1 Introduction

In this paper, we consider the following MHD system in a bounded domain \(\Omega \subset {\mathbb {R}}^3\) with the smooth boundary \(\partial \Omega \):

$$\begin{aligned} \left\{ \begin{array}{l} u_t-\Delta u+(u \cdot \nabla )u-(B \cdot \nabla )B+\nabla (p+\vert B \vert ^2/2)=0 \quad \text {in}\ \Omega \times (0,T),\\ B_t-\Delta B+(u \cdot \nabla )B-(B \cdot \nabla )u=0 \quad \text {in}\ \Omega \times (0,T),\\ \textrm{div} \ u= \textrm{div} \ B = 0 \quad \text {in}\ \Omega \times (0,T),\\ u \vert _{\partial \Omega }=0,\ B\cdot \nu \vert _{\partial \Omega }=0,\ ({\textrm{rot}}\ B)\times \nu \vert _{\partial \Omega } =0,\\ u(0)=u_0,\ B(0)=B_0, \end{array} \right. \end{aligned}$$
(1.1)

where \(u=u(x,t) = (u_1(x, t), u_2(x, t), u_3(x, t)),\) \(B=B(x,t) = (B_1(x, t), B_2(x, t), B_3(x, t))\) and \(p = p(x, t)\) denote the unknown velocity, magnetic field and the pressure, respectively, while \(u_0=u_0(x) = (u_{0, 1}(x), u_{0,2}(x), u_{0,3}(x))\) and \(B_0=B_0(x) = (B_{0, 1}(x), B_{0,2}(x), B_{0,3}(x))\) denote the given initial velocity and magnetic field, respectively. Notice that \(\nu \) is the unit outer normal to \(\partial \Omega .\) For smooth solutions u and B of (1.1), the following energy equality holds:

$$\begin{aligned} \frac{1}{2}\Vert u(t)\Vert ^2_2+\frac{1}{2}\Vert B(t)\Vert ^2_2+\int ^t_0\Vert \nabla u\Vert ^2_2ds+\int ^t_0\Vert \nabla B\Vert ^2_2ds=\frac{1}{2}\Vert u_0\Vert ^2_2+\frac{1}{2}\Vert B_0\Vert ^2_2. \end{aligned}$$
(1.2)

Our aim is to show the energy equality for a larger class of solutions.

The energy conservation law of MHD system has been developed together with that of the Navier–Stokes equations since both equations have a similar non-linear structure. Let us first recall previous results on the Navier–Stokes equations. The pioneer paper work by Shinbrot [11] showed that the energy equality holds for the solution in \(L^p(0,T; L^q(\Omega ))\) for \(2/p+2/q \leqq 1\) with \(4 \leqq q.\) It should be noted that such a class given by Shinbrot is larger than that of the scaling invariant space given by Serrin [10]. Recently, Cheskidov–Constantin–Friedlander–Shvydkoy [3] obtained an elegant result in \({\mathbb {R}}^3\) in the largest class \(L^3(0,T; B^{1/3}_{3,\infty }({\mathbb {R}}^3)).\) From this result and the interpolations with \(L^\infty (0,T; L^2({\mathbb {R}}^3)) \cap L^2(0,T;H^1({\mathbb {R}}^3)),\) the energy equality holds for the solution in the following class:

$$\begin{aligned}{} & {} \begin{array}{c} L^p (0,T;B^\alpha _{q, \infty }{({\mathbb {R}}^3)})\ \ \text {for}\ \ \dfrac{2}{p}+\dfrac{3}{q}-1-\alpha = 1+\alpha -\dfrac{3}{q}\\ \text {with}\ \ 3\leqq p\ \ \text {and}\ \ 1\leqq q\leqq p \leqq \infty , \end{array}\end{aligned}$$
(1.3)
$$\begin{aligned}{} & {} \begin{array}{c} L^p (0,T;B^\alpha _{q, \infty }{({\mathbb {R}}^3)})\ \ \text {for}\ \ \dfrac{2}{p}+\dfrac{3}{q}-1-\alpha = \dfrac{1}{5}-\dfrac{\alpha }{5}+\dfrac{3}{5q}\\ \text {with}\ \ 1\leqq \dfrac{1}{p}+\dfrac{2}{q},\ 1\leqq p \leqq 3\ \ \text {and}\ \ 1\leqq q\leqq \infty , \end{array}\end{aligned}$$
(1.4)
$$\begin{aligned}{} & {} \begin{array}{c} L^p (0,T; B^\alpha _{q, \infty }{({\mathbb {R}}^3)})\ \ \text {for}\ \ \dfrac{2}{p}+\dfrac{3}{q}-1-\alpha = \dfrac{1}{q}\\ \text {with}\ \ \dfrac{1}{p}+\dfrac{2}{q}\leqq 1\ \ \text {and}\ \ 1\leqq p \leqq q \leqq \infty , \end{array} \end{aligned}$$
(1.5)

where \(B^\alpha _{q,\infty }{({\mathbb {R}}^3)}\) denotes the inhomogeneous Besov space in \({\mathbb {R}}^3.\) The criteria (1.3)–(1.5) and the interpolation method are mentioned in Cheskidov–Luo [4, Lemma 2.1]). Berselli–Chiodaroli [1] showed the energy equality under the following conditions which are included in (1.3)–(1.5) for exponents pq but deal with a bounded domain:

$$\begin{aligned}{} & {} L^p(0,T;W^{1,q}(\Omega ))\quad \text {for}\ \frac{2}{p}+\frac{3}{q}-2= 2-\frac{3}{q}\ \text {with}\ \frac{3}{2}<q<\frac{9}{5},\end{aligned}$$
(1.6)
$$\begin{aligned}{} & {} L^p(0,T;W^{1,q}(\Omega ))\quad \text {for}\ \frac{2}{p}+\frac{3}{q}-2= \frac{3}{5q}\ \text {with}\ \frac{9}{5}\leqq q \leqq 3,\end{aligned}$$
(1.7)
$$\begin{aligned}{} & {} L^p (0,T;W^{1,q}(\Omega ))\quad \text {for}\ \frac{2}{p}+\frac{3}{q}-2= \frac{3}{q}-\frac{4}{q+2}\ \text {with}\ 3 < q \leqq \infty . \end{aligned}$$
(1.8)

As for another extension of these results, to enlarge the space in [3], Cheskidov and Luo [4] introduced the weak \(L^p\)-space including \(0<p<1\) as follows:

$$\begin{aligned}{} & {} \begin{array}{c} L^p (0, T; B^\alpha _{q,\infty }{({\mathbb {R}}^3)})\quad \text {for} \dfrac{2}{p}+\dfrac{3}{q}-1-\alpha = \dfrac{1}{5}-\dfrac{\alpha }{5}+\dfrac{3}{5q}\\ \text {with}\ 1\leqq \dfrac{1}{p}+\dfrac{2}{q},\ 0<p<1\ \text {and}\ 1\leqq q\leqq \infty , \end{array}\end{aligned}$$
(1.9)
$$\begin{aligned}{} & {} \begin{array}{c} L^{p}_{w}(0, T; B^\alpha _{q,\infty }{({\mathbb {R}}^3)})\quad \text {for}\ \ \dfrac{2}{p}+\dfrac{3}{q}-1-\alpha = \dfrac{1}{q}\\ \text {with}\ \dfrac{2}{q}+\dfrac{1}{p}<1\ \text {and}\ 1\leqq p<q\leqq \infty , \end{array} \end{aligned}$$
(1.10)

where \(L^p_w\) denote the weak \(L^p\) space on (0, T). Let us introduce the class

$$\begin{aligned} u \in L^p(0, T; W^{\alpha , q}) \end{aligned}$$
(1.11)

for \(2/p+3/q-1-\alpha =\varepsilon (p,q,\alpha ).\) In the case \(\varepsilon (p,q,\alpha )=0,\) the class (1.11) coincides with the scaling invariant space. From viewpoint of the scaling invariant space, it is easy to prove (1.2) in the case \(\varepsilon (p,q,\alpha ) \leqq 0.\) Hence, the essential problem for the validity of the energy equality is to take \(\varepsilon (p,q,\alpha )>0.\)

Concerning ideal MHD system, the result corresponding to Constantin–E–Titi [5] in the Euler equation was handled by Caflisch–Klapper–Steele [2]. According to the paper, the energy equality holds if \(u \in C([0,T];B^{\alpha _1}_{3,\infty }({\mathbb {T}}^3))\) and \(B \in C([0,T];B^{\alpha _2}_{3,\infty }({\mathbb {T}}^3)),\) where the Hölder exponents \(\alpha _1\) and \(\alpha _2\) satisfy \(\alpha _1>1/3\) and \(\alpha _1 + 2 \alpha _2 > 1.\) In the whole space \({\mathbb {R}}^3,\) the same result on the critical case of Caflisch et al. [2] was obtained by Kang–Lee [8]. Namely, the energy equality for the ideal MHD system holds if \(u \in L^3([0,T];B^{\alpha _1}_{3,c({\mathbb {N}})}({\mathbb {R}}^3))\) and \(B \in L^3([0,T];B^{\alpha _2}_{3,c({\mathbb {N}})}({\mathbb {R}}^3)),\) where the Hölder exponents \(\alpha _1\) and \(\alpha _2\) satisfy \(\alpha _1 \geqq 1/3\) and \(\alpha _1 + 2 \alpha _2 \geqq 1.\) Recently, in the case of the bounded domain \(\Omega ,\) the result corresponding to Caflisch et al. [2] was obtained by Wang–Zuo [15].

Concerning MHD system with viscosity and magnetic diffusion, He–Xin [6] showed the validity of the energy equality for u in the Serrin class even though B is in the class \(L^\infty (0, T; L^2({\mathbb {R}}^3)) \cap L^2(0, T; H^1({\mathbb {R}}^3))\) given by Leray [9] and Hopf [7]. The same result on the critical case \(u \in L^\infty (0,T; L^3({\mathbb {R}}^3))\) of Serrin’s was obtained by Yong–Jiu [16]. On the other hand, the critical case \(L^2_w(0,T;BMO_{loc}({\mathbb {R}}^3))\) of Serrin’s was handled by Tan–Wu [12]. They introduce the function space \(L^2_w(0,T;BMO_{loc}({\mathbb {R}}^3))\) which is bigger than \(L^2(0,T;L^\infty ({\mathbb {R}}^3))\) but they assume that B must also belong to \(L^2_w(0,T;BMO_{loc}({\mathbb {R}}^3)).\) Recently, Wang–Zuo [14] dealt with the corresponding result to He–Xin [6] in the bounded domain \(\Omega .\) They also dealt with the critical case of Shinbrot [11]. The result corresponding to Shinbrot [11] and Taniuchi [13] was also handled by Zeng [17]. However, in the Shinbrot–Taniuchi condition dealt in Wang–Zuo [14] and Zeng [17], they assumed that B belongs to the same class of u. Since the nonlinear structures of u and B in the MHD system are the same, it is expected that if B belongs to the same class of u satisfying the energy equality of the Navier–Stokes equations, then we have (1.2). Hence, it seems to be an interesting question to find a larger class of B than that of u so that we may obtain (1.2).

In this paper, we relax the condition on B in such a way that the energy equality holds for B in a larger class than that of Berselli–Chiodaroli [1] for the bounded domain \(\Omega \) and Cheskidov–Constantin–Friedlander–Shvydkoy [3] for the whole space \({\mathbb {R}}^3.\)

This paper is organized as follows. In Sect. 2, we state our main results. Sections 3 and 4 are devoted to the proof of our main theorems.

2 Main results

Theorem 2.1

Let u and B be weak solutions of (1.1) in \(L^\infty (0,T;L^2(\Omega )) \cap L^2(0,T;H^1(\Omega )).\)

  1. (i)

    Case \(3/2 < q \leqq 9/5{:}\) If u and B satisfy

    $$\begin{aligned}{} & {} {\textrm{rot}}\ u \in L^p(0,T; L^q(\Omega ))\quad \text {for}\ \frac{2}{p}+\frac{3}{q}-2=2-\frac{3}{q},\end{aligned}$$
    (2.1)
    $$\begin{aligned}{} & {} \begin{array}{c} {\textrm{rot}}\ B \in L^\sigma (0,T;L^\tau (\Omega ))\quad \text {for}\ \dfrac{2}{\sigma }+\dfrac{3}{\tau }-2=2-\dfrac{3}{\tau },\\ \quad \text {with}\ \dfrac{3}{2} < \tau \leqq \dfrac{9}{5}+\varepsilon _1(q), \end{array} \end{aligned}$$
    (2.2)

    then the energy equality (1.2) holds, where

    $$\begin{aligned} 0\leqq \varepsilon _1(q)=\frac{6q}{5q-3}-\frac{9}{5}. \end{aligned}$$

    Notice that \(\varepsilon _1(9/5) = 0\) and \(0<\varepsilon _1(q)\) for \(3/2< q < 9/5.\)

  2. (ii)

    Case \(9/5 \leqq q < 3{:}\) If u and B satisfy

    $$\begin{aligned}{} & {} {\textrm{rot}}\ u \in L^p(0,T; L^q(\Omega ))\quad \text {for}\ \frac{2}{p}+\frac{3}{q}-2=\frac{3}{5q},\end{aligned}$$
    (2.3)
    $$\begin{aligned}{} & {} \begin{array}{c} {\textrm{rot}}\ B \in L^\sigma (0,T;L^\tau (\Omega ))\quad \text {for}\ \dfrac{2}{\sigma }+\dfrac{3}{\tau }-2=\dfrac{3}{5\tau },\\ \quad \text {with}\ \dfrac{9}{5}-\varepsilon _2(q) \leqq \tau < 3, \end{array} \end{aligned}$$
    (2.4)

    then the energy equality (1.2) holds,  where

    $$\begin{aligned} 0 \leqq \varepsilon _2(q)=\frac{9}{5}-\frac{6q}{5q-3}. \end{aligned}$$

Notice that \(\varepsilon _2(9/5) = 0\) and \(0<\varepsilon _2(q)\) for \(9/5< q < 3.\)

Since \(q=3\) exhibits a threshold exponent for validity of (1.2), it is suitable to introduce the Besov space to obtain a new criteria.

Theorem 2.2

Let u and B be weak solutions of (1.1) in \(L^\infty (0,T;L^2({\mathbb {R}}^3)) \cap L^2(0,T;H^1({\mathbb {R}}^3)).\) Assume that \(3 \leqq q <\infty .\) If u and B satisfy

$$\begin{aligned}{} & {} u \in L^p(0,T; B^1_{q,\infty }({\mathbb {R}}^3))\quad \text {for}\ \frac{2}{p}+\frac{3}{q}-2=\frac{3}{5q},\end{aligned}$$
(2.5)
$$\begin{aligned}{} & {} \begin{array}{c} B \in L^\sigma (0,T;B^1_{\tau ,c({\mathbb {N}})}({\mathbb {R}}^3)) \quad \text {for}\ \dfrac{2}{\sigma }+\dfrac{3}{\tau }-2=\dfrac{3}{5\tau },\\ \quad \text {with}\ \dfrac{9}{5}-\varepsilon _2(q)< \tau < \infty , \end{array} \end{aligned}$$
(2.6)

then the energy equality (1.2) holds,  where \(0 < \varepsilon _2(q).\)

Remark 1

1. The classes (2.1), (2.3) and (2.5) of u are same as those in recent works [1, 3] for the Navier–Stokes equations. However, our classes (2.2), (2.4) and (2.6) of B are larger than those of them. Especially, if we take u in the scaling invariant spaces \(L^1(0,T;W^{1,\infty })\) and \(L^\infty (0,T; W^{1,3/2}),\) then B reaches the Leray-Hopf class. Hence, we may regard our Theorems 2.1 and 2.2 as a generalization of He–Xin [6].

2. The classes (2.1) and (2.2) correspond to (1.6). The classes (2.3) and (2.4) also correspond to (1.7). Notice that the classes (2.5) and (2.6) dose not relate to (1.8), but corresponds to (1.4) in the case \(3<q.\) More precisely, (1.8) represents the nonlinear relation between quantities 2/p and 3/q by means to the inhomogeneous term \(4/(q+2)\) on the RHS. On the other hand, our advantage of (2.5) and (2.6) exhibit the linear relation between 2/p and 3/q,  which enables us to obtain (1.2) in the Leray-Hopf class for B when u belongs to the Serrin class. In (2.5) and (2.6), we should also emphasize the advantage to make use of the Besov space, which is suitable for us to handle freely derivatives of the nonlinear convection term.

3. The class in our Theorems 2.1 and 2.2 are not embedded in the Kang–Lee’s class which deal with the ideal MHD in \({\mathbb {R}}^3\) by the interpolations with the Leray–Hopf class. Indeed, when u belongs to the class near the critical cases \(L^1(0,T;B^1_{\infty ,\infty }({\mathbb {R}}^3))\) or \(L^\infty (0,T; W^{1,3/2}({\mathbb {R}}^3)),\) our results are not embedded in the class of Kang–Lee [8]. For instance, we consider the critical case \(u \in L^1(0,T;B^1_{\infty ,\infty }({\mathbb {R}}^3))\) and \(B \in L^\infty (0,T; B^1_{6/5, \infty }({\mathbb {R}}^3))\) in Theorem 2.2. By the interpolation method mentioned in Cheskidov–Luo [4], we see that \(\alpha _2=1/6\) is the largest exponent \(\alpha _2\) in which the following embedding holds:

$$\begin{aligned} \begin{aligned} L^\infty (0,T; B^1_{6/5, \infty }({\mathbb {R}}^3))&\cap L^\infty (0,T;L^2({\mathbb {R}}^3)) \cap L^2(0,T&;H^1({\mathbb {R}}^3))\\&\quad \hookrightarrow L^3(0,T; B^{\alpha _2}_{3,\infty }({\mathbb {R}}^3)). \end{aligned} \end{aligned}$$

Then, we choose the smallest exponent \(\alpha _1 = 2/3\) satisfying \(1 \leqq \alpha _1 + 2 \alpha _2.\) Now, we consider the following imbedding:

$$\begin{aligned} \begin{aligned} L^1 (0,T; B^1_{\infty , \infty }({\mathbb {R}}^3)) \cap L^\infty (0&,T;L^2({\mathbb {R}}^3)) \cap L^2(0,T;H^1({\mathbb {R}}^3))\\&\hookrightarrow L^3(0,T; B^s_{p,\infty }({\mathbb {R}}^3)) \hookrightarrow L^3(0,T; B^{\alpha _2}_{3,\infty }({\mathbb {R}}^3)), \end{aligned} \end{aligned}$$

with the conditions

$$\begin{aligned}{} & {} \alpha _1 = 2/3, \ 1 \leqq p \leqq 3, \ s-\frac{3}{p} \geqq \frac{2}{3} - 1, \ 0 \leqq \theta _1,\ \theta _2 \leqq 1, \ \theta _1 + \theta _2 \leqq 1,\\{} & {} s = \theta _1 + \theta _2, \ \frac{1}{p} = \frac{\theta _1}{\infty } + \frac{\theta _2}{2} + \frac{1 - \theta _1 - \theta _2}{2}, \ \frac{1}{3} = \frac{\theta _1}{1} + \frac{\theta _2}{\infty } + \frac{1 - \theta _1 - \theta _2}{2}. \end{aligned}$$

However, there exist no exponents sp\(\theta _1\) and \(\theta _2\) satisfying the imbedding and conditions above. Indeed, these conditions imply \(1/3 \geqq \theta _1\) and \(\theta _1 \geqq 1,\) which is a contradiction. Hence, our class in Theorem 2.2 is not embedded in the Kang–Lee’s class by the interpolations with the Leray–Hopf class. In the case of \(u \in L^\infty (0,T; W^{1,3/2}({\mathbb {R}}^3))\) and \(B \in L^2(0,T; W^{1,2}({\mathbb {R}}^3))\) in Theorem 2.1, our class is also not embedded in the class in Kang–Lee [8] by the interpolations with the Leray–Hopf class. Moreover, our conditions can deal with not only the whole space \({\mathbb {R}}^3\) but also the bounded domain \(\Omega .\)

3 Preliminaries

3.1 Notations

In this paper, \(p'\) and \(p^*\) denote the Hölder and Sobolev exponents defined by

$$\begin{aligned}{} & {} p'=p/(p-1) \quad \text {for } 1 \leqq p \leqq \infty ,\\{} & {} p^*=3p/(3-p) \quad \text {for } 1 \leqq p <3. \end{aligned}$$

We also use the inner product \((\cdot ,\cdot )\) and norms \(\Vert f\Vert _{L^p L^q}\) defined by

$$\begin{aligned}{} & {} (f,g) = \int _{\Omega } f(x)g(x) dx,\\{} & {} \Vert f\Vert _{L^p L^q}=\Vert f\Vert _{L^p(0,T;L^q)}=\left\{ \int ^T_0 \Vert f(t)\Vert ^p_q dt \right\} ^{\frac{1}{p}}. \end{aligned}$$

Next, let us recall the Littlewood–Paley decomposition and Besov spaces, for which we refer to [3] or [4]. Let \(\chi \in C^\infty _0(B_1(0))\) be a symmetric cut-off function satisfying \(\chi =1\) in \(B_{\frac{3}{4}}(0)\) and \(\chi =0\) in \(B_1(0)^c.\) We define \( \lambda _n=2^n, \ \phi (\xi ) = \chi (\lambda _1^{-1}\xi )-\chi (\xi )\) and \(\phi _n(\xi )=\phi (\lambda _n^{-1}\xi ).\) Then the inhomogeneous Besov space is denoted by \(B^s_{p,q}\) with the norm \(\Vert u\Vert _{B^s_{p,q}}=\Vert \lambda ^s_n\Vert \Delta _n u\Vert _p\Vert _{l^q},\) where \(\Delta _n u=\check{\phi _n}*u\ \text {for}\ 0 \leqq n\ \text {and}\ \Delta _{-1}u={\check{\chi }}*u.\) We also define \(B^s_{p,c({\mathbb {N}})}\) as

$$\begin{aligned} B^s_{p,c({\mathbb {N}})}=\{ u \in B^s_{p, \infty }; \lim _{n \rightarrow \infty } \lambda ^s_n \Vert \Delta _n u\Vert _p = 0 \}. \end{aligned}$$

The function space \(B^s_{p,c({\mathbb {N}})}\) is introduced in Cheskidov–Constantin–Friedlander–Shvydkoy [3] and is also written as \(B^s_{p,\infty -}.\) Here, we note that for any \(1 \leqq q < \infty \) following imbedding holds;

$$\begin{aligned} B^s_{p,q} \hookrightarrow B^s_{p,c({\mathbb {N}})}. \end{aligned}$$

We use notations \(u_{\leqq N},\ {{\bar{h}}},\ {{\bar{h}}}_N\) and \(K_{(\alpha )}\) defined as follows:

$$\begin{aligned} u_{\leqq N}= & {} \sum _{n=-1}^{N} \Delta _n u,\\ {{\bar{h}}}_N(y)= & {} \lambda _{N+1}^3 {{\check{\chi }}} (\lambda _{N+1}y),\\ K_{(\alpha )}(n)= & {} \left\{ \begin{array}{l} \lambda ^{-(1-\alpha )}_n\ (0\leqq n)\\ \lambda ^{\alpha }_n\ (n<0) \end{array} \right. \quad \text {for}\ 0\leqq \alpha \leqq 1. \end{aligned}$$

3.2 Estimates for non-linear terms

The proofs of our theorems are basically derived from the standard approximation argument for the energy equality as [1, 3]. Roughly speaking, since the estimates of linear terms are established in the Leray-Hopf class \(L^\infty (0,T;L^2) \cap L^2(0,T;H^1),\) the estimates of non-linear terms in following Lemmas 3.1 and 3.2 play an essential role for the proof of our theorems.

Lemma 3.1

Let \(f,\ g_1,\ g_2 \in L^\infty (0,T;L^2(\Omega )) \cap L^2(0,T;H^1(\Omega ))\) with \(f \vert _{\partial \Omega }=g_1\cdot \nu \vert _{\partial \Omega }=g_2\cdot \nu \vert _{\partial \Omega }=0.\)

  1. (i)

    Case \(3/2 < q \leqq 9/5{:}\) Suppose that f\(g_1\) and \(g_2\) satisfy

    $$\begin{aligned} \nabla f \in L^p(0,T;L^q(\Omega ))\quad \text {for}\ \frac{2}{p}+\frac{3}{q}-2=2-\frac{3}{q}, \\ \begin{array}{c} \nabla g_1,\ \nabla g_2 \in L^\sigma (0,T;L^\tau (\Omega ))\quad \text {for}\ \dfrac{2}{\sigma }+\dfrac{3}{\tau }-2=2-\dfrac{3}{\tau },\\ \text {with}\ \dfrac{3}{2} < \tau \leqq \dfrac{9}{5}+\varepsilon _1(q). \end{array} \end{aligned}$$
  2. (i-i)

    We take \(\alpha \) and \(\beta \) so that

    $$\begin{aligned} \frac{1}{6}< \frac{1}{\alpha } \leqq \frac{1}{q^*},\ \frac{1}{6} < \frac{1}{\beta } \leqq \frac{1}{\tau ^*} \text { and } \frac{1}{\tau '} = \frac{1}{\alpha } + \frac{1}{\beta }. \end{aligned}$$

    For such \(\alpha \) and \(\beta ,\) we take \(0 < \theta ,\ \tilde{\theta }\leqq 1\) so that

    $$\begin{aligned} \frac{1}{\alpha } = \frac{\theta }{q^*} + \frac{1-\theta }{6},\ \frac{1}{\beta } = \frac{{{\tilde{\theta }}}}{\tau ^*} + \frac{1-\tilde{\theta }}{6}. \end{aligned}$$

    Then we have the estimate

    $$\begin{aligned} \left| \int ^t_0 ((f \cdot \nabla )g_1,g_2) ds \right| \leqq C \Vert \nabla f\Vert ^{\theta }_{L^pL^q} \Vert \nabla f\Vert ^{1-\theta }_{L^2L^2} \Vert \nabla g_1\Vert _{L^\sigma L^\tau } \Vert \nabla g_2\Vert ^{{{\tilde{\theta }}}}_{L^\sigma L^\tau } \Vert \nabla g_2\Vert ^{1-\tilde{\theta }}_{L^2 L^2}, \end{aligned}$$

    where \(C = C(q,\tau ,\alpha ,\beta ).\)

  3. (i-ii)

    We also take \(0 < {{\tilde{\theta }}} \leqq 1\) so that

    $$\begin{aligned} \frac{1}{2q'}=\frac{{{\tilde{\theta }}}}{\tau ^*} + \frac{1-\tilde{\theta }}{6}. \end{aligned}$$

    Then we have the estimate

    $$\begin{aligned} \left| \int ^t_0 ((g_1 \cdot \nabla ) f,g_2)ds \right| \leqq C \Vert \nabla g_1\Vert ^{{{\tilde{\theta }}}}_{L^\sigma L^\tau } \Vert \nabla g_1\Vert ^{1 - {{\tilde{\theta }}}}_{L^2 L^2} \Vert \nabla f\Vert _{L^p L^q} \Vert \nabla g_2\Vert ^{{{\tilde{\theta }}}}_{L^\sigma L^\tau } \Vert \nabla g_2\Vert ^{1 - \tilde{\theta }}_{L^2 L^2}, \end{aligned}$$

    where \(C = C(q,\tau ).\)

  4. (ii)

    Case \(9/5 \leqq q < 3\): Suppose that f\(g_1\) and \(g_2\) satisfy

    $$\begin{aligned}{} & {} \nabla f \in L^p(0,T;L^q(\Omega ))\quad \text {for}\ \frac{2}{p}+\frac{3}{q}-2=\frac{3}{5q}, \\{} & {} \begin{array}{c} \nabla g_1,\ \nabla g_2 \in L^\sigma (0,T;L^\tau (\Omega ))\quad \text {for}\ \dfrac{2}{\sigma } + \dfrac{3}{\tau }-2=\dfrac{3}{5\tau },\\ \text {with}\ \dfrac{9}{5}-\varepsilon _2(q) \leqq \tau < 3. \end{array} \end{aligned}$$
  5. (ii-i)

    We take \(\alpha \) and \(\beta \) so that

    $$\begin{aligned} \frac{1}{q^*} \leqq \frac{1}{\alpha }<\frac{1}{2},\ \frac{1}{\tau ^*} \leqq \beta <\frac{1}{2} \text { and } \frac{1}{\tau '}=\frac{1}{\alpha }+\frac{1}{\beta }. \end{aligned}$$

    For such \(\alpha \) and \(\beta ,\) we take \(0 \leqq \theta ,\ \tilde{\theta }< 1\) so that

    $$\begin{aligned} \frac{1}{\alpha }=\frac{\theta }{2}+\frac{1-\theta }{q^*},\ \frac{1}{\beta }=\frac{{{\tilde{\theta }}}}{2}+\frac{1-\tilde{\theta }}{\tau ^*}. \end{aligned}$$

    Then we have the estimate

    $$\begin{aligned} \left| \int ^t_0 ((f \cdot \nabla )g_1,g_2) ds \right| \leqq C \Vert f\Vert ^{\theta }_{L^\infty L^2} \Vert \nabla f\Vert ^{1-\theta }_{L^p L^q} \Vert \nabla g_1\Vert _{L^\sigma L^\tau } \Vert g_2\Vert ^{{\tilde{\theta }}}_{L^\infty L^2} \Vert \nabla g_2\Vert ^{1 - \tilde{\theta }}_{L^\sigma L^\tau }, \end{aligned}$$

    where \(C = C(q,\tau ,\alpha ,\beta ).\)

  6. (ii-ii)

    We also take \(0 \leqq {\tilde{\theta }} < 1\) so that

    $$\begin{aligned} \frac{1}{2q'} = \frac{{{\tilde{\theta }}}}{2} + \frac{1-\tilde{\theta }}{\tau ^*}. \end{aligned}$$

    Then we have the estimate

    $$\begin{aligned} \left| \int ^t_0 ((g_1 \cdot \nabla ) f,g_2) ds \right| \leqq C \Vert g_1\Vert ^{{\tilde{\theta }}}_{L^\infty L^2} \Vert \nabla g_1\Vert ^{1 - {{\tilde{\theta }}}}_{L^\sigma L^\tau } \Vert \nabla f\Vert _{L^p L^q} \Vert g_2\Vert ^{{\tilde{\theta }}}_{L^\infty L^2} \Vert \nabla g_2\Vert ^{1 - \tilde{\theta }}_{L^\sigma L^\tau }, \end{aligned}$$

    where \(C = C(q,\tau ).\)

Proof of Lemma 3.1

In the case of (i-i), by the convex interpolation, we have

$$\begin{aligned} \begin{aligned} \left| \int ^t_0 ((f \cdot \nabla )g_1,g_2)ds\right| \leqq&\int ^t_0 \Vert f\Vert _\alpha \Vert \nabla g_1\Vert _\tau \Vert g_2\Vert _\beta ds\\ \leqq&\int ^t_0 \Vert f\Vert ^{\theta }_{q^*}\Vert f\Vert ^{1-\theta }_6 \Vert \nabla g_1\Vert _\tau \Vert g_2\Vert ^{{\tilde{\theta }}}_{\tau ^*} \Vert g_2\Vert ^{1-{\tilde{\theta }}}_{6} ds\\ \leqq&C \int ^t_0 \Vert \nabla f\Vert ^{\theta }_q \Vert \nabla f\Vert ^{1-\theta }_2 \Vert \nabla g_1\Vert _\tau \Vert \nabla g_2\Vert ^{{\tilde{\theta }}}_\tau \Vert \nabla g_2\Vert ^{1-{\tilde{\theta }}}_{2} ds. \end{aligned} \end{aligned}$$

Now, take the exponents \(1 \leqq a_1,\ a_2,\ b_1,\ b_2 \leqq \infty \) as

$$\begin{aligned} a_1=\frac{p}{\theta },\ a_2=\frac{2}{1-\theta },\ b_1= \frac{\sigma }{{\tilde{\theta }}},\ b_2=\frac{2}{1-{\tilde{\theta }}}. \end{aligned}$$

Then we see that

$$\begin{aligned} 1 = \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{\sigma } + \frac{1}{b_1} + \frac{1}{b_2}. \end{aligned}$$

Hence, it follows from the Hölder inequality in time that

$$\begin{aligned} \left| \int ^t_0 ((f \cdot \nabla )g_1,g_2)ds\right| \leqq C \Vert \nabla f\Vert ^{\theta }_{L^pL^q}\Vert \nabla f\Vert ^{1-\theta }_{L^2L^2} \Vert \nabla g_1\Vert _{L^\sigma L^\tau } \Vert \nabla g_2\Vert ^{\tilde{\theta }}_{L^\sigma L^\tau }\Vert \nabla g_2\Vert ^{1- {{\tilde{\theta }}}}_{L^2 L^2}. \end{aligned}$$

In the case of (i-ii), by the convex interpolation, we have

$$\begin{aligned} \begin{aligned} \left| \int ^t_0 ((g_1\cdot \nabla )f,g_2) ds\right|&\leqq \int ^t_0\Vert g_1\Vert _{2q'} \Vert \nabla f\Vert _q \Vert g_2\Vert _{2q'} ds\\&\leqq \int ^t_0 \Vert g_1\Vert ^{{\tilde{\theta }}}_{\tau ^*}\Vert g_1\Vert ^{1-{\tilde{\theta }}}_6 \Vert \nabla f\Vert _q \Vert g_2\Vert ^{{\tilde{\theta }}}_{\tau ^*} \Vert g_2\Vert ^{1-{\tilde{\theta }}}_6 ds\\&\leqq C \int ^t_0 \Vert \nabla g_1\Vert ^{{\tilde{\theta }}}_\tau \Vert \nabla g_1\Vert ^{1-{\tilde{\theta }}}_2 \Vert \nabla f\Vert _q \Vert \nabla g_2\Vert ^{{\tilde{\theta }}}_\tau \Vert \nabla g_2\Vert ^{1-{\tilde{\theta }}}_2 ds. \end{aligned} \end{aligned}$$

Here, it should be noted that \(\tau ^* \leqq 2q' < 6\) holds by the assumptions \(3/2 < q\) and \(\tau \leqq 9/5+\varepsilon _1(q).\) Now, take the exponents \(1\leqq b_1,\) \(b_2 \leqq \infty \) as

$$\begin{aligned} b_1=\frac{\sigma }{{\tilde{\theta }}},\ b_2=\frac{2}{1-{\tilde{\theta }}}. \end{aligned}$$

Then we see that

$$\begin{aligned} 1=\frac{1}{b_1}+\frac{1}{b_2}+\frac{1}{p}+\frac{1}{b_1}+\frac{1}{b_2}. \end{aligned}$$

Hence, the Hölder inequality gives

$$\begin{aligned} \left| \int ^t_0 ((g_1\cdot \nabla )f,g_2) ds\right| \leqq C \Vert \nabla g_1\Vert ^{{{\tilde{\theta }}}}_{L^\sigma L^\tau } \Vert \nabla g_1\Vert ^{1-{{\tilde{\theta }}}}_{L^2 L^2} \Vert \nabla f\Vert _{L^p L^q} \Vert \nabla g_2\Vert ^{{{\tilde{\theta }}}}_{L^\sigma L^\tau } \Vert \nabla g_2\Vert ^{1-\tilde{\theta }}_{L^2 L^2}. \end{aligned}$$

In the case of (ii-i), it follows from the convex interpolation that

$$\begin{aligned} \begin{aligned} \left| \int ^t_0 ((f\cdot \nabla )g_1,g_2)ds \right|&\leqq \int ^t_0 \Vert f\Vert _\alpha \Vert \nabla g_1\Vert _\tau \Vert g_2\Vert _\beta ds \\&\leqq C \int ^t_0 \Vert f\Vert ^{\theta }_2 \Vert \nabla f\Vert ^{1-\theta }_q \Vert \nabla g_1\Vert _\tau \Vert g_2\Vert ^{{\tilde{\theta }}}_2 \Vert \nabla g_2\Vert ^{1-{\tilde{\theta }}}_\tau ds. \end{aligned} \end{aligned}$$

Now, take the exponents \(1 \leqq a,\) \(b \leqq \infty \) as

$$\begin{aligned} a=\frac{p}{1-\theta },\ b=\frac{\sigma }{1-{{\tilde{\theta }}}}. \end{aligned}$$

Then we see that

$$\begin{aligned} 1=\frac{1}{a} + \frac{1}{\sigma } + \frac{1}{b}. \end{aligned}$$

Hence, the Hölder inequality gives

$$\begin{aligned} \left| \int ^t_0 ((f\cdot \nabla )g_1,g_2)ds \right| \leqq C \Vert f\Vert ^{\theta }_{L^\infty L^2} \Vert \nabla f\Vert ^{1-\theta }_{L^p L^q} \Vert \nabla g_1\Vert _{L^\sigma L^\tau } \Vert g_2\Vert ^{{\tilde{\theta }}}_{L^\infty L^2} \Vert \nabla g_2\Vert ^{1-{{\tilde{\theta }}}}_{L^\sigma L^\tau }. \end{aligned}$$

In the case of (ii-ii), it follows from the convex interpolation that

$$\begin{aligned} \begin{aligned} \left| \int ^t_0 ((g_1\cdot \nabla )f,g_2) ds\right| \leqq&\int ^t_0 \Vert g_1\Vert _{2q'} \Vert \nabla f\Vert _q \Vert g_2\Vert _{2q'} ds \\ \leqq&C \int ^t_0 \Vert g_1\Vert ^{{{\tilde{\theta }}}}_2 \Vert \nabla g_1\Vert ^{1-\tilde{\theta }}_\tau \Vert \nabla f\Vert _q \Vert g_2\Vert ^{{{\tilde{\theta }}}}_2 \Vert \nabla g_2\Vert ^{1-{{\tilde{\theta }}}}_\tau ds. \end{aligned} \end{aligned}$$

Here, it should be noted that \(2 < 2q' \leqq \tau ^*\) holds by the assumption \(9/5 - \varepsilon _2(q) \leqq \tau .\) Now, take the exponent \(1< b < \infty \) as

$$\begin{aligned} b = \frac{\sigma }{1-{{\tilde{\theta }}}}. \end{aligned}$$

Then we see that

$$\begin{aligned} 1=\frac{1}{b} +\frac{1}{p} + \frac{1}{b}. \end{aligned}$$

Hence, the Hölder inequality gives

$$\begin{aligned} \left| \int ^t_0 ((g_1\cdot \nabla )f,g_2) ds\right| \leqq C \Vert g_1\Vert ^{{{\tilde{\theta }}}}_{L^\infty L^2} \Vert \nabla g_1\Vert ^{1-\tilde{\theta }}_{L^\sigma L^\tau } \Vert \nabla f\Vert _{L^p L^q} \Vert g_2\Vert ^{\tilde{\theta }}_{L^\infty L^2} \Vert \nabla g_2\Vert ^{{1-{{\tilde{\theta }}}}}_{L^\sigma L^\tau }. \end{aligned}$$

This completes the proof of Lemma 3.1.

Now, we remark on the existence of the exponents \(\alpha \) and \(\beta \) in the proof. \(\square \)

Remark 2

1. Let \(3/2 < q \leqq 9/5\) and \(3/2 < \tau \leqq 9/5+\varepsilon _1(q).\) Let \(1/6 < 1/\beta \leqq { 1/\tau ^*}.\) Set \(\alpha \) as \(\tau ' \leqq \alpha \) and \(1/\tau '=1/\alpha +1/\beta .\) Then we have

$$\begin{aligned} L_1 \leqq \frac{1}{\beta }< L_2 \Rightarrow \frac{1}{6} < {\frac{1}{\alpha }} \leqq \frac{1}{q^*}, \end{aligned}$$

where

$$\begin{aligned} L_1=\frac{4}{3}-\frac{1}{\tau }-\frac{1}{q},\ L_2=\frac{5}{6}-\frac{1}{\tau }. \end{aligned}$$

Moreover, we can choose \(\beta \) satisfying \(1/6 < 1/\beta \leqq 1/\tau ^*\) and \(L_1 \leqq 1/\beta < L_2\) because we see that \((1/6, 1/\tau ^*] \cap [L_1, L_2) \ne \text{\O }\) by the assumption \(3/2 < \tau \leqq 9/5+\varepsilon _1(q).\)

2. Let \(9/5 \leqq q < 3\) and \(9/5- \varepsilon _2(q) \leqq \tau < 3.\) Let \(1/\tau ^* \leqq 1/\beta < 1/2.\) Set \(\alpha \) as \(\tau ' \leqq \alpha \) and \(1/\tau '=1/\alpha +1/\beta .\) Then we have

$$\begin{aligned} L_3< \frac{1}{\beta } \leqq L_1 \Rightarrow \frac{1}{q^*} \leqq \frac{1}{\alpha } < \frac{1}{2}, \end{aligned}$$

where

$$\begin{aligned} L_3=\frac{1}{2}-\frac{1}{\tau } \end{aligned}$$

Moreover, we can choose \(\beta \) satisfying \(1/\tau ^* \leqq 1/\beta < 1/2\) and \(L_3 < 1/\beta \leqq L_1\) because we see that \([1/\tau ^*,1/2) \cap (L_3,L_1] \ne \text{\O }\) by the assumption \(9/5 - \varepsilon _2(q) \leqq \tau .\)

Lemma 3.2

Let \(3 \leqq q < \infty \) and \(f,\ g_1,\ g_2 \in L^\infty (0,T;L^2({\mathbb {R}}^3)) \cap L^2(0,T;H^1({\mathbb {R}}^3))\) with \(f \vert _{\partial \Omega } = g_1 \cdot \nu \vert _{\partial \Omega } = g_2 \cdot \nu \vert _{\partial \Omega } = 0.\) Let us define \(\Pi _{N,f,g_1,g_2}\) by

$$\begin{aligned}{} & {} \Pi _{N,f,g_1,g_2}=\Pi ^{(1)}_{N,f,g_1,g_2}+\Pi ^{(2)}_{N,f,g_1,g_2}, \\{} & {} \Pi ^{(1)}_{N,f,g_1,g_2}=(r_N(f,g_1);\nabla (g_2)_{\leqq N}), \\{} & {} \Pi ^{(2)}_{N,f,g_1,g_2}=((f-f_{\leqq N})\otimes (g_1-(g_1)_{\leqq N}); \nabla (g_2)_{\leqq N}), \\{} & {} r_N(f,g_1)=\int \bar{h}_N(y)(f(x-y)-f(x))\otimes (g_1(x-y)-g_1(x))dy. \end{aligned}$$

Suppose that f\(g_1\) and \(g_2\) satisfy

$$\begin{aligned}{} & {} f \in L^p(0,T;B^1_{q,\infty })\quad \text {for} \dfrac{2}{p}+\dfrac{3}{q}-2=\dfrac{3}{5q}, \\{} & {} \begin{array}{c} g_1,\ g_2 \in L^\sigma (0,T;B^1_{\tau ,c({\mathbb {N}})})\quad \text {for}\ \dfrac{2}{\sigma } + \dfrac{3}{\tau }-2=\dfrac{3}{5\tau },\\ \text {with}\ \dfrac{9}{5} - \varepsilon _2(q)< \tau < \infty . \end{array} \end{aligned}$$

(i) Case \(3 \leqq q < \infty \) and \(9/5-\varepsilon _2(q) < \tau \leqq 2 q'\): We take \(0< \rho < 1\) so that

$$\begin{aligned} 0<\rho<\textrm{min} \left\{ \frac{2q}{5q-6},\ \left( \frac{5\tau - 6}{2\tau }-\frac{3}{2q} \right) \frac{4q}{5q - 6}\right\} <1. \end{aligned}$$

For such \(\rho ,\) we take \(0< \alpha ,\ \beta < 1\) and \(0< {{\tilde{\rho }}} < 1\) so that

$$\begin{aligned} \alpha =1-\rho \frac{5q - 6}{2q},\ \beta =\frac{1-\alpha }{2}=\rho \cdot \frac{5q-6}{4q} \text { and } {{\tilde{\rho }}} = \frac{2\tau }{5\tau -6}\left( \rho \cdot \frac{5q-6}{4q} + \frac{3}{2q} \right) . \end{aligned}$$

Then we have the estimate

$$\begin{aligned}{} & {} \left| \int ^t_0 \right. \Pi _{N,f,g_1,g_2} ds \biggr \vert + \left| \int ^t_0 \Pi _{N,g_1,g_2,f} ds \right| \nonumber \\{} & {} \quad \leqq C \sum _{-1\leqq n,m,l} K_{(\alpha )}(N-n) K_{(\beta )}(N-m)K_{(\beta )}(N-l)\nonumber \\{} & {} \qquad \times \Vert \Delta _n f\Vert ^{\rho }_{L^\infty L^2} \left\{ \int ^T_0 \lambda ^{p}_n \Vert \Delta _n f\Vert ^{p}_q ds \right\} ^{\frac{1-\rho }{p}} \nonumber \\{} & {} \qquad \times \Vert \Delta _m g_1\Vert ^{1-{{\tilde{\rho }}}}_{L^\infty L^2} \left\{ \int ^T_0 \lambda ^\sigma _m\Vert \Delta _m g_1\Vert ^{\sigma }_\tau ds \right\} ^{\frac{{{\tilde{\rho }}}}{\sigma }}\nonumber \\{} & {} \qquad \times \Vert \Delta _l g_2\Vert ^{1-{{\tilde{\rho }}}}_{L^\infty L^2} \left\{ \int ^T_0 \lambda ^\sigma _l\Vert \Delta _l g_2\Vert ^{\sigma }_\tau ds \right\} ^{\frac{{{\tilde{\rho }}}}{\sigma }}, \end{aligned}$$
(3.1)

where \(C = C(q,\tau ,\rho ).\)

(ii) Case \(3 \leqq q < \infty \) and \(2q'< \tau < \infty \): We take \(0< \rho < 1\) so that

$$\begin{aligned} 0<\rho<\textrm{min} \left\{ \frac{2q}{5q-6},\ \frac{\tau }{\tau -2} \cdot \frac{4}{5q-6} \right\} <1. \end{aligned}$$

For such \(\rho ,\) we take \(0 < \alpha ,\) \(\beta < 1\) and \(0< \tilde{\rho }< 1\) so that

$$\begin{aligned}{} & {} \alpha = 1-\rho \frac{5q - 6}{2q},\ \beta = \frac{1 - \alpha }{2}=\rho \cdot \frac{5q-6}{4q} \text { and }\\{} & {} {{\tilde{\rho }}} = \frac{\sigma }{2 \left( \frac{p}{1-\rho } \right) ' } = \frac{1}{2} \cdot \frac{5 \tau }{5 \tau - 6} \cdot \frac{5 q - (1 - \rho )(5 q - 6)}{5 q}. \end{aligned}$$

Then we have the estimate

$$\begin{aligned}{} & {} \left| \int ^t_0 \right. \Pi _{N,f,g_1,g_2} ds\biggr \vert +\left| \int ^t_0 \Pi _{N,g_1,g_2,f} ds\right| \nonumber \\{} & {} \quad \leqq C \sum _{-1\leqq n,m,l} K_{(\alpha )}(N-n) K_{(\beta )}(N-m)K_{(\beta )}(N-l)\nonumber \\{} & {} \qquad \times \Vert \Delta _n f\Vert ^{\rho }_{L^\infty L^2} \left\{ \int ^T_0 \lambda ^{p}_n\Vert \Delta _n f\Vert ^{p}_q ds \right\} ^{\frac{1-\rho }{p}} \nonumber \\{} & {} \qquad \times \Vert \Delta _m g_1\Vert ^{1-{{\tilde{\rho }}}}_{L^\infty L^2} \left\{ \int ^T_0 \lambda ^\sigma _m \Vert \Delta _m g_1\Vert ^{\sigma }_\tau ds \right\} ^{\frac{{{\tilde{\rho }}}}{\sigma }}\nonumber \\{} & {} \qquad \times \Vert \Delta _l g_2\Vert ^{1-{{\tilde{\rho }}}}_{L^\infty L^2} \left\{ \int ^T_0 \lambda ^\sigma _l \Vert \Delta _l g_2\Vert ^{\sigma }_\tau ds \right\} ^{\frac{{{\tilde{\rho }}}}{\sigma }}, \end{aligned}$$
(3.2)

where \(C = C(q, \tau , \rho ).\)

We first prove a following proposition.

Proposition 3.3

Let f\(g_1\) and \(g_2\) be as in Lemma 3.2. Let the exponents \(2 \leqq a,\ b,\ c < \infty \) and \(0< \alpha ,\ \beta ,\ \gamma < 1\) satisfy \(1=1/a+1/b+1/c\) and \(1=\alpha + \beta + \gamma ,\) respectively. Then we have

$$\begin{aligned} \vert \Pi _{N,f,g_1,g_2} \vert\leqq & {} C \sum _{-1\leqq n,m,l} K_{(\alpha )}(N-n)\lambda ^\alpha _n\Vert \Delta _n f\Vert _a\\{} & {} \times K_{(\beta )}(N-m)\lambda ^\beta _m\Vert \Delta _m g_1\Vert _b K_{(\gamma )}(N-l)\lambda ^\gamma _l\Vert \Delta _l g_2\Vert _c, \end{aligned}$$

where C is a constant independent of \(N,\ f,\ g_1,\ g_2.\)

Proof of Proposition 3.3

Let us use the Bernstein inequality and imbedding \(B^0_{a,2} \hookrightarrow L^a\) to estimate

$$\begin{aligned}{} & {} \Vert f(\cdot -y)-f(\cdot )\Vert ^2_a\leqq C \left( \sum _{n\leqq N} \vert y \vert ^2\lambda ^2_n \Vert \Delta _n f\Vert ^2_a+\sum _{N<n} \Vert \Delta _n f\Vert ^2_a \right) \nonumber \\{} & {} \quad = C \left( \lambda ^{2(1-\alpha )}_N \vert y\vert ^2 \sum _{n\leqq N} \lambda ^{-2(1-\alpha )}_{N-n} (\lambda ^\alpha _n \Vert \Delta _n f\Vert _a)^2+\lambda ^{-2\alpha }_N \sum _{N<n} \lambda ^{2\alpha }_{N-n}(\lambda ^\alpha _n \Vert \Delta _n f\Vert _a)^2 \right) \nonumber \\{} & {} \quad \leqq C (\lambda ^{2(1-\alpha )}_N \vert y \vert ^2+\lambda ^{-2\alpha }_N) \sum _{-1\leqq n} (K_{(\alpha )}(N-n))^2 (\lambda ^\alpha _n \Vert \Delta _n f\Vert _a)^2. \end{aligned}$$
(3.3)

By the estimate (3.3), we have

$$\begin{aligned}{} & {} \int \vert {{\bar{h}}}_N (y)\vert \Vert f(\cdot -y)-f(\cdot )\Vert ^2_a dy \\{} & {} \quad \leqq C \left( \lambda ^{2(1-\alpha )}_N\int \vert {{\bar{h}}}_N (y)\vert \vert y\vert ^2 dy +\lambda ^{-2\alpha }_N \right) \sum _{-1\leqq n} (K_{(\alpha )}(N-n))^2 (\lambda ^\alpha _n \Vert \Delta _n f\Vert _a)^2\\{} & {} \quad \leqq C \lambda ^{-2\alpha }_N \sum _{-1\leqq n} (K_{(\alpha )}(N-n))^2 (\lambda ^\alpha _n\Vert \Delta _n f\Vert _a)^2. \end{aligned}$$

Hence we obtain

$$\begin{aligned}{} & {} \vert \Pi ^{(1)}_{N,f,g_1,g_2} \vert \\{} & {} \quad \leqq C \left\{ \int \vert {{\bar{h}}}_N (y)\vert \Vert f(\cdot -y)-f(\cdot )\Vert ^2_a dy \right\} ^{\frac{1}{2}} \left\{ \int \vert {{\bar{h}}}_N (y) \vert \Vert g_1(\cdot -y)-g_1(\cdot )\Vert ^2_b dy \right\} ^{\frac{1}{2}}\\{} & {} \qquad \times \left\{ \sum _{l\leqq N} \lambda ^2_l \Vert \Delta _l g_2\Vert ^2_c \right\} ^{\frac{1}{2}} \\{} & {} \quad \leqq C \lambda ^{1-(\alpha +\beta +\gamma )}_N \sum _{-1\leqq n,m,l} K_{(\alpha )}(N-n)\lambda ^\alpha _n\Vert \Delta _n f\Vert _a K_{(\beta )}(N-m)\lambda ^\beta _m\Vert \Delta _m g_1\Vert _b\\{} & {} \qquad \times K_{(\gamma )}(N-l)\lambda ^\gamma _l\Vert \Delta _l g_2\Vert _c. \end{aligned}$$

Analogously, we have

$$\begin{aligned} \vert \Pi ^{(2)}_{N,f,g_1,g_2} \vert\leqq & {} C \left\{ \sum _{N<n} \Vert \Delta _n f\Vert ^2_a \right\} ^{\frac{1}{2}} \left\{ \sum _{N<m} \Vert \Delta _m g_1\Vert ^2_b \right\} ^{\frac{1}{2}} \left\{ \sum _{l \leqq N} \lambda ^2_l \Vert \Delta _l g_2\Vert ^2_c \right\} ^{\frac{1}{2}}\\\leqq & {} C \sum _{-1\leqq n,m,l} K_{(\alpha )}(N-n)\lambda ^\alpha _n\Vert \Delta _n f\Vert _a K_{(\beta )}(N-m)\lambda ^\beta _m\Vert \Delta _m g_1\Vert _b \\{} & {} \times K_{(\gamma )}(N-l)\lambda ^\gamma _l\Vert \Delta _l g_2\Vert _c. \end{aligned}$$

\(\square \)

Proof of Lemma 3.2

In the case of (i), it follows from the Bernstein inequality that

$$\begin{aligned} \lambda ^\alpha _n \Vert \Delta _n f\Vert _q\leqq & {} C \lambda _n^{\alpha +3\rho \left( \frac{1}{2} - \frac{1}{q} \right) } \Vert \Delta _n f\Vert ^\rho _2 \Vert \Delta _n f\Vert ^{1-\rho }_q\nonumber \\= & {} C \lambda _n^{1-\rho } \Vert \Delta _n f\Vert ^{1-\rho }_q \Vert \Delta _n f\Vert ^\rho _2,\end{aligned}$$
(3.4)
$$\begin{aligned} \lambda ^\beta _m \Vert \Delta _m g_1\Vert _{2q'}\leqq & {} C \lambda ^{\beta +\left( \frac{3}{\tau } - \frac{3}{2} + \frac{3}{2q} \right) {{\tilde{\rho }}} + \frac{3}{2q}(1-{{\tilde{\rho }}})}_m \Vert \Delta _m g_1\Vert ^{{{\tilde{\rho }}}}_\tau \Vert \Delta _m g_1\Vert ^{1-{{\tilde{\rho }}}}_2\nonumber \\= & {} C \lambda ^{{{\tilde{\rho }}}}_m \Vert \Delta _m g_1\Vert ^{{{\tilde{\rho }}}}_\tau \Vert \Delta _m g_1\Vert ^{1-{{\tilde{\rho }}}}_2. \end{aligned}$$
(3.5)

Note the facts that

$$\begin{aligned}{} & {} 0<\rho< \frac{2q}{5q-6} \Rightarrow 0<\alpha ,\ \beta<1, \\{} & {} 0<\rho<\left( \frac{5\tau - 6}{2\tau }-\frac{3}{2q} \right) \frac{4q}{5q - 6} \Rightarrow 0<{{\tilde{\rho }}}<1. \end{aligned}$$

Hence, combining (3.4), (3.5) and Proposition 3.3, we have

$$\begin{aligned}{} & {} \int ^t_0 \vert \Pi _{N,f,g_1,g_2} \vert ds\\{} & {} \quad \leqq C \sum _{-1\leqq n,m,l} K_{(\alpha )}(N-n) K_{(\beta )}(N-m)K_{(\beta )}(N-l)\\{} & {} \qquad \times \int ^T_0 \lambda ^\alpha _n\Vert \Delta _n f\Vert _q \lambda ^\beta _m \Vert \Delta _m g_1\Vert _{2q'} \lambda ^\beta _l \Vert \Delta _l g_2\Vert _{2q'} ds\\{} & {} \quad \leqq C \sum _{-1\leqq n,m,l} K_{(\alpha )}(N-n) K_{(\beta )}(N-m)K_{(\beta )}(N-l)\\{} & {} \qquad \times \Vert \Delta _n f\Vert ^{\rho }_{L^\infty L^2} \left\{ \int ^T_0 \lambda ^{p}_n \Vert \Delta _n f\Vert ^{p}_q ds \right\} ^{\frac{1}{r}}\\{} & {} \qquad \times \Vert \Delta _m g_1\Vert ^{1-{{\tilde{\rho }}}}_{L^\infty L^2} \left\{ \int ^T_0 \lambda ^{2r' {{\tilde{\rho }}}}_m \Vert \Delta _m g_1\Vert ^{2r' {{\tilde{\rho }}}}_\tau ds \right\} ^{\frac{1}{2r'}} \\{} & {} \qquad \times \Vert \Delta _l g_2\Vert ^{1-{{\tilde{\rho }}}}_{L^\infty L^2} \left\{ \int ^T_0 \lambda ^{2r' {{\tilde{\rho }}}}_l \Vert \Delta _l g_2\Vert ^{2r' \tilde{\rho }}_\tau ds \right\} ^{\frac{1}{2r'}}, \end{aligned}$$

where \(1<r<\infty \) is given by

$$\begin{aligned} r=\frac{p}{1-\rho }. \end{aligned}$$

By \(2r'{{\tilde{\rho }}}=\sigma ,\) the term \(\int ^t_0 \vert \Pi _{N,f,g_1,g_2} \vert ds\) is bounded by the RHS of the inequality (3.1). Similarly, we can also estimate the term \(\int ^t_0 \vert \Pi _{N,g_1,g_2,f} \vert ds.\) Hence, we obtain (3.1).

In the case of (ii), it follows from the Bernstein inequality that

$$\begin{aligned} \lambda ^\alpha _n \Vert \Delta _n f\Vert _q \leqq C \lambda _n^{1-\rho } \Vert \Delta _n f\Vert ^{1-\rho }_q \Vert \Delta _n f\Vert ^\rho _2. \end{aligned}$$
(3.6)

We set \(0<{{\tilde{\theta }}} < 1\) as

$$\begin{aligned} \frac{1}{2q'} = \frac{{{\tilde{\theta }}}}{2} + \frac{1-\tilde{\theta }}{\tau }. \end{aligned}$$

Then, we have

$$\begin{aligned} \Vert \Delta _m g_1\Vert _{2q'} \leqq \Vert \Delta _m g_1\Vert ^{{{\tilde{\theta }}}}_2 \Vert \Delta _m g_1\Vert ^{1-{{\tilde{\theta }}}}_\tau . \end{aligned}$$
(3.7)

Hence, combining (3.6), (3.7) and Proposition 3.3, we have

$$\begin{aligned}{} & {} \int ^t_0 \vert \Pi _{N,f,g_1,g_2} \vert ds\nonumber \\{} & {} \quad \leqq C \sum _{-1\leqq n,m,l} K_{(\alpha )}(N-n) K_{(\beta )}(N-m)K_{(\beta )}(N-l)\nonumber \\{} & {} \qquad \times \int ^T_0 \lambda ^\alpha _n\Vert \Delta _n f\Vert _q \lambda ^\beta _m \Vert \Delta _m g_1\Vert _{2q'} \lambda ^\beta _l \Vert \Delta _l g_2\Vert _{2q'} ds\nonumber \\{} & {} \quad \leqq C \sum _{-1\leqq n,m,l} K_{(\alpha )}(N-n) K_{(\beta )}(N-m)K_{(\beta )}(N-l)\nonumber \\{} & {} \qquad \times \int ^T_0 \lambda ^{1-\rho }_n \Vert \Delta _n f\Vert ^{1-\rho }_q \Vert \Delta _n f\Vert ^\rho _2 \lambda ^\beta _m \Vert \Delta _m g_1\Vert ^{\tilde{\theta }}_2 \Vert \Delta _m g_1\Vert ^{1-{{\tilde{\theta }}}}_\tau \lambda ^\beta _l \Vert \Delta _l g_2\Vert ^{{{\tilde{\theta }}}}_2 \Vert \Delta _l g_2\Vert ^{1-\tilde{\theta }}_\tau ds\nonumber \\{} & {} \quad \leqq C \sum _{-1\leqq n,m,l} K_{(\alpha )}(N-n) K_{(\beta )}(N-m)K_{(\beta )}(N-l) \Vert \Delta _n f\Vert ^{\rho }_{L^\infty L^2}\nonumber \\{} & {} \qquad \times \left\{ \int ^T_0 \lambda ^{p}_n \Vert \Delta _n f\Vert ^{p}_q ds \right\} ^{\frac{1}{r}} \left\{ \int ^T_0 \lambda ^{2r'\beta }_m \Vert \Delta _m g_1\Vert ^{2 r' {{\tilde{\theta }}}}_2 \Vert \Delta _m g_1\Vert ^{2 r' (1 - {{\tilde{\theta }}})}_\tau ds \right\} ^{\frac{1}{2r'}}\nonumber \\{} & {} \qquad \times \left\{ \int ^T_0 \lambda ^{2r'\beta }_l \Vert \Delta _l g_2\Vert ^{2 r' {{\tilde{\theta }}}}_2 \Vert \Delta _l g_2\Vert ^{2 r' (1 - {{\tilde{\theta }}})}_\tau ds \right\} ^{\frac{1}{2r'}}, \end{aligned}$$
(3.8)

where the exponents r is given by

$$\begin{aligned} r=\frac{p}{1-\rho }. \end{aligned}$$

Here, note the fact

$$\begin{aligned} \rho< \frac{\tau }{\tau -2} \cdot \frac{4}{5q-6} \Rightarrow \sigma < 2r'(1-{{\tilde{\theta }}}). \end{aligned}$$

Then the Bernstein inequality gives

$$\begin{aligned}{} & {} \lambda ^{2r'\beta }_m \Vert \Delta _m g_1\Vert ^{2r' {{\tilde{\theta }}}}_2 \Vert \Delta _m g_1\Vert ^{2r'(1-{{\tilde{\theta }}})}_\tau \nonumber \\{} & {} \quad =\lambda ^{2r'\beta }_m \Vert \Delta _m g_1\Vert ^{2r' {{\tilde{\theta }}}}_2 \Vert \Delta _m g_1\Vert ^{2r'(1-{{\tilde{\theta }}})-\sigma }_\tau \Vert \Delta _m g_1\Vert ^\sigma _\tau \nonumber \\{} & {} \quad \leqq C \lambda ^{2r' \beta + \left( \frac{3}{2} - \frac{3}{\tau } \right) (2r'(1- {{\tilde{\theta }}})-\sigma )}_m \Vert \Delta _m g_1\Vert ^{2r'-\sigma }_2 \Vert \Delta _m g_1\Vert ^\sigma _\tau \nonumber \\{} & {} \quad = C \lambda ^\sigma _m \Vert \Delta _m g_1\Vert ^\sigma _\tau \Vert \Delta _m g_1\Vert ^{2r'-\sigma }_2. \end{aligned}$$
(3.9)

Combining (3.8) and (3.9), the term \(\int ^t_0 \vert \Pi _{N,f,g_1,g_2} \vert ds\) is bounded by the RHS of the inequality (3.2). Similarly, we can also estimate the term \(\int ^t_0 \vert \Pi _{N,g_1,g_2,f} \vert ds.\) Hence, we obtain (3.2). This completes the proof of Lemma 3.2. \(\square \)

4 Proof of main results

4.1 Proof of Theorem 2.1

Let \(X^{1,q}\) and \(Y^{1,\tau }\) be function spaces given by

$$\begin{aligned} X^{1,q}= & {} \{f;\ {\textrm{rot}}\ f \in L^q(\Omega ),\ \textrm{div} f = 0 \text { in } \Omega ,\ f \vert _{\partial \Omega } = 0\}, \\ Y^{1,\tau }= & {} \{g;\ {\textrm{rot}}\ g \in L^\tau (\Omega ),\ \textrm{div} g = 0 \text { in } \Omega ,\ g \cdot \nu \vert _{\partial \Omega } = 0\}. \end{aligned}$$

In the same way as [1], let us take approximating functions \(\{u_m\}^\infty _{m=1}\) and \(\{B_m\}^\infty _{m=1}\) of u and B with \(\textrm{div}u_m = 0,\) \(u_m \vert _{\partial \Omega }=0,\) \(\textrm{div}B_m = 0\) and \(B_m \cdot \nu \vert _{\partial \Omega }=0\) in the norm of \(L^\infty (0,T;L^2(\Omega )) \cap L^2(0,T;H^1(\Omega )) \cap L^p(0,T;X^{1,q})\) and \(L^\infty (0,T; L^2(\Omega ))\cap L^2(0,T;H^1(\Omega )) \cap L^\sigma (0,T;Y^{1,\tau }),\) respectively. Let \(f^\varepsilon \) donate the mollifier of f in time, i.e.,

$$\begin{aligned} \begin{array}{c} f^\varepsilon (t,x)= \int ^T_0 \rho _\varepsilon (t-s) f (s,x) ds\\ \text { with } \rho \in C^\infty _0([-1,1]),\ \int ^\infty _{-\infty } \rho (s) ds = 1 \text { and } \rho _\varepsilon (t) = \dfrac{1}{\varepsilon } \rho \left( \dfrac{t}{\varepsilon } \right) . \end{array} \end{aligned}$$

We take \((u_m)^\varepsilon \) and \((B_m)^\varepsilon \) as the test functions of the first and second equation in (1.1), respectively. Then adding the two identities, we obtain

$$\begin{aligned} L_u+L_B=-h_1+h_2, \end{aligned}$$

where

$$\begin{aligned} L_u= & {} (u(t),(u_m)^\varepsilon (t))-(u_0,(u_m)^\varepsilon (0))\\{} & {} -\int ^t_0 (u,\partial _t (u_m)^\varepsilon )ds+\int ^t_0(\nabla u,\nabla (u_m)^\varepsilon )ds,\\ L_B= & {} (B(t),(B_m)^\varepsilon (t))-(B_0,(B_m)^\varepsilon (0))\\{} & {} -\int ^t_0 (B,\partial _t (B_m)^\varepsilon )ds+\int ^t_0(\nabla B,\nabla (B_m)^\varepsilon )ds,\\ h_1= & {} h(u,u,(u_m)^\varepsilon )+h(u,B,(B_m)^\varepsilon ), \\ h_2= & {} h(B,B,(u_m)^\varepsilon )+h(B,u,(B_m)^\varepsilon ), \\ h(\phi ,\psi ,\chi )= & {} \int ^t_0((\phi \cdot \nabla )\psi ,\chi )ds. \end{aligned}$$

For the proof of the energy equality (1.2), it suffices to show that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} \left( \lim _{m \rightarrow \infty } h_1\right) =0,\ \lim _{\varepsilon \rightarrow 0} \left( \lim _{m \rightarrow \infty } h_2\right) =0. \end{aligned}$$

To show this, we first split \(h_1\) into following six terms:

$$\begin{aligned} h_1= & {} h(u,u,(u_m)^\varepsilon -u^\varepsilon )+h(u,u,u^\varepsilon -u)+h(u,u,u)\nonumber \\{} & {} +h(u,B,(B_m)^\varepsilon -B^\varepsilon )+h(u,B,B^\varepsilon -B)+h(u,B,B). \end{aligned}$$
(4.1)

Moreover, we can show that h(uuu) and h(uBB) are exactly zero since following facts holds by the same method as [1]:

$$\begin{aligned}{} & {} h(u,u_m,u_m) \rightarrow h(u,u,u)\ as\ m\rightarrow \infty ,\ h(u,u_m,u_m)=0\quad \text {for}\ \text {all}\ m, \\{} & {} h(u,B_m,B_m) \rightarrow h(u,B,B)\ as\ m\rightarrow \infty ,\ h(u,B_m,B_m)=0\quad \text {for}\ \text {all}\ m. \end{aligned}$$

In \(h_2,\) to control the error of the solution, we decompose \(h_2\) into the following forms:

$$\begin{aligned} h_2= & {} h(B,B-B_m,(u_m)^\varepsilon ) + h(B,B_m,(u_m)^\varepsilon )\nonumber \\{} & {} + h(B,u-u_m,(B_m)^\varepsilon ) + h(B,u_m,(B_m)^\varepsilon )\nonumber \\= & {} h(B,B-B_m,(u_m)^\varepsilon ) + h(B,u-u_m,(B_m)^\varepsilon )\nonumber \\{} & {} + h(B,B_m,(u_m)^\varepsilon ) - h(B,(B_m)^\varepsilon ,u_m)\nonumber \\= & {} h(B,B-B_m,(u_m)^\varepsilon ) + h(B,u-u_m,(B_m)^\varepsilon )\nonumber \\{} & {} + h(B,B_m,(u_m)^\varepsilon -u^\varepsilon ) + h(B,B_m,u^\varepsilon -u)\nonumber \\{} & {} + h(B,B_m-B,u) + h(B,B,u)\nonumber \\{} & {} - h(B,(B_m)^\varepsilon -B^\varepsilon ,u_m) - h(B,B^\varepsilon -B,u_m)\nonumber \\{} & {} - h(B,B,u_m-u) - h(B,B,u). \end{aligned}$$
(4.2)

Notice that all trilinear forms on the RHS of \(h_1\) and \(h_2\) in (4.1) and (4.2) are written in the following three types:

$$\begin{aligned} I_1=h(f_1,f_2,f_3),\ I_2=h(f, g_1,g_2),\ I_3=h(g_1,f,g_2), \end{aligned}$$

where

$$\begin{aligned}{} & {} f,\ f_i \in L^\infty (0,T;L^2(\Omega )) \cap L^2(0,T;H^1(\Omega )) \cap L^p(0,T;X^{1,q}),\quad i=1,2,3,\nonumber \\{} & {} g,\ g_i \in L^\infty (0,T;L^2(\Omega )) \cap L^2(0,T;H^1(\Omega )) \cap L^\sigma (0,T;Y^{1,\tau }),\quad i=1,2,\nonumber \\{} & {} {\textrm{div}}\ f=\textrm{div}\ f_i=\textrm{div}\ g=\textrm{div}\ g_i=0\ \text {in}\ (0,T) \times \Omega , \end{aligned}$$
(4.3)
$$\begin{aligned}{} & {} f \vert _{\partial \Omega }=f_i \vert _{\partial \Omega }=g \cdot \nu \vert _{\partial \Omega }=g_i \cdot \nu \vert _{\partial \Omega }=0. \end{aligned}$$
(4.4)

Here, by the virtue of (4.3) and (4.4), the following inequalities from the vector analysis hold:

$$\begin{aligned}{} & {} \Vert \nabla f\Vert _q \leqq C (\Vert {\textrm{rot}}\ f\Vert _q+\Vert f\Vert _q),\ \Vert \nabla f_i\Vert _q\leqq C(\Vert {\textrm{rot}}\ f_i\Vert _q+\Vert f_i\Vert _q),\quad i=1,2,3,\qquad \qquad \end{aligned}$$
(4.5)
$$\begin{aligned}{} & {} \Vert \nabla g\Vert _\tau \leqq C(\Vert {\textrm{rot}}\ g\Vert _\tau +\Vert g\Vert _\tau ),\ \Vert \nabla g_i\Vert _\tau \leqq C(\Vert {\textrm{rot}}\ g_i\Vert _\tau +\Vert g_i\Vert _\tau ),\quad i=1,2. \end{aligned}$$
(4.6)

Combining (4.5), (4.6) and Lemma 3.1, we see that the non-linear terms \(I_2\) and \(I_3\) are estimated in \(L^\infty (0,T;L^2(\Omega )) \cap L^2(0,T;H^1(\Omega )) \cap L^p(0,T;X^{1,q})\) and \(L^\infty (0,T;L^2(\Omega )) \cap L^2(0,T;H^1(\Omega )) \cap L^p(0,T;Y^{1,\tau }).\) Here, by \(\theta ,\ {{\tilde{\theta }}} \ne 1\) in the Lemma 3.1(ii), we can prove the convergence of the error term in \(L^2(0,T;H^1(\Omega )) \cap L^p(0,T;X^{1,q})\) and \(L^2(0,T;H^1(\Omega )) \cap L^p(0,T;Y^{1,\tau }).\) Notice that the terms \(\Vert f,f_i\Vert _{L^p L^q},\) \(i=1,2,3\) and \(\Vert g,g_i\Vert _{L^\sigma L^\tau },\) \(i=1,2\) appear because of RHS of (4.5) and (4.6). But it is easy to show that these terms are bounded by the norm of these functions in the Leray–Hopf class (see Remark 3 for details) . Moreover, it follows from the previous work [1] and (4.5) that the nonlinear term \(I_1\) is estimated in \(L^\infty (0,T;L^2(\Omega )) \cap L^2(0,T;H^1(\Omega )) \cap L^p(0,T;X^{1,q}).\) Hence, the solutions u and B satisfy the energy equality (1.2).

In the case \((0,T) \times {\mathbb {R}}^3,\) it should be noted that Theorem 2.1 is immediately established by the Lemma 3.1 because the terms \(\Vert f,f_i\Vert _{L^p L^q},\) \(i=1,2,3\) and \(\Vert g,g_i\Vert _{L^\sigma L^\tau },\) \(i=1,2\) in the RHS in (4.5) and (4.6) vanish. \(\Box \)

Remark 3

1. For \(2/r+3/s=3/2\) and \(0 < 1/r \leqq 1/2,\) the following imbedding holds

$$\begin{aligned} L^\infty (0,T;L^2(\Omega )) \cap L^2(0,T;H^1(\Omega )) \hookrightarrow L^r(0,T; L^s(\Omega )) \end{aligned}$$

with the estimate

$$\begin{aligned} \Vert f \Vert _{L^r L^s} \leqq C \Vert f \Vert _{L^\infty L^2}^\gamma \Vert f \Vert _{L^2H^1}^{1-\gamma }, \end{aligned}$$
(4.7)

for some \(0 \leqq \gamma <1.\) Here, the constant C is independent of f. Note that \(0< 1/r \leqq 1/2\) is equivalent to \(1/6 \leqq 1/s <1/2.\)

2. For \(2/\sigma +3/\tau -2=2-3/\tau \) with \(1/2< 1/\tau < 2/3,\) it follows from (4.7) that

$$\begin{aligned} L^\infty (0,T;L^2(\Omega )) \cap L^2(0,T;H^1(\Omega )) \hookrightarrow L^\sigma (0,T; L^s(\Omega )) \hookrightarrow L^\sigma (0,T; L^\tau (\Omega )), \end{aligned}$$

where \(2/\sigma +3/s = 3/2.\) Here, it should be noted that \(1/2< 1/\tau < 2/3\) is equivalent to \(0< 1/r = 1/\sigma <1/2.\)

3. For \(2/\sigma + 3/\tau -2 = 3/5\tau ,\) in the case \(5/12< 1/\tau < 2/3,\) it follows that

$$\begin{aligned} L^\infty (0,T;L^2(\Omega )) \cap L^2(0,T;H^1(\Omega )) \hookrightarrow L^\sigma (0,T; L^s(\Omega )) \hookrightarrow L^\sigma (0,T; L^\tau (\Omega )), \end{aligned}$$

where \(2/\sigma +3/s = 3/2.\) Here, it should be noted that \(5/12< 1/\tau < 2/3\) is equivalent to \(1/5< 1/r = 1/\sigma <1/2.\) In the case \(1/3 < 1/\tau \leqq 5/12,\) it follows that

$$\begin{aligned} L^\infty (0,T;L^2(\Omega )) \cap L^2(0,T;H^1(\Omega )) \hookrightarrow L^r(0,T; L^\tau (\Omega )) \hookrightarrow L^\sigma (0,T; L^\tau (\Omega )), \end{aligned}$$

where \(2/r+3/\tau = 3/2.\) Here, it should be noted that \(1/6< 1/3< 1/\tau = 1/s< 5/12 < 1/2.\)

4. By \(1-\gamma \ne 0,\) we can prove the convergence of the error terms in \(L^2(0,T; H^1(\Omega )).\)

4.2 Proof of Theorem 2.2

In the same way as [4], we take \((u_{\leqq N})_{\leqq N}\) and \((B_{\leqq N})_{\leqq N}\) as the test functions of the first and second equality in (1.1), respectively. Then we obtain

$$\begin{aligned}{} & {} \frac{1}{2}\Vert u_{\leqq N}(t)\Vert ^2_2-\frac{1}{2}\Vert (u_0)_{\leqq N}\Vert ^2_2+\int ^t_0\Vert \nabla u_{\leqq N}\Vert ^2_2ds\\{} & {} \qquad +\frac{1}{2}\Vert B_{\leqq N}(t)\Vert ^2_2-\frac{1}{2}\Vert (B_0)_{\leqq N}\Vert ^2_2+\int ^t_0\Vert \nabla B_{\leqq N}\Vert ^2_2ds\\{} & {} \quad =-\int ^t_0 \Pi _{N,u,u,u}ds+\int ^t_0 \Pi _{N,B,B,u}ds-\int ^t_0 \Pi _{N,u,B,B}ds+\int ^t_0 \Pi _{N,B,u,B}ds. \end{aligned}$$

By Lemma 3.2, we see that

$$\begin{aligned} \varlimsup _{N \rightarrow \infty } \int ^t_0 \vert \Pi _{N,B,B,u} \vert ds = \varlimsup _{N \rightarrow \infty } \int ^t_0 \vert \Pi _{N,u,B,B} \vert ds = \varlimsup _{N \rightarrow \infty } \int ^t_0 \vert \Pi _{N,B,u,B}\vert ds = 0. \end{aligned}$$

Moreover, the estimate of the term \(\int ^t_0 \Pi _{N,u,u,u}ds\) was proved in [3]. Hence, u and B satisfy the energy equality (1.2). \(\square \)