If I have a set of polynomials defined on six points of a triangle, such that $\phi_i(p_j) = \delta_{ij}$, how do I use an affine transformation to get new polynomials so that $\bar{\phi_i}(\bar{p}_j) = \delta_{ij}$ for the corresponding points in the triangle.
For example, I have a triangle defined by the points { (0, 0), (.5, 0), (1, 0), (.5, .5), (0, 1), (0, .5)}
The polynomial defined for (0, 0) is $1 – 3x + 2x^2 – 3y + 4xy + 2y^2$. Let's say I want to get the corresponding polynomial for a triangle with vertices { (1, 2), (0, 2), (1, 1) }. I know the transformation to map to the new triangle is:
$$ \left[ \begin{array}{ccc}
\bar{x} \\
\bar{y}
\end{array} \right] = \left[ \begin{array}{ccc}
x_2 – x_1 & x_3 – x_1 \\
y_2 – y_1 & x_3 – x_1
\end{array} \right] \left[ \begin{array}{ccc}
x \\
y
\end{array} \right] + \left[ \begin{array}{ccc}
x_1 \\
y_1
\end{array} \right]
$$
How do I use this matrix to get a new set of polynomials?
Best Answer
The transformation map is (I think there is a typo in your question): $$ \begin{pmatrix}\hat{x}\\ \hat{y}\end{pmatrix} = \begin{pmatrix}x_2 - x_1& x_3 - x_1 \\ y_2 - y_1& y_3 - y_1\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} + \begin{pmatrix}x_1\\ y_1\end{pmatrix}. $$ This just means a point $(x,y)$ in the reference triangle will get transformed into the new triangle with orientation preserved. In your case: $$ \begin{pmatrix}\hat{x}\\ \hat{y}\end{pmatrix} = \begin{pmatrix}-1& 0 \\ 0& -1\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} + \begin{pmatrix}1\\ 2\end{pmatrix} = \begin{pmatrix}1 - x\\ 2 - y\end{pmatrix}.\tag{1} $$ The orientation is preserved in the sense that: say $p(x,y)= x$ in the reference triangle the convex hull of $\{(0,0),(1,0),(0,1)\} = \{P_1,P_2,P_3\}$: $$p|_{P_1} = 0,\; p|_{P_2} = 1,\;p_{P_3} = 0,$$ and the new $\hat{p}(x) = 1-x$, in the new triangle $\{(1,2),(0,2),(1,1)\} = \{\hat{P}_1,\hat{P}_2,\hat{P}_3\}$: $$\hat{p}|_{\hat{P}_1} = 0,\; \hat{p}|_{\hat{P}_2} = 1,\;\hat{p}_{\hat{P}_3} = 0.$$ Similar things happes to $y\mapsto 2-y$. Therefore, for $$ f(x,y) = 1 - 3x + 2x^2 - 3y + 4xy + 2y^2, $$ the new polynomial should be: $$ \hat{f}(x,y) = 1 - 3(1-x) + 2(1-x)^2 -3(2-y) + 4(1-x)(2-y) + 2(2-y)^2, $$ and you can check that this new polynomial, preserves the values of the original polynomial on the vertices and midpoints of the reference triangle.