Using the characteristics to get the canonical form of a pde

Question

I've been asked to consider this parabolic equation.

$ 3\frac{∂^2u}{∂x^2} + 6\frac{∂^2u}{∂x∂y} +3\frac{∂^2u}{∂y^2} - \frac{∂u}{∂x} - 4\frac{∂u}{∂y} + u = 0$

I calculated the characteristic coordinates to be $ξ = y - x, η = x$. The question then asks to transform the equation to the canonical form. I've got the method in other questions but can't seem to work out how to transfer the method from those examples to this one.

Answer 1

Hint

We have$${\partial^2 u\over \partial x^2}={\partial^2 u\over \partial \eta^2}+{\partial^2 u\over \partial ξ^2}-2{\partial^2 u\over \partial \eta\partial ξ}\\{\partial u\over \partial x}={\partial u\over \partial \eta}-{\partial u\over \partial ξ}\\{\partial u\over \partial y}={\partial u\over \partial ξ}\\{\partial^2 u\over \partial y^2}={\partial \over \partial y}{\partial u\over \partial ξ}={\partial^2 u\over \partial ξ^2}\\{\partial^2 u\over \partial x\partial y}={\partial^2 u\over \partial ξ\partial \eta}-{\partial^2 u\over \partial ξ^2}$$Can you finish now?

Answer 2

I'm using subscripts as partials to save time

\begin{align} u_x &= u_\xi\xi_x + u_\eta\eta_x = - u_\xi + u_\eta \\ u_y &= u_\xi\xi_y + u_\eta\eta_y = u_\xi \end{align}

Using the chain rule again

\begin{align} u_{xx} &= (-u_\xi + u_\eta)_x = -u_{\xi\xi}\xi_x - u_{\xi\eta}\eta_x + u_{\eta\xi}\xi_x + u_{\eta\eta}\eta_x = u_{\xi\xi} - 2u_{\xi\eta} + u_{\eta\eta} \\ u_{yy} &= (u_\xi)_y = u_{\xi\xi}\xi_y + u_{\xi\eta}\eta_y = u_{\xi\xi} \\ u_{xy} &= (u_\xi)_x = u_{\xi\xi}\xi_x + u_{\xi\eta}\eta_x = -u_{\xi\xi} + u_{\eta\xi} \end{align}

Combining everything:

$$ 3u_{xx} + 6u_{xy} + 3u_{yy} - u_x - 4u_y + u = 3u_{\eta\eta} - 3u_\xi - u_\eta + u = 0 $$

which is indeed parabolic