Linear Water Waves

Question

When solving , $\displaystyle {\partial^2 \tilde\phi\over\partial x^2}+{\partial^2 \tilde\phi\over\partial y^2}=0$

Why is it that $-h<y<0$

as opposed to $-h<y<\delta\tilde S(x,t)$ ?

Answer 1

The interface is not just arbitrarily set to $y=0$. The solution is being developed as a perturbation expansion

$$\phi(x,y,t) = \delta \phi^{(1)} + \delta^2 \phi^{(2)} + \ldots, \\ S(x,t) = \delta S^{(1)} + \delta^2 S^{(2)} + \ldots $$

You are simply solving for the leading-order solution for small $\delta$.

When you apply a boundary condition at the interface $y = S(x,t)$, for example a continuity condition for the derivative of the velocity potential, you expand in a Taylor series around $y=0$:

$$\frac{\partial \phi}{\partial y}[x,S(x,t))] \approx \frac{\partial \phi}{\partial y}(x,0) + \frac{\partial^2 \phi}{\partial y^2}(x,0)S(x,t)+ \ldots=\delta\frac{\partial \phi^{(1)}}{\partial y}(x,0) + \delta^2 \frac{\partial^2 \phi^{(1)}}{\partial y^2}(x,0)S^{(0)}(x,t)+ \ldots$$.

The $O(\delta)$ solution then depends only on boundary terms of the same order and involves a condition applied at $y=0$.

There are other wave problems where the second-order solution might be relevant.