这是一份warwick华威大学PX264的成功案例

$\mathbf{v}=\mathbf{0}$ when $r=a$. Assume the solution takes the form $p=C z, C$ constant, $v_{z}=v_{z}(r)$, and $v_{r}=v_{\theta}=0$. obtain
$$
C=\mu \Delta v_{z}=\mu\left(\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial v_{z}}{\partial r}\right)\right)
$$
Integration yields
$$
v_{z}=-\frac{C}{4 \mu} r^{2}+A \log r+B
$$
where $A, B$ are constants. Because we require that the solution be bounded, $A$ must be 0 , because $\log r \rightarrow-\infty$ as $r \rightarrow 0$. Use the no-slip condition to determine $B$ and obtain
$$
v_{z}=\frac{C}{4 \mu}\left(a^{2}-r^{2}\right)
$$

PX264 COURSE NOTES ：

Let the velocity of $\partial D$ be specified as $\mathbf{V}$, so
$$
\mathbf{u} \cdot \mathbf{n}=\mathbf{V} \cdot \mathbf{n}
$$
Thus, $\varphi$ solves the Neumann problem:
$$
\Delta \varphi=0, \quad \frac{\partial \varphi}{\partial n}=\mathbf{V} \cdot \mathbf{n}
$$
If $\varphi$ is a solution, then $\mathbf{u}=\operatorname{grad} \varphi$ is a solution of the stationary homogeneous Euler equations, i.e.,
$$
\begin{gathered}
\rho(\mathbf{u} \cdot \nabla) \mathbf{u}=-\operatorname{grad} p \
\operatorname{div} \mathbf{u}=0 \
\mathbf{u} \cdot \mathbf{n}=\mathbf{V} \cdot \mathbf{n} \quad \text { on } \partial D
\end{gathered}
$$