Find the direction of the gradient on a finite element mesh

Question

Suppose we have a triangular mesh of a two dimensional shape $\Omega$, and on this mesh we define a P1 finite element structure. I know that given $u,v$ by their values at the vertices of the triangles, we can compute $\int_\Omega \nabla u \cdot \nabla v$ using the stiffness matrix $A = (\int_\Omega \nabla \varphi_i\cdot \nabla\varphi_j)$ (where $\varphi_i$ is the P1 basis). This makes it possible to find expressions of the form $\int_\Omega |\nabla u|^2$ if $u$ is known at the triangle vertices.

How can we compute the approximation of the gradient $\nabla u$, given the values of $u$ on the points?

I know that the gradient of $\varphi_i$ is constant on each triangle, but how do we need to combine these gradients for heighboring triangles, such that the value we obtain in the end is relevant.

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The purpose of doing this: I would like to integrate a quantity of the form $\int_\Omega \varphi(\nabla u)^2$, where $\varphi$ is a norm, different from the euclidean one. Now that I think about it, I guess it is necessary to compute the matrix $A = \int_\Omega b(\nabla \varphi_i,\nabla \varphi_j)$, where $b$ is the scalar product associated to $\varphi$.

Answer 1

Your approach with the matrix won't work unless the norm is quadratic. Here is the general approach:

• Your solution is defined as $u_h(x) = \sum_j U_j \phi_j(x)$.

• If you want to compute $\int_\Omega \varphi(\nabla u_h(x)) dx$ (i.e., a number), then you need to replace the integral with a quadrature formula. If you use piecewise linear elements, then the one-point midpoint formula is sufficient, i.e., $$ \int_\Omega \varphi(\nabla u_h(x)) dx = \sum_K |K| \; \varphi(\nabla u_h(x_K)) $$ where $x_K$ is the midpoint (or, in fact, any point in the interior) of cell $K$.

• So you only need to evaluate $\nabla u_h(x_K)$. Using the definition of $u_h$, you get that this equals $$ \nabla u_h(x_K) = \sum_j U_j \nabla \phi_j(x_K). $$ On the other hand, because all shape functions are zero on cell $K$ except the ones defined at the three vertices of $K$, you have that in fact $$ \nabla u_h(x_K) = \sum_{j_K=1}^3 U_{j(j_K)} \nabla \phi_{j(j_K)}(x_K) = \sum_{j_K=1}^3 U_{j(j_K)} \nabla \phi^K_{j_K}(x_K). $$ Here, $j(j_K)$ is the global index of the $j_K$-th shape function on cell $K$ where $j_K=1..3$, and $\phi^K_{j_K}$ are the three shape functions defined on cell $K$.

So, if you can evaluate the gradients of the three shape functions on cell $K$, then you can compute $\nabla u_h(x_K)$ and then $\varphi(\nabla u_h(x_K)$ and from there the whole integral.