Kelvin-Helmholtz Instabilitys
There are a total of four boundary conditions $$ \phi_1 \to 0 \quad \text{as } z \to +\infty, \qquad \phi_2 \to 0 \quad \text{as } z \to -\infty $$ $$ \overbrace{ \underbrace{\mathbf{n}}_{\text{local normal}} \cdot \nabla \tilde{\phi}_1 = \underbrace{\mathbf{n}}_{\text{local normal}} \cdot \underbrace{\mathbf{u}_s}_{\text{velocity of the interface}} = \underbrace{\mathbf{n}}_{\text{local normal}} \cdot \nabla \tilde{\phi}_2 \quad \text{on } z = \zeta }^{\text{Kinematic boundary condition}} $$ $$ \overbrace{ \underbrace{p_1, p_2}_{\text{pressures above/below}} : \quad p_1 = p_2 \quad \text{on } z = \zeta }^{\text{Dynamic boundary condition}} $$

Kinematic boundary condition can be rewritten $$ \mathbf{n} \cdot \left\{ \frac{\partial \tilde{\phi}_1}{\partial x}\mathbf{e}_x + \frac{\partial \tilde{\phi}_1}{\partial z}\mathbf{e}_z \right\} = \mathbf{n} \cdot \left\{ \frac{\partial \zeta}{\partial t}\mathbf{e}_z \right\} = \mathbf{n} \cdot \left\{ \frac{\partial \tilde{\phi}_2}{\partial x}\mathbf{e}_x + \frac{\partial \tilde{\phi}_2}{\partial z}\mathbf{e}_z \right\} \quad \text{on } z = \zeta $$ The normal vector $$ \mathbf{n} = \nabla f / |\nabla f| = [-(\partial \zeta/\partial x)\mathbf{e}_x + \mathbf{e}_z] / \sqrt{1 + (\partial \zeta/\partial x)^2} $$ when $f(x,z,t) = z - \zeta(x,t) = 0$ defines the interface, and $ \mathbf{u}_s = (\partial \zeta / \partial t)\mathbf{e}_z $ can be considered purely vertical $$ \frac{ -\left(U_1 + \frac{\partial \phi_1}{\partial x}\right)\frac{\partial \zeta}{\partial x} + \frac{\partial \phi_1}{\partial z} }{ \cancel{\sqrt{1+\left(\frac{\partial \zeta}{\partial x}\right)^2}}} = \frac{\frac{\partial \zeta}{\partial t}}{\cancel{\sqrt{1+\left(\frac{\partial \zeta}{\partial x}\right)^2}}} = \frac{ -\left(U_2 + \frac{\partial \phi_2}{\partial x}\right)\frac{\partial \zeta}{\partial x} + \frac{\partial \phi_2}{\partial z} }{ \cancel{\sqrt{1+\left(\frac{\partial \zeta}{\partial x}\right)^2}}} $$ The normal vector reduces to $$ -\left(U_1 + \frac{\partial \phi_1}{\partial x}\right)\frac{\partial \zeta}{\partial x} + \frac{\partial \phi_1}{\partial z} = \frac{\partial \zeta}{\partial t} = -\left(U_2 + \frac{\partial \phi_2}{\partial x}\right)\frac{\partial \zeta}{\partial x} + \frac{\partial \phi_2}{\partial z} \quad \text{on } z = \zeta $$ This condition can be linearized by applying it at $z = 0$ instead of at $z = \zeta$ and by neglecting quadratic terms. Thus, the simplified version of the kinematic boundary condition is $$ - U_1 \frac{\partial \zeta}{\partial x} + \frac{\partial \phi_1}{\partial z} = \frac{\partial \zeta}{\partial t} = - U_2 \frac{\partial \zeta}{\partial x} + \frac{\partial \phi_2}{\partial z} \quad \text{on } z = 0 $$