 Solution

The wall position is $x_w(t)=\mathrm{Re}(c{e}^{-i\omega t})$, so its velocity is $\mathrm{Re}(-i\omega c{e}^{-i\omega t})$. The kinematic boundary condition equates the normal fluid velocity to the normal wall velocity. Linearizing the evaluation point to the mean wall position givesThe other wall and bottom retain $f_x(L,z)=0$ and $f_z(x,-h)=0$, while the free surface gives $f_z(x,0)=({\omega }^2/g)f(x,0)$.