Xenos and A. C. Felias (Department of Mathematics, University of Ioannina, Greece)"
]
}
]
},
{
"type": "Heading",
"id": "S3.SSx6.SSSx1",
"depth": 3,
"content": [
"Solution by the proposers"
]
},
{
"type": "Paragraph",
"content": [
"Notice that, for ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "∇p=𝐅=𝟎",
"meta": {
"altText": "\\nabla p=\\mathbf{F}=\\mathbf{0}"
}
},
", equation (",
{
"type": "Cite",
"target": "S3-E20",
"content": [
"20"
]
},
") becomes"
]
},
{
"type": "MathBlock",
"id": "S3.E23",
"mathLanguage": "mathml",
"text": "ρ(∂𝐮∂t+𝐮⋅∇𝐮)=μ∇2𝐮.",
"meta": {
"altText": "\\rho\\Bigl(\\frac{\\partial\\mathbf{u}}{\\partial t}+\\mathbf{u}\\cdot\\nabla\\mathbf{u}\\Bigr)=\\mu\\nabla^{2}\\mathbf{u}."
}
},
{
"type": "Paragraph",
"content": [
"Since, for an incompressible fluid, ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "ρ",
"meta": {
"altText": "\\rho"
}
},
" is a nonzero constant, one can divide both sides of equation (",
{
"type": "Cite",
"target": "S3-E23",
"content": [
"23"
]
},
") by ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "ρ",
"meta": {
"altText": "\\rho"
}
},
" and thus obtain equation (",
{
"type": "Cite",
"target": "S3-E21",
"content": [
"21"
]
},
").\nNow consider the motion of a one-dimensional viscous fluid with fluid velocity ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "u",
"meta": {
"altText": "u"
}
},
" along the ",
{
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"mathLanguage": "mathml",
"text": "x",
"meta": {
"altText": "x"
}
},
"-axis as time passes, ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "u=u(x,t)",
"meta": {
"altText": "u=u(x,t)"
}
},
".\nIn this case, equation (",
{
"type": "Cite",
"target": "S3-E21",
"content": [
"21"
]
},
") transforms into equation (",
{
"type": "Cite",
"target": "S3-E22",
"content": [
"22"
]
},
").\nIntroduce the transformation of ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "u",
"meta": {
"altText": "u"
}
},
" given by"
]
},
{
"type": "MathBlock",
"id": "S3.E24",
"mathLanguage": "mathml",
"text": "0,\\,\\lambda\\neq 0,\\end{aligned}\\right.\" class=\"ltx_math_unparsed\" display=\"block\">{u(x,t)=u(ζ),ζ=μ(x−λt),μ>0,λ≠0,",
"meta": {
"altText": "\\left\\{\\begin{aligned} u(x,t)&=u(\\zeta),\\\\\n\\zeta&=\\mu(x-\\lambda t),\\quad\\mu>0,\\,\\lambda\\neq 0,\\end{aligned}\\right."
}
},
{
"type": "Paragraph",
"content": [
"with ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "μ",
"meta": {
"altText": "\\mu"
}
},
" representing the wave number and ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "λ",
"meta": {
"altText": "\\lambda"
}
},
" the velocity.\nThen transformation (",
{
"type": "Cite",
"target": "S3-E24",
"content": [
"24"
]
},
") reduces equation (",
{
"type": "Cite",
"target": "S3-E22",
"content": [
"22"
]
},
") to the following ODE for ",
{
"type": "MathFragment",
"mathLanguage": "mathml",
"text": "u(ζ)",
"meta": {
"altText": "u(\\zeta)"
}
},
":"
]
},
{
"type": "MathBlock",
"id": "S3.E25",
"mathLanguage": "mathml",
"text": "−λu′(ζ)+u(ζ)u′(ζ)−νμu′′(ζ)=0.",
"meta": {
"altText": "-\\lambda u^{\\prime}(\\zeta)+u(\\zeta)u^{\\prime}(\\zeta)-\\nu\\mu u^{\\prime\\prime}(\\zeta)=0."
}
},
{
"type": "Paragraph",
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"25"
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"text": "−λu(ζ)+12u2(ζ)−νμu′(ζ)=0.",
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"altText": "-\\lambda u(\\zeta)+\\frac{1}{2}u^{2}(\\zeta)-\\nu\\mu u^{\\prime}(\\zeta)=0."
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"content": [
"The idea behind the Tanh method uses a key property of the functional derivatives all being written in terms of the Tanh function [",
{
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"content": [
"2"
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"3"
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"].\nThe following identity is used:"
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"text": "sech2ζ=1−tanh2ζ,ζ∈ℝ.",
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"type": "Cite",
"target": "S3-E25",
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"25"
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},
{
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"text": "y=tanhζ,",
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{
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"content": [
"solution(s) can be sought in the form"
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{
"type": "MathBlock",
"id": "S3.E29",
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"text": "u(y)=∑n=0Nanyn.",
"meta": {
"altText": "u(y)=\\sum_{n=0}^{N}a_{n}y^{n}."
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},
{
"type": "Paragraph",
"content": [
"Chain differentiation yields"
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},
{
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"id": "S3.E30",
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"text": "ddζ=ddydydζ=sech2ζddy=(30)(1−tanh2ζ)ddy=(31)(1−y2)ddy.",
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"altText": "\\begin{split}\\frac{d}{d\\zeta}&=\\frac{d}{dy}\\frac{dy}{d\\zeta}=\\textrm{sech}^{2}\\zeta\\frac{d}{dy}\\\\\n&\\underset{(30)}{=}(1-\\textrm{tanh}^{2}\\zeta)\\frac{d}{dy}\\underset{(31)}{=}(1-y^{2})\\frac{d}{dy}.\\end{split}"
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{
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"type": "Paragraph",
"content": [
"Different right-moving, kink-shaped solutions of the viscous Burgers equation, for ",
{
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"mathLanguage": "mathml",
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"29"
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"id": "S3.Ex67",
"mathLanguage": "mathml",
"text": "{a122+a1νμ=0,a0a1−a1λ=0,a022−a0λ−a1νμ=0,",
"meta": {
"altText": "\\left\\{\\begin{aligned} \\frac{a_{1}^{2}}{2}+a_{1}\\nu\\mu&=0,\\\\\na_{0}a_{1}-a_{1}\\lambda&=0,\\\\\n\\frac{\\smash[t]{a_{0}^{2}}}{2}-a_{0}\\lambda-a_{1}\\nu\\mu&=0,\\end{aligned}\\right."
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"altText": "\\left\\{\\begin{aligned} \\mu&=\\frac{\\lambda}{2\\nu},\\quad\\lambda>0,\\\\\na_{0}&=\\lambda,\\\\\na_{1}&=-\\lambda.\\end{aligned}\\right."
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"target": "S3-E31",
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"31"
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") and (",
{
"type": "Cite",
"target": "S3-E29",
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"29"
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},
") and using expression (",
{
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"target": "S3-E28",
"content": [
"28"
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},
"), one obtains"
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},
{
"type": "MathBlock",
"id": "S3.Ex68",
"mathLanguage": "mathml",
"text": "u(ζ)=λ(1−Tanhζ),",
"meta": {
"altText": "u(\\zeta)=\\lambda(1-\\operatorname{Tanh}\\zeta),"
}
},
{
"type": "Paragraph",
"content": [
"and finally"
]
},
{
"type": "MathBlock",
"id": "S3.Ex69",
"mathLanguage": "mathml",
"text": "u(x,t)=λ[1−Tanh(λ2ν(x−λt))].",
"meta": {
"altText": "u(x,t)=\\lambda\\Bigl[1-\\operatorname{Tanh}\\Bigl(\\frac{\\lambda}{2\\nu}(x-\\lambda t)\\Bigr)\\Bigr]."
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},
{
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"content": [
"Figure ",
{
"type": "Cite",
"target": "S3-F2",
"content": [
"2"
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},
" displays right-moving analytical solutions of the viscous Burgers equation for different values of the kinematic viscosity ",
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"."
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},
{
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"id": "p2",
"content": [
{
"type": "Emphasis",
"content": [
"We wait to receive your solutions to the proposed problems and ideas on the open problems.\nSend your solutions to Michael Th. Rassias by email to ",
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"type": "Link",
"target": "mailto:mthrassias@yahoo.com",
"content": [
"mthrassias@yahoo.com"
]
},
"."
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}
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"id": "p3",
"content": [
{
"type": "Emphasis",
"content": [
"We also solicit your new problems with their solutions for the next “Solved and unsolved problems” column, which will be devoted to ",
{
"type": "Emphasis",
"content": [
"Probability Theory"
]
},
"."
]
}
]
}
]
}