- #1

mathmari

Gold Member

MHB

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I have to find the Normal form of the 2.order partial differential equation. I am not sure if my solution is correct..

The differential equation is:

$ u_{xx}-4u_{xy}+4_{yy}-6u_x+12u_y-9u=0$

$a=1, b=-2, c=4$

$b^2-ac=4-4=0 \Rightarrow $ parabolic

$\frac{dy}{dx}=\frac{1}{a}(b \pm \sqrt{b^2-ac})$

$\frac{dy}{dx}=-2 \Rightarrow dy=-2dx \Rightarrow y=-2x+c \Rightarrow \xi=y+2x$

To find $ \eta$ ,we know that the determinant of the Jacobi-Matrix should be different from $0$ .

$J=\begin{vmatrix}

\xi_x & \xi_y\\

\eta_x & \eta_y

\end{vmatrix}=\xi_x \eta_y-\xi_y \eta_x \neq 0$

$ 2 \eta_y-n_x \neq 0$

So we could take $\eta=y$, couldn't we?

In this case it is $ 2 \neq 0$.

$\partial_x=2\partial_{\xi}$

$\partial_y=\partial_{\xi}+\partial_{\eta}$

$\partial_{xx}=4\partial_{\xi \xi}$

$\partial_{yy}=\partial_{\xi \xi}+2 \partial_{\xi \eta}+\partial_{\eta \eta}$

$\partial_{xy}=2\partial_{\xi \xi}+2 \partial_{\xi \eta}$

By replacing this in the differential equation we have:

$u_{\eta \eta}+3u_{\eta}-\frac{9}{4}u=0$

Could you tell me if my result is correct?

Shouldn't it be in the form $ u_{\xi \xi}=f(u,u_{\xi}, u_{\eta})$ ?

My result is in the form $ u_{\eta \eta}=f(u,u_{\xi}, u_{\eta})$.. (Thinking)