Parabolic 2. order partial differential equation

[mathmari]

Hey!

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)

 

[I like Serena]

Hey!

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)

Hey!

The calculations look good. :)

But I think you should assign the constant equation to the 2nd variable $\eta$.
As a result $\eta = C$ will correspond to the tangent line.

Consider that for a regular upright parabola $y=x^2$ the tangent line at the top is $y=0$.

You might use for instance:
\begin{cases}
\xi = x \\
\eta = 2x + y
\end{cases}

 

[mathmari]

But I think you should assign the constant equation to the 2nd variable $\eta$.
As a result $\eta = C$ will correspond to the tangent line.

Consider that for a regular upright parabola $y=x^2$ the tangent line at the top is $y=0$.

You might use for instance:
\begin{cases}
\xi = x \\
\eta = 2x + y
\end{cases}

Using \begin{cases}
\xi = x \\
\eta = 2x + y
\end{cases} I found the following:$\partial_x=\partial_{\xi}+2 \partial_{\eta}$

$\partial_y=\partial_{\eta}$

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

$\partial_{yy}=\partial_{\eta \eta}$

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

So substituting these at the differential equation we get:
$$u_{\xi \xi}-6u_{\xi}-9u=0$$

That is the form that it should be! (Yes)

Thank you for your help! (Smirk)