Entire solutions of hydrodynamical equations with exponential dissipation

# Entire solutions of hydrodynamical equations with exponential dissipation

## Abstract

We consider a modification of the three-dimensional Navier–Stokes equations and other hydrodynamical evolution equations with space-periodic initial conditions in which the usual Laplacian of the dissipation operator is replaced by an operator whose Fourier symbol grows exponentially as at high wavenumbers . Using estimates in suitable classes of analytic functions, we show that the solutions with initially finite energy become immediately entire in the space variables and that the Fourier coefficients decay faster than for any . The same result holds for the one-dimensional Burgers equation with exponential dissipation but can be improved: heuristic arguments and very precise simulations, analyzed by the method of asymptotic extrapolation of van der Hoeven, indicate that the leading-order asymptotics is precisely of the above form with . The same behavior with a universal constant is conjectured for the Navier–Stokes equations with exponential dissipation in any space dimension. This universality prevents the strong growth of intermittency in the far dissipation range which is obtained for ordinary Navier–Stokes turbulence. Possible applications to improved spectral simulations are briefly discussed.

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## 1 Introduction

More than a quarter of a millenium after the introduction by Leonhard Euler of the equations of incompressible fluid dynamics the question of their well-posedness in three dimensions (3D) with sufficiently smooth initial data is still moot (1); (2); (3); (4) (see also many papers in (5) and references therein). Even more vexing is the fact that switching to viscous flow for the solution of the Navier–Stokes equations (NSE) barely improves the situation in 3D (6); (7); (8); (9); (10). Finite-time blow up of the solution to the NSE can thus not be ruled out, but there is no numerical evidence that this happens.

In contrast, there is strong numerical evidence that for analytic spatially periodic initial data both the 3D Euler and NSE have complex space singularities. Indeed, when such equations are solved by (pseudo-)spectral techniques the Fourier transforms of the solution display an exponential decrease at high wavenumbers, which is a signature of complex singularities (11). This behavior was already conjectured by von Neumann (12) who pointed out on p. 461 that the solution should be analytic with an exponentially decreasing spectrum. Recently Li and Sinai used a Renormalization Group method to prove that for certain complex-valued initial data the 3D NSE display finite-time blow up in the real domain (and, as a trivial corollary, also in the complex domain) (13).

For some PDEs in lower space dimensions explicit information about the position and type of complex singularities may be available. For example, complex singularities can sometimes be related to poles of elliptic functions in connection with the reaction diffusion equation (14) and 2D incompressible Euler equations in Lagrangian coordinates (15). The best understood case is that of the 1D Burgers equation with ordinary (Laplacian) dissipation:1 its singularities are poles located at the zeroes of the solutions of the heat equation to which it can be mapped by using the Hopf–Cole transformation (see, e.g., (16); (17) and references therein).

We now return to the 3D NSE with real analytic data. It is known that blow up in the real domain can be avoided altogether by modifying the dissipative operator, whose Fourier-space symbol is , to a higher power of the Laplacian with symbol () (6); (18). The numerical evidence is however that complex singularities cannot be avoided by this “hyperviscous” procedure, frequently used in geophysical simulations (see, for example, (19)).

Actually, we are unaware of any instance of a nonlinear space-time PDE, with the property that the Cauchy problem is well posed in the complex space domain for at least some time and which is guaranteed never to have any complex-space singularities at a finite distance from the real domain. In other words the solution stays or becomes entire for all . Here we shall show that solutions of the Cauchy problem are entire for a fairly large class of pseudo-differential nonlinear equations, encompassing variants of the 3D NSE, which possess “exponential dissipation”, that is dissipation with a symbol growing exponentially as with the ratio of the wavenumber to a reference wavenumber .

The paper is organized as follows. In Section 2 we consider the forced 3D incompressible NSE in a periodic domain with exponential dissipation. The initial conditions are assumed just to have finite energy. The main theorem is established using classes of analytic functions whose norms contain exponentially growing weights in the Fourier space (20); (21). In Section 3 we show that the Fourier transform of the solution decays at high wavenumbers faster than for any . Here, is the nondimensionalised wavenumber. In Section 4 we briefly present extensions of the result to other instances: different space dimensions and dissipation rates, problems formulated in the whole space and on a sphere, and different equations. In Section 5 we then turn to the 1D Burgers equation with a dissipation growing exponentially at high wavenumbers, for which the same bounds hold as for the 3D Navier–Stokes case. However in the Burgers case, simple heuristic considerations (Section 5.1) and very accurate numerical simulations performed by two different techniques (Sections 5.2 and 5.3), indicate that the leading-order asymptotic decay is precisely . We observe that the heuristic approach, which involves a dominant balance argument applied in spatial Fourier space, is also applicable to the 3D Navier–Stokes case with exactly the same prediction regarding the asymptotic decay. In the concluding Section 6 we discuss open problems and a possible application.

## 2 Proof that the solution is entire

We consider the following 3D spatially periodic Navier–Stokes equations with an exponential dissipation (expNSE)

 ∂u∂t+u⋅∇u=−∇p−μDu+f,∇⋅u=0, (2.1) u(x,0)=u0(x). (2.2)

Here, is the (pseudo-differential) operator whose Fourier space symbol is , that is a dissipation rate varying exponentially with the wavenumber , is the initial condition, is a prescribed driving force and and are prescribed positive coefficients. The problem is formulated in a periodic domain (for simplicity of notation we take ). The driving force is assumed to be a divergence-free trigonometric polynomial in the spatial coordinates. For technical convenience we use in the statements and proofs of mathematical results, while the use of the reference wavenumber is preferred when discussing the results.

The initial condition is taken to be a divergence-free periodic vector field with a finite norm (finite energy).

As usual the problem is rewritten as an abstract ordinary differential equation in a suitable function space, namely

 dudt+μe2σA1/2u+B(u,u)=f; (2.3) u(0)=u0; (2.4)

where and is a suitable quadratic form which takes into account the nonlinear term, the pressure term and the incompressibility constraint (see, e.g. (6); (7); (9)). Note that the Fourier symbol of is .

The problem is formulated in the space is periodic, , . Here, for any , the Fourier symbol of the operator is given by , where .

To prove the entire character, with respect to the spatial variables, of the solution of expNSE for , it suffices to show that its Fourier coefficients decrease faster than exponentially with the wavenumber . This will be done by showing that, for any , the norm of , the solution with an exponential weight in Fourier space, is finite. As usual, we here denote the norm of a real space-periodic function by . Moreover, will be the usual Sobolev space of index (i.e., functions which have up to space derivatives in ).

The main result (Theorem 2.1) will make use of the following Proposition which was inspired by (20) (see also (21))
Proposition 2.1    Let , , and . Then

 ∣∣eαA1/2B(φ,φ)∣∣ ≤CA(lκ(β))a(κ)∣∣eαA1/2φ∣∣2−a(κ)  ∣∣e(α+β)A1/2φ∣∣a(κ), (2.5)

where is a universal constant and

 lκ(β)\coloneqqsup0≤x<∞xκe−βx=(κβ)κe−κ. (2.6)

Notation   In Proposition 2.1 and also in the sequel we use the following notation (to avoid fractions in exponents):

 a(κ)\coloneqq52κ,d(κ)\coloneqq2κ+52κ−5 e(κ)\coloneqq4κ−52κ−5,f(κ)\coloneqq2κ2κ−5. (2.7)

Proof   Let . By using the Fourier representations and and Parseval’s theorem, we have

 1(2π)3(eαA1/2B(φ,φ),w)=∑k∈Z3k≠0  eα|k|⎛⎝∑l+m=k;l,m≠0(^φl⋅im)^φm⎞⎠⋅^w∗k, (2.8)

where the means complex conjugation.

Since , when , we can estimate the absolute value of the right-hand side from above as

 ≤∑k≠0∑l+m=k, l,m≠0eα|l||^ϕl||m|eα|m||^ϕm||^wk|=1(2π)3∫Φ(x)Ψ(x)W(x)dx≤1(2π)3∥Ψ∥L∞|Φ||W|, (2.9)

where the functions , and are given by

 Φ(x)=∑l≠0eα|l||^φl|eil⋅x, (2.10)
 Ψ(x)=∑m≠0|m|eα|m||^φm|eim⋅x, (2.11)

and

 W(x)=∑k≠0|^wk|eik⋅x, (2.12)

and the last inequality follows from the Cauchy–Schwarz inequality.

 ∥Ψ∥L∞≤CA∥Ψ∥12H1∥Ψ∥12H2≤CA|A1/2Ψ|12|AΨ|12=CA∣∣AeαA1/2φ∣∣∣∣A3/2eαA1/2φ∣∣, (2.13)

where is a universal constant. By using (2.10), (2.13) and the fact that , we obtain

 ∣∣(eαA1/2B(φ,φ),w)∣∣≤CA∣∣eαA1/2φ∣∣∣∣AeαA1/2φ∣∣12∣∣∣A32eαA1/2φ∣∣∣12|w|. (2.14)

And by using the interpolation inequality between and , where , we obtain2

 ∣∣(eαA1/2B(φ,φ),w)∣∣≤CA∣∣eαA1/2φ∣∣2−a(κ)∣∣Aκ2eαA1/2φ∣∣a(κ)|w|. (2.15)

Now, to obtain the inequality in Proposition 2.1, we just need to estimate the operator norm

 ∥∥Aκ2e−βA1/2∥∥L(H) ≤sup0≤x<∞xκe−βx=(κβ)κe−κ=lκ(β). (2.16)

This concludes the proof of Proposition 2.1.

Next, we state and present the proof of the main theorem. The steps of the proof are made in a formal way, however, they can be justified rigorously by establishing them first for a Galerkin approximation system and using the usual Aubin compactness theorem to pass to the limit (see, e.g. (6); (7); (9)). Furthermore, we do not assume that the initial condition is entire; it is only assumed to be square integrable, although it will become entire for any . This is why in estimating norms of the solution with exponential weights we have to stay clear of .

Theorem 2.1   Let , fix and let be an entire function with respect to the spatial variable . Then for every there exist constants and which depend on , and on the norm

 ∫T0∣∣e(n−1)σA1/2f(s)∣∣2ds, (2.17)

moreover there exists integers such that

 ∣∣enσA1/2u(t)∣∣2≤Kntpn+Cn,for all t∈(0,T] (2.18)

and

 ∫Tt|e(n+1)σA1/2u(s)|2ds≤¯Kntqn+¯Cn,for allt∈(0,T], (2.19)

where is the solution of (2.3)-(2.4).

Corollary 2.1    Let and let be an entire function with respect to the spatial variable such that for every we have . Then, the solution of (2.3)-(2.4) is an entire function with respect to the spatial variable for all , and satisfies the estimates (2.18) and (2.19) in Theorem 2.1 for any .

Proof of Corollary 2.1    Consider the Fourier series representation

 u(x,t)=∑keik⋅x^u(k,t). (2.20)

From (2.18) and Parseval’s theorem, we have, for any

 ∑ke2nσ|k||^u(k,t)|2<∞, (2.21)

for . In (2.20) we change to a complex location and obtain

 u(x+iy,t) = ∑keik⋅(x+iy)^u(k,t) (2.22) = ∑k[eik⋅xe−|k|][e−k⋅y+|k|^u(k,t)]. (2.23)

For any , the series (2.23) of complex analytic functions converges uniformly in the strip . This is because the sum in (2.23) is shown to be bounded, for any , by use of the Cauchy–Schwarz inequality applied to the two bracketed expressions and use of (2.21) with . Hence the Fourier series representation converges in the whole complex domain. This concludes the proof of the entire character of the solution with respect to the spatial variables.
Remark    This corollary just expresses the most obvious part of the Paley–Wiener Theorem.

Proof of Theorem 2.1    The proof of the theorem proceeds by mathematical induction.

Step    We prove the statement of the theorem for . We take the inner product of (2.3) with and use the fact that to obtain (when there is no ambiguity we shall henceforth frequently denote by )

 12ddt|u|2+μ∣∣eσA1/2u∣∣2 = (f,u)=(e−σA1/2f,eσA1/2u) (2.24) ≤ ∣∣e−σA1/2f∣∣∣∣eσA1/2u∣∣ ≤ |e−σA1/2f|22μ+μ2|eσA1/2u|2, (2.25)

where Young’s inequality has been used to obtain the third line. Therefore

 ddt|u|2+μ∣∣eσA1/2u∣∣2≤|e−σA1/2f|2μ. (2.26)

Integrating the above from to , we obtain

 |u(t)|2+μ∫T0∣∣eσA1/2u(s)∣∣2ds≤C0\coloneqq|u0|2+1μ∫T0∣∣e−σA1/2f(s)∣∣2ds. (2.27)

Hence

 |u(t)|2≤C0, (2.28)

and

 Missing or unrecognized delimiter for \right (2.29)

From (2.28) and (2.29) we obtain (2.18) and (2.19) for the case . Here is given by (2.27), and . Notice that since there is no need to determine the integers and ; however, for the sake of initializing the induction process we chose .

Step Assume that (2.18) and (2.19) are true up to and we would like to prove them for . Let us take the inner product of (2.3) with and obtain

 12ddt∣∣e(m+1)σA1/2u∣∣2 + μ∣∣e(m+2)σA1/2u∣∣2 ≤ ∣∣(f,e2(m+1)σA1/2u)∣∣+∣∣(B(u,u),e2(m+1)σA1/2u)∣∣ ≤ ∣∣emσA1/2f∣∣∣∣e(m+2)σA1/2u∣∣+∣∣emσA1/2B(u,u)∣∣∣∣e(m+2)σA1/2u∣∣.

Now we use Proposition 2.1 to majorize the previous expression by

 Missing \left or extra \right (2.30)

By Young’s inequality we have

 CA[lκ(β)]a(κ)∣∣emσA1/2u∣∣2−a(κ)∣∣e(m+2)σA1/2u∣∣1+a(κ)≤2κ−54κμ−d(κ)C2f(κ)A(lκ(β))2a(κ)f(κ)(2κ+5κ)d(κ)∣∣emσA1/2u∣∣2e(κ)+μ4∣∣e(m+2)σA1/2u∣∣2. (2.31)

It follows that

 ddt∣∣e(m+1)σA1/2u∣∣2+μ∣∣e(m+2)σA1/2u∣∣2≤2μ∣∣emσA1/2f∣∣2+2κ−52κμ−d(κ)C2f(κ)A(lκ(β))2a(κ)f(κ)(2κ+5κ)d(κ)∣∣emσA1/2u∣∣2e(κ). (2.32)

Now we integrate this inequality on the interval , obtaining

 ∣∣e(m+1)σA1/2u(t)∣∣2+μ∫ts∣∣e(m+2)σA1/2u(s′)∣∣2ds′≤∣∣e(m+1)σA1/2u(s)∣∣2+2μ∫ts∣∣emσA1/2f(s′)∣∣2ds′+C′∫ts∣∣emσA1/2u(s′)∣∣2e(κ)ds′≤∣∣e(m+1)σA1/2u(s)∣∣2+2μ∫T0∣∣emσA1/2f(s′)∣∣2ds′+C′∫ts∣∣emσA1/2u(s′)∣∣2e(κ)ds′, (2.33)

where we have set for brevity

 C′=C(μ,β,κ)\coloneqq2κ−52κμ−d(κ)C2f(κ)A(lκ(β))2a(κ)f(κ)(2κ+5κ)d(κ), (2.34)

and where is given by (2.6).

Now we come to the point where we use the actual induction assumptions. We use (2.18) and the midpoint convexity to estimate the integrand in the last integral:

 ∣∣emσA1/2u(t)∣∣2e(κ)≤(Kmtpm+Cm)e(κ)≤2f(κ)(Kmtpm)e(κ)+2f(κ)Ce(κ)m. (2.35)

Whence it follows that

 C′∫ts∣∣emσA1/2u(s′)∣∣2e(κ)ds′≤2f(κ)C′1e(κ)pm−1Ke(κ)m(1se(κ)pm−1−1te(κ)pm−1)+2f(κ)C′Ce(κ)m(t−s)≤2f(κ)C′1e(κ)pm−1Ke(κ)m1se(κ)pm−1+2f(κ)C′Ce(κ)m(t−s). (2.36)

Discarding the positive term in (2.33), we obtain from (2.33) and (2.36)

 ∣∣e(m+1)σA1/2u(t)∣∣2≤∣∣e(m+1)σA1/2u(s)∣∣2+2μ∫T0∣∣emσA1/2f(s′)∣∣2ds′+2f(κ)C′ 1e(κ)pm−1Ke(κ)m1se(κ)pm−1+2f(κ)C′Ce(κ)m(t−s). (2.37)

Integrating this inequality with respect to over we get

 ∣∣e(m+1)σA1/2u(t)∣∣2≤2t∫tt/2∣∣e(m+1)σA1/2u(s)∣∣2ds+2μ∫T0∣∣emσA1/2f(s′)∣∣2ds′+2f(κ)C′Ke(κ)m(e(κ)pm−1)(e(κ)pm−2)(2t)e(κ)pm−1+2f(κ)C′Ce(κ)mt4. (2.38)

Note that implies that

 e(κ)pm−2>0. (2.39)

By using (2.19), we have

 ∣∣e(m+1)σA1/2u(t)∣∣2≤¯¯¯¯¯Km(2t)qm+1+¯¯¯¯Cm2t+2μ∫T0∣∣emσA1/2f(s′)∣∣2ds′+2f(κ)C′Ke(κ)m(e(κ)pm−1)(e(κ)pm−2)(2t)e(κ)pm−1+2f(κ)C′Ce(κ)mT4. (2.40)

From this relation follows that (2.18) holds for with

 pm+1=max{e(κ)pm−1,qm+1}, (2.41) qm+1=max{e(κ)pm−1,qm+1}. (2.42)

By the induction assumption we use (2.33) to estimate

 ∫Tt/2∣∣e(m+1)σA1/2u(s)∣∣2ds≤2qm¯Kmtqm+¯Cm. (2.43)

From this estimate and the above we conclude the existence of the constants and the integer such that (2.18) holds for . Using the estimate that we have just established in (2.18) for , and substituting this in (2.33), we immediately obtain the estimate (2.19) for . This concludes the proof of Theorem 2.1.

## 3 Rate of decay of the Fourier coefficients

The purpose of this section is to specify the behavior of various constants appearing in the preceding section to obtain the rate of decay with the wavenumber of the Fourier coefficients for . We again consider the 3D case in the periodic domain. Since the decay may depend on the rate of decay of the Fourier transform of the forcing term , for simplicity we assume zero external forcing, which we expect to behave as the case with sufficiently rapidly decaying forcing. The adaptation to sufficiently regular forced cases, for example a trigonometric polynomial, is similar but more technical.3 Furthermore, it is enough to prove the decay result up to a time such that , where and are a typical length scale and velocity of the initial data. Extending the results to later times is easy (by propagation of regularity).

We shall show that the bound for the square of the norm of the velocity weighted by is a double exponential in . Specifically, we have

Theorem 3.1   Let be the solution of (2.3)-(2.4) in with and . Then for every and , there exists a number , depending on and , such that, for all integer

 ∣∣enσA1/2u(t)∣∣2 ≤ (ΛLUt)an,t∈(0,T], (3.1) ∫Tt∣∣e(n+1)σA1/2u(s)∣∣2ds ≤ (ΛLUt)an,t∈(0,T] (3.2) wherean = ((1+δ)4κ−52κ−5)n. (3.3)

Corollary 3.1    For any the function of (2.3)-(2.4) is an entire function in the space variable and its (spatial) Fourier coefficients tend to zero in the following faster-than-exponential way: there exists a constant such that, for any , we have

 |^u(k,t)|≤e−σ(1−ε)βκ,δ|k|ln|k|, for all|k|≥(√ΛLUt)βκ,δεσ, (3.4)

where

 βκ,δ=ln((1+δ)4κ−52κ−5). (3.5)

Proof of Corollary 3.1   Since we are dealing with a Fourier series, the modulus of any Fourier coefficient of the function cannot exceed its norm, hence it is bounded by (3.1). Thus, discarding a factor , we have for all and

 |^u(k,t)|≤e−nσ|k|(√ΛLUt)((1+δ)e(κ))n=exp(ln√ΛLUtenβκ,δ−nσ|k|), (3.6)

where is defined in (2.7). Now choosing

 ln|k|≥1εβκ,δσln√ΛLUt, (3.7)

we obtain with the following estimate

 |^u(k,t)|≤exp[−σβκ,δ|k|ln|k|(1−βκ,δln√ΛL/Utσln|k|)]≤e−σ(1−ε)βκ,δ|k|ln|k|. (3.8)

Remark 3.1    Since and can be chosen arbitrarily small and arbitrarily large, Corollary 3.1 implies that, in terms of the dimensionless wavenumber , the Fourier amplitude has a bound (at high enough ) of the form for any . We shall see that the upper bound for the constant can probably be improved to .

Proof of Theorem 3.1    We proceed again by induction. We assume that the following inequalities hold

 ∣∣enσA1/2u(t)∣∣2≤Kntan, (3.9)

and

 ∫Tt∣∣e(n+1)σA1/2u(s)∣∣2ds≤Kntan, (3.10)

where and are still to be determined. Starting from expNSE (2.3), we take the inner product with . Then we obtain from Proposition 2.1 with and

 Missing \left or extra \right (3.11)

Then it follows that

 ddt∣∣e(n+1)σA1/2u∣∣2+μ∣∣e(n+2)σA1/2u∣∣2≤2κ−52κμ−d(κ)C2f(κ)A(lκ(β))2a(κ)f(κ)(2κ+52κ)d(κ)∣∣enσA1/2u∣∣2e(κ). (3.12)

By using the induction assumption we obtain

 ddt∣∣e(n+1)σA1/2u∣∣2+μ∣∣e(n+2)σA1/2u∣∣2≤C′′(Kntan)e(κ), (3.13)

where we have set

 C′′=2κ−52κμ−d(κ)C2f(κ)A(lκ(β))2a(κ)f(κ)(2κ+52κ)d(κ). (3.14)

Renaming the time variable in (3.13) from to and integrating over from to (with ) we obtain

 ∣∣e(n+1)σA1/2u(t)