Navier–Stokes equations: a new estimate of a possible gap related to the energy equality of a suitable weak solution

The paper is concerned with the IBVP of the Navier-Stokes equations. The result of the paper is in the wake of analogous results obtained by the authors in previous articles Crispo et al. (Ricerche Mat 70:235–249, 2021). The goal is to estimate the possible gap between the energy equality and the energy inequality deduced for a weak solution.


Introduction
This note concerns the 3D-Navier-Stokes initial boundary value problem: (1) In system (1) Ω ⊆ R 3 is assumed bounded or exterior, and its boundary is assumed smooth.
In the two recent papers [4,5] the authors look for an energy equality for suitable weak solutions.Here, the term suitable is meant in the sense that a new solution is exhibited and not that an improvement is obtained to the one given in [3].Actually, the crucial result of papers [4,5], and it seems the first, is the strong convergence in L p (0, T ; W 1,2 (Ω)) ∩ L 2 (0, T ; L 2 (Ω)), for all T > 0 and p ∈ [1, 2), of a sequence {v m } of smooth solutions to the "Leray's approximating Navier-Stokes Cauchy problem" (see (4) below), [7].
Since the strong convergence is not in L 2 (0, T ; W 1,2 (Ω)), the authors attempt to obtain the energy equality employing the (differential and integral) energy equality of the approximating solutions and some auxiliary functions.Actually, the approaches used so far allow to prove an energy equality which involves other quantities.Here it is proved that a suitable weak solution * 1 Dipartimento di Matematica e Fisica, Università degli Studi della Campania "L.Vanvitelli", via Vivaldi 43, 81100 Caserta, Italy.francesca.crispo@unicampania.it , paolo.maremonti@unicampania.it 2 Dipartimento di Matematica Università di Pisa, via Buonarroti 1/c, 56127 Pisa, Italy.carlo.romano.grisanti@unipi.itexists and satisfies the following relation where is of full measure in (0, T ), and where J m (α) is the union of, at most, a countable sequence (N(α, m)) of disjoint intervals (s h , t h ) ⊂ (s, t) and the following holds: Instead in the case of s = 0, one obtains where −M (0, T ) := lim Roughly speaking the above intervals seem to contain the possible singular points S of the weak solution that, as is known, has H 1 2 (S) = 0 (H a Hausdorff's measure), [11].Of course, independently of the meaning of the conjecture for the intervals, from a physical view point the energy relation (2) would add a dissipative quantity which is not justifiable.If this is a necessary consequence of an initial datum only in L 2 , then from a physical point of view it is a right reason to reject the L 2 -class as a class of existence.
However, the validity of an energy equality, without requiring extra conditions 1 , is interesting to better delimit the case of validity of possible counterexamples.
Actually, in the papers [2] and [1] two examples of non-uniqueness are furnished.
1 In this connection in paper [9], the so called Prodi-Serrin condition for the energy equality for a weak solution is not required on the whole interval of existence, but just on (ε, T ), that is L 4 (ε, T ; L 4 (Ω)), for all ε > 0. This means that no extra assumption on the initial datum in L 2 is needed for the validity of the energy equality.
Following [6], under the same quoted assumption, the same result of energy equality holds in the set of very-weak solutions.
The former works for very-weak solutions, which are continuous in L 2 -norm, but do not verify an energy inequality of the kind given by Leray-Hopf, in other words neglecting the term M (s, t) with ≤ 0. Further, in the case of Leray-Hopf weak solutions their counterexample does not work.
The latter works with a homogeneous initial datum.Actually, the non-uniqueness is exhibited for solutions corresponding to a suitable data force, that, among other things, allows an energy equality.
The plan of the paper is the following.In sect. 2 some preliminary lemmas are recalled and some new results of strong convergence are furnished.In sect. 3 the statement and the proof of the chief result are performed.

Preliminary results
We set J 1,2 (Ω):=completion of C 0 (Ω) in W 1,2 -norm, where C 0 (Ω) is the set of the test functions of the hydrodynamics.Definition 1.For weak solution to the IBVP (1) we mean a field v : (0, ∞) → R 3 such that for all T > 0

the field v solves the integral equation
For our goals we consider a mollified Navier-Stokes system.Hence problem (1) becomes where Lemma 1.For all m ∈ N there exists a unique solution to problem (4) such that for all T > 0 Moreover, the sequence {v m } is strong convergent to a limit v in L p (0, T ; W 1,2 (Ω))∩L 2 (0, T ; L 2 (Ω)), for all p ∈ [1, 2), and the limit v is a weak solution to problem (1) with (v(t), ϕ) ∈ C([0, T )), for all ϕ ∈ J 2 (Ω).
Proof.This lemma for data force f = 0 is Theorem 6.1.1 proved in [4].It is not difficult to image that the proof can be modified without difficulty assuming f = 0.So that we consider as achieved the proof of the lemma.
Lemma 3.For any T > 0, there exists a constant M > 0, not depending on m, such that where v m is the solution of problem (4) stated in Lemma 1.
Proof.By virtue of the regularity of (v m , π m ) stated in (5), we multiply equation ( 4) Integrating by parts on Ω, and applying the Hölder inequality, we get Applying inequality (6) with r = ∞ and q = 6, by virtue of the Sobolev inequality, we obtain By inequalities (7) and ( 8), we get for all m ∈ N and a.e. in t > 0 .Substituting in inequality (9) the identity and dividing by (1 + ||∇v m (t)|| 2 2 ) 2 , we get the following estimate Integrating on (0, T ) we have Using the identity (10) we get Using once again identity (10) we get Lemma 4. Let {h m (t)} be a sequence of non-negative functions bounded in L 1 (0, T ).Also, assume that h m (t) → h(t) a.e. in t ∈ (0, T ) with h(t) ∈ L 1 (0, T ).Let be g : (0, α 0 ) −→ R a continuous and strictly increasing function such that lim •) is weakly decreasing and lim ρ→+∞ p(α, ρ) = 0 for any α ∈ (0, α 0 ).Then we get Proof.We have We fix α ∈ (0, α 0 ) and we consider the first integral.For any ε ∈ (0, α 0 − α) we set Hence we have By (12) we get | hence, by the dominated convergence theorem, we have Since p(α, •) is decreasing, we get Using the boundedness of the sequence (h m ) in L 1 we obtain that

The chief result
We recall the definition where {v m } is the sequence of solutions to problem (4).By virtue of the strong convergence stated in Lemma 1, the set T is certainly not empty and, as matter of fact, it is of full measure in (0, T ) .
Theorem 1.Let v be the weak solution stated in Lemma 1. Then v satisfies the relation with Proof.We consider the sequence {v m } of solutions to problem (4) whose existence is ensured by Lemma 1.For all m ∈ N the Reynolds-Orr equation holds: We set ρ m (t) := ||∇v m (t)|| 2 2 , and we consider Let be and fix s, t ∈ T , with s < t .Let α 1 be such that , for all α ∈ (α 1 , 1) .
Hence, by virtue of the pointwise convergence, we claim the existence of m 0 such that , for all m ≥ m 0 and α ∈ (α 1 , 1) .
We set A m := max ρ m (t).We denote by , there exists the minimum s > s such that ρ m (s) = tan α π 2 , as well, being ρ m (t) < tan α π 2 , there exists the maximum t < t such that ρ m (t) = tan α π 2 .Thus, if J m (α) is a non-empty set, by the regularity of ρ m (t), we get that J m (α) is at most the union of a sequence of open interval (s h , t h ) such that We justify the claim.The set J m (α) is an open set, hence it is at most the countable union of maximal intervals (s h , t h ).We set where we took the energy relation (17) into account and the strong convergence of the right-hand side too.Estimate (20) leads to (16).Recalling the definition of p(α, ρ m (t)), we have where we took into account that, for all α ∈ (0, 1), function p is Lipschitz's function in ρ m , and ρ m (t) is a regular function in t.Hence, we get p(α, ρ m (t)) is Lipschitz's function with respect to t.We multiply equation (17) for p(α, ρ m (τ )), with α > α 1 , and we integrate by parts on (s, t): where we set ||v m (τ )|| where we set Recalling the properties of J m (α), for all α and m, integrating by parts, we get Hence we arrive at h∈N(α,m) We estimate the last integral.Let be It results that On the complement of the set T we can set ρ = 0, since the value on a null measure set does not Since ρ ∈ L 1 and, by (25), Concerning the force term we have Using algebraic manipulation we obtain the following relation:  (f, v m ) dτ.
uniformly with respect to m .(16) Moreover, if s = 0, the relation (15) holds setting s = 0 in the left-hand side, and with the right-hand side replaced by lim k M (s k , t) where {s k } is any sequence in T converging to 0.