Category Archives: Navier-Stokes Eqns

Smoothness for small initial data

$\displaystyle \begin{cases}u_t+u\cdot\nabla u-\nu\Delta u+\nabla p=0\\div u=0\\u(x,0)=u_0\end{cases}\text{ on }\mathbb{R}^3$

Thm(Kato): If ${u_0\in \dot{H}^{1/2}}$ is small, then the solution with initial data ${u_0}$ is global smooth.

Proof: ${\Lambda u=\sqrt{-\Delta}u}$ is a well defined operator on ${\dot{H}^{1/2}}$. Multiplying ${\Lambda u}$ on both sides, we get

$\displaystyle \frac{1}{2}\frac{d}{dt}||u||_{\dot{H}^{1/2}}^2+||u||_{\dot{H}^{3/2}}^2\leq \int |u||\nabla u||\Lambda u|\leq ||u||_{L^{3}}||\nabla u||_{L^3}||\Lambda u||_{L^3}$

By the Sobolev embedding,

$\displaystyle ||u||_{L^3}\leq C||u||_{\dot{H}^{1/2}}$

So we can get

$\displaystyle ||u||_{L^{3}}||\nabla u||_{L^3}||\Lambda u||_{L^3}\leq C||u||_{\dot{H}^{1/2}}||u||_{\dot{H}^{3/2}}^2$

Let ${G=||u||_{\dot{H}^{3/2}}^2}$, ${N=||u||_{\dot{H}^{1/2}}^2}$,

$\displaystyle \frac{1}{2}\frac{d}{dt}N+G(1-C\sqrt{N})\leq 0$

If ${N(0)<1/C^2}$ is small enough, then ${N'(t)}$ will be negative thus ${N(t)<1/C^2}$ and small enough. Consequently, ${||u||_{L^3}\leq C||u||_{\dot{H}^{1/2}}}$ will also be small. While from the NS equation,

$\displaystyle \frac{1}{2}\frac{d}{dt}|\nabla u|_{L^2}^2+|\Delta u|_{L^2}^2\leq\int|u||\nabla u||\Delta u|\leq C||u||_{L^3}||\Delta u||_{L^2}^2$

When ${||u||_{L^3}}$ small enough, the right hand side can be absorded by left hand side. Thus ${|\nabla u|_{L^2}^2}$ will be bounded. And from the standard regularity theory, ${u}$ must be smooth. $\Box$

Remark: Follows from my note of lectures given by Peter Constantin.

Finite blow up of NS solution

$\displaystyle \begin{cases}u_t+u\cdot\nabla u-\nu\Delta u+\nabla p=0\\div u=0\\u(x,0)=u_0\end{cases}\text{ on }\mathbb{R}^3$

Thm: If NS 3-D has an initial data leading to an infinite time blow up then it also has an initial data leading to finite blow up.

Proof: Suppose ${u\in C([0,\infty);W^{1,2})}$ is a weak solution of NS 3-D. Assume ${u}$ blows up at infinity, then ${\exists\,t_n\rightarrow \infty}$, ${||u(t_n)||_{H^1}\rightarrow \infty}$. NS equation implies

$\displaystyle \frac{1}{2}\frac{d}{dt}|u|_{L^2}^2+|u|_{\dot{H}^1}^2\leq 0$

where ${\dot{H}^1}$ is the homogeneous ${H^1}$ space, ${|u|_{\dot{H}^1}^2=\int |\nabla u|^2dx}$. Denote ${E=\int_{\mathbb{R}^3}|u_0|^2dx}$, then

$\displaystyle |u|_{L^2}^2\leq E$

$\displaystyle \int_t^{t+1}|u|_{\dot{H}^1}^2\leq 2E\quad \forall\, t>0$

Then ${\exists\, \tau_n\in (t_n-1,t_n)}$ such that ${|u(\tau_n)|_{\dot{H}^1}\leq \sqrt{2E}}$. Consider ${\tilde{u}(t)=u(t+\tau_n)}$, note that ${|\tilde{u}_n(0)|_{H^1}\leq |u(\tau_n)|_{H^1}\leq C}$. Then there exists subsequence converging weakly in ${H^1}$, suppose ${\tilde{u}_n(0)\rightarrow \tilde{u}_0}$ in ${L^2}$ and converge weakly in ${H^1}$.

Claim: Start NS with initial data ${\tilde{u}_0}$, the solution must blow up in finite time. Suppose not. Let ${v}$ be the solution with initial data ${\tilde{u}_0}$ and it is nice in ${[0,1]}$. Let ${w=v-\tilde{u}_n}$, then

$\displaystyle \begin{cases}w_t-\nu\Delta w+v\cdot\nabla w+w\cdot\nabla v+w\cdot\nabla w+\nabla p=0 \\ div w=0\\w(x,0)\text{ small in } L^2\end{cases}\\$

This equation implies

$\displaystyle \frac{1}{2}\frac{d}{dt}|w|_{L^2}^2+|w|_{\dot{H}^1}^2\leq C|w|^2_{L^2}$

Since ${|w|_{L^2}}$ can arbitrarily small, we can find ${t_0\in (0,1)}$ such that ${|w(0)|_{{H}^1}^2\ll 1}$. Multiplying the equation of ${w}$ by ${\Delta w}$, one can get

$\displaystyle \frac{d}{dt}|w|_{H^1}^2\leq C(|u|^3_{H^1}+|u|^2_{H^1})$

This implies ${|w(t)|_{H^1}\ll 1}$ for ${t\in [t_0,1]}$. This contradicts ${|u(t_n)|_{H^1}\rightarrow \infty}$. $\Box$

Remark: Follows from my note of lectures given by Peter Constantin. As he said, this proof is done by Foias and Temam.

One regularity of NS 3-D

$\displaystyle \begin{cases}u_t+u\cdot\nabla u-\nu\Delta u+\nabla p=0\\div u=0\\u(x,0)=u_0\end{cases}\text{ on }\mathbb{R}^3$

Thm: If ${u}$ solves NS 3-D and ${u\in C([0,T];L^3(\mathbb{R}^3))}$, then ${u}$ is regular.

Proof: Denote

$\displaystyle E=\int_{\mathbb{R}^3}|u_0|^2dx$

Note that ${\int_{\mathbb{R}^3}|u(x,t)|^2dx\leq E}$ for all ${t>0}$.

Fix ${M>0}$, which will be determined later, letting ${G_M=\{x|u(x,t)\leq M\}}$, by Chebyshev’s inequality

$\displaystyle M^2\left|G^c_M\right|\leq \int_{\mathbb{R}^3}|u(x,t)|^2dx\leq E$

$\displaystyle \left|G^c_M\right|\leq \frac{E}{M^2}.$

Since ${u\in C([0,T];L^3)}$, then ${u(\cdot,t)}$ is compact in ${L^3}$, in particular, ${u(\cdot,t)}$ is uniformly integrable, namely

$\displaystyle \sup_t\left|\int_B|u(x,t)|^3dx\right|\leq \varepsilon\text{ whenever }|B|\leq \delta(\varepsilon),t\in [0,T]$

Multiplying the NS equation by ${\Delta u}$, we get

$\displaystyle \frac{1}{2}\frac{d}{dt}\int |\nabla u|^2+\nu\int |\Delta u|^2dx\leq \int |u||\nabla u||\Delta u|$

$\displaystyle \leq \int_{G_M}|u||\nabla u||\Delta u|dx+\int_{G_M^c}|u||\nabla u||\Delta u|dx$

$\displaystyle \leq M\int |\nabla u||\Delta u|dx+\left(\int_{G_M^c} u^3dx\right)^{1/3}||\nabla u||_{L^6}||\Delta u||_{L^2}$

Choose ${M}$ such that ${\int_{G_M^c} u^3dx}$ small enough and use the Sobolev inequality ${||\nabla u||_{L^6}||\leq C||\Delta u||_{L^2}}$,

$\displaystyle \frac{1}{2}\frac{d}{dt}\int |\nabla u|^2+\frac{\nu}{2}\int |\Delta u|^2dx\leq C||\nabla u||_{L^2}||\Delta u||_{L^2}$

By Young’s inequality,

$\displaystyle \frac{1}{2}\frac{d}{dt}\int |\nabla u|^2\leq C||\nabla u||_{L^2}^2$

So ${\nabla u\in L^\infty([0,T]; L^2)}$. From the regularity theory, ${u}$ must be a regular solution. $\Box$

Remark: Follows from my note of lectures given by peter constantin.

Counterexample in the plane for exterior domain

On page 75 of one Amick’s paper said,

Indeed, one can easily construct solenoidal velocity fields in $\Omega$ with finite Dirichlet norm which become unbounded like $(\ln(r))^\alpha$ with $0<\alpha<\frac{1}{2}$ as $r\to \infty$.

Actually the example here can be

$\psi=x\ln ^\alpha r$

$w=curl\psi=(\partial_y\psi,-\partial_x\psi)$

Apprently $w$ is a divergence free vector field. One can verify that the Dirichlet norm $\int_{|x|>1}|\nabla w|^2dx<\infty$ exactly when $\alpha<\frac{1}{2}$. Howver, $w$ grows like $\ln^\alpha r$ at infinity.

Remark: Charles J. Amick, On Leray’s problem of steady Navier-Stokes flow past a body in the plane.

Hardy inequality in dimesion 2

Suppose ${u}$ is a smooth function defined in ${B^c=\{|x|>1\}}$ in the plane ${\mathbb{R}^2}$, assume ${u=0}$ on ${\partial B^c}$ and also has compact support, then

$\displaystyle \int_{|x|> 1}\frac{u(x)^2}{|x|^2\ln^2|x|}dx\leq 4\int_{|x|>1}|\nabla u|^2dx$

There is a way presented by my advisor to explain why the strange function ${|x|^2\ln^2|x|}$ pops up here. First we transform LHS to polar coordinates

$\displaystyle \int_{|x|> 1}\frac{u(x)^2}{|x|^2\ln^2|x|}dx=\int_{0}^{2\pi}\int_{1}^\infty\frac{u(r,\theta)}{r^2\ln^2 r}rdrd\theta$

which leads us to start with a very general ${f(r)}$

$\displaystyle \int_{0}^{2\pi}\int_{1}^\infty\frac{u(r,\theta)}{f(r)}rdrd\theta$

Suppose ${F'(r)=r/f(r)}$, then

$\displaystyle \int_{0}^{2\pi}\int_{1}^\infty\frac{u(r,\theta)}{f(r)}rdrd\theta=\int_{0}^{2\pi}\int_{1}^\infty u(r,\theta)dF(r)d\theta$

$\displaystyle =\int_{0}^{2\pi}\left(u(r,\theta)F(r)|_1^\infty-\int_{1}^\infty F(r)\partial_r u(r,\theta)dr\right)d\theta$

$\displaystyle =-\int_{0}^{2\pi}\int_{1}^\infty F(r)\partial_r u(r,\theta)drd\theta=-\int_{|x|>1} \frac{F(r)\partial_r u(x)}{r}dx$

$\displaystyle \leq 2\left(\int_{|x|>1}\frac{u(x)^2}{f(r)}dx\right)^{1/2}\left(\int_{|x|>1}\frac{f(r)F^2(r)}{r^2}|\nabla u|^2dx\right)^{1/2}$

So we only need to find ${f(r)}$ and ${F(r)}$ such that

$\displaystyle \frac{f(r)F^2(r)}{r^2}\leq C$

Actually one can solve the ODE

$\displaystyle \frac{f(r)F^2(r)}{r^2}=1, \quad F'(r)=\frac{r}{f(r)}$

to get ${F(r)=\frac{-1}{\ln r}}$, ${f(r)=r^2\ln^2 r}$. Plugging in this function back to the above proof gives you the desired inequality.

Bound of norm on gradient implies bound on function

$\mathbf{Problem:}$ Suppose $p\in L^2_{loc}(B_1)$, $\nabla p\in L^2(B_1)$, prove that $p\in L^2(B_1)$ where $B_1\subset\mathbb{R}^n$ is a ball with radius 1.

We are going to use the following lemma.

Lemma 7.16 on GT’s book. p162

Let $\Omega$ be convex and $u\in W^{1,1}(\Omega)$. Then

$\displaystyle |u(x)-u_S|\leq \frac{d^n}{n|S|}\int_{\Omega}\frac{|Du(y)|}{|x-y|^{n-1}}dy$,   $a.e. \Omega$

where $\displaystyle u_S=\frac{1}{|S|}\int_S udx$, $d=diam\,\Omega$.

$\mathbf{Proof:}$ For any $B_r\subset B_1$ concentric with $B_1$, $p\in W^{1,1}(B_r)$, apply the lemma 7.6

$\displaystyle |p(x)-p_S|\leq \frac{r^n}{n|S|}\int_{B_r}\frac{|\nabla p(y)|}{|x-y|^{n-1}}dy\leq \frac{1}{n|S|}\int_{B_1}\frac{|\nabla p(y)|^2}{|x-y|^{n-1}}dy\int_{B_1}\frac{1}{|x-y|^{n-1}}dy$

Since $\displaystyle \int_{B_1}\frac{1}{|x-y|^{n-1}}dy$ is uniformly bounded for $\forall x\in\mathbb{R}^n$, then the above inequality implies that

$\displaystyle |p(x)-p_S|\leq C\int_{B_1}\frac{|\nabla p(y)|^2}{|x-y|^{n-1}}dy$

$\displaystyle \int_{B_r}|p(x)-p_S|^2dx\leq C\int_{B_1}\int_{B_1}\frac{|\nabla p(y)|^2}{|x-y|^{n-1}}dydx\leq C\int_{B_1}|\nabla p|^2dy$

Let $r\to 1^+$, we get $||p||_{L^2}\leq C(p_S+||\nabla p||_{L^2})$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ For any bounded domain with locally lipschitz boundary, this proposition is true. Because the locally boundary enable the domain to have locally star-shaped property, we can apply this proposition locally.

Uniqueness and nonuniqueness of steady Stokes equations

Simply take $u$ as $\overset{\rightharpoonup}{u}$, $u$ is understood as a vector function.

$\displaystyle D(\Omega)=\{\overset{\rightharpoonup}{u}\in C^\infty_c(\Omega)| \nabla\cdot \overset{\rightharpoonup}{u}=0 \}$

$\displaystyle H(\Omega)$ is  the closure of $D(\Omega)$ under the norm of $H^1_0(\Omega)$

$\langle\displaystyle \overset{\rightharpoonup}{u}, \overset{\rightharpoonup}{v}\rangle_{H(\Omega)}=\int_{\Omega}\nabla\overset{\rightharpoonup}{u}\cdot \nabla\overset{\rightharpoonup}{v}dx$

$H(\Omega)$ is defined for arbitrary domain $\Omega\subset \mathbb{R}^3$ or $\mathbb{R}^2$, unless $\Omega=\mathbb{R}^2$

$\displaystyle \widehat{H}(\Omega)=\{\overset{\rightharpoonup}{u}|\overset{\rightharpoonup}{u} \in H^1_0(\Omega), \nabla\cdot \overset{\rightharpoonup}{u}=0\}$.

$\displaystyle (1)\begin{cases}\nu\Delta \overset{\rightharpoonup}{v}=\nabla p-\overset{\rightharpoonup}{f}\\ \nabla\cdot \overset{\rightharpoonup}{v}=0\\ \overset{\rightharpoonup}{v}|_{\partial \Omega}=0\end{cases}$

Reasonable solution of this problem is $\overset{\rightharpoonup}{v}_x\in L^2(\Omega)$ from physical reason. So it is also “natural” to require $\overset{\rightharpoonup}{v}\in W^{1,2}(\Omega)$. However, if the $\partial\, \Omega$ is too bad(say not lipschitz), we can not talk about the $\overset{\rightharpoonup}{v}|_{\partial \Omega}$ in a right sense. So we have to require $\overset{\rightharpoonup}{v}$ is in a reasonable space and $\partial \Omega$ is has some regularities. In order to overcome this difficulty, we force $\overset{\rightharpoonup}{v}\in H(\Omega)$ for any domain, bounded or unbounded, except $\Omega=\mathbb{R}^2$.

Formulation I: A generalized solution is $\overset{\rightharpoonup}{v}\in H(\Omega)$ and $\displaystyle \nu\langle\overset{\rightharpoonup}{v},\overset{\rightharpoonup}{u}\rangle=\int_{\Omega}\overset{\rightharpoonup}{f}\cdot\overset{\rightharpoonup}{u} dx$, for all $\overset{\rightharpoonup}{u}\in H(\Omega)$.

In this formulation, we force $\overset{\rightharpoonup}{v}\in H(\Omega)$. If $\overset{\rightharpoonup}{f}\in H'(\Omega)$, which means

$\displaystyle \int_{\Omega}\overset{\rightharpoonup}{f}\cdot\overset{\rightharpoonup}{u}dx\leq C||u||_{H(\Omega)}$

then $(1)$ is uniquely solved. Additionally, we do not need to make any assumption on $p(x)$ near the boundary of $\Omega$.

Formulation II: A generalized solution is $\overset{\rightharpoonup}{v}\in \widehat{H}(\Omega)$ and $\displaystyle \nu\langle\overset{\rightharpoonup}{v},\overset{\rightharpoonup}{u}\rangle=\int_{\Omega}\overset{\rightharpoonup}{f}\cdot\overset{\rightharpoonup}{u} dx$, for all $\overset{\rightharpoonup}{u}\in \widehat{H}(\Omega)$.

In this formulation, in order to get $\displaystyle \nu\langle\overset{\rightharpoonup}{v},\overset{\rightharpoonup}{u}\rangle=\int_{\Omega}\overset{\rightharpoonup}{f}\cdot\overset{\rightharpoonup}{u} dx$ from $(1)$, we need $\int_{\Omega}\nabla p\cdot \overset{\rightharpoonup}{u}dx=0$, $\forall \overset{\rightharpoonup}{u}\in \widehat{H}(\Omega)$. But this require $p(x)\in L^2(\Omega)$. We can not isolate finding of $\overset{\rightharpoonup}{v}$ from the of $p$. Moreover, it is not natural to require $p\in L^2(\Omega)$ for unbounded domain from the physical point of view.

Formulation III: A generalized solution is $\overset{\rightharpoonup}{v}\in \widehat{H}(\Omega)$ and $\displaystyle \nu\langle\overset{\rightharpoonup}{v},\overset{\rightharpoonup}{u}\rangle=\int_{\Omega}\overset{\rightharpoonup}{f}\cdot\overset{\rightharpoonup}{u} dx$, for all $\overset{\rightharpoonup}{u}\in {H}(\Omega)$.(see[2])

In this formulation, we are able to isolate $\overset{\rightharpoonup}{v}$ and $p$. But this may results in $(1)$ has more than one solutions. According to the paper of Heywood, the number of solution is the number of cosets $\widehat{H}(\Omega)/H(\Omega)$. He actually gives an example such that $\widehat{H}(\Omega)/H(\Omega)$ is one-dimensional.

So the real question is whether $H(\Omega)$ coincides with $\widehat{H}(\Omega)$. [1] says that if $\Omega$ is “star shaped” bounded domains, or $\Omega$ has a compact lipschitz boundary.

[2] proves that if $\Omega$ is the exterior domain with $C^2$ boundaries, then they are the same space.

Application:

$\displaystyle (2)\begin{cases}\nu\Delta \overset{\rightharpoonup}{v}=\nabla p-\overset{\rightharpoonup}{f}\\ \nabla\cdot \overset{\rightharpoonup}{v}=0\\ \overset{\rightharpoonup}{v}|_{\partial \Omega}=\alpha(x)\end{cases}$

When one try to solve $(2)$, first one takes some solenoidal extension $\overset{\rightharpoonup}{a}$  of $\alpha(x)$ inside of $\Omega$. If one forces $\overset{\rightharpoonup}{a}\in W^{1,2}(\Omega)$, one can find a solution of $(2)$ in the form of $\overset{\rightharpoonup}{a}+\overset{\rightharpoonup}{v}$, where $\overset{\rightharpoonup}{v}\in H(\Omega)$. However, there is question here, if we find another extension $\overset{\rightharpoonup}{a}$ of $\alpha(x)$, are we necessarily find the same solution?

In another way, we can find solution of $(2)$ by hydrodynamic-potentials method, is this solution the same with what we have obtained before?

All of these question is related to $\widehat{H}(\Omega)/H(\Omega)$.

$\textbf{Remark:}$

[1]O.A. Ladyzhenskaya, V.A. Solonnikov. Some problems for vector analysis and generalized formulations of boundary-value problems for the navier-stokes equations.

[2]J.G. Heywood. On uniqueness questions in the theory of viscous flow.

Completeness of H(\Omega)

$\displaystyle D(\Omega)=\{\overset{\rightharpoonup}{u}\in C^\infty_c(\Omega)| \nabla\cdot \overset{\rightharpoonup}{u}=0 \}$

$\displaystyle H(\Omega)$ is  the closure of $D(\Omega)$ under the norm of $H^1_0(\Omega)$

$\langle\displaystyle \overset{\rightharpoonup}{u}, \overset{\rightharpoonup}{v}\rangle_{H(\Omega)}=\int_{\Omega}\nabla\overset{\rightharpoonup}{u}\cdot \nabla\overset{\rightharpoonup}{v}dx$

$H(\Omega)$ is defined for arbitrary domain $\Omega\subset \mathbb{R}^3$ or $\mathbb{R}^2$, unless $\Omega=\mathbb{R}^2$

Let us prove $H(\Omega)$ is complete.

Simply take $u$ as $\overset{\rightharpoonup}{u}$, $u$ is understood as a vector function.

$\textbf{Proof:}$ $n=3$. Recall the Hardy inequality, for all $u\in C^\infty_c(\Omega)$

$\displaystyle \int_{\mathbb{R}^3}\frac{u^2(x)}{|x-y|^2}dx\leq 4\int_{\mathbb{R}^3}|\nabla u|^2dx$, $\forall y\in \mathbb{R}^3$, $\quad (1)$

Suppose $u_n$ is cauchy sequence in $H(\Omega)$, namely $\int_{\Omega}|\nabla u_n-\nabla u_m|^2dx\to 0$ as $n,m\to \infty$.

Since $D(\Omega)$ is dense in $H(\Omega)$, WLOG assume $u_n\in D(\Omega)$, $\forall n$.

Then apply $(1)$, we get,

$\displaystyle \int_{\Omega}\frac{(u_n(x)-u_m(x))^2}{|x-y|^2}dx\leq 4\int_{\Omega}|\nabla u_n-\nabla u_m|^2dx\to \infty$, $y$ is fixed.

This means $\displaystyle \frac{u_n(x)}{|x-y|}\to \frac{u(x)}{|x-y|}$ in $L^2(\Omega)$.

Apparently, $\nabla u_n$ should converge to some $v$ in $L^2(\Omega)$. Let us prove $\nabla u=v$.

For any $w\in D(\Omega)$

$\displaystyle \int_{\Omega}u_n\cdot\nabla wdx=-\int_{\Omega}\frac{u_n}{|x-y|}\cdot w\nabla|x-y| dx+\int_{\Omega}\frac{u_n}{|x-y|}\cdot\nabla(|x-y|w)$  $(2)$

Since $w\nabla|x-y|$ and $\nabla(|x-y|w)$ are both $C^\infty_c(\Omega)$ functions except a singular point $y$,

$\displaystyle \int_{\Omega}\frac{u_n}{|x-y|}\cdot w\nabla|x-y| dx\to \int_{\Omega}\frac{u}{|x-y|}\cdot w\nabla|x-y| dx$ as $n\to \infty$

$\displaystyle \int_{\Omega}\frac{u_n}{|x-y|}\cdot\nabla(|x-y|w)\to \int_{\Omega}\frac{u}{|x-y|}\cdot\nabla(|x-y|w)$ as $n\to \infty$

$(2)$ becomes

$\displaystyle \int_{\Omega}u_n\cdot\nabla wdx\to \int_{\Omega}u\cdot\nabla wdx$

However $\displaystyle \int_{\Omega}u_n\cdot\nabla wdx=-\int_{\Omega}\nabla u_n\cdot wdx \to -\int_{\Omega}v\cdot wdx$, which means

$\displaystyle \int_{\Omega}u\cdot\nabla wdx\to -\int_{\Omega}v\cdot wdx$

Then $\nabla u=v$ and it is easy to prove $v\in H(\Omega)$. And $u_n\to v$ in $H(\Omega)$

When $n=2$, use the same method by

$\displaystyle \int_{|x-y|>1}\frac{u^2(x)}{|x-y|^2\ln^2|x-y|}dx\leq 4\int_{|x-y|>1}|\nabla u|^2dx$, $\forall y\in \mathbb{R}^3$

Extend the boundary value to a solenoidal vector field whole domain

$\mathbf{Problem:}$ Suppose $\Omega\subset \mathbb{R}^2$ or $\mathbb{R}^3$ has smooth boundary $\partial \Omega$. If $\boldsymbol{f}$ is smooth vector field on $\partial\Omega$, satisfies

$\displaystyle \int_{\partial \Omega}\boldsymbol{f}\cdot\boldsymbol{n}=0$

Then there exists a soleniodal vector field $\boldsymbol{u}$ takes $\boldsymbol{f}$ on the boundary. That is

$\displaystyle \begin{cases}div\boldsymbol{u}=0\\ \boldsymbol{u}|_{\partial \Omega}=\boldsymbol{f}\end{cases}$

$\mathbf{Proof:}$  Decompose $\boldsymbol{f}=a(x)\boldsymbol{n}+\boldsymbol{g}$, where $a(x)=\boldsymbol{f}\cdot \boldsymbol{n}$. We can solve the Neumann problem

$\displaystyle\Delta \phi=0, \quad \frac{\partial \phi}{\partial n}=a(x)$

So if we can find $\boldsymbol{u}$ takes $\boldsymbol{g}$ on $\partial \Omega$, $\boldsymbol{u}+\nabla \phi$ is the solution.

(1) Suppose $\Omega\subset\mathbb{R}^2$. We will construct $\psi$, $\displaystyle \boldsymbol{u}=curl \psi=(\frac{\partial \psi}{\partial x_2}, -\frac{\partial \psi}{\partial x_1})$, then $div \boldsymbol{u}=0$

Since $\partial \Omega$ is locally a curve in $\mathbb{R}^2$, we assume it has parameter representation $(x_1(t),x_2(t))$, $t\in (-\epsilon,\epsilon)$.

If we are forcing $curl\psi=\boldsymbol{g}=(g_1,g_2)$, then

$\displaystyle \psi(t)=\psi(0)+\int_0^t \frac{\partial \psi}{\partial x_1}x'_1(t)+ \frac{\partial \psi}{\partial x_2}x'_2(t)dt=\int_0^t -g_2x'_1(t)+g_2x'_2(t) dt=\psi(0)$

since $\boldsymbol{g}\cdot \boldsymbol{n}=0$. So we can let $\psi=cons.$ and $\nabla\psi=(-g_2,g_1)$ on $\partial \Omega$. Such $\psi$ can be extended to the whole $\Omega$.

then $curl\psi+\nabla\phi$ is the vector we need. In particular case, if $\Omega$ is simply connected, $u$ can be represented by a single $curl \psi$.

(2) If $\Omega\subset \mathbb{R}^3$. Then there exists a partition of unity, namely

$\displaystyle I=\sum\limits_{k=1}^K \xi_k$

each $\xi_k$ supports in a small domain $U_k$. Let $\boldsymbol{g_k}=\boldsymbol{g}\xi_k$. We will construct a $\boldsymbol{v_k}$ for each $\boldsymbol{g_k}$ such that $curl \boldsymbol{v_k}=\boldsymbol{g_k}$ on $U_k\cap \partial \Omega$. Then $\sum curl \boldsymbol{v_k}$ is the vector field we want to find. For one $\boldsymbol{g_k}$, for simplicity, ignore the subindex as $\boldsymbol{g}$. Change the coordinates such that $\partial \Omega\cap U_k$ can be represented as

$\displaystyle y_3=f(y_1,y_2)$

Then $\boldsymbol{v}=(g_2(y_3-f(y_1,y_2)), -g_1(y_3-f(y_1,y_2)),0)$. Verify that $curl\boldsymbol{v}=\boldsymbol{g}$ by the fact that $\boldsymbol{g}\cdot \boldsymbol{n}=0$ where $\boldsymbol{n}=(f_{y_1},f_{y_2},-1)$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Refer to Olga Aleksandrovna Ladyzhenskaya’s book. The second part was taken from the lecture notes of Yanyan, Li.