We investigate into the local well-posedness of incompressible fluid equations. $\require{extpfeil}\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ First we consider the incompressible Navier-Stokes system in $\Omega\subset \R^n:$
\begin{equation}
\label{NS}
\begin{split}
&u_t+u\cdot \nabla u+\nabla p = \nu \Delta u+f,\\
&\nabla \cdot u= 0,\\
\end{split}
\tag{NS}
\end{equation}Here $u $ is the velocity, $p$ is the pressure and $f$ is the exterior force. The constant $\nu>0$ denote the viscosity.
The incompressible Euler equations arise as the varnishing viscosity limit $\nu\to 0$:
\begin{equation}
\label{Eul}
\begin{split}
&u_t+u\cdot \nabla u+\nabla p = f,\\
&\nabla \cdot u= 0,\\
\end{split}
\tag{Eul}
\end{equation}The method we adopt in this lecture is different from mild solution, since the dissipation does not occur in the Euler flow.
Smooth Approximation
Now we will prove the local well-posedness of smooth solution of Navier-Stokes/Euler equation by the regular approximation and Picard contraction. Consider the mollified equations:
\begin{equation}
\label{mNS}
\begin{split}
& u_t^\epsilon + M_\epsilon (M_\epsilon u^\epsilon \cdot\nabla M_\epsilon u^\epsilon )=-\nabla p^\epsilon +\nu M_\epsilon (M_\epsilon \Delta u^\epsilon )\\
& \nabla \cdot u^\epsilon =0.
\end{split}
\tag{mNS}
\end{equation}Here $M_\epsilon$ is the standard mollification operator (see Appendix Mollification). Applying the Leray projection $P$ to the both side, we can get that
\begin{equation}
\label{trans}
u_t^\epsilon=\nu M_\epsilon (M_\epsilon \Delta u^\epsilon )-P M_\epsilon (M_\epsilon u^\epsilon \cdot\nabla M_\epsilon u^\epsilon )\eqqcolon F_\epsilon^1(u^\epsilon )-F_\epsilon^2(u^\epsilon )\eqqcolon F_\epsilon (u^\epsilon ).
\end{equation}The first claim is the global existence for regularized system:
- For $u\in H^k_\sigma$, it is clear that $\nabla \cdot F_\epsilon (u)=0$ since $P$ maps into divergence-free vector fields. Moreover, we can compute by Proposition 5 that
The detailed calculation for the last term is that: $$\norm{M_\epsilon u\otimes M_\epsilon u }_{H^k}\le C\norm{M_\epsilon u }_{L^\infty} \norm{u }_{H^k}\le C\epsilon^{-d/2} \norm{u }_{L^2}\norm{u}_{H^k}\le C C\epsilon^{-d/2}\norm{u}_{H^k}^2. $$\begin{equation*}
\begin{split}
\norm{F_\epsilon (u ) }_{H^k}=& \nu\norm{M_\epsilon^2 \Delta u }_{H^k}+\norm{PM_\epsilon \br{\nabla \cdot (M_\epsilon u \otimes M_\epsilon u ) } }_{H^k}\\
\le & \nu \norm{M_\epsilon^2 u }_{H^{k+2}}+C\epsilon^{-1}\norm{M_\epsilon u\otimes M_\epsilon u }_{H^k}\\
\le & C\br{\nu \epsilon^{-2}\norm{u}_{H^k}+\epsilon^{-\frac d2-1}\norm{u}_{H^k}^2 }.
\end{split}
\end{equation*}Then it implies $F_\epsilon (u)\in H^k_\sigma$ immediately. - We show that $F_\epsilon$ is locally continuous in any open ball
\begin{equation*}
B_R\coloneqq \set{u\in H^k_\sigma| \norm{u}_{H^k}\le R }.
\end{equation*} Given $u_1,u_2\in B_R$, a direct result is $\norm{F^1_\epsilon (u_1)-F^1_\epsilon(u_2) }_{H^k}\le C\nu\epsilon^{-2}\norm{u_1-u_2}_{H^k} $. As for the second part, we have that
\begin{equation*}
\begin{split}
\norm{F^2_\epsilon(u_1 )-F^2_\epsilon (u_2) }_{H^k}\le & C\norm{M_\epsilon (u_1-u_2)\cdot \nabla M_\epsilon u_1 }_{H^k}+C\norm{M_\epsilon u_2\cdot \nabla M_\epsilon (u_1-u_2) }_{H^k}\\
\le & C (\norm{\nabla M_\epsilon u_1 }_{L^\infty} \norm{u_1-u_2}_{H^k}+\norm{\nabla M_\epsilon u_1 }_{H^k}\norm{M_\epsilon (u_1-u_2)}_{L^\infty}\\
&+\norm{M_\epsilon u_2 }_{L^\infty} \norm{\nabla M_\epsilon (u_2-u_1) }_{H^k}+\norm{u_2}_{H^k}\norm{\nabla M_\epsilon (u_1-u_2 ) }_{L^\infty})\\
\le & C \epsilon^{-(\frac d2 +1 )}\norm{u_1}_{L^2}\norm{u_1-u_2 }_{H^k }+C \epsilon^{-(1+k+\frac d2 ) }\norm{u_1}_{L^2}\norm{u_1-u_2 }_{L^2 }\\
&+C\epsilon^{-(\frac d2+1 )}\norm{ u_2}_{L^2} \norm{u_1-u_2 }_{H^k }+ C \epsilon^{-(1+k+\frac d2 ) }\norm{u_2}_{L^2}\norm{u_1-u_2 }_{L^2 } \\
\le & C\epsilon^{-(\frac d2+1+k )}\br{\norm{u_1}_{L^2}+\norm{u_2}_{L^2} }\norm{u_1-u_2 }_{H^k}.
\end{split}
\end{equation*} The final result is $\norm{F_\epsilon (u_1)-F_\epsilon (u_2)}_{H^k}\le C(\epsilon,d,k,\norm{u_i}_{L^2} )\norm{u_1-u_2}_{H^k}$, so $F_\epsilon$ is locally Lipschitz in $H^k_\sigma$.
In conclusion, the Picard theorem grants a unique solution $u^\epsilon\in C^1\br{[0,T_\epsilon);H^k_\sigma}$ for some $T_\epsilon>0$. Then we attempt to extend the existence interval to infinity. It is enough to show that for any solution $u^\epsilon $, $\norm{u^\epsilon }_{H^k}$ is bounded by $\norm{u_0 }_{H^k}$. Indeed, we can show
$$\sup_{0\le t\le T} \norm{u^\epsilon (t) }_{L^2}\le \norm{u_0 }_{L^2} $$ for any $T\in [0,T_\epsilon)$, by multiplying $u^\epsilon $ to the both side of \eqref{mNS}. Furthermore, we set $u_1=u^\epsilon, u_2=0$, then by the above estimates
\begin{equation*}
\frac{d}{dt} \norm{u^\epsilon (t) }_{H^k}=\norm{F_\epsilon^2 (u^\epsilon ) }_{H^k} \le C\br{\epsilon,d, k, \norm{u^\epsilon }_{L^2} } \norm{u^\epsilon }_{H^k}\le C\br{\epsilon,d, k,\norm{u_0 }_{L^2} } \norm{u^\epsilon }_{H^k}.
\end{equation*} Grönwall's inequality implies $\norm{u^\epsilon (t) } _{H^k}\le \norm{u_0 }_{H^k}e^{Ct} $ immediately.
Besides, we prove the following high-order energy estimate for the regularized solution $u^\epsilon$, which will be applied in further discussion.
\begin{equation*}
\frac 12 \frac d{dt}\norm{u^\epsilon }_{H^k}^2+\nu \norm{ \nabla M_\epsilon u^\epsilon }_{H^k}^2\le C_k \norm{\nabla M_\epsilon u^\epsilon }_{L^\infty }\norm{u^\epsilon }_{H^k }^2.
\end{equation*}
\begin{equation*}
\frac d{dt} \norm{u^\epsilon }_{H^k }\le C_k \norm{ \nabla M_\epsilon u^\epsilon }_{L^\infty } \norm{u^\epsilon }_{H^k}\le C_k \norm{u^\epsilon }_{H^k}^2.
\end{equation*} And hence for any $\epsilon $,
\begin{equation}
\label{appr bound}
\sup_{0\le t\le T } \norm{u^\epsilon }_{H^k}\le \frac{\norm{u_0 }_{H^k} }{ 1-C_k T\norm{u_0 }_{H^k } }.
\end{equation} It is notable that the above estimates is independent of viscosity constant $\nu$.
\begin{equation*}
\begin{split}
\frac 12 \frac{d}{dt}\norm{\partial^\alpha u^\epsilon }_{L^2 }^2 =&\nu \br{ M_\epsilon^2 \Delta \partial^\alpha u^\epsilon, \partial^\alpha u^\epsilon }- \br{\partial^\alpha PM_\epsilon (M_\epsilon u^\epsilon \cdot \nabla M_\epsilon u^\epsilon ), \partial^\alpha u^\epsilon } \\
=&-\nu \norm{\nabla M_\epsilon \partial^\alpha u^\epsilon }_{L^2}^2-\br{ PM_\epsilon (M_\epsilon u^\epsilon \cdot \nabla \partial^\alpha M_\epsilon u^\epsilon ), \partial^\alpha u^\epsilon }\\
&+ \br{\partial^\alpha PM_\epsilon (M_\epsilon u^\epsilon \cdot \nabla M_\epsilon u^\epsilon )-PM_\epsilon (M_\epsilon u^\epsilon \cdot \nabla \partial^\alpha M_\epsilon u^\epsilon) , \partial^\alpha u^\epsilon }.
\end{split}
\end{equation*} Here the divergence-free condition $\nabla \cdot u^\epsilon=0 $ implies:
\begin{equation*}
\br{ PM_\epsilon (M_\epsilon u^\epsilon \cdot \nabla \partial^\alpha M_\epsilon u^\epsilon ), \partial^\alpha u^\epsilon }=\br{ M_\epsilon u^\epsilon , \nabla \br{\partial^\alpha M_\epsilon u^\epsilon }^2 }=0.
\end{equation*} Summing over $|\alpha |\le k$, by Lemma 9 we have
\begin{equation*}
\begin{split}
&\frac 12\frac {d}{dt}\norm{ u^\epsilon }_{H^k}^2 +\nu \norm{\nabla M_\epsilon u^\epsilon }_{H^k}^2\\
& \le \norm{u^\epsilon }_{H^k} \sum_{|\alpha |\le k}\norm{\partial^\alpha (M_\epsilon u^\epsilon \cdot \nabla M_\epsilon u^\epsilon )- (M_\epsilon u^\epsilon \cdot \nabla \partial^\alpha M_\epsilon u^\epsilon) }_{L^2} \\
& \le C_k\norm{u^\epsilon }_{H^k}\br{ \norm{\nabla M_\epsilon u^\epsilon }_{L^\infty }\norm{ \partial^{m-1} M_\epsilon u^\epsilon }_{L^2}+\norm{\nabla M_\epsilon u^\epsilon }_{L^\infty}\norm{\partial^m M_\epsilon u^\epsilon }_{L^2} }\\
& \le C_k\norm{ \nabla M_\epsilon u^\epsilon }_{L^\infty } \norm{u^\epsilon }_{H^k}^2.
\end{split}
\end{equation*} Finally we get the high-order energy estimate.
With the above regularized result, we now show the local existence of the Euler/Navier-Stokes equations \eqref{NS} provided the initial data with enough regularity.
$$u^\nu\in C\br{[0,T];C^2 (\R^d ) }\cap C^1([0,T];C(\R^d ) ) $$ to Euler/Naiver-Stokes equations \eqref{NS}. Moreover, we have uniform energy estimate
$$\sup_{0\le t\le T}\norm{u^\nu }_{H^k }\le \frac{\norm{u_0 }_{H^k} }{1-C_kT \norm{u_0 }_{H^k} } .$$
Indeed, the significant observation is that $u^\nu$ is the limit of $u^\epsilon$ of corresponding regularized system (in subsequence). The strategy for the proof of Theorem 3 is to establish the energy estimate for $u^\epsilon$, then we show a contraction in $L^2-$norm and apply an interpolation inequality to prove convergence as $\epsilon\to 0 $.
\begin{equation*}
\begin{split}
\norm{u_t^\epsilon }_{H^{k-2} }=& \norm{F_\epsilon (u^\epsilon ) }_{H^{k-2} }\\
\le & \nu \norm{\Delta M_\epsilon u^\epsilon }_{H^{k-2} }+\norm{\nabla \cdot (M_\epsilon u^\epsilon \otimes M_\epsilon u^\epsilon ) }_{H^{k-2}}\\
\le & \nu \norm{u^\epsilon }_{H^k}+\norm{M_\epsilon u^\epsilon }_{L^\infty }\norm{M_\epsilon u^\epsilon }_{H^{k-1} }\\
\le & \nu \norm{u^\epsilon }_{H^k }+C\norm{u^\epsilon }_{H^k }^2.
\end{split}
\end{equation*} Consequently, since $H^{k}\overset{K}{\hookrightarrow} H^{k-2}\hookrightarrow H^{k-2}$ in bounded domain, we see $\set{ u^\epsilon}$ is precompact in $C_t (H^{k-2}_{loc} )$, i.e., there exists $u^\epsilon \xrightarrow{C_t H^{k-2}_{loc} } u^\nu$ in subsequence sense. Now we show $u^{\nu}$ is indeed the solution of following equation:
\begin{equation*}
u^\nu_t = F_\nu (u^\nu ), F_\nu (u)=\nu \Delta u+P (u\cdot \nabla u ).
\end{equation*} Indeed, we can show by term splitting and mollification properties that2
\begin{equation*}
\begin{split}
\frac{d}{dt} \norm{u^\epsilon -u^\nu }_{L^2}\le C \br{\epsilon +\norm{u^\epsilon -u^\nu }_{L^2} },
\end{split}
\end{equation*} and thus
\begin{equation*}
\sup_{0\le t\le T} \norm{u^\epsilon -u^\nu }_{L^2 }\le e^{CT}\br{\epsilon +\norm{u_0^\epsilon-u_0 }_{L^2} }-\epsilon \le C(T )\epsilon .
\end{equation*} The final step is to recover the pressure.
$$\lim_{t\to T^* } \norm{u^\nu (t) }_{H^k }=\infty. $$
Appendix
We introduce the standard bump function $\rho:\R^d\mapsto \R$ as
\begin{equation*}
\rho(x)=\begin{cases}
\frac 1c e^{-\frac {1}{1-|x|^2} } & |x|\le 1 \\
0, |x|>1,
\end{cases}
\end{equation*}where $c=\int_{B_1}e^{-\frac{1}{1-|x|^2} }dx $. It can be check that $\rho\in \Di(\R^n )$ and $\int \rho dx=1$. Let $\rho_\epsilon(x)\coloneqq \epsilon^{-d}\rho(x/\epsilon)$, we define the $\epsilon-$mollification of $u\in L^1_{loc}(\Omega )$ as
\begin{equation*}
M_\epsilon u=\rho_\epsilon * u=\epsilon^{-d} \int \rho\br{\frac{\cdot -y}{\epsilon} } u(y)dy.
\end{equation*}Some important properties about mollification operator is listed as following.
- $M_\epsilon u\in \Ei(\R^n)$ and $\supp M_\epsilon u\in \Omega_\epsilon, \partial^\alpha M_\epsilon u=u* \partial^\alpha \rho_\epsilon$;
- If $u\in L^p_{loc}(\Omega )$, then $M_\epsilon u\to u$ in $L^p_{loc}(\Omega )$;
- If $u\in C(\Omega)$, then $M_\epsilon u\to u$ in $C_{loc}(\Omega)$;
- If $u\in H^s(\R^d),s\in \R$, then $M_\epsilon u\to u$ in $H^s$. Particularly, $M_\epsilon u\to u$ in $H^{s-1}$ at rate $O(\epsilon )$, i.e.
\begin{equation*}
\norm{M_\epsilon u-u }_{H^{s-1}}\le C\epsilon \norm{u}_{H^s};
\end{equation*} - If $u\in H^{k}(\R^d), k\in \N$, then for $\epsilon>0$,
\begin{equation*}
\begin{split}
\norm{M_\epsilon u }_{H^{k+m}}\le & \frac{C_{m,k}}{\epsilon^m}\norm{u }_{H^k};\\
\norm{M_\epsilon \partial^m u }_{L^\infty}\le & \frac{C_k}{\epsilon^{d/2+k} }\norm{u}_{L^2}.
\end{split}
\end{equation*} - $M_\epsilon, P$ and $\partial^\alpha$ are commutative with each other.
\begin{equation*}
\begin{split}
\norm{M_\epsilon u-u }_{H^{s-1}}^2=&\int ( 1+|\xi |^2 )^{s-1}(\hat \rho_\epsilon (\xi )-1)^2 \hat u(\xi)^2 d\xi\\
=& \int \frac{(\hat \rho_\epsilon (\xi )-1)^2}{(1+ |\xi|^2 ) } ( 1+|\xi |^2 )^{s} \hat u(\xi)^2 d\xi \\
\le & C\norm{\frac{(\hat \rho_\epsilon (\xi )-1)^2}{(1+ |\xi|^2 ) } }_{L^\infty_\xi }\norm{u }_{H^s}^2.
\end{split}
\end{equation*} Then our work is to estimate the rate of $\norm{\frac{(\hat \rho_\epsilon (\xi )-1)^2}{(1+ |\xi|^2 ) } }_{L^\infty_\xi }$.
\begin{equation}
\label{Compactness}
\norm{u }_Y\le \eta \norm{v }_{X }+C_\eta \norm{v }_Z.
\end{equation} Indeed, if there is a $\eta_0>0$ and sequence $\set{u_n } $ such that
\begin{equation*}
\norm{u_n }_Y>\eta_0 \norm{u_n }_X+n\norm{u_n}_Z,\forall n\in \N.
\end{equation*} Let $w_n=u_n/\norm{u_n}_X$, then the compact embedding $X\overset{K}{\hookrightarrow} Y$ implies $\set{w_n }$ is precompact in $Y$, i.e., there is a subsequence $w_{n_k}\to w_0\not=0$ in $Y$. This implies also $\norm{w_{n_k}}_Y\le C$ and then $w_{n_k}\to 0$ in $Z$, contradiction to $w_0\not=0$.
Now we show the continuity in time: we assume $\norm{u_n (t) }_{X}\le C_1,\forall n\in \N, t\in [0,T] $. Then for any $\epsilon>0,t_0\in [0,T] $, there exists $\delta (\epsilon,t_0 )>0 $ such that for any $|t-t_0|\sl\delta $, we have
\begin{equation*}
\norm{u_n(t)-u_n(t_0) }_Z\le \epsilon/(2 C_\eta),\forall n\in \N.
\end{equation*} Let $\eta=\epsilon/ (4C_1)$, then \eqref{Compactness} implies that
\begin{equation*}
\begin{split}
\norm{u_n(t)-u_n(t_0) }_Y\le & \frac{\epsilon }{4C_1}\norm{u_n(t)-u_n(t_0) }_X+C_{\frac \epsilon {4C_1} } \norm{u_n (t)-u_n(t_0) }_Z\\
\le & \frac{\epsilon }{4C_1}\cdot 2C_1+ C_{\frac \epsilon{4C_1} }\cdot \frac{\epsilon }{2C_{\frac \epsilon{4C_1}} }=\epsilon.
\end{split}
\end{equation*} Consequently, we see $u_n\in C([0,T];Y )$ for any $n\in \N$. Moreover, since $X\hookrightarrow Z$, then $\set{u_n} $ is uniformly bounded in $Z$. Together with equi-continuity on $C([0,T];Z)$, Arzela-Ascoli lemma implies $\set{u_n}$ is precompact in $C([0,T];Z)$, i,e, $\epsilon>0,\exists N>0,$ s.t.
\begin{equation*}
\sup_{t\in [0,T]} \norm{u_{n_k}(t) -u_{n_l} (t) }_Z\le \frac{\epsilon}{C_{\epsilon/4C_1 } },\forall k,l>N.
\end{equation*} Apply the inequality \eqref{Compactness} with $\eta=\epsilon/(4C_1)$ again, we have
\begin{equation*}
\begin{split}
\norm{u_{n_k} (t)-u_{n_l} (t) }_{Y}\le \frac{\epsilon }{4C}\norm{u_{n_k}(t)-u_{n_l}(t) }_X+C_{\frac \epsilon {4C} } \norm{u_{n_k} (t)-u_{n_l}(t) }_Z\le \epsilon.
\end{split}
\end{equation*} Take supremum for $t$ at both hand, we derive the precompactness in $C([0,T];Y)$.
\begin{equation*}
\frac 1p=\frac jd+\br{\frac 1r-\frac kd }\alpha+\frac{1-\alpha}{q}, \frac jk\le \alpha \le 1.
\end{equation*} then there exists a constant $C$ depending on $k,d,j,q,r,\alpha$ such that
\begin{equation*}
\norm{\partial^j u}_{L^p}\le C\norm{\partial^k u }_{L^r}^\alpha \norm{u }_{L^q}^{1-\alpha}.
\end{equation*}
\begin{equation*}
\norm{\partial^j u }_{L^p}\le C_1\norm{\partial^k u }_{L^r}^\alpha\norm{u}_{L^q}^{1-\alpha}+C_2\norm{u}_{L^s}
\end{equation*} for arbitrary $s\ge 1$, where the constants $C_1,C_2$ depend on $\Omega$ and $s$ in addition to the other parameters.
\begin{equation*}
\norm{uv}_{H^s}\le C\norm{u}_{H^s}\norm{v}_{H^s}.
\end{equation*}
\begin{equation*}
\norm{\hat u}_{L^1}=\int \ip{\xi}^{-s}\ip{\xi}^s|\hat u(\xi) |d\xi\le \int \ip{\xi}^{-2s}d\xi \cdot \norm{u}_{H^s}\le C\norm{u}_{H^s}
\end{equation*} provided $s>d/2$. Consequently, we have
\begin{equation*}
\begin{split}
\norm{uv}_{H^s}=&\norm{ \ip{\xi}^s|\widehat{uv}(\xi) | }_{L^2_\xi}\le \norm{\int \ip{\xi}^s\abs{\hat u(\xi-\eta)\hat v(\eta)}d\eta }_{L^2_\xi}\\
\le &C\norm{\int \ip{\xi-\eta}^s \abs{\hat u(\xi-\eta)\hat v(\eta) } d\eta}_{L^2_\xi}+C\norm{\int \ip{\eta}^s \abs{\hat u(\xi-\eta)\hat v(\eta) } d\eta}_{L^2_\xi}\\
\le & C\norm{ \ip{\cdot }^s |\hat u| *|\hat v |(\xi)}_{L^2_\xi}+C\norm{|\hat u |*\ip{\cdot }^s|\hat v|(\xi) }_{L^2_\xi}\\
\le & C\norm{\ip{\cdot }^s|\hat u | }_{L^s}\norm{\hat v }_{L^1}+C\norm{\hat u}_{L^1}\norm{\ip{\cdot }^s|\hat v | }_{L^s}\\
\le & C\norm{u}_{H^s}\norm{\hat v}_{L^1}+\norm{v }_{H^s}\norm{\hat u}_{L^1}\le C\norm{u}_{H^s}\norm{v}_{H^s.}
\end{split}
\end{equation*} Here in the third step, we use the fact: For $r>0,$
\begin{equation*}
\begin{split}
(1+|\xi |^2 )^r\le & (1+2|\xi-\eta |^2+2|\eta|^2 )^r\\
\le & 2^r\br{1+|\xi-\eta |^2+1+|\eta|^2 }^r\\
\le & C\br{1+|\xi-\eta |^2 }^r +C(1+|\eta |^2 )^r
\end{split}
\end{equation*} where $C=\max\{2^r,2^{2r-1}\}$, so $\ip{\xi}^ s\le C\ip{\xi-\eta }^s+C\ip{\eta}^s$.
\begin{equation*}
\norm{uv}_{H^k}\le C\br{\norm{u}_{L^\infty}\norm{\partial^k v}_{L^2}+\norm{\partial^k u}_{L^2}\norm{v}_{L^\infty} },
\end{equation*} \begin{equation*}
\sum_{|\alpha |\le k } \norm{\partial^\alpha (uv )-u\partial^\alpha v }_{L^2}\le C\br{ \norm{ \nabla u}_{L^2 }\norm{\partial^{k-1}v }_{L^2}+\norm{\partial^k u}_{L^2}\norm{v}_{L^\infty} }.
\end{equation*}
\begin{equation*}
\norm{uv}_{H^k}\le C\br{\norm{u}_{L^\infty}\norm{v}_{H^s}+\norm{ u}_{H^s}\norm{v}_{L^\infty} }.
\end{equation*} See Tao's blog 254A, Notes 1, Appendix for details (proved via Littlewood-Paley theory).
\begin{equation*}
\norm{F(u_1)-F(u_2)}_X\le L\norm{u_1-u_2}_{X},\forall u_1,u_2\in U.
\end{equation*} Then for any $u_0\in O$, there exists a time $T$ such that the ODE
\begin{equation*}
\begin{cases}
\frac{d u}{dt}=F(u)\\
u|_{t=0}=u_0
\end{cases}
\end{equation*} has a unique (local) solution $u\in C^1\br{[0,T); O}$.
Notice that $\rho_\epsilon * u \cdot v=\int \rho_\epsilon (y-x)u(x)dy\ \cdot v(x)=\int \rho_\epsilon (y-x)v(x)dy\ \cdot u(x)=u \cdot \rho_\epsilon * v. $ ↩
You can see similar proof in Majda-Bertozzi-Ogawa Lemma 3.7. ↩