D.5 2D coordinate transformation derivation

This note gives a more detailed description where the expressions (18.22) for the coefficients $a'$, $b'$, $c'$ and $d'$ comes from.

In terms of the notations for the general case, you have

$$a=a_{11} \qquad b=a_{12}=a_{21} \qquad c=a_{22} \qquad x=x_1 \qquad y=x_2$$

and

$$a'=a'_{11} \qquad b'=a'_{12}=a'_{21} \qquad c'=a'_{22} \qquad \xi=x_1 \qquad \eta=x_2$$

The introduction noted that the new coefficients can be found from

$$a'_{kl} = \sum_{i=1}^n \sum_{j=1}^n a_{ij} \frac{\part... ..._i\partial x_j} \right) \frac{\partial u}{\partial \xi_k}$$

As an example, let’s find the value for $a'$. In terms of the notations in the general case, $a'$ is the 1,1 element of matrix $A'$: $a'$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a'_{11}$. To get $a'$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a'_{11}$, put $k$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 in

$$a'_{kl} = \sum_{i=1}^n \sum_{j=1}^n a_{ij} \frac{\partial \xi_k}{\partial x_i} \frac{\partial \xi_l}{\partial x_j}.$$

That turns $\xi_k$ and $\xi_l$ into $\xi_1$, or $\xi$ for short:

$$a' = a'_{11} = \sum_{i=1}^2 \sum_{j=1}^2 a_{ij} \frac{\partial\xi}{\partial x_i} \frac{\partial\xi}{\partial x_j}$$

If we write out the four terms of the double sum explicitly, that becomes:

$$a' = a_{11}\frac{\partial\xi}{\partial x_1} \frac{\parti... ...c{\partial\xi}{\partial x_2} \frac{\partial\xi}{\partial x_2}$$

Now note that by definition $a_{11}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a$, $a_{12}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a_{21}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $b$, $a_{22}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $c$, $x_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $x$, and $x_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $y$, and you get the expression for $a'$ claimed:

$$a' = a \left(\xi_x\right)^2 + 2b \left(\xi_x\right)\left(\xi_y\right) + c \left(\xi_y\right)^2$$

The expressions for $b'$, $c'$ and $d'$ may be verified similarly.