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.