where $f=2 \Omega$. We may align the $x$-axis in the direction of the velocity at infinity, so that the boundary condition at $z \rightarrow \infty$ is simply
$$
\begin{aligned}
&u=U, \
&r=0, \quad z \rightarrow \infty \
&w=0
\end{aligned}
$$
At the rigid surface $z=0$ we suppose that the friction, in direct analogy with the role of molecular viscosity, inhibits the fluid motion to such an extent that both the normal and tangential velocities vanish on $z=0$, i.e.,
$$
u=v=w=0 \quad(z=0) .
$$
The crucial condition is the condition of no slip on the tangential volocities $u$ and $v$. In the absence of retarding frictional forces, this boundary condition would be absent. Then, the uniform velocity $U$ would be an exact solution of $(4.3 .1 \mathrm{a}, \mathrm{b}, \mathrm{c})$, i.e.,
$$
\begin{aligned}
&u=U=-\frac{1}{\rho f} \frac{\hat{\partial p}}{\partial y}, \
&v=w=0,
\end{aligned}
$$
MATH3974 COURSE NOTES ：
Since the fluid is homogeneous, it follows from that
or
$$
\begin{gathered}
\hat{\mathcal{c}}|\hat{\rho} p / \partial x| \
\hat{\partial z}|\hat{\hat{p} p / \partial y}|
\end{gathered}=0,
$$
so that the horizontal pressure gradient must be independent of $z$. Now as $z \rightarrow \infty$ both $u$ and $v$ become constant. so that for very large $z$, using
$$
\begin{gathered}
0=-\frac{1}{\rho} \frac{\hat{\partial} p}{\partial x} \
f U=-\frac{1 \hat{c} p}{\rho \hat{c} y}
\end{gathered}
$$