# Elementary matrix of Raviart-Thomas elements

We can use the $$RT0$$ to solve the Darcy equation, i.e. $$k^{-1}\mathbf{u}+\nabla p = 0, \text{ in } \Omega,$$ $$-\nabla \cdot \mathbf{u} = 0, \text{ in } \Omega,$$ $$p = p_D \text{ on } \partial\Omega,$$ $$\mathbf{u \cdot n} = 0 \text{ on } \partial\Omega, \tag{1}$$ where $$\mathbf{u}$$ is the velocity, $$k$$ is conductivity coefficient, and $$p$$ is the pressure.

With the incorporation of test functions, we have: $$\int_{T}k^{-1}\mathbf{u}\cdot\mathbf{v}\mathrm{d}T - \int_{T}p\nabla\cdot \mathbf{v}\mathrm{d}T=-\int_{\partial T}p_D(\mathbf{v\cdot n})\mathrm{d}(\partial T),$$ $$-\int_{T}\nabla\cdot \mathbf{u}q\mathrm{d}T=0,\tag{2}$$ where $$\mathbf{v}$$ and $$q$$ are basis functions (they can be replaced with specific basis functions, like $$RT0$$), and $$T$$ is an elementary triangle in a mesh. The Eq. (2) can be rewritten as $$\left[\begin{matrix}\mathbf{A}&\mathbf{B}\\\mathbf{B}^{TPS}&0\end{matrix}\right]\left[\begin{matrix}\mathbf{u}\\p\end{matrix}\right] = \left[\begin{matrix}\mathbf{f}\\0\end{matrix}\right], \tag{3}$$ which is a local (elementary) matrix.

The basis function $$\mathbf{\hat{\phi}}$$ of $$RT0$$ is written as $$\hat{\phi_1} = \sqrt{2}\left(\hat{x_1} \quad \hat{x_2}\right)^{T}, \hat{\phi_2} = \left(-1+\hat{x_1} \quad \hat{x_2}\right)^{T}, \text{ and } \hat{\phi_3} = \left(\hat{x_1} \quad -1+\hat{x_2}\right)^{T},\tag{4}$$ where $$\hat{}$$ is a reminder of reference triangle, $$\hat{x_i}$$ is coordinates of corners of the reference triangle, i.e. 0 or 1.

The transformation from $$T$$ to $$\hat{T}$$ is $$$$\left[\begin{matrix}x_1\\x_2\end{matrix}\right]=B_k\left[\begin{matrix}\hat{x_1}\\ \hat{x_2}\end{matrix}\right]+b=\left[\begin{matrix}x_1^2-x_1^1&x_1^3-x_1^1\\x_2^2-x_2^1&x_2^3-x_2^1\end{matrix}\right]\left[\begin{matrix}\hat{x_1}\\ \hat{x_2}\end{matrix}\right] + \left[\begin{matrix}{x_1}\\ {x_2}\end{matrix}\right]$$,\tag{5}$$ where $$x_i$$ is coordinates of nodes of $$T$$.

Now let's approximate $$\mathbf{A}$$ in Eq. (3) with $$\mathbf{\hat{\phi}}$$: $$\mathbf{A} = \alpha_{ij}^T = \frac{1}{|\text{det}B_k|}\int_{\hat{T}}k^{-1}[s_i][s_j](B_k\hat{\phi_i})\cdot(B_k\hat{\phi_j})\mathbf{d}\hat{T},$$ where $$[s_i]$$ is +1 or -1 if the normal direction of on edge $$i$$ of reference triangle is identical or not identical to the prescribed global normal direction of edge $$i$$. And we may use Gauss Quadrature Rule to approximate the integral, and the coordinates of Gauss points are $$\hat{x_i}$$ which can be input to $$\hat{\phi_i}$$ ($$i = 1, 2, 3$$ means the three edges).

My question is: since $$B_k$$ is $$2 \times 2$$ and $$\hat{\phi_i}$$ is $$2\times 1$$, the $$B_k\hat{\phi_i}$$ is $$2\times 1$$. Then the result of the dot product,i.e. $$(B_k\hat{\phi_i})\cdot(B_k\hat{\phi_j})$$, is a scalar. Thus, $$\mathbf{A}$$ is $$3 \times 3$$. However, $$\mathbf{u}$$ is $$6 \times 1$$, i.e. $$[u_{x1}, u_{x2}, u_{x3}, u_{y1}, u_{y2}, u_{y3}]^{TPS}$$ (if $$\Omega$$ is in 2D). They are not consistent in the number of rows. Why? Is my understanding incorrect?

The Raviart-Thomas element has only 3 degrees of freedom on each triangle, namely the normal components of the vectors at the midpoints of the three edges. So the root of your misunderstanding is that $$\mathbf u$$ is a vector of size 3, matching the element matrix size.
• ohhh. thanks a million! Prof. Wolfgang. But how it is that possible? The $\mathbf{u}$ at a point must have two or three components in reality. Maybe $\mathbf{u}$ is only a magnitude along the normal direction??? Nov 24 '21 at 6:41