Lateral boundary conditions

Let $ G$ denote the lateral boundary of the model domain with the closed land boundary $ G^c$ and the open boundary $ G^o$ such that $ G^c \cup G^o=G$ and $ G^c \cap G^o=\emptyset$. Let further $ \vec u=(u,v)$ denote the horizontal velocity vector and $ \vec u_n=(-v,u)$ its normal vector. At closed boundaries, the flow must be parallel to the boundary:

$\displaystyle \vec u_n \cdot \vec\nabla G^c = 0$ (10)

with $ \vec\nabla=(\partial_x,\partial_y)$ being the gradient operator.

For an eastern or a western closed boundary with $ \vec\nabla G^c=(0,1)$ this has the consequence that $ u=0$ and, equivalently, for a southern or a northern closed boundary with $ \vec\nabla G^c=(1,0)$ this has the consequence that $ v=0$.

At open boundaries, the velocity gradients across the boundary vanish:

$\displaystyle \vec\nabla_n u \cdot \vec\nabla G^o = 0, \quad \vec\nabla_n v \cdot \vec\nabla G^o = 0, \quad$ (11)

with $ \vec\nabla_n=(-\partial_y,\partial_x)$ being the operator normal to the gradient operator.

For an eastern or a western open boundary with this has the consequence that $ \partial_x u=\partial_x v =0$ and, equivalently, for a southern or a northern open boundary this has the consequence that $ \partial_y u=\partial_y v =0$.

At so-called forced open boundaries, the sea surface elevation $ \zeta$ is prescribed. At passive open boundaries, it is assumed that the curvature of the surface elevation normal to the boundary is zero, with the consequence that the spatial derivatives of the surface slopes normal to the boundaries vanish.

kklingbe 2017-10-02