Mild solution and semigroup theory

Mild solution is an important object to investigate the strong solution of hydrodynamic flows with dissipation. The investigation will apply semigroup theory and some other functional tools.

Introduction

Consider Naiver-Stokes equation with zero-force
\begin{equation}
\label{Zero Force} \tag{NS}
\begin{cases}
\partial_t v-\Delta v+v\cdot \nabla v+\nabla p=0\\
\nabla \cdot v=0, v_{|t=0}=v_0.
\end{cases}
\end{equation}Take Helmholtz projection $P$ at both hand of \eqref{Zero Force}, and remind the divergence free condition, then the pressure is eliminated: $\partial_t v+Av+P (v\cdot \nabla )v=0$, where $A=-P\Delta $ is the Stokes operator((For more information about $A$ and the semigroup $e^{-tA}$, see the appendix.)). Now solve formally as ODE with initial data $v_0$, it figure as:
\begin{equation}
\label{Integral Equation} \tag{2}
v(t)=e^{-tA}v_0-\int_0^t e^{-(t-\tau )A}P \nabla (v\otimes v)(\tau )d\tau \eqqcolon v_1(t)+B(v,v)(t).
\end{equation}Here $B(u,v)\coloneqq -\int_0^t e^{-(t-s)A}P \nabla (u\otimes v)(s)ds$. Then $v$ is a mild solution of \eqref{Zero Force} in $X(\Omega_T)$ with initial data $v_0$, if $v=v_1+B(v,v)$ holds in $X(\Omega_T )$.

Fix structure $X$ and spatial domain $\Omega$, moreover, we say $v$ is a mild solution in $ [t_0,t_1) $ if $u(t)=v(t+t_0)$ is a mild solution in $X(\Omega_{t_1-t_0} )$ with initial data $v(t_0)$, i.e.
\begin{equation*}
\label{Restriction Mild Solution} \tag{3}
\begin{split}
v(t)=&e^{-(t-t_0)A}v(t_0)-\int_{t_0}^t e^{-(t-\tau) A }P\nabla (v\otimes v )(\tau )d\tau,\forall t\in [t_0,t_1).\\
\end{split}
\end{equation*}The following claim tells that the restriction preserves a mild solution.

If $v$ is a mild solution in $[0,T)$ with initial data $v_0$, then for any $[t_0,t_1)\subset [0,T) $, $v$ is also a mild solution in $[t_0,t_1)$ with inital data $v(t_0)$.(($v(t_0)$ need to be well-defined.))

Denote $P \nabla (v\otimes v)(t)=F(t) $ for convenience, then
\begin{equation*}
\begin{split}
v(t)=& e^{-tA}v_0-\int_0^t e^{-(t-\tau )A}F(\tau )d\tau\\
=& e^{-(t-t_0)A} e^{-t_0A} v_0-\int_0^{t_0} e^{-(t-t_0)}e^{-(t_0-\tau)} F(\tau )d\tau
-\int_{t_0}^t e^{-(t-\tau )A}F(\tau )d\tau\\
=& e^{-(t-t_0)A} \br{e^{-t_0 A}+ \int_0^{t_0} e^{-(t_0-\tau)}F(\tau )d\tau }- \int_{t_0}^t e^{-(t-\tau )A}F(\tau )d\tau\\
=& e^{-(t-t_0 ) A} v(t_0)-\int_{t_0}^t e^{-(t-\tau)A }F(\tau )d\tau
\end{split}
\end{equation*}for any $t\in [t_0,t_1)$.
Clearly it corresponds to the characterization \eqref{Restriction Mild Solution}.

Existence with $L^q(q>3)$ and $L^3$ data

Now we discuss about the existence of mild solution in $L^q(q>3)$ and $L^3$ case respectively. The main tool is Banach fixed-point theorem, where we consider the following mapping:
\begin{equation}
\label{Integral Transform} \tag{4}
\Psi: v\mapsto v_1+ B(v,v), v\in X .
\end{equation}To make it a contraction, different spaces $X$ are constructed for different cases.

Suppose $v_0\in L^q_\sigma(q>3)$, then with initial data $v_0$ there exists a mild solution
$v\in C([0, T_1]; L^q_\sigma )$ for equation \eqref{Zero Force}, where
$$ T_1\sim \norm{v_0}_q^{-s }, \dfrac{2}{s}+\dfrac{3}{p}=1.$$

Denote $X=C_tL^q_\sigma$. It is necessary to give out the estimate for $B(u,v)$ and $e^{-tA}v_0 $ first. Suppose $u,v\in X$ and notice $\frac 12-\frac 3{2q}>0$, then
\begin{equation*}
\begin{split}
\norm{B(u,v)(t)}_q \le & \int_0^t \norm{e^{-(t-\tau)A}P\nabla }_{\frac q2\to q }\norm{u(\tau ) }_q\norm{v(\tau ) }_qd\tau\\
\le & C\br{\int_0^t(t-s)^{- \frac 3{2q}-\frac 12 }ds} \norm{u}_{X}\norm{v}_{X}\\
\le & C_0 t^{\frac 12-\frac 3{2q} }\norm{u}_{X}\norm{v}_{X};
\end{split}
\end{equation*}Moreover, $\norm{e^{-tA}v_0 }_q\le C_q\norm{v_0}_q$ by \ref{}. Consequently $B(u,v), e^{-tA}v_0\in X$ as $e^{-tA}$ is a strong continuous semigroup in $L^q_\sigma$.Now we consider the mapping $\Psi$ and show that it is a contraction in
$$B_1=\{v\in X| \norm{v}_{X}\le 2 C_q \norm{v_0}_q \}$$for proper time length $T_1>0$.

$\Psi$ stablize $B_1$((That is say, $\Psi(B)\subset B$)). The former estimates show that $\Psi:X\to X$, now for any $\norm{v}_{X}\le 2C_q\norm{v_0}_q$, it comes
\begin{equation*}
\begin{split}
\norm{\Psi(v) }_{X}\le & \norm{e^{-tA}v_0 }_{X}+\norm{B(v,v) }_{X}\\
\le & C_q\norm{v_0}_q+ C_0T_1^{\frac 12-\frac 3{2q} }\cdot (4C_q^2 \norm{v_0}^2_q)\\
\le & 2C_q \norm{v_0}_q
\end{split}
\end{equation*}once we let $T_1\le (4C_0C_q\norm{v_0}_q)^{-s} $;
$\Psi$ contracts $B_1$. Indeed, $\forall v_1,v_2\in X ,$
\begin{equation*}
\begin{split}
\norm{\Psi(v_1)-\Psi(v_2) }_{X}=& \norm{B(v_1,v_1)-B(v_2,v_2) }_{X}=\norm{B(v_1,v_1-v_2)+B(v_1-v_2,v_2) }_{X}\\
\le & C_0 T_1^{\frac 12-\frac 3{2q} }(\norm{v_1}_{X}+\norm{v_2}_{X})\norm{v_1-v_2}_{X}\\
\le & C_0 T^{\frac 12-\frac 3{2q} }\cdot{4C_q\norm{v_0}_q \cdot} \norm{v_1-v_2}_{X}\le \frac 12\norm{v_1-v_2 }_{X}
\end{split}
\end{equation*}if $T_1\le (8C_0C_q\norm{v_0}_q )^{-s}$.

In conclusion, $\Psi$ is a contraction in $B_1$ for $T_1=(8C_0C_q\norm{v_0}_q)^{-s}\sim \norm{v_0}_q^{-s}$. In this case, there is a unique $v$ s.t. $v(t)=e^{-tA}v_0+B(v,v)(t)$ holds in $B_1$ (and thus in $X$).

The above proof only grants the uniqueness in class $\norm{v}_X\le 2C_q\norm{v_0}_q $ but not in the whole $X$. Thus we need to verify the uniqueness respectively:

Suppose $0\sl T\le T',$$ v\in C([0,T], L^q_\sigma), u\in C([0,T'], L^q_\sigma)$ are two mild solution of \eqref{Zero Force} with same initial data $v_0\in L^q_\sigma$, then $v=u$ in $[0, T]$.

Suppose $v,u$ diverge at $t_0$, we show that $t_0=T$. Denote $w=u-v$ on $[0,T]$, then $w=0$ on $[0,t_0]$. For $t_1\in (t_0, T]$. We observe that
\begin{equation*}
\begin{split}
\norm{w(t)}_q\le & \norm{B(u,w)(t)+B(w,v)(t) }_q\\
\le & \int_{t_0}^t \norm{e^{-(t-\tau)A}P\nabla } \norm{w(\tau)}_q(\norm{u(\tau )}_q+\norm{v(\tau)}_q )d\tau\\
\le & C_0 (t-t_0)^{\frac 12-\frac 3{2q} }(\norm{ u}_{X}+\norm{v}_{X} )\norm{w}_{X}.
\end{split}
\end{equation*}Then we can choose $t'$ small enough that for $t\in (t_0,t_0+t'),$
$$C_0 (t-t_0)^{\frac 12-\frac 3{2q} }(\norm{ u}_X+\norm{v}_X )\sl 1\Longrightarrow w(t)=0, $$which contracts to the selection of $t_0$.

That is say, two mild solution must coincident on their common interval. So it is proper to prolong the interval.

Next we consider the existence for $L^3$ data. To reach it, we introduce a space for $p\ge 3:$
\begin{equation*}
\norm{v}_{Y^p}=\norm{ \norm{v }_{Z^p}(t) }_{L^\infty_t}= \norm{t^{\frac 12-\frac 3{2q} }\norm{v(t)}_p }_{L^\infty_t},
\end{equation*}and set
$$Y^p_0=\{v\in Y^p| \norm{v}_{Z^p}(t)\xrightarrow{t\to \infty} 0 \}, Y^3_0=Y^3.$$Denote $Y^{p,q}=Y^p\cap Y^q, Y^{p,q}_0=Y^p_0\cap Y^q_0$. See appendix for more information about $Y^p$. Now we claim the existence as

Suppose $v_0\in L^3_\sigma$, then for any $q\in (3,\infty)$, there exists $T_2>0$ such that $v\in Y^{3,q}(\Omega_{T_2} ) $ is a mild solution of \eqref{Zero Force} with initial data $v_0$.

Set $\alpha \in (3,q)$. The about corollary indicates for $u\in Y^\alpha,$\
\begin{equation*}
v\in Y^{3,q}/Y^{3,q}_0\Longrightarrow B(u,v)\in Y^{3,q}/Y^{3,q}_0.
\end{equation*}Now we consider $\Psi$ defined by \eqref{Integral Transform} in space
$$B_2=\{v\in Y^{3,q}_0| \norm{v}_{Y^\alpha}\le 2\norm{e^{-tA}v_0 }_{Y^\alpha} \} $$and find the proper time length $T_2$.

$\Psi$ stablize $B$. First we prove $\Psi: Y^{3,q}_0\to Y^{3,q}_0$. Indeed,
\begin{equation*}
\norm{e^{-tA}v_0}_{3}\le C_3\norm{v_0}_3, \norm{e^{-tA}v_0}_q\le C t^{-\frac 32 (\frac 13-\frac 1q )}\norm{v_0}_3\Longrightarrow e^{-tA}v_0\in Y^{3,q}.
\end{equation*}Moreover, by the asymptomatic behavior of $t^{\frac 12-\frac 3{2q} } \norm{e^{-tA}v_0 }_q$ in \ref{Asymp Behavior} we see $e^{-tA}v_0\in Y^{3,q}_0$. As for $B(v,v)$, $v\in Y^{3,q}_0\Longrightarrow B(v,v)\in Y^{3,q}_0$ by \ref{Bilinear Y}.Now for $\norm{v}_{Y^\alpha}\le 2\norm{e^{-tA}v_0}_{Y^\alpha}$, we have
\begin{equation*}
\begin{split}
\norm{\Psi(v)}_{Y^\alpha}\le & \norm{e^{-tA}v_0 }_{Y^\alpha}+\norm{B(v,v) }_{Y^\alpha}\\
\le & \norm{e^{-tA}v_0 }_{Y^\alpha}+4C\norm{e^{-tA}v_0}_{Y^\alpha}^2\\
\le &2 \norm{e^{-tA}v_0 }_{Y^\alpha},
\end{split}
\end{equation*}provide $\norm{ e^{-tA}v_0}_{Y^\alpha}$ is small enough.
$\Psi$ contracts $B$. Take $u,v\in B$, then
\begin{equation*}
\begin{split}
\norm{\Psi(u)-\Psi(v) }_{Y^{3,q}}=& \norm{B(u, u-v )+B(u-v,v) }_{Y^{3,q}}\\
\le & C(\norm{u}_{Y^\alpha}+\norm{v}_{Y^\alpha} )\norm{u-v}_{Y^{3,q}}\\
\le & 4C \norm{e^{-tA}v_0 }_\alpha \norm{u-v }_{Y^{3,q}},
\end{split}
\end{equation*}which contracts also for small $\norm{e^{-tA}v_0}_{Y^\alpha}$.

The criterion for above is $\norm{e^{-tA}v_0 }_{Y^\alpha}\le (8C)^{-1} $, which can be reached when

$\norm{v_0}_3$ is small, since $\norm{e^{-tA}v_0 }_{\alpha}\le C_\alpha t^{-(\frac 12-\frac{3 }{2\alpha } ) }\norm{v_0}_3 $;
$T_2$ is small, since $\norm{e^{-tA}v_0 }_{Z^\alpha}\to 0$ as $t\to 0$ by \ref{}.

Consequently, if $\norm{v_0}_3\le (8CC_\alpha)^{-1}$, there exists unique global mild solution under structure $Y^{3,q}_0 $;
if not, there exists unique mild solution in $[0,T_2]$ under $Y^{3,q}_0$, for $T_2$ the critical time satisfying $\norm{e^{-tA}v_0 }_{Z^\alpha}\le (8C)^{-1},\forall t\in [0, T_2]$.

Suppose $0\sl T\le T', v\in Y^{3,q}_0(\Omega_T ), u\in Y^{3,q}_0(\Omega_{T'} )$ are two mild solution of \eqref{Zero Force} with same initial data $v_0\in L^3_\sigma$, then $v=u$ in $[0, T]$.

Similarly we define $w=u-v$ and $t_1$ the diverging time. Then
\begin{equation*}
\begin{split}
\norm{w}_\alpha \le & \int_{t_1}^t \norm{e^{-(t-\tau)A}P\nabla }_{\frac \alpha 2\to \alpha }(\norm{u(\tau)}_{\alpha}+\norm{v(\tau ) }_\alpha )\norm{w(\tau)}_\alpha d\tau\\
\le & C \int_{t_1}^t (t-\tau)^{-\frac 12-\frac{3}{2\alpha}} \tau^{-1+\frac 3\alpha }d\tau (\norm{u}_{Y^\alpha}+\norm{v}_{Y^\alpha})\norm{w}_{Y^\alpha(\Omega_t )}\\
\le & C t^{-\frac 12+\frac 3{2\alpha} } \br{ \int^1_{\frac{t_1}{t}} (1-\theta )^{-\frac 12-\frac 3{2\alpha } }\theta^{-1+\frac{3}{\alpha}} d\theta} (\norm{u}_{Y^\alpha}+\norm{v}_{Y^\alpha})\norm{w}_{Y^\alpha(\Omega_t )}\\
\le & Ct^{-\frac 12+\frac 3{2\alpha } }\br{1-\frac {t_1} t}^{-\frac 12+\frac 3{2\alpha } }(\norm{u }_{Y^\alpha }+\norm{v}_{Y^\alpha } )\norm{w }_{Y^\alpha(\Omega_t ) }\\
\Longrightarrow \norm{w}_{Y^\alpha(\Omega_t )}\le & C (1-\frac{t_1}{t} )^{-\frac 12+\frac 3{2\alpha } }\br{\norm{u}_{Y^\alpha}+\norm{v }_{Y^\alpha } }\norm{w }_{Y^\alpha(\Omega_t ) }.
\end{split}
\end{equation*}Then $w(t)=0$ for $t$ large than $t_1$.

Regularity of Mild Solution

We show that the mild solution obtained above is also a strong solution (in probably smaller interval), and it is .

The above mild solution lays in $L^sL^q$ for any $\frac{2}{s}+\frac 3q=1, s\in (3,\infty)$. Moreover, if $v_0\in L^2_\sigma\cap L^3_\sigma,$ then $v$ lays in $L^\infty L^2\cap L^2\dot H^1$.

We will set $B_3, B_4, B_5$ in $L^sL^q,L^\infty L^2, L^2 \dot H^1 $ respectively, and show $\Psi$ is a contraction in $B_2\cap B_3\cap B_4\cap B_5$ for proper small time length $T$.

$L^sL^q$: First we give out $L^sL^q$ estimate for $v_1, B(v,v)$ first. The estimate \ref{LsLq v1} gives out $\norm{v_1 }_{L^sL^q}\le C\norm{v_0}_3 $. As for $B(v,v)$, HLS \eqref{HLS} implies
\begin{equation*}
\begin{split}
\norm{B(u,v)}_{L^sL^q}\le & C\norm{ \int_0^t |t-\tau|^{-\frac 12-\frac 3{2q} }\norm{u(\tau )}_q\norm{v(\tau )}_q d\tau}_{L^s}\\
\le & C\norm{\frac{1}{t^{\frac 12+\frac 3{2q} }}*\br{ \norm{u(t)}_q\norm{v(t)}_q} }_{L^{s}_t}\\
\le & C\norm{\norm{u(t)}_q\norm{v(t)}_q }_{L^{s/2}_t}\\
\le & C\norm{u}_{L^sL^q}\norm{v}_{L^sL^q}.
\end{split}
\end{equation*}Now we show $\Psi$ is a contraction in $B_{3}=\{v\in L^sL^q| \norm{v}_{L^sL^q}\le 1/(4C) \}$ for small $T$.
1. $\Psi$ stablizes $B_3$. Since $v_1\in {L^sL^q(\Omega_{\infty})}$$\Longrightarrow \norm{v_1}_{L^sL^q(\Omega_t ) }\to 0$ as $t\to 0$. Then
  \begin{equation*}
  \begin{split}
  \norm{\Psi (v)}_{L^sL^q}\le & \norm{ e^{-tA}v_0 }_{L^sL^q}+C\norm{v}_{L^sL^q}^2\\
  \le & \frac{1}{8C}+ \frac{1}{16C}\le \frac{1}{4C},
  \end{split}
  \end{equation*}if $T$ is small enough.
2. $\Psi$ contracts $B_3$:
  \begin{equation*}
  \begin{split}
  \norm{\Psi(u)-\Psi(v)}_{L^sL^q}= & \norm{B(u,u-v)+B(u-v,v) }_{L^sL^q}\\
  \le & C\br{\norm{u}_{L^sL^q}+\norm{u}_{L^sL^q}}\norm{u-v}_{L^sL^q}\\
  \le & \frac{1}{2} \norm{u-v}_{L^sL^q}.
  \end{split}
  \end{equation*}
$L^2\dot H^1:$ Apply \eqref{Stokes Identity} and HLS inequality again, we have
\begin{equation*}
\norm{\nabla e^{-tA}v_0 }_{L^2L^2}\le \norm{v_0}_2;
\end{equation*}\begin{equation*}
\begin{split}
\norm{\nabla B(u,v)(t) }_{L^2L^2}\le & \norm{\int_0^t \norm{\nabla e^{-(t-\tau )A}P }_{\frac{1}{1/q+ 1/2} \to 2}\norm{u(\tau)}_{q}\norm{\nabla v(\tau )}_2 d\tau}_{L^2_t} \\
\le &\norm{ \int_0^t \abs{t-\tau }^{-\frac 12-\frac 3{2q} }\norm{u(\tau)}_q\norm{\nabla v(\tau)}_2 d\tau}_{L^2_t}\\
\le & \norm{ t^{-(\frac 12+\frac 3{2q} )} * \norm{u(t)}_q\norm{\nabla v}_2}_2\\
\le & C\norm{u }_{L^sL^q}\norm{\nabla v }_{L^2L^2}.
\end{split}
\end{equation*}Set $B_5=\{v\in L^2\dot H^1| \norm{v}_{L^2\dot H^1}\le 2\norm{v_0}_2 \}$. Then for stabilization and contraction,
\begin{equation*}
\begin{split}
\norm{\Psi(v)}_{L^2\dot H^1 }\le \norm{v_0}_2+C\norm{v}_{L^sL^q}\norm{v}_{L^2\dot H^1 }\le 2\norm{v_0};
\end{split}
\end{equation*}\begin{equation*}
\begin{split}
\norm{\Psi(u)-\Psi(v) }_{L^2\dot H^1}\le & C\br{\norm{v}_{L^sL^q}+\norm{u }_{L^sL^q} }\norm{u-v}_{L^2 \dot H^1} \\
\le & \frac 12\norm{u-v}_{L^2\dot H^1}
\end{split}
\end{equation*}when $u,v\in B_5\cap B_3$ and $ T$ small enough.
$L^\infty L^2:$ The estimate is given by:
\begin{equation*}
\norm{v_1(t)}_{L^2}\le \norm{v_0}_2;
\end{equation*}\begin{equation*}
\begin{split}
\norm{B(u,v)(t)}_{L^2}\le & \int_0^t \norm{e^{-(t-\tau)A} P\nabla }_{\frac 1{1/\alpha +1/2}\to 2 }\norm{u(\tau) }_\alpha\norm{v(\tau) }_2 d\tau\\
\le & \int_0^t |t-\tau |^{-\frac 12-\frac 3{2\alpha}}\cdot \tau^{-\frac{1}{2}+\frac{3}{2\alpha} } \norm{u(\tau )}_{Z^\alpha} \norm{v(\tau) }_2 d\tau\\
\le &C\norm{u }_{Y^\alpha}\norm{v}_{L^\infty L^2}.
\end{split}
\end{equation*}Set $B_4=\{v\in L^\infty L^2|\norm{v}_{L^\infty L^2 }\le \norm{v_0}_2 \}$, then the contraction in $B_2\cap B_4$ is promised:
\begin{equation*}
\begin{split}
\norm{\Psi(v)}_{L^2L^\infty}\le & \norm{e^{-tA}v_0}_{L^\infty L^2}+\norm{B(v,v) }_{L^\infty L^2}\\
\le & \norm{v_0}_{2}+C\norm{v}_{Y^\alpha}\norm{v}_{L^\infty L^2}\le 2\norm{v_0};
\end{split}
\end{equation*}\begin{equation*}
\begin{split}
\norm{\Psi(u)-\Psi(v) }_{L^2L^\infty}\le & C (\norm{u}_{Y^\alpha}+\norm{v}_{Y^\alpha})\norm{u-v}_{L^\infty L^2}\\
\le & \frac 12 \norm{u-v}_{L^\infty L^2}
\end{split}
\end{equation*}when $u,v\in B_2\cap B_4$ and $T$ small enough.

The above three arguments shows $\Psi$ is a contraction in $B_2\cap B_3\cap B_4\cap B_5$ for small $T$. Then $v$ obtained in \ref{L^3 Mild Solution} lays in $L^sL^q$ locally. moreover, it lays in $L^\infty L^2\cap L^2\dot H$ if $v_0\in L^2_\sigma\cap L^3_\sigma$, which implies a strong solution.

Appendix

For $3\le p\le r\le q$, we have$Y^{p,q}\hookrightarrow Y^r, Y^{p,q}_0\hookrightarrow Y^r_0.$

By the index relation, we can always take $\theta \in [0,1]$ such that $\frac 1r=\frac \theta p+\frac {(1-\theta)}q$, then it comes
\begin{equation}
\begin{split}
\norm{v}_{Z^r}(t)=& t^{\frac 12-\frac 3{2r}} \norm{v(t)}_r
\le t^{\frac \theta 2-\frac{3\theta}{2p} } \cdot t^{\frac{1-\theta}{2}-\frac{1-\theta}{2q} }\norm{v(t)}_p^\theta\norm{v(t)}_q^{1-\theta}\\
=& \br{t^{\frac 12-\frac{3}{2p} }\norm{v(t)}_p }^\theta\cdot \br{t^{\frac 12-\frac{3}{2q} }\norm{v(t)}_q }^{1-\theta }\\
= & \norm{v }_{Z^p}^{\theta}(t) \norm{v }_{Z^q}^{1-\theta}(t).
\end{split}
\end{equation}Consequently, it comes that $\norm{v}_{Y^r}\le \norm{v}_{Y^p}^\theta\norm{v}_{Y^q}^{1-\theta}\Longrightarrow Y^{p,q}\hookrightarrow Y^r$. Apply the asymptomatic behavior we also get $Y^{p,q}_0\hookrightarrow Y^r_0$.

Suppose $\frac {1}{\gamma}=\frac 1 \alpha+\frac 1\beta, \alpha,\beta\ge 3$. If $u\in Y^\alpha, v\in Y^\beta/ Y^\beta_0$, then $B(u,v)\in Y^\beta/Y^\beta_0$.

The estimate below apply inequality \eqref{Rearrangement}:
\begin{equation*}
\begin{split}
\norm{B(u,v)(t) }_\beta=& \norm{\int_0^t e^{-(t-\tau)A}P\nabla (u\otimes v )(\tau )d\tau }_\beta\\
\le & \int_0^t \norm{e^{-(t-\tau )A}P\nabla }_{\gamma \to \beta }\norm{u(\tau ) }_\alpha\norm{v(\tau)}_\beta d\tau\\
\le & \int_0^t (t-\tau )^{-\frac 12-\frac 3{2\alpha} }\norm{u(\tau)}_\alpha\norm{v(\tau) }_\beta d\tau\\
= & \int_0^t (t-\tau)^{-\frac 12-\frac 3{2\alpha}}\tau^{-(1-\frac{3}{2\alpha}-\frac 3{2\beta} )}\cdot \br{\tau^{\frac 12-\frac 3{2\alpha} }\norm{u}_\alpha }\br{\tau^{\frac 12-\frac 3{2\beta} }\norm{v}_\beta} d\tau\\
\le & C t^{-\frac 12+ \frac 3{2\beta}}\norm{u}_{Z^\alpha}(t)\norm{v}_{Z^\beta}(t).
\end{split}
\end{equation*}Consequently, we see $\norm{B(u,v)}_{Z^\beta}(t)\le C\norm{u }_{Z^\alpha}(t)\norm{v }_{Z^\beta}(t) $. The claim follows immediately.

$\int_0^t (t-\tau)^r\tau^sd\tau\le C t^{r+s+1 }$ if $r+s>-1$.

\begin{equation*}
\begin{split}
\int_0^t(t-\tau )^r \tau^sd\tau\le & t^{r+s+1 } \int_0^t \br{1-\frac{\tau }{t} }^r\br{\frac \tau t }^s d\frac{\tau}{t} \\
\le & t^{r+s+1}\int_0^1 \br{1-\theta }^r\theta^s d\theta\\
\le & t^{r+s+1} \int_0^1 \theta^{r+s}d\theta\le \frac {t^{r+s+1} }{r+s+1}.
\end{split}
\end{equation*}

[Hardy-Littlewood-Sobolev Inequality]
For $p>1,\alpha=n(1-\frac{1}{p}+\frac 1q )$, we have
\begin{equation*}
\norm{\frac{1}{x^{\alpha }}* f }_q\lesssim \norm{f}_p,\forall f\in L^p.
\end{equation*}

Suppose $v\in L^3$, then $\norm{e^{-tA}Pv}_{L^sL^q}\le C\norm{v}_{3}$ for $\frac{2}{s}+\frac 3q=1, s\in (3,\infty]$.

The case $s=\infty$ is directly from estimate. For others, we consider following mapping:
$$T: L^r(\Omega; \R^3)\to L^s_w(0,\infty), v\mapsto \norm{e^{-tA}Pv }_{L^q}. $$Notice $\norm{e^{-tA}Pv }_q\le Ct^{-\frac 1s}\norm{v}_3$, we can show that $T$ is weak-$(3,s)$ for $s\in [3,\infty)$ as(($\norm{t^{-\frac 1s } }_{L^s_w}=\sup_{\lambda}\{\lambda |\{|t^{-\frac 1s} |> \lambda \} |^\frac 1s \}$$=\sup_\lambda \{\lambda \cdot (2\lambda)^{-s\cdot \frac 1s } \}=\frac 12. $))
\begin{equation*}
\begin{split}
\norm{Tv}_{L^s_w}\le C\norm{ t^{-\frac 1s} }_{L^s_w} \norm{v}_3\le C\norm{v}_3.
\end{split}
\end{equation*}Then Marcinkiewicz interpolation tells $T$ is strong $(3,s)$ for any $s\in (3,\infty)$.

$v_1(t)=e^{-tA}v_0 $ is a classic solution of Stokes system
\begin{equation*}
\begin{cases}
\partial_t v-\Delta v+\nabla p=0,\\
\nabla\cdot v=0,v(0)=v_0.
\end{cases}
\end{equation*}Then it satisfies following energy identity:
\begin{equation}
\label{Stokes Identity} \tag{6}
\norm{v_1(t)}_2^2+2\int_0^t \norm{\nabla v_1 }_2^2=\norm{v_0}_2.
\end{equation}

Stokes Operator

For $L^q(\Omega)$ where $q\in (1,\infty)$ and $\Omega\subset \R^n$ is $C^2$/half/whole space, we have Holmholtz decomposition $L^q =L^p_\sigma\oplus G_q$, where((See Galdi's III-1,2. Indeed, the second holds only for $u\in C^1$.))
\begin{equation*}
\begin{split}
G_q(\Omega)\coloneqq & \{u\in L^q| \exists p\in W^{1,q}_{loc}, \nabla p=u \},\\
L^q_\sigma(\Omega)\coloneqq \ol{\Di_\sigma}^{q}&= \{u\in L^q| \int u\cdot \varphi=0, \forall \varphi\in G_{q'} \}\\
&= \{u\in L^q| \nabla \cdot u=0, u\cdot n_{|\partial \Omega}=0 \}.
\end{split}
\end{equation*}We define the Holmholtz projection as $P_q: L^{q}\mapsto L^q_{\sigma}$, and furthermore
$$A_q\coloneqq -P_q\Delta: W^{1,q}_{0,\sigma}\cap W^{2,q}\to L^q_\sigma, $$here $W^{1,q}_{0,\sigma}=\ol{\Di_\sigma}^{1,q}=\{u\in W^{1,q}| \nabla \cdot u=0,$$ u_{|\partial \Omega}=0 \}$. $A_q$ is a densely defined closed operator in $L^q_\sigma$ and self-adjoint.((See https://www.math.colostate.edu/~pauld/M646/StokesOperator.pdf. )) Now we attempt to find the resolvent estimate of $-A_q$, before this we introduce the fan-shaped area: ((Especially, $\sum_{\pi ,0}=\C\setminus (-\infty, 0]$.))
$$\Sigma_{\theta, \epsilon}=\{\lambda \in \C| |\arg \lambda|\sl \theta, |\lambda |>\epsilon \} ,$$especially, we denote $\Sigma_{\theta,0}=\Sigma_{\theta}$.

For any $q\in (1,\infty),\theta\in (\frac \pi 2,\pi ),\lambda\in \Sigma_{\theta,0}$, there exists $C_{\theta, q}$ such that
\begin{equation}
\label{Resolvent Estimate} \tag{Resolvent Estimate}
\norm{(\lambda +A_q )^{-1} f}_{q}\le \frac{C_{\theta, q }}{\lambda}\norm{f}_q,\forall f\in L^{q}_\sigma.
\end{equation}

This problem can be reduced to estimate of the solution of following Stokes equations:
\begin{equation}
\label{Stokes} \tag{Stokes}
\begin{cases}
(\lambda-\Delta) u+\nabla p=f,\\
\nabla \cdot u=0,\\
u_{|\partial \Omega}=0.
\end{cases}
\end{equation}Consider the above function in $L^q$ sense, the last two conditions implies $u\in W^{1,q}_{0,\sigma}=D(A_q)$, then we can function $P_q$ on the both side of first equation to get((Note $D(A_q), L^q_\sigma$ both implies $\nabla \cdot v=0$, thus $P_q$ functions null.))
\begin{equation*}
(\lambda +A_q) u=f\Longrightarrow u=(\lambda+A_q )^{-1}f.
\end{equation*}Then the resolvent estimate of $-A_q$ is transformed to $(p,p)$ estimate of \eqref{Stokes}, which is a elliptic system. Thus we are able to reach it by Potential method: recall that for Laplace equation $-\Delta u=f$, we take Fourier transform on both side to get:
\begin{equation*}
\xi^2 \hat u=\hat f\Longrightarrow u= \Fi^{-1}\br{\frac{1}{\xi^2}}* f=\Gamma *f.
\end{equation*}and then use this representation to get the estimate. The similar fact holds for \eqref{Stokes},((See Giga.)) where $u=K*f$ and
\begin{equation*}
\hat K= \frac{\delta_{ij}-\frac{\xi_i\xi_j}{|\xi|^2} }{\lambda+|\xi|^2}.
\end{equation*}And we have

Let $p\in (1,\infty)$. Then for any $\theta\sl \pi $ there exists $C_{\theta,q}$ such that
\begin{equation*}
\norm{u}_q\le \frac{C_{\theta, q }}{|\lambda|} \norm{f}_q,\forall f\in L^q.
\end{equation*}

The above estimate implies \eqref{Resolvent Estimate} directly.

Before this, we introduce the semigroup estimate for Stokes operator first:

Suppose $1\sl p\le q$, then we have
\begin{equation*}
\label{SE1} \tag{SE1}
\begin{split}
\norm{e^{-tA} P v }_q\le & Ct^{-\frac{3}{2}(\frac 1p-\frac 1q ) } \norm{v}_p;\\
\norm{e^{-tA} P\nabla v }_q\le & Ct^{ -\frac 12-\frac{3}{2}(\frac 1p-\frac 1q ) } \norm{v}_p;\\
\norm{\nabla e^{-tA} P v }_q\le & Ct^{-\frac 12 -\frac{3}{2}(\frac 1p-\frac 1q ) } \norm{v}_p;\\
\end{split}
\end{equation*}

We prove it later.

If $p\sl q, v\in L^p$, then we have
\begin{equation*}
t^{\frac 32( \frac 1p-\frac 1q) }\norm{e^{-tA}P v }_q\to
\begin{cases}
\norm{Pv }_p, & \text{as } p=q,t\to 0;\\
0, & \text{as } p\sl q, t\to 0;\\
0, & \text{as } p=q, t\to \infty ;\\
0, & \text{as } p\sl q, t\to \infty .
\end{cases}
\end{equation*}

Clearly $\norm{e^{-tA}P v }_p \xrightarrow{t\to 0} \norm{Pv}_p$ holds since $e^{-tA} $ is a strong continuous semigroup. For the remained case, we consider $u\in L^p\cap L^r$, where $r\in (1,q]$. Then
\begin{equation*}
\begin{split}
t^{\frac 32(\frac 1p-\frac 1q ) }\norm{e^{-tA} Pv }_q\le& t^{\frac 32(\frac 1p-\frac 1q ) }\norm{e^{-tA} P(v-u) }_q+t^{\frac 32(\frac 1p-\frac 1q ) }\norm{e^{-tA} Pu }_q\\
\le &C\norm{v-u }_p+ t^{\frac 32(\frac 1p-\frac 1q ) }\cdot C t^{-\frac 32(\frac 1r-\frac 1q ) }\norm{u}_r\\
\le & C\norm{v-u}_p+t^{\frac 32(\frac 1p-\frac 1r ) }\norm{u}_r.
\end{split}
\end{equation*}Since $L^p\cap L^r$ is dense in $L^p$, we can always set $\norm{v-u}_q\to 0 $. As for the latter,

when $t\to 0, p\sl q$: we choose $r\in (p,q]$, then $\frac 32(\frac 1p-\frac 1r )>0\Longrightarrow t^{\frac 32(\frac 1p-\frac 1r ) }$$\norm{u}_r\xrightarrow{t\to 0 } 0; $
when $t\to \infty, p\le q$: we choose $r\in (1,p)$, then $\frac 32(\frac 1p-\frac 1r )\sl 0\Longrightarrow t^{\frac 32(\frac 1p-\frac 1r ) }$$\norm{u}_r\xrightarrow{t\to \infty } 0.$

In conclusion, $t^{\frac 32( \frac 1p-\frac 1q) }\norm{e^{-tA}P v }_q\to 0 $ holds for the remained three cases.

Analytic Semigroup

Suppose $\{T_{z}\}\subset L(X),z\in \Sigma_{\theta}\cup\{0\}, $, we say $T_z$ is a analytic semigroup of angle $\theta$ if

$T_z$ satisfies semigroup properties: $T_0=0, T_{z_1}T_{z_2}=T_{z_1+z_2}$;
$T_z$ is analytic in $z$ in the sense of uniform operator topology.

Suppose $A$ is densely defined and closed on $X$, then $A$ generates an analytic semigroup of angle $\theta(\theta\in (0,\frac \pi 2 ))$ iff

$\rho(A)\supset \Sigma_{\frac \pi 2+\theta }$;
$\norm{R(\lambda, A ) }\le \dfrac{C }{|\lambda | },\forall \lambda \in \Sigma_{\frac \pi 2+\theta } $.

In this case, the semigroup is figured as
\begin{equation*}
e^{zA}=\frac 1{2\pi i }\int_\gamma e^{\lambda z }R(\lambda ,A )d\lambda, z\in \Sigma_{\theta}.
\end{equation*}$\gamma$ is an admissible curve defined by: $$\gamma= - e^{-i(\frac \pi 2+ \eta)}[r, \infty)+ r e^{i(- \frac \pi 2-\eta, \frac \pi 2 +\eta )} +e^{i(\frac \pi 2+ \eta)}[r, \infty), \eta\in (\frac {|\arg z|+\theta}{2}, \theta ), r\sl |z|.$$((Details see this lecture and Wikipedia.))

Consequently, the resolvent estimate implies $A_q$ generates a analytic semigroup $e^{zA_q}$ of angle $\theta$, for any $\theta \in (0, \frac \pi 2)$.