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Let's consider the triangle $\Delta$ in $\mathbb{R}^2$ with the vertices $e_1=(0,0), e_2=(1,0)$ and $e_3=(0,1)$. Let $\mathbb{P}$ denote the set of bivariate polynomials of degree $\leq 1$.

In the first step I am asked to find a basis $S_1,S_2,S_3$ of the space $\mathbb{P}$ such that $S_i(e_j)=\delta_{i,j}$ holds for all $i,j=1,2,3$ .
Since each $S_i$ has to be of the form $\alpha+\beta \xi_1 +\gamma \xi_2$ we can just solve the following system of equations:

$\left ( \begin{array}{ccccc} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 &1 \end{array}\right ) \left( \begin{array}{c}\alpha \\ \beta \\ \gamma \end{array}\right ) = \left( \begin{array}{c} 1 \\0 \\0\end{array} \right);\left( \begin{array}{c} 0 \\1 \\0\end{array} \right); \left( \begin{array}{c} 0 \\0 \\1\end{array} \right) $.

The vectors on the right side correspond to $S_1,S_2,S_3$ respectively. That gives me the following solution:
$S_1(\xi_1,\xi_2)=1-\xi_1-\xi_2$
$S_2(\xi_1,\xi_2)=\xi_1$
$S_3(\xi_2,\xi_2)=\xi_2$.

In the next step I would like find basis functions $\phi_1,\phi_2,\phi_3$ for arbitrary vertices $x_1,x_2,x_3$ of the triangle $\mathbb{T}$, i.e. find $\phi_1,\phi_2,\phi_3$ such that $\phi_i(x_j)=\delta_{i,j}$. I was thinking about trying to generalize the result from above using the coordinate transformation

$\Delta \rightarrow \mathbb{P}: x=(\xi_1,\xi_2) \mapsto x_1+\xi_1(x_2-x_1)+\xi_2(x_3-x_1)$.

Is this the right idea? I'm having trouble to rewrite the functions $S_1,S_2,S_3$ in the new coordinate system. Could someone help me with that?

Also, I'm asked to determine the mass matrix $M$ given by \begin{equation} M_{i,j}=\int_{\mathbb{T}} \phi_i \phi_j dx \end{equation} Here I had the same idea as above. I first determined the mass matrix for the fuctions $S_1,S_2,S_3$ which gave: $M_{S_1,S_2,S_3}=\left ( \begin{array}{ccc} 1/12 & 1/24 & 1/24 \\ 1/24 & 1/12 & 1/24 \\ 1/24 & 1/24 &1/12 \end{array}\right ) $

I wonder if there is a way to generalize this result to the functions $\phi_1,\phi_2,\phi_3$ using coordinate transformation?

Andreas804
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1 Answers1

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Introducing $f : \xi \mapsto x_1 + \xi_1(x_2 - x_1) + \xi_2(x_3 - x_1)$ is the good method. Indeed, $\mathbb{P}$ is stable by composition with $f$ and $f^{-1}$ (which exists unless $\mathbb{T}$ is flat) and $f(e_i) = x_i$ for all $i$, so $\phi_i(x_j) = \delta_{ij}$ for all $i,j$ is equivalent to $\phi_i(f(e_j)) = \delta_{ij}$ for all $i,j$, hence $\phi_i \circ f = S_i$ so $\phi_i = S_i \circ f^{-1}$. And for the matrix, you can use a change of variable to deduce that, $$ M_{ij} = \int_{\mathbb{T}} \phi_i\phi_j \, dx = \int_{f(\Delta)} (S_iS_j) \circ f^{-1} \, dx = \int_\Delta S_iS_j|\det(df)| \, d\xi. $$ Now, notice that $\det(df) = \det(x_2 - x_1,x_3 - x_1)$ is constant so $M = 2AM^0$ where $M^0$ is the matrix you obtain with the $S_i$ and $A = \frac{1}{2}|\det(x_2 - x_1,x_3 - x_1)|$ is the area of $\mathbb{T}$.

Cactus
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  • Thank you! Could I do the same if I had to evaluate $A:=\int_{\mathbb{T}}\nabla \phi_i \nabla \phi_j dx=\int_{\mathbb{T}}\nabla ( S_i f^{-1} )\nabla (S_j f^{-1} )dx$. Can I put $f^{-1}$ outside of the gradient? How would $A$ differ from $A^0$? – Andreas804 Jun 12 '23 at 14:47
  • I'm not sure it is gonna work this time because $\nabla(S_i \circ f^{-1})(x) = d(f^{-1})(x)^\top\nabla S_i(f^{-1}(x)) = d(f^{-1})^\top\nabla S_i(f^{-1}(x))$ ($d(f^{-1})$ is constant). Therefore, $\nabla(S_i f^{-1})(x) \cdot \nabla(S_j f^{-1})(x) = \nabla S_i(f^{-1})(x)^\top d(f^{-1})^\top d(f^{-1})\nabla S_j(f^{-1}(x))$ so you can do again the same change of variable but I don't think you can get rif of the matrix $d(f^{-1})^\top d(f^{-1})$ that appears in the middle of your integral. – Cactus Jun 13 '23 at 07:48
  • Thanks! Do you mean the Jacobian of $f^{-1}$ by $d(f^{-1})$ ?And do I get a scalar at the end if there is a matrix in my integral? – Andreas804 Jun 13 '23 at 14:20
  • Yes it is the Jacobian, and yes you obtain a scalar because $\nabla S_i(f^{-1}(x))^T$ is $1 \times 3$, $d(f^{-1})^\top d(f^{-1})$ is $3 \times 3$ and $\nabla S_j(f^{-1}(x))$ is $3 \times 1$. – Cactus Jun 13 '23 at 14:41
  • But this means I can't put the Jacobian outside of the integral even though it's constant, right? – Andreas804 Jun 13 '23 at 14:43
  • Yes, it would have been possible if it were not in the middle of the intergand – Cactus Jun 13 '23 at 16:42