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Solved and unsolved problems

  • Michael Th. Rassias

    Hellenic Military Academy, Greece

The present column is devoted to Number Theory.

I Six new problems – solutions solicited

Solutions will appear in a subsequent issue.


Consider two positive integers n1n\geq 1 and a2a\geq 2 such that


is a prime. Prove that nn is a power of 33.

Dorin Andrica and George Cătălin Ţurcaş (Babeş-Bolyai University, Cluj-Napoca, Romania)


The Collatz map is defined as follows:

Col(n):={n/2if n is even,3n+1if n is odd.\operatorname{Col}(n):=\begin{cases}n/2&\textrm{if}\ n\ \textrm{is even},\\ 3n+1&\textrm{if}\ n\ \textrm{is odd}.\end{cases}


tm,xmin(n>0:Colm(n)x).t_{m,x}≔\min(n>0:\operatorname{Col}^{m}(n)\geq x).

That is, tm,xt_{m,x} is the smallest integer such that, if we apply the Collatz map mm times, the result is larger than xx.

(a) Find t3,1000t_{3,1000} and t4,1000t_{4,1000}.

(b) Show that, for xx large enough (larger than (say) 10001000), we have

t4,x3mod4ort4,x6mod8.t_{4,x}\equiv 3\bmod 4\quad\textrm{or}\quad t_{4,x}\equiv 6\bmod 8.

(c) In general, for mm odd and xx large enough, there exists a constant Xm,xX_{m,x} such that tm,xt_{m,x} is the smallest n>Xm,xn>X_{m,x} such that ncmmodMmn\equiv c_{m}\bmod M_{m}. Find MmM_{m} and relate cmc_{m} to cm1c_{m-1}.

Christopher Lutsko (Department of Mathematics, Rutgers University, Piscataway, USA)


The light-bulb problem: Alice and Bob are in jail for trying to divide by 00. The jailer proposes the following game to decide their freedom: Alice will be shown an n×nn\times n grid of light bulbs. The jailer will point to a light bulb of his choice and Alice will decide whether it should be on or off. Then the jailer will point to another bulb of his choice and Alice will decide on/off. This continues until the very last bulb, when the jailer will decide whether this bulb is on or off. So the jailer controls the order of the selection, and the state of the final bulb. Alice is now removed from the room, and Bob is brought in. Bob’s goal is to choose nn bulbs such that his selection includes the final bulb (the one determined by the jailer).

Is there a strategy that Alice and Bob can use to guarantee success? What 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)?

Christopher Lutsko (Department of Mathematics, Rutgers University, Piscataway, USA)


Let pp and qq be coprime integers greater than or equal to 22. Let invq(p)\operatorname{inv}_{q}(p) and invp(q)\operatorname{inv}_{p}(q) denote the modular inverse of pmodqp\bmod q and qmodpq\bmod p, respectively. That is, invq(p)p1modq\operatorname{inv}_{q}(p)p\equiv 1\bmod q and invp(q)q1modp\operatorname{inv}_{p}(q)q\equiv 1\bmod p.

(a) Show that

invp(q)p2if and only ifinvq(p)>q2.\operatorname{inv}_{p}(q)\leq\frac{p}{2}\quad\textrm{if and only if}\quad\operatorname{inv}_{q}(p)>\frac{q}{2}.

(b) Show by providing an example that, if 1u<v1\leq u<v are coprime integers and αu/v\alpha≔u/v, then the statement

invp(q)αp if and only if invq(p)>(1α)q\operatorname{inv}_{p}(q)\leq\alpha p\quad\textrm{ if and only if }\quad\operatorname{inv}_{q}(p)>(1-\alpha)q

is not necessarily true.

(c) What additional assumption should pp and/or qq satisfy so that the equivalence (1) holds?

Athanasios Sourmelidis (Institut für Analysis und Zahlentheorie, Technische Universität Graz, Austria)


Let cn(k)c_{n}(k) denote the Ramanujan sum defined as the sum of kkth powers of the primitive nnth roots of unity. Show that, for any integer m1m\geq 1,


where the sum is over all ordered pairs (n,k)(n,k) of positive integers n,kn,k such that their lcm is mm, and φ\varphi is Euler’s totient function.

László Tóth (University of Pécs, Hungary)


Show that, for every integer n1n\geq 1, we have the polynomial identity

k=1(k,n)=1n(x(k1,n)1)=dnΦd(x)φ(n)/φ(d),\prod_{\begin{subarray}{c}k=1\\ (k,n)=1\end{subarray}}^{n}(x^{(k-1,n)}-1)=\prod_{d\mid n}\Phi_{d}(x)^{\varphi(n)/\varphi(d)},

where Φd(x)\Phi_{d}(x) are the cyclotomic polynomials and φ\varphi denotes Euler’s totient function.

László Tóth (Department of Mathematics, University of Pécs, Hungary)

II Open problem

275*. Chowla’s conjecture and its relatives

by Terence Tao (UCLA, Department of Mathematics, Los Angeles, USA)

Let λ ⁣:N{1,+1}\lambda\colon\mathbb{N}\to\{-1,+1\} denote the Liouville function. In [2 S. Chowla, The Riemann hypothesis and Hilbert’s tenth problem. Mathematics and its Applications 4, Gordon and Breach Science Publishers, New York (1965) ], Chowla conjectured that

nxλ(n+h1)λ(n+hk)=o(x)\sum_{n\leq x}\lambda(n+h_{1})\cdots\lambda(n+h_{k})=o(x)

as xx\to\infty, for any distinct natural numbers h1,,hkh_{1},\ldots,h_{k} (in fact, Chowla made the more general conjecture that

nxλ(P(n))=o(x)\sum_{n\leq x}\lambda(P(n))=o(x)

whenever PP is a square-free polynomial mapping from N\mathbb{N} to N\mathbb{N}). Chowla’s conjecture was extended to other bounded multiplicative functions by Elliott [3 P. D. T. A. Elliott, On the correlation of multiplicative functions. Notas Soc. Mat. Chile 11, 1–11 (1992) ] (see also a technical correction to the conjecture in [8 K. Matomäki, M. Radziwiłł and T. Tao, An averaged form of Chowla’s conjecture. Algebra Number Theory 9, 2167–2196 (2015) ]).

One can view (2) as a less difficult cousin of the notorious Hardy–Littlewood prime tuples conjecture [4 G. H. Hardy and J. E. Littlewood, Some problems of ‘Partitio numerorum’; III: On the expression of a number as a sum of primes. Acta Math. 44, 1–70 (1923) ], which conjectures an asymptotic of the form

nxΛ(n+h1)Λ(n+hk)=Sx+o(x)\sum_{n\leq x}\Lambda(n+h_{1})\cdots\Lambda(n+h_{k})=\mathfrak{S}x+o(x)

where the singular seriesS\mathfrak{S} is an explicit product over primes of factors involving the numbers h1,,hkh_{1},\ldots,h_{k}. For k=1k=1, both conjectures follow readily from the prime number theorem, but they remain open for higher kk. However, the analogue of (2) (and (3) for k2k\leq 2) were recently established in certain function fields [11 W. Sawin and M. Shusterman, On the Chowla and twin primes conjectures over 𝔽q⁢[T]. Ann. of Math. (2) 196, 457–506 (2022) ], and are also known to hold in the presence of a Siegel zero [1 J. Chinis, Siegel zeros and Sarnak’s conjecture, preprint, arXiv:2105.14653 (2021) , 5 D. R. Heath-Brown, Prime twins and Siegel zeros. Proc. London Math. Soc. (3) 47, 193–224 (1983) , 15 T. Tao, J. Teräväinen, The Hardy–Littlewood–Chowla conjecture in the presence of a Siegel zero, preprint, arXiv:2109.06291 (2021) , 6 K. Matomäki, J. Merikoski, Siegel zeros, twin primes, Goldbach’s conjecture, and primes in short intervals, preprint, arXiv:2112.11412 (2021) ]. The conjecture (2) is also known if one performs enough averaging in the h1,,hkh_{1},\dots,h_{k} variables [8 K. Matomäki, M. Radziwiłł and T. Tao, An averaged form of Chowla’s conjecture. Algebra Number Theory 9, 2167–2196 (2015) ].

The logarithmically averaged version

nxλ(n+h1)λ(n+hk)n=o(logx)\sum_{n\leq x}\frac{\lambda(n+h_{1})\cdots\lambda(n+h_{k})}{n}=o(\log x)

turns out to be more tractable, as it can be analyzed by the “entropy decrement method” [12 T. Tao, The logarithmically averaged Chowla and Elliott conjectures for two-point correlations. Forum Math. Pi 4, Paper No. e8 (2016) ], which has been successfully used to establish the conjecture for k=2k=2 (see [12 T. Tao, The logarithmically averaged Chowla and Elliott conjectures for two-point correlations. Forum Math. Pi 4, Paper No. e8 (2016) ], building upon the breakthrough work [7 K. Matomäki and M. Radziwiłł, Multiplicative functions in short intervals. Ann. of Math. (2) 183, 1015–1056 (2016) ]) and for odd kk (see [14 T. Tao and J. Teräväinen, Odd order cases of the logarithmically averaged Chowla conjecture. J. Théor. Nombres Bordeaux 30, 997–1015 (2018) , 16 T. Tao and J. Teräväinen, The structure of logarithmically averaged correlations of multiplicative functions, with applications to the Chowla and Elliott conjectures. Duke Math. J. 168, 1977–2027 (2019) ]). The conjecture (4) for arbitrary kk is also known to be equivalent [13 T. Tao, Equivalence of the logarithmically averaged Chowla and Sarnak conjectures. In Number Theory—Diophantine Problems, Uniform Distribution and Applications, Springer, Cham, 391–421 (2017) ] to the (logarithmically averaged) Sarnak conjecture [10 P. Sarnak, Mobius randomness and dynamics. Not. S. Afr. Math. Soc. 43, 89–97 (2012) ], which asserts that

nxλ(n)F(Tnx0)n=o(logx)\sum_{n\leq x}\frac{\lambda(n)F(T^{n}x_{0})}{n}=o(\log x)

whenever T ⁣:XXT\colon X\to X is a compact dynamical system of zero entropy, x0x_{0} is a point in XX, and F ⁣:XCF\colon X\to\mathbb{C} is continuous. Many special cases of this conjecture are known, unfortunately too many to list here.

The conjecture (4) would also follow from a higher-order local Fourier uniformity conjecture [12 T. Tao, The logarithmically averaged Chowla and Elliott conjectures for two-point correlations. Forum Math. Pi 4, Paper No. e8 (2016) ], which is somewhat complicated to state in full generality here; however, the first unsolved special case of this conjecture asserts that

X2XsupαR/Zxnx+Hλ(n)e(αn)dx=o(XH)as X\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

whenever 1H=H(X)X1\leq H=H(X)\leq X is such that H(X)H(X)\to\infty as XX\to\infty, where e(θ)e2πiθe(\theta)≔e^{2\pi i\theta}. This is currently only established in the regime HXεH\geq X^{\varepsilon} for a fixed ε>0\varepsilon>0 (see [9 K. Matomäki, M. Radziwiłł and T. Tao, Fourier uniformity of bounded multiplicative functions in short intervals on average. Invent. Math. 220, 1–58 (2020) ]).

III Solutions


Let f ⁣:[0,)Rf\colon[0,\infty)\to\mathbb{R} be a C1C^{1}-differentiable and convex function with f(0)=0f(0)=0.

(i) Prove that, for every x[0,)x\in[0,\infty), the following inequality holds:


(ii) Determine all functions ff for which we have equality.

Dorin Andrica (“Babeş-Bolyai” University, Cluj-Napoca, Romania) and Mihai Piticari (“Dragoş Vodă” National College, Câmpulung Moldovenesc, Romania)

Solution by the proposers

(i) We have


By the Mean Value Theorem,


for some ct(0,t)[0,x]c_{t}\in(0,t)\subset[0,x]. Since ff is convex, it follows that ff^{\prime} is increasing; hence f(ct)f(x)f^{\prime}(c_{t})\leq f^{\prime}(x). Therefore, f(t)=tf(ct)tf(x)f(t)=tf^{\prime}(c_{t})\leq tf^{\prime}(x), and we obtain


(ii) We have to find all the solutions to the equation



g(x)=0xf(t)dt,x[0,),g(x)=\mathop{\smash[t]{\int_{0}^{x}}}f(t)\,dt,\quad x\in[0,\infty),

the above equation is equivalent to the second-order differential equation

g(x)=x22g(x),x[0,).g(x)=\smash[t]{\frac{x^{2}}{2}}g^{\prime\prime}(x),\quad x\in[0,\infty).

Note that if gg is a solution, then




It follows that


and so

x22g(x)xg(x)=C1,x[0,),\smash[t]{\frac{x^{2}}{2}}g^{\prime}(x)-xg(x)=C_{1},\quad x\in[0,\infty),

where C1C_{1} is an arbitrary constant. This last equation is equivalent to

x2g(x)2xg(x)x4=2C1x4,x(0,),\smash[b]{\frac{x^{2}g^{\prime}(x)-2xg(x)}{x^{4}}=\frac{2C_{1}}{x^{4}}},\quad x\in(0,\infty),


(g(x)x2)=2C1x4,x(0,).\smash{\Bigl(\frac{g(x)}{x^{2}}\Bigr)^{\prime}=\frac{2C_{1}}{x^{4}}},\quad x\in(0,\infty).


g(x)x2=2C13x3+C2,x(0,),\smash[b]{\frac{g(x)}{x^{2}}=\frac{-2C_{1}}{3x^{3}}}+C_{2},\quad x\in(0,\infty),

and we get

g(x)=2C13x+C2x2,x(0,).g(x)=\smash[t]{\frac{-2C_{1}}{3x}}+C_{2}x^{2},\quad x\in(0,\infty).

On the other hand, gg is continuous and g(0)=0g(0)=0. This implies that C1=0C_{1}=0 and g(x)=C2x2g(x)=C_{2}x^{2}. We conclude that the sought-for functions are necessarily of the form f(x)=g(x)=2C2xf(x)=g^{\prime}(x)=2C_{2}x, i.e., f(x)=Axf(x)=Ax, where AA is an arbitrary real constant.


Let y(x)y(x) be the unknown function of the following fractional-order derivative Cauchy problem:

{Dαy=f(x,y),0<α<1,y(0)=y.\left\{\begin{aligned} D^{\alpha}y&=f(x,y),\quad 0<\alpha<1,\\ y(0)&=y^{*}.\end{aligned}\right.

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.

Carlo Cattani (Engineering School, DEIM, University “La Tuscia”, Viterbo, Italy)

Solution by the proposer

Before we give the solution of (5), let us make some preliminary remarks about the most popular definitions of fractional derivatives.

The Riemann–Liouville integral of fractional order ν0\nu\geq 0 of a function f(x)f(x) is defined as

(Jνf)(t)={1Γ(ν)0t(tτ)ν1f(τ)dτ,ν>0,f(t),ν=0.(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] f(t),&\nu=0.\end{cases}

The corresponding Riemann–Liouville fractional derivative of order α>0\alpha>0 is defined as

DRLαf(t)=dndtnJnαf(t),nN,n1<αn.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.

The main problem with this derivative is that it assigns a nonzero value to a constant function. To avoid this issue, people often use the so-called order-α\alpha Caputo fractional derivative, defined as

DCαf(x)={dnf(x)dxn,0<αN,1Γ(nα)0xf(n)(τ)(xτ)αn+1dτ,t>0,0n1<α<n,D^{\alpha}_{\mathrm{C}}f(x)=\begin{cases}\dfrac{d^{n}f(x)}{dx^{n}},&0<\alpha\in\mathbb{N},\\[3.0pt] \displaystyle\frac{1}{\Gamma(n-\alpha)}\int_{0}^{x}\frac{f^{(n)}(\tau)}{(x-\tau)^{\alpha-n+1}}\,d\tau,&t>0,\\[-3.0pt] &0\leq n-1<\alpha<n,\end{cases}

where nn is an integer, x>0x>0, and fCnf\in\mathcal{C}^{n}.

Riemann–Liouville (RL) and Caputo (C) derivatives are the most popular and have been used in many applications; nevertheless, they both have some drawbacks. For this reason, many authors have introduced some more flexible fractional operators. The most general fractional derivative with a given kernel K(x,α)K(x,\alpha) is defined as

Dαf(x)={dnf(x)dxn,0<αN,0xf(n)(τ)K(xτ,α)dτ,t>0,0n1<α<n.D^{\alpha}f(x)=\begin{cases}\dfrac{d^{n}f(x)}{dx^{n}},&0<\alpha\in\mathbb{N},\\[6.45831pt] \displaystyle\int_{0}^{x}f^{(n)}(\tau)K(x-\tau,\alpha)\,d\tau,&t>0,\\[-3.0pt] &0\leq n-1<\alpha<n.\end{cases}

The kernel should be chosen in a such a way that at least the following two conditions are satisfied:

limα0K(xτ,α)=1andlimα1K(xτ,α)=δ(xτ).\lim_{\alpha\to 0}K(x-\tau,\alpha)=1\quad\textrm{and}\quad\lim_{\alpha\to 1}K(x-\tau,\alpha)=\delta(x-\tau).

Moreover, to ensure that one is dealing with a non-singular kernel, one requires that

limxτK(xτ,α)0α.\lim_{x\to\tau}K(x-\tau,\alpha)\neq 0\quad\forall\alpha.

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. This issue can be avoided by using the following simple definition, which is based on an intuitive interpolation.

Limiting ourselves to the case n=1n=1, the general structure of the Caputo-type fractional derivative

DCαf(x)=0xf(τ)K(xτ,α)dτ,0<α<1,D^{\alpha}_{\mathrm{C}}f(x)=\int_{0}^{x}f^{\prime}(\tau)K(x-\tau,\alpha)\,d\tau,\quad 0<\alpha<1,

is based on the kernel K(xτ,α)K(x-\tau,\alpha), which is a positive function that decays at infinity (to ensure convergence), while according to (6), the general structure of the Riemann–Liouville first-order derivative is

DRLαf(x)=1Γ(1α)ddx0xf(τ)(xτ)αdτ,0<α1.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.

Usually, to find the solution of (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. In any case, the solution will depend not only on the independent variable xx and the initial condition yy^{*}, but also on the kernel and on the fractional-order parameter:


The dependence on the fractional parameter α\alpha is essential in solving fractional-order problems. However, 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. Because of this lack of uniqueness, fractional calculus is missing a strong mathematical motivation. On 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. Therefore, 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. In 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 (5).

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 (7) and the accompanying choice of kernel. This 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. Thus, we set


so that the initial value problem (5) becomes


Now, for 0<α<10<\alpha<1, we easily obtain the following ordinary differential problem that is equivalent to (5):

{dydx(11α)y=(11α)y+1αf(x,y),y(0)=y0.\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),\\ y(0)&=y_{0}.\end{aligned}\right.

For the moment, we can suppose that the initial conditions of (5) and (8) do not necessarily coincide, i.e., yy0y^{*}\neq y_{0}. However, some further assumptions can be made when the function f(x,y)f(x,y) is given explicitly. In particular, to achieve a perfect equivalence between (5) and (8), we can set y=(1α)τ+αy0τRy^{*}=(1-\alpha)\tau+\alpha y_{0}\,\forall\tau\in\mathbb{R}, and then let α1\alpha\to 1.


Let y(x)y(x) be the unknown function of the following Bernoulli fractional-order Cauchy problem:

{Dαy=g(x)yβ,0<α<1,β0,1,y(0)=y,\left\{\begin{aligned} D^{\alpha}y&=g(x)y^{\beta},\quad 0<\alpha<1,\,\beta\neq 0,1,\\ y(0)&=y^{*},\end{aligned}\right.

where g(x)g(x) is a continuous function in the interval I=[0,)I=[0,\infty).

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.

Carlo Cattani (Engineering School, DEIM, University “La Tuscia”, Viterbo, Italy)

Solution by the proposer

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.

Thus, we set


so that (9) becomes


Then taking 0<α<10<\alpha<1 and using equations (10) and (11), we easily get the following ordinary differential problem equivalent to (9):

{dydx(11α)y=(11α)y+1αg(x)yβ,β{0,1},y(0)=y0.\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] &\hskip 40.00006pt\beta\neq\{0,1\},\\[-3.0pt] y(0)&=y_{0}.\end{aligned}\right.

For the moment, we search a general solution of (9) by assuming that yy0y^{*}\neq y_{0}. Solving separately the two equations


we obtain the respective solutions

y(x)=y+k1e(11α)x,\displaystyle y(x)=y^{*}+k_{1}e^{(1-\frac{1}{\alpha})x},
y(x)1β=e(1β)(11α)x[k2+1βα0xg(ξ)e1βαξdξ],β{0,1}.\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] &\hskip 40.00006pt\beta\neq\{0,1\}.\end{split}

Consequently, the solution of problem (12), which is also the solution of problem (9), takes the forms listed below.

1. Let y0yy_{0}\neq y^{*} and β{0,1}\beta\neq\{0,1\}. Then

y(x)=y+12(y0y)e(11α)x+e(1β)(11α)x[12(y0y)+1βα0xg(ξ)e1βαξdξ].\begin{split}y(x)&=y^{*}+\frac{1}{2}(y_{0}-y^{*})e^{(1-\frac{1}{\alpha})x}\\[-3.0pt] &\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}

2. Let y0=yy_{0}=y^{*} and β{0,1}\beta\neq\{0,1\}. Then


3. Let y0=0y_{0}=0, y0y^{*}\neq 0 and β{0,1}\beta\neq\{0,1\}. In this case, note that, in order to solve the given fractional-order Cauchy problem (9) via an equivalent ordinary differential problem, we can simply set y0=0y_{0}=0 in (12) and thus obtain the solution

y(x)=y12ye(11α)x+e(1β)(11α)x[12y+1βα0xg(ξ)e1βαξdξ].\begin{split}y(x)&=y^{*}-\frac{1}{2}y^{*}e^{(1-\frac{1}{\alpha})x}\\[-3.0pt] &\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}

4. Let y0=y=0y_{0}=y^{*}=0 and β{0,1}\beta\neq\{0,1\}. Then



Let gg be a real-valued C2C^{2}-function defined on (0,)(0,\infty), strictly increasing, such that g(x)>1g(x)>1 for all x(0,)x\in(0,\infty) and g(2)<4g(2)<4. Consider the boundary value problem

y=g(x)y,y(0)=1,y(0)=0.y^{\prime\prime}=-g(x)y,\quad y(0)=1,\quad y^{\prime}(0)=0.

Prove that the solution yy has exactly one zero in (0,π/2)(0,\pi/2), i.e., there exists a unique point x0(0,π/2)x_{0}\in(0,\pi/2) such that y(x0)=0y(x_{0})=0, and give a positive lower bound for x0x_{0}.

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)

Solution by the proposer

First suppose that y(x)>0y(x)>0 for x(0,π/2)x\in(0,\pi/2). The function z(x)=cosxz(x)=\cos x is the solution to the auxiliary initial value problem

z=z,z(0)=1,z(0)=0.z^{\prime\prime}=-z,\quad z(0)=1,\quad z^{\prime}(0)=0.



Integrating this equality over the interval (0,π/2)(0,\pi/2), we obtain

0π/2(1g(x))zydx=0π/2(yzyz)dx=(yzyz)0π/2=y(π/2),\begin{split}\int_{0}^{\pi/2}(1-g(x))zy\,dx&=\int_{0}^{\pi/2}(y^{\prime\prime}z-yz^{\prime\prime})\,dx\\ &=(y^{\prime}z-yz^{\prime})|_{0}^{\pi/2}=y(\pi/2),\end{split}

and y(π/2)0y(\pi/2)\geq 0 by the continuity of yy. But

(1g(x))zy<0in (0,π/2),(1-g(x))zy<0\quad\textrm{in}\ (0,\pi/2),

so we reached a contradiction. Thus, yy has at least one zero in (0,π/2)(0,\pi/2).

Next, observe that π/2<2\pi/2<2 and by assumption g(2)<4g(2)<4. Consider the function w(x)=cos(2x)w(x)=\cos(2x), which is a solution to the second auxiliary initial value problem

w=4w,w(0)=1,w(0)=0w^{\prime\prime}=-4w,\quad w(0)=1,\quad w^{\prime}(0)=0

and satisfies w(x)>0w(x)>0 for x(0,π/4)x\in(0,\pi/4). Suppose that yy has (at least) one zero in (0,π/4)(0,\pi/4). Denote the smallest such zero by x1x_{1}. Then y(x)y(x) is positive for x(0,x1)x\in(0,x_{1}) (recall that y(0)=1y(0)=1) and y(x1)=0y(x_{1})=0; hence y(x1)0y^{\prime}(x_{1})\leq 0. An argument analogous to the one above shows that

0x1(g(x)4)wydx=0π/2(wywy)dx=(wywy)0x1=w(x1)y(x1).\begin{split}\int_{0}^{x_{1}}(g(x)-4)wy\,dx&=\int_{0}^{\pi/2}(w^{\prime\prime}y-wy^{\prime\prime})\,dx\\ &=(w^{\prime}y-wy^{\prime})|_{0}^{x_{1}}=-w(x_{1})y^{\prime}(x_{1}).\end{split}

Note that w(x1)y(x1)0-w(x_{1})y^{\prime}(x_{1})\geq 0, but the integrand (g(x)4)wy<0(g(x)-4)wy<0, so again we reached a contradiction. Thus, yy has no zero in (0,π/4)(0,\pi/4), so π/4\pi/4 is a positive lower bound for the zeros of yy.

Finally, suppose that yy has more than one zero in (π/4,π/2)(\pi/4,\pi/2), namely, there exist at least two points x2,x3(π/4,π/2)x_{2},x_{3}\in(\pi/4,\pi/2) such that x2<x3x_{2}<x_{3} and y(x2)=y(x3)=0y(x_{2})=y(x_{3})=0. Take the function


where a,ba,b are chosen in such a way that v(x2)=0v(x_{2})=0 and v(x)v(x) is negative for x(x2,x3)x\in(x_{2},x_{3}); see Figure 1.

Figure 1. Assuming yy has more than one zero in (π/4,π/2)(\pi/4,\pi/2)

We have v=4vv^{\prime\prime}=-4v, so yvyv=(4g(x))yvy^{\prime\prime}v-yv^{\prime\prime}=(4-g(x))yv, which is positive on the interval (x2,x3)(x_{2},x_{3}). Therefore,


On the other hand,


and the right-hand side is negative since y(x3)=0y(x_{3})=0 and y(x)<0y(x)<0 for x(x2,x3)x\in(x_{2},x_{3}).


We propose an interesting stochastic-source scattering problem in acoustics. The 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. In 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.

Consider the following approximation problem of a stochastic-source Helmholtz equation:

Δu+k2u=fin D,\displaystyle\Delta u+{k^{2}}u=f\quad\textrm{in}\ D,
u=0,xD,\displaystyle u=0,\quad x\in\partial D,

where f=afaHaf=\sum_{a}f_{a}H_{a} is a generalized stochastic source. For the stochastic problem (13), we use the equations

u=auaHa,f=afaHa,u=\sum_{a}u_{a}H_{a},\quad f=\sum_{a}f_{a}H_{a},

and we get the collection of deterministic problems

Δua+k2ua=fain D,\displaystyle\Delta u_{a}+k^{2}u_{a}=f_{a}\quad\textrm{in}\ D,
ua=0,xD.\displaystyle u_{a}=0,\quad x\in\partial D.

Assume that uaH01(D)u_{a}\in H_{0}^{1}(D) solves problem (14). Then prove that, for all vH01(D)v\in H_{0}^{1}(D), the solution uaH01(D)u_{a}\in H_{0}^{1}(D) satisfies

Duavdx+Dk2uavdx=Dfavdx.-\int_{D}\nabla u_{a}\cdot\nabla v\,dx+\int_{D}k^{2}u_{a}v\,dx=\int_{D}f_{a}v\,dx.

George Kanakoudis, Konstantinos G. Lallas and Vassilios Sevroglou (Department of Statistics and Insurance Science, University of Piraeus, Greece)

Solution by the proposers

We decompose our problem into a hierarchy of deterministic evolution (BVPs), and we give their corresponding variational formulations.

For a=0\lvert a\rvert=0, we get

{Δu0+k2u0=f0in D,u0=0on D.\left\{\begin{aligned} \Delta u_{0}+k^{2}u_{0}&=f_{0}&&\quad\textrm{in}\ D,\\ u_{0}&=0&&\quad\textrm{on}\ \partial D.\end{aligned}\right.

The estimation of a solution of problem (15) is

u0H1(D)c0f0L2(D).\lVert u_{0}\rVert_{H^{1}(D)}\leq c_{0}\lVert f_{0}\rVert_{L^{2}(D)}.

For a=1\lvert a\rvert=1, we get

Δu1+k2u1=f1in D,\displaystyle\Delta u_{1}+k^{2}u_{1}=f_{1}\quad\textrm{in}\ D,
u1=0on D.\displaystyle u_{1}=0\quad\textrm{on}\ \partial D.

We take an arbitrary vH01(D)v\in H_{0}^{1}(D) and multiply equation (16) by vv. Then we get

(Δu1)v+k2u1v=f1v(\Delta u_{1})v+k^{2}u_{1}v=f_{1}v

and integrate over DD. Every term is integrable since uaH01(D)u_{a}\in H_{0}^{1}(D), and hence we have Δu1H01(D)\Delta u_{1}\in H_{0}^{1}(D) and vH01(D)v\in H_{0}^{1}(D), so

(Δu1)vH01(D),k2L(D),u1H01(D),vH01(D).(\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).

Therefore, k2u1vH01(D)k^{2}u_{1}v\in H_{0}^{1}(D) and f1L2(D)f_{1}\in L^{2}(D), so f1vH01(D)f_{1}v\in H_{0}^{1}(D). We obtain

D(Δu1)vdx+Dk2u1vdx=Df1vdx.\int_{D}(\Delta u_{1})v\,dx+\int_{D}k^{2}u_{1}v\,dx=\int_{D}f_{1}v\,dx.

We use Green’s formula according to which

D(Δu1)vdx=Du1vdx+Dγ1(u1)γ0(v)dΓ\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

since vH01(D)v\in H_{0}^{1}(D) is equivalent to γ0(v)=0\gamma_{0}(v)=0.

Let H01(D)H_{0}^{1}(D) be a stochastic Hilbert space. If we now assume the bilinear form on H01(D)×H01(D)H_{0}^{1}(D)\times H_{0}^{1}(D),

a(u1,v)=D(u1v+k2u1v)dx,a(u_{1},v)=\int_{D}(-\nabla u_{1}\cdot\nabla v+k^{2}u_{1}v)\,dx,

and the linear functional on H01(D)H_{0}^{1}(D),


then the variational formulation of problem (17) is

a(u1,v)=(v)vH01(D).a(u_{1},v)=\ell(v)\quad\forall v\in H_{0}^{1}(D).

The estimation of a solution of problem (18) is

u1H1(D)c1f1L2(D).\lVert u_{1}\rVert_{H^{1}(D)}\leq c_{1}\lVert f_{1}\rVert_{L^{2}(D)}.

For a=n\lvert a\rvert=n, we get

{Δun+k2un=fnin D,un=0on D.\left\{\begin{aligned} \Delta u_{n}+k^{2}u_{n}&=f_{n}&&\quad\textrm{in}\ D,\\ u_{n}&=0&&\quad\textrm{on}\ \partial D.\end{aligned}\right.

The estimation of a solution of problem (19) is

unH1(D)cnfnL2(D).\lVert u_{n}\rVert_{H^{1}(D)}\leq c_{n}\lVert f_{n}\rVert_{L^{2}(D)}.

Via the above variational formulations and taking into account u=auaHau=\sum_{a}u_{a}H_{a}, we can prove that the solution uu of the stochastic boundary value problem (13) satisfies the following inequality:

uH1(D)c0f0L2(D)+c1f1L2(D)++cnfnL2(D),\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)},\\

where cic_{i}, i=0,1,,ni=0,1,\ldots,n, are considered to be in agreement with appropriate built-in weights. The solution uu belongs to the space {H1(D),Ω,F,μ}\{H^{1}(D),\Omega,F,\mu\} which is a stochastic Hilbert space with μ\mu the probability measure defined by Ha(ω)H_{a}(\omega).


For a Newtonian incompressible fluid, the Navier–Stokes momentum equation, in vector form, reads [4]

ρ(ut+uu)=p+μ2u+F,\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},
u=u(x,t),u ⁣:Rn×(0,)Rn.\displaystyle\mathbf{u}=\mathbf{u}(\mathbf{x},t),\quad\mathbf{u}\colon\mathbf{R}^{n}\times(0,\infty)\to\mathbf{R}^{n}.

Here, ρ\rho is the fluid density, u\mathbf{u} is the velocity vector field, pp is the pressure, μ\mu is the viscosity, and F\mathbf{F} is an external force field.

(i) Assuming that both the pressure drop p\nabla p and the external field F\mathbf{F} are negligible, it is easy to show that equation (20) reduces to

ut+uu=ν2u,\frac{\partial\mathbf{u}}{\partial t}+\mathbf{u}\cdot\nabla\mathbf{u}=\nu\nabla^{2}\mathbf{u},

and finally to equation (22), where ν=μρ\nu=\frac{\mu}{\rho} is the so-called kinematic viscosity [5].

(ii) Regarding the one-dimensional viscous Burgers equation

ut+uux=ν2ux2,u=u(x,t),\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=\nu\frac{\partial^{2}u}{\partial x^{2}},\quad u=u(x,t),

prove that an analytical solution can be obtained by means of the Tanh Method [2, 3, 5] as

u(x,t)=λ[1tanh(λ2ν(xλt))],λ>0.u(x,t)=\lambda\Bigl[1-\operatorname{tanh}\Bigl(\frac{\lambda}{2\nu}(x-\lambda t)\Bigr)\Bigr],\quad\lambda>0.

M. A. Xenos and A. C. Felias (Department of Mathematics, University of Ioannina, Greece)

Solution by the proposers

Notice that, for p=F=0\nabla p=\mathbf{F}=\mathbf{0}, equation (20) becomes

ρ(ut+uu)=μ2u.\rho\Bigl(\frac{\partial\mathbf{u}}{\partial t}+\mathbf{u}\cdot\nabla\mathbf{u}\Bigr)=\mu\nabla^{2}\mathbf{u}.

Since, for an incompressible fluid, ρ\rho is a nonzero constant, one can divide both sides of equation (23) by ρ\rho and thus obtain equation (21). Now consider the motion of a one-dimensional viscous fluid with fluid velocity uu along the xx-axis as time passes, u=u(x,t)u=u(x,t). In this case, equation (21) transforms into equation (22