Evaluate integral on Boundary - FEM

Question

I am implementing a MatLab program to solve the equation given in this paper, which involves solving integrals coming from the variational formulation of the problem. One of them is $$\int_{\partial \Omega} f\left(\sum_{i=1}^Nu_i\phi_i\right) \cdot \phi_j d\Omega$$ where $\phi_j$ is a function such that, in the node $(x_j, y_j)$ of the triangular partition of $\Omega$, its value is 1 and 0 elsewhere. That is, $\{\phi_i\}$ is the standard nodal basis defined for a triangular partition on FEM.

My confusion is how to evaluate this integral. Here is what I think I need to do:

At each iteration, we want to evaluate $u^{(k)} = (u_1^{(k)}, ..., u_N^{(k)})^T$ an approximation of $u$ at each node from our triangular partition with $N$ nodes. This integral is only evaluated at the nodes which lie on the boundary $\partial \Omega$. Suppose, for example, that $(x_1, y_1)$ is one of those nodes and we want the value of $u$ at this node (ie, we want u_1 at some iteration). Then, we should have to evaluate:

$$\int_{\partial \Omega} f\left(\sum_{i=1}^Nu_i\phi_i(x_1, y_1)\right) \cdot \phi_1(x_1, y_1) d\Omega = \int_{\partial \Omega} f(u_1) d\Omega$$

I am really confused if this is correct, and if it is, how do I calculate this integral?

I am sorry if the question is confusing, but the paper can give a clear idea of what should be done, if needed.

Thanks!

Answer 1

So taking your first expression, and giving it a name to make an equation out of it, and substituting a quadrature rule says: $$ \big[F(u)\big]_{j} \approx \sum_q w_q f\Big(\sum_i u_i \phi_i(x_q)\Big)\phi_j(x_q) $$ where $w_q$ and $x_q$ are the weights and points of your quadrature rule. If this is the RHS of your overall system, then $\phi_j$ should probably not appear since this would make a matrix for all $i$ and $j$ in your domain. If this is in the typical left-hand side, then you have additional contribution to the $i,j$ entry of your stiffness matrix. If $x_q$ is not in any of the elements surrounding $x_i$, then you can skip that term in the inner sum since $\phi_i(x_q)=0$ in that case and just add $f(0)$ to your partial sum for that $i$. It might be good if you edited to include the whole equation you're trying to solve, e.g. $Lu=F(u)$, for a linear operator $L$, or $F(u)={\rm something else}$ to help clarify. Either way, you're probably looking to form a matrix equation to solve for each $u_i$. If $u_i$ is unknown and $f$ is non-linear, then forming a matrix out of this system to solve for $u_i$ is probably impossible.

Due to this inner sum inside $f$ which is presumably non-linear, this is going to be an expensive term to compute. You might be able to quickly build the full $N_qN_i$ table if both $N$s are of reasonable size and $f$ is cheap.

Answer 2

You seem to be confused with the definition of the basis $\{\phi_i\}.$ You are right that $\phi_i(x_i,y_i)=1$ and $\phi_i(x_j,y_j)=0$ for all $i\neq j.$ What you probably miss is that $\phi_i$ is linear, i.e. it is non-zero on all triangular elements which touch $(x_i,y_i).$

The paper describes a standard Galerkin discretisation method. The integral in the left-hand side of your expression can be evaluated analytically, since the function $f(u)=-u(u-a)(u-b)$ is a polynomial.