I am slogging through FEM self study leaning heavily on the 1st Zienkiewicz text. I think the book is great but I cannot seem wrap my head around switching between the element of interest and the reference element. I found a paper that talks about implementing a code in MatLab. The paper does some hand holding on the assembly but I am still having some gaps. Hopefully the community can give me a helping hand.
So I pick up with the definition of triangular element
$$
\hat{v}_1 = (1,0), \hat{v}_2=(0,1) \:\mathrm{and}\: \hat{v}_3=(0,0)
$$
So the text says we can generate an affine map translating the reference triangle to any triangle in the mesh by say matching up the verities. I am ok there…
$$
F:\hat{\tau}\rightarrow\tau\;\mathrm{where}\\B=\left[\begin{matrix}x_1-x_3& y_1-y_3 \\x_2-x_3 & y_2-y_3 \end{matrix} \right]and \;c=(x_3,y_3)^T
$$
Things go downhill from here when this gets folded back into integrating over the element
$$
\int_\tau \nabla\lambda_i \cdot\nabla\lambda_jdxdy=\int_\tau(B^{-1}\hat{\nabla}\hat{\lambda_i})\cdot(B^{-1}\hat{\nabla}\hat{\lambda_J})|det(B)|d\hat{x}d\hat{y}
$$
So I know the hat is in terms of the reference element and they also state $\hat{u}(\hat{\mathbf{x}})=u(F(\hat{\mathbf{x}}))$ which explains where B gets introduced. The inclusion of the determinant of B confuses me. Where did that come from?
The next line is that
$$
\int_\tau \nabla\lambda_i \cdot\nabla\lambda_jdxdy= \frac{1}{2}|det(B)|(B^{-1}\hat{\nabla}\hat{\lambda_i})\cdot(B^{-1}\hat{\nabla}\hat{\lambda_j})
$$
I am not clear what took place with the integration. I read that barycentric coordinates add up to 1 if they are homogenous, is that what happened?
If someone could fill in some misconceptions it would be much appreciated.
Best Answer
In the first equation you are basically doing change of coordinates, similar to cartesian to e.g. polar. The determinant of the change of variables Jacobian needs to be included, since you are changing from $\boldsymbol{x} $ to $ \widehat{\boldsymbol{x}}$ through $\boldsymbol{x} = B \widehat{\boldsymbol{x}}$. Thus, $$\frac{\mathrm d \boldsymbol{x} }{\mathrm d \widehat{\boldsymbol{x}}} = B \Rightarrow \mathrm d \boldsymbol{x} = B \mathrm d \widehat{\boldsymbol{x}}$$ For the formalization see e.g. this Wikipedia entry for a start, although there are probably lots of better resources out there, like this.
Coming to the second equality, there is no exact integration, but instead Gaussian Quadrature (the standard for FEM), i.e., numerical integration going on. Since only one summand is used, it looks like the order one method is used. This explains also the factor $1/2$ (the weight) in front. All terms are then evaluated at $\hat x = \hat y = 1/3$. This resource might be a good starting point for looking into Gaussian Quadrature.