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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.

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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

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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?

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  • $\begingroup$ My previous examples seem to suggest that for this question ∂u/∂y = ∂u/∂ξ + ∂u/∂η. Might be me misunderstanding the examples. $\endgroup$ – User19098 Feb 25 at 13:42
  • $\begingroup$ Your $u_x$ is not correct $\endgroup$ – Dylan Feb 25 at 16:11
  • $\begingroup$ @Dylan thanks.... $\endgroup$ – Mostafa Ayaz Feb 25 at 18:34
  • $\begingroup$ @User19098 but we have $${\partial \eta\over \partial y}=0$$ $\endgroup$ – Mostafa Ayaz Feb 25 at 18:34
  • $\begingroup$ I think one of your second derivatives is not right either. $\endgroup$ – Dylan Feb 25 at 19:22

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