## Tag Archives: solenoidal vector field

### 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.

### 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.