2440180 [math.GM] 21 Oct 2018

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Oct 21, 2018 - arXiv:submit/2440180 [math.GM] 21 Oct 2018. Reduction of modern problems of mathematics to the classical Riemann-Poincar-Hilbert problem.
arXiv:submit/2440180 [math.GM] 21 Oct 2018

Reduction of modern problems of mathematics to the classical Riemann-Poincar-Hilbert problem Durmagambetov A.A1 a 010000,

Institute of Information and Computational Technologies, International Science Complex Astana, Kazakhstan

Abstract Using the example of such a complicated problem as the Cauchy problem for the Navier-Stokes equation, we show how the Poincar-Riemann-Hilbert boundary value problem enables us to construct effective estimates of solutions for this case. The apparatus of the three-dimensional inverse problem of quantum scattering theory is developing for this. In which it is shown that the unitary scattering operator can be studied as a solution of the Poincar-Riemann-Hilbert boundary-value problem. This allows us to go on to study the potential in the Schrodinger equation, which we consider as a velocity component in the NavierStokes equation. The same scheme of reduction of Riemann integral equations for the zeta-function to the Poincar-Riemann-Hilbert boundary-value problem allows us to construct effective estimates that describe the behavior of the zeros of the zeta function very well. Keywords: Schr¨odinger’s equation; potential; scattering amplitude; Cauchy problem; Navier–Stokes equations; Fourier transform, the global solvability and uniqueness of the Cauchy problem,the loss of smoothness,The Millennium Prize Problems,Euler product, Dirichlet, Riemann, Hilbert, Poincar, Riemann hypothesis, zeta function, Hadamard, Landau, Walvis, Estarmann, Chernoff

Email address: [email protected] (Durmagambetov A.A) URL: (Durmagambetov A.A) 1

Preprint submitted to Journal of LATEX Templates

October 21, 2018

1. INTRODUCTION Using the example of such a complicated problem as the Cauchy problem for the Navier-Stokes equation, we show how the Poincar-Riemann-Hilbert boundary value problem enables us to construct effective estimates of solutions for this case. The apparatus of the three-dimensional inverse problem of quantum scattering theory is developing for this. In which it is shown that the unitary scattering operator can be studied as a solution of the Poincar-Riemann-Hilbert boundary-value problem. This allows us to go on to study the potential in the Schrodinger equation, which we consider as a velocity component in the NavierStokes equation. The same scheme of reduction of Riemann integral equations for the zeta-function to the Poincar-Riemann-Hilbert boundary-value problem allows us to construct effective estimates that describe the behavior of the zeros of the zeta function very well.

2. Results for the one-dimensional case Let us consider one-dimensional function f and its Fourier transformation f˜. Using notions of module and phase, we write Fourier transformation in the following form f˜ = |f˜| exp(iΨ) , where Ψ is phase. To cite Plancherel equality: ||f ||L2 = Const||f˜||L2 . Here we can see that a phase is not contributed

to determination of X norm. To estimate a maximum we have a simple estimate as max|f |2 ≤ 2||f ||L2 ||∇f ||L2 .Now we have an estimate of the function maximum in which a phase is not involved. Let us consider a behavior of a progressing wave running with a constant velocity of v = a described by function F (x, t) = f (x + at). For its Fourier transformation along x variable we have F˜ = f˜exp(iatk). Again in this case we can see that when we will be studying a module of the Fourier transformation, we will not obtain major physical information about the wave, such as its velocity and location of the wave crest because of |F˜ | = |f˜| . These two examples show w eaknesses of studying Fourier transformation. On the other hand, many researchers focus on the study of functions using embedding theorem, but in the embedding theorems main object of 2

the study is module of function. But as we have seen in given examples, a phase is a main physical characteristic of a process, and as we can see in the mathematical studies, which use embedding theorems with energy estimates, the phase disappears. Along with phase, all reasonable information about physical process disappears, as demonstrated by Terence Tao [1] and other research considerations. In fact, he built progressing waves that are not followed energy estimates. Let us proceed with more essential analysis of influence of the phase on behavior of functions. Theorem 1. There are functions of W21 (R) with a constant rate of the norm for a gradient catastrophe of which a phase change of its Fourier transformation is sufficient. Proof: To prove this, we consider a sequence of testing functions f˜n = ∆/(1+k 2 ), ∆ = (i−k)n /(i+k)n . it is obvious that |f˜n | = 1/(1+k 2 ). max|fn |2 ≤ 2||fn ||L2 ||∇fn ||L2 ≤ Const.. Calculating the Fourier transformation of these testing functions, we obtain: fn (x) = x(−1)(n−1) 2π exp(−x)L1(n−1) (2x)if x > 0, fn (x) = 0 if x ≤ 0

(1)

where L1(n−1) (2x) is a Laguerre polynomial . Now we see that the functions are equibounded and derivatives of these functions will grow with the growth of n. Thus, we have built an example of a sequence of the bounded functions of W21 (R) which have a constant norm W21 (R) and this sequence converges to a discontinuous function. Theorem 2. There are functions of W21 (R) with a constant rate of the norm for a gradient catastrophe of which a phase change of its Fourier transformation is sufficient. Proof: To prove this, we consider a sequence of testing functions f˜n = ∆/(1+k 2 ), ∆ = (i−k)n /(i+k)n . it is obvious that |f˜n | = 1/(1+k 2 ). max|fn |2 ≤ 2||fn ||L2 ||∇fn ||L2 ≤ Const.. Calculating the Fourier transformation of these testing functions, we obtain: fn (x) = x(−1)(n−1) 2π exp(−x)L1(n−1) (2x)if x > 0, fn (x) = 0 if x ≤ 0 3

(2)

where L1(n−1) (2x) is a Laguerre polynomial . Now we see that the functions are equibounded and derivatives of these functions will grow with the growth of n. Thus, we have built an example of a sequence of the bounded functions of W21 (R) which have a constant norm W21 (R) and this sequence converges to a discontinuous function. 3. Results for the three-dimensional case Consider Schrodinger’sequation: −∆x Ψ + qΨ = k 2 Ψ, k ∈ C

(3)

Let Ψ+ (k, θ, x) be a solution of ( 3 )with the following asympotic behavior:   ′ eik|x| 1 , |x| → ∞, (4) A(k, θ , θ) + 0 Ψ+ (k, θ, x) = Ψ0 (k, θ, x) + |x| |x| ′



where A(k, θ , θ) is the scattering amplitude and θ = {Imk ≥ 0} Ψ0 (k, θ, x) = eik(θ,x) ′

A(k, θ , θ) = −

1 4π

Z

x |x| , θ

∈ S 2 for k ∈ C¯ + =



q(x)Ψ+ (k, θ, x)e−ikθ x dx.

R3

Solutions ( 3 )- ( 4 ) are obtained by solving the integral equation Z e+ik|x−y| Ψ+(k, θ, y)dy = G(qΨ+ ) Ψ+ (k, θ, x) = Ψ0 (k, θ, x) + q(y) |x − y| R3 which is called the Lippman-Schwinger equation. Let inroduce ′

θ, θ ∈ S 2 , Df = k

Z







A(k, θ , θ)f (k, θ )dθ ,

S2

Let us also define the solution Ψ− (k, θ, x) for k ∈ C¯ − = {Imk ≤ 0} as Ψ− (k, θ, x) = Ψ+ (−k, −θ, x) . As is well known[8] : k Ψ+ (k, θ, x) − Ψ− (k, θ, x) = − 4π

Z



S2





A(k, θ , θ)Ψ− (k, θ , x)dθ , k ∈ R.

(5)

This equation is the key to solving the inverse scattering problem, and was first used by Newton [8,9] and Somersalo et al. [10]. 4

Definition 1. The set of measurable functions R with the norm, defined by Z q(x)q(y) ||q||R = dxdy < ∞ 2 R6 |x − y| is recognized as being of Rollnik class. Equation (5) is equivalent to the following: Ψ+ = SΨ− , where S is a scattering operator with the kernel S(k, l), Z Ψ+ (k, x)Ψ∗− (l, x)dx , S(k, l) = R3

. The following theorem was stated in [9]: Theorem 3. (The energy and momentum conservation laws) Let q ∈ R. Then, SS ∗ = I, S ∗ S = I, where I is a unitary operator. Corollary 1. SS ∗ = I, S ∗ S = I, yeild ′



A(k, θ , θ) − A(k, θ, θ )∗ =

ik 2π

Z

′′



′′

A(k, θ, θ )A(k, θ , θ )∗ dθ

′′

S2

Theorem 4. (Birmann–Schwinger estimation). Let q ∈ R. Then, the number of discrete eigenvalues can be estimated as: Z Z 1 q(x)q(y) N (q) ≤ dxdy. 2 (4π) R3 R3 |x − y|2  Lemma 1. Let |q|L1 (R3 ) + 4π|q|L2 (R3 ) < α < 1/2. Then,  |q|L1 (R3 ) + 4π|q|L2 (R3 ) α < kΨ+ kL∞ ≤ 1−α 1 − |q|L1 (R3 ) + 4π|q|L2 (R3 )



∂(Ψ+ − Ψ0 ) |q|L1 (R3 ) + 4π|q|L2 (R3 ) α

< ≤

∂k 1 − α 3 3 1 − |q| + 4π|q| L1 (R ) L2 (R ) L∞

Proof. By Lippman-Schwinger equation’s

|Ψ+ − Ψ0 | ≤ |GqΨ+ | , 5

|Ψ+ − Ψ0 |L∞ ≤ |Ψ+ − Ψ0 |L∞ |Gq| + |Gq| finaly

 |q|L1 (R3 ) + 4π|q|L2 (R3 )  |Ψ+ − Ψ0 | ≤ 1 − |q|L1 (R3 ) + 4π|q|L2 (R3 )

By Lippman-Schwinger equation’s ∂ (Ψ+ − Ψ0 ) ∂Gq ∂ (Ψ+ − Ψ0 ) ≤ + |Gq| ∂k Ψ+ + Gq ∂k ∂k ∂(Ψ+ − Ψ0 )  ≤ |q|L (R3 ) + 4π|q|L (R3 ) 1 2 ∂k

∂(Ψ+ − Ψ0 ) |q|L1 (R3 ) + 4π|q|L2 (R3 )

 ≤

∂k 1 − |q|L1 (R3 ) + 4π|q|L2 (R3 ) L∞

proof completes

Let us introduce the following notations: Z ′ ′ q(x)eik(θ−θ )x dx, K(s) = s, X(x) = x. Q(k, θ, θ ) = R3

T+ Q =

Z

+∞

−∞



Q(s, θ, θ ) ds, T− Q = s − t − i0

Z

+∞

−∞



Q(s, θ, θ ) ds. s − t + i0

3

Lemma 2. Let q ∈ R ∩ L1 (R ), kqkL1 + 4π|q|L2 (R3 ) < α < 1/2. Then, kA+ kL∞ < α +

α . 1−α



∂A+ α

0, T− f = lim s−z 2πi Imz→0

−∞

Z∞

−∞

Tf =

f (s) ds, Im z < 0, s−z

1 (T+ + T− )f. 2

Consider the Riemann problem of finding a function Φ, that is analytic in the complex plane with a cut along the real axis.Values of Φ on the sides of the cut are denoted as Φ+ , Φ− .The following presents the results of [12]: Lemma 3. TT =

1 1 1 1 1 I, T T+ = T+ , T T− = − T− , T+ = T + I, T− = T − I, T− T− = −T− 4 2 2 2 2

Denote Φ+ (k, θ, x) = Ψ+ (k, θ, x)−Ψ0 (k, θ, x), Φ− (k, θ, x) = Ψ− (k, −θ, x)−Ψ0 (k, θ, x), g(k, θ, x) = Φ+ (k, θ, x) − Φ− (k, θ, x) Lemma 4. Let q ∈ R, N (q) < 1 g+ = g(k, θ, x), g− = g(k, −θ, x).. Then, Φ+ (k, θ, x) = T+ g+ + eikθx , Φ− (k, θ, x) = T− g+ + eikθx . Proof. The proof of the above follows from the classic results for the Riemann problem. Lemma 5. Let q ∈ R, N (q) < 1 g+ = g(k, θ, x), g− = g(k, −θ, x), ). Then, Ψ+ (k, θ, x) = (T+ g+ + eikθx ), Ψ− (k, θ, x) = (T− g− + e−ikθx ). Proof. The proof of the above follows from the definitions of g, Φ± , Ψ± . Lemma 6. Let

∞ Z Z ′ pA(p, θ , θ) αdθ < 1/2 dp < α, sup 4π(p − k + i0) k S2 −∞

Then

Z ∞ kj A(kj , θ′ , θk ) j kj dkj dθkj ≤ 2−n 4π(k − k + i0) j+1 j S2 −∞

Y Z

0≤j 1 + δ, δ > 0.

(21)

Hadamard was the first to apply P (s) in the study of the zeta function [28]. Chernoff made significant progress in the Riemann hypothesis using P (s) [29]. In the present study, modifications of Chernoff’s results are obtained. Specifically, his study on the pseudo zeta function is completed. Chernoff obtained an equivalent formulation of the Riemann hypothesis in terms of a pseudo zeta function as follows.

22

THEOREM. (Chernoff) Let C(s) =

Y

n>1

1−

1 (nln(n))s

−1

. Then, C(s) continues analytically into the critical strip and has no zeros there. The significance of this theorem is that if the primes were distributed more regularly (i.e., if pn ≡ n log n), then the Riemann hypothesis would be trivially true. In an effort to further develop the work of Chernoff and Hadamard, the following question naturally arises: Does the pseudo zeta function P (s) continues analytically into the critical strip? It should be noted that analytic extensions of P (s) were first studied by E. Landau and A. Walvis [30] and T. Estarmann [31], [32]; however, no satisfactory estimates for P (s) were obtained, and the present study is concerned with this question. THEOREM. ( E. Landau, A. Walvis, T. Estarmann) Let µ(n) − f unctionM o¨bius . Then, P (s) =

X µ(n)lnζ(ns) as Re(s) > 1 + δ, δ > 0, n

n≥1

X µ(n)lnζ(ns) − meromorphic f unction as Re(s) > δ, δ > 0. n

n≥1

The section is organized as follows. Intermediate estimates are first obtained for the lnζ(s). Subsequently, the sets where the logarithm of the zeta function is uniquely determined are defined. These sets are composed of rectangles in which the zeta function has no roots, and they cover the entire critical strip except for the rectangular regions in which the zeros of the zeta functions are located. In the rectangles in which there are no zeros of the zeta function, the real value of its logarithm can be defined, and in these sets, the mirrorsymmetric equation that arises by taking the logarithm on both sides of the Riemann functional equation is investigated. Then, the Fourier transform is applied to it, and it is multiplied by a regulating factor. Thus, a Riemann– Hilbert boundary value problem is obtained for the lnζ(s) The properties of the solution to the Riemann–Hilbert boundary value problem are expressed in 23

terms of the Hilbert integral transform. In the rectangles in which the zeta function has no roots, the Hilbert transform can be used to obtain exact lower bounds for the zeta function in the critical strip.

8. RESULTS As mentioned in Introduction, certain simple intermediate estimates are first obtained. The rectangles in which the zeta function has zeros are first introduced as follows: D(n) = (s|0.1 < Re(s) < 0.9, Im(s) 6= Im(sn ), Im(sn )−dn ≤ Im(s) ≤ Im(sn )+dn D(n) = (s|0.1 < Re(s) < 0.9, , Im(−s) 6= Im(−sn ), −Im(sn )−dn ≤ Im(s) ≤ −Im(sn )+dn , where ζ(sn+1 ) = 0, ζ(sn ) = 0, ζ(1 − sn ) = 0, ζ(1 − sn+1 ) = 0, ζ(1 − sn ) = 0, dn = (Im(sn+1 ) − Im(sn ))/2 The sets of D(n), is shown in the figure below. D(n)

Im(s)

0

1/2−ǫ

1 −ǫ

1/2+ǫ

Re(s)

D(n)

Theorem 7. Let s ∈ Dn ∪ D(n) , F (s) = 2s ln(π) − ln(Γ(s/2)) − ln(Γ(1 − s)/2)). Then,

sup s∈Dn ∪D(n)

|F (τ + iα)| +

sup s∈Dn ∪D(n)

24

|

dF (τ + iα) | < CCn dτ

1−s 2 ln(π)

+

Proof. As 0.1 < τ < 0.9 implies that F - is holomorphic which completes the proof. As mentioned in Introduction, a Riemann–Hilbert boundary value problem should be obtained. To this end, an equation should be derived that determines the difference between the boundary values of the analytic functions in the upper plane and the lower plane. Denote Q(s) = Re(ln(ζ(s))) = ln(|ζ(s)|). Theorem 8. Let s ∈ D(n) or s ∈ D(n) and F (s) =

1−s s ln(π) − ln(Γ(s/2)) − ln(π) + ln(Γ(1 − s)/2)). 2 2

Then, Q(s) = Q(1 − s) + Re(F (s)) Proof. The Riemann equation (20) can be logarithmized for s ∈ D(n) or s ∈ D(n), and an equation for Q(s) may be obtained as an equation for the real part. As the resulting equation is not suitable for the formulation of a Riemann– Hilbert boundary value problem, a number of transformations should be performed. Certain preliminary arguments suggest that the Fourier transform is an appropriate choice. In Riemann–Hilbert boundary value problems, the asymptotic behavior of the unknown functions is highly important. To ensure this behavior, it is necessary to have estimates for the derivatives. The evaluation of the derivatives is usually carried out by integration by parts. Thus, the following

25

functions are introduced:   νǫ (s) = 0, Re(s) < 2ǫ;        νǫ (s) = 1, ǫ < Re(s) < 1/2 − 2ǫ;      νǫ (s) = 0, 1/2 − 2ǫ < Re(s) < 1/2 + 2ǫ;        νǫ (s) = 1, 1/2 + 2ǫ < Re(s) < 1 − 2ǫ,

νǫ (s) = 0, Re(s) > 1 − 2ǫ;      1 1 R1   ψ(t) = σe t2 −1 , t2 < 1; σ1 = e t2 −1 dt    −1     2  ψ(t) = 0, t ≥ 1;      µ (x) = R ψ(s/ǫ)ν (x − t)dt/ǫ = R ψ(x − t/ǫ)ν (t)dt/ǫ; ǫ ǫ ǫ

The graph of νǫ is shown in the figure below. νǫ

0

1/2−ǫ

1 −ǫ

1/2+ǫ

s

As the equation is multiplied by µǫ , it should be verified that the symmetry of the equation for Q is not affected. Lemma 16. For νǫ and µǫ , we have νǫ (x) = νǫ (1 − x), µǫ (x) = µǫ (1 − x). Proof. We have νǫ (x) = νǫ (1 − x) by definition. Moreover, Z Z µǫ (x) = ψ(s/ǫ)νǫ (x − s)ds/ǫ = ψ(s/ǫ)νǫ (1 − x + s)ds/ǫ = µǫ (1 − x).

Lemma 17. Let 1/2 + 3ǫ < x < 1 − 3ǫ. Then, µǫ (x) = 1. Proof. µǫ (x) =

Z

ψ(s/ǫ)νǫ (x − s)ds/ǫ =

Z

ǫ

−ǫ

ψ(s/ǫ)νǫ (x − s)ds/ǫ =

26

Z

ǫ

−ǫ

ψ(s/ǫ)ds/ǫ = 1.

To compute the Fourier transform of the equation for Q, the functions R(k), Qǫ (s), Qǫ (1 − s), Re(Fǫ (s)), the Fourier transform of Qǫ (s), the Fourier transform of Qǫ (1 − s), the Fourier transform of Fǫ (s), are introduced as follows: Qǫ (s) = Q(s)µǫ (Re(s)), Qǫ (1 − s) = Q(1 − s)µǫ (1 − Re(s)) e−ik R(k) = + 1 , Fǫ (s) = Re(F (s))µǫ (Re(s)); k − ia Z 0.9 Z 1 e−ik 0.9 √ Qǫ (1 − τ − iα)e−ikτ dτ = √ µǫ (1 − Re(s))Qǫ (τ − iα)eikτ dτ ; 2π 0.1 2π 0.1 Z 0.9 fǫ (k, α) = √1 Qǫ (τ + iα)e−ikτ dτ, Q 2π 0.1 Z 0.9 1 f Re(Fǫ (τ + iα)e−ikτ )ds. Fǫ (k, α) = √ 2π 0.1 1 Jǫ (k, α) = √ 2π

Z

0.9

0.1

1 Qǫ (τ − iα)eikτ dτ , Iǫ (k, α) = √ 2π

Z

0.9

Qǫ (τ + iα)e−ikτ dτ. (22)

0.1

To obtain the Riemann-Hilbert boundary value problem , the following lemma is required. Lemma 18. Let a > 2 then ind(R) = 0 Proof. By definition ind(R) =

1 2πi

Z

+∞

−∞

R(k)′ dk R(k)

As Im(z) < 0, |e−iz | < 1 and |z − ia| > 2 yield

R(k)′ R(k)

have nothing pole. Latest statement and Lemma of Jordan yield ind(R) = 0.

To obtain the necessary asymptotics, the following lemma is required. Lemma 19. Let a>2 then ln(R(k)) is single-valued analitical fuction in lower half plane.

27

Proof. As Im(k) ≤ 0 yeild Re(R(k)) = 1 + Re



 eik >0 k − ia

which completes proof. Denote Ω(s) = Re(s − sn ), ω(s) = (s − sn )(s − 1 + sn ) To obtain the necessary asymptotics, the following lemma is required. Lemma 20. Let γ = 1/4 and |Ω(1/2)| = ǫn > 0, dn = (Im(sn+1 ) − Im(sn ))/2 Then, we have the following estimate as ǫ = 0.1ǫn : Z 0.9 2 sup |Qǫ (τ + iα)| + |Qǫ (τ + iα)| dτ < Cn Cǫn Cγ .

Imsn −dn ≤α≤Imsn +dn

0.1

sup

Z

−Imsn −dn ≤α≤−Imsn +dn

sup Imsn −dn ≤α≤Imsn +dn

0.9

0.1

Z

0.9

−Imsn −dn ≤α≤−Imsn +dn

Z

2

|Fǫ (τ + iα)| + |Fǫ (τ + iα)| dτ < Cn Cǫn Cγ .

0.1

sup

2

|Qǫ (τ − iα)| + |Qǫ (τ − iα)| dτ < Cn Cǫn Cγ .

0.9 0.1

2

|Fǫ (τ − iα)| + |Fǫ (τ − iα)| dτ < Cn Cǫn Cγ .

Proof. by definion Qǫ (s), IQ =

Z

0.9

0.1

2

|Qǫ (τ + iα)| +|Qǫ (τ + iα)| dτ =

Z

0.9

0.1

|ln|ζ(τ + iα)||2 +|ln |ζ(τ + iα)|| dτ ≤

2 2 0.9 ζ(τ + iα) 1 1 + ln + ln ln ζ(τ + iα) + ln ω(τ + iα) ω(τ + iα) ω(τ + iα) ω(τ + iα) dτ 0.1

Z

Denote

Lmax =

max s∈D(n)∪D(n)

ζ(s) ω(s) , Lmin = 28

min s∈D(n)∪D(n)

ζ(s) ω(s)

2 Z 0.9 1 2γ 1 γ 1 1 IQ ≤ ln Lmax + + + ln L + + max ω(s) ω(s) dτ Lmin Lmin 0.1

which completes the proof.

The previous constructions allow the calculation of the asymptotics as follows. Lemma 21. Let (3/4 + iα) ∈ D(n).Then

lim

Im(k)→−∞

Iǫ (k, α) = 0,

lim

Im(k)→∞

Jǫ (k, α) = 0.

(23)

and as Im(k)=0 lim

Re(k)→∞

Iǫ (k, α) = 0,

lim

Re(k)→∞

Jǫ (k, α) = 0.

(24)

Proof. Lemma 20 yield Z |Iǫ (k, α)| =

0.9

0.1

Z Qǫ (τ + iα)e−ikτ dτ ≤

0.9

0.1

 1/2 |Qǫ (τ + iα)|2 dτ

1 |Im(k)|1/2

A similar argument is used for the function Z 0.9 1 Jǫ (k, α) = √ Qǫ (τ − iα)eikτ dτ. 2π 0.1

As Im(k) > 0 , Jǫ (k, α) can be estimated using the last expression and Lemma 5 as follows: |Jǫ (k, α)|
0, T− f = lim s−z 2πi Imz→0

Z∞

−∞

Tf =

f (s) ds, Im z < 0. s−z

1 (T+ + T− )f. 2

These operators are closely related to the Hilbert transform, whose isometric properties were studied by Poincar´e. The following result is from [33]. Lemma 22. TT =

1 1 1 1 1 I, T T+ = T+ , T T− = − T− , T+ = T + I, T− = T − I, 4 2 2 2 2

where I is the identity operator If = f . The reduction to a Riemann–Hilbert boundary value problem can now be formulated as follows. Theorem 9. Let (3/4 + iα) ∈ D(n), a > 2 Z ∞ 1 ln(R(t))dt Γ+ (k) = − 2πi −∞ t − k − i0 Z ∞ 1 ln(R(t))dt Γ− (k) = − 2πi −∞ t − k + i0 X (k) − X+ (k) = eΓ+ (k) , X− (k) = eΓ− (k) , R(k) = , Gǫ (k, α) = Jǫ (k, α). X+ (k)

(26) (27) (28) (29)

Then, Z



dt Gǫ Gǫ (t, α) = X+ (k)T+ X− −∞ X− (t) t − k − i0 Z ∞ fǫ (k, α) dt Iǫ (k, α) F Gǫ X− (k) Gǫ (t, α) . − =− dt = X− (k)T− k − ia k − ia 2πi X (t) t − k + i0 X − − −∞ Jǫ (k, α) = −

X+ (k) 2πi

30

Proof. By Theorem 8 and Lemma 16 we have Qǫ (s) = Qǫ (1 − s) + Fǫ (s).

(30)

Using the Fourier transform, we obtain fǫ (k, α). Iǫ (k, α) = e−ik Jǫ (k, α) + F

Multiplying this equation by

1 k−ia

(31)

we get

fǫ (k, α) e−ik Jǫ (k, α) F Iǫ (k, α) = + . k − ia k − ia k − ia

(32)

fǫ (k, α) Iǫ (k, α) F − = R(k)Jǫ (k, α) + Jǫ (k, α). k − ia k − ia

(33)

Rewriting latest equation

Ψ− (k, α) =

fǫ (k, α) Iǫ (k, α) F − k − ia k − ia Ψ+ (k, α) = Jǫ (k, α)

(35)

Gǫ (t, α) = Jǫ (k, α).

(36)

(34)

Using Lemma 21, the following Riemann-Hilbert boundary value problem is obtained regarding the definition of an analytic function from its boundary values on the real line:

lim

Im(k)→∞

Ψ− (k, α) = R(k)Ψ+ (k, α) + Gǫ (k, α),

(37)

Ψ+ (k, α) = 0,

(38)

lim

Im(k)→−∞

Ψ+ (k, α) = 0.

Hilbert’s formula and Lemma 20, 21 gives the solution to the RiemannHilbert boundary value problem (37),(38) Z ∞ dt X+ (k) Gǫ (t, α) Ψ+ (k, α) = − 2πi X (t) t − k − i0 − −∞ Z ∞ dt Gǫ (t, α) X− (k) Ψ− (k, α) = − 2πi X (t) t − k + i0 − −∞ 31

(39) (40)

Denote

Jǫ (k, α)R(k) X− (k) fǫ (k, α) Ψ− (k, α) Iǫ (k, α) − F Φ− (k, α) = − X− (k) X− (k)(k − ia) Φ+ (k, α) = Ψ+ (k, α) −

(41) (42)

For Φ+ (k, α), Φ− (k, α) we have new boundary-value problem of Riemann-Hilbert :

lim

Im(k)→∞

Φ+ (k, α) = 0,

Φ− (k, α) = Φ+ (k, α)

(43)

lim

(44)

Im(k)→−∞

Φ− (k, α) = 0.

Liouville theorem implies that Φ− (k, α) = 0 , Φ+ (k, α) = 0.

(45)

which completes the proof.

9. DISCUSSION Our computations led to a new definition of the functions Iǫ (k), Jǫ (k), which we obtained from the Riemann-Hilbert boundary-value problem. From the uniqueness of the solution of the Riemann-Hilbert boundary value problem functions Iǫ (k), Jǫ (k), defined earlier in (6) and obtained from the Hilbert formula are equal! To obtain the final estimates for the zeta function, the isometric properties of the integral Hilbert transform will be used. Theorem 10. Let (3/4 + iα) ∈ D(n) and a > 2. Then, C −1 < |X− (t)| < C, C −1 < |X+ (t)| < C. ||Ψ+ ||L2 ≤ Cǫ , ||Ψ− ||L2 ≤ Cǫ , Proof. By Lemma 18,19,22 we get Z ∞ ln(R(t))dt 1 = T− ln(R) = ln(R) Γ− (k) = 2πi −∞ t − k + i0 32

X− (t) = R(t), X+ (t) = 1 latest estimates implies C −1 < |X− (t)| < C, |X+ (t)| = 1.

Using Theorem 9 we obtain ||Ψ− ||2L2

+

||Ψ+ ||2L2

=

Z

+∞

−∞

2 Z +∞ I (k, α) F fǫ (k, α) ǫ 2 |Jǫ (k, α)| dk ≤ Cn Cǫ − dk + k − ia k − ia −∞

Lemma 23. Let βn , φn satisfies equations eβ =

p (2πn + φ)2 + (β − a)2 ,

φ = π − arg (2πn + φ + i(−a + β)) Then tn = 2πn + π + iβn + φn , n ≥ 0 root of equation R(k) = 0 and βn =

ln(2πn) + o(1), φn = π + O(ln(n)/n), n ≥ 1 2

Proof. R(tn ) =

e−itn e−i2πn+β−iφ +1= +1= tn − ia 2πn + φ + i(βn − a)

e−i2πn+βn −iφ p +1= ei arg(2πn+φ+i(−a+βn )) (2πn + φ)2 + (βn − a)2 −eβn p + 1 = −1 + 1 = 0 (2πn + φn )2 + (βn − a)2

take β = ln(nπ) + γn then

33

e

γn

=

s

1+

(ln(nπ) + γn − a)2 , (2πn + φ)2

for φ we have φn = π − arctan



(−a + β) 2πn + φn



and we get φn = π + O(ln(n)/n) proof complete. Theorem 11. Let s ∈ D(m) ∪ D(m) and a > 2, with 1/2 + 3ǫ < Re(s) < 1 − 3ǫ. Then, |Q(s)| < Cm Cǫ . Proof. By Theorem 9, Z ∞ fǫ (k, α) dt X− (k) Gǫ (t, α) Iǫ (k, α) F − =− . Ψ− (k, α) = k − ia k − ia 2πi X (t) t − k + i0 − −∞ Denote I1 =

Z

∞ −∞

Gǫ (t, α) dt, I2 = (k − ia)I1 X− (t)(t − k + i0)

Lemma 22 and Lemma of Jordan yield I1 =

Z



−∞



X Gǫ (t, α) Gǫ (tn , α) dt = ′ (t )(t − k + i0) X− (t)(t − k + i0) X n − n 0

Iǫ (k, α) = Fǫ (k, α) + X− (k)(k − ia)

∞ X 0

Gǫ (tn , α) ′ (t )(t − k + i0) X− n n

As k ∈ (−N, N ) uniformly-convergent series yields Z

+N

e

itk

−N

d2 dk 2



Iǫ (k, α) X− (k)



dk =

Z

+N

e −N

itk

d2 dk 2



Fǫ (k, α) X− (k)



dk+

Z

+N

−N

eitk

d2 ((k − ia)I1 ) dk dk 2

By definition I1 : Z

+N

−N

eitk

d2 ((k − ia)I1 ) dk = dk 2

Z

N +N X

−N

1

34

eitk

d2 dk 2



(k − ia)Gǫ (tn , α) ′ (t )(t − k + i0) X− n n



dk

+

Z

∞ +N X

−N

W1 = W2 =

Z

Z

N



+N

−N

Denote W10 (k)

(k − ia)Gǫ (tn , α) ′ (t )(t − k + i0) X− n n



eitk dk = W1 + W2

N +1

 N X = 2 1

W20 (k) =



 Gǫ (tn , α) (k − ia)Gǫ (tn , α) 2 eitk dk ′ (t )(t − k + i0)2 + X ′ (t )(t − k + i0)3 X n n n n − − 1   ∞ X Gǫ (tn , α) (k − ia)Gǫ (tn , α) 2 eitk dk ′ (t )(t − k + i0)2 + X ′ (t )(t − k + i0)3 X− n n n − n

N +N X

−N

d2 dk 2

∞ X

2

N +1

Gǫ (tn , α) (k − ia)Gǫ (tn , α) + ′ ′ 2 X− (tn )(tn − k + i0) X− (tn )(tn − k + i0)3



(k − ia)Gǫ (tn , α) Gǫ (tn , α) ′ (t )(t − k + i0)2 + X ′ (t )(t − k + i0)3 X− n n n − n



eitk



eitk

Consider the integration of the functions W10 , W20 around the square contour S with vertices functions

W10 , W20

±N , ±N − iN and oriented positively. Analyticity of the yields Z

S

W10 dS

= 0,

Z

S

W20 dS = 0

and W1 =

Z

−iN

0

W2 =

Z

0

W10 (−N

+ iτ )idτ +

Z

0

−iN

−iN

W20 (−N + iτ )idτ +

Z

W10 (N

+ iτ )idτ +

0

−iN

W20 (N + iτ )idτ +

Z

−N

N

Z

W10 (−iN + τ )dτ

−N

N

W20 (−iN + τ )dτ

Latest integrals yields Z Z Z −N −iN 0 |W1 | ≤ W10 (−N + iτ )idτ + W10 (N + iτ )idτ + W10 (−iN + τ )dτ = 0 −iN N R11 + R12 + R13

Enough to calculate R11 , the remaining calculations are done in the same way. R11

Z N N X Cǫ Cǫ 1 dτ ≤ ǫ ≤ 0.1 2 + (b − τ )2 n (2πn + φ − N ) N n 0 1

We will perform the same calculations for W2 Z Z Z −N −iN 0 |W2 | ≤ W20 (−N + iτ )idτ + W20 (N + iτ )idτ + W20 (−iN + τ )dτ = 0 −iN N 35

R21 + R22 + R23 Also enough to calculate R21 . Z N ∞ X 1 Cǫ Cǫ R21 ≤ dτ ≤ ǫ n0.1 0 (2πn + φ − N )2 + (bn − τ )2 N N +1

We perform simple calculations to form a final estimate.     Z +N Z +N +N 2 2 1 Iǫ (k, α) d itk d itk d itk d e e dk−2 dk e Iǫ (k, α)dk = (Iǫ (k, α)) dk 2 dk 2 X− (k) dk dk X− (k) −N −N −N   Z +N Z +N Z +N d2 d2 1 1 d2 eitk 2 Iǫ dk− dk+ eitk 2 (Iǫ (k, α)) dk eitk Iǫ (k, α) 2 − dk X (k) dk dk X (k) − − −N −N −N

Z

Theorem 9 and the estimates of W1 , W2 yields Z +N 2 itk d e I dk ≤ Cǫ Cm ǫ −N dk 2

By Lemma 17, as 1/2 + 3ǫ < Re(s) < 1 − 3ǫ, the last estimate yields |Q(s)| < Cm Cǫ .,

which completes the proof, As mentioned in Introduction, the values of the zeta function in adjacent rectangles should be compared. This will be carried out in the following theorem. Theorem 12. The Riemann’s function has nontrivial zeros only on the line Re(s) = 1/2. Proof. Let it be assumed that there is a root of the zeta function with sn = 1/2 + δn + i ∗ αn , where δn > 0. Let sn−1 = 1 + δn−1 + iαn−1 , where δn−1 > 0 be another root nearest to it. Then, the following sets corresponding to sn are constructed: D(n) = (s|0.1 < Re(s) < 0.9, Im(s) 6= Im(sn ), Im(sn )−dn ≤ Im(s) ≤ Im(sn )+dn D(n) = (s|0.1 < Re(s) < 0.9, Im(−s) 6= Im(−sn ), −Im(sn )−dn ≤ Im(s) ≤ −Im(sn )+dn , where ζ(sn+1 ) = 0, ζ(sn ) = 0, ζ(1 − sn ) = 0, ζ(1 − sn+1 ) = 0, ζ(1 − sn ) = 0, 36

dn = (Im(sn+1 ) − Im(sn ))/2 where ǫ = δn /10. As 1/2 < Re(s) < 0.9 and s ∈ D(n) ∪ D(n), thus, we have the equation for Q. Theorem 11 now yields |ln(|ζ(1/2 + δn + iαn − iδ)|) = |Q(1/2 + δn − iαn − iδ)| < 2Cn Cǫ Furthermore, lim |ln(|ζ(1/2 + δn + iαn − iδ)|)| = ∞.

δ→0

These estimates for |Q(s)| , |f (s)| imply that the function does not have zeros on the half plane Re(s) > 1/2. By the integral representation (3), these results are extended to the half plane Re(s) < 1/2. Thus, Riemann’s hypothesis has been proved.

10. Conclusions As can be seen from the results obtained, the genius of such Poincar scholars, Riemann and Hilbert, still illuminates the path of modern scholars. An example of such a complicated problem as the Cauchy problem for the Navier-Stokes equation shows how the Poincar-Riemann-Hilbert boundary value problem allows us to construct effective estimates of solutions for this case. For this, the apparatus of the three-dimensional inverse problem of the theory of quantum scattering develops. It is shown that the unitary scattering operator can be investigated as a solution of the Poincar-Riemann-Hilbert boundary-value problem. This allows us to continue studying the potential in the Schrodinger equation, which we consider as a velocity component in the Navier-Stokes equation. The same scheme of reduction of Riemann integral equations for the zeta function to the Poincar-Riemann-Hilbert boundary value problem enables us to construct effective estimates that describe the behavior of the zeros of the zeta function well. Summarizing it is possible to tell these outstanding scientists the problems have been formulated and all modern methods of their decision are put in pawn. 37

11. ACKNOWLEDGEMENTS The author thanks the National Engineering Academy of the Republic of Kazakhstan, in particular, Academician NAS RK B.Zhumagulov for constant attention and support. Moreover, the author thanks the Mathematics seminar at the Kazakhstan branch of the Moscow State University for attention and valuable comments, as well as Professors B. Kanguzhin and M. Otelbaev, and the organizers of Automorphicformsworkshop.org/AFW2018 for their detailed review and valuable comments.The author is especially grateful to Professor Steven Miller for a thorough analysis of the work and detailed recommendations that have significantly improved the paper.The author is especially grateful to Professor Plotnikov P.I for a thorough analysis of the work and detailed recommendations that have significantly improved the paper.The author is especially grateful to Professor Alexander Mednykh for a thorough analysis of the work and detailed recommendations that have significantly improved the paper.

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