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.

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