1 Introduction

In this paper we study the following Navier–Stokes equations with damping

$$\begin{aligned} {\left\{ \begin{array}{ll} u_t-\Delta u+(u\cdot \nabla )u+\alpha |u|^{\beta -1}u+\nabla p=0, \quad (x,t)\in \Omega \times (0,\infty ),\\ \text {div} u=0, \quad (x,t)\in \Omega \times (0, \infty ),\\ u|_{t=0}=u_0, \quad x\in \Omega ,\\ u|_{\partial \Omega }=0, \quad t\in (0,\infty ), \end{array}\right. } \end{aligned}$$
(1.1)

where \(\alpha >0\) and \(\beta \ge 1\) are constants and the unknown functions u(x, t) and p(x, t) are velocity and pressure of the fluid, respectively. Let \(\Omega =\mathbb {R}^3\) or \(\Omega \subset \mathbb {R}^3\) be a bounded domain with a sufficiently smooth boundary. When \(\Omega \) is the whole space, the boundary condition of (1.1) is replaced by \(|u(x)|\rightarrow 0 (|x|\rightarrow \infty )\).

The Navier–Stokes equations with damping is a modified Navier–Stokes equations and describe the flow affected by the resistance to the motion of fluid such as porous media flow and the flow with drag or friction effect (see [1] and its references). The damping term \(\alpha |u|^{\beta -1}u\) in (1.1) represents the resistance force, which for instance, appears in the flow of cerebrospinal fluid inside the porous brain tissues [8]. Mathematically, (1.1) is a regularization of the Naver-Stokes equations, which means that the weak solutions of (1.1) converges to a weak solution of the Navier–Stokes equations as \(\alpha \rightarrow 0\) (see [6] for more details). For the problem (1.1), finding the restrictions on parameters \(\alpha \) and \(\beta \) to guarantee the existence of global strong solutions and uniqueness of weak solutions is one of the most interesting things.

The problem (1.1) has been studied intensively in recent years.

In case of the bounded domains, [11] asserted that (1.1) has a global strong solution if \(\beta \ge 7/2\) and \(u_0 \in H^1 \cap L^{\beta +1}\), but we wonder if the proof of this result is correct. Indeed, in a priori estimates multiplying \(-\Delta u\), they thought \(-(\Delta u, \nabla p)\) is zero although the normal component of \(-\Delta u\) does not vanish on \(\partial \Omega \), when \(\Omega \) is a bounded domain. Even if we multiply Au, the pressure could be vanished but on the contrary the product \((|u|^{\beta -1}u, Au)\) becomes an obstacle and it should be estimated by \((|u|^{\beta -1}u, Au)\le \Vert u\Vert ^{\beta }_{2\beta } \Vert Au\Vert \). Considering these points, [5] proved the global existence of strong solutions for \(\beta =3, \alpha >1/4\) or \(3<\beta <5\) with \(u_0\in H^1\). After that, they proved the global \(L^2-\)boundedness of space and time derivatives of the weak solutions, under the restriction \(u_0\in H^2\) with \(\beta =3, \alpha >1/4\) or \(\beta >3\) in [6]. On the other hand, the uniqueness of weak solutions was proved for \(\beta \ge 4\) in [9] and finally, proved for \(\beta =3, \alpha \ge 1/4\) or \(\beta >3\) in [6].

In case of the whole space, [1, 12] presented the results for the existence of global strong solutions with the restrictions \(\beta \ge 7/2, u_0\in H^1 \cap L^{\beta +1}\)([1]), \(\beta >3, u_0\in H^1 \cap L^{\beta +1}\) ([12]) and [13] showed that local strong solutions could be extended globally when \(\beta \ge 3, \alpha =1, u_0\in H^1\). Their results are based on the estimations for the smooth solutions([1, 13]) or the Galerkin approximations([12]) taking \(-\Delta u\) as a test function. Here, note that we aren’t able to construct the Galerkin basis with the eigenfunctions of \(-\Delta u\) in the whole space, so we should recognize that a prior estimates of [12] is only applicable to the smooth solutions not to the Galerkin approximations. In conclusion, it is understood that in [1, 12, 13] they proved the possibility of extension for the local strong solution. Then, the question of whether there exist local strong solutions, is naturally raised. The local existence of the strong solutions of (1.1) couldn’t be found in any literature and we don’t think it is trivial. In fact, to prove the local existence of strong solutions to (1.1), considering the damping term as the external force and following the same argument with the Navier–Stokes equations, we encounter with the difficulties because the damping term is not contained in \(L^2(0,T;L^2)\) when \(\beta \) is large.

Besides, global regularity for the 3D inhomogeneous damped Navier-Stokes equations was studied in [7].

To the best of our knowledge, it seems that there is no result for the existence of local strong solutions when \(\beta \ge 5\). Also in the bounded domains, the restriction \(\beta <5\) is essential for the existence of the strong solutions due to the problems occurred by the damping term(see [5] for details). So we are going to study the local regularity of weak solutions to complete the proof for the global existence of strong solutions in the whole space as well as to improve the restriction \(\beta <5\) for the global existence of strong solutions in bounded domains. We employ the approach used in [3, 4], which is associated with the Stokes semigroup, however, to treat the damping term is a difficulty for our case, when \(\beta \) is large.

\(\Vert \cdot \Vert _X\), \(\Vert \cdot \Vert _{X \rightarrow Y}\), \(\Vert \cdot \Vert _q\), \(\Vert \cdot \Vert _{q,s;T}\) and \(\Vert \cdot \Vert _{q,s;a, b}\) denote the norms of the Banach space X, linear operator from X to Y, \(L^q(\Omega )^n\), \(L^s(0,T;L^q(\Omega )^n)\) and \(L^s(a,b;L^q(\Omega )^n)\), respectively and \(\Vert \cdot \Vert =\Vert \cdot \Vert _2\).

We define the following function spaces

$$\begin{aligned}{} & {} C_{0,\sigma }^\infty (\Omega ):=\{u\in C_0^\infty (\Omega )^3, \text {div}u=0\},\\{} & {} V:=\overline{C_{0,\sigma }^\infty (\Omega )}^{\Vert \cdot \Vert _{W^{1,2}(\Omega )^3}},\\{} & {} L_\sigma ^q(\Omega ):=\overline{C_{0,\sigma }^\infty (\Omega )}^{\Vert \cdot \Vert _q}, 1< q<\infty \end{aligned}$$

and H denotes \(L^2_\sigma (\Omega )\).

\(P_q:L^q(\Omega )^3\rightarrow L_\sigma ^q(\Omega )\) \((1<q<\infty )\) denotes the Helmholtz projection and let \(A_q:=-P_q\Delta \) be the Stokes operator with domain \(D(A_q)\), \(A_q^{1/2}\) be the square root of \(A_q\), and \(S_q(t):=e^{-tA_q}\) be the semigroup generated by \(-A_q\). We may write \(P_q=P\), \(A_q=A\), and \(S_q=S\), if there is no misunderstanding. \(\mathbb {N}\) is the set of natural numbers and \(\mathbb {N}_0=\mathbb {N}\cup \{0\}\).

Definition 1.1

Let \(0<T<\infty \). A function \(u\in L^\infty (0,T;H)\cap L^2(0,T;V)\cap L^{\beta +1}(0,T;L^{\beta +1}(\Omega )^3)\) is called a weak solution of (1.1) on [0, T], if it satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{d}{dt}(u,v)+(\nabla u, \nabla v)+((u\cdot \nabla ) u, v)+\alpha (|u|^{\beta -1}u, v)=0, \\ \quad \quad \quad \quad \forall v\in V\cap L^{\beta +1}(\Omega )^3, t\in (0, T),\\ u(0)=u_0, \end{array}\right. } \end{aligned}$$
(1.2)

where \((f, g)=\int _\Omega f\cdot g dx\). And u is called a strong solution of (1.1) on [0, T], when u is a weak solution of (1.1) on [0, T] and \(u \in L^\infty (0,T;V \cap L^{1.5(\beta -1)}(\Omega )^3)\cap L^2(0, T; D(A_2))\).

Our main results are the following theorems and corollary.

Theorem 1.1

Let \(\beta \ge 3\) and \(u_0 \in H \cap L^{1.5(\beta -1)}_\sigma (\Omega )\). Then, there exists an interval [0, T], only depending on \(\Omega , \alpha , \beta \) and \(u_0\), with the following property: There is a unique weak solution u of (1.1) on [0, T], satisfying

$$\begin{aligned} u\in C([0,T]; H\cap L_\sigma ^{1.5(\beta -1)}(\Omega )), \end{aligned}$$
(1.3)

with \(u(0)=u_0\) and

$$\begin{aligned}{} & {} t^{1/2}A^{1/2}u\in C([0,T];H\cap L_\sigma ^{1.5(\beta -1)}(\Omega )), \end{aligned}$$
(1.4)
$$\begin{aligned}{} & {} t^{\frac{\beta (2kp-3)-2kp+2p+3}{2p(\beta -1)}}A^{k} u\in C([0, T]; L^p_\sigma (\Omega )), 1.5(\beta -1)< p <\infty , 0\le k\le 1/2,\nonumber \\ \end{aligned}$$
(1.5)

with values zero at \(t=0\). Moreover,

$$\begin{aligned} u\in L^2(\epsilon , T; D(A_2))\cap C([\epsilon , T]; V\cap L_\sigma ^{1.5(\beta -1)}(\Omega )) \end{aligned}$$
(1.6)

for any \(0<\epsilon <T\).

Theorem 1.2

Let \(\beta \ge 3\) and \(u_0\in V\cap L_\sigma ^{\bar{\beta }}(\Omega )\), where \({\bar{\beta }}=1.5(\beta -1)\) in case of \(\beta \ne 5\) and \(6<{\bar{\beta }}<\infty \) in case of \(\beta = 5\). Then, there exist an interval [0, T], only depending on \(\Omega , \alpha , \beta \) and \(u_0\), and a strong solution u of (1.1) on [0, T] satisfying

$$\begin{aligned} u\in L^2(0, T;D(A_2))\cap C([0, T]; V\cap L^{\bar{\beta }}_\sigma (\Omega )). \end{aligned}$$
(1.7)

Corollary 1.1

Let \(\beta =3, \alpha \ge 1/4\) or \(\beta>3, \alpha >0\).

a) Let \(\Omega =\mathbb {R}^3\) and \(u_0\in V\cap L^{\bar{\beta }}(\Omega )\), where \({\bar{\beta }}=1.5(\beta -1)\) in case of \(\beta \ne 5\) and \(6<{\bar{\beta }}<\infty \) in case of \(\beta = 5\). Then there exists a unique strong solution u of (1.1), such that

$$\begin{aligned} u\in L^2(0,T;D(A_2))\cap C([0,T]; V \cap L_\sigma ^{\bar{\beta }}(\Omega )) \end{aligned}$$
(1.8)

for all \(0<T<\infty \).

b) Let \(\Omega \) be a bounded domain of \(\mathbb {R}^3\) with a sufficiently smooth boundary and \(\beta =5, u_0 \in V\cap L_\sigma ^{\bar{\beta }}(\Omega )(6<{\bar{\beta }}<\infty )\). Then there exists a unique strong solution u of (1.1), such that

$$\begin{aligned} u\in L^2(0,T;D(A_2))\cap C([0,T]; V\cap L^{\bar{\beta }}_\sigma (\Omega )) \end{aligned}$$
(1.9)

for all \(0<T<\infty \).

Remark 1.1

Corollary 1.1 extends the upper restriction on \(\beta \), from \(\beta <5\)([5]) to \(\beta \le 5\), for the existence of global strong solutions to (1.1) in the bounded domains. However, the problem of whether or not (1.1) has a global strong solution in case \(\Omega \) is a bounded domain and \(\beta >5\), still remains open.

2 Proof of Theorems 1.1 and 1.2

First, we recall some previous results on the properties of the Stokes semigroup and regularity and uniqueness of the weak solutions to (1.1).

Lemma 2.1

[2] Let \(\Omega \) be the whole space or a bounded domain of \(\mathbb {R}^3\) with sufficiently smooth boundary and let \(k\ge 0\), \(1<p\le q<\infty \). Then, \(S(t)(t\ge 0)\) is an analytic semigroup and

$$\begin{aligned}{} & {} \Vert t^{(2k+3/p-3/q)/2}A^kS(t)u\Vert _q\le C \Vert u\Vert _p, u\in L_\sigma ^p(\Omega ),t>0, \end{aligned}$$
(2.1)
$$\begin{aligned}{} & {} \lim _{t\rightarrow 0+} \Vert t^{(2k+3/p-3/q)/2} A^k S(t)u\Vert _q=0, u\in L_\sigma ^p(\Omega ), 0<(2k+3/p-3/q)/2\le 1\nonumber \\ \end{aligned}$$
(2.2)

are satisfied, where \(C>0\) is only depends on \(\Omega , k, p\) and q.

Lemma 2.2

[6, (i) of Theorem 1.2] Let \(\Omega \) be a bounded domain of \(\mathbb {R}^3\) with sufficiently smooth boundary and let \(\beta =3, \alpha \ge 1/4\) or \(\beta>3, \alpha >0\). If \(u_0\in D(A_2)\), then (1.1) has a weak solution such that

$$\begin{aligned} u\in W_{loc}^{1, \infty }(0, \infty ;H)\cap W_{loc}^{1, 2}(0, \infty ; V). \end{aligned}$$
(2.3)

Remark 2.1

In [6, (i) of Theorem 1.2], the restriction on \(\alpha \) and \(\beta \), was \(\beta =3, \alpha >1/4\) or \(\beta>3, \alpha >0\), but \(\beta =3, \alpha =1/4\) is also possible for (2.3).

Lemma 2.3

[6, Corollary 2.1] Let \(\Omega \) be the whole space or a bounded domain of \(\mathbb {R}^3\) with sufficiently smooth boundary and let \(\beta =3, \alpha \ge 1/4\) or \(\beta>3, \alpha >0\). If \(u_0\in H\), then the weak solution of (1.1) is unique.

Consider the following approximations

$$\begin{aligned}{} & {} u^{(0)}(t):=S(t)u_0, \end{aligned}$$
(2.4)
$$\begin{aligned}{} & {} u^{(n)}(t):=u^{(0)}+Gu^{(n-1)}(t)+\alpha Bu^{(n-1)} (t), n\in \mathbb {N} \end{aligned}$$
(2.5)

where

$$\begin{aligned} Gu(t):= & {} -\int _0^t{S(t-s)P((u\cdot \nabla ) u(s))ds}, \end{aligned}$$
(2.6)
$$\begin{aligned} Bu(t):= & {} -\int _0^t{S(t-s)P(|u|^{\beta -1}u(s))ds}. \end{aligned}$$
(2.7)

The following lemmas are useful to prove the Theorem 1.1 and note that the constant C in the inequalities below, only depends on \(\Omega , \alpha \) and \(\beta \).

Lemma 2.4

Let \(\beta \ge 3\) and \(u_0 \in H\cap L_\sigma ^{1.5(\beta -1)}(\Omega )\). Then, for any \(K>0\), there exists \(0<T<1\) such that

$$\begin{aligned}{} & {} \begin{aligned}&t^{1/2}A^{1/2}u^{(n)}\in C([0, T]; L_\sigma ^p(\Omega )),\\&{\Vert t^{1/2}A^{1/2}u^{(n)}\Vert _p} \le K, 2\le p \le 1.5(\beta -1), \end{aligned} \end{aligned}$$
(2.8)
$$\begin{aligned}{} & {} \begin{aligned}&t^\frac{(2p-3)\beta +3}{2p\beta (\beta -1)} u^{(n)} \in C([0,T]; L^{p\beta }_\sigma (\Omega )), \\&\Vert t^\frac{(2p-3)\beta +3}{2p\beta (\beta -1)} u^{(n)}\Vert _{p\beta } \le K, 2\le p\le 1.5(\beta -1), \end{aligned} \end{aligned}$$
(2.9)

with values zero at \(t=0\), for all \(n\in \mathbb {N}_0\).

Proof

It is sufficient to prove this lemma when \(K>0\) is small enough. From (2.2) and \(u_0 \in H\cap L_\sigma ^{1.5(\beta -1)}(\Omega )\), we can choose some \(0<T<1\) such that

$$\begin{aligned} \Vert t^{1/2}A^{1/2}S(t)u_0\Vert _p \le K/2, 2\le p\le 1.5(\beta -1) \end{aligned}$$
(2.10)

and

$$\begin{aligned} \Vert t^\frac{(2p-3)\beta +3}{2p\beta (\beta -1)} S(t)u_0\Vert _{p\beta } \le K/2, 2\le p\le 1.5(\beta -1) \end{aligned}$$
(2.11)

are fulfilled for all \(0 \le t \le T\).

For the same T, we will show that

$$\begin{aligned} {\Vert t^{1/2}A^{1/2}u^{(n)}(t)\Vert _p} \le K, 2\le p \le 1.5(\beta -1), 0\le t\le T \end{aligned}$$
(2.12)

and

$$\begin{aligned} \Vert t^\frac{(2p-3)\beta +3}{2p\beta (\beta -1)} u^{(n)}(t)\Vert _{p\beta } \le K, 2\le p\le 1.5(\beta -1), 0\le t\le T \end{aligned}$$
(2.13)

are satisfied for all \(n\in \mathbb {N}_0\), by induction.

If \(n=0\), (2.12) and (2.13) are trivial by (2.10) and (2.11).

Next, assume that (2.12) and (2.13) are satisfied for \(n\in \mathbb {N}_0\), and prove they are true for \(n+1\) by estimating \(\Vert A^kGu^{(n)}(t)\Vert _p, \Vert Gu^{(n)}(t)\Vert _{p\beta }, \Vert A^k Bu^{(n)}(t)\Vert _p\) and \(\Vert Bu^{(n)}(t)\Vert _{p\beta }\) (\(k=0\) or \(1/2, 2 \le p \le 1.5(\beta -1), 0 \le t \le T\)) using Hölder’s inequality, (2.1), (2.12) and (2.13) for n.

Setting \(\frac{1}{q}=\frac{1}{3\beta }+\frac{1}{p}\),

$$\begin{aligned} \Vert A^kGu^{(n)}(t)\Vert _p&\le \int _0^t {\Vert A^k S(t-s)P((u^{(n)} \cdot \nabla ) u^{(n)}(s))\Vert _pds} \nonumber \\&\le C \int _0^t (t-s)^{-(2k+\frac{3}{q}-\frac{3}{p})/2} \Vert (u^{(n)} \cdot \nabla ) u^{(n)}(s)\Vert _q ds\nonumber \\&\le C \int _0^t (t-s)^{-(2k+\frac{3}{q}-\frac{3}{p})/2} \Vert u^{(n)}(s)\Vert _{3\beta } \Vert A^{1/2}u^{(n)}(s)\Vert _p ds\nonumber \\&\le C \int _0^t (t-s)^{-k-\frac{1}{2\beta }} s^{-\frac{\beta +1}{2\beta (\beta -1)}-\frac{1}{2}} K^2 ds\nonumber \\&\le C t^{-k + \frac{\beta -3}{2(\beta -1)}}K^2 \end{aligned}$$
(2.14)

holds. And if \(\frac{1}{q}=\frac{1}{3}+\frac{1}{p\beta }\), then we have

$$\begin{aligned} \Vert Gu^{(n)}(t)\Vert _{p\beta }&\le \int _0^t \Vert S(t-s) P((u^{(n)}\cdot \nabla ) u^{(n)} (s))\Vert _{p\beta } ds\nonumber \\&\le C \int _0^t (t-s)^{-(\frac{3}{q}-\frac{3}{p\beta })/2} \Vert u^{(n)} (s)\Vert _{p\beta } \Vert A^{1/2} u^{(n)}(s)\Vert _3 ds\nonumber \\&\le C \int _0^t (t-s)^{-\frac{1}{2}} s^{- \frac{(2p-3)\beta +3}{2p\beta (\beta -1)} -\frac{1}{2}}K^2 ds\nonumber \\&\le C t^{-\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}}K^2. \end{aligned}$$
(2.15)

We obtain

$$\begin{aligned} \Vert A^k B u^{(n)} (t) \Vert _p&\le \int _0^t \Vert A^k S(t-s) P (| u^{(n)}|^{\beta -1} u^{(n)} (s)) \Vert _p ds\nonumber \\&\le C \int _0^t (t-s)^{-(2k+\frac{3}{q}-\frac{3}{p})/2} \Vert u^{(n)}(s)\Vert _{q\beta }^{\beta } ds\nonumber \\&\le C \int _0^t (t-s)^{-(2k+\frac{3}{q}-\frac{3}{p})/2} s^{-\frac{(2q-3)\beta +3}{2q(\beta -1)}} K^\beta ds\nonumber \\&\le C K^\beta t^{-k+\frac{3\beta -2p-3}{2p(\beta -1)}}, \end{aligned}$$
(2.16)

where \(q=\frac{3p+4}{p+3}\). Finally,

$$\begin{aligned} \Vert Bu^{(n)}(t) \Vert _{p\beta }&\le \int _0^t \Vert S(t-s)P(|u^{(n)}|^{\beta -1}u^{(n)}(s))\Vert _{p\beta } ds\nonumber \\&\le C \int _0^t (t-s)^{-(\frac{3}{2}-\frac{3}{p\beta })/2} \Vert u^{(n)}(s)\Vert _{2\beta }^{\beta }ds\nonumber \\&\le C \int _0^t (t-s)^{-\frac{3}{4}+\frac{3}{2p\beta }} s^{-\frac{\beta +3}{4(\beta -1)}} K^\beta ds\nonumber \\&\le C K^\beta t^{-\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}} \end{aligned}$$
(2.17)

is satisfied.

When \(2\le p \le 1.5(\beta -1)\) and \(0 \le t \le T\), the inequalities (2.10), (2.14) and (2.16) with \(k=1/2\) lead to

$$\begin{aligned} \Vert t^{1/2}&A^{1/2}u^{(n+1)}(t)\Vert _p\nonumber \\&\le \Vert t^{1/2}A^{1/2}S(t)u_0\Vert _p +\Vert t^{1/2}A^{1/2}Gu^{(n)}(t)\Vert _p+|\alpha |\Vert t^{1/2}A^{1/2}Bu^{(n)}(t)\Vert _p\nonumber \\&\le K/2 +C(t^{\frac{\beta -3}{2(\beta -1)}}K^2+t^{\frac{3\beta -2p-3}{2p(\beta -1)}}K^\beta )\nonumber \\&\le K/2+C(K^2+K^\beta )\nonumber \\&\le K. \end{aligned}$$
(2.18)

Also, (2.11), (2.15) and (2.17) imply

$$\begin{aligned} \Vert t^{\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}}&u^{(n+1)} (t)\Vert _{p\beta }\nonumber \\&\le \Vert t^{\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}}S(t)u_0\Vert _{p\beta }+ \Vert t^{\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}}Gu^{(n)}(t)\Vert _{p\beta }\nonumber \\&\qquad +|\alpha |\Vert t^{\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}}Bu^{(n)}(t)\Vert _{p\beta }\nonumber \\&\le K/2+ C(K^2+K^\beta )\nonumber \\&\le K, \end{aligned}$$
(2.19)

for \(2\le p \le 1.5(\beta -1)\) and \(0 \le t \le T\). In (2.18) and (2.19), we use the fact \(C(K^2+K^\beta )\le K/2\) because K is small enough. (2.18) and (2.19) complete the induction.

The continuities

$$\begin{aligned}{} & {} t^{1/2}A^{1/2}u^{(n)}(t)\in C([0, T]; L_\sigma ^p(\Omega )), 2\le p\le 1.5(\beta -1),\\{} & {} t^\frac{(2p-3)\beta +3}{2p\beta (\beta -1)} u^{(n)}(t) \in C([0,T]; L^{p\beta }_\sigma (\Omega )), 2\le p \le 1.5(\beta -1) \end{aligned}$$

with values zero at \(t=0\), are proved by the same argument as [3] using (2.12) and (2.13). \(\square \)

To show the uniformly convergence of \(\{u^{(n)}\}\), estimate the difference

$$\begin{aligned} w^{(n)}(t):=u^{(n)}(t)-u^{(n-1)}(t), n\in \mathbb {N}. \end{aligned}$$
(2.20)

Lemma 2.5

Let \(\beta \ge 3\) and \(u_0 \in H\cap L_\sigma ^{1.5(\beta -1)}(\Omega )\). Then for any \(K>0\), there exists \(0<T<1\) such that

$$\begin{aligned}{} & {} \sup _{2\le p \le 1.5(\beta -1), 0 \le t \le T} \Vert w^{(n)}(t) \Vert _p \le (C_0K)^n, \end{aligned}$$
(2.21)
$$\begin{aligned}{} & {} \sup _{2\le p \le 1.5(\beta -1), 0 \le t \le T} \Vert t^{1/2}A^{1/2}w^{(n)}(t)\Vert _p \le (C_0K)^n, \end{aligned}$$
(2.22)
$$\begin{aligned}{} & {} \sup _{2\le p \le 1.5(\beta -1), 0 \le t \le T} \Vert t^{\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}} w^{(n)}(t) \Vert _{p\beta }\le (C_0K)^n, \end{aligned}$$
(2.23)

for all \(n\in \mathbb {N}\), where \(C_0>0\) only depends on \(\Omega , \alpha \) and \(\beta \).

Proof

We may assume that \(K>0\) is sufficiently small. By Lemma 2.4 there exists \(0<T<1\) satisfying (2.12) and (2.13). For the same T, prove that (2.21)–(2.23) are fulfilled for all \(n\in \mathbb {N}\) by induction. Note that \(C_0\) is determined in process of the proof.

If \(n=1\), \(w^{(1)}(t)=Gu^{(0)}(t)+\alpha Bu^{(0)}(t)\) satisfies

$$\begin{aligned} \Vert t^{k}A^{k}w^{(1)}(t)\Vert _p&\le \Vert t^{k}A^{k} Gu^{(0)}(t)\Vert _p+|\alpha | \Vert t^{k}A^{k} Bu^{(0)} (t)\Vert _p\nonumber \\&\le C(t^{\frac{\beta -3}{2(\beta -1)}}K^2+t^{\frac{3\beta -2p-3}{2p(\beta -1)}}K^\beta )\nonumber \\&\le C_0K, \end{aligned}$$
(2.24)

by (2.14) and (2.16) with \(k=0\) or \(k=1/2\). Also,

$$\begin{aligned} \Vert t^{\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}}w^{(1)} (t)\Vert _{p\beta }&\le \Vert t^{\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}} G^{(0)}(t)\Vert _{p\beta } + |\alpha | \Vert t^{\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}} B^{(0)}(t)\Vert _{p\beta }\nonumber \\&\le C(K^2+K^\beta )\nonumber \\&\le C_0K \end{aligned}$$
(2.25)

holds by (2.15) and (2.17), for any \(2\le p \le 1.5(\beta -1)\) and \(0 \le t\le T\), so (2.21)–(2.23) are true.

Next, assume that (2.21)–(2.23) are satisfied for \(n\in \mathbb {N}\), and prove them for \(n+1\). For this purpose, we need to estimate \(\Vert A^k(G^{(n)}(t)-G^{(n-1)}(t)) \Vert _p, \Vert Gu^{(n)}(t)-Gu^{(n-1)}(t)\Vert _{p\beta }, \Vert A^k(Bu^{(n)}(t)-B^{(n-1)}(t))\Vert _p\) and \(\Vert Bu^{(n)}(t)-Bu^{(n-1)}(t)\Vert _{p\beta }\), where \(k=0\) or \(k=1/2\), \(2 \le p \le 1.5(\beta -1)\) and \(0 \le t \le T\). In estimating them, we use (2.1), (2.12), (2.13), (2.21)–(2.23) for n and Höder’s inequality, and note that q is chosen carefully so that the integral like \(\int _0^t (t-s)^{a} s^{b}ds\), obtained finally in calculation, is finite.

Setting \(\frac{1}{q}=\frac{1}{2\beta }+\frac{1}{p}\), we have

$$\begin{aligned}{} & {} \Vert A^k (Gu^{(n)}(t)-Gu^{(n-1)}(t))\Vert _p \nonumber \\{} & {} \quad \le \int _0^t \Vert A^k S(t-s) P((u^{(n)}\cdot \nabla ) w^{(n)}(s)+(w^{(n)}\cdot \nabla ) u^{(n-1)}(s))\Vert _p ds\nonumber \\{} & {} \quad \le C \int _0^t (t-s)^{-(2k+\frac{3}{q}-\frac{3}{p})/2}(\Vert (u^{(n)}\cdot \nabla ) w^{(n)}(s)\Vert _q+\Vert (w^{(n)}\cdot \nabla ) u^{(n-1)}(s)\Vert _q) ds\nonumber \\{} & {} \quad \le C \int _0^t (t-s)^{-k-\frac{3}{4\beta }} (\Vert u^{(n)}(s)\Vert _{2\beta } \Vert A^{1/2}w^{(n)}(s)\Vert _p+ \Vert w^{(n)}(s)\Vert _{2\beta } \Vert A^{1/2}u^{(n-1)}(s)\Vert _p)ds\nonumber \\{} & {} \quad \le C \int _0^t (t-s)^{-k -\frac{3}{4\beta }} s^{-\frac{\beta +3}{4\beta (\beta -1)}-\frac{1}{2}} (C_0K)^n Kds\nonumber \\{} & {} \quad \le C K (C_0K)^n t^{-k+\frac{\beta -3}{2(\beta -1)}}. \end{aligned}$$
(2.26)

And \(\frac{1}{q}=\frac{1}{p\beta }+\frac{1}{3}\) leads to

$$\begin{aligned} \Vert G&u^{(n)}(t)-Gu^{(n-1)}(t) \Vert _{p\beta } \nonumber \\&\le \int _0^t \Vert S(t-s)P((u^{(n)}\cdot \nabla ) w^{(n)}(s)+(w^{(n)}\cdot \nabla ) u^{(n-1)}(s))\Vert _{p\beta }ds\nonumber \\&\le C \int _0^t (t-s)^{-(\frac{3}{q}-\frac{3}{p\beta })/2} (\Vert (u^{(n)}\cdot \nabla ) w^{(n)}(s)\Vert _q+\Vert (w^{(n)}\cdot \nabla ) u^{(n-1)}(s)\Vert _q )ds\nonumber \\&\le C \int _0^t (t-s)^{-\frac{1}{2}} (\Vert u^{(n)}(s)\Vert _{p\beta } \Vert A^{1/2}w^{(n)}(s)\Vert _3+\Vert w^{(n)}(s)\Vert _{p\beta }\Vert A^{1/2}u^{(n-1)}(s)\Vert _3 ds\nonumber \\&\le C \int _0^t (t-s)^{-\frac{1}{2}} s^{-\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}-\frac{1}{2}} (C_0K)^n K ds\nonumber \\&\le CK (C_0K)^n t^{-\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}}. \end{aligned}$$
(2.27)

If \(\frac{1}{q}=\frac{1}{p}+\frac{\beta -1}{3\beta }\),

$$\begin{aligned} \Vert A^k&(Bu^{(n)}(t)-Bu^{(n-1)}(t))\Vert _p \nonumber \\&\le \int _0^t \Vert A^k S(t-s) P(|u^{(n)}|^{\beta -1}u^{(n)}(s)-|u^{(n-1)}|^{\beta -1}u^{(n-1)}(s))\Vert _p ds\nonumber \\&\le C \int _0^t (t-s)^{-(2k+\frac{3}{q}-\frac{3}{p})/2} \Vert |u^{(n)}|^{\beta -1}u^{(n)}(s)-|u^{(n-1)}|^{\beta -1}u^{(n-1)}(s)\Vert _q ds\nonumber \\&\le C \int _0^t (t-s)^{-k-\frac{\beta -1}{2\beta }} \Vert w^{(n)}(s)\Vert _p (\Vert u^{(n)}(s)\Vert _{3\beta }^{\beta -1}+\Vert u^{(n-1)}(s)\Vert _{3\beta }^{\beta -1}) ds\nonumber \\&\le C \int _0^t (t-s)^{-k-\frac{\beta -1}{2\beta }} s^{-\frac{\beta +1}{2\beta }} (C_0K)^n K^{\beta -1} ds\nonumber \\&\le CK^{\beta -1}(C_0K)^n t^{-k} \end{aligned}$$
(2.28)

holds and if \(\frac{1}{q}=\frac{1}{p\beta }+\frac{\beta -1}{2\beta }\)

$$\begin{aligned} \Vert B&u^{(n)}(t)-Bu^{(n-1)}(t)\Vert _{p\beta }\nonumber \\&\le \int _0^t \Vert S(t-s)P(|u^{(n)}|^{\beta -1}u^{(n)}(s)-|u^{(n-1)}|^{\beta -1}u^{(n-1)}(s))\Vert _{p\beta }ds\nonumber \\&\le C \int _0^t (t-s)^{-(\frac{3}{q}-\frac{3}{p\beta })/2} \Vert |u^{(n)}|^{\beta -1}u^{(n)}(s)-|u^{(n-1)}|^{\beta -1}u^{(n-1)}(s))\Vert _q ds\nonumber \\&\le C \int _0^t (t-s)^{-(\frac{3}{q}-\frac{3}{p\beta })/2} \Vert w^{(n)}\Vert _{p\beta }(\Vert u^{(n)}(s)\Vert _{2\beta }^{\beta -1}+\Vert u^{(n-1)}(s)\Vert _{2\beta }^{\beta -1})ds\nonumber \\&\le C\int _0^t (t-s)^{-\frac{3(\beta -1)}{4\beta }} s^{-\frac{\beta +3}{4\beta }-\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}} (C_0K)^n K^{\beta -1} ds\nonumber \\&\le CK^{\beta -1}(C_0K)^n t^{-\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}} \end{aligned}$$
(2.29)

is fulfilled, where we used the inequality

$$\begin{aligned} \Vert |u_1|^{\beta -1}u_1-|u_2|^{\beta -1}u_2\Vert _r \le \frac{\beta }{2} \Vert u_1-u_2\Vert _{r_1} (\Vert u_1\Vert _{r_2(\beta -1)}^{\beta -1}+\Vert u_2\Vert _{r_2(\beta -1)}^{\beta -1})\qquad \end{aligned}$$
(2.30)

with \(1< r, r_1, r_2 <\infty , \frac{1}{r}=\frac{1}{r_1}+\frac{1}{r2}\), easily obtained by

$$\begin{aligned} ||x|^{\beta -1}x-|y|^{\beta -1}y| \le \frac{\beta }{2} |x-y|(|x|^{\beta -1}+|y|^{\beta -1}),\quad x, y \in \mathbb {R}^3 \end{aligned}$$
(2.31)

and Hölder’s inequality. By (2.26) and (2.28), we get

$$\begin{aligned} \Vert t^k&A^k w^{(n+1)}(t)\Vert _p\nonumber \\&\le \Vert t^k A^k(Gu^{(n)}(t)-Gu^{(n-1)}(t))\Vert _p+|\alpha | \Vert t^k A^k(Bu^{(n)}(t)-Bu^{(n-1)}(t))\Vert _p\nonumber \\&\le CK(C_0K)^n(t^{\frac{\beta -3}{2(\beta -1)}}+K^{\beta -2})\nonumber \\&\le (C_0K)^{n+1}. \end{aligned}$$
(2.32)

Also, (2.27) and (2.29) imply

$$\begin{aligned} \Vert t&^{\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}} w^{(n+1)}(t)\Vert _{p\beta }\nonumber \\&\le \Vert t^{\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}} (Gu^{(n)}(t)-Gu^{(n-1)}(t)) \Vert _{p\beta }\nonumber \\&\quad + |\alpha | \Vert t^{\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}} (Bu^{(n)}(t)-Bu^{(n-1)}(t))\Vert _{p\beta }\nonumber \\&\le CK (C_0K)^n (1+K^{\beta -2})\nonumber \\&\le (C_0K)^{n+1}. \end{aligned}$$
(2.33)

The inequalities (2.32) and (2.33) complete the proof.

Proof of Theorem 1.1

(Existence) By Lemma 2.5 when \(K=\frac{1}{2C_0}\), there exists \(0<T<1\) satisfying

$$\begin{aligned}{} & {} \sup _{2\le p \le 1.5(\beta -1), 0\le t\le T} \Vert w^{(n)}(t)\Vert _p\le (1/2)^n, \end{aligned}$$
(2.34)
$$\begin{aligned}{} & {} \sup _{2\le p \le 1.5(\beta -1), 0\le t\le T} \Vert t^{1/2}A^{1/2}w^{(n)}(t)\Vert _p\le (1/2)^n, \end{aligned}$$
(2.35)
$$\begin{aligned}{} & {} \sup _{2\le p \le 1.5(\beta -1), 0\le t\le T} \Vert t^{\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}}w^{(n)}(t)\Vert _{p\beta } \le (1/2)^n, \end{aligned}$$
(2.36)

for all \(n\in \mathbb {N}\). Then \(\{u^{(n)}\}\) uniformly converges to some \(u\in C([0,T];H \cap L^{1.5(\beta -1)}_\sigma (\Omega ))\) in [0, T], because \(u^{(n)}\in C([0, T]; H\cap L^{1.5(\beta -1)}(\Omega ))\) with

$$\begin{aligned} \sum _{j=0}^{n} \Vert u^{(j+1)}(t)-u^{(j)}(t)\Vert _p \le \sum _{j=0}^{n} \frac{1}{2^{j+1}} \le 1 \end{aligned}$$
(2.37)

got by (2.34) for \(2\le p \le 1.5(\beta -1)\) and \(0\le t\le T\). Similarly \(\{t^{1/2}A^{1/2}u^{(n)}\}\) and \(\{t^{\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}} u^{(n)}\}\) uniformly converge to \(t^{1/2}A^{1/2} u \in C([0, T]; L^p_\sigma (\Omega ))\) and \( t^{\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}} u\in C([0, T]; L^{p\beta }_\sigma (\Omega ))\), respectively when \(2 \le p \le 1.5(\beta -1)\). Thus, we have

$$\begin{aligned} u\in C([0,T];H \cap L^{1.5(\beta -1)}_\sigma (\Omega )), \end{aligned}$$
(2.38)

with \(u(0)=u_0\) and

$$\begin{aligned}{} & {} t^{1/2}A^{1/2}u\in C([0, T]; H\cap L^{1.5(\beta -1)}_\sigma (\Omega )), \end{aligned}$$
(2.39)
$$\begin{aligned}{} & {} t^{\frac{(2p-3)\beta +3}{2p\beta (\beta -1)}} u\in C([0, T]; L^{p\beta }_\sigma (\Omega )), 2\le p\le 1.5(\beta -1) \end{aligned}$$
(2.40)

with values zero at \(t=0\).

First, prove that u is a weak solution of (1.1). Uniform convergence of \(\{u^{n}\}\) and following the calculations of (2.26) and (2.28) with replacing \(u^{(n-1)}\) by u, lead to uniform convergence of \(\{Gu^{(n)}\}\) to Gu as well as \(\{Bu^{(n)}\}\) to Bu, in C([0, T]; H). Thus u satisfies

$$\begin{aligned} u(t)=S(t)u_0+Gu(t)+\alpha Bu(t), 0\le t\le T \end{aligned}$$
(2.41)

by (2.5) as \(n\rightarrow \infty \).

By using Hölder’s inequality, (2.38), (2.39) and (2.1), we have

$$\begin{aligned} \Vert A^{1/2}Gu(t)\Vert _2&\le \int _0^t \Vert A^{1/2}S(t-s)P((u \cdot \nabla ) u(s))\Vert _2ds\nonumber \\&\le C\int _0^t (t-s)^{-(1+\frac{3}{1.5}-\frac{3}{2})/2} \Vert u(s)\Vert _3\Vert A^{1/2}u(s)\Vert _3 ds\nonumber \\&\le C\Vert u\Vert _{3,\infty ;T}\Vert t^{1/2}A^{1/2}u\Vert _{3,\infty ;T}\int _0^t (t-s)^{-3/4}s^{-1/2} ds\nonumber \\&\le C\Vert u\Vert _{3,\infty ;T}\Vert t^{1/2}A^{1/2}u\Vert _{3,\infty ;T} t^{-1/4} \end{aligned}$$
(2.42)

which implies

$$\begin{aligned} \Vert A^{1/2}Gu\Vert _{2,2;T}&\le C\Vert u\Vert _{3,\infty ;T}\Vert t^{1/2}A^{1/2}u\Vert _{3,\infty ;T} (\int _0^T t^{-1/2} ds)^{1/2}\nonumber \\&\le C\Vert u\Vert _{3,\infty ;T}\Vert t^{1/2}A^{1/2}u\Vert _{3,\infty ;T} T^{1/4}\nonumber \\&<\infty . \end{aligned}$$
(2.43)

Also, Hölder’s inequality, (2.40) and (2.1) lead to

$$\begin{aligned} \Vert A^{1/2} Bu(t)\Vert _2&\le \int _0^t \Vert A^{1/2}S(t-s)P(|u|^{\beta -1}u(s))\Vert _2 ds\nonumber \\&\le C \int _0^t (t-s)^{-1/2} \Vert u(s)\Vert _{2\beta }^\beta ds\nonumber \\&\le C \Vert t^{\frac{\beta +3}{4\beta (\beta -1)}} u\Vert _{2\beta , \infty ;T} \int _0^t (t-s)^{-1/2} s^{-\frac{\beta +3}{4(\beta -1)}} ds\nonumber \\&\le C \Vert t^{\frac{\beta +3}{4\beta (\beta -1)}} u\Vert _{2\beta , \infty ;T} t^\frac{\beta -5}{4(\beta -1)} \end{aligned}$$
(2.44)

and

$$\begin{aligned} \Vert A^{1/2}Bu\Vert _{2,2;T}&\le C\Vert t^{\frac{\beta +3}{4\beta (\beta -1)}} u\Vert _{2\beta , \infty ;T} (\int _0^T t^{\frac{\beta -5}{2(\beta -1)}}ds)^{1/2}\nonumber \\&\le C \Vert t^{\frac{\beta +3}{4\beta (\beta -1)}} u\Vert _{2\beta , \infty ;T} T^{\frac{3\beta -7}{4(\beta -1)}}\nonumber \\&<\infty . \end{aligned}$$
(2.45)

Applying Lemma [10, Chap.IV, Lemma 1.5.3], obtains

$$\begin{aligned} \Vert A^{1/2} S(t)u_0\Vert _{2,2;T} \le \Vert u_0\Vert _2. \end{aligned}$$
(2.46)

The equation (2.41) and the inequalities (2.43), (2.45) and (2.46) imply

$$\begin{aligned} A^{1/2}u\in L^2(0,T; H). \end{aligned}$$
(2.47)

On the other hand,

$$\begin{aligned} \Vert u\Vert _{\beta +1, \beta +1;T}&\le C \Vert u\Vert _{2,\infty ;T}^{\frac{1}{\beta +1}} \Vert u\Vert _{2\beta , \beta ; T}^{\frac{\beta }{\beta +1}} \nonumber \\&\le C\Vert u\Vert _{2,\infty ;T}^{\frac{1}{\beta +1}}\Vert t^{\frac{\beta +3}{4\beta (\beta -1)}} u\Vert _{2\beta , \infty ; T}^{\frac{\beta }{\beta +1}} (\int _0^T t^{-\frac{\beta +3}{4(\beta -1)}} ds)^{1/(\beta +1)}\nonumber \\&\le C\Vert u\Vert _{2,\infty ;T}^{\frac{1}{\beta +1}}\Vert t^{\frac{\beta +3}{4\beta (\beta -1)}} u\Vert _{2\beta , \infty ; T}^{\frac{\beta }{\beta +1}} T^{\frac{3\beta -7}{4(\beta +1)(\beta -1)}}\nonumber \\&<\infty \end{aligned}$$
(2.48)

leads to

$$\begin{aligned} u\in L^{\beta +1}(0, T; L_\sigma ^{\beta +1}(\Omega )). \end{aligned}$$
(2.49)

Combining (2.38), (2.47) and (2.49), we get

$$\begin{aligned} u\in L^2(0,T; V)\cap L^\infty (0,T; H)\cap L^{\beta +1}(0, T; L_\sigma ^{\beta +1}(\Omega )), \end{aligned}$$
(2.50)

which shows that u is a weak solution of (1.1).

Next, prove that u satisfies (1.3)–(1.5). Obviously, u satisfies (1.3) and (1.4) by (2.38) and (2.39).

When \(k=0, 1.5(\beta -1)<p<\infty \), and \(0\le t\le T\), using (2.1) and Hölder’s inequality, we have

$$\begin{aligned} \Vert u^{(0)}(t)\Vert _p= & {} \Vert S(t)u_0\Vert _p \le C t^{-\frac{-3\beta +2p+3}{2p(\beta -1)}} \Vert u_0\Vert _{1.5(\beta -1)}, \end{aligned}$$
(2.51)
$$\begin{aligned} \Vert Gu(t)\Vert _p\le & {} \int _0^t {\Vert S(t-s)P((u \cdot \nabla ) u(s))\Vert _pds} \nonumber \\\le & {} C \int _0^t (t-s)^{-(\frac{3}{1.5(\beta -1)/2}-\frac{3}{p})/2} \Vert u(s)\Vert _{1.5(\beta -1)} \Vert A^{1/2}u(s)\Vert _{1.5(\beta -1)} ds\nonumber \\\le & {} C \Vert u \Vert _{1.5(\beta -1), \infty ; t} \Vert s^{1/2}A^{1/2} u\Vert _{1.5(\beta -1), \infty ;t} \int _0^t (t-s)^{-\frac{2}{\beta -1}+\frac{3}{2p}} s^{-\frac{1}{2}}ds\nonumber \\\le & {} C \Vert u \Vert _{1.5(\beta -1), \infty ; t} \Vert s^{1/2}A^{1/2} u\Vert _{1.5(\beta -1), \infty ; t} t^{(\frac{1}{2}-\frac{1}{\beta -1})-\frac{-3\beta +2p+3}{2p(\beta -1)}}\nonumber \\\le & {} C \Vert u \Vert _{1.5(\beta -1), \infty ; t} \Vert s^{1/2}A^{1/2} u\Vert _{1.5(\beta -1), \infty ; t} t^{-\frac{-3\beta +2p+3}{2p(\beta -1)}}, \end{aligned}$$
(2.52)
$$\begin{aligned} \Vert B u (t) \Vert _p\le & {} \int _0^t \Vert S(t-s) P (| u^{(n)}|^{\beta -1} u(s)) \Vert _p ds\nonumber \\\le & {} C \int _0^t (t-s)^{-(\frac{3}{2}-\frac{3}{p})/2} \Vert u(s)\Vert _{2\beta }^{\beta } ds\nonumber \\\le & {} C \Vert s^{\frac{\beta +3}{4\beta (\beta -1)}} u\Vert _{2\beta , \infty ; t}^\beta \int _0^t (t-s)^{-\frac{3}{4}+\frac{3}{2p}} s^{-\frac{\beta +3}{4(\beta -1)}} ds\nonumber \\\le & {} C \Vert s^{\frac{\beta +3}{4\beta (\beta -1)}} u\Vert _{2\beta , \infty ; t}^\beta t^{-\frac{-3\beta +2p+3}{2p(\beta -1)}}, \end{aligned}$$
(2.53)

thus, \(t^{ \frac{-3\beta +2p+3}{2p(\beta -1)}} u \in C([0, T]; L_\sigma ^p(\Omega ))\) with value zero at \(t=0\).

In case of \(0< k \le 1/2, 1.5(\beta -1)<p<\infty \) and \(0\le t \le T\), the estimates

$$\begin{aligned} \Vert A^k u^{(0)}(t)\Vert _p= & {} \Vert A^k S(t)u_0\Vert _p \le C t^{-\frac{\beta (2kp-3)-2kp+2p+3}{2p(\beta -1)}} \Vert u_0\Vert _{1.5(\beta -1)}, \end{aligned}$$
(2.54)
$$\begin{aligned} \Vert A^kGu(t)\Vert _p\le & {} \int _0^t {\Vert A^k S(t-s)P((u \cdot \nabla ) u(s))\Vert _pds} \nonumber \\\le & {} C \int _0^t (t-s)^{-(2k+3(\frac{1}{3}+\frac{1}{2p})-\frac{3}{p})/2} \Vert u(s)\Vert _{2p} \Vert A^{1/2}u(s)\Vert _3 ds\nonumber \\\le & {} C \Vert s^{1/2}A^{1/2}u\Vert _{3, \infty ; t} \Vert s^\frac{-3\beta +4p+3}{4p(\beta -1)} u\Vert _{2p, \infty ; t}\nonumber \\{} & {} \int _0^t (t-s)^{-k-\frac{1}{2}+\frac{3}{4p}} s^{-\frac{-3\beta +4p+3}{4p(\beta -1)}-\frac{1}{2}} ds\nonumber \\\le & {} C \Vert s^{1/2}A^{1/2}u\Vert _{3, \infty ;t} \Vert s^\frac{-3\beta +4p+3}{4p(\beta -1)} u\Vert _{2p, \infty ; t} t^{-\frac{\beta (2kp-3)-2kp+2p+3}{2p(\beta -1)}}\nonumber \\ \end{aligned}$$
(2.55)

and

$$\begin{aligned} \Vert A^k B u^{(n)} (t) \Vert _p&\le \int _0^t \Vert A^k S(t-s) P (| u^{(n)}|^{\beta -1} u^{(n)} (s) )\Vert _p ds\nonumber \\&\le C \int _0^t (t-s)^{-(2k+\frac{3}{q}-\frac{3}{p})/2} \Vert u^{(n)}(s)\Vert _{q\beta }^{\beta } ds\nonumber \\&\le C \Vert s^{\frac{(2q-3)\beta +3}{2q\beta (\beta -1)}} u\Vert _{q\beta , \infty ; t}^\beta \int _0^t (t-s)^{-k-\frac{3}{2q}+\frac{3}{2p}} s^{-\frac{(2q-3)\beta +3}{2q(\beta -1)}} ds\nonumber \\&\le C \Vert s^{\frac{(2q-3)\beta +3}{2q\beta (\beta -1)}} u\Vert _{q\beta , \infty ; t}^\beta t^{-\frac{\beta (2kp-3)-2kp+2p+3}{2p(\beta -1)}}, \end{aligned}$$
(2.56)

where \(q=\frac{3p+6}{p+3}\), imply \(t^{\frac{\beta (2kp-3)-2kp+2p+3}{2p(\beta -1)}}A^ku\in C([0, T]; L^p_\sigma (\Omega ))\) with value zero at \(t=0\). Thus, (1.5) is proved.

Finally, show that u satisfies (1.6) for any \(0<\epsilon <T\). It is obvious that \(u\in C([\epsilon , T]; V\cap L^{1.5(\beta -1)}_\sigma (\Omega ))\), by (1.3) and (1.4).

We have \((u\cdot \nabla ) u\in L^2(\epsilon , T; L^2(\Omega )^3)\) from \(\Vert (u\cdot \nabla ) u\Vert _2 \le C\Vert u\Vert _6 \Vert A^{1/2} u\Vert _3 \le C\Vert A^{1/2} u\Vert _2 \Vert A^{1/2}u\Vert _3\) and (1.4). Also \(|u|^{\beta -1}u \in L^2(\epsilon , T; L^2(\Omega )^3)\) is obtained from (1.5), so

$$\begin{aligned} (u\cdot \nabla ) u +\alpha |u|^{\beta -1}u \in L^2(\epsilon , T; L^2(\Omega )^3). \end{aligned}$$
(2.57)

Applying [10, Chap.IV, Theorem 2.5.4] with \(u(\epsilon )\in V\) and (2.57), implies \(u\in L^2(\epsilon , T; D(A_2))\), which leads to (1.6) and completes the proof of this lemma. \(\square \)

(Uniqueness) The proof of uniqueness is the same as [4], so we omit it here.

Proof of Theorem 1.2

By the same way as Theorem 1.1, with \(u_0\in V \cap L^{\bar{\beta }}_\sigma (\Omega ) \subset H \cap L^{\max \{6, \bar{\beta }\}}_\sigma (\Omega )\), we can prove that there exist \(0<T<1\) and a unique weak solution of (1.1) on [0, T], satisfying

$$\begin{aligned}{} & {} u\in C([0, T]; H\cap L_\sigma ^{\max \{6, \bar{\beta }\}}(\Omega )), \end{aligned}$$
(2.58)
$$\begin{aligned}{} & {} t^{1/2}A^{1/2}u\in C([0, T]; H\cap L_\sigma ^{\max \{6, \bar{\beta }\}}(\Omega )), \end{aligned}$$
(2.59)
$$\begin{aligned}{} & {} t^{\frac{2kp{\max \{6, \bar{\beta }\}}+3p-3{\max \{6, \bar{\beta }\}}}{2p{\max \{6, \bar{\beta }\}}}} A^k u\in C([0, T]; L_\sigma ^p(\Omega )), {\max \{6, \bar{\beta }\}}<p<\infty , 0\le k\le 1/2.\nonumber \\ \end{aligned}$$
(2.60)

And the condition \(u_0\in V\) provides more regularity

$$\begin{aligned} t^{\frac{3p-6}{4p}} A^{1/2}u\in C([0, T]; L^p_\sigma (\Omega )), 2\le p\le 3. \end{aligned}$$
(2.61)

In fact, when \(2\le p\le 3, 0\le t\le T\), the estimates

$$\begin{aligned} \Vert A^{1/2} S(t)u_0\Vert _p&= \Vert S(t) A^{1/2} u_0\Vert _p \le C t^{-\frac{3p-6}{4p}}\Vert A^{1/2} u_0\Vert , \end{aligned}$$
(2.62)
$$\begin{aligned} \Vert A^{1/2} Gu(t)\Vert _p&\le \int _0^t {\Vert A^{1/2} S(t-s)P((u \cdot \nabla ) u(s))\Vert _p ds} \nonumber \\&\le C \int _0^t (t-s)^{-1/2} \Vert (u \cdot \nabla ) u(s)\Vert _p ds\nonumber \\&\le C \int _0^t (t-s)^{-1/2} \Vert u(s)\Vert _{2p} \Vert A^{1/2}u(s)\Vert _{2p} ds\nonumber \\&\le C \Vert u\Vert _{2p,\infty ;t} \Vert s^{1/2} A^{1/2}u\Vert _{2p, \infty ; t} \int _0^t (t-s)^{-1/2} s^{-1/2}ds\nonumber \\&\le C \Vert u\Vert _{2p,\infty ;t} \Vert s^{1/2} A^{1/2}u\Vert _{2p, \infty ; t}, \end{aligned}$$
(2.63)
$$\begin{aligned} \Vert A^{1/2} B u (t) \Vert _p&\le \int _0^t \Vert A^{1/2} S(t-s) P (| u|^{\beta -1} u (s) )\Vert _p ds\nonumber \\&\le C \int _0^t (t-s)^{-1/2} \Vert u(s)\Vert _{p\beta }^{\beta } ds\nonumber \\&\le C \Vert s^{\frac{3p\beta -3\max \{6, \bar{\beta }\}}{2p\beta \max \{6, \bar{\beta }\}}} u\Vert _{p\beta , \infty ; t}^\beta \int _0^t (t-s)^{-1/2} s^{-\frac{3p\beta -3\max \{6, \bar{\beta }\}}{2p\max \{6, \bar{\beta }\}}} ds\nonumber \\&\le C \Vert s^{\frac{3p\beta -3\max \{6, \bar{\beta }\}}{2p\beta \max \{6, \bar{\beta }\}}} u\Vert _{p\beta , \infty ; t}^\beta t^{-\frac{3p-6}{4p}+\frac{5\max \{6, \bar{\beta }\}-6\beta }{4\max \{6, \bar{\beta }\}}}\nonumber \\&\le C \Vert s^{\frac{3p\beta -3\max \{6, \bar{\beta }\}}{2p\beta \max \{6, \bar{\beta }\}}} u\Vert _{p\beta , \infty ; t}^\beta t^{-\frac{3p-6}{4p}}, \end{aligned}$$
(2.64)

with (2.58), (2.59) and (2.60), imply (2.61).

Now, show that u is a strong solution of (1.1) on [0, T] satisfying (1.7). It is obvious that \(u\in C([0, T]; V\cap L^{\bar{\beta }}_\sigma (\Omega ))\) from (2.58) and (2.61). By (2.58), (2.61) and Hölder’s inequality,

$$\begin{aligned} \Vert (u\cdot \nabla ) u\Vert _{2,2;T}&\le C\Vert u\Vert _{6,\infty ;T} \Vert A^{1/2}u\Vert _{3,2;T}\nonumber \\&\le C\Vert u\Vert _{6,\infty ;T}\Vert t^{1/4}A^{1/2}u\Vert _{3,\infty ;T} (\int _0^T t^{-1/2} ds)^{1/2}\nonumber \\&\le C \Vert u\Vert _{6,\infty ;T}\Vert t^{1/4}A^{1/2}u\Vert _{3,\infty ;T} T^{1/4}\nonumber \\&<\infty \end{aligned}$$
(2.65)

is satisfied. Also,

$$\begin{aligned} \Vert |u|^{\beta -1}u\Vert _{2,2;T}&= \Vert u\Vert _{2\beta , 2\beta ;T}^\beta \nonumber \\&\le C\Vert t^{\frac{6\beta -3\max \{6, \bar{\beta }\}}{4\beta \max \{6, \bar{\beta }\}}}u\Vert _{2\beta , \infty ; T}^\beta (\int _0^T t^{-\frac{6\beta -3\max \{6, \bar{\beta }\}}{2\max \{6, \bar{\beta }\}}} ds)^{1/2}\nonumber \\&\le C \Vert t^{\frac{6\beta -3\max \{6, \bar{\beta }\}}{4\beta \max \{6, \bar{\beta }\}}}u\Vert _{2\beta , \infty ; T}^\beta T ^{\frac{5{\max \{6, \bar{\beta }\}}-6\beta }{4{\max \{6, \bar{\beta }\}}}}\nonumber \\&< \infty \end{aligned}$$
(2.66)

holds by (2.60). Applying [10, Chap.IV, Theorem 2.5.4] with \(u_0\in V\), (2.65) and (2.66), leads to \(u\in L^2(0, T;D(A_2))\). \(\square \)

Proof of Corollary 1.1

a) By Theorem 1.2 there exists a local strong solution \(u\in L^2(0, T_1;D(A_2))\cap C([0, T_1]; V\cap L^{\bar{\beta }}(\Omega )) (0<T_1<\infty )\).

Let

$$\begin{aligned} T^*=sup\{T\ge T_1| u \in L^2(0, T;D(A_2))\cap C([0, T]; V\cap L^{\bar{\beta }}(\Omega ))\}. \end{aligned}$$
(2.67)

If \(T^*<\infty \), then we will prove \(u \in L^2(t_0, T^*;D(A_2)) \cap C([t_0, T^*]; V \cap L^{\bar{\beta }}_\sigma (\Omega )) \) on a certain interval \([t_0, T^*)(0<t_0\le T_1)\) from a priori estimates for u. Then, there exist \(T>0\) and a strong solution \(\tilde{u}\in L^2(T^*, T^*+T;D(A_2))\cap C([T^*, T^*+T]; V\cap L^{\bar{\beta }}(\Omega ))\) with initial condition \(\tilde{u}(T^*)=u(T^*)\) by Theorem 1.2. Moreover, \(\tilde{u}\) coincides with u by Lemma 2.3, which implies \(u \in L^2(0, T^*+T;D(A_2))\cap C([0, T^*+T]; V\cap L^{\bar{\beta }}(\Omega ))\) and contradicts to (2.67). Thus, \(T^*=\infty \) and u satisfies (1.8) for all \(0<T<\infty \). The uniqueness is obvious by Lemma 2.3.

Assuming \(T^*<\infty \), prove that \(u \in L^2(t_0, T^*;D(A_2)) \cap C ([t_0, T^*]; V \cap L^{\bar{\beta }}_\sigma (\Omega ))(0<t_0\le T_1)\).

First, multiply the first equation of (1.1) by u and integrate by parts on \(\Omega \) using \(((u \cdot \nabla )u,u)=0\), we have

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\Vert u(t)\Vert ^2+\Vert \nabla u\Vert +\alpha \Vert u\Vert _{\beta +1}^{\beta +1}=0, 0\le t<T^* \end{aligned}$$
(2.68)

and integrating (2.68) leads to

$$\begin{aligned} \Vert u\Vert _{2,\infty ;T^*} \le \Vert u_0\Vert <\infty . \end{aligned}$$
(2.69)

Next, for the boundedness of \(\Vert A^{1/2}u(t)\Vert \), multiply the first equation of (1.1) by \(-\Delta u\) and integrate by parts on \(\Omega \) using the relation \(-(\nabla p, \Delta u)=0\) and

$$\begin{aligned} (|u|^{\beta -1}u, -\Delta u)=\Vert |u|^\frac{\beta -1}{2} \nabla u\Vert ^2+\frac{\beta -1}{4(\beta +1)^2} \Vert \nabla |u|^\frac{\beta +1}{2}\Vert ^2, \end{aligned}$$
(2.70)

to get

$$\begin{aligned} \frac{1}{2}\frac{d}{dt} \Vert A^{1/2}u(t)\Vert ^2&+\Vert Au\Vert ^2+\alpha \Vert |u|^\frac{\beta -1}{2} \nabla u\Vert ^2+\frac{\beta -1}{4(\beta +1)^2} \Vert \nabla |u|^\frac{\beta +1}{2}\Vert ^2 \nonumber \\&=((u\cdot \nabla )u, \Delta u), 0\le t<T^*. \end{aligned}$$
(2.71)

From \(\beta =3, \alpha \ge 1/4\) or \(\beta>3, \alpha >0\), the following inequality

$$\begin{aligned} ((u\cdot \nabla ) u, \Delta u)&\le \Vert Au\Vert ^2+\frac{1}{4}\Vert |u| \nabla u \Vert ^2\nonumber \\&\le \Vert Au\Vert ^2+\alpha \Vert |u|^\frac{\beta -1}{2}\nabla u \Vert ^2+C\Vert A^{1/2} u\Vert ^2 \end{aligned}$$
(2.72)

holds by Hölder’s inequality and \(\frac{1}{4}|x|^2 \le \alpha |x|^{\beta -1}+C, x\in \mathbb {R}^3\). Adding (2.72) to (2.71) implies

$$\begin{aligned} \frac{d}{dt}\Vert A^{1/2}u(t)\Vert ^2\le C\Vert A^{1/2} u\Vert ^2, 0\le t<T^* \end{aligned}$$
(2.73)

and applying Gronwall’s lemma to (2.73) gives

$$\begin{aligned} \Vert A^{1/2}u\Vert _{2,\infty ;T^*} \le \Vert A^{1/2}u_0\Vert \text {exp}(CT^*)<\infty . \end{aligned}$$
(2.74)

Finally, prove the continuity of u(t) in \(V\cap L^{\bar{\beta }}_\sigma (\Omega )\). Since \(u\in L^2(0, T_1;D(A_2))\), there exists some \(0<t_0<T_1\) such that \(u(t_0)\in D(A_2)\). Differentiating the first equation of (1.1) with respect t and taking the inner product with \(u_t\) in H, we have

$$\begin{aligned} \frac{1}{2} \frac{d}{dt} \Vert u_t(t)\Vert ^2 +\Vert \nabla u_t\Vert ^2 +\alpha ((|u|^{\beta -1}u)_t, u_t) = -((u\cdot \nabla u)_t, u_t), t_0\le t<T^*. \end{aligned}$$
(2.75)

Using the inequalities

$$\begin{aligned} \alpha ((|u|^{\beta -1}u)_t, u_t)&=\alpha (|u|^{\beta -1}u_t, u_t)+\frac{\alpha (\beta -1)}{4} \int _\Omega |u|^{\beta -3} | \frac{\partial }{\partial t} |u|^2|^2 dx\nonumber \\&\ge \alpha \Vert |u|^{(\beta -1)/2} u_t \Vert ^2, \end{aligned}$$
(2.76)
$$\begin{aligned} -((u\cdot \nabla u)_t, u_t)&=-((u_t\cdot \nabla )u, u_t)-((u \cdot \nabla )u_t, u_t)\nonumber \\&=-((u_t\cdot \nabla )u, u_t)\nonumber \\&=((u_t\cdot \nabla )u_t, u)\nonumber \\&\le \Vert \nabla u_t \Vert ^2 +\frac{1}{4} \Vert |u|u_t\Vert ^2\nonumber \\&\le \Vert \nabla u_t \Vert ^2 +\alpha \Vert |u|^{(\beta -1)/2} u_t \Vert ^2 +C \Vert u_t\Vert ^2, \end{aligned}$$
(2.77)

(2.75) becomes

$$\begin{aligned} \frac{d}{dt} \Vert u_t(t)\Vert ^2 \le C\Vert u_t\Vert ^2, t_0 \le t<T^*. \end{aligned}$$
(2.78)

By \(\Vert u\Vert _\infty \le C \Vert Au \Vert ^{1/2} \Vert A^{1/2} u \Vert ^{1/2}\), \(\Vert u\Vert _{2\beta }^{\beta } \le C\Vert Au\Vert ^\frac{3(\beta -1)}{4} \Vert u \Vert ^\frac{\beta +3}{4}\), \(u(t_0)\in D(A_2)\) and (1.1), we have

$$\begin{aligned} \Vert u_t (t_0) \Vert&\le \Vert Au(t_0)\Vert +\Vert P((u\cdot \nabla ) u(t_0)) \Vert +\alpha \Vert P |u|^{\beta -1} u(t_0)\Vert \nonumber \\&\le \Vert Au(t_0)\Vert +C\Vert A^{1/2} u(t_0)\Vert \Vert u(t_0)\Vert _\infty +C\alpha \Vert Au(t_0)\Vert ^\frac{3(\beta -1)}{4} \Vert u(t_0)\Vert ^\frac{\beta +3}{4}\nonumber \\&\le \Vert Au(t_0)\Vert +C\Vert A^{1/2} u(t_0)\Vert ^{3/2} \Vert Au(t_0)\Vert ^{1/2} +C\alpha \Vert Au(t_0)\Vert ^\frac{3(\beta -1)}{4} \Vert u(t_0)\Vert ^\frac{\beta +3}{4}\nonumber \\&< \infty . \end{aligned}$$
(2.79)

So, applying Gronwall’s inequality to (2.78) implies

$$\begin{aligned} \Vert u_t\Vert _{2,\infty ;t_0,T^*} \le \Vert u_t(t_0)\Vert \text {exp} (C(T^*-t_0)) <\infty . \end{aligned}$$
(2.80)

Multiplying the first equation of (1.1) by \(-\Delta u\) and using the inequality

$$\begin{aligned} ((u\cdot \nabla u, \Delta u)&\le \Vert Au\Vert \Vert A^{1/2} u\Vert \Vert u\Vert _\infty \\&\le C \Vert Au \Vert ^{3/2} \Vert A^{1/2} u\Vert ^{3/2}\\&\le \frac{1}{4} \Vert Au\Vert ^2 +C\Vert A^{1/2} u\Vert ^ 6 \end{aligned}$$

and

$$\begin{aligned} (u_t, -\Delta u)\le \Vert u_t \Vert ^2 +\frac{1}{4} \Vert Au\Vert ^2, \end{aligned}$$

we obtain

$$\begin{aligned} \frac{1}{2}\Vert Au\Vert ^2 \le \Vert u_t\Vert ^2+C\Vert A^{1/2}u\Vert ^6, t_0\le t<T^*, \end{aligned}$$
(2.81)

thus, get

$$\begin{aligned} \Vert Au\Vert _{2,\infty ;t_0,T^*} \le C(\Vert u \Vert _{2,\infty ;t_0, T^*}+\Vert A^{1/2} u\Vert _{2, \infty ;t_0, T^*}^3)<\infty \end{aligned}$$
(2.82)

from (2.74), (2.80). Gagliardo-Nirenberg inequality \(\Vert u\Vert _{\bar{\beta }} \le C \Vert Au\Vert ^{\frac{3({\bar{\beta }}-2)}{4{\bar{\beta }}}} \Vert u\Vert ^{\frac{{\bar{\beta }}+6}{4{\bar{\beta }}}}\), (2.69) and (2.82) give

$$\begin{aligned} \Vert u \Vert _{{\bar{\beta }}, \infty ;t_0, T^*} < \infty . \end{aligned}$$
(2.83)

Also, the inequalities

$$\begin{aligned} \Vert (|u|^{\bar{\beta }-1} u)_t \Vert _1&\le C \Vert |u|^{\bar{\beta }-1} |u_t|\Vert _1\nonumber \\&\le C \Vert u_t\Vert \Vert u\Vert _{2(\bar{\beta }-1)}^{\bar{\beta }-1}\nonumber \\&\le C \Vert u_t\Vert \Vert Au\Vert ^{\frac{3(\bar{\beta }-2)}{4}}\Vert u\Vert ^{\frac{\bar{\beta }+2}{4}}, \end{aligned}$$
(2.84)

(2.69), (2.80) and (2.82) lead to

$$\begin{aligned} \Vert (|u|^{\bar{\beta }-1} u)_t\Vert _{1, \infty ;t_0, T^*} <\infty .\nonumber \\ \end{aligned}$$
(2.85)

Using the estimate

$$\begin{aligned} -((u\cdot \nabla u)_t, u_t)&=((u_t\cdot \nabla )u_t, u)\nonumber \\&\le \frac{1}{2}\Vert \nabla u_t \Vert ^2 +\frac{1}{2} \Vert u\Vert _\infty ^2 \Vert u_t\Vert ^2\nonumber \\&\le \frac{1}{2}\Vert \nabla u_t \Vert ^2 +C \Vert Au\Vert ^2\Vert A^{1/2}u\Vert ^2\Vert u_t\Vert ^2 \end{aligned}$$
(2.86)

and \(((|u|^{\beta -1}u)_t, u_t)\ge 0\), (2.75) implies

$$\begin{aligned} \frac{d}{dt} \Vert u_t(t)\Vert ^2 +\Vert A^{1/2} u_t\Vert ^2 \le C\Vert Au\Vert ^2\Vert A^{1/2}u\Vert ^2\Vert u_t\Vert ^2, t_0\le t<T^*. \end{aligned}$$
(2.87)

Integrate (2.87) considering (2.74), (2.80) and (2.82), to get

$$\begin{aligned} \Vert A^{1/2} u_t\Vert _{2, 2;t_0, T^*}< \infty . \end{aligned}$$
(2.88)

Combining (2.69), (2.74), (2.80), (2.82), (2.83), (2.85) and (2.88), we have

$$\begin{aligned} u \in L^2(t_0, T^*;D(A_2)) \cap C([t_0, T^*]; V \cap L^{\bar{\beta }}_\sigma (\Omega )). \end{aligned}$$
(2.89)

b) We employ the same continuation argument as a) of Corollary 1.1. The local strong solution \(u\in L^2(0, T_1;D(A_2))\cap C([0, T_1]; V\cap L^{\bar{\beta }}(\Omega )) (0<T_1<\infty )\) obtained by Theorem 1.2, satisfies (1.9) for all \(0<T<\infty \) and uniqueness is guaranteed by Lemma 2.3. In fact, let \(T^*\) be defined by (2.67) and assume that \(T^*<\infty \) and \(u(t_0)\in D(A_2)(0<t_0<T_1)\). Applying Lemma 2.2 with \(u(t_0)\in D(A_2)\) and using the uniqueness of weak solutions (Lemma 2.3), we have \(u\in W^{1,2}(t_0, 2T^*; V)\) therefore, \(u \in C([t_0, 2T^*]; V) \subset C([t_0, 2T^*]; L^6_\sigma (\Omega )=L^{1.5(\beta -1)}_\sigma (\Omega ))\). So there exists \(t_0<T'<T^*\) and a strong solution \(\tilde{u}\in L^2((T'+T^*)/2, 2T^*-T'; D(A_2))\cap C([(T'+T^*)/2, 2T^*-T']; V\cap L_\sigma ^{\bar{\beta }}(\Omega ))\) with initial condition \(\tilde{u}((T'+T^*)/2)=u((T'+T^*)/2)\) by (1.5) and (1.6) of Theorem 1.1. By Lemma 2.3, \(\tilde{u}\) coincides with u, which means that \(u\in L^2(0, 2T^*-T';D(A_2))\cap C([0, 2T^*-T']; V\cap L^{\bar{\beta }}(\Omega ))\) and contradicts (2.67). \(\square \)