Solution
= Solution
Integrating the normal momentum equation from the free surface gives the hydrostatic pressure
$$
\boxed{
p(x,y,t)=p_{\rm atm}
+\rho g\cos\alpha\,[h(x,t)-y]}.
$$
Thus $p_x=\rho g\cos\alpha\,h_x$. Twice integrating the downslope equation and applying $u(0)=0$ and $u_y(h)=0$ gives
$$
\boxed{
u(x,y,t)
=\frac{\rho g}{\mu}
\left(\sin\alpha-\cos\alpha\,h_x\right)
\left(hy-\frac{y^2}{2}\right)}.
$$