Interestingly, even with the specification of the field equation \(\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial z^2} = 0\) and the three boundary conditions, \(w = \frac{\partial \phi}{\partial z} = 0 \text{ on } z = -H\), \(\left( \frac{\partial \phi}{\partial z} \right)_{z=0} \cong \frac{\partial \eta}{\partial t}\), and \(\left( \frac{\partial \phi}{\partial t} + \frac{p}{\rho} + gz \right)_{z = \eta} \cong \left( \frac{\partial \phi}{\partial t} \right)_{z=0} + g \eta \cong 0 \text{ or } \left( \frac{\partial \phi}{\partial t} \right)_{z=0} \cong -g \eta\), the overall linear surface-wave problem is not fully defined without initial conditions for the surface shape

For simplicity, choose \( \eta(x,t = 0) = a \cos(kx) \) since it matches the simple sinusoidal wave \(\eta(x, t) = a \cos[kx - \omega t]\), which now becomes a foundational part of the solution. To produce a cosine dependence for \( \eta \) on the phase \( (kx - \omega t) \) in \(\eta(x, t) = a \cos[kx - \omega t]\), \(\left( \frac{\partial \phi}{\partial z} \right)_{z=0} \cong \frac{\partial \eta}{\partial t}\), and \(\left( \frac{\partial \phi}{\partial t} + \frac{p}{\rho} + gz \right)_{z = \eta} \cong \left( \frac{\partial \phi}{\partial t} \right)_{z=0} + g \eta \cong 0 \text{ or } \left( \frac{\partial \phi}{\partial t} \right)_{z=0} \cong -g \eta\) require \( \phi \) to be a sine function of \( (kx - \omega t) \)
\(\therefore\) A solution for \( \phi \) in the form \(\phi(x, z, t) = \boxed{f(z)} \sin (kx - \boxed{\omega(k)}t)\)
\(\therefore\) Substite into the Laplace equation \(\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial z^2} = 0\) \(\Rightarrow d^2f/dz^2 - k^2 f = 0\)
\(\because\) Has the general solution \( f(z) = Ae^{kz} + Be^{-kz} \), where \( A \) and \( B \) are constants
\(\therefore\) \(\phi(x, z, t) = (Ae^{kz} + Be^{-kz}) \sin(kx - \omega t)\)
\(\therefore\) The constants \( A \) and \( B \) can be determined by substituting \(\phi(x, z, t) = (Ae^{kz} + Be^{-kz}) \sin(kx - \omega t)\) into \(w = \frac{\partial \phi}{\partial z} = 0 \text{ on } z = -H\) \(\Rightarrow k(Ae^{-kH} - Be^{kH}) \sin(kx - \omega t) = 0 \text{ or } B = Ae^{-2kH}\)
\(\therefore\) Substituting \(\eta(x, t) = a \cos[kx - \omega t]\) and \(\phi(x, z, t) = (Ae^{kz} + Be^{-kz}) \sin(kx - \omega t)\) into \(\left( \frac{\partial \phi}{\partial z} \right)_{z=0} \cong \frac{\partial \eta}{\partial t}\) \(\Rightarrow k(A - B) \sin(kx - \omega t) = \omega a \sin(kx - \omega t) \text{ or } k(A - B) = \omega a\)
\(\because\) Solving \(\begin{cases} k\left(Ae^{-kH} - Be^{kH}\right) \sin(kx - \omega t) = 0 & \text{ or } B = Ae^{-2kH} \\ k(A - B) \sin(kx - \omega t) = \omega a \sin(kx - \omega t) & \text{ or } k(A - B) = \omega a \end{cases}\) for \( A \) and \( B \)
\(\therefore\) \(A = \frac{\omega a}{k(1 - e^{-2kH})} \quad \text{and} \quad B = \frac{\omega a e^{-2kH}}{k(1 - e^{-2kH})}\)
\(\because\) \( \cosh(x) = \dfrac{e^x + e^{-x}}{2} \), \( \sinh(x) = \dfrac{e^x - e^{-x}}{2} \), \( e^{kz} + e^{-2kH} e^{-kz} = 2e^{-kH} \cosh(k(z + H)) \text{ and } \frac{1}{\sinh(kH)} = \frac{2e^{-kH}}{1 - e^{-2kH}}\)
\(\therefore\) The velocity potential \(\phi(x, z, t) = (Ae^{kz} + Be^{-kz}) \sin(kx - \omega t)\) then becomes \(\phi = \frac{\omega a}{k} \frac{\cosh(k(z + H))}{\sinh(kH)} \sin(kx - \omega t)\)
\(\Rightarrow\) The fluid velocity components \(u = \omega a \frac{\cosh(k(z + H))}{\sinh(kH)} \cos(kx - \omega t) \quad \text{and} \quad w = \omega a \frac{\sinh(k(z + H))}{\sinh(kH)} \sin(kx - \omega t)\)

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