How Can You Reduce This PDE to Its Canonical Form?

[Julio1]

Let the PDE $u_{xx}-4u_{xy}+4u_{yy}=0.$ Reduce to the canonical form.Good Morning MHB :). My problem is find the canonical form of the PDE know an variable change. But how I can transform the equation? Thanks.

 

[evinda]

Let the PDE $u_{xx}-4u_{xy}+4u_{yy}=0.$ Reduce to the canonical form.Good Morning MHB :). My problem is find the canonical form of the PDE know an variable change. But how I can transform the equation? Thanks.

(Wave)

$u_{xx}-4u_{xy}+4u_{yy}=0 \Rightarrow \left(\frac{\partial^2}{\partial{x}^2} -4 \frac{\partial^2}{\partial{xy}}+4 \frac{\partial^2}{\partial{y}^2} \right) u=0 $

Find the discriminant of $\frac{\partial^2}{\partial{x}^2} -4 \frac{\partial^2}{\partial{xy}}+4 \frac{\partial^2}{\partial{y}^2}=0$ in order to find the characteristic equations.

 

[Julio1]

(Wave)

$u_{xx}-4u_{xy}+4u_{yy}=0 \Rightarrow \left(\frac{\partial^2}{\partial{x}^2} -4 \frac{\partial^2}{\partial{xy}}+4 \frac{\partial^2}{\partial{y}^2} \right) u=0 $

Find the discriminant of $\frac{\partial^2}{\partial{x}^2} -4 \frac{\partial^2}{\partial{xy}}+4 \frac{\partial^2}{\partial{y}^2}=0$ in order to find the characteristic equations.

Thanks evinda :).

But why the discriminant? I don't understand this :(.

One question: What's up if does the change $\xi=\xi(x,y)$ and $\eta=\eta(x,y)$?

 

[evinda]

Thanks evinda :).

But why the discriminant? I don't understand this :(.

One question: What's up if does the change $\xi=\xi(x,y)$ and $\eta=\eta(x,y)$?

You will set $\xi$ and $\eta$ equal to the solutions of the equation of which you find the discriminant.

 

[Julio1]

Thanks Evinda, I can solve in this form...

As $y(x)=-2x+C$ is the characteristic curve, we have that $y+2x=C.$ Let $\xi(x,y)=y+2x$ and $\eta=x$, it follow that:

$\dfrac{\partial \xi}{\partial x}=2, \quad \dfrac{\partial \eta}{\partial x}=1.$

$\dfrac{\partial \xi}{\partial y}=1, \quad \dfrac{\partial \eta}{\partial y}=0.$

Then,

$\dfrac{\partial}{\partial x}=\dfrac{\partial}{\partial \xi}\dfrac{\partial \xi}{\partial x}+\dfrac{\partial}{\partial \eta}\dfrac{\partial \eta}{\partial x}=2\dfrac{\partial}{\partial \xi}+\dfrac{\partial}{\partial \eta}$

$\dfrac{\partial}{\partial y}=\dfrac{\partial}{\partial \xi}\dfrac{\partial \xi}{\partial y}+\dfrac{\partial}{\partial \eta}\dfrac{\partial \eta}{\partial y}=\dfrac{\partial}{\partial \xi}.$

$\begin{eqnarray*}
\dfrac{\partial^2}{\partial x^2}&=&\dfrac{\partial}{\partial x}\left(\dfrac{\partial}{\partial x}\right)\\
&=&\left(2\dfrac{\partial}{\partial \xi}+\dfrac{\partial}{\partial \eta}\right) \left(2\dfrac{\partial}{\partial \xi}+\dfrac{\partial}{\partial \eta}\right)\\
&=&4\dfrac{\partial^2}{\partial \xi^2}+4\dfrac{\partial^2}{\partial \eta\partial \xi}+\dfrac{\partial^2}{\partial \eta^2}.
\end{eqnarray*}
$

$\dfrac{\partial^2}{\partial y^2}=\dfrac{\partial^2}{\partial \xi^2}$

$\dfrac{\partial^2}{\partial y\partial x}=2\dfrac{\partial^2}{\partial \xi^2}+\dfrac{\partial^2}{\partial \xi \partial \eta}$

Thus, we have that:

$\left(\dfrac{\partial^2}{\partial x^2}-4\dfrac{\partial^2}{\partial y\partial x}+4\dfrac{\partial^2}{\partial y^2}\right)u=0\implies u_{\eta \eta}=0.$

This is correct? :)

 

[Julio1]

Since nobody ratified if I was okay, I give terminate.

Thanks :).