= Solution
{c}
The two wall <boundary conditions> are
$$
\boxed{\hat u=0,\qquad \hat w=0\quad(z=0).}
$$
Define the <capillary parameter for a falling film>
$$
\boxed{S=\frac{2\gamma}{3\rho gL^2\sin\alpha}.}
$$
Retain $S=O(1)$ in the <long-wave approximation>. Thus <surface tension> is retained in the leading <normal stress> balance even though geometric slope corrections are small. For $h_x=O(1)$ in the rescaled coordinates, the dimensional surface slope is $\varepsilon h_x$.
The three <free surface> <boundary conditions>, accurate through $O(\varepsilon)$, are
$$
\boxed{\begin{aligned}
h_t+(\bar u+\hat u)h_x&=\hat w,\\
\bar u_z+\hat u_z&=0,\\
\hat p&=(h-1)\cot\alpha-\frac32S h_{xx}+\varepsilon\hat w_z,
\end{aligned}\qquad z=h(x,t).}
$$
The first is the <kinematic boundary condition>; the second is the zero tangential <traction> condition. For the third, the dimensional liquid-side <pressure> is, to the required order,
$$
p-p_a=-\gamma\frac{h_0}{L^2}h_{xx}
+\frac{2\mu\varepsilon U_0}{h_0}w_z.
$$
Division by $\rho gh_0\sin\alpha$ gives $p=-3S h_{xx}/2+\varepsilon w_z$ in dimensionless variables. Subtracting $\bar p(h)=(1-h)\cot\alpha$ gives the stated perturbation condition. The additional slope-times-shear contribution to <normal stress> vanishes at this order because the tangential <traction> is zero. The omitted geometric and tangential-stress corrections are $O(\varepsilon^2)$ under this ordering.
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