Interpolation at user-defined locations

All the quantities deals with in SWAN are located at the vertices. Hence, due to the user-defined locations of the wave parameters, interpolations are required. Let parameter $\varphi_j = \varphi(\vec{x}_j)$ and Cartesian coordinates $\vec{x}_j = (x_j, y_j)$, with $j \in \{1,2,3\}$ indicating the vertices of cell $i$, be given. The vertices 1, 2 and 3 are ordered in a counterclockwise direction, see Figure 8.2. The associated 3 edges are denoted as 12, 23 and 31.


Linear interpolation, with $\vec{x}_0$ inside cell $i$ and $\varphi_0 = \varphi(\vec{x}_0)$, is given by

  $$\varphi(\vec{x}) = \varphi_0 + \nabla \varphi \cdot (\vec{x} - \vec{x}_0)$$ (8.17)

where $\nabla \varphi$ is a constant vector inside cell $i$. We apply Green-Gauss reconstruction, i.e.,

  $$\nabla \varphi \approx \frac{1}{A_i} \int_{\triangle i} \nabla \v... ...e i} \varphi \vec{n} d\Gamma \approx \frac{1}{A_i} \sum_e \varphi_e \vec{n}_e$$ (8.18)

where $A_i$ is the area of cell $i$ and the summation runs over the 3 edges $e \in \{12,23,31\}$ of cell $i$. The values $\varphi_e$ at edges are taken as averages:

  $$\varphi_{12} = \frac{1}{2} (\varphi_1+\varphi_2)\, , \quad \varp... ...phi_2+\varphi_3)\, , \quad \varphi_{31} = \frac{1}{2} (\varphi_3+\varphi_1)\$$ (8.19)

Furthermore, $\vec{n}_e$ is the outward pointing normal at edge $e$ and is obtained by rotating the edge over 90$^o$ in the clockwise direction. Hence,

  $$\vec{n}_{12} = R\vec{t}_{12}\, , \quad R = \left( \begin{array}{l... ... \\ -1 & 0 \end{array} \right) \, , \quad \vec{t}_{12} = \vec{x}_2 - \vec{x}_1$$ (8.20)

We also need the following identity

  $$\vec{n}_{12} + \vec{n}_{23} + \vec{n}_{31} = 0$$ (8.21)

It is not difficult to show that

$$\nabla \varphi = \frac{1}{2A_i} \left [ \vec{n}_{12} (\varphi_1 - \varphi_3) + \vec{n}_{31} (\varphi_1 - \varphi_2) \right ]$$

$$= -\frac{1}{2A_i} \left [ \varphi_1 \vec{n}_{23} + \varphi_2 \vec{n}_{31} + \varphi_3 \vec{n}_{12} \right ]$$ (8.22)

or

  $$\frac{\partial \varphi}{\partial x} = \frac{1}{2A_i} \left [ \varphi_1 (y_2 - y_3) + \varphi_2 (y_3 - y_1) + \varphi_3 (y_1 - y_2) \right]$$ (8.23)

and

  $$\frac{\partial \varphi}{\partial y} = \frac{1}{2A_i} \left [ \varphi_1 (x_3 - x_2) + \varphi_2 (x_1 - x_3) + \varphi_3 (x_2 - x_1) \right]$$ (8.24)

The area $A_i$ of cell $i$ is given by $\vert\vec{t}_{12} \cdot \vec{n}_{13}\vert/2$. Hence, with

  $$\vec{n}_{13} = R\vec{t}_{13} = \left( \begin{array}{l} y_3 - y_1 \\ x_1 - x_3 \end{array} \right)$$ (8.25)

we have

  $$A_i = \frac{1}{2} \vert (x_2 - x_1)(y_3 - y_1) - (x_3 - x_1)(y_2 - y_1) \vert$$ (8.26)

Alternatively, we may interpolate using the following relation

  $$\varphi(\vec{x}) = \sum_k \varphi_k \lambda_k(\vec{x}) = \varphi_1 \lambda_1 + \varphi_2 \lambda_2 +\varphi_3 \lambda_3$$ (8.27)

where $\lambda_k$ is a linear shape function with the following properties:

1. $\lambda_k$ is linear per cell and
2. $\lambda_k(\vec{x}_j) = \delta_{kj}$ with $\delta_{kj}$ is the Kronecker delta.

We choose the following shape function

  $$\lambda_k(\vec{x}) = a_0^k + a_x^k x + a_y^k y$$ (8.28)

and the coefficients $a$ follow from solving

  $$\left( \begin{array}{lll} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ ... ...egin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$$ (8.29)