## 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 $\Omega=\mathbb{R}^2$

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

Consider the steady Stokes problem

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

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.