I am trying to find a transformation that maps each internal point of my first quadrilateral domain into each point of my second quadrilateral domain. Please see the following image:
Each domain is defined by four points and my aim is to find a matrix or any other way that enables the transformation of a point of domain 1 (for example a0_1 into a1_1).
This is similar to affine transformations but not quite.
Any hints would be greatly appreciated.
Here's a vectorial approach that will work in $\mathbb R^3$ for any pair of quadrilaterals that are not even co-planar. Let the vertices of each quad be represented by $x-$, $y-$, and $z-$ coordinates such that the position vectors at these points are
$$ \vec{P_{01}} = [x_{01},y_{01},z_{01}] ,\hspace 0.25 cm \vec{P_{02}} = [x_{02},y_{02},z_{02}] ,\hspace 0.25 cm \vec{P_{03}} = [x_{03},y_{03},z_{03}] ,\hspace 0.25 cm \vec{P_{04}} = [x_{04},y_{04},z_{04}] $$
for the first quadrilateral (index $0$), and likewise for the second quadrilateral (index 1)
$$ \vec{P_{11}} = [x_{11},y_{11},z_{11}] ,\hspace 0.25 cm \vec{P_{12}} = [x_{12},y_{12},z_{12}] ,\hspace 0.25 cm \vec{P_{13}} =[x_{13},y_{13},z_{13}] ,\hspace 0.25 cm \vec{P_{14}} = [x_{14},y_{14},z_{14}] $$
The position vector at any given point within the first quadrilateral (index $0$) can be represented by the bilinear parametrisation
$$ \vec{P_{0}} =\vec{A_{0}}+\vec{B_{0}}\xi +\vec{C_{0}}\eta+\vec{D_{0}}\xi\eta $$
where
$$ \vec{A_{0}}=\vec{P_{01}}\\ \vec{B_{0}}=\vec{P_{02}}-\vec{P_{01}}\\ \vec{C_{0}}=\vec{P_{03}}-\vec{P_{01}}\\ \vec{D_{0}}=\vec{P_{04}}-\vec{P_{02}}-\vec{P_{03}}+\vec{P_{01}}\\ $$
The objective is to find the normalised $\xi$ and $\eta$ coordinates corresponding to this point. This is achieved by first defining two vectors $\vec u$ orthogonal to $\vec{D_{0}}$ and $\vec v$ orthogonal to $\vec{C_{0}}$ as follows
$$ \vec u=[\vec{D_{0}} \times \vec{C_{0}}] \times\vec{D_{0}} $$ and $$ \vec v=[\vec{D_{0}} \times \vec{C_{0}}] \times\vec{C_{0}} $$
The inner products of the position vector $\vec{P_{0}}$ with $\vec u$ and $\vec v$ yield the following set of equations for $\xi$ and $\eta$
$$ a=b\xi+c\eta\hspace 2 cm \text{[EQ. 1]}\\ f=g\xi+h\xi\eta\hspace 2 cm \text{[EQ. 2]} $$
where
$$ a=(\vec{P_{0}}-\vec{A_{0}})\cdot \vec u\\ b=\vec{B_{0}}\cdot \vec u\\ c=\vec{C_{0}}\cdot \vec u $$ and $$ f=(\vec{P_{0}}-\vec{A_{0}})\cdot \vec v\\ g=\vec{B_{0}}\cdot \vec v\\ h=\vec{D_{0}}\cdot \vec v $$
From EQ. 1, we have
$$ \eta=\frac{a-b\xi}{c}\hspace 2 cm \text{[EQ. 3]} $$
Substituting EQ. 3 in EQ. 2 yields a quadratic in $\xi$ and only the root lying in the interval $[0,1]$ is relevant. Once this root is calculated, $\eta$ can be calculated from Eq. 3.
The corresponding point $\vec{P_{1}}$ in the second quadrilateral (with index 1) can now be evaluated as
$$ \vec{P_{1}} =\vec{A_{1}}+\vec{B_{1}}\xi +\vec{C_{1}}\eta+\vec{D_{1}}\xi\eta $$
where
$$ \vec{A_{1}}=\vec{P_{11}}\\ \vec{B_{1}}=\vec{P_{12}}-\vec{P_{11}}\\ \vec{C_{1}}=\vec{P_{13}}-\vec{P_{11}}\\ \vec{D_{1}}=\vec{P_{14}}-\vec{P_{12}}-\vec{P_{13}}+\vec{P_{11}}\\ $$
If the first quadrilateral (index $0$) is a rectangle, then $\vec D_0$ (and hence $\vec u$ and $\vec v$) will be the null vector. In this case, set $\vec u=\vec B_0$ and $\vec v=\vec C_0$ whereupon EQs. 1 and 2 will be replaced by a set of simultaneous linear equations in $\xi$ and $\eta$.
Hope this helps.