How can I compute the difference between shape function and dual solution in dwr?

Question

I am trying to find error estimation based on weighted residual technique:

$$Q(e_h)=\sum_{k \in {\cal T}}\eta_k, $$ where $$\eta_k=\int_kR(z-v_h)d\Omega+\int_{\partial k}J(z-v_h)ds,$$ or in the weak form $$\eta_k=\int_k l(z-v_h)d\Omega+\int_k \nabla u_h.\nabla(z-v_h)d\Omega,$$ where $u_h$ is the solution of primal problem $a(u,v)=l(v),$ and $z$ is the solution of the dual problem $a(z,v)=Q(v),$ also $Q(v)=|\Omega|^{(-1)}(1,v),$ where $v$ is the test function. Moreover $R$ and $J$ are residual and jump respectively. Using approximation $z_h=\sum_i \phi_i*z_i$ (instead of z) at Gaussian quadrature point leads to a scalar where using $\phi$ as test function at any point gives a vector, but $(z-v_h)$ should be a scalar not a vector. How can I solve this problem?

Answer 1

You seem to be confused about what is a vector and what is a scalar. $z_h$ and $\phi_i$ are both scalars if evaluates at quadrature points. The fact that the $z_i$ form a vector really has no relevance since you are computing integrals which you approximate via quadrature for which you need $z_h(x)$, not the individual components $z_i$ of the solution vector $Z$.

I'm going to add something to the answer that you did not ask, but should know anyway. If you approximate $z$ by a finite element solution $z_h$ that uses the same finite element space as $u_h$, then you will be out of luck because you could then choose $v_h=z_h$ and your entire error estimate would be zero. In other words, approximating $z$ by $z_h$ is not a good choice.

There are numerous solutions to this problem. I might in fact refer you to the book about dual weighted residual error estimators, namely W. Bangerth and R. Rannacher, "Adaptive Finite Element Methods for Solving Differential Equations", Birkhäuser, 2003. In the interest of full disclosure, I will mention the obvious: I'm the first author of this book.