Get 20M+ Full-Text Papers For Less Than $1.50/day. Start a 14-Day Trial for You or Your Team.

Learn More →

Well posedness of magnetohydrodynamic equations in 3D mixed-norm Lebesgue space

Well posedness of magnetohydrodynamic equations in 3D mixed-norm Lebesgue space 1IntroductionUnlike a usual Lebesgue space [1,2,3], the mixed-norm Lebesgue space allows its norm decay to zero with different rates as ∣x∣→∞| x| \to \infty in different spatial directions [4]. On the basis of the mixed-norm Lebesgue space feature, we investigate the following magnetohydrodynamic (MHD) equations in R3{{\mathbb{R}}}^{3}[1]: (1.1)∂tu−1ReΔu+u⋅∇u−S(∇×b)×b+∇p˜=0,inR3×(0,+∞),∂tb−∇×(u×b)+1Rm∇×(∇×b)=0,inR3×(0,+∞),u(x,0)=u0(x),b(x,0)=b0(x),inR3,\left\{\begin{array}{ll}{\partial }_{t}u-\frac{1}{{\rm{Re}}}\Delta u+u\cdot \nabla u-S\left(\nabla \times b)\times b+\nabla \tilde{p}=0,& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ {\partial }_{t}b-\nabla \times \left(u\times b)+\frac{1}{{\rm{Rm}}}\nabla \times \left(\nabla \times b)=0,& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ u\left(x,0)={u}_{0}\left(x),\hspace{1em}b\left(x,0)={b}_{0}\left(x),& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3},\end{array}\right.where uu, bb, u0{u}_{0}, and b0{b}_{0}satisfy (1.2)divu(x,t)=divb(x,t)=0,divu0=divb0=0,{\rm{div}}\hspace{0.33em}u\left(x,t)={\rm{div}}\hspace{0.33em}b\left(x,t)=0,\hspace{1em}{\rm{div}}\hspace{0.33em}{u}_{0}={\rm{div}}\hspace{0.33em}{b}_{0}=0,S=M2ReRmS=\frac{{M}^{2}}{{\rm{ReRm}}}, Re>0{\rm{Re}}\gt 0is the Reynolds number, Rm>0{\rm{Rm}}\gt 0is the magnetic Reynolds number, MMis the Hartman number. u:R3×(0,+∞)→R3u:{{\mathbb{R}}}^{3}\times \left(0,+\infty )\to {{\mathbb{R}}}^{3}denote the velocity of the fluid, b:R3×(0,+∞)→R3b:{{\mathbb{R}}}^{3}\times \left(0,+\infty )\to {{\mathbb{R}}}^{3}denote the magnetic field, and p˜=p˜(x,t)∈R\tilde{p}=\tilde{p}\left(x,t)\in {\mathbb{R}}denote the pressure.The main purpose of this paper is to study the well posedness of the solution for the equations (1.1) in 3D mixed-norm Lebesgue spaces. First, we review some of the relevant work of the MHD equations. If b=0b=0, then the equations (1.1) can be reduced to the incompressible Navier-Stokes equations: ∂tu−1ReΔu+u⋅∇u+∇p˜=0,inR3×(0,+∞),u(x,0)=u0(x),inR3.\left\{\begin{array}{ll}{\partial }_{t}u-\frac{1}{{\rm{Re}}}\Delta u+u\cdot \nabla u+\nabla \tilde{p}=0,& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ u\left(x,0)={u}_{0}\left(x),& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}.\end{array}\right.The mathematical theory of the Navier-Stokes equation has been much studied in recent years. For example, Leray first introduced the weak solution [5], and later, Hopf gained the existence of global weak solutions with u0∈Hs(RN){u}_{0}\in {H}^{s}\left({{\mathbb{R}}}^{N})[6], and Fujita and Kato demonstrated that the well posedness of the Cauchy problem when u0∈Hs(RN)(N≥2){u}_{0}\in {H}^{s}\left({{\mathbb{R}}}^{N})\left(N\ge 2)[7]. In addition, there are many monographs that study the Navier-Stokes equation, for example, Temam [8], Lions [9], and Constantin and Foias [10]. The mild and self-similar solutions in R3{{\mathbb{R}}}^{3}are obtained in references [11,12]. Especially, Jia and Šverák derived that the homogeneous classical Cauchy problem with initial value has a global scale invariant solution, and the solution is smooth in positive time [13].For the MHD system, the coupling between uuand bbmakes the situation more complicated. Because it describes abundant natural phenomena, as well as physical importance and mathematical challenges, the MHD system has become the subject of the study by physicists and mathematicians. Duraut and Lions derived the global weak solution and the local strong solution of the initial boundary value problem of equations (1.1) and derived the existence of the global strong solution in the case of the small initial value [14]. Nevertheless, it is still a challenging open problem whether a unique local solution of the exists globally when the initial value is large. Furthermore, Sermange and Temam derived the regularity of the weak solution (u,b)∈L∞([0,T];H1(R3))\left(u,b)\in {L}^{\infty }\left(\left[0,T];\hspace{0.33em}{H}^{1}\left({{\mathbb{R}}}^{3}))[3]. Kozono derived the existence of classical solutions in bounded domain Ω⊂R3\Omega \subset {{\mathbb{R}}}^{3}for equations (1.1) [15]. For the appropriately weak solution, He and Xin gained different local regularity results [16]. Cao and Wu obtained the global well posedness of the MHD system for any initial value in H2(R2){H}^{2}\left({{\mathbb{R}}}^{2}), but it needs a condition of mixed partial dissipation and additional magnetic diffusion in R2{{\mathbb{R}}}^{2}[2].Since the coefficients in the equations have no critical influence on the subsequent analysis, we can simply take Re=Rm=S=1{\rm{Re}}={\rm{Rm}}=S=1. Futhermore, we can obtain the following equations [1]: (∇×b)×b=(b⋅∇)b−b⋅(∇b),∇×∇×b=∇divb−Δb,∇×(u×b)=(b⋅∇)u−(u⋅∇)b+udivb−bdivu.\begin{array}{rcl}\left(\nabla \times b)\times b& =& \left(b\cdot \nabla )b-b\cdot \left(\nabla b),\\ \hspace{0.35em}\nabla \times \nabla \times b& =& \nabla {\rm{div}}\hspace{0.33em}b-\Delta b,\\ \hspace{0.35em}\nabla \times \left(u\times b)& =& \left(b\cdot \nabla )u-\left(u\cdot \nabla )b+u\hspace{0.33em}{\rm{div}}\hspace{0.33em}b-b\hspace{0.33em}{\rm{div}}\hspace{0.33em}u.\end{array}Then, (1.1) can be rewritten as follows: (1.3)∂tu−Δu+(u⋅∇)u−(b⋅∇)b+b⋅(∇b)+∇p˜,inR3×(0,+∞),∂tb−Δb+(u⋅∇)b−(b⋅∇)u=0,inR3×(0,+∞),u(x,0)=u0(x),b(x,0)=b0(x),inR3.\left\{\begin{array}{ll}{\partial }_{t}u-\Delta u+\left(u\cdot \nabla )u-\left(b\cdot \nabla )b+b\cdot \left(\nabla b)+\nabla \tilde{p},& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ {\partial }_{t}b-\Delta b+\left(u\cdot \nabla )b-\left(b\cdot \nabla )u=0,& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ u\left(x,0)={u}_{0}\left(x),\hspace{1.0em}b\left(x,0)={b}_{0}\left(x),& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}.\end{array}\right.In 2D and 3D cases, Ai et al. applied the semi-Galerkin approximation method to obtain the existence of weak solutions [1]. In 2D case, Ai et al. proved the global existence of strong solutions to (1.1), the continuous dependence of initial-boundary data, and the uniqueness of weak-strong solutions. On this basis, they also proved the existence of a uniform attractor for (1.1). Inspired by [1,4], the main purpose of this paper is to study the well posedness of the solution to (1.1) in 3D mixed-norm Lebesgue spaces. Specifically, in Section 2, we state some mixed-norm legesgue spaces and related properties. In Section 3, we prove the existence, uniqueness, and stability of the solution.2PreliminaryWe define the 3D mixed-norm Lebesgue spaces as follows [4]: Lp1p2p3(R3)=ff:R3→R,∥f∥Lp1p2p3(R3)=∫∫∫∣f∣p1dx1p2p1dx2p3p2dx31p3<+∞,{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})=\left\{f\left|f:{{\mathbb{R}}}^{3}\to {\mathbb{R}},{\parallel f\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}={\left(\int {\left(\int {\left(\int {| f| }^{{p}_{1}}{\rm{d}}{x}_{1}\right)}^{\frac{{p}_{2}}{{p}_{1}}}{\rm{d}}{x}_{2}\right)}^{\frac{{p}_{3}}{{p}_{2}}}{\rm{d}}{x}_{3}\right)}^{\tfrac{1}{{p}_{3}}}\right.\lt +\infty \right\},where p1,p2,p3∈[1,+∞){p}_{1},{p}_{2},{p}_{3}\in {[}1,+\infty ).Let PLp1p2p3(R3){{\rm{PL}}}_{{p}_{1}{p}_{2}{p}_{3}}\left({{\mathbb{R}}}^{3})as follows: PLp1p2p3(R3)={f∈Lp1p2p3(R3):divf=0}.{{\rm{PL}}}_{{p}_{1}{p}_{2}{p}_{3}}\left({{\mathbb{R}}}^{3})=\{f\in {L}_{{p}_{1}{p}_{2}{p}_{3}}\left({{\mathbb{R}}}^{3}):{\rm{div}}\hspace{0.33em}f=0\}.All the spaces that appear in this paper are invariant with respect to the scaling f(⋅)→λf(λ⋅)f\left(\cdot )\to \lambda f\left(\lambda \cdot ), λ>0\lambda \gt 0. The mixed-norm space Lp1p2p3(R3){L}_{{p}_{1}{p}_{2}{p}_{3}}\left({{\mathbb{R}}}^{3})is invariant if and only if 1p1+1p2+1p3=1\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}+\frac{1}{{p}_{3}}=1[4]. For given T∈(0,∞]T\in (0,\infty ], 1p1+1p2+1p3=1\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}+\frac{1}{{p}_{3}}=1, 1q1+1q2+1q3=δ∈(0,1)\frac{1}{{q}_{1}}+\frac{1}{{q}_{2}}+\frac{1}{{q}_{3}}=\delta \in \left(0,1), pk∈(1,+∞){p}_{k}\in \left(1,+\infty ), qk∈(pk,+∞){q}_{k}\in \left({p}_{k},+\infty ), k=1,2,3k=1,2,3, and f:R3×[0,∞)→R3f:{{\mathbb{R}}}^{3}\times {[}0,\infty )\to {{\mathbb{R}}}^{3}denote the measurable vector field functions, we denote Xp,q,T{{\mathcal{X}}}_{p,q,T}as follows [4]: Xp,q,T=f:g(x,t)≔t1−δ2f(x,t),g˜(x,t)≔t12Dxf(x,t),(x,t)∈R3×(0,T),{{\mathcal{X}}}_{p,q,T}=\left\{f:g\left(x,t):= {t}^{\tfrac{1-\delta }{2}}f\left(x,t),\hspace{1em}\tilde{g}\left(x,t):= {t}^{\tfrac{1}{2}}{D}_{x}f\left(x,t),\hspace{1.0em}\left(x,t)\in {{\mathbb{R}}}^{3}\times \left(0,T)\right\},then g∈C([0,T],PLq1q2q3(R3)),g˜∈C([0,T],PLp1p2p3(R3)).g\in C\left(\left[0,T],{{\rm{PL}}}_{{q}_{1}{q}_{2}{q}_{3}}\left({{\mathbb{R}}}^{3})),\hspace{1em}\tilde{g}\in C\left(\left[0,T],{{\rm{PL}}}_{{p}_{1}{p}_{2}{p}_{3}}\left({{\mathbb{R}}}^{3})).Moreover, g(x,0)=0g\left(x,0)=0, g˜(x,0)=0\tilde{g}\left(x,0)=0, and the norm ‖f‖χp,q,T=supt∈(0,T)[‖g(⋅,t)‖Lq1q2q3(R3)+‖g˜‖Lp1p2p3(R3)]<+∞.\hspace{-24.6em}\Vert f\Vert {\chi }_{p,q,T}=\mathop{\sup }\limits_{t\in \left(0,T)}{[}\Vert g\left(\cdot ,\hspace{0.33em}t){\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}+\Vert \tilde{g}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}]\lt +\infty .We denote Yp,T{{\mathcal{Y}}}_{p,T}as follows [4]: Yp,T=f:f∈C([0,T],PLp1p2p3(R3)),t12Dxf∈C([0,T],PLp1p2p3(R3)),{{\mathcal{Y}}}_{p,T}=\left\{f:f\in C\left(\left[0,T],{{\rm{PL}}}_{{p}_{1}{p}_{2}{p}_{3}}\left({{\mathbb{R}}}^{3})),\hspace{1em}{t}^{\tfrac{1}{2}}{D}_{x}f\in C\left(\left[0,T],{{\rm{PL}}}_{{p}_{1}{p}_{2}{p}_{3}}\left({{\mathbb{R}}}^{3}))\right\},and the norm ‖f‖Yp,T=supt∈(0,T)[‖f(t)‖Lp1p2p3(R3)+t1/2∥Dxf(t)∥Lp1p2p3(R3)]<+∞.\Vert f{\Vert }_{{{\mathcal{Y}}}_{p,T}}=\mathop{\sup }\limits_{t\in \left(0,T)}{[}\Vert f\left(t){\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+{t}^{1\text{/}2}{\parallel {D}_{x}f\left(t)\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}]\lt +\infty .We state the following result on Young’s inequality in mixed-norm Lebesgue spaces [4].Lemma 2.1[4] Let pk,rk{p}_{k},{r}_{k}, and qk{q}_{k}be given numbers in [1,+∞]\left[1,+\infty ]that satisfy1pk+1=1qk+1rk,k=1,2,3.\frac{1}{{p}_{k}}+1=\frac{1}{{q}_{k}}+\frac{1}{{r}_{k}},\hspace{1em}k=1,2,3.Then, (2.1)∥f∗g∥Lp1p2p3(R3)=∥f∥Lq1q2q3(R3)∥g∥Lr1r2r3(R3),{\parallel f\ast g\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}={\parallel f\parallel }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel g\parallel }_{{L}_{{r}_{1}{r}_{2}{r}_{3}}({{\mathbb{R}}}^{3})},for all f∈Lq1q2q3(R3)f\in {L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})and g∈Lr1r2r3(R3)g\in {L}_{{r}_{1}{r}_{2}{r}_{3}}({{\mathbb{R}}}^{3}).We state the following results on heat equations in mixed-norm Lebesgue spaces. First, we see the Cauchy problem for the heat equations: (2.2)ut−Δu=0,inR3×(0,+∞),bt−Δb=0,inR3×(0,+∞),u(x,0)=u0(x),b(x,0)=b0(x),inR3.\left\{\begin{array}{ll}{u}_{t}-\Delta u=0,& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ {b}_{t}-\Delta b=0,& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ u\left(x,0)={u}_{0}\left(x),b\left(x,0)={b}_{0}\left(x),& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}.\end{array}\right.We see that (2.2) can be written as follows: (2.3)Ut−ΔU=0,inR3×(0,+∞),U(x,0)=U0(x),inR3,\left\{\begin{array}{ll}{U}_{t}-\Delta U=0,& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ U\left(x,0)={U}_{0}\left(x),& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3},\end{array}\right.\hspace{6.65em}where U=ubU=\left(\begin{array}{l}u\\ b\end{array}\right), U0=u0b0{U}_{0}=\left(\begin{array}{l}{u}_{0}\\ {b}_{0}\end{array}\right). It is well known that a solution of (2.3) is (2.4)U(x,t)=eΔtU0(x)=(Gt*U0)(x,t),(x,t)∈R3×(0,+∞),U\left(x,t)={e}^{\Delta t}{U}_{0}\left(x)=\left({G}_{t}* {U}_{0})\left(x,t),\hspace{1.0em}\left(x,t)\in {{\mathbb{R}}}^{3}\times \left(0,+\infty ),where Gt(x)=1(4πt)n2e−∣x∣24t,(x,t)∈R3×(0,+∞).{G}_{t}\left(x)=\frac{1}{{\left(4\pi t)}^{\tfrac{n}{2}}}{e}^{-\tfrac{{| x| }^{2}}{4t}},\hspace{1.0em}\left(x,t)\in {{\mathbb{R}}}^{3}\times \left(0,+\infty ).Next, we state the following fundamental results of the solution of the heat equation in the mixed-norm Lebesgue space.Lemma 2.2[4] Let 1≤pk≤qk≤+∞1\le {p}_{k}\le {q}_{k}\le +\infty . There exists a positive constant NNdepending only on p1{p}_{1}, p2{p}_{2}, p3{p}_{3}, q1{q}_{1}, q2{q}_{2}, and q3{q}_{3}such that for every solution U(x,t)=eΔtU0(x)U\left(x,t)={e}^{\Delta t}{U}_{0}\left(x)defined in (2.4) of the Cauchy problem (2.3) with U0∈Lq1q2q3(R3){U}_{0}\in {L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3}), then for t>0t\gt 0(2.5)∥U(x,t)∥Lp1p2p3(R3)≤Nt−12∑k=1n=31qk−1pk∥U0∥Lq1q2q3(R3),{\parallel U\left(x,t)\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le N{t}^{-\tfrac{1}{2}\mathop{\sum }\limits_{k=1}^{n=3}\left(\tfrac{1}{{q}_{k}}-\tfrac{1}{{p}_{k}}\right)}{\parallel {U}_{0}\parallel }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})},(2.6)∥DxU(x,t)∥Lp1p2p3(R3)≤Nt−12−12∑k=1n=31qk−1pk∥U0∥Lq1q2q3(R3).{\parallel {D}_{x}U\left(x,t)\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le N{t}^{-\tfrac{1}{2}-\tfrac{1}{2}\mathop{\sum }\limits_{k=1}^{n=3}\left(\tfrac{1}{{q}_{k}}-\tfrac{1}{{p}_{k}}\right)}{\parallel {U}_{0}\parallel }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}.Lemma 2.3[4] Let p1,p2,p3∈(1,∞){p}_{1},{p}_{2},{p}_{3}\in \left(1,\infty ), and U0∈Lp1p2p3(R3){U}_{0}\in {L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3}). Let U(x,t)=eΔtU0(x)U\left(x,t)={e}^{\Delta t}{U}_{0}\left(x)be the solution of the heat equation (2.3) defined in (2.4). Then, U∈C([0,∞),Lp1p2p3(R3))U\in C\left({[}0,\infty ),{L}_{{p}_{1}{p}_{2}{p}_{3}}\left({{\mathbb{R}}}^{3}))and(2.7)limt→0+∥U(x,t)−U0∥Lp1p2p3(R3)=0.\mathop{\mathrm{lim}}\limits_{t\to {0}^{+}}{\parallel U\left(x,t)-{U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}=0.The following consequence illustrates the boundedness of Helmholtz-Leray projection P{\mathbb{P}}in mixed-norm Lebesgue spaces.Lemma 2.4[4] Let P=Id−∇Δ−1∇⋅{\mathbb{P}}={\rm{Id}}-\nabla {\Delta }^{-1}\nabla \cdot be the Helmholtz-Leray projection onto the divergence-free vector fields. Let p1,p2,p3∈(1,∞){p}_{1},{p}_{2},{p}_{3}\in \left(1,\infty ). Then, one has(2.8)∥P(f)∥Lp1p2p3(R3)≤N∥f∥Lp1p2p3(R3),{\parallel {\mathbb{P}}(f)\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le N{\parallel f\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},for all f∈(Lp1p2p3(R3))3f\in {({L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3}))}^{3}, where N=N(p1,p2,p3)N=N\left({p}_{1},{p}_{2},{p}_{3})is a positive constant.3MHD equations in 3D mixed-norm Lebesgue spaceWe apply P{\mathbb{P}}on the system (1.3), and then (1.3) can be expressed as follows: (3.1)Ut+AU+F(U,U)=0,inR3×(0,+∞),U(x,0)=U0(x),inR3,\left\{\begin{array}{ll}{U}_{t}+{\mathcal{A}}U+F\left(U,U)=0,& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ U\left(x,0)={U}_{0}\left(x),& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3},\end{array}\right.as P∇p˜=0{\mathbb{P}}\nabla \tilde{p}=0, where U=ub,U0=u0b0,A=−PΔ00−PΔ,F(U,U)=P(u⋅∇)u−P(b⋅∇)b+Pb⋅(∇b)P(u⋅∇)b−P(b⋅∇)u.\begin{array}{l}\phantom{\rule[-1.5em]{}{0ex}}U=\left(\begin{array}{l}u\\ b\end{array}\right),\hspace{1em}{U}_{0}=\left(\begin{array}{l}{u}_{0}\\ {b}_{0}\end{array}\right),\hspace{1em}{\mathcal{A}}=\left(\begin{array}{ll}-{\mathbb{P}}\Delta & 0\\ 0& -{\mathbb{P}}\Delta \end{array}\right),\\ F\left(U,U)=\left(\begin{array}{l}{\mathbb{P}}\left(u\cdot \nabla )u-{\mathbb{P}}\left(b\cdot \nabla )b+{\mathbb{P}}b\cdot \left(\nabla b)\\ {\mathbb{P}}\left(u\cdot \nabla )b-{\mathbb{P}}\left(b\cdot \nabla )u\end{array}\right).\end{array}By the Duhamel’s principle, the system (3.1) can be transformed into the following integral equations: (3.2)U=U1+G(U,U),U={U}_{1}+G\left(U,U),where U1=e−AtU0(x)=eΔtu0(x)eΔtb0(x){U}_{1}={e}^{-{\mathcal{A}}t}{U}_{0}\left(x)=\left(\begin{array}{l}{e}^{\Delta t}{u}_{0}\left(x)\\ {e}^{\Delta t}{b}_{0}\left(x)\end{array}\right)and (3.3)G(U,U)=−∫0te−(t−s)AF(U,U)ds\begin{array}{rcl}G\left(U,U)& =& -\underset{0}{\overset{t}{\displaystyle \int }}{e}^{-\left(t-s){\mathcal{A}}}F\left(U,U){\rm{d}}s\end{array}=−∫0te(t−s)Δ(P(u⋅∇)u−P(b⋅∇)b+Pb⋅(∇b))ds−∫0te(t−s)Δ(P(u⋅∇)b−P(b⋅∇)u)ds.\begin{array}{rcl}& =& \left(\begin{array}{l}-\underset{0}{\overset{t}{\displaystyle \int }}{e}^{\left(t-s)\Delta }({\mathbb{P}}\left(u\cdot \nabla )u-{\mathbb{P}}\left(b\cdot \nabla )b+{\mathbb{P}}b\cdot \left(\nabla b)){\rm{d}}s\\ -\underset{0}{\overset{t}{\displaystyle \int }}{e}^{\left(t-s)\Delta }({\mathbb{P}}\left(u\cdot \nabla )b-{\mathbb{P}}\left(b\cdot \nabla )u){\rm{d}}s\end{array}\right).\end{array}The main results in the paper are as follows.Theorem 3.1Let pk∈(1,+∞){p}_{k}\in \left(1,+\infty ), qk∈[pk,+∞){q}_{k}\in {[}{p}_{k},+\infty ), k=1,2,3,k=1,2,3,and1p1+1p2+1p3=1,1q1+1q2+1q3=δ∈(0,1).\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}+\frac{1}{{p}_{3}}=1,\hspace{1em}\frac{1}{{q}_{1}}+\frac{1}{{q}_{2}}+\frac{1}{{q}_{3}}=\delta \in \left(0,1).Then, there are a sufficiently small constant λ0>0{\lambda }_{0}\gt 0and a number N>0N\gt 0, depending on p1{p}_{1}, p2{p}_{2}, p3{p}_{3}, q1{q}_{1}, q2{q}_{2}, and q3{q}_{3}, such that the following results hold. (i)For all U0∈Lp1p2p3(R3){U}_{0}\in {L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})with divU0=0{\rm{div}}\hspace{0.33em}{U}_{0}=0, if ∥U0∥Lp1p2p3(R3)≤λ0{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le {\lambda }_{0}, then (3.1) has an unique global time solution U∈Xp,q,∞∩Yp,∞U\in {{\mathcal{X}}}_{p,q,\infty }\cap {{\mathcal{Y}}}_{p,\infty }with‖U‖χp,q,∞≤N∥U0∥Lp1p2p3(R3),‖U‖Yp,∞≤N(∥U0∥Lp1p2p3(R3)+∥U0∥Lp1p2p3(R3)2).\begin{array}{lcl}\Vert U{\Vert }_{{\chi }_{p,q,\infty }}& \le & N{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},\\ \Vert U{\Vert }_{{{\mathcal{Y}}}_{p,\infty }}& \le & N({\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}^{2}).\end{array}\hspace{0.55em}(ii)For all U0∈Lp1p2p3(R3){U}_{0}\in {L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})with divU0=0{\rm{div}}\hspace{0.33em}{U}_{0}=0, there is a sufficiently small T0>0{T}_{0}\gt 0depending on p1{p}_{1}, p2{p}_{2}, p3{p}_{3}, q1{q}_{1}, q2{q}_{2}, and q3{q}_{3}such that (3.1) has an unique local time solution U∈Xp,q,T0∩Yp,T0U\in {{\mathcal{X}}}_{p,q,{T}_{0}}\cap {{\mathcal{Y}}}_{p,{T}_{0}}with‖U‖χp,q,T0≤N∥U0∥Lp1p2p3(R3),‖U‖Yp,T0≤N(∥U0∥Lp1p2p3(R3)+∥U0∥Lp1p2p3(R3)2).\begin{array}{lcl}\Vert U{\Vert }_{{\chi }_{p,q,{T}_{0}}}& \le & N{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},\\ \Vert U{\Vert }_{{{\mathcal{Y}}}_{p,{T}_{0}}}& \le & N\left({\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}^{2}).\end{array}\hspace{0.2em}To prove Theorem 3.1, we need the following lemmas.Lemma 3.1[4] Let p1{p}_{1}, p2{p}_{2}, p3{p}_{3}, q1{q}_{1}, q2{q}_{2}, and q3{q}_{3}be given numbers and 1<pk≤qk<∞1\lt {p}_{k}\le {q}_{k}\lt \infty . Also, let σ≥0\sigma \ge 0be defined by σ=∑k=1n=31pk−1qk\sigma ={\sum }_{k=1}^{n=3}\left(\frac{1}{{p}_{k}}-\frac{1}{{q}_{k}}\right). (i)There exists a number NNdepending only on p1{p}_{1}, p2{p}_{2}, p3{p}_{3}, q1{q}_{1}, q2{q}_{2}, and q3{q}_{3}such that(3.4)∥e−AtPf∥Lq1q1q3(R3)≤Nt−σ2‖f‖Lp1P2p3(R3),{\parallel {e}^{-{\mathcal{A}}t}{\mathbb{P}}f\parallel }_{{L}_{{q}_{1}{q}_{1}{q}_{3}}({{\mathbb{R}}}^{3})}\le N{t}^{-\tfrac{\sigma }{2}}\Vert f{\Vert }_{{L}_{{p}_{1}{P}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},(3.5)∥Dxe−AtPf∥Lq1q1q3(R3)≤Nt−12(1+σ)‖f‖Lp1p2p3(R3),{\parallel {D}_{x}{e}^{-{\mathcal{A}}t}{\mathbb{P}}f\parallel }_{{L}_{{q}_{1}{q}_{1}{q}_{3}}({{\mathbb{R}}}^{3})}\le N{t}^{-\tfrac{1}{2}\left(1+\sigma )}\Vert f{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},for all f∈Lp1p2p3(R3)3f\in {L}_{{p}_{1}{p}_{2}{p}_{3}}{({{\mathbb{R}}}^{3})}^{3}.(ii)For all f∈Lp1p2p3(R3)3f\in {L}_{{p}_{1}{p}_{2}{p}_{3}}{({{\mathbb{R}}}^{3})}^{3}, the following assertions hold: if σ>0\sigma \gt 0, then(3.6)limt→0+tσ2∥e−AtPf∥Lq1q1q3(R3)=0,\mathop{\mathrm{lim}}\limits_{t\to {0}^{+}}{t}^{\tfrac{\sigma }{2}}{\parallel {e}^{-{\mathcal{A}}t}{\mathbb{P}}f\parallel }_{{L}_{{q}_{1}{q}_{1}{q}_{3}}({{\mathbb{R}}}^{3})}=0,(3.7)limt→0+∥[e−AtPf]−Pf∥Lp1P1p3(R3)=0,\mathop{\mathrm{lim}}\limits_{t\to {0}^{+}}{\parallel {[}{e}^{-{\mathcal{A}}t}{\mathbb{P}}f]-{\mathbb{P}}f\parallel }_{{L}_{{p}_{1}{P}_{1}{p}_{3}}({{\mathbb{R}}}^{3})}=0,(3.8)limt→0+t−12(1+σ)∥Dxe−AtPf∥Lq1q1q3(R3)=0.\mathop{\mathrm{lim}}\limits_{t\to {0}^{+}}{t}^{-\tfrac{1}{2}\left(1+\sigma )}{\parallel {D}_{x}{e}^{-{\mathcal{A}}t}{\mathbb{P}}f\parallel }_{{L}_{{q}_{1}{q}_{1}{q}_{3}}({{\mathbb{R}}}^{3})}=0.Lemma 3.2Let pk∈(1,∞),αk,βk,γk∈(0,1]{p}_{k}\in \left(1,\infty ),{\alpha }_{k},{\beta }_{k},{\gamma }_{k}\in (0,1]be given numbers satisfying γk≤αk+βk<pk{\gamma }_{k}\le {\alpha }_{k}+{\beta }_{k}\lt {p}_{k}, k=1,2,3.k=1,2,3.Letα=α1p1+α2p2+α2p2,β=β1p1+β2p2+β2p2,γ=γ1p1+γ2p2+γ2p2.\alpha =\frac{{\alpha }_{1}}{{p}_{1}}+\frac{{\alpha }_{2}}{{p}_{2}}+\frac{{\alpha }_{2}}{{p}_{2}},\hspace{1em}\beta =\frac{{\beta }_{1}}{{p}_{1}}+\frac{{\beta }_{2}}{{p}_{2}}+\frac{{\beta }_{2}}{{p}_{2}},\hspace{1em}\gamma =\frac{{\gamma }_{1}}{{p}_{1}}+\frac{{\gamma }_{2}}{{p}_{2}}+\frac{{\gamma }_{2}}{{p}_{2}}.Then, (3.9)‖G(U,U)‖Lp1γ1p2γ2p3γ3(R3)≤N∫0t(t−s)−α+β−γ2‖u‖Lp1α1p2α2p3α3(R3)∥Dxu∥Lp1β1p2β2p3β3(R3)+2‖b‖Lp1α1p2α2p3α3(R3)∥Dxb∥Lp1β1p2β2p3β3(R3)+‖u‖Lp1α1p2α2p3α3(R3)∥Dxb∥Lp1β1p2β2p3β3(R3)+‖b‖Lp1α1p2α2p3α3(R3)∥Dxu∥Lp1β1p2β2p3β3(R3)ds,\begin{array}{l}\Vert G\left(U,U){\Vert }_{{L}_{\frac{{p}_{1}}{{\gamma }_{1}}\frac{{p}_{2}}{{\gamma }_{2}}\frac{{p}_{3}}{{\gamma }_{3}}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\le N\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{\alpha +\beta -\gamma }{2}}\left(\Vert u{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}+2\Vert b{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}\right.\\ \hspace{2.0em}\left.+\Vert u{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}+\Vert b{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}\right){\rm{d}}s,\end{array}\hspace{0.65em}(3.10)‖DxG(U,U)‖Lp1γ1p2γ2p3γ3(R3)≤N∫0t(t−s)−1+α+β−γ2‖u‖Lp1α1p2α2p3α3(R3)∥Dxu∥Lp1β1p2β2p3β3(R3)+2‖b‖Lp1α1p2α2p3α3(R3)∥Dxb∥Lp1β1p2β2p3β3(R3)+‖u‖Lp1α1p2α2p3α3(R3)∥Dxb∥Lp1β1p2β2p3β3(R3)+‖b‖Lp1α1p2α2p3α3(R3)∥Dxu∥Lp1β1p2β2p3β3(R3)ds,\begin{array}{l}\Vert {D}_{x}G\left(U,U){\Vert }_{{L}_{\frac{{p}_{1}}{{\gamma }_{1}}\frac{{p}_{2}}{{\gamma }_{2}}\frac{{p}_{3}}{{\gamma }_{3}}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\le N\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1+\alpha +\beta -\gamma }{2}}\left(\Vert u{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}+2\Vert b{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}\right.\\ \hspace{2.0em}\left.+\Vert u{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}+\Vert b{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}\right){\rm{d}}s,\end{array}where N>0N\gt 0is a constant depending on pk,αk,βk,γk,k=1,2,3{p}_{k},{\alpha }_{k},{\beta }_{k},{\gamma }_{k},k=1,2,3.ProofWe first prove (3.9) in Lemma 3.2. For γk≤αk+βk<pk{\gamma }_{k}\le {\alpha }_{k}+{\beta }_{k}\lt {p}_{k}, we can gain pkγk≥pkαk+βk,∑k=1n=3αk+βkpk−γkpk=α+β−γ.\frac{{p}_{k}}{{\gamma }_{k}}\ge \frac{{p}_{k}}{{\alpha }_{k}+{\beta }_{k}},\hspace{1.0em}\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{{\alpha }_{k}+{\beta }_{k}}{{p}_{k}}-\frac{{\gamma }_{k}}{{p}_{k}}\right)=\alpha +\beta -\gamma .By (3.4), we can obtain (3.11)‖G(U,U)‖Lp1γ1p2γ2p3γ3(R3)≤N∫0t(t−s)−α+β−γ2‖F1(U,U)‖Lp1α1+β1p2α2+β2p3α3+β13(R3)ds,\Vert G\left(U,U){\Vert }_{{L}_{\frac{{p}_{1}}{{\gamma }_{1}}\frac{{p}_{2}}{{\gamma }_{2}}\frac{{p}_{3}}{{\gamma }_{3}}}({{\mathbb{R}}}^{3})}\le N\underset{0}{\overset{t}{\int }}{\left(t-s)}^{-\tfrac{\alpha +\beta -\gamma }{2}}\Vert {F}_{1}\left(U,U){\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{13}}}({{\mathbb{R}}}^{3})}{\rm{d}}s,where F1(U,U)=(u⋅∇)u−(b⋅∇)b+b⋅(∇b)(u⋅∇)b−(b⋅∇)u{F}_{1}\left(U,U)=\left(\begin{array}{l}\left(u\cdot \nabla )u-\left(b\cdot \nabla )b+b\cdot \left(\nabla b)\\ \left(u\cdot \nabla )b-\left(b\cdot \nabla )u\end{array}\right).By using Hölder’s inequality repeatedly, we can find that ‖(u⋅∇)b‖Lp1α1+β1p2α2+β2p3α3+β3(R3)=∫∫∫∣u⋅(∇b)∣p1α1+β1dx1p2(α1+β1)p1(α2+β2)dx2p3(α2+β2)p2(α3+β3)dx3(α3+β3)p3≤∫∫∫∣u∣p1α1dx1α1p1∫∣Dxb∣p1β1dx1β1p1p2(α2+β2)dx2p3(α2+β2)p2(α3+β3)dx3(α3+β3)p3≤∫∫∫∣u∣p1α1dx1p2α1p1α2dx2α2p2∫∫∣Dxb∣p1β1dx1p2β1p1β2dx2β2p2p3(α3+β3)dx3(α3+β3)p3\begin{array}{l}\Vert \left(u\cdot \nabla )b{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{3}}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}={\left[\displaystyle \int {\left[\displaystyle \int {\left[\displaystyle \int {| u\cdot \left(\nabla b)| }^{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}}{\rm{d}}{x}_{1}\right]}^{\frac{{p}_{2}\left({\alpha }_{1}+{\beta }_{1})}{{p}_{1}\left({\alpha }_{2}+{\beta }_{2})}}{\rm{d}}{x}_{2}\right]}^{\frac{{p}_{3}\left({\alpha }_{2}+{\beta }_{2})}{{p}_{2}\left({\alpha }_{3}+{\beta }_{3})}}{\rm{d}}{x}_{3}\right]}^{\tfrac{\left({\alpha }_{3}+{\beta }_{3})}{{p}_{3}}}\\ \hspace{1.0em}\le {\left[\displaystyle \int {\left[\displaystyle \int {\left[{\left(\displaystyle \int {| u| }^{\frac{{p}_{1}}{{\alpha }_{1}}}{\rm{d}}{x}_{1}\right)}^{\frac{{\alpha }_{1}}{{p}_{1}}}{\left(\displaystyle \int {| {D}_{x}b| }^{\frac{{p}_{1}}{{\beta }_{1}}}{\rm{d}}{x}_{1}\right)}^{\frac{{\beta }_{1}}{{p}_{1}}}\right]}^{\frac{{p}_{2}}{\left({\alpha }_{2}+{\beta }_{2})}}{\rm{d}}{x}_{2}\right]}^{\frac{{p}_{3}\left({\alpha }_{2}+{\beta }_{2})}{{p}_{2}\left({\alpha }_{3}+{\beta }_{3})}}{\rm{d}}{x}_{3}\right]}^{\tfrac{\left({\alpha }_{3}+{\beta }_{3})}{{p}_{3}}}\\ \hspace{1.0em}\le {\left[\displaystyle \int {\left[{\left[\displaystyle \int {\left[\displaystyle \int {| u| }^{\frac{{p}_{1}}{{\alpha }_{1}}}{\rm{d}}{x}_{1}\right]}^{\frac{{p}_{2}{\alpha }_{1}}{{p}_{1}{\alpha }_{2}}}{\rm{d}}{x}_{2}\right]}^{\frac{{\alpha }_{2}}{{p}_{2}}}{\left[\displaystyle \int {\left[\displaystyle \int {| {D}_{x}b| }^{\frac{{p}_{1}}{{\beta }_{1}}}{\rm{d}}{x}_{1}\right]}^{\frac{{p}_{2}{\beta }_{1}}{{p}_{1}{\beta }_{2}}}{\rm{d}}{x}_{2}\right]}^{\frac{{\beta }_{2}}{{p}_{2}}}\right]}^{\frac{{p}_{3}}{\left({\alpha }_{3}+{\beta }_{3})}}{\rm{d}}{x}_{3}\right]}^{\tfrac{\left({\alpha }_{3}+{\beta }_{3})}{{p}_{3}}}\end{array}\hspace{0.35em}≤∫∫∫∣u∣p1α1dx1p2α1p1α2dx2p3α2p2α3dx3α3p3∫∫∫∣Dxb∣p1β1dx1p2β1p1β2dx2p3β2p2β3dx3β3p3=‖u‖Lp1α1p2α2p3α3(R3)‖Dxb‖Lp1β1p2β2p3β3(R3).\begin{array}{l}\hspace{1.0em}\le {\left[\displaystyle \int {\left[\displaystyle \int {\left[\displaystyle \int | u{| }^{\frac{{p}_{1}}{{\alpha }_{1}}}{\rm{d}}{x}_{1}\right]}^{\frac{{p}_{2}{\alpha }_{1}}{{p}_{1}{\alpha }_{2}}}{\rm{d}}{x}_{2}\right]}^{\frac{{p}_{3}{\alpha }_{2}}{{p}_{2}{\alpha }_{3}}}{\rm{d}}{x}_{3}\right]}^{\tfrac{{\alpha }_{3}}{{p}_{3}}}{\left[\displaystyle \int {\left[\displaystyle \int {\left[\displaystyle \int {| {D}_{x}b| }^{\frac{{p}_{1}}{{\beta }_{1}}}{\rm{d}}{x}_{1}\right]}^{\frac{{p}_{2}{\beta }_{1}}{{p}_{1}{\beta }_{2}}}{\rm{d}}{x}_{2}\right]}^{\frac{{p}_{3}{\beta }_{2}}{{p}_{2}{\beta }_{3}}}{\rm{d}}{x}_{3}\right]}^{\tfrac{{\beta }_{3}}{{p}_{3}}}\\ \hspace{1.0em}=\Vert u{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}\Vert {D}_{x}b{\Vert }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}.\end{array}\hspace{2.75em}Then, ‖F1(U,U)‖Lp1α1+β1p2α2+β2p3α3+β13(R3)≤∥∣(u⋅∇)u∣+∣(b⋅∇)b∣+∣b⋅(∇b)∣+∣(u⋅∇)b∣+∣(b⋅∇)u∣∥Lp1α1+β1p2α2+β2p3α3+β13(R3)≤∥(u⋅∇)u∥Lp1α1+β1p2α2+β2p3α3+β13(R3)+∥(b⋅∇)b∥Lp1α1+β1p2α2+β2p3α3+β13(R3)+∥b⋅(∇b)∥Lp1α1+β1p2α2+β2p3α3+β13(R3)+∥(u⋅∇)b∥Lp1α1+β1p2α2+β2p3α3+β13(R3)+∥(b⋅∇)u∥Lp1α1+β1p2α2+β2p3α3+β13(R3)≤‖u‖Lp1α1p2α2p3α3(R3)‖Dxu‖Lp1β1p2β2p3β3(R3)+2‖b‖Lp1α1p2α2p3α3(R3)‖Dxb‖Lp1β1p2β2p3β3(R3)+‖u‖Lp1α1p2α2p3α3(R3)‖Dxb‖Lp1β1p2β2p3β3(R3)+‖b‖Lp1α1p2α2p3α3(R3)‖Dxu‖Lp1β1p2β2p3β3(R3).\begin{array}{l}\Vert {F}_{1}\left(U,U){\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{13}}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\le \parallel | \left(u\cdot \nabla )u| +| \left(b\cdot \nabla )b| +| b\cdot \left(\nabla b)| +| \left(u\cdot \nabla )b| +| \left(b\cdot \nabla )u| {\parallel }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{13}}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\le \parallel \left(u\cdot \nabla )u{\parallel }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{13}}}({{\mathbb{R}}}^{3})}+\parallel \left(b\cdot \nabla )b{\parallel }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{13}}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\hspace{1.0em}+\parallel b\cdot \left(\nabla b){\parallel }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{13}}}({{\mathbb{R}}}^{3})}+\parallel \left(u\cdot \nabla )b{\parallel }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{13}}}({{\mathbb{R}}}^{3})}+\parallel \left(b\cdot \nabla )u{\parallel }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{13}}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\le \Vert u{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}\Vert {D}_{x}u{\Vert }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}+2\Vert b{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}\Vert {D}_{x}b{\Vert }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\hspace{1.0em}+\Vert u{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}\Vert {D}_{x}b{\Vert }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}+\Vert b{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}\Vert {D}_{x}u{\Vert }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}.\end{array}By substituting the aforementioned formula into (3.11), we can obtain (3.9). Similarly, (3.10) can be proved by (3.5).□Lemma 3.3[4] Let XXbe a Banach space with norm ‖⋅‖X\Vert \cdot {\Vert }_{X}. Let G:X×X→XG:X\times X\to Xbe a bilinear map such that there is N0>0{N}_{0}\gt 0so that‖G(U,V)‖X≤N0‖U‖X‖V‖X,∀U,V∈X.\Vert G\left(U,V){\Vert }_{X}\le {N}_{0}\Vert U{\Vert }_{X}\Vert V{\Vert }_{X},\hspace{1em}\forall U,V\in X.Then, for all U1∈X{U}_{1}\in Xwith 4N0‖U1‖X<14{N}_{0}\Vert {U}_{1}{\Vert }_{X}\lt 1, the equationU=U1+G(U,U)U={U}_{1}+G\left(U,U)has an unique solution UUwith‖U‖X≤2‖U1‖X.\Vert U{\Vert }_{X}\le 2\Vert {U}_{1}{\Vert }_{X}.Proof of Theorem 3.1We now prove (i). First, we start from the proof that U∈χp,q,∞U\in {\chi }_{p,q,\infty }. From Lemma 2.2 and σ=∑k=1n=31pk−1qk=1−δ\sigma ={\sum }_{k=1}^{n=3}\left(\frac{1}{{p}_{k}}-\frac{1}{{q}_{k}}\right)=1-\delta , we have ∥U1∥Lq1q1q3(R3)≤N1t−1−δ2‖U0‖Lp1P2p3(R3),∥DxU1∥Lq1q2q3(R3)≤N1t−12‖U0‖Lp1p2p3(R3),\begin{array}{rcl}{\parallel {U}_{1}\parallel }_{{L}_{{q}_{1}{q}_{1}{q}_{3}}({{\mathbb{R}}}^{3})}& \le & {N}_{1}{t}^{-\tfrac{1-\delta }{2}}\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{P}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},\\ {\parallel {D}_{x}{U}_{1}\parallel }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}& \le & {N}_{1}{t}^{-\tfrac{1}{2}}\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},\end{array}where N1>0{N}_{1}\gt 0is a constant depending on p1{p}_{1}, p2{p}_{2}, p3{p}_{3}, q1{q}_{1}, q2{q}_{2}, and q3{q}_{3}.Furthermore, according to Lemma 3.1, we know that when t→0t\to 0, t−1−σ2e−AtP→0{t}^{-\tfrac{1-\sigma }{2}}{e}^{-{\mathcal{A}}t}{\mathbb{P}}\to 0, and when t=0t=0, t−1−σ2U1=0{t}^{-\tfrac{1-\sigma }{2}}{U}_{1}=0. Hence, t−1−σ2e−AtP:Lp1p2p3(R3)→PLq1q2q3(R3){t}^{-\tfrac{1-\sigma }{2}}{e}^{-{\mathcal{A}}t}{\mathbb{P}}:{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})\to P{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})is uniformly bounded. Similarly, when t→0t\to 0, t−12Dxe−AtP→0{t}^{-\tfrac{1}{2}}{D}_{x}{e}^{-{\mathcal{A}}t}{\mathbb{P}}\to 0, and when t=0t=0, t−12DxU1=0{t}^{-\tfrac{1}{2}}{D}_{x}{U}_{1}=0. Hence, t−1−σ2Dxe−AtP:Lp1p2p3(R3)→PLp1p2p3(R3){t}^{-\tfrac{1-\sigma }{2}}{D}_{x}{e}^{-{\mathcal{A}}t}{\mathbb{P}}:{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})\to P{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})is also uniformly bounded. Hence, we have U0∈χp,q,∞{U}_{0}\in {\chi }_{p,q,\infty }and (3.12)‖U1‖χp,q,∞≤N1∥U0∥Lp1p2p3(R3).\Vert {U}_{1}{\Vert }_{{\chi }_{p,q,\infty }}\le {N}_{1}{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}.Now let’s prove that bilinear G:χp,q,∞×χp,q,∞→χp,q,∞G:{\chi }_{p,q,\infty }\times {\chi }_{p,q,\infty }\to {\chi }_{p,q,\infty }is bounded.Let βk=1{\beta }_{k}=1, γk=αk=pkqk∈(0,1]{\gamma }_{k}={\alpha }_{k}=\frac{{p}_{k}}{{q}_{k}}\in (0,1], we obtain pkγk=qk\frac{{p}_{k}}{{\gamma }_{k}}={q}_{k}, −α+β−γ2=−12∑k=1n=3αkpk+βkpk−γkpk=−12∑k=1n=31qk+1pk−1qk=−12∑k=1n=31pk=−12.-\frac{\alpha +\beta -\gamma }{2}=-\frac{1}{2}\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{{\alpha }_{k}}{{p}_{k}}+\frac{{\beta }_{k}}{{p}_{k}}-\frac{{\gamma }_{k}}{{p}_{k}}\right)=-\frac{1}{2}\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{1}{{q}_{k}}+\frac{1}{{p}_{k}}-\frac{1}{{q}_{k}}\right)=-\frac{1}{2}\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{1}{{p}_{k}}\right)=-\frac{1}{2}.By using 1p1+1p2+1p3=1\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}+\frac{1}{{p}_{3}}=1, 1q1+1q2+1q3=δ∈(0,1)\frac{1}{{q}_{1}}+\frac{1}{{q}_{2}}+\frac{1}{{q}_{3}}=\delta \in \left(0,1), (3.9) and the definition of Xp,q,T{{\mathcal{X}}}_{p,q,T}, and applying (3.3), we can obtain ‖G(U,U)‖Lq1q2q3(R3)≤N∫0t(t−s)−12(‖u‖Lq1q2q3(R3)∥Dxu∥Lp1p2p3(R3)+2‖b‖Lq1q2q3(R3)∥Dxb∥Lp1p2p3(R3)+‖u‖Lq1q2q3(R3)∥Dxb∥Lp1p2p3(R3)+‖b‖Lq1q2q3(R3)∥Dxu∥Lp1p2p3(R3))ds≤N∫0t(t−s)−12s1−δ2‖u‖Lq1q2q3(R3)s12∥Dxu∥Lp1p2p3(R3)s−1−δ2s−12+2s1−δ2‖b‖Lq1q2q3(R3)s12∥Dxb∥Lp1p2p3(R3)s−1−δ2s−12+s1−δ2‖u‖Lq1q2q3(R3)s12×∥Dxb∥Lp1p2p3(R3)s−1−δ2s−12+s1−δ2‖b‖Lq1q2q3(R3)s12∥Dxu∥Lp1p2p3(R3)s−1−δ2s−12ds≤N(∥u∥χp,q,∞2+∥b∥χp,q,∞2+2‖u‖χp,q,∞‖b‖χp,q,∞)∫0t(t−s)−12s−1+δ2ds,\begin{array}{rcl}\Vert G\left(U,U){\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}& \le & N\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1}{2}}(\Vert u{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+2\Vert b{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\\ & & +\Vert u{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+\Vert b{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}){\rm{d}}s\\ & \le & N\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1}{2}}\left({s}^{\tfrac{1-\delta }{2}}\Vert u{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{s}^{\tfrac{1}{2}}{\parallel {D}_{x}u\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}{s}^{-\tfrac{1-\delta }{2}}{s}^{-\tfrac{1}{2}}\right.\\ & & +2{s}^{\tfrac{1-\delta }{2}}\Vert b{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{s}^{\tfrac{1}{2}}{\parallel {D}_{x}b\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}{s}^{-\tfrac{1-\delta }{2}}{s}^{-\tfrac{1}{2}}+{s}^{\tfrac{1-\delta }{2}}\Vert u{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{s}^{\tfrac{1}{2}}\\ & & \times {\parallel {D}_{x}b\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}{s}^{-\tfrac{1-\delta }{2}}{s}^{-\tfrac{1}{2}}\left.+{s}^{\tfrac{1-\delta }{2}}\Vert b{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{s}^{\tfrac{1}{2}}{\parallel {D}_{x}u\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}{s}^{-\tfrac{1-\delta }{2}}{s}^{-\tfrac{1}{2}}\right){\rm{d}}s\\ & \le & N({\parallel u\parallel }_{{\chi }_{p,q,\infty }}^{2}+{\parallel b\parallel }_{{\chi }_{p,q,\infty }}^{2}+2\Vert u{\Vert }_{{\chi }_{p,q,\infty }}\Vert b{\Vert }_{{\chi }_{p,q,\infty }})\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1}{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s,\end{array}and ∫0t(t−s)−12s−1+δ2ds=∫0t2(t−s)−12s−1+δ2ds+∫t2t(t−s)−12s−1+δ2ds≤t2−12∫0t2s−1+δ2ds+t2−1+δ2∫t2t(t−s)−12ds≤Nt2−12t2δ2+Nt2−1+δ2t212≤Nt−1−δ2.\hspace{-29em}\begin{array}{rcl}\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1}{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s& =& \underset{0}{\overset{\frac{t}{2}}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1}{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s+\underset{\frac{t}{2}}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1}{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s\\ & \le & {\left(\frac{t}{2}\right)}^{-\tfrac{1}{2}}\underset{0}{\overset{\frac{t}{2}}{\displaystyle \int }}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s+{\left(\frac{t}{2}\right)}^{-1+\tfrac{\delta }{2}}\underset{\frac{t}{2}}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1}{2}}{\rm{d}}s\\ & \le & N{\left(\frac{t}{2}\right)}^{-\tfrac{1}{2}}{\left(\frac{t}{2}\right)}^{\tfrac{\delta }{2}}+N{\left(\frac{t}{2}\right)}^{-1+\tfrac{\delta }{2}}{\left(\frac{t}{2}\right)}^{\tfrac{1}{2}}\\ & \le & N{t}^{-\tfrac{1-\delta }{2}}.\end{array}\hspace{12.2em}So, we have (3.13)‖G(U,U)‖Lq1q2q3(R3)≤Nt−1−δ2(∥u∥χp,q,∞2+2∥b∥χp,q,∞2+2‖u‖χp,q,∞‖b‖χp,q,∞).\Vert G\left(U,U){\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}\le N{t}^{-\tfrac{1-\delta }{2}}({\parallel u\parallel }_{{\chi }_{p,q,\infty }}^{2}+2{\parallel b\parallel }_{{\chi }_{p,q,\infty }}^{2}+2\Vert u{\Vert }_{{\chi }_{p,q,\infty }}\Vert b{\Vert }_{{\chi }_{p,q,\infty }}).\hspace{2.1em}Similarly, let βk=γk=1{\beta }_{k}={\gamma }_{k}=1, αk=pkqk∈(0,1]{\alpha }_{k}=\frac{{p}_{k}}{{q}_{k}}\in (0,1], and pkγk=qk\frac{{p}_{k}}{{\gamma }_{k}}={q}_{k}, −1+α+β−γ2=−121+∑k=1n=3αkpk+βkpk−γkpk=−121+∑k=1n=31qk+1pk−1pk=−121+∑k=1n=31qk=−1+δ2.-\frac{1+\alpha +\beta -\gamma }{2}=-\frac{1}{2}\left(1+\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{{\alpha }_{k}}{{p}_{k}}+\frac{{\beta }_{k}}{{p}_{k}}-\frac{{\gamma }_{k}}{{p}_{k}}\right)\right)=-\frac{1}{2}\left(1+\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{1}{{q}_{k}}+\frac{1}{{p}_{k}}-\frac{1}{{p}_{k}}\right)\right)=-\frac{1}{2}\left(1+\mathop{\sum }\limits_{k=1}^{n=3}\frac{1}{{q}_{k}}\right)=-\frac{1+\delta }{2}.By using 1p1+1p2+1p3=1\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}+\frac{1}{{p}_{3}}=1, 1q1+1q2+1q3=δ∈(0,1)\frac{1}{{q}_{1}}+\frac{1}{{q}_{2}}+\frac{1}{{q}_{3}}=\delta \in \left(0,1), and (3.10), we also obtain ‖DxG(U,U)‖Lp1p2p3(R3)≤N∫0t(t−s)−1+δ2(‖u‖Lq1q2q3(R3)∥Dxu∥Lp1p2p3(R3)+2‖b‖Lq1q2q3(R3)∥Dxb∥Lp1p2p3(R3)+‖u‖Lq1q2q3(R3)∥Dxb∥Lp1p2p3(R3)+‖b‖Lq1q2q3(R3)∥Dxu∥Lp1p2p3(R3))ds≤N(∥u∥χp,q,∞2+∥b∥χp,q,∞2+2‖u‖χp,q,∞‖b‖χp,q,∞)∫0t(t−s)−1+δ2s−1+δ2ds,\begin{array}{l}\Vert {D}_{x}G\left(U,U){\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\le N\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1+\delta }{2}}(\Vert u{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+2\Vert b{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\hspace{1.0em}+\Vert u{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+\Vert b{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}){\rm{d}}s\\ \hspace{1.0em}\le N({\parallel u\parallel }_{{\chi }_{p,q,\infty }}^{2}+{\parallel b\parallel }_{{\chi }_{p,q,\infty }}^{2}+2\Vert u{\Vert }_{{\chi }_{p,q,\infty }}\Vert b{\Vert }_{{\chi }_{p,q,\infty }})\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1+\delta }{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s,\end{array}and ∫0t(t−s)−1+δ2s−1+δ2ds=∫0t2(t−s)−1+δ2s−1+δ2ds+∫t2t(t−s)−1+δ2s−1+δ2ds≤t2−1+δ2∫0t2s−1+δ2ds+t2−1+δ2∫t2t(t−s)−1+δ2ds≤Nt2−1+δ2t2δ2+Nt2−1+δ2t21+δ2≤Nt−12.\hspace{6.25em}\begin{array}{l}\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1+\delta }{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s\\ \hspace{1.0em}=\underset{0}{\overset{\frac{t}{2}}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1+\delta }{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s+\underset{\frac{t}{2}}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1+\delta }{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s\\ \hspace{1.0em}\le {\left(\frac{t}{2}\right)}^{-\tfrac{1+\delta }{2}}\underset{0}{\overset{\frac{t}{2}}{\displaystyle \int }}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s+{\left(\frac{t}{2}\right)}^{-1+\tfrac{\delta }{2}}\underset{\frac{t}{2}}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1+\delta }{2}}{\rm{d}}s\\ \hspace{1.0em}\le N{\left(\frac{t}{2}\right)}^{-\tfrac{1+\delta }{2}}{\left(\frac{t}{2}\right)}^{\tfrac{\delta }{2}}+N{\left(\frac{t}{2}\right)}^{-1+\tfrac{\delta }{2}}{\left(\frac{t}{2}\right)}^{\tfrac{1+\delta }{2}}\\ \hspace{1.0em}\le N{t}^{-\tfrac{1}{2}}.\end{array}\hspace{12.5em}So, we have (3.14)‖DxG(U,U)‖Lp1p2p3(R3)≤Nt−12(∥u∥χp,q,∞2+2∥b∥χp,q,∞2+2‖u‖χp,q,∞‖b‖χp,q,∞).\Vert {D}_{x}G\left(U,U){\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le N{t}^{-\tfrac{1}{2}}({\parallel u\parallel }_{{\chi }_{p,q,\infty }}^{2}+2{\parallel b\parallel }_{{\chi }_{p,q,\infty }}^{2}+2\Vert u{\Vert }_{{\chi }_{p,q,\infty }}\Vert b{\Vert }_{{\chi }_{p,q,\infty }}).From the estimates (3.13), (3.14), and Lemma 3.1, we can prove that t1−δ2G(U,U):[0,∞)→PLq1q2q3(R3){t}^{\tfrac{1-\delta }{2}}G\left(U,U):{[}0,\infty )\to {\mathbb{P}}{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})is continuous and when t→0t\to 0, t1−δ2G(U,U)→0{t}^{\tfrac{1-\delta }{2}}G\left(U,U)\to 0. Similarly, we can also prove that t12DxG(U,U):[0,∞)→PLp1p2p3(R3){t}^{\tfrac{1}{2}}{D}_{x}G\left(U,U):{[}0,\infty )\to {\mathbb{P}}{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})is continuous and when t→0t\to 0, t12DxG(U,U)→0{t}^{\tfrac{1}{2}}{D}_{x}G\left(U,U)\to 0. Therefore, we obtain G(U,U)∈χp,q,∞G\left(U,U)\in {\chi }_{p,q,\infty }and for all U∈χp,q,∞U\in {\chi }_{p,q,\infty }: (3.15)‖G(U,U)‖χp,q,∞≤N(∥u∥χp,q,∞2+2∥b∥χp,q,∞2+2‖u‖χp,q,∞‖b‖χp,q,∞)≤N‖U‖χp,q,∞‖U‖χp,q,∞≤N2∥U∥χp,q,∞2,∀U∈χp,q,∞,\begin{array}{rcl}\Vert G\left(U,U)\Vert {\chi }_{p,q,\infty }& \le & N({\parallel u\parallel }_{{\chi }_{p,q,\infty }}^{2}+2{\parallel b\parallel }_{{\chi }_{p,q,\infty }}^{2}+2\Vert u{\Vert }_{{\chi }_{p,q,\infty }}\Vert b{\Vert }_{{\chi }_{p,q,\infty }})\\ & \le & N\Vert U{\Vert }_{{\chi }_{p,q,\infty }}\Vert U{\Vert }_{{\chi }_{p,q,\infty }}\\ & \le & {N}_{2}{\parallel U\parallel }_{{\chi }_{p,q,\infty }}^{2},\hspace{1em}\forall U\in {\chi }_{p,q,\infty },\end{array}where N2{N}_{2}is a constant depending on p1{p}_{1}, p2{p}_{2}, p3{p}_{3}, q1{q}_{1}, q2{q}_{2}, and q3{q}_{3}. That is, the bilinear G:χp,q,∞×χp,q,∞→χp,q,∞G:{\chi }_{p,q,\infty }\times {\chi }_{p,q,\infty }\to {\chi }_{p,q,\infty }is bounded.Next, let us choose a sufficiently small constant λ0{\lambda }_{0}so that 4N1N2λ0<1,4{N}_{1}{N}_{2}{\lambda }_{0}\lt 1,where N1{N}_{1}is defined in (3.12) and N2{N}_{2}is defined in (3.15). If ∥U0∥Lp1p2p3(R3)≤λ0{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le {\lambda }_{0}, then it can be obtained by (3.12) 4N2‖U1‖χp,q,∞≤4N1N2∥U0∥Lp1p2p3(R3)≤4N1N2λ0<1.4{N}_{2}\Vert {U}_{1}{\Vert }_{{\chi }_{p,q,\infty }}\le 4{N}_{1}{N}_{2}{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le 4{N}_{1}{N}_{2}{\lambda }_{0}\lt 1.By this and Lemma 3.3, we can gain a unique time solution U∈χp,q,∞U\in {\chi }_{p,q,\infty }of (3.2) such that (3.16)‖U‖χp,q,∞≤2∥U1∥χp,q,∞≤2N1∥U0∥Lp1p2p3(R3).\Vert U{\Vert }_{{\chi }_{p,q,\infty }}\le 2{\parallel {U}_{1}\parallel }_{{\chi }_{p,q,\infty }}\le 2{N}_{1}{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}.Now, we need to prove that U∈Yp,∞U\in {{\mathcal{Y}}}_{p,\infty }. As U=U1+G(U,U)U={U}_{1}+G\left(U,U), we have (3.17)‖U‖Lp1p2p3(R3)≤‖U1‖Lp1p2p3(R3)+‖G(U,U)‖Lp1p2p3(R3),\Vert U{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le \Vert {U}_{1}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+\Vert G\left(U,U){\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},\hspace{0.95em}(3.18)‖DxU‖Lp1p2p3(R3)≤‖DxU1‖Lp1p2p3(R3)+‖DxG(U,U)‖Lp1p2p3(R3).\Vert {D}_{x}U{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le \Vert {D}_{x}{U}_{1}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+\Vert {D}_{x}G\left(U,U){\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}.Then, by Lemma 3.1, we see that (3.19)∥U1∥Lp1p1p3(R3)≤N‖U0‖Lp1p2p3(R3),{\parallel {U}_{1}\parallel }_{{L}_{{p}_{1}{p}_{1}{p}_{3}}({{\mathbb{R}}}^{3})}\le N\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},(3.20)∥DxU1∥Lp1p2p3(R3)≤Nt−12‖U0‖Lp1p2p3(R3).{\parallel {D}_{x}{U}_{1}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le N{t}^{-\tfrac{1}{2}}\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}.Let βk=γk=1{\beta }_{k}={\gamma }_{k}=1, αk=pkqk∈(0,1]{\alpha }_{k}=\frac{{p}_{k}}{{q}_{k}}\in (0,1], and pkγk=pk\frac{{p}_{k}}{{\gamma }_{k}}={p}_{k}, −α+β−γ2=−12∑k=1n=3αkpk+βkpk−γkpk=−12∑k=1n=31qk+1pk−1pk=−12∑k=1n=31qk=−δ2.-\frac{\alpha +\beta -\gamma }{2}=-\frac{1}{2}\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{{\alpha }_{k}}{{p}_{k}}+\frac{{\beta }_{k}}{{p}_{k}}-\frac{{\gamma }_{k}}{{p}_{k}}\right)=-\frac{1}{2}\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{1}{{q}_{k}}+\frac{1}{{p}_{k}}-\frac{1}{{p}_{k}}\right)=-\frac{1}{2}\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{1}{{q}_{k}}\right)=-\frac{\delta }{2}.By using (3.9), we can infer that ‖G(U,U)‖Lp1p2p3(R3)≤N∫0t(t−s)−δ2(‖u‖Lq1q2q3(R3)∥Dxu∥Lp1p2p3(R3)+2‖b‖Lq1q2q3(R3)∥Dxb∥Lp1p2p3(R3)+‖u‖Lq1q2q3(R3)∥Dxb∥Lp1p2p3(R3)+‖b‖Lq1q2q3(R3)∥Dxu∥Lp1p2p3(R3))ds≤N(∥u∥χp,q,∞2+∥b∥χp,q,∞2+2‖u‖χp,q,∞‖b‖χp,q,∞)∫0t(t−s)−δ2s−1+δ2ds.\begin{array}{l}\Vert G\left(U,U){\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\le N\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{\delta }{2}}(\Vert u{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+2\Vert b{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\hspace{1.0em}+\Vert u{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+\Vert b{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}){\rm{d}}s\\ \hspace{1.0em}\le N({\parallel u\parallel }_{{\chi }_{p,q,\infty }}^{2}+{\parallel b\parallel }_{{\chi }_{p,q,\infty }}^{2}+2\Vert u{\Vert }_{{\chi }_{p,q,\infty }}\Vert b{\Vert }_{{\chi }_{p,q,\infty }})\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{\delta }{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s.\end{array}Since ∫0t(t−s)−δ2s−1+δ2ds=∫0t2(t−s)−δ2s−1+δ2ds+∫t2t(t−s)−δ2s−1+δ2ds≤t2−δ2∫0t2s−1+δ2ds+t2−1+δ2∫t2t(t−s)−δ2ds≤N1t2−δ2t2δ2+N2t2−1+δ2t2−δ2+1≤N,\hspace{-27.1em}\begin{array}{rcl}\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{\delta }{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s& =& \underset{0}{\overset{\frac{t}{2}}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{\delta }{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s+\underset{\frac{t}{2}}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{\delta }{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s\\ & \le & {\left(\frac{t}{2}\right)}^{-\tfrac{\delta }{2}}\underset{0}{\overset{\frac{t}{2}}{\displaystyle \int }}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s+{\left(\frac{t}{2}\right)}^{-1+\tfrac{\delta }{2}}\underset{\frac{t}{2}}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{\delta }{2}}{\rm{d}}s\\ & \le & {N}_{1}{\left(\frac{t}{2}\right)}^{-\tfrac{\delta }{2}}{\left(\frac{t}{2}\right)}^{\tfrac{\delta }{2}}+{N}_{2}{\left(\frac{t}{2}\right)}^{-1+\tfrac{\delta }{2}}{\left(\frac{t}{2}\right)}^{-\tfrac{\delta }{2}+1}\\ & \le & N,\end{array}by using (3.16), we obtain that (3.21)‖G(U,U)‖Lp1p2p3(R3)≤N(∥u∥χp,q,∞2+2∥b∥χp,q,∞2+2‖u‖χp,q,∞‖b‖χp,q,∞)≤N∥U∥χp,q,∞2≤N∥U0∥Lp1p2p3(R3)2.\begin{array}{rcl}\Vert G\left(U,U){\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}& \le & N({\parallel u\parallel }_{{\chi }_{p,q,\infty }}^{2}+2{\parallel b\parallel }_{{\chi }_{p,q,\infty }}^{2}+2\Vert u{\Vert }_{{\chi }_{p,q,\infty }}\Vert b{\Vert }_{{\chi }_{p,q,\infty }})\\ & \le & N{\parallel U\parallel }_{{\chi }_{p,q,\infty }}^{2}\le N{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}^{2}.\end{array}Similarly, by using (3.14) and (3.16), we obtain that (3.22)‖DxG(U,U)‖Lp1p2p3(R3)≤Nt−12(∥u∥χp,q,∞2+2∥b∥χp,q,∞2+2‖u‖χp,q,∞‖b‖χp,q,∞)≤Nt−12∥U∥χp,q,∞2≤Nt−12∥U0∥Lp1p2p3(R3)2.\begin{array}{rcl}\Vert {D}_{x}G\left(U,U){\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}& \le & N{t}^{-\tfrac{1}{2}}({\parallel u\parallel }_{{\chi }_{p,q,\infty }}^{2}+2{\parallel b\parallel }_{{\chi }_{p,q,\infty }}^{2}+2\Vert u{\Vert }_{{\chi }_{p,q,\infty }}\Vert b{\Vert }_{{\chi }_{p,q,\infty }})\\ & \le & N{t}^{-\tfrac{1}{2}}{\parallel U\parallel }_{{\chi }_{p,q,\infty }}^{2}\le N{t}^{-\tfrac{1}{2}}{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}^{2}.\end{array}\hspace{0.1em}From estimates (3.17), (3.19), and (3.21), we see that ‖U‖Lp1p2p3(R3)≤N(‖U0‖Lp1p2p3(R3)+‖U0‖Lp1p2p3(R3)2).\Vert U{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le N(\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}^{2}).\hspace{2.5em}From the estimates (3.18), (3.20), and (3.22), we see that ‖DxU‖Lp1p2p3(R3)≤Nt−12(‖U0‖Lp1p2p3(R3)+‖U0‖Lp1p2p3(R3)2).\Vert {D}_{x}U{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le N{t}^{-\tfrac{1}{2}}(\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}^{2}).\hspace{2.5em}It is known that ‖U0‖Lp1p2p3(R3)\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}is sufficiently small, we can gain ‖U‖Yp,∞≤N0(‖U0‖Lp1p2p3(R3)+‖U0‖Lp1p2p3(R3)2).\Vert U{\Vert }_{{{\mathcal{Y}}}_{p,\infty }}\le {N}_{0}(\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}^{2}).\hspace{0.35em}Now, we have to prove (ii). Similar to the proof of (3.12), we found that U1∈χp,q,∞{U}_{1}\in {\chi }_{p,q,\infty }, χp,q,∞{\chi }_{p,q,\infty }is continuous, and when t→0t\to 0, t1−δ2U1→0{t}^{\tfrac{1-\delta }{2}}{U}_{1}\to 0, t12DxU1→0{t}^{\tfrac{1}{2}}{D}_{x}{U}_{1}\to 0. There is a sufficiently small constant T0>0{T}_{0}\gt 0depending on p1{p}_{1}, p2{p}_{2}, p3{p}_{3}, q1{q}_{1}, q2{q}_{2}, q3{q}_{3}, and U0{U}_{0}such that ‖U1‖χp,q,T0≤λ0.\Vert {U}_{1}{\Vert }_{{\chi }_{p,q,{T}_{0}}}\le {\lambda }_{0}.\hspace{5.75em}Furthermore, similar to the proof of (3.15), we also found that the bilinear GG: χp,q,T0×χp,q,T0→χp,q,T0{\chi }_{p,q,{T}_{0}}\times {\chi }_{p,q,{T}_{0}}\to {\chi }_{p,q,{T}_{0}}is bounded and ‖G(U,U)‖χp,q,T0≤N‖U‖χp,q,T0‖U‖χp,q,T0,∀U∈χp,q,T0.\Vert G\left(U,U){\Vert }_{{\chi }_{p,q,{T}_{0}}}\le N\Vert U{\Vert }_{{\chi }_{p,q,{T}_{0}}}\Vert U{\Vert }_{{\chi }_{p,q,{T}_{0}}},\hspace{1em}\forall U\in {\chi }_{p,q,{T}_{0}}.Then, by Lemma 3.3, we can obtain a unique local time solution U∈χp,q,T0U\in {\chi }_{p,q,{T}_{0}}of (3.2) with ‖U‖χp,q,T0≤2N1∥U0∥Lp1p2p3(R3).\Vert U{\Vert }_{{\chi }_{p,q,{T}_{0}}}\le 2{N}_{1}{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}.\hspace{4.1em}As in proof (i), we can gain U∈Yp,T0U\in {{\mathcal{Y}}}_{p,{T}_{0}}and ‖U‖Yp,T0≤N(∥U0∥Lp1p2p3(R3)+∥U0∥Lp1p2p3(R3)2).□\hspace{12em}\Vert U{\Vert }_{{{\mathcal{Y}}}_{p,{T}_{0}}}\le N({\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}^{2}).\hspace{14.25em}\square http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Open Mathematics de Gruyter

Well posedness of magnetohydrodynamic equations in 3D mixed-norm Lebesgue space

Open Mathematics , Volume 20 (1): 11 – Jan 1, 2022

Loading next page...
 
/lp/de-gruyter/well-posedness-of-magnetohydrodynamic-equations-in-3d-mixed-norm-jAmipHt5Pg
Publisher
de Gruyter
Copyright
© 2022 Yongfang Liu and Chaosheng Zhu, published by De Gruyter
ISSN
2391-5455
eISSN
2391-5455
DOI
10.1515/math-2022-0014
Publisher site
See Article on Publisher Site

Abstract

1IntroductionUnlike a usual Lebesgue space [1,2,3], the mixed-norm Lebesgue space allows its norm decay to zero with different rates as ∣x∣→∞| x| \to \infty in different spatial directions [4]. On the basis of the mixed-norm Lebesgue space feature, we investigate the following magnetohydrodynamic (MHD) equations in R3{{\mathbb{R}}}^{3}[1]: (1.1)∂tu−1ReΔu+u⋅∇u−S(∇×b)×b+∇p˜=0,inR3×(0,+∞),∂tb−∇×(u×b)+1Rm∇×(∇×b)=0,inR3×(0,+∞),u(x,0)=u0(x),b(x,0)=b0(x),inR3,\left\{\begin{array}{ll}{\partial }_{t}u-\frac{1}{{\rm{Re}}}\Delta u+u\cdot \nabla u-S\left(\nabla \times b)\times b+\nabla \tilde{p}=0,& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ {\partial }_{t}b-\nabla \times \left(u\times b)+\frac{1}{{\rm{Rm}}}\nabla \times \left(\nabla \times b)=0,& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ u\left(x,0)={u}_{0}\left(x),\hspace{1em}b\left(x,0)={b}_{0}\left(x),& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3},\end{array}\right.where uu, bb, u0{u}_{0}, and b0{b}_{0}satisfy (1.2)divu(x,t)=divb(x,t)=0,divu0=divb0=0,{\rm{div}}\hspace{0.33em}u\left(x,t)={\rm{div}}\hspace{0.33em}b\left(x,t)=0,\hspace{1em}{\rm{div}}\hspace{0.33em}{u}_{0}={\rm{div}}\hspace{0.33em}{b}_{0}=0,S=M2ReRmS=\frac{{M}^{2}}{{\rm{ReRm}}}, Re>0{\rm{Re}}\gt 0is the Reynolds number, Rm>0{\rm{Rm}}\gt 0is the magnetic Reynolds number, MMis the Hartman number. u:R3×(0,+∞)→R3u:{{\mathbb{R}}}^{3}\times \left(0,+\infty )\to {{\mathbb{R}}}^{3}denote the velocity of the fluid, b:R3×(0,+∞)→R3b:{{\mathbb{R}}}^{3}\times \left(0,+\infty )\to {{\mathbb{R}}}^{3}denote the magnetic field, and p˜=p˜(x,t)∈R\tilde{p}=\tilde{p}\left(x,t)\in {\mathbb{R}}denote the pressure.The main purpose of this paper is to study the well posedness of the solution for the equations (1.1) in 3D mixed-norm Lebesgue spaces. First, we review some of the relevant work of the MHD equations. If b=0b=0, then the equations (1.1) can be reduced to the incompressible Navier-Stokes equations: ∂tu−1ReΔu+u⋅∇u+∇p˜=0,inR3×(0,+∞),u(x,0)=u0(x),inR3.\left\{\begin{array}{ll}{\partial }_{t}u-\frac{1}{{\rm{Re}}}\Delta u+u\cdot \nabla u+\nabla \tilde{p}=0,& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ u\left(x,0)={u}_{0}\left(x),& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}.\end{array}\right.The mathematical theory of the Navier-Stokes equation has been much studied in recent years. For example, Leray first introduced the weak solution [5], and later, Hopf gained the existence of global weak solutions with u0∈Hs(RN){u}_{0}\in {H}^{s}\left({{\mathbb{R}}}^{N})[6], and Fujita and Kato demonstrated that the well posedness of the Cauchy problem when u0∈Hs(RN)(N≥2){u}_{0}\in {H}^{s}\left({{\mathbb{R}}}^{N})\left(N\ge 2)[7]. In addition, there are many monographs that study the Navier-Stokes equation, for example, Temam [8], Lions [9], and Constantin and Foias [10]. The mild and self-similar solutions in R3{{\mathbb{R}}}^{3}are obtained in references [11,12]. Especially, Jia and Šverák derived that the homogeneous classical Cauchy problem with initial value has a global scale invariant solution, and the solution is smooth in positive time [13].For the MHD system, the coupling between uuand bbmakes the situation more complicated. Because it describes abundant natural phenomena, as well as physical importance and mathematical challenges, the MHD system has become the subject of the study by physicists and mathematicians. Duraut and Lions derived the global weak solution and the local strong solution of the initial boundary value problem of equations (1.1) and derived the existence of the global strong solution in the case of the small initial value [14]. Nevertheless, it is still a challenging open problem whether a unique local solution of the exists globally when the initial value is large. Furthermore, Sermange and Temam derived the regularity of the weak solution (u,b)∈L∞([0,T];H1(R3))\left(u,b)\in {L}^{\infty }\left(\left[0,T];\hspace{0.33em}{H}^{1}\left({{\mathbb{R}}}^{3}))[3]. Kozono derived the existence of classical solutions in bounded domain Ω⊂R3\Omega \subset {{\mathbb{R}}}^{3}for equations (1.1) [15]. For the appropriately weak solution, He and Xin gained different local regularity results [16]. Cao and Wu obtained the global well posedness of the MHD system for any initial value in H2(R2){H}^{2}\left({{\mathbb{R}}}^{2}), but it needs a condition of mixed partial dissipation and additional magnetic diffusion in R2{{\mathbb{R}}}^{2}[2].Since the coefficients in the equations have no critical influence on the subsequent analysis, we can simply take Re=Rm=S=1{\rm{Re}}={\rm{Rm}}=S=1. Futhermore, we can obtain the following equations [1]: (∇×b)×b=(b⋅∇)b−b⋅(∇b),∇×∇×b=∇divb−Δb,∇×(u×b)=(b⋅∇)u−(u⋅∇)b+udivb−bdivu.\begin{array}{rcl}\left(\nabla \times b)\times b& =& \left(b\cdot \nabla )b-b\cdot \left(\nabla b),\\ \hspace{0.35em}\nabla \times \nabla \times b& =& \nabla {\rm{div}}\hspace{0.33em}b-\Delta b,\\ \hspace{0.35em}\nabla \times \left(u\times b)& =& \left(b\cdot \nabla )u-\left(u\cdot \nabla )b+u\hspace{0.33em}{\rm{div}}\hspace{0.33em}b-b\hspace{0.33em}{\rm{div}}\hspace{0.33em}u.\end{array}Then, (1.1) can be rewritten as follows: (1.3)∂tu−Δu+(u⋅∇)u−(b⋅∇)b+b⋅(∇b)+∇p˜,inR3×(0,+∞),∂tb−Δb+(u⋅∇)b−(b⋅∇)u=0,inR3×(0,+∞),u(x,0)=u0(x),b(x,0)=b0(x),inR3.\left\{\begin{array}{ll}{\partial }_{t}u-\Delta u+\left(u\cdot \nabla )u-\left(b\cdot \nabla )b+b\cdot \left(\nabla b)+\nabla \tilde{p},& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ {\partial }_{t}b-\Delta b+\left(u\cdot \nabla )b-\left(b\cdot \nabla )u=0,& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ u\left(x,0)={u}_{0}\left(x),\hspace{1.0em}b\left(x,0)={b}_{0}\left(x),& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}.\end{array}\right.In 2D and 3D cases, Ai et al. applied the semi-Galerkin approximation method to obtain the existence of weak solutions [1]. In 2D case, Ai et al. proved the global existence of strong solutions to (1.1), the continuous dependence of initial-boundary data, and the uniqueness of weak-strong solutions. On this basis, they also proved the existence of a uniform attractor for (1.1). Inspired by [1,4], the main purpose of this paper is to study the well posedness of the solution to (1.1) in 3D mixed-norm Lebesgue spaces. Specifically, in Section 2, we state some mixed-norm legesgue spaces and related properties. In Section 3, we prove the existence, uniqueness, and stability of the solution.2PreliminaryWe define the 3D mixed-norm Lebesgue spaces as follows [4]: Lp1p2p3(R3)=ff:R3→R,∥f∥Lp1p2p3(R3)=∫∫∫∣f∣p1dx1p2p1dx2p3p2dx31p3<+∞,{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})=\left\{f\left|f:{{\mathbb{R}}}^{3}\to {\mathbb{R}},{\parallel f\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}={\left(\int {\left(\int {\left(\int {| f| }^{{p}_{1}}{\rm{d}}{x}_{1}\right)}^{\frac{{p}_{2}}{{p}_{1}}}{\rm{d}}{x}_{2}\right)}^{\frac{{p}_{3}}{{p}_{2}}}{\rm{d}}{x}_{3}\right)}^{\tfrac{1}{{p}_{3}}}\right.\lt +\infty \right\},where p1,p2,p3∈[1,+∞){p}_{1},{p}_{2},{p}_{3}\in {[}1,+\infty ).Let PLp1p2p3(R3){{\rm{PL}}}_{{p}_{1}{p}_{2}{p}_{3}}\left({{\mathbb{R}}}^{3})as follows: PLp1p2p3(R3)={f∈Lp1p2p3(R3):divf=0}.{{\rm{PL}}}_{{p}_{1}{p}_{2}{p}_{3}}\left({{\mathbb{R}}}^{3})=\{f\in {L}_{{p}_{1}{p}_{2}{p}_{3}}\left({{\mathbb{R}}}^{3}):{\rm{div}}\hspace{0.33em}f=0\}.All the spaces that appear in this paper are invariant with respect to the scaling f(⋅)→λf(λ⋅)f\left(\cdot )\to \lambda f\left(\lambda \cdot ), λ>0\lambda \gt 0. The mixed-norm space Lp1p2p3(R3){L}_{{p}_{1}{p}_{2}{p}_{3}}\left({{\mathbb{R}}}^{3})is invariant if and only if 1p1+1p2+1p3=1\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}+\frac{1}{{p}_{3}}=1[4]. For given T∈(0,∞]T\in (0,\infty ], 1p1+1p2+1p3=1\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}+\frac{1}{{p}_{3}}=1, 1q1+1q2+1q3=δ∈(0,1)\frac{1}{{q}_{1}}+\frac{1}{{q}_{2}}+\frac{1}{{q}_{3}}=\delta \in \left(0,1), pk∈(1,+∞){p}_{k}\in \left(1,+\infty ), qk∈(pk,+∞){q}_{k}\in \left({p}_{k},+\infty ), k=1,2,3k=1,2,3, and f:R3×[0,∞)→R3f:{{\mathbb{R}}}^{3}\times {[}0,\infty )\to {{\mathbb{R}}}^{3}denote the measurable vector field functions, we denote Xp,q,T{{\mathcal{X}}}_{p,q,T}as follows [4]: Xp,q,T=f:g(x,t)≔t1−δ2f(x,t),g˜(x,t)≔t12Dxf(x,t),(x,t)∈R3×(0,T),{{\mathcal{X}}}_{p,q,T}=\left\{f:g\left(x,t):= {t}^{\tfrac{1-\delta }{2}}f\left(x,t),\hspace{1em}\tilde{g}\left(x,t):= {t}^{\tfrac{1}{2}}{D}_{x}f\left(x,t),\hspace{1.0em}\left(x,t)\in {{\mathbb{R}}}^{3}\times \left(0,T)\right\},then g∈C([0,T],PLq1q2q3(R3)),g˜∈C([0,T],PLp1p2p3(R3)).g\in C\left(\left[0,T],{{\rm{PL}}}_{{q}_{1}{q}_{2}{q}_{3}}\left({{\mathbb{R}}}^{3})),\hspace{1em}\tilde{g}\in C\left(\left[0,T],{{\rm{PL}}}_{{p}_{1}{p}_{2}{p}_{3}}\left({{\mathbb{R}}}^{3})).Moreover, g(x,0)=0g\left(x,0)=0, g˜(x,0)=0\tilde{g}\left(x,0)=0, and the norm ‖f‖χp,q,T=supt∈(0,T)[‖g(⋅,t)‖Lq1q2q3(R3)+‖g˜‖Lp1p2p3(R3)]<+∞.\hspace{-24.6em}\Vert f\Vert {\chi }_{p,q,T}=\mathop{\sup }\limits_{t\in \left(0,T)}{[}\Vert g\left(\cdot ,\hspace{0.33em}t){\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}+\Vert \tilde{g}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}]\lt +\infty .We denote Yp,T{{\mathcal{Y}}}_{p,T}as follows [4]: Yp,T=f:f∈C([0,T],PLp1p2p3(R3)),t12Dxf∈C([0,T],PLp1p2p3(R3)),{{\mathcal{Y}}}_{p,T}=\left\{f:f\in C\left(\left[0,T],{{\rm{PL}}}_{{p}_{1}{p}_{2}{p}_{3}}\left({{\mathbb{R}}}^{3})),\hspace{1em}{t}^{\tfrac{1}{2}}{D}_{x}f\in C\left(\left[0,T],{{\rm{PL}}}_{{p}_{1}{p}_{2}{p}_{3}}\left({{\mathbb{R}}}^{3}))\right\},and the norm ‖f‖Yp,T=supt∈(0,T)[‖f(t)‖Lp1p2p3(R3)+t1/2∥Dxf(t)∥Lp1p2p3(R3)]<+∞.\Vert f{\Vert }_{{{\mathcal{Y}}}_{p,T}}=\mathop{\sup }\limits_{t\in \left(0,T)}{[}\Vert f\left(t){\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+{t}^{1\text{/}2}{\parallel {D}_{x}f\left(t)\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}]\lt +\infty .We state the following result on Young’s inequality in mixed-norm Lebesgue spaces [4].Lemma 2.1[4] Let pk,rk{p}_{k},{r}_{k}, and qk{q}_{k}be given numbers in [1,+∞]\left[1,+\infty ]that satisfy1pk+1=1qk+1rk,k=1,2,3.\frac{1}{{p}_{k}}+1=\frac{1}{{q}_{k}}+\frac{1}{{r}_{k}},\hspace{1em}k=1,2,3.Then, (2.1)∥f∗g∥Lp1p2p3(R3)=∥f∥Lq1q2q3(R3)∥g∥Lr1r2r3(R3),{\parallel f\ast g\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}={\parallel f\parallel }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel g\parallel }_{{L}_{{r}_{1}{r}_{2}{r}_{3}}({{\mathbb{R}}}^{3})},for all f∈Lq1q2q3(R3)f\in {L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})and g∈Lr1r2r3(R3)g\in {L}_{{r}_{1}{r}_{2}{r}_{3}}({{\mathbb{R}}}^{3}).We state the following results on heat equations in mixed-norm Lebesgue spaces. First, we see the Cauchy problem for the heat equations: (2.2)ut−Δu=0,inR3×(0,+∞),bt−Δb=0,inR3×(0,+∞),u(x,0)=u0(x),b(x,0)=b0(x),inR3.\left\{\begin{array}{ll}{u}_{t}-\Delta u=0,& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ {b}_{t}-\Delta b=0,& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ u\left(x,0)={u}_{0}\left(x),b\left(x,0)={b}_{0}\left(x),& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}.\end{array}\right.We see that (2.2) can be written as follows: (2.3)Ut−ΔU=0,inR3×(0,+∞),U(x,0)=U0(x),inR3,\left\{\begin{array}{ll}{U}_{t}-\Delta U=0,& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ U\left(x,0)={U}_{0}\left(x),& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3},\end{array}\right.\hspace{6.65em}where U=ubU=\left(\begin{array}{l}u\\ b\end{array}\right), U0=u0b0{U}_{0}=\left(\begin{array}{l}{u}_{0}\\ {b}_{0}\end{array}\right). It is well known that a solution of (2.3) is (2.4)U(x,t)=eΔtU0(x)=(Gt*U0)(x,t),(x,t)∈R3×(0,+∞),U\left(x,t)={e}^{\Delta t}{U}_{0}\left(x)=\left({G}_{t}* {U}_{0})\left(x,t),\hspace{1.0em}\left(x,t)\in {{\mathbb{R}}}^{3}\times \left(0,+\infty ),where Gt(x)=1(4πt)n2e−∣x∣24t,(x,t)∈R3×(0,+∞).{G}_{t}\left(x)=\frac{1}{{\left(4\pi t)}^{\tfrac{n}{2}}}{e}^{-\tfrac{{| x| }^{2}}{4t}},\hspace{1.0em}\left(x,t)\in {{\mathbb{R}}}^{3}\times \left(0,+\infty ).Next, we state the following fundamental results of the solution of the heat equation in the mixed-norm Lebesgue space.Lemma 2.2[4] Let 1≤pk≤qk≤+∞1\le {p}_{k}\le {q}_{k}\le +\infty . There exists a positive constant NNdepending only on p1{p}_{1}, p2{p}_{2}, p3{p}_{3}, q1{q}_{1}, q2{q}_{2}, and q3{q}_{3}such that for every solution U(x,t)=eΔtU0(x)U\left(x,t)={e}^{\Delta t}{U}_{0}\left(x)defined in (2.4) of the Cauchy problem (2.3) with U0∈Lq1q2q3(R3){U}_{0}\in {L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3}), then for t>0t\gt 0(2.5)∥U(x,t)∥Lp1p2p3(R3)≤Nt−12∑k=1n=31qk−1pk∥U0∥Lq1q2q3(R3),{\parallel U\left(x,t)\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le N{t}^{-\tfrac{1}{2}\mathop{\sum }\limits_{k=1}^{n=3}\left(\tfrac{1}{{q}_{k}}-\tfrac{1}{{p}_{k}}\right)}{\parallel {U}_{0}\parallel }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})},(2.6)∥DxU(x,t)∥Lp1p2p3(R3)≤Nt−12−12∑k=1n=31qk−1pk∥U0∥Lq1q2q3(R3).{\parallel {D}_{x}U\left(x,t)\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le N{t}^{-\tfrac{1}{2}-\tfrac{1}{2}\mathop{\sum }\limits_{k=1}^{n=3}\left(\tfrac{1}{{q}_{k}}-\tfrac{1}{{p}_{k}}\right)}{\parallel {U}_{0}\parallel }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}.Lemma 2.3[4] Let p1,p2,p3∈(1,∞){p}_{1},{p}_{2},{p}_{3}\in \left(1,\infty ), and U0∈Lp1p2p3(R3){U}_{0}\in {L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3}). Let U(x,t)=eΔtU0(x)U\left(x,t)={e}^{\Delta t}{U}_{0}\left(x)be the solution of the heat equation (2.3) defined in (2.4). Then, U∈C([0,∞),Lp1p2p3(R3))U\in C\left({[}0,\infty ),{L}_{{p}_{1}{p}_{2}{p}_{3}}\left({{\mathbb{R}}}^{3}))and(2.7)limt→0+∥U(x,t)−U0∥Lp1p2p3(R3)=0.\mathop{\mathrm{lim}}\limits_{t\to {0}^{+}}{\parallel U\left(x,t)-{U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}=0.The following consequence illustrates the boundedness of Helmholtz-Leray projection P{\mathbb{P}}in mixed-norm Lebesgue spaces.Lemma 2.4[4] Let P=Id−∇Δ−1∇⋅{\mathbb{P}}={\rm{Id}}-\nabla {\Delta }^{-1}\nabla \cdot be the Helmholtz-Leray projection onto the divergence-free vector fields. Let p1,p2,p3∈(1,∞){p}_{1},{p}_{2},{p}_{3}\in \left(1,\infty ). Then, one has(2.8)∥P(f)∥Lp1p2p3(R3)≤N∥f∥Lp1p2p3(R3),{\parallel {\mathbb{P}}(f)\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le N{\parallel f\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},for all f∈(Lp1p2p3(R3))3f\in {({L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3}))}^{3}, where N=N(p1,p2,p3)N=N\left({p}_{1},{p}_{2},{p}_{3})is a positive constant.3MHD equations in 3D mixed-norm Lebesgue spaceWe apply P{\mathbb{P}}on the system (1.3), and then (1.3) can be expressed as follows: (3.1)Ut+AU+F(U,U)=0,inR3×(0,+∞),U(x,0)=U0(x),inR3,\left\{\begin{array}{ll}{U}_{t}+{\mathcal{A}}U+F\left(U,U)=0,& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3}\times \left(0,+\infty ),\\ U\left(x,0)={U}_{0}\left(x),& \hspace{0.1em}\text{in}\hspace{0.1em}\hspace{0.33em}{{\mathbb{R}}}^{3},\end{array}\right.as P∇p˜=0{\mathbb{P}}\nabla \tilde{p}=0, where U=ub,U0=u0b0,A=−PΔ00−PΔ,F(U,U)=P(u⋅∇)u−P(b⋅∇)b+Pb⋅(∇b)P(u⋅∇)b−P(b⋅∇)u.\begin{array}{l}\phantom{\rule[-1.5em]{}{0ex}}U=\left(\begin{array}{l}u\\ b\end{array}\right),\hspace{1em}{U}_{0}=\left(\begin{array}{l}{u}_{0}\\ {b}_{0}\end{array}\right),\hspace{1em}{\mathcal{A}}=\left(\begin{array}{ll}-{\mathbb{P}}\Delta & 0\\ 0& -{\mathbb{P}}\Delta \end{array}\right),\\ F\left(U,U)=\left(\begin{array}{l}{\mathbb{P}}\left(u\cdot \nabla )u-{\mathbb{P}}\left(b\cdot \nabla )b+{\mathbb{P}}b\cdot \left(\nabla b)\\ {\mathbb{P}}\left(u\cdot \nabla )b-{\mathbb{P}}\left(b\cdot \nabla )u\end{array}\right).\end{array}By the Duhamel’s principle, the system (3.1) can be transformed into the following integral equations: (3.2)U=U1+G(U,U),U={U}_{1}+G\left(U,U),where U1=e−AtU0(x)=eΔtu0(x)eΔtb0(x){U}_{1}={e}^{-{\mathcal{A}}t}{U}_{0}\left(x)=\left(\begin{array}{l}{e}^{\Delta t}{u}_{0}\left(x)\\ {e}^{\Delta t}{b}_{0}\left(x)\end{array}\right)and (3.3)G(U,U)=−∫0te−(t−s)AF(U,U)ds\begin{array}{rcl}G\left(U,U)& =& -\underset{0}{\overset{t}{\displaystyle \int }}{e}^{-\left(t-s){\mathcal{A}}}F\left(U,U){\rm{d}}s\end{array}=−∫0te(t−s)Δ(P(u⋅∇)u−P(b⋅∇)b+Pb⋅(∇b))ds−∫0te(t−s)Δ(P(u⋅∇)b−P(b⋅∇)u)ds.\begin{array}{rcl}& =& \left(\begin{array}{l}-\underset{0}{\overset{t}{\displaystyle \int }}{e}^{\left(t-s)\Delta }({\mathbb{P}}\left(u\cdot \nabla )u-{\mathbb{P}}\left(b\cdot \nabla )b+{\mathbb{P}}b\cdot \left(\nabla b)){\rm{d}}s\\ -\underset{0}{\overset{t}{\displaystyle \int }}{e}^{\left(t-s)\Delta }({\mathbb{P}}\left(u\cdot \nabla )b-{\mathbb{P}}\left(b\cdot \nabla )u){\rm{d}}s\end{array}\right).\end{array}The main results in the paper are as follows.Theorem 3.1Let pk∈(1,+∞){p}_{k}\in \left(1,+\infty ), qk∈[pk,+∞){q}_{k}\in {[}{p}_{k},+\infty ), k=1,2,3,k=1,2,3,and1p1+1p2+1p3=1,1q1+1q2+1q3=δ∈(0,1).\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}+\frac{1}{{p}_{3}}=1,\hspace{1em}\frac{1}{{q}_{1}}+\frac{1}{{q}_{2}}+\frac{1}{{q}_{3}}=\delta \in \left(0,1).Then, there are a sufficiently small constant λ0>0{\lambda }_{0}\gt 0and a number N>0N\gt 0, depending on p1{p}_{1}, p2{p}_{2}, p3{p}_{3}, q1{q}_{1}, q2{q}_{2}, and q3{q}_{3}, such that the following results hold. (i)For all U0∈Lp1p2p3(R3){U}_{0}\in {L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})with divU0=0{\rm{div}}\hspace{0.33em}{U}_{0}=0, if ∥U0∥Lp1p2p3(R3)≤λ0{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le {\lambda }_{0}, then (3.1) has an unique global time solution U∈Xp,q,∞∩Yp,∞U\in {{\mathcal{X}}}_{p,q,\infty }\cap {{\mathcal{Y}}}_{p,\infty }with‖U‖χp,q,∞≤N∥U0∥Lp1p2p3(R3),‖U‖Yp,∞≤N(∥U0∥Lp1p2p3(R3)+∥U0∥Lp1p2p3(R3)2).\begin{array}{lcl}\Vert U{\Vert }_{{\chi }_{p,q,\infty }}& \le & N{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},\\ \Vert U{\Vert }_{{{\mathcal{Y}}}_{p,\infty }}& \le & N({\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}^{2}).\end{array}\hspace{0.55em}(ii)For all U0∈Lp1p2p3(R3){U}_{0}\in {L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})with divU0=0{\rm{div}}\hspace{0.33em}{U}_{0}=0, there is a sufficiently small T0>0{T}_{0}\gt 0depending on p1{p}_{1}, p2{p}_{2}, p3{p}_{3}, q1{q}_{1}, q2{q}_{2}, and q3{q}_{3}such that (3.1) has an unique local time solution U∈Xp,q,T0∩Yp,T0U\in {{\mathcal{X}}}_{p,q,{T}_{0}}\cap {{\mathcal{Y}}}_{p,{T}_{0}}with‖U‖χp,q,T0≤N∥U0∥Lp1p2p3(R3),‖U‖Yp,T0≤N(∥U0∥Lp1p2p3(R3)+∥U0∥Lp1p2p3(R3)2).\begin{array}{lcl}\Vert U{\Vert }_{{\chi }_{p,q,{T}_{0}}}& \le & N{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},\\ \Vert U{\Vert }_{{{\mathcal{Y}}}_{p,{T}_{0}}}& \le & N\left({\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}^{2}).\end{array}\hspace{0.2em}To prove Theorem 3.1, we need the following lemmas.Lemma 3.1[4] Let p1{p}_{1}, p2{p}_{2}, p3{p}_{3}, q1{q}_{1}, q2{q}_{2}, and q3{q}_{3}be given numbers and 1<pk≤qk<∞1\lt {p}_{k}\le {q}_{k}\lt \infty . Also, let σ≥0\sigma \ge 0be defined by σ=∑k=1n=31pk−1qk\sigma ={\sum }_{k=1}^{n=3}\left(\frac{1}{{p}_{k}}-\frac{1}{{q}_{k}}\right). (i)There exists a number NNdepending only on p1{p}_{1}, p2{p}_{2}, p3{p}_{3}, q1{q}_{1}, q2{q}_{2}, and q3{q}_{3}such that(3.4)∥e−AtPf∥Lq1q1q3(R3)≤Nt−σ2‖f‖Lp1P2p3(R3),{\parallel {e}^{-{\mathcal{A}}t}{\mathbb{P}}f\parallel }_{{L}_{{q}_{1}{q}_{1}{q}_{3}}({{\mathbb{R}}}^{3})}\le N{t}^{-\tfrac{\sigma }{2}}\Vert f{\Vert }_{{L}_{{p}_{1}{P}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},(3.5)∥Dxe−AtPf∥Lq1q1q3(R3)≤Nt−12(1+σ)‖f‖Lp1p2p3(R3),{\parallel {D}_{x}{e}^{-{\mathcal{A}}t}{\mathbb{P}}f\parallel }_{{L}_{{q}_{1}{q}_{1}{q}_{3}}({{\mathbb{R}}}^{3})}\le N{t}^{-\tfrac{1}{2}\left(1+\sigma )}\Vert f{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},for all f∈Lp1p2p3(R3)3f\in {L}_{{p}_{1}{p}_{2}{p}_{3}}{({{\mathbb{R}}}^{3})}^{3}.(ii)For all f∈Lp1p2p3(R3)3f\in {L}_{{p}_{1}{p}_{2}{p}_{3}}{({{\mathbb{R}}}^{3})}^{3}, the following assertions hold: if σ>0\sigma \gt 0, then(3.6)limt→0+tσ2∥e−AtPf∥Lq1q1q3(R3)=0,\mathop{\mathrm{lim}}\limits_{t\to {0}^{+}}{t}^{\tfrac{\sigma }{2}}{\parallel {e}^{-{\mathcal{A}}t}{\mathbb{P}}f\parallel }_{{L}_{{q}_{1}{q}_{1}{q}_{3}}({{\mathbb{R}}}^{3})}=0,(3.7)limt→0+∥[e−AtPf]−Pf∥Lp1P1p3(R3)=0,\mathop{\mathrm{lim}}\limits_{t\to {0}^{+}}{\parallel {[}{e}^{-{\mathcal{A}}t}{\mathbb{P}}f]-{\mathbb{P}}f\parallel }_{{L}_{{p}_{1}{P}_{1}{p}_{3}}({{\mathbb{R}}}^{3})}=0,(3.8)limt→0+t−12(1+σ)∥Dxe−AtPf∥Lq1q1q3(R3)=0.\mathop{\mathrm{lim}}\limits_{t\to {0}^{+}}{t}^{-\tfrac{1}{2}\left(1+\sigma )}{\parallel {D}_{x}{e}^{-{\mathcal{A}}t}{\mathbb{P}}f\parallel }_{{L}_{{q}_{1}{q}_{1}{q}_{3}}({{\mathbb{R}}}^{3})}=0.Lemma 3.2Let pk∈(1,∞),αk,βk,γk∈(0,1]{p}_{k}\in \left(1,\infty ),{\alpha }_{k},{\beta }_{k},{\gamma }_{k}\in (0,1]be given numbers satisfying γk≤αk+βk<pk{\gamma }_{k}\le {\alpha }_{k}+{\beta }_{k}\lt {p}_{k}, k=1,2,3.k=1,2,3.Letα=α1p1+α2p2+α2p2,β=β1p1+β2p2+β2p2,γ=γ1p1+γ2p2+γ2p2.\alpha =\frac{{\alpha }_{1}}{{p}_{1}}+\frac{{\alpha }_{2}}{{p}_{2}}+\frac{{\alpha }_{2}}{{p}_{2}},\hspace{1em}\beta =\frac{{\beta }_{1}}{{p}_{1}}+\frac{{\beta }_{2}}{{p}_{2}}+\frac{{\beta }_{2}}{{p}_{2}},\hspace{1em}\gamma =\frac{{\gamma }_{1}}{{p}_{1}}+\frac{{\gamma }_{2}}{{p}_{2}}+\frac{{\gamma }_{2}}{{p}_{2}}.Then, (3.9)‖G(U,U)‖Lp1γ1p2γ2p3γ3(R3)≤N∫0t(t−s)−α+β−γ2‖u‖Lp1α1p2α2p3α3(R3)∥Dxu∥Lp1β1p2β2p3β3(R3)+2‖b‖Lp1α1p2α2p3α3(R3)∥Dxb∥Lp1β1p2β2p3β3(R3)+‖u‖Lp1α1p2α2p3α3(R3)∥Dxb∥Lp1β1p2β2p3β3(R3)+‖b‖Lp1α1p2α2p3α3(R3)∥Dxu∥Lp1β1p2β2p3β3(R3)ds,\begin{array}{l}\Vert G\left(U,U){\Vert }_{{L}_{\frac{{p}_{1}}{{\gamma }_{1}}\frac{{p}_{2}}{{\gamma }_{2}}\frac{{p}_{3}}{{\gamma }_{3}}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\le N\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{\alpha +\beta -\gamma }{2}}\left(\Vert u{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}+2\Vert b{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}\right.\\ \hspace{2.0em}\left.+\Vert u{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}+\Vert b{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}\right){\rm{d}}s,\end{array}\hspace{0.65em}(3.10)‖DxG(U,U)‖Lp1γ1p2γ2p3γ3(R3)≤N∫0t(t−s)−1+α+β−γ2‖u‖Lp1α1p2α2p3α3(R3)∥Dxu∥Lp1β1p2β2p3β3(R3)+2‖b‖Lp1α1p2α2p3α3(R3)∥Dxb∥Lp1β1p2β2p3β3(R3)+‖u‖Lp1α1p2α2p3α3(R3)∥Dxb∥Lp1β1p2β2p3β3(R3)+‖b‖Lp1α1p2α2p3α3(R3)∥Dxu∥Lp1β1p2β2p3β3(R3)ds,\begin{array}{l}\Vert {D}_{x}G\left(U,U){\Vert }_{{L}_{\frac{{p}_{1}}{{\gamma }_{1}}\frac{{p}_{2}}{{\gamma }_{2}}\frac{{p}_{3}}{{\gamma }_{3}}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\le N\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1+\alpha +\beta -\gamma }{2}}\left(\Vert u{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}+2\Vert b{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}\right.\\ \hspace{2.0em}\left.+\Vert u{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}+\Vert b{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}\right){\rm{d}}s,\end{array}where N>0N\gt 0is a constant depending on pk,αk,βk,γk,k=1,2,3{p}_{k},{\alpha }_{k},{\beta }_{k},{\gamma }_{k},k=1,2,3.ProofWe first prove (3.9) in Lemma 3.2. For γk≤αk+βk<pk{\gamma }_{k}\le {\alpha }_{k}+{\beta }_{k}\lt {p}_{k}, we can gain pkγk≥pkαk+βk,∑k=1n=3αk+βkpk−γkpk=α+β−γ.\frac{{p}_{k}}{{\gamma }_{k}}\ge \frac{{p}_{k}}{{\alpha }_{k}+{\beta }_{k}},\hspace{1.0em}\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{{\alpha }_{k}+{\beta }_{k}}{{p}_{k}}-\frac{{\gamma }_{k}}{{p}_{k}}\right)=\alpha +\beta -\gamma .By (3.4), we can obtain (3.11)‖G(U,U)‖Lp1γ1p2γ2p3γ3(R3)≤N∫0t(t−s)−α+β−γ2‖F1(U,U)‖Lp1α1+β1p2α2+β2p3α3+β13(R3)ds,\Vert G\left(U,U){\Vert }_{{L}_{\frac{{p}_{1}}{{\gamma }_{1}}\frac{{p}_{2}}{{\gamma }_{2}}\frac{{p}_{3}}{{\gamma }_{3}}}({{\mathbb{R}}}^{3})}\le N\underset{0}{\overset{t}{\int }}{\left(t-s)}^{-\tfrac{\alpha +\beta -\gamma }{2}}\Vert {F}_{1}\left(U,U){\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{13}}}({{\mathbb{R}}}^{3})}{\rm{d}}s,where F1(U,U)=(u⋅∇)u−(b⋅∇)b+b⋅(∇b)(u⋅∇)b−(b⋅∇)u{F}_{1}\left(U,U)=\left(\begin{array}{l}\left(u\cdot \nabla )u-\left(b\cdot \nabla )b+b\cdot \left(\nabla b)\\ \left(u\cdot \nabla )b-\left(b\cdot \nabla )u\end{array}\right).By using Hölder’s inequality repeatedly, we can find that ‖(u⋅∇)b‖Lp1α1+β1p2α2+β2p3α3+β3(R3)=∫∫∫∣u⋅(∇b)∣p1α1+β1dx1p2(α1+β1)p1(α2+β2)dx2p3(α2+β2)p2(α3+β3)dx3(α3+β3)p3≤∫∫∫∣u∣p1α1dx1α1p1∫∣Dxb∣p1β1dx1β1p1p2(α2+β2)dx2p3(α2+β2)p2(α3+β3)dx3(α3+β3)p3≤∫∫∫∣u∣p1α1dx1p2α1p1α2dx2α2p2∫∫∣Dxb∣p1β1dx1p2β1p1β2dx2β2p2p3(α3+β3)dx3(α3+β3)p3\begin{array}{l}\Vert \left(u\cdot \nabla )b{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{3}}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}={\left[\displaystyle \int {\left[\displaystyle \int {\left[\displaystyle \int {| u\cdot \left(\nabla b)| }^{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}}{\rm{d}}{x}_{1}\right]}^{\frac{{p}_{2}\left({\alpha }_{1}+{\beta }_{1})}{{p}_{1}\left({\alpha }_{2}+{\beta }_{2})}}{\rm{d}}{x}_{2}\right]}^{\frac{{p}_{3}\left({\alpha }_{2}+{\beta }_{2})}{{p}_{2}\left({\alpha }_{3}+{\beta }_{3})}}{\rm{d}}{x}_{3}\right]}^{\tfrac{\left({\alpha }_{3}+{\beta }_{3})}{{p}_{3}}}\\ \hspace{1.0em}\le {\left[\displaystyle \int {\left[\displaystyle \int {\left[{\left(\displaystyle \int {| u| }^{\frac{{p}_{1}}{{\alpha }_{1}}}{\rm{d}}{x}_{1}\right)}^{\frac{{\alpha }_{1}}{{p}_{1}}}{\left(\displaystyle \int {| {D}_{x}b| }^{\frac{{p}_{1}}{{\beta }_{1}}}{\rm{d}}{x}_{1}\right)}^{\frac{{\beta }_{1}}{{p}_{1}}}\right]}^{\frac{{p}_{2}}{\left({\alpha }_{2}+{\beta }_{2})}}{\rm{d}}{x}_{2}\right]}^{\frac{{p}_{3}\left({\alpha }_{2}+{\beta }_{2})}{{p}_{2}\left({\alpha }_{3}+{\beta }_{3})}}{\rm{d}}{x}_{3}\right]}^{\tfrac{\left({\alpha }_{3}+{\beta }_{3})}{{p}_{3}}}\\ \hspace{1.0em}\le {\left[\displaystyle \int {\left[{\left[\displaystyle \int {\left[\displaystyle \int {| u| }^{\frac{{p}_{1}}{{\alpha }_{1}}}{\rm{d}}{x}_{1}\right]}^{\frac{{p}_{2}{\alpha }_{1}}{{p}_{1}{\alpha }_{2}}}{\rm{d}}{x}_{2}\right]}^{\frac{{\alpha }_{2}}{{p}_{2}}}{\left[\displaystyle \int {\left[\displaystyle \int {| {D}_{x}b| }^{\frac{{p}_{1}}{{\beta }_{1}}}{\rm{d}}{x}_{1}\right]}^{\frac{{p}_{2}{\beta }_{1}}{{p}_{1}{\beta }_{2}}}{\rm{d}}{x}_{2}\right]}^{\frac{{\beta }_{2}}{{p}_{2}}}\right]}^{\frac{{p}_{3}}{\left({\alpha }_{3}+{\beta }_{3})}}{\rm{d}}{x}_{3}\right]}^{\tfrac{\left({\alpha }_{3}+{\beta }_{3})}{{p}_{3}}}\end{array}\hspace{0.35em}≤∫∫∫∣u∣p1α1dx1p2α1p1α2dx2p3α2p2α3dx3α3p3∫∫∫∣Dxb∣p1β1dx1p2β1p1β2dx2p3β2p2β3dx3β3p3=‖u‖Lp1α1p2α2p3α3(R3)‖Dxb‖Lp1β1p2β2p3β3(R3).\begin{array}{l}\hspace{1.0em}\le {\left[\displaystyle \int {\left[\displaystyle \int {\left[\displaystyle \int | u{| }^{\frac{{p}_{1}}{{\alpha }_{1}}}{\rm{d}}{x}_{1}\right]}^{\frac{{p}_{2}{\alpha }_{1}}{{p}_{1}{\alpha }_{2}}}{\rm{d}}{x}_{2}\right]}^{\frac{{p}_{3}{\alpha }_{2}}{{p}_{2}{\alpha }_{3}}}{\rm{d}}{x}_{3}\right]}^{\tfrac{{\alpha }_{3}}{{p}_{3}}}{\left[\displaystyle \int {\left[\displaystyle \int {\left[\displaystyle \int {| {D}_{x}b| }^{\frac{{p}_{1}}{{\beta }_{1}}}{\rm{d}}{x}_{1}\right]}^{\frac{{p}_{2}{\beta }_{1}}{{p}_{1}{\beta }_{2}}}{\rm{d}}{x}_{2}\right]}^{\frac{{p}_{3}{\beta }_{2}}{{p}_{2}{\beta }_{3}}}{\rm{d}}{x}_{3}\right]}^{\tfrac{{\beta }_{3}}{{p}_{3}}}\\ \hspace{1.0em}=\Vert u{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}\Vert {D}_{x}b{\Vert }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}.\end{array}\hspace{2.75em}Then, ‖F1(U,U)‖Lp1α1+β1p2α2+β2p3α3+β13(R3)≤∥∣(u⋅∇)u∣+∣(b⋅∇)b∣+∣b⋅(∇b)∣+∣(u⋅∇)b∣+∣(b⋅∇)u∣∥Lp1α1+β1p2α2+β2p3α3+β13(R3)≤∥(u⋅∇)u∥Lp1α1+β1p2α2+β2p3α3+β13(R3)+∥(b⋅∇)b∥Lp1α1+β1p2α2+β2p3α3+β13(R3)+∥b⋅(∇b)∥Lp1α1+β1p2α2+β2p3α3+β13(R3)+∥(u⋅∇)b∥Lp1α1+β1p2α2+β2p3α3+β13(R3)+∥(b⋅∇)u∥Lp1α1+β1p2α2+β2p3α3+β13(R3)≤‖u‖Lp1α1p2α2p3α3(R3)‖Dxu‖Lp1β1p2β2p3β3(R3)+2‖b‖Lp1α1p2α2p3α3(R3)‖Dxb‖Lp1β1p2β2p3β3(R3)+‖u‖Lp1α1p2α2p3α3(R3)‖Dxb‖Lp1β1p2β2p3β3(R3)+‖b‖Lp1α1p2α2p3α3(R3)‖Dxu‖Lp1β1p2β2p3β3(R3).\begin{array}{l}\Vert {F}_{1}\left(U,U){\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{13}}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\le \parallel | \left(u\cdot \nabla )u| +| \left(b\cdot \nabla )b| +| b\cdot \left(\nabla b)| +| \left(u\cdot \nabla )b| +| \left(b\cdot \nabla )u| {\parallel }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{13}}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\le \parallel \left(u\cdot \nabla )u{\parallel }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{13}}}({{\mathbb{R}}}^{3})}+\parallel \left(b\cdot \nabla )b{\parallel }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{13}}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\hspace{1.0em}+\parallel b\cdot \left(\nabla b){\parallel }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{13}}}({{\mathbb{R}}}^{3})}+\parallel \left(u\cdot \nabla )b{\parallel }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{13}}}({{\mathbb{R}}}^{3})}+\parallel \left(b\cdot \nabla )u{\parallel }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}+{\beta }_{1}}\frac{{p}_{2}}{{\alpha }_{2}+{\beta }_{2}}\frac{{p}_{3}}{{\alpha }_{3}+{\beta }_{13}}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\le \Vert u{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}\Vert {D}_{x}u{\Vert }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}+2\Vert b{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}\Vert {D}_{x}b{\Vert }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\hspace{1.0em}+\Vert u{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}\Vert {D}_{x}b{\Vert }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}+\Vert b{\Vert }_{{L}_{\frac{{p}_{1}}{{\alpha }_{1}}\frac{{p}_{2}}{{\alpha }_{2}}\frac{{p}_{3}}{{\alpha }_{3}}}({{\mathbb{R}}}^{3})}\Vert {D}_{x}u{\Vert }_{{L}_{\frac{{p}_{1}}{{\beta }_{1}}\frac{{p}_{2}}{{\beta }_{2}}\frac{{p}_{3}}{{\beta }_{3}}}({{\mathbb{R}}}^{3})}.\end{array}By substituting the aforementioned formula into (3.11), we can obtain (3.9). Similarly, (3.10) can be proved by (3.5).□Lemma 3.3[4] Let XXbe a Banach space with norm ‖⋅‖X\Vert \cdot {\Vert }_{X}. Let G:X×X→XG:X\times X\to Xbe a bilinear map such that there is N0>0{N}_{0}\gt 0so that‖G(U,V)‖X≤N0‖U‖X‖V‖X,∀U,V∈X.\Vert G\left(U,V){\Vert }_{X}\le {N}_{0}\Vert U{\Vert }_{X}\Vert V{\Vert }_{X},\hspace{1em}\forall U,V\in X.Then, for all U1∈X{U}_{1}\in Xwith 4N0‖U1‖X<14{N}_{0}\Vert {U}_{1}{\Vert }_{X}\lt 1, the equationU=U1+G(U,U)U={U}_{1}+G\left(U,U)has an unique solution UUwith‖U‖X≤2‖U1‖X.\Vert U{\Vert }_{X}\le 2\Vert {U}_{1}{\Vert }_{X}.Proof of Theorem 3.1We now prove (i). First, we start from the proof that U∈χp,q,∞U\in {\chi }_{p,q,\infty }. From Lemma 2.2 and σ=∑k=1n=31pk−1qk=1−δ\sigma ={\sum }_{k=1}^{n=3}\left(\frac{1}{{p}_{k}}-\frac{1}{{q}_{k}}\right)=1-\delta , we have ∥U1∥Lq1q1q3(R3)≤N1t−1−δ2‖U0‖Lp1P2p3(R3),∥DxU1∥Lq1q2q3(R3)≤N1t−12‖U0‖Lp1p2p3(R3),\begin{array}{rcl}{\parallel {U}_{1}\parallel }_{{L}_{{q}_{1}{q}_{1}{q}_{3}}({{\mathbb{R}}}^{3})}& \le & {N}_{1}{t}^{-\tfrac{1-\delta }{2}}\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{P}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},\\ {\parallel {D}_{x}{U}_{1}\parallel }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}& \le & {N}_{1}{t}^{-\tfrac{1}{2}}\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},\end{array}where N1>0{N}_{1}\gt 0is a constant depending on p1{p}_{1}, p2{p}_{2}, p3{p}_{3}, q1{q}_{1}, q2{q}_{2}, and q3{q}_{3}.Furthermore, according to Lemma 3.1, we know that when t→0t\to 0, t−1−σ2e−AtP→0{t}^{-\tfrac{1-\sigma }{2}}{e}^{-{\mathcal{A}}t}{\mathbb{P}}\to 0, and when t=0t=0, t−1−σ2U1=0{t}^{-\tfrac{1-\sigma }{2}}{U}_{1}=0. Hence, t−1−σ2e−AtP:Lp1p2p3(R3)→PLq1q2q3(R3){t}^{-\tfrac{1-\sigma }{2}}{e}^{-{\mathcal{A}}t}{\mathbb{P}}:{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})\to P{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})is uniformly bounded. Similarly, when t→0t\to 0, t−12Dxe−AtP→0{t}^{-\tfrac{1}{2}}{D}_{x}{e}^{-{\mathcal{A}}t}{\mathbb{P}}\to 0, and when t=0t=0, t−12DxU1=0{t}^{-\tfrac{1}{2}}{D}_{x}{U}_{1}=0. Hence, t−1−σ2Dxe−AtP:Lp1p2p3(R3)→PLp1p2p3(R3){t}^{-\tfrac{1-\sigma }{2}}{D}_{x}{e}^{-{\mathcal{A}}t}{\mathbb{P}}:{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})\to P{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})is also uniformly bounded. Hence, we have U0∈χp,q,∞{U}_{0}\in {\chi }_{p,q,\infty }and (3.12)‖U1‖χp,q,∞≤N1∥U0∥Lp1p2p3(R3).\Vert {U}_{1}{\Vert }_{{\chi }_{p,q,\infty }}\le {N}_{1}{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}.Now let’s prove that bilinear G:χp,q,∞×χp,q,∞→χp,q,∞G:{\chi }_{p,q,\infty }\times {\chi }_{p,q,\infty }\to {\chi }_{p,q,\infty }is bounded.Let βk=1{\beta }_{k}=1, γk=αk=pkqk∈(0,1]{\gamma }_{k}={\alpha }_{k}=\frac{{p}_{k}}{{q}_{k}}\in (0,1], we obtain pkγk=qk\frac{{p}_{k}}{{\gamma }_{k}}={q}_{k}, −α+β−γ2=−12∑k=1n=3αkpk+βkpk−γkpk=−12∑k=1n=31qk+1pk−1qk=−12∑k=1n=31pk=−12.-\frac{\alpha +\beta -\gamma }{2}=-\frac{1}{2}\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{{\alpha }_{k}}{{p}_{k}}+\frac{{\beta }_{k}}{{p}_{k}}-\frac{{\gamma }_{k}}{{p}_{k}}\right)=-\frac{1}{2}\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{1}{{q}_{k}}+\frac{1}{{p}_{k}}-\frac{1}{{q}_{k}}\right)=-\frac{1}{2}\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{1}{{p}_{k}}\right)=-\frac{1}{2}.By using 1p1+1p2+1p3=1\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}+\frac{1}{{p}_{3}}=1, 1q1+1q2+1q3=δ∈(0,1)\frac{1}{{q}_{1}}+\frac{1}{{q}_{2}}+\frac{1}{{q}_{3}}=\delta \in \left(0,1), (3.9) and the definition of Xp,q,T{{\mathcal{X}}}_{p,q,T}, and applying (3.3), we can obtain ‖G(U,U)‖Lq1q2q3(R3)≤N∫0t(t−s)−12(‖u‖Lq1q2q3(R3)∥Dxu∥Lp1p2p3(R3)+2‖b‖Lq1q2q3(R3)∥Dxb∥Lp1p2p3(R3)+‖u‖Lq1q2q3(R3)∥Dxb∥Lp1p2p3(R3)+‖b‖Lq1q2q3(R3)∥Dxu∥Lp1p2p3(R3))ds≤N∫0t(t−s)−12s1−δ2‖u‖Lq1q2q3(R3)s12∥Dxu∥Lp1p2p3(R3)s−1−δ2s−12+2s1−δ2‖b‖Lq1q2q3(R3)s12∥Dxb∥Lp1p2p3(R3)s−1−δ2s−12+s1−δ2‖u‖Lq1q2q3(R3)s12×∥Dxb∥Lp1p2p3(R3)s−1−δ2s−12+s1−δ2‖b‖Lq1q2q3(R3)s12∥Dxu∥Lp1p2p3(R3)s−1−δ2s−12ds≤N(∥u∥χp,q,∞2+∥b∥χp,q,∞2+2‖u‖χp,q,∞‖b‖χp,q,∞)∫0t(t−s)−12s−1+δ2ds,\begin{array}{rcl}\Vert G\left(U,U){\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}& \le & N\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1}{2}}(\Vert u{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+2\Vert b{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\\ & & +\Vert u{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+\Vert b{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}){\rm{d}}s\\ & \le & N\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1}{2}}\left({s}^{\tfrac{1-\delta }{2}}\Vert u{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{s}^{\tfrac{1}{2}}{\parallel {D}_{x}u\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}{s}^{-\tfrac{1-\delta }{2}}{s}^{-\tfrac{1}{2}}\right.\\ & & +2{s}^{\tfrac{1-\delta }{2}}\Vert b{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{s}^{\tfrac{1}{2}}{\parallel {D}_{x}b\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}{s}^{-\tfrac{1-\delta }{2}}{s}^{-\tfrac{1}{2}}+{s}^{\tfrac{1-\delta }{2}}\Vert u{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{s}^{\tfrac{1}{2}}\\ & & \times {\parallel {D}_{x}b\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}{s}^{-\tfrac{1-\delta }{2}}{s}^{-\tfrac{1}{2}}\left.+{s}^{\tfrac{1-\delta }{2}}\Vert b{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{s}^{\tfrac{1}{2}}{\parallel {D}_{x}u\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}{s}^{-\tfrac{1-\delta }{2}}{s}^{-\tfrac{1}{2}}\right){\rm{d}}s\\ & \le & N({\parallel u\parallel }_{{\chi }_{p,q,\infty }}^{2}+{\parallel b\parallel }_{{\chi }_{p,q,\infty }}^{2}+2\Vert u{\Vert }_{{\chi }_{p,q,\infty }}\Vert b{\Vert }_{{\chi }_{p,q,\infty }})\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1}{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s,\end{array}and ∫0t(t−s)−12s−1+δ2ds=∫0t2(t−s)−12s−1+δ2ds+∫t2t(t−s)−12s−1+δ2ds≤t2−12∫0t2s−1+δ2ds+t2−1+δ2∫t2t(t−s)−12ds≤Nt2−12t2δ2+Nt2−1+δ2t212≤Nt−1−δ2.\hspace{-29em}\begin{array}{rcl}\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1}{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s& =& \underset{0}{\overset{\frac{t}{2}}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1}{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s+\underset{\frac{t}{2}}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1}{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s\\ & \le & {\left(\frac{t}{2}\right)}^{-\tfrac{1}{2}}\underset{0}{\overset{\frac{t}{2}}{\displaystyle \int }}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s+{\left(\frac{t}{2}\right)}^{-1+\tfrac{\delta }{2}}\underset{\frac{t}{2}}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1}{2}}{\rm{d}}s\\ & \le & N{\left(\frac{t}{2}\right)}^{-\tfrac{1}{2}}{\left(\frac{t}{2}\right)}^{\tfrac{\delta }{2}}+N{\left(\frac{t}{2}\right)}^{-1+\tfrac{\delta }{2}}{\left(\frac{t}{2}\right)}^{\tfrac{1}{2}}\\ & \le & N{t}^{-\tfrac{1-\delta }{2}}.\end{array}\hspace{12.2em}So, we have (3.13)‖G(U,U)‖Lq1q2q3(R3)≤Nt−1−δ2(∥u∥χp,q,∞2+2∥b∥χp,q,∞2+2‖u‖χp,q,∞‖b‖χp,q,∞).\Vert G\left(U,U){\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}\le N{t}^{-\tfrac{1-\delta }{2}}({\parallel u\parallel }_{{\chi }_{p,q,\infty }}^{2}+2{\parallel b\parallel }_{{\chi }_{p,q,\infty }}^{2}+2\Vert u{\Vert }_{{\chi }_{p,q,\infty }}\Vert b{\Vert }_{{\chi }_{p,q,\infty }}).\hspace{2.1em}Similarly, let βk=γk=1{\beta }_{k}={\gamma }_{k}=1, αk=pkqk∈(0,1]{\alpha }_{k}=\frac{{p}_{k}}{{q}_{k}}\in (0,1], and pkγk=qk\frac{{p}_{k}}{{\gamma }_{k}}={q}_{k}, −1+α+β−γ2=−121+∑k=1n=3αkpk+βkpk−γkpk=−121+∑k=1n=31qk+1pk−1pk=−121+∑k=1n=31qk=−1+δ2.-\frac{1+\alpha +\beta -\gamma }{2}=-\frac{1}{2}\left(1+\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{{\alpha }_{k}}{{p}_{k}}+\frac{{\beta }_{k}}{{p}_{k}}-\frac{{\gamma }_{k}}{{p}_{k}}\right)\right)=-\frac{1}{2}\left(1+\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{1}{{q}_{k}}+\frac{1}{{p}_{k}}-\frac{1}{{p}_{k}}\right)\right)=-\frac{1}{2}\left(1+\mathop{\sum }\limits_{k=1}^{n=3}\frac{1}{{q}_{k}}\right)=-\frac{1+\delta }{2}.By using 1p1+1p2+1p3=1\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}+\frac{1}{{p}_{3}}=1, 1q1+1q2+1q3=δ∈(0,1)\frac{1}{{q}_{1}}+\frac{1}{{q}_{2}}+\frac{1}{{q}_{3}}=\delta \in \left(0,1), and (3.10), we also obtain ‖DxG(U,U)‖Lp1p2p3(R3)≤N∫0t(t−s)−1+δ2(‖u‖Lq1q2q3(R3)∥Dxu∥Lp1p2p3(R3)+2‖b‖Lq1q2q3(R3)∥Dxb∥Lp1p2p3(R3)+‖u‖Lq1q2q3(R3)∥Dxb∥Lp1p2p3(R3)+‖b‖Lq1q2q3(R3)∥Dxu∥Lp1p2p3(R3))ds≤N(∥u∥χp,q,∞2+∥b∥χp,q,∞2+2‖u‖χp,q,∞‖b‖χp,q,∞)∫0t(t−s)−1+δ2s−1+δ2ds,\begin{array}{l}\Vert {D}_{x}G\left(U,U){\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\le N\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1+\delta }{2}}(\Vert u{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+2\Vert b{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\hspace{1.0em}+\Vert u{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+\Vert b{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}){\rm{d}}s\\ \hspace{1.0em}\le N({\parallel u\parallel }_{{\chi }_{p,q,\infty }}^{2}+{\parallel b\parallel }_{{\chi }_{p,q,\infty }}^{2}+2\Vert u{\Vert }_{{\chi }_{p,q,\infty }}\Vert b{\Vert }_{{\chi }_{p,q,\infty }})\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1+\delta }{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s,\end{array}and ∫0t(t−s)−1+δ2s−1+δ2ds=∫0t2(t−s)−1+δ2s−1+δ2ds+∫t2t(t−s)−1+δ2s−1+δ2ds≤t2−1+δ2∫0t2s−1+δ2ds+t2−1+δ2∫t2t(t−s)−1+δ2ds≤Nt2−1+δ2t2δ2+Nt2−1+δ2t21+δ2≤Nt−12.\hspace{6.25em}\begin{array}{l}\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1+\delta }{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s\\ \hspace{1.0em}=\underset{0}{\overset{\frac{t}{2}}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1+\delta }{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s+\underset{\frac{t}{2}}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1+\delta }{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s\\ \hspace{1.0em}\le {\left(\frac{t}{2}\right)}^{-\tfrac{1+\delta }{2}}\underset{0}{\overset{\frac{t}{2}}{\displaystyle \int }}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s+{\left(\frac{t}{2}\right)}^{-1+\tfrac{\delta }{2}}\underset{\frac{t}{2}}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{1+\delta }{2}}{\rm{d}}s\\ \hspace{1.0em}\le N{\left(\frac{t}{2}\right)}^{-\tfrac{1+\delta }{2}}{\left(\frac{t}{2}\right)}^{\tfrac{\delta }{2}}+N{\left(\frac{t}{2}\right)}^{-1+\tfrac{\delta }{2}}{\left(\frac{t}{2}\right)}^{\tfrac{1+\delta }{2}}\\ \hspace{1.0em}\le N{t}^{-\tfrac{1}{2}}.\end{array}\hspace{12.5em}So, we have (3.14)‖DxG(U,U)‖Lp1p2p3(R3)≤Nt−12(∥u∥χp,q,∞2+2∥b∥χp,q,∞2+2‖u‖χp,q,∞‖b‖χp,q,∞).\Vert {D}_{x}G\left(U,U){\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le N{t}^{-\tfrac{1}{2}}({\parallel u\parallel }_{{\chi }_{p,q,\infty }}^{2}+2{\parallel b\parallel }_{{\chi }_{p,q,\infty }}^{2}+2\Vert u{\Vert }_{{\chi }_{p,q,\infty }}\Vert b{\Vert }_{{\chi }_{p,q,\infty }}).From the estimates (3.13), (3.14), and Lemma 3.1, we can prove that t1−δ2G(U,U):[0,∞)→PLq1q2q3(R3){t}^{\tfrac{1-\delta }{2}}G\left(U,U):{[}0,\infty )\to {\mathbb{P}}{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})is continuous and when t→0t\to 0, t1−δ2G(U,U)→0{t}^{\tfrac{1-\delta }{2}}G\left(U,U)\to 0. Similarly, we can also prove that t12DxG(U,U):[0,∞)→PLp1p2p3(R3){t}^{\tfrac{1}{2}}{D}_{x}G\left(U,U):{[}0,\infty )\to {\mathbb{P}}{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})is continuous and when t→0t\to 0, t12DxG(U,U)→0{t}^{\tfrac{1}{2}}{D}_{x}G\left(U,U)\to 0. Therefore, we obtain G(U,U)∈χp,q,∞G\left(U,U)\in {\chi }_{p,q,\infty }and for all U∈χp,q,∞U\in {\chi }_{p,q,\infty }: (3.15)‖G(U,U)‖χp,q,∞≤N(∥u∥χp,q,∞2+2∥b∥χp,q,∞2+2‖u‖χp,q,∞‖b‖χp,q,∞)≤N‖U‖χp,q,∞‖U‖χp,q,∞≤N2∥U∥χp,q,∞2,∀U∈χp,q,∞,\begin{array}{rcl}\Vert G\left(U,U)\Vert {\chi }_{p,q,\infty }& \le & N({\parallel u\parallel }_{{\chi }_{p,q,\infty }}^{2}+2{\parallel b\parallel }_{{\chi }_{p,q,\infty }}^{2}+2\Vert u{\Vert }_{{\chi }_{p,q,\infty }}\Vert b{\Vert }_{{\chi }_{p,q,\infty }})\\ & \le & N\Vert U{\Vert }_{{\chi }_{p,q,\infty }}\Vert U{\Vert }_{{\chi }_{p,q,\infty }}\\ & \le & {N}_{2}{\parallel U\parallel }_{{\chi }_{p,q,\infty }}^{2},\hspace{1em}\forall U\in {\chi }_{p,q,\infty },\end{array}where N2{N}_{2}is a constant depending on p1{p}_{1}, p2{p}_{2}, p3{p}_{3}, q1{q}_{1}, q2{q}_{2}, and q3{q}_{3}. That is, the bilinear G:χp,q,∞×χp,q,∞→χp,q,∞G:{\chi }_{p,q,\infty }\times {\chi }_{p,q,\infty }\to {\chi }_{p,q,\infty }is bounded.Next, let us choose a sufficiently small constant λ0{\lambda }_{0}so that 4N1N2λ0<1,4{N}_{1}{N}_{2}{\lambda }_{0}\lt 1,where N1{N}_{1}is defined in (3.12) and N2{N}_{2}is defined in (3.15). If ∥U0∥Lp1p2p3(R3)≤λ0{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le {\lambda }_{0}, then it can be obtained by (3.12) 4N2‖U1‖χp,q,∞≤4N1N2∥U0∥Lp1p2p3(R3)≤4N1N2λ0<1.4{N}_{2}\Vert {U}_{1}{\Vert }_{{\chi }_{p,q,\infty }}\le 4{N}_{1}{N}_{2}{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le 4{N}_{1}{N}_{2}{\lambda }_{0}\lt 1.By this and Lemma 3.3, we can gain a unique time solution U∈χp,q,∞U\in {\chi }_{p,q,\infty }of (3.2) such that (3.16)‖U‖χp,q,∞≤2∥U1∥χp,q,∞≤2N1∥U0∥Lp1p2p3(R3).\Vert U{\Vert }_{{\chi }_{p,q,\infty }}\le 2{\parallel {U}_{1}\parallel }_{{\chi }_{p,q,\infty }}\le 2{N}_{1}{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}.Now, we need to prove that U∈Yp,∞U\in {{\mathcal{Y}}}_{p,\infty }. As U=U1+G(U,U)U={U}_{1}+G\left(U,U), we have (3.17)‖U‖Lp1p2p3(R3)≤‖U1‖Lp1p2p3(R3)+‖G(U,U)‖Lp1p2p3(R3),\Vert U{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le \Vert {U}_{1}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+\Vert G\left(U,U){\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},\hspace{0.95em}(3.18)‖DxU‖Lp1p2p3(R3)≤‖DxU1‖Lp1p2p3(R3)+‖DxG(U,U)‖Lp1p2p3(R3).\Vert {D}_{x}U{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le \Vert {D}_{x}{U}_{1}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+\Vert {D}_{x}G\left(U,U){\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}.Then, by Lemma 3.1, we see that (3.19)∥U1∥Lp1p1p3(R3)≤N‖U0‖Lp1p2p3(R3),{\parallel {U}_{1}\parallel }_{{L}_{{p}_{1}{p}_{1}{p}_{3}}({{\mathbb{R}}}^{3})}\le N\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})},(3.20)∥DxU1∥Lp1p2p3(R3)≤Nt−12‖U0‖Lp1p2p3(R3).{\parallel {D}_{x}{U}_{1}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le N{t}^{-\tfrac{1}{2}}\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}.Let βk=γk=1{\beta }_{k}={\gamma }_{k}=1, αk=pkqk∈(0,1]{\alpha }_{k}=\frac{{p}_{k}}{{q}_{k}}\in (0,1], and pkγk=pk\frac{{p}_{k}}{{\gamma }_{k}}={p}_{k}, −α+β−γ2=−12∑k=1n=3αkpk+βkpk−γkpk=−12∑k=1n=31qk+1pk−1pk=−12∑k=1n=31qk=−δ2.-\frac{\alpha +\beta -\gamma }{2}=-\frac{1}{2}\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{{\alpha }_{k}}{{p}_{k}}+\frac{{\beta }_{k}}{{p}_{k}}-\frac{{\gamma }_{k}}{{p}_{k}}\right)=-\frac{1}{2}\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{1}{{q}_{k}}+\frac{1}{{p}_{k}}-\frac{1}{{p}_{k}}\right)=-\frac{1}{2}\mathop{\sum }\limits_{k=1}^{n=3}\left(\frac{1}{{q}_{k}}\right)=-\frac{\delta }{2}.By using (3.9), we can infer that ‖G(U,U)‖Lp1p2p3(R3)≤N∫0t(t−s)−δ2(‖u‖Lq1q2q3(R3)∥Dxu∥Lp1p2p3(R3)+2‖b‖Lq1q2q3(R3)∥Dxb∥Lp1p2p3(R3)+‖u‖Lq1q2q3(R3)∥Dxb∥Lp1p2p3(R3)+‖b‖Lq1q2q3(R3)∥Dxu∥Lp1p2p3(R3))ds≤N(∥u∥χp,q,∞2+∥b∥χp,q,∞2+2‖u‖χp,q,∞‖b‖χp,q,∞)∫0t(t−s)−δ2s−1+δ2ds.\begin{array}{l}\Vert G\left(U,U){\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\le N\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{\delta }{2}}(\Vert u{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+2\Vert b{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\\ \hspace{1.0em}\hspace{1.0em}+\Vert u{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}b\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+\Vert b{\Vert }_{{L}_{{q}_{1}{q}_{2}{q}_{3}}({{\mathbb{R}}}^{3})}{\parallel {D}_{x}u\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}){\rm{d}}s\\ \hspace{1.0em}\le N({\parallel u\parallel }_{{\chi }_{p,q,\infty }}^{2}+{\parallel b\parallel }_{{\chi }_{p,q,\infty }}^{2}+2\Vert u{\Vert }_{{\chi }_{p,q,\infty }}\Vert b{\Vert }_{{\chi }_{p,q,\infty }})\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{\delta }{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s.\end{array}Since ∫0t(t−s)−δ2s−1+δ2ds=∫0t2(t−s)−δ2s−1+δ2ds+∫t2t(t−s)−δ2s−1+δ2ds≤t2−δ2∫0t2s−1+δ2ds+t2−1+δ2∫t2t(t−s)−δ2ds≤N1t2−δ2t2δ2+N2t2−1+δ2t2−δ2+1≤N,\hspace{-27.1em}\begin{array}{rcl}\underset{0}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{\delta }{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s& =& \underset{0}{\overset{\frac{t}{2}}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{\delta }{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s+\underset{\frac{t}{2}}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{\delta }{2}}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s\\ & \le & {\left(\frac{t}{2}\right)}^{-\tfrac{\delta }{2}}\underset{0}{\overset{\frac{t}{2}}{\displaystyle \int }}{s}^{-1+\tfrac{\delta }{2}}{\rm{d}}s+{\left(\frac{t}{2}\right)}^{-1+\tfrac{\delta }{2}}\underset{\frac{t}{2}}{\overset{t}{\displaystyle \int }}{\left(t-s)}^{-\tfrac{\delta }{2}}{\rm{d}}s\\ & \le & {N}_{1}{\left(\frac{t}{2}\right)}^{-\tfrac{\delta }{2}}{\left(\frac{t}{2}\right)}^{\tfrac{\delta }{2}}+{N}_{2}{\left(\frac{t}{2}\right)}^{-1+\tfrac{\delta }{2}}{\left(\frac{t}{2}\right)}^{-\tfrac{\delta }{2}+1}\\ & \le & N,\end{array}by using (3.16), we obtain that (3.21)‖G(U,U)‖Lp1p2p3(R3)≤N(∥u∥χp,q,∞2+2∥b∥χp,q,∞2+2‖u‖χp,q,∞‖b‖χp,q,∞)≤N∥U∥χp,q,∞2≤N∥U0∥Lp1p2p3(R3)2.\begin{array}{rcl}\Vert G\left(U,U){\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}& \le & N({\parallel u\parallel }_{{\chi }_{p,q,\infty }}^{2}+2{\parallel b\parallel }_{{\chi }_{p,q,\infty }}^{2}+2\Vert u{\Vert }_{{\chi }_{p,q,\infty }}\Vert b{\Vert }_{{\chi }_{p,q,\infty }})\\ & \le & N{\parallel U\parallel }_{{\chi }_{p,q,\infty }}^{2}\le N{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}^{2}.\end{array}Similarly, by using (3.14) and (3.16), we obtain that (3.22)‖DxG(U,U)‖Lp1p2p3(R3)≤Nt−12(∥u∥χp,q,∞2+2∥b∥χp,q,∞2+2‖u‖χp,q,∞‖b‖χp,q,∞)≤Nt−12∥U∥χp,q,∞2≤Nt−12∥U0∥Lp1p2p3(R3)2.\begin{array}{rcl}\Vert {D}_{x}G\left(U,U){\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}& \le & N{t}^{-\tfrac{1}{2}}({\parallel u\parallel }_{{\chi }_{p,q,\infty }}^{2}+2{\parallel b\parallel }_{{\chi }_{p,q,\infty }}^{2}+2\Vert u{\Vert }_{{\chi }_{p,q,\infty }}\Vert b{\Vert }_{{\chi }_{p,q,\infty }})\\ & \le & N{t}^{-\tfrac{1}{2}}{\parallel U\parallel }_{{\chi }_{p,q,\infty }}^{2}\le N{t}^{-\tfrac{1}{2}}{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}^{2}.\end{array}\hspace{0.1em}From estimates (3.17), (3.19), and (3.21), we see that ‖U‖Lp1p2p3(R3)≤N(‖U0‖Lp1p2p3(R3)+‖U0‖Lp1p2p3(R3)2).\Vert U{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le N(\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}^{2}).\hspace{2.5em}From the estimates (3.18), (3.20), and (3.22), we see that ‖DxU‖Lp1p2p3(R3)≤Nt−12(‖U0‖Lp1p2p3(R3)+‖U0‖Lp1p2p3(R3)2).\Vert {D}_{x}U{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}\le N{t}^{-\tfrac{1}{2}}(\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}^{2}).\hspace{2.5em}It is known that ‖U0‖Lp1p2p3(R3)\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}is sufficiently small, we can gain ‖U‖Yp,∞≤N0(‖U0‖Lp1p2p3(R3)+‖U0‖Lp1p2p3(R3)2).\Vert U{\Vert }_{{{\mathcal{Y}}}_{p,\infty }}\le {N}_{0}(\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+\Vert {U}_{0}{\Vert }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}^{2}).\hspace{0.35em}Now, we have to prove (ii). Similar to the proof of (3.12), we found that U1∈χp,q,∞{U}_{1}\in {\chi }_{p,q,\infty }, χp,q,∞{\chi }_{p,q,\infty }is continuous, and when t→0t\to 0, t1−δ2U1→0{t}^{\tfrac{1-\delta }{2}}{U}_{1}\to 0, t12DxU1→0{t}^{\tfrac{1}{2}}{D}_{x}{U}_{1}\to 0. There is a sufficiently small constant T0>0{T}_{0}\gt 0depending on p1{p}_{1}, p2{p}_{2}, p3{p}_{3}, q1{q}_{1}, q2{q}_{2}, q3{q}_{3}, and U0{U}_{0}such that ‖U1‖χp,q,T0≤λ0.\Vert {U}_{1}{\Vert }_{{\chi }_{p,q,{T}_{0}}}\le {\lambda }_{0}.\hspace{5.75em}Furthermore, similar to the proof of (3.15), we also found that the bilinear GG: χp,q,T0×χp,q,T0→χp,q,T0{\chi }_{p,q,{T}_{0}}\times {\chi }_{p,q,{T}_{0}}\to {\chi }_{p,q,{T}_{0}}is bounded and ‖G(U,U)‖χp,q,T0≤N‖U‖χp,q,T0‖U‖χp,q,T0,∀U∈χp,q,T0.\Vert G\left(U,U){\Vert }_{{\chi }_{p,q,{T}_{0}}}\le N\Vert U{\Vert }_{{\chi }_{p,q,{T}_{0}}}\Vert U{\Vert }_{{\chi }_{p,q,{T}_{0}}},\hspace{1em}\forall U\in {\chi }_{p,q,{T}_{0}}.Then, by Lemma 3.3, we can obtain a unique local time solution U∈χp,q,T0U\in {\chi }_{p,q,{T}_{0}}of (3.2) with ‖U‖χp,q,T0≤2N1∥U0∥Lp1p2p3(R3).\Vert U{\Vert }_{{\chi }_{p,q,{T}_{0}}}\le 2{N}_{1}{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}.\hspace{4.1em}As in proof (i), we can gain U∈Yp,T0U\in {{\mathcal{Y}}}_{p,{T}_{0}}and ‖U‖Yp,T0≤N(∥U0∥Lp1p2p3(R3)+∥U0∥Lp1p2p3(R3)2).□\hspace{12em}\Vert U{\Vert }_{{{\mathcal{Y}}}_{p,{T}_{0}}}\le N({\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}+{\parallel {U}_{0}\parallel }_{{L}_{{p}_{1}{p}_{2}{p}_{3}}({{\mathbb{R}}}^{3})}^{2}).\hspace{14.25em}\square

Journal

Open Mathematicsde Gruyter

Published: Jan 1, 2022

Keywords: magnetohydrodynamic equations; local well posedness; global well posedness; Primary 35D30; Secondary 35Q30; 76W05; 76D06

There are no references for this article.