Suppose I have the PDE:
$3f_{x_1x_1} + 3f_{x_1x_2} + 3f_{x_2x_2} = 0$
And I want to transform this PDE into:
$\alpha_1f_{z_1z_1} + \alpha_2f_{z_2z_2} = 0$
where $(z_1, z_2)^T = A(x_1, x_2)^T$
How can I figure out the matrix $A$ that will accomplish the desired result?
It's an elliptic equation so it's posible write it in its canonical form with a suitable variable change. Following the general idea here I've found the following new variables (constant coefficients help :):
$$z_1=x_1-x_2$$ $$z_2=\frac{x_1}{\sqrt 3}+\frac{x_2}{\sqrt 3}$$
And in matrix notation $$\left(\begin{matrix} z_1\\ z_2 \end{matrix}\right) = \left(\begin{matrix} 1 & -1\\ \frac{1}{\sqrt 3} & \frac{1}{\sqrt 3} \end{matrix}\right) \left(\begin{matrix} x_1\\ x_2 \end{matrix}\right)$$
I've tried to find a more, so said, simetric transformation without success. Anyway, we obtain the two dimension Laplace equation:
$$f_{z_1z_1}+f_{z_2z_2}=0$$
So, if you wish $\alpha_1f_{z_1z_1}+\alpha_2f_{z_2z_2}=0$ you only need to rescale the variables:
$$\left(\begin{matrix} z_1\\ z_2 \end{matrix}\right) = \left(\begin{matrix} \sqrt\alpha_1 & 0\\ 0 & \sqrt\alpha_2 \end{matrix}\right) \left(\begin{matrix} 1 & -1\\ \frac{1}{\sqrt 3} & \frac{1}{\sqrt 3} \end{matrix}\right) \left(\begin{matrix} x_1\\ x_2 \end{matrix}\right)$$ So: $$A= \left(\begin{matrix} \sqrt\alpha_1 & -\sqrt\alpha_1\\ \frac{\sqrt\alpha_2}{\sqrt 3} & \frac{\sqrt\alpha_2}{\sqrt 3} \end{matrix}\right)$$