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.
\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.
$v\in C([0, T_1]; L^q_\sigma )$ for equation \eqref{Zero Force}, where
$$ T_1\sim \norm{v_0}_q^{-s }, \dfrac{2}{s}+\dfrac{3}{p}=1.$$
\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:
\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
\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 $\norm{e^{-tA}v_0 }_{Z^\alpha}\le (8C)^{-1},\forall t\in [0, T_2]$.
\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 .
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 $T$.- $\Psi$ stablizes $B_3$. Since $v_1\in {L^sL^q(\Omega_{\infty})}$$\Longrightarrow \norm{v_1}_{L^sL^q(\Omega_t ) }\to 0$ as $t\to 0$. Then
\begin{equation*}
\begin{split}
\norm{\Psi (v)}_{L^sL^q}\le & \norm{ e^{-tA}v_0 }_{L^sL^q}+C\norm{v}_{L^sL^q}^2\\
\le & \frac{1}{8C}+ \frac{1}{16C}\le \frac{1}{4C},
\end{split}
\end{equation*}if $T$ is small enough. - $\Psi$ contracts $B_3$:
\begin{equation*}
\begin{split}
\norm{\Psi(u)-\Psi(v)}_{L^sL^q}= & \norm{B(u,u-v)+B(u-v,v) }_{L^sL^q}\\
\le & C\br{\norm{u}_{L^sL^q}+\norm{u}_{L^sL^q}}\norm{u-v}_{L^sL^q}\\
\le & \frac{1}{2} \norm{u-v}_{L^sL^q}.
\end{split}
\end{equation*}
- $\Psi$ stablizes $B_3$. Since $v_1\in {L^sL^q(\Omega_{\infty})}$$\Longrightarrow \norm{v_1}_{L^sL^q(\Omega_t ) }\to 0$ as $t\to 0$. Then
- $L^2\dot H^1:$ 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
\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 $Y^{p,q}_0\hookrightarrow Y^r_0$.
\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.
\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*}
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*}
$$T: L^r(\Omega; \R^3)\to L^s_w(0,\infty), v\mapsto \norm{e^{-tA}Pv }_{L^q}. $$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 \}$$=\sup_\lambda \{\lambda \cdot (2\lambda)^{-s\cdot \frac 1s } \}=\frac 12. $))
\begin{equation*}
\begin{split}
\norm{Tv}_{L^s_w}\le C\norm{ t^{-\frac 1s} }_{L^s_w} \norm{v}_3\le C\norm{v}_3.
\end{split}
\end{equation*}Then Marcinkiewicz interpolation tells $T$ is strong $(3,s)$ for any $s\in (3,\infty)$.
\begin{equation*}
\begin{cases}
\partial_t v-\Delta v+\nabla p=0,\\
\nabla\cdot v=0,v(0)=v_0.
\end{cases}
\end{equation*}Then it satisfies following energy identity:
\begin{equation}
\label{Stokes Identity} \tag{6}
\norm{v_1(t)}_2^2+2\int_0^t \norm{\nabla v_1 }_2^2=\norm{v_0}_2.
\end{equation}
Stokes Operator
For $L^q(\Omega)$ where $q\in (1,\infty)$ and $\Omega\subset \R^n$ is $C^2$/half/whole space, we have Holmholtz decomposition $L^q =L^p_\sigma\oplus G_q$, where((See Galdi's III-1,2. Indeed, the second holds only for $u\in C^1$.))
\begin{equation*}
\begin{split}
G_q(\Omega)\coloneqq & \{u\in L^q| \exists p\in W^{1,q}_{loc}, \nabla p=u \},\\
L^q_\sigma(\Omega)\coloneqq \ol{\Di_\sigma}^{q}&= \{u\in L^q| \int u\cdot \varphi=0, \forall \varphi\in G_{q'} \}\\
&= \{u\in L^q| \nabla \cdot u=0, u\cdot n_{|\partial \Omega}=0 \}.
\end{split}
\end{equation*}We define the Holmholtz projection as $P_q: L^{q}\mapsto L^q_{\sigma}$, and furthermore
$$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 $\Sigma_{\theta,0}=\Sigma_{\theta}$.
\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
\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:
\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*}
\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*}
\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 $\theta \in (0, \frac \pi 2)$.
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