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

Abstract

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.

1 Introduction

This note concerns the 3D-Navier–Stokes initial boundary value problem:

$$\begin{aligned} \begin{array}{l}v_t+v\cdot \nabla v+\nabla \pi _v=\Delta v+f,\;\nabla \cdot v=0, \text{ in } (0,T)\times \Omega ,\\ \displaystyle v=0 \text{ on } (0,T)\times \partial \Omega , v(0,x)=v_0(x) \text{ on } \{0\}\times \Omega .\end{array} \end{aligned}$$

(1)

In system (1) $\Omega \subseteq {\mathbb {R}}^{3}$ is assumed bounded or exterior, and its boundary is assumed smooth.

In the two recent papers [5, 6] 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 [5, 6] is the strong convergence in $L^p(0,T;W^{1,2}(\Omega ))$, for all $T>0$ and $p\in [1,2)$, of a sequence $\{v^m\}$ of smooth solutions to the “Leray’s approximating Navier–Stokes Cauchy problem” (see (4) below), [11].

Since the strong convergence is not in $L^2(0,T;W^{1,2}(\Omega ))$, 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 exists and satisfies the following relation

$$\begin{aligned}{} & {} \Vert v(t)\Vert _2^2+2 \int \limits _{{s}}^{{t}}\Vert \nabla v(\tau )\Vert _2^2d\tau +M(s,t)\nonumber \\{} & {} \quad =\Vert v(s)\Vert _2^2+\int \limits _{{s}}^{{t}}(f,v)d\tau \text{ for } \text{ all } 0<s<t\in {\mathcal {T}}, \end{aligned}$$

(2)

where, thanks to the result of strong convergence in $L^p(0,T;W^{1,2}(\Omega ))$, $p\in [1,2)$ (see Lemma 1),

$$\begin{aligned} {\mathcal {T}}:=\left\{ t\in (0,T): \Vert v^m(t)\Vert _{1,2}\rightarrow \Vert v(t)\Vert _{1,2}\right\} \end{aligned}$$

is of full measure in (0, T) for all $T>0$, and

$$\begin{aligned}{} & {} M(s,t):=2\lim _{\alpha \rightarrow 1^-}\overline{\lim _m}\int \limits _{{J^m(\alpha )}}\Vert \nabla v^m(\tau )\Vert _2^2\,d\tau \nonumber \\{} & {}\hskip1.45cm=\lim _{\alpha \rightarrow 1^-}\overline{\lim _m} {\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }} \Big [ \Vert v^m(s_h)\Vert _2^2-\Vert v^m(t_h)\Vert _2^2\Big ]\, \end{aligned}$$

where $J^m(\alpha )$ is the union of, at most, a countable sequence (${\mathbb {N}}(\alpha ,m)$) of disjoint intervals $(s_h,t_h)\subset (s,t)$ and the following holds:

$$\begin{aligned}{} & {} \lim _{\alpha \rightarrow 1^{-}}\frac{|J^m(\alpha )|}{1-\alpha }\le \frac{1}{\pi }\Vert v_0\Vert _2^2 +\frac{2}{\pi }\int \limits _{{0}}^{{t}}(f,v)d\tau \,,\;\nonumber \\{} & {} \hskip4cm \text{ uniformly } \text{ in } m\in {\mathbb {N}}. \end{aligned}$$

Instead in the case of $s=0$, one obtains

$$\begin{aligned}{} & {} \Vert v(t)\Vert _2^2+2 \int \limits _{{0}}^{{t}}\Vert \nabla v(\tau )\Vert _2^2d\tau +M(0,t)\nonumber \\{} & {} \hskip2cm =\Vert v_0\Vert _2^2+\int \limits _{{0}}^{{t}}(f,v)d\tau \text{ for } \text{ all } t\in {\mathcal {T}}\,, \end{aligned}$$

(3)

where

$$\begin{aligned} M(0,t):=\lim _{s_k\rightarrow 0}M(s_k,t)\,, \text{ for } \text{ any } \{s_k\}\subset {\mathcal {T}}\,. \end{aligned}$$

Roughly speaking the above intervals seem to contain the possible singular points S of the weak solution that, as is known, has ${\mathcal {H}}^\frac{1}{2}(S)=0$ (${\mathcal {H}}^a$ Hausdorff’s measure), [16]. 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.

Also in [15] the author considers the possibility to add a further dissipative term to the right hand side of the classical energy inequality, but, as already stressed in [5], our result is different, since we obtain the equality (2) with M(s, t) expressed only in terms of energy quantities (“kinetic or dissipated”). We think that this difference is of a special interest.

The proof of our result is based on a new existence theorem, where our weak solution is the limit of the sequence $\{v^m\}$ of solutions to problem (4). In addition to the usual weak convergences of $\{v^m\}$, there is the peculiarity that our weak solution is strong limit in $L^p(0,T;W^{1,2}(\Omega ))$, for all $T>0$ and $p\in [1,2)$. This result, proved for the first time in [5] (as far as we know it is also the unique known proof), is obtained under the minimal assumption of $v_0\in L^2(\Omega )$ and divergence free. As already said, it is important in order to obtain that $\lim _m\Vert \nabla v^m(t)-\nabla v(t)\Vert _2=0$ almost every where in $t>0$. This is a main difference with other results of existence of weak solutions, classical or more recent, as the ones furnished in [8] and in [9], obtained with stronger assumptions on the initial datum $v_0$.

By making the minimal requirement on $v_0$, from one hand we match the result¹ obtained in [13], and from another hand we better match the questions of counterexamples, as we remark below.

The validity of an energy equality, without requiring extra conditions, 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.

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 $\ge 0$. 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.

2 Preliminary results

We set $J^{1,2}(\Omega )$:=completion of ${\mathscr {C}}_0(\Omega )$ in $W^{1,2}$-norm, where ${\mathscr {C}}_0(\Omega )$ is the set of the test functions of the hydrodynamics.

Definition 1

For weak solution to the IBVP (1) we mean a field $v:(0,\infty )\times \Omega \rightarrow {{\mathbb {R}}^{3}}$ such that for all $T>0$

• 1. $v\in L^\infty (0,T;L^2(\Omega ))\cap L^2(0,T;J^{1,2}(\Omega )) ,$

• 2. the field v solves the integral equation

  $\int \limits _{{s}}^{{t}}\Big [(v,\varphi _\tau )-(\nabla v,\nabla \varphi )+(v\cdot \nabla \varphi ,v)+(\pi _v,\nabla \cdot \varphi )\Big ]d\tau +(v(s),\varphi (s))=(v(t),\varphi (t)),$

  for all $\varphi \in C^1_0([0,T)\times \Omega ),$

• 3. $\displaystyle \lim _{t\rightarrow 0}\Vert v(t)-v_0\Vert _2=0\,.$

For our goals we consider a mollified Navier–Stokes system. Hence problem (1) becomes

$$\begin{aligned} \begin{array}{l}v_t^m+J_m[v^m]\cdot \nabla v^m+\nabla \pi _{v^m}=\Delta v^m+f,\;\nabla \cdot v^m=0, \text{ in } (0,T)\times \Omega ,\\ \displaystyle v^m=0 \text{ on } (0,T)\times \partial \Omega , v^m(0,x)=v_0^m(x) \text{ on } \{0\}\times \Omega ,\end{array} \end{aligned}$$

(4)

where $f\in L^2(0,T,L^2(\Omega ))$, $\{v_0^m\}\subset J^{1,2}(\Omega )$ converges to $v_0$ in $J^2(\Omega )$ and $J_m[\cdot ]\equiv {{\widetilde{J}}}_{\frac{1}{m}}[\cdot ]$ where ${{\widetilde{J}}}_{\frac{1}{m}}[\cdot ]$ is Friedrichs’ (spatial) mollifier and we suppose that $v^m$ is extended to zero in ${\mathbb {R}}^3-\Omega$.

Lemma 1

For all $m\in {\mathbb {N}}$ there exists a unique solution to problem (4) such that for all $T>0$

$$\begin{aligned}{} & {} \displaystyle \Vert v^m(t)\Vert _2^2+2\int \limits _{{0}}^{{t}}\Vert \nabla v^m(\tau )\Vert _2^2 \nonumber \\{} & {} \quad =\Vert v^m_0\Vert _2^2+2\int \limits _{{0}}^{{t}} (f(\tau ),v^m(\tau ))d\tau \,, \text{ for } \text{ all } t>0\,, \nonumber \\{} & {} \quad v^m\in C([0,T);J^{1,2}(\Omega ))\cap L^2(0,T;W^{2,2}(\Omega ))\,, \nonumber \\{} & {} \quad \displaystyle v^m_t,\nabla \pi ^m\in L^2(0,T;L^2(\Omega ))\,.\end{aligned}$$

(5)

Moreover, the sequence $\{v^m\}$ is strong convergent to a limit v in $L^p(0,T;W^{1,2}(\Omega ))\cap L^2(0,T;L^2(\Omega ))$, for all $p\in [1,2)$, and the limit v is a weak solution to problem (1) with $(v(t),\varphi )\in C([0,T))$, for all $\varphi \in J^2(\Omega )$.

Proof

This lemma for data force $f=0$ is Theorem 6.1.1 proved in [5]. It is not difficult to image that the proof can be modified without difficulty assuming $f\ne 0$. So that we consider as achieved the proof of the lemma.$\square$

Lemma 2

Let $\Omega \subseteq {\mathbb {R}}^n$ and let $u\in W^{2,2}(\Omega )\cap J^{1,2}(\Omega )$. Then there exists a constant c independent of u such that

$$\begin{aligned}{} & {} \Vert u\Vert _r\!\le \! c\Vert P\Delta u\Vert _2^a\Vert u\Vert _q^{1-a}, \hskip0.5cm a\big ({\frac{1}{2}-\frac{2}{n}}\big )+(1-a){\frac{1}{q}}={\textstyle \frac{1}{r}}, \end{aligned}$$

(6)

provided that $a\in [0,1)$.

Proof

See [12, 14] .$\square$

The following lemma furnishes an integrability property of derivatives with respect to t of the sequence $\{\Vert \nabla v^m\Vert _2\}$. This is made following the approach given in paper [5]. However, there are similar results directly concerning weak solutions. For the sake of completeness, we give the following references [4, 7, 17]. In any case, our proof is different from those given in the quoted papers.

Lemma 3

For any $T>0$, there exists a constant $M>0$, not depending on m, such that

$$\begin{aligned} \int _0^T\frac{\left| \frac{d}{dt}\Vert \nabla v^m(t)\Vert _2^2\right| }{\left( 1+\Vert \nabla v^m\Vert _2^2\right) ^2}\,dt\le M \end{aligned}$$

where $v^m$ is the solution of problem (4) stated in Lemma 1.

Proof

By virtue of the regularity of $(v^m,\pi ^m)$ stated in (5), we multiply Eq. (4)$_1$ by $P\Delta v^m-v^m_t$. Integrating by parts on $\Omega$, and applying the Hölder inequality, we get

$$\begin{aligned} \Vert P\Delta v^m-v^m_t\Vert _2^2\le 2\Vert J_m[v^m]\cdot \nabla v^m\Vert _2^2+2\Vert f\Vert _2^2\,, \text{ a.e. } \text{ in } t>0\,. \end{aligned}$$

(7)

Applying inequality (6) with $r=\infty$ and $q=6$, by virtue of the Sobolev inequality, we obtain

$$\begin{aligned}{} & {} \Vert J_m[v^m]\cdot \nabla v^m\Vert _2\le \Vert v^m\Vert _\infty \Vert \nabla v^m\Vert _2\nonumber \\{} & {}\hskip3cm\le c\Vert P\Delta v^m\Vert _2^\frac{1}{2}\Vert \nabla v^m\Vert _2^\frac{3}{2}. \end{aligned}$$

(8)

By inequalities (7) and (8), we get

$$\begin{aligned}{} & {} \displaystyle \quad \Vert P\Delta v^m-v^m_t\Vert _2^2\le c\Vert P\Delta v^m\Vert _2\Vert \nabla v^m\Vert _2^3+2\Vert f\Vert _2^2 \displaystyle \nonumber \\{} & {} \hskip3.1cm \le \frac{1}{2}\Vert P\Delta v^m\Vert _2^2+ c\Vert \nabla v^m\Vert _2^6+2\Vert f\Vert _2^2, \end{aligned}$$

(9)

for all $m\in {\mathbb {N}}$ and a.e. in $t>0\,$. Substituting in inequality (9) the identity

$$\begin{aligned} \displaystyle \quad {\frac{d}{dt}}\Vert \nabla v^m\Vert _2^2+\Vert P\Delta v^m\Vert _2^2+\Vert v^m_t\Vert _2^2=\Vert P\Delta v^m-v^m_t\Vert _2^2 \end{aligned}$$

(10)

and dividing by $(1+\Vert \nabla v^m(t)\Vert _2^2)^2$, we get the following estimate

$$\begin{aligned} \frac{{\dot{\rho }}_m}{(1+\rho _m)^2 } +\frac{\frac{1}{2}\Vert P\Delta v^m\Vert _2^2+\Vert v_t^m\Vert _2^2}{(1+ \rho _m)^2}\le c\rho _m+\frac{2\Vert f\Vert _2^2}{\left( 1+\rho _m\right) ^2}\,, \end{aligned}$$

where we set $\rho _m(t):=\Vert \nabla v^m(t)\Vert _2^2$. Integrating on (0, T) we have

$$\begin{aligned} \begin{aligned}&\frac{1}{1+ \Vert \nabla v^m_0\Vert _2^2}-\frac{1}{1+ \Vert \nabla v^m(T)\Vert ^2_2}\\&\hskip3cm +\int \limits _0^T\frac{\frac{1}{2}\Vert P\Delta v^m\Vert _2^2+\Vert v_t^m\Vert _2^2}{(1+ \rho _m)^2}\,dt\\&\quad \le c \int \limits _0^T\rho _m\,dt+2\int _0^T\frac{2\Vert f\Vert _2^2}{\left( 1+\rho _m\right) ^2}\,dt\le C. \end{aligned} \end{aligned}$$

It follows that

$$\begin{aligned} \int \limits _0^T\frac{\Vert P\Delta v^m\Vert _2^2}{(1+ \rho _m)^2}\,dt\le 2C+2, \quad \int \limits _0^T\frac{\Vert v_t^m\Vert _2^2}{(1+ \rho _m)^2}\,dt\le C+1. \end{aligned}$$

Using the identity (10) we get

$$\begin{aligned} \begin{aligned}&\int \limits _0^T\frac{\Vert P\Delta v^m-v^m_t\Vert _2^2}{(1+ \rho _m)^2}\,dt=\int \limits _0^T\frac{{\frac{d}{dt}}\rho _m}{(1+\rho _m)^2}\,dt+\int \limits _0^T\frac{\Vert P\Delta v^m\Vert _2^2}{(1+ \rho _m)^2}\,dt\\&\hskip7cm + \int \limits _0^T\frac{\Vert v_t^m\Vert _2^2}{(1+ \rho _m)^2}\,dt\\& \hskip0.5cm\le \frac{1}{1+ \Vert \nabla v^m_0\Vert _2^2} -\frac{1}{1+ \Vert \nabla v^m(T)\Vert ^2_2} +3C+3\le 3C+4. \end{aligned} \end{aligned}$$

Using once again identity (10) we get

$$\begin{aligned} \begin{aligned}&\int \limits _0^T\frac{\left| {\frac{d}{dt}}\rho _m\right| }{(1+\rho _m)^2}\,dt \le \int \limits _0^T\frac{\Vert P\Delta v^m-v^m_t\Vert _2^2}{(1+ \rho _m)^2}\,dt\\&\hskip2cm +\int \limits _0^T\frac{\Vert P\Delta v^m\Vert _2^2}{(1+ \rho _m)^2}\,dt+\int \limits _0^T\frac{\Vert v_t^m\Vert _2^2}{(1+ \rho _m)^2}\,dt\\&\hskip3cm \le 6C+7=:M. \end{aligned} \end{aligned}$$

$\square$

Lemma 4

Let $\{h_m(t)\}$ be a sequence of non-negative functions bounded in $L^1(0,T)$. Also, assume that $h_m(t)\rightarrow h(t)$ a.e. in $t\in (0,T)$ with $h(t)\in L^1(0,T)$. Let be $g:(0,\alpha _0)\longrightarrow {\mathbb {R}}$ a continuous and strictly increasing function such that $\lim \limits _{\alpha \rightarrow \alpha _0}g(\alpha )=+\infty$ and $p:[0,1)\times [0,\infty )\longrightarrow [0,1]$ a continuous function such that $p(\alpha ,\rho )=1$ if $0\le \rho \le g(\alpha )$, $p(\alpha ,\cdot )$ is weakly decreasing and $\lim \limits _{\rho \rightarrow +\infty }p(\alpha ,\rho )=0$ for any $\alpha \in (0,\alpha _0)$.

Then we get

$$\begin{aligned} \lim _{\alpha \rightarrow \alpha _0}\lim _m\int \limits _{{0}}^{{T}}h_m(t)p(\alpha ,h_m(t))dt=\int \limits _{{0}}^{{T}}h(t)dt\,, \end{aligned}$$

(11)

Proof

We have

$$\begin{aligned}{} & {} \displaystyle \int \limits _{{0}}^{{T}}h_m(t)p(\alpha ,h_m(t))dt\\{} & {} \quad \displaystyle =\int \limits _{{0}}^{{T}}(h_m(t)-h(t))p(\alpha ,h_m(t))dt\\{} & {} \hskip3cm + \int \limits _{{0}}^{{T}}h(t)p(\alpha ,h_m(t))\\{} & {} \quad =:I_1(\alpha ,m)+I_2(\alpha ,m)\,. \end{aligned}$$

We fix $\alpha \in (0,\alpha _0)$ and we consider the first integral. For any $\varepsilon \in (0,\alpha _0-\alpha )$ we set

$$\begin{aligned}{} & {} J^-_m(\varepsilon )=\{t:h_m(t)\le g(\alpha _0-\varepsilon )\},\nonumber \\{} & {} \qquad J^+_m(\varepsilon )=\{t:g(\alpha _0-\varepsilon )<h_m(t)\}. \end{aligned}$$

(12)

Hence we have

$$\begin{aligned} \begin{aligned}&I_1(\alpha ,m)=\int \limits _{{0}}^{{T}}\chi _{J^-_m(\varepsilon )}(t)(h_m(t)-h(t))p(\alpha ,h_m(t)) dt\\&\quad +\int \limits _{{0}}^{{T}}\chi _{J_m^+(\varepsilon )}(t)(h_m(t)-h(t))p(\alpha ,h_m(t))dt\,\\&\quad =:I_1^-(\alpha ,m,\varepsilon )+I_1^+(\alpha ,m,\varepsilon ). \end{aligned} \end{aligned}$$

By (12) we get

$$\begin{aligned} |\chi _{J^-_m(\varepsilon )}(t)(h_m(t)-h(t))p(\alpha ,h_m(t))|\le g(\alpha _0-\varepsilon )+|h(t)| \end{aligned}$$

hence, by the dominated convergence theorem, we have

$$\begin{aligned} \lim _m I_1^-(\alpha ,m,\varepsilon )=0, \qquad \forall \,\alpha ,\varepsilon . \end{aligned}$$

(13)

Since $p(\alpha ,\cdot )$ is decreasing, we get

$$\begin{aligned}{} & {} \left| \chi _{J_m^+(\varepsilon )}(t)(h_m(t)-h(t))p(\alpha ,h_m(t))\right| \\{} & {} \quad \le p(\alpha ,g(\alpha _o-\varepsilon ))\left( |h_m(t)|+|h(t)|\right) . \end{aligned}$$

Using the boundedness of the sequence $\{h_m\}$ in $L^1$ we obtain that

$$\begin{aligned} |I_1^+(\alpha ,m,\varepsilon )|\le c p(\alpha ,g(\alpha _0-\varepsilon )),\qquad \forall m\in {\mathbb {N}}. \end{aligned}$$

(14)

By (13) and (14) we get

$$\begin{aligned} 0\le \overline{\lim _m} |I_1(\alpha ,m)|\le c p(\alpha ,g(\alpha _0-\varepsilon )),\qquad \forall \,\alpha ,\varepsilon . \end{aligned}$$

Since $\lim \limits _{\varepsilon \rightarrow 0}p(\alpha ,g(\alpha _0-\varepsilon ))=0$ we have that

$$\begin{aligned} \lim _m I_1(\alpha ,m)=0,\qquad \forall \,\alpha . \end{aligned}$$

Now we consider the integral $I_2(\alpha ,m)$. Since $\left| p(\alpha ,h_m(t))h(t)\right| \le 1$ and $\lim \limits _m h_m(t)=h(t)$ a.e. in $t\in (0,T)$, by the dominated convergence theorem, we get

$$\begin{aligned} \lim _mI_2(\alpha ,m)=\int \limits _{{0}}^{{T}}h(t)p(\alpha ,h(t))dt. \end{aligned}$$

Finally, since $\lim \limits _{\alpha \rightarrow \alpha _0}p(\alpha ,h(t))=1$ we have that

$$\begin{aligned} \lim _{\alpha \rightarrow \alpha _0}\lim _mI_2(\alpha ,m)=\int \limits _{{0}}^{{T}}h(t)dt, \end{aligned}$$

and this completes the proof. $\square$

3 The chief result

We recall the definition

$$\begin{aligned} {\mathcal {T}}:=\left\{ t\in (0,T): \Vert v^m(t)\Vert _{1,2}\rightarrow \Vert v(t)\Vert _{1,2}\right\} , \end{aligned}$$

(15)

where $\{v^m\}$ is the sequence of solutions to problem (4). By virtue of the strong convergence stated in Lemma 1, the set ${\mathcal {T}}$ is certainly not empty and, as matter of fact, it is of full measure in (0, T) for all $T>0$.

Theorem 1

Let v be the weak solution and $\{v^m\}$ the related approximating sequence stated in Lemma 1. Then, for all $t,s\in {\mathcal {T}}$, v satisfies the relation

$$\begin{aligned}{} & {} \Vert v(t)\Vert _2^2+2 \int \limits _{{s}}^{{t}}\Vert \nabla v(\tau )\Vert _2^2d\tau +M(s,t) \nonumber \\{} & {} \quad =\Vert v(s)\Vert _2^2+\int \limits _{{s}}^{{t}}(f,v)d\tau \,, \end{aligned}$$

(16)

with

$$\begin{aligned}{} & {} M(s,t):=2\lim _{\alpha \rightarrow 1^-}\overline{\lim _m}\int \limits _{{J^m(\alpha )}}\Vert \nabla v^m(\tau )\Vert _2^2\,d\tau \\{} & {} \quad =\lim _{\alpha \rightarrow 1^-}\overline{\lim _m}{\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\Big [ \Vert v^m(s_h)\Vert _2^2-\Vert v^m(t_h)\Vert _2^2\Big ] \end{aligned}$$

where, for a suitable positive $\alpha _0$ depending on (s, t), for all $\alpha \in (\alpha _0,1)$, $J^m(\alpha )\equiv {\underset{i\in {\mathbb {N}}(\alpha ,m)}{\cup }}(s_i,t_i)$ with ${\mathbb {N}}(\alpha ,m)$ which is, at most, a sequence of integers, and for all $i\in {\mathbb {N}}(\alpha ,m)$ $(s_i,t_i)\subset (s,t)$ with $(s_i,t_i)\cap (s_j,t_j)=\emptyset$ for any $i\not =j$, and

$$\begin{aligned}{} & {} \displaystyle \lim _{\alpha \rightarrow 1^-}\frac{|J^m(\alpha )|}{1-\alpha }\le \frac{1}{\pi }\Vert v_0\Vert _2^2\nonumber \\{} & {} \quad +\frac{2}{\pi }\int \limits _{{0}}^{{t}}(f,v)d\tau \,, \text { uniformly with respect to }m\,. \end{aligned}$$

(17)

Moreover, if $s=0$, the relation (16) holds with $M(0,t)=\lim \limits _k M(s_k,t)$ where $\{s_k\}$ is any sequence in ${\mathcal {T}}$ converging to 0.

Proof

We consider the sequence $\{v^m\}$ of solutions to problem (4) whose existence is ensured by Lemma 1. For all $m\in {\mathbb {N}}$ the Reynolds-Orr equation holds:

$$\begin{aligned} \frac{d}{d\tau }\Vert v^m(\tau )\Vert _2^2+2\Vert \nabla v^m(\tau )\Vert _2^2=(f,v^m)\,. \end{aligned}$$

(18)

We set $\rho _m(t):=\Vert \nabla v^m(t)\Vert _2^2$ , and we consider

$$\begin{aligned} \alpha \in (0,1)\,,\;p(\alpha ,\rho _m):=\left\{ \begin{array}{ll}1&{}\text{ if } \rho _m\in [0,\tan \alpha \frac{\pi }{2}]\\ \frac{\frac{\pi }{2}-\arctan \rho _m}{(1-\alpha )\frac{\pi }{2}}&{}\text { if }\rho _m\in (\tan \alpha \frac{\pi }{2}\,,\infty )\,\end{array}\right. . \end{aligned}$$

(19)

Fix $s,t\in {\mathcal {T}}$, with $s<t$ , ${\mathcal {T}}$ given in (15) . Let $\alpha _1$ be such that

$$\begin{aligned} \max \{\Vert \nabla v(s)\Vert _2^2,\Vert \nabla v(t)\Vert _2^2\}<\tan \alpha \frac{\pi }{2}\,, \text{ for } \text{ all } \alpha \in (\alpha _1,1)\,. \end{aligned}$$

Hence, by virtue of the pointwise convergence, we claim the existence of $m_0$ such that

$$\begin{aligned} \max \{\Vert \nabla v^m(s)\Vert _2^2,\Vert \nabla v^m(t)\Vert _2^2\}<\tan \alpha \frac{\pi }{2}\,, \text{ for } \text{ all } m\ge m_0 \text{ and } \alpha \in (\alpha _1,1)\,. \end{aligned}$$

(20)

We set $\displaystyle A^m:=\max _{[s,t]}\rho _m(t)$. We denote by

$$\begin{aligned} J^m(\alpha ):=\{\tau :\rho _m(\tau )\in (\tan \alpha \frac{\pi }{2},A^m]\}. \end{aligned}$$

If $A_m\le \tan \alpha \frac{\pi }{2}$, then $J^m(\alpha )$ is an empty set. If $A_m>\tan \alpha \frac{\pi }{2}$ holds, since $\rho _m(s)<\tan \alpha \frac{\pi }{2}$, there exists the minimum $\overline{s}>s$ such that $\rho _m(\overline{ s})=\tan \alpha \frac{\pi }{2}$ , as well, being $\rho _m(t)<\tan \alpha \frac{\pi }{2}$, there exists the maximum $\overline{ t}<t$ such that $\rho _m(\overline{ t})=\tan \alpha \frac{\pi }{2}$. Thus, if $J^m(\alpha )$ is a non-empty set, by the regularity of $\rho _m(t)$, we get that $J^m(\alpha )$ is at most the union of a sequence of open interval $(s_h,t_h)$ such that $\rho _m(s_h)=\rho _m(t_h)=\tan \alpha \frac{\pi }{2}$. We justify the claim.

The set $J^m(\alpha )$ is an open set, hence it is at most the countable union of maximal intervals $(s_h,t_h)$. We set $E^m:=(s, t)-\overline{{\underset{h\in {\mathbb {N}}}{\cup }}(s_h,t_h)}$.

For all $\tau \in E^m$ we have $\rho _m(\tau )\le \tan \alpha \frac{\pi }{2}$, thus, by continuity of $\rho _m$, we get $\rho _m(s_h) =\tan \alpha \frac{\pi }{2}=\rho _m(t_h)$ for all $h\in {\mathbb {N}}$. For the measure of $J^m(\alpha )$ we get

$$\begin{aligned} |J^m(\alpha )|\tan \alpha \frac{\pi }{2} \le \int \limits _{{J^m(\alpha )}}\rho _m(\tau )d\tau <\frac{1}{2}\Vert v(s)\Vert _2^2+\int \limits _{{s}}^{{t}}(f,v)d\tau \,, \end{aligned}$$

(21)

where we took the energy relation (18) into account and the strong convergence of the right-hand side too. Estimate (21) leads to (17). Recalling the definition of $p(\alpha ,\rho _m(t))$, we have

$$\begin{aligned} \frac{d}{d\tau }p(\alpha ,\rho _m(\tau ))=\left\{ \begin{array}{ll}0&{}\text{ a.e. } \text{ in } \tau \in E^m\,, \\ \displaystyle \frac{-2}{(1-\alpha )\pi }\frac{{\dot{\rho }}_m(\tau )}{1+(\rho _m(\tau ))^2}&{}\text{ for } \text{ all } \tau \in J^m(\alpha )\,,\end{array}\right. \end{aligned}$$

(22)

where we took into account that, for all $\alpha \in (0,1)$, function p is a Lipschitz’s function in $\rho _m$, and $\rho _m(t)$ is a regular function in t. Hence, we get $p(\alpha ,\rho _m(t))$ is a Lipschitz’s function with respect to t. We multiply Eq. (18) for $p(\alpha ,\rho _m(\tau ))$, with $\alpha >\alpha _1$, and we integrate by parts on (s, t):

$$\begin{aligned} \Vert v^m(t)\Vert _2^2+2\int \limits _{{s}}^{{t}} p(\alpha ,\rho _m(\tau ))\Vert \nabla v^m(\tau )\Vert _2^2d\tau +\frac{2A(t,s,m,\alpha )}{(1-\alpha )\pi }=\Vert v^m(s)\Vert _2^2+\int \limits _{{s}}^{{t}}(f,v^m)p(\alpha ,\rho _m(\tau ))d\tau \,, \end{aligned}$$

where we set

$$\begin{aligned} A(t,s,m,\alpha ):= \int \limits _{{J^m(\alpha )}}\frac{\Vert v^m(\tau )\Vert _2^2}{1+\rho _m^2(\tau )} {{\dot{\rho }}_m(\tau )}d\tau \end{aligned}$$

where we took (20) and definition of p into account. Letting $m\rightarrow \infty$ and $\alpha \rightarrow 1$, by virtue of the pointwise convergence in s and in t, and Lemma 4, we arrive at

$$\begin{aligned} \Vert v(t)\Vert _2^2+ M(s,t)+2 \int \limits _{{s}}^{{t}}\Vert \nabla v(\tau )\Vert _2^2d\tau =\Vert v(s)\Vert _2^2+\int \limits _{{s}}^{{t}}(f,v)d\tau \,, \end{aligned}$$

(23)

where we set

$$\begin{aligned} M(s,t):=\lim _{\alpha \rightarrow 1^-}\overline{\lim _m} \frac{2}{(1-\alpha )\pi }\int \limits _{{J^m(\alpha )}}\frac{\Vert v^m(\tau )\Vert _2^2}{1+(\rho _m(\tau ))^2}{\dot{\rho }}_m(\tau )d\tau \,. \end{aligned}$$

Recalling the properties of $J^m(\alpha )$, for all $\alpha$ and m, integrating by parts, we get

$$\begin{aligned} \begin{array}{ll} \displaystyle \int \limits _{{J^m(\alpha )}}\!\!\frac{\Vert v^m\Vert _2^2}{1+\rho _m^2}{\dot{\rho }}_md\tau \hskip-0.4cm&{}\displaystyle ={\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\int \limits _{{s_h}}^{{t_h}}\frac{\Vert v^m\Vert _2^2}{1+\rho _m^2}{\dot{\rho }}_md\tau \\ &{}\displaystyle =\frac{\tan \alpha \frac{\pi }{2}}{1+\tan ^2\!\alpha \frac{\pi }{2} } {\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\Big [ \Vert v^m(t_h)\Vert _2^2-\Vert v^m(s_h)\Vert _2^2\Big ]\\ {} &{}\displaystyle+\hskip-0.3cm{\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\int \limits _{{s_h}}^{{t_h}}\hskip-0.15cm\frac{2\rho _m^2}{1+\rho _m^2}d\tau \displaystyle -2\hskip-0.3cm\sum _{h\in {\mathbb {N}}(\alpha ,m)}\int \limits _{s_h}^{t_h}\hskip-0.15cm\frac{\rho _m(f,v_m)}{1+\rho _m^2}\,d\tau\\ {} &{}\displaystyle\hskip3cm +2\hskip-0.2cm{\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\int \limits _{{s_h}}^{{t_h}}\frac{\Vert v^m\Vert _2^2\rho _m^2}{(1+\rho _m^2)^2 } {\dot{\rho }}_md\tau \,.\end{array} \end{aligned}$$

Hence, we arrive at

$$\begin{aligned}{} & {} \displaystyle {\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\int \limits _{{s_h}}^{{t_h}}\frac{\Vert v^m\Vert _2^2}{1+\rho _m^2}{\dot{\rho }}_md\tau \nonumber \\{} & {} \quad -2{\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\int \limits _{{s_h}}^{{t_h}}\frac{\Vert v^m\Vert _2^2\rho _m^2}{(1+\rho _m^2)^2 } {\dot{\rho }}_md\tau +2\sum _{h\in {\mathbb {N}}(\alpha ,m)}\int \limits _{s_h}^{t_h}\frac{\rho _m(f,v_m)}{1+\rho _m^2}\,d\tau \nonumber \\{} & {} \quad \displaystyle \hskip1cm=\frac{\tan \alpha \frac{\pi }{2}}{1+\tan ^2\!\alpha \frac{\pi }{2}}{\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\Big [ \Vert v^m(t_h)\Vert _2^2-\Vert v^m(s_h)\Vert _2^2\Big ] \nonumber \\{} & {} \hskip6cm +{\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\int \limits _{{s_h}}^{{t_h}}\frac{2\rho _m^2}{1+\rho _m^2}d\tau \,. \end{aligned}$$

(24)

We estimate the last integral. Let be

$$\begin{aligned} {{\widetilde{J}}}(\alpha ):=\overline{\lim _m}\, J^m(\alpha )=\bigcap _{j=0}^\infty \bigcup _{m=j}^\infty J^m(\alpha ). \end{aligned}$$

(25)

It results that

$$\begin{aligned}{} & {} \tau \in {{\widetilde{J}}}(\alpha )\iff \exists \, m_k\rightarrow \infty \text{ s.t. } \tau \in J^{m_k}(\alpha )\ \\{} & {} \quad \forall k\in {\mathbb {N}}\iff \overline{\lim _m}\, \chi _{J^m(\alpha )}(\tau )=1. \end{aligned}$$

Hence, if $\tau \in {{\widetilde{J}}}(\alpha )\cap {\mathcal {T}}$ we get that

$$\begin{aligned} \rho (\tau )=\lim _{k\rightarrow \infty }\rho _{m_k}(\tau )\ge \tan \frac{\alpha \pi }{2}. \end{aligned}$$

(26)

On the complement of the set ${\mathcal {T}}$ we can set $\rho =0$, since the value on a null measure set does not change the estimates. Since $0\le \chi _{J^m(\alpha )}\frac{\rho _m^2}{1+\rho _m^2}\le 1$, by Fatou’s lemma, it follows that

$$\begin{aligned}\begin{aligned}&\frac{1}{1-\alpha }\overline{\lim _m} \int \limits _{J^m(\alpha )}\frac{\rho _m^2}{1+\rho _m^2}\,d\tau =\frac{1}{1-\alpha }\overline{\lim _m} \,\int \limits _s^t\chi _{J^m(\alpha )}\frac{\rho _m^2}{1+\rho _m^2}\,d\tau \\&\quad \le \frac{1}{1-\alpha }\int \limits _s^t \overline{\lim _m}\, \chi _{J^m(\alpha )}\frac{\rho _m^2}{1+\rho _m^2}\,d\tau \\&\quad =\frac{1}{1-\alpha }\int \limits _s^t\chi _{{{\widetilde{J}}}(\alpha )}\frac{\rho ^2}{1+\rho ^2}\,d\tau =\frac{1}{1-\alpha }\int \limits _{{{\widetilde{J}}}(\alpha )}\frac{\rho ^2}{1+\rho ^2}\,d\tau \\&\quad \le \frac{1}{1-\alpha }\frac{1}{\tan \frac{\alpha \pi }{2}}\int \limits _{{{\widetilde{J}}}(\alpha )}\rho (\tau )\,d\tau \end{aligned} \end{aligned}$$

Since $\rho \in L^1$ and, by (26),

$$\begin{aligned} \left| {{\widetilde{J}}}(\alpha )\right| \le \frac{\Vert \rho \Vert _1}{\tan \frac{\alpha \pi }{2}} \end{aligned}$$

(27)

the last integral vanishes as $\alpha$ tends to $1^-$. Moreover

$$\begin{aligned} \lim _{\alpha \rightarrow 1^-}\frac{1}{(1-\alpha )\tan \alpha \frac{\pi }{2}}=\frac{\pi }{2}\,\end{aligned}$$

(28)

hence

$$\begin{aligned} \lim _{\alpha \rightarrow 1^-}\overline{\lim _m}\frac{1}{1-\alpha }\int \limits _{J^m(\alpha )}\frac{\rho _m^2}{1+\rho _m^2}\,d\tau =0. \end{aligned}$$

(29)

Concerning the force term we have

$$\begin{aligned} \begin{aligned}&\left|\; \int \limits _{J^m(\alpha )}\frac{\rho _m(f,v^m)}{1+\rho _m^2}\,d\tau \right| \le \left(\; \int \limits _{J^m(\alpha )}\frac{\rho _m^2}{(1+\rho _m^2)^2}\,d\tau \right) ^{\frac{1}{2}}\\&\hskip5cm \left(\; \int \limits _{J^m(\alpha )}\Vert f(\tau )\Vert _2^2\Vert v^m(\tau )\Vert _2^2\,d\tau \right) ^{\frac{1}{2}}\\{} & {} \quad \le \left( \frac{1}{1+(\tan \frac{\alpha \pi }{2})^2}\right) ^{\frac{1}{2}}\left(\; \int \limits _{J^m(\alpha )}\frac{\rho _m^2}{1+\rho _m^2}\,d\tau \right) ^{\frac{1}{2}}\\&\hskip4cm \sup _{m,t}\Vert v^m(t)\Vert _2\left(\; \int \limits _{J^m(\alpha )}\Vert f(\tau )\Vert _2^2\,d\tau \right) ^{\frac{1}{2}}\\& \le \frac{C}{\tan \frac{\alpha \pi }{2}}\left[ \!\frac{(\tan \frac{\alpha \pi }{2})^2}{1+(\tan \frac{\alpha \pi }{2})^2} |J^m(\alpha )|\right] ^{\!\frac{1}{2}} \!\!\le \frac{C}{\tan \frac{\alpha \pi }{2}}\left[ \!\frac{C}{\tan \frac{\alpha \pi }{2}}\right] ^{\!\frac{1}{2}}\!. \end{aligned} \end{aligned}$$

It follows that

$$\begin{aligned} \lim _{\alpha \rightarrow 1^-}\frac{1}{1-\alpha }\overline{\lim _m}\int \limits _{J^m(\alpha )}\frac{\rho _m(f,v^m)}{1+\rho _m^2}\,d\tau =0. \end{aligned}$$

(30)

Using algebraic manipulation we obtain the following relation:

$$\begin{aligned} \begin{array}{l} \displaystyle {\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\int \limits _{{s_h}}^{{t_h}}\frac{\Vert v^m\Vert _2^2}{1+\rho _m^2}{\dot{\rho }}_md\tau -2{\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\int \limits _{{s_h}}^{{t_h}}\frac{\Vert v^m\Vert _2^2\rho _m^2}{(1+\rho _m^2)^2}{\dot{\rho }}_md\tau \\ \displaystyle =-{\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\int \limits _{{s_h}}^{{t_h}}\frac{\Vert v^m\Vert _2^2}{1+\rho _m^2}{\dot{\rho }}_md\tau +2{\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\int \limits _{{s_h}}^{{t_h}}\frac{\Vert v^m\Vert _2^2}{(1+\rho _m^2)^2} {\dot{\rho }}_md\tau \,.\end{array} \end{aligned}$$

Substituting the above relation in Eq. (24) we get

$$\begin{aligned} \begin{aligned}&\displaystyle -{\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\int \limits _{{s_h}}^{{t_h}}\frac{\Vert v^m\Vert _2^2}{1+\rho _m^2}{\dot{\rho }}_md\tau +2{\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\int \limits _{{s_h}}^{{t_h}}\frac{\Vert v^m\Vert _2^2}{(1+\rho _m^2)^2 } {\dot{\rho }}_md\tau \\&\hskip1cm +\sum _{h\in {\mathbb {N}}(\alpha ,m)}\int \limits _{s_h}^{t_h}\frac{\rho _m(f,v_m)}{1+\rho _m^2}\,d\tau \\&\quad \displaystyle =\frac{\tan \alpha \frac{\pi }{2}}{1+\tan ^2\!\alpha \frac{\pi }{2}}{\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\Big [ \Vert v^m(t_h)\Vert _2^2-\Vert v^m(s_h)\Vert _2^2\Big ]\\&\hskip5cm +{\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }} \int \limits _{{s_h}}^{{t_h}}\frac{2\rho _m^2}{1+\rho _m^2}d\tau \end{aligned} \end{aligned}$$

(31)

At last we estimate the integral

$$\begin{aligned} \begin{aligned}{} & {} \left|\;\, \int \limits _{J^m(\alpha )}\frac{\Vert v^m\Vert _2^2}{(1+\rho _m^2)^2 } {\dot{\rho }}_md\tau\, \right| \le \int \limits _{J^m(\alpha )}\frac{\Vert v^m\Vert _2^2}{(1+\rho _m^2)^2 } |{\dot{\rho }}_m|\,d\tau \\{} & {} \quad \le \sup _{t,m}\Vert v^m(t)\Vert _2^2\int \limits _{J^m(\alpha )}\frac{|{\dot{\rho }}_m|}{(1+\rho _m^2)^2 }\,d\tau \\{} & {} \quad \le c\,\frac{1}{1+(\tan \frac{\alpha \pi }{2})^2}\int \limits _{J^m(\alpha )}\frac{|{\dot{\rho }}_m|}{1+\rho _m^2 }\,d\tau \\{} & {}\le \frac{2c}{1+(\tan \frac{\alpha \pi }{2})^2} \int \limits _{J^m(\alpha )}\frac{|{\dot{\rho }}_m|}{(1+\rho _m)^2 }\,d\tau \le \frac{2cM}{1+(\tan \frac{\alpha \pi }{2})^2} \end{aligned} \end{aligned}$$

(32)

where the last inequality follows by Lemma 3. Hence, by (28), we get

$$\begin{aligned}{} & {} \lim _{\alpha \rightarrow 1^-}\overline{\lim _m} \, \left| \frac{2}{1-\alpha }\int \limits _{J^m(\alpha )}\frac{\Vert v^m\Vert _2^2}{(1+\rho _m^2)^2 } {\dot{\rho }}_md\tau \right| \\{} & {} \hskip2cm \le \lim _{\alpha \rightarrow 1^-}\frac{2}{1-\alpha }\frac{2cM}{1+(\tan \frac{\alpha \pi }{2})^2}=0. \end{aligned}$$

Multiplying Eq. (31) by $\frac{2}{(1-\alpha )\pi }$ and passing to the limit using (32), (30) and (29), we get

$$\begin{aligned} M(s,t)=\lim _{\alpha \rightarrow 1^-}\overline{\lim _m}{\underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }}\Big [ \Vert v^m(s_h)\Vert _2^2-\Vert v^m(t_h)\Vert _2^2\Big ]. \end{aligned}$$

By Eq. (18) we get

$$\begin{aligned}{} & {} \underset{h\in {\mathbb {N}}(\alpha ,m)}{\sum }\Big [ \Vert v^m(t_h)\Vert _2^2-\Vert v^m(s_h)\Vert _2^2\Big ]=\\{} & {} \hskip1.5cm -2\int \limits _{J^m(\alpha )}\Vert \nabla v^m(\tau )\Vert _2^2\,d\tau +\int \limits _{J^m(\alpha )}(f,v^m)\,d\tau . \end{aligned}$$

Let us consider the last integral. Since $\left| \left( f(\tau ),v^m(\tau )\right) \right| \le \Vert f(\tau )\Vert _2\Vert v^m(\tau )\Vert _2\le c\Vert f(\tau )\Vert _2$ we can apply the Fatou’s lemma to get

$$\begin{aligned} \begin{aligned}{} & {} \overline{\lim _m} \left|\;\, \int \limits _{J^m(\alpha )}(f,v^m)\,d\tau \,\right| =\overline{\lim _m}\left|\; \int \limits _s^t\chi _{J^m(\alpha )}(f,v^m)\,d\tau \,\right| \\{} & {} \quad \le \int \limits _s^t\overline{\lim _m}\,\left| \chi _{J^m(\alpha )}(f,v^m)\,\right| \,d\tau \\{} & {} \quad \le c\int \limits _s^t\Vert f\Vert _2\overline{\lim _m} \,\chi _{J^m(\alpha )}\,d\tau \le c\int \limits _s^t\Vert f\Vert _2\chi _{{{\widetilde{J}}}(\alpha )}\,d\tau \end{aligned} \end{aligned}$$

with ${{\widetilde{J}}}(\alpha )$ defined in (25). Since $\Vert f(\tau )\Vert _2$ is summable, considering (27), we get

$$\begin{aligned} \lim _{\alpha \rightarrow 1^-}\overline{\lim _m} \left|\;\, \int \limits _{J^m(\alpha )}(f,v^m)\,d\tau \,\right| =0 \end{aligned}$$

and this completes the proof in the case of $s,t\in {\mathcal {T}}$. In order to complete the proof of the theorem, we limit ourselves to remark that, letting $s\rightarrow 0$, the left-hand side tends to values in 0, in particular on any sequence $\{s_k\}\subset {\mathcal {T}}$ letting to 0, and as a consequence the limit on $\{s_k\}$ of the right hand side is well posed. $\square$