MHB Normal form of the differential equation

[mathmari]

Hey!

I have the following exercise:
Write in the normal form the differential equation
$$u_{xx}+\frac{2y}{x}u_{xy}+\frac{y^2}{x^2}[(1+y^2)u_{yy}+2yu_y]=0$$
Hint: You can suppose that the one new variable is given by $\xi=x$

I have done the following:

$a=1, b=\frac{y}{x}, c=\frac{y^2}{x^2}(1+y^2)$

$b^2-ac=\frac{y^2}{x^2}-\frac{y^2}{x^2}(1+y^2)=-\frac{y^4}{x^2}=(\frac{y^2}{x}i)^2<0$

So the differential equation is elliptic.

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

$\frac{dy}{dx}=\frac{y}{x} \pm \frac{y^2}{x}i$

• $$\frac{dy}{dx}=\frac{y+y^2i}{x} \Rightarrow \frac{dy}{y(1+yi)}=\frac{dx}{x} \Rightarrow \frac{1}{y(1+yi)}=\frac{1}{x}+c_1 \Rightarrow \frac{1}{y(1+y^2)}-\frac{1}{x}-\frac{1}{1+y^2}i=c_1$$
• $$\frac{dy}{dx}=\frac{y-y^2i}{x} \Rightarrow \frac{dy}{y(1+yi)}=\frac{dx}{x} \Rightarrow \frac{1}{y(1-yi)}=\frac{1}{x}+c_2 \Rightarrow \frac{1}{y(1+y^2)}-\frac{1}{x}+\frac{1}{1+y^2}i=c_2$$

Is it correct so far?

How can I use the hint?

 

[I like Serena]

Hey!

I have the following exercise:
Write in the normal form the differential equation
$$u_{xx}+\frac{2y}{x}u_{xy}+\frac{y^2}{x^2}[(1+y^2)u_{yy}+2yu_y]=0$$
Hint: You can suppose that the one new variable is given by $\xi=x$

Hi! :)

To be honest, I'm not really familiar with elliptic partial differential equations.
And Google isn't helping much.
But let's see what we can do.

$$\frac{dy}{dx}=\frac{y+y^2i}{x} \Rightarrow \frac{dy}{y(1+yi)}=\frac{dx}{x} \Rightarrow \frac{1}{y(1+yi)}=\frac{1}{x}+c_1 \Rightarrow \frac{1}{y(1+y^2)}-\frac{1}{x}-\frac{1}{1+y^2}i=c_1$$

This does not look right.
If you integrate left and right, you should have:
\begin{aligned}
\frac{dy}{y(1+yi)}=\frac{dx}{x}
&\Rightarrow \frac{dy}{y} - \frac{i\ dy}{1+yi}=\frac{dx}{x} \\
&\Rightarrow \ln y - \ln(1+yi)=\ln x + \text{Constant} \\
&\Rightarrow \frac y {1+yi}=c_1 x
\end{aligned}

As for the hint, I think you are supposed to switch coordinates from $(x,y)$ to $(\xi,\eta)$.
That is, introduce functions $\xi(x,y)$ and $\eta(x,y)$, that are intended to normalize to the form:
$$u_{\xi\xi} + u_{\eta\eta} + \text{lower order terms} = 0$$

 

[mathmari]

This does not look right.
If you integrate left and right, you should have:
\begin{aligned}
\frac{dy}{y(1+yi)}=\frac{dx}{x}
&\Rightarrow \frac{dy}{y} - \frac{i\ dy}{1+yi}=\frac{dx}{x} \\
&\Rightarrow \ln y - \ln(1+yi)=\ln x + \text{Constant} \\
&\Rightarrow \frac y {1+yi}=c_1 x
\end{aligned}

Oh yes.. (Tmi) you're right! (Smile)

As for the hint, I think you are supposed to switch coordinates from $(x,y)$ to $(\xi,\eta)$.
That is, introduce functions $\xi(x,y)$ and $\eta(x,y)$, that are intended to normalize to the form:
$$u_{\xi\xi} + u_{\eta\eta} + \text{lower order terms} = 0$$

At an other exercise I was also supposed to switch coordinates from $(x,y)$ to $(\xi,\eta)$, and I found the new variables as followed:
$\cdots$
$\frac{dy}{dx}=\frac{3}{2} \pm \frac{1}{2}i$
$y-(\frac{3}{2}+\frac{1}{2}i)x=a$
$y-(\frac{3}{2}-\frac{1}{2}i)x=b$

So $\xi =Re(a)=Re(b) \Rightarrow \xi=y-\frac{3}{2}x$
and $\eta=Im(b) \Rightarrow \eta =\frac{1}{2}x$

But in this case, $\frac y {1+yi}=c_1 x \Rightarrow c_1=\frac{y}{x(1+y^2)}-\frac{y^2}{x(1+y^2)}i$.

$\frac{dy}{y(1-yi)}=\frac{dx}{x} \Rightarrow (\frac{1}{y}+\frac{i}{1-yi})dy=\frac{1}{x}dx \Rightarrow lny+ln(1-yi)=lnx+c_2 \Rightarrow ln (y(1-yi))=lnx+c_2 \Rightarrow y(1-yi)=xc_2 \Rightarrow \frac{y}{x}-\frac{y^2}{x}i=c_2$.
But we cannot take $\xi=Re(c_1)=Re(c_2)$, since $Re(c_1)=\frac{y}{x(1+y^2)} \text{ and }Re(c_2)=\frac{y}{x}$.
Or have I done something wrong with the calculations?? (Thinking)

 

[I like Serena]

$\frac{dy}{y(1-yi)}=\frac{dx}{x} \Rightarrow (\frac{1}{y}+\frac{i}{1-yi})dy=\frac{1}{x}dx \Rightarrow lny+ln(1-yi)=lnx+c_2 \Rightarrow ln (y(1-yi))=lnx+c_2 \Rightarrow y(1-yi)=xc_2 \Rightarrow \frac{y}{x}-\frac{y^2}{x}i=c_2$.
But we cannot take $\xi=Re(c_1)=Re(c_2)$, since $Re(c_1)=\frac{y}{x(1+y^2)} \text{ and }Re(c_2)=\frac{y}{x}$.
Or have I done something wrong with the calculations?? (Thinking)

Hmm, maybe you've got a $+$ and $-$ mixed up? (Wasntme)

 

[mathmari]

Hmm, maybe you've got a $+$ and $-$ mixed up? (Wasntme)

(Nod) (Blush)

$\frac{dy}{y(1-yi)}=\frac{dx}{x} \Rightarrow (\frac{1}{y}+\frac{i}{1-yi})dy=\frac{1}{x}dx \Rightarrow lny-ln(1-yi)=lnx+c_2 \Rightarrow ln (\frac{y}{1-yi})=lnx+c_2 \Rightarrow \frac{y}{1-yi}=xc_2 \Rightarrow \frac{y(1+yi)}{1+y^2}=xc_2 \Rightarrow \frac{y}{x(1+y^2)}+\frac{y^2}{x(1+y^2)}i=c_2$.

So now $Re(c_1)=Re(c_2)= \frac{y}{x(1+y^2)}$.

But I haven't understood yet how to use the hint.. (Doh)

 

[I like Serena]

(Nod) (Blush)

$\frac{dy}{y(1-yi)}=\frac{dx}{x} \Rightarrow (\frac{1}{y}+\frac{i}{1-yi})dy=\frac{1}{x}dx \Rightarrow lny-ln(1-yi)=lnx+c_2 \Rightarrow ln (\frac{y}{1-yi})=lnx+c_2 \Rightarrow \frac{y}{1-yi}=xc_2 \Rightarrow \frac{y(1+yi)}{1+y^2}=xc_2 \Rightarrow \frac{y}{x(1+y^2)}+\frac{y^2}{x(1+y^2)}i=c_2$.

So now $Re(c_1)=Re(c_2)= \frac{y}{x(1+y^2)}$.

Yep.

But I haven't understood yet how to use the hint.. (Doh)

Your method finds expressions that are way too complicated.

Let's go back to basics.
Suppose we pick $\xi = x$ and we accept that we don't know $\eta$ yet.

Then we can set up the equations referring to $\eta_x, \eta_y, \eta_{xx}, \eta_{xy}, \eta_{yy}$ such that the $u_{\xi\eta}$ term disappears and the $u_{\eta\eta}$ gets a coefficient of 1.
From there we can find 2 equations with only $\eta_x$ and $\eta_y$.

Solving it leads to a simple expression for $\eta$. (Mmm)

 

[mathmari]

Let's go back to basics.
Suppose we pick $\xi = x$ and we accept that we don't know $\eta$ yet.

Then we can set up the equations referring to $\eta_x, \eta_y, \eta_{xx}, \eta_{xy}, \eta_{yy}$ such that the $u_{\xi\eta}$ term disappears and the $u_{\eta\eta}$ gets a coefficient of 1.
From there we can find 2 equations with only $\eta_x$ and $\eta_y$.

Solving it leads to a simple expression for $\eta$. (Mmm)

So I will try it...

For $\xi =x$ we have the following:

$$\partial_x=\partial_{\xi}+{\eta}_x \partial_{\eta}$$

$$\partial_y={\eta}_y \partial_{\eta}$$

$$\partial_{xx}=\partial_{\xi \xi}+2 {\eta}_x\partial_{\xi \eta}+({\eta}_x)^2 \partial_{\eta \eta}$$

$$\partial_{yy}=({\eta}_y)^2 \partial_{\eta \eta}$$

$$\partial_{xy}={\eta}_y \partial_{\xi \eta}+{\eta}_x {\eta}_y\partial_{\eta \eta}$$

Substituting these at the differential equation we get:

$$(\partial_{\xi \xi}+(2 {\eta}_x+\frac{2y}{x} \eta_y)\partial_{\xi \eta}+(({\eta}_x)^2+ \frac{ 2y }{ x } {\eta}_x {\eta}_y+\frac{y^2}{x^2}(1+y^2)({\eta}_y)^2) \partial_{\eta \eta}+\frac{2y^3}{x^2} {\eta}_y \partial_{\eta})u=0$$

It should be: $$2 {\eta}_x+\frac{2y}{x} {\eta}_y=0 \Rightarrow x n_x=-y n_y$$

So we could take: $n_x=-y$ and $n_y=x$, couldn't we?

$n_x=-y \Rightarrow n=-yx+g(y) \Rightarrow n_y=-x+g'(y)=x \Rightarrow g'(y)=2x \Rightarrow g(y)=2xy+c \text{ for } c=0 \text{ we have } g(y)=2xy \Rightarrow n=-yx+2xy \Rightarrow n=xy$

So,
$$\partial_x=\partial_{\xi}+y \partial_{\eta}$$

$$\partial_y=x\partial_{\eta}$$

$$\partial_{xx}=\partial_{\xi \xi}+2 y\partial_{\xi \eta}+y^2 \partial_{\eta \eta}$$

$$\partial_{yy}=x^2 \partial_{\eta \eta}$$

$$\partial_{xy}=x \partial_{\xi \eta}+xy\partial_{\eta \eta}$$And the differential equation becomes:
$$(\partial_{\xi \xi}+(y^2+\frac{2y}{x} yx+\frac{y^2}{x^2}(1+y^2)x^2)\partial_{\eta \eta}+\frac{2y^3}{x^2}x\partial_{\eta})u=0 \Rightarrow $$

$$(\partial_{\xi \xi}+(4y^2+y^4)\partial_{\eta \eta}+\frac{2y^3}{x}\partial_{\eta})u=0$$

Do I have to substitute $x=\xi$ and $y=\frac{\eta}{\xi}$?

So it would be:

$$(\partial_{\xi \xi}+(4(\frac{\eta}{\xi})^2+(\frac{\eta}{\xi})^4) \partial_{\eta \eta}+\frac{2{\eta}^3}{{\xi}^4}\partial_{\eta})u=0 \Rightarrow u_{\xi \xi}+(4(\frac{\eta}{\xi})^2+(\frac{\eta}{\xi})^4) u_{\eta \eta}+\frac{2{\eta}^3}{{\xi}^4}u_{\eta}=0$$

Is this correct?

 

[I like Serena]

So I will try it...

For $\xi =x$ we have the following:

$$\partial_x=\partial_{\xi}+{\eta}_x \partial_{\eta}$$

$$\partial_y={\eta}_y \partial_{\eta}$$

$$\partial_{xx}=\partial_{\xi \xi}+2 {\eta}_x\partial_{\xi \eta}+({\eta}_x)^2 \partial_{\eta \eta}$$

$$\partial_{yy}=({\eta}_y)^2 \partial_{\eta \eta}$$

$$\partial_{xy}={\eta}_y \partial_{\xi \eta}+{\eta}_x {\eta}_y\partial_{\eta \eta}$$

Substituting these at the differential equation we get:

$$(\partial_{\xi \xi}+(2 {\eta}_x+\frac{2y}{x} \eta_y)\partial_{\xi \eta}+(({\eta}_x)^2+ \frac{ 2y }{ x } {\eta}_x {\eta}_y+\frac{y^2}{x^2}(1+y^2)({\eta}_y)^2) \partial_{\eta \eta}+\frac{2y^3}{x^2} {\eta}_y \partial_{\eta})u=0$$

It should be: $$2 {\eta}_x+\frac{2y}{x} {\eta}_y=0 \Rightarrow x n_x=-y n_y$$

Yep.
That's also what I have. (Nod)

So we could take: $n_x=-y$ and $n_y=x$, couldn't we?

$n_x=-y \Rightarrow n=-yx+g(y) \Rightarrow n_y=-x+g'(y)=x \Rightarrow g'(y)=2x \Rightarrow g(y)=2xy+c \text{ for } c=0 \text{ we have } g(y)=2xy \Rightarrow n=-yx+2xy \Rightarrow n=xy$

Hmm, something is not right here.
If we pick $\eta = xy$ we get $\eta_x = y$, but that does not match! (Sadface)

So,
$$\partial_x=\partial_{\xi}+y \partial_{\eta}$$

$$\partial_y=x\partial_{\eta}$$

$$\partial_{xx}=\partial_{\xi \xi}+2 y\partial_{\xi \eta}+y^2 \partial_{\eta \eta}$$

$$\partial_{yy}=x^2 \partial_{\eta \eta}$$

$$\partial_{xy}=x \partial_{\xi \eta}+xy\partial_{\eta \eta}$$And the differential equation becomes:
$$(\partial_{\xi \xi}+(y^2+\frac{2y}{x} yx+\frac{y^2}{x^2}(1+y^2)x^2)\partial_{\eta \eta}+\frac{2y^3}{x^2}x\partial_{\eta})u=0 \Rightarrow $$

$$(\partial_{\xi \xi}+(4y^2+y^4)\partial_{\eta \eta}+\frac{2y^3}{x}\partial_{\eta})u=0$$

Here we have an extra confirmation that something is wrong.
The coefficient of $\partial_{\eta \eta} \ne 1$. (Sweating)

That's actually the extra equation that needs to be satisfied with the choice for $\eta$.

 

[mathmari]

Hmm, something is not right here.
If we pick $\eta = xy$ we get $\eta_x = y$, but that does not match! (Sadface)

Could you give me a hint how to find $\eta$? I got stuck right now.. (Worried)

 

[I like Serena]

Substituting these at the differential equation we get:
$$(\partial_{\xi \xi}+(2 {\eta}_x+\frac{2y}{x} \eta_y)\partial_{\xi \eta}+(({\eta}_x)^2+ \frac{ 2y }{ x } {\eta}_x {\eta}_y+\frac{y^2}{x^2}(1+y^2)({\eta}_y)^2) \partial_{\eta \eta}+\frac{2y^3}{x^2} {\eta}_y \partial_{\eta})u=0$$

It should be: $$2 {\eta}_x+\frac{2y}{x} {\eta}_y=0 \Rightarrow x n_x=-y n_y$$

Could you give me a hint how to find $\eta$? I got stuck right now.. (Worried)

You can write that equation as:
$$\eta_y = - \frac x y \eta_x$$

Since the coefficient of $\partial_{\eta \eta}$ should be 1, you get the additional equation:
$$({\eta}_x)^2+ \frac{ 2y }{ x } {\eta}_x {\eta}_y+\frac{y^2}{x^2}(1+y^2)({\eta}_y)^2 = 1$$

Substitute $\eta_y$? (Wondering)

 

[mathmari]

For $\xi =x$ we have the following:

$$\partial_x=\partial_{\xi}+{\eta}_x \partial_{\eta}$$

$$\partial_y={\eta}_y \partial_{\eta}$$

$$\partial_{xx}=\partial_{\xi \xi}+2 {\eta}_x\partial_{\xi \eta}+({\eta}_x)^2 \partial_{\eta \eta}$$

$$\partial_{yy}=({\eta}_y)^2 \partial_{\eta \eta}$$

$$\partial_{xy}={\eta}_y \partial_{\xi \eta}+{\eta}_x {\eta}_y\partial_{\eta \eta}$$

Substituting these at the differential equation we get:

$$(\partial_{\xi \xi}+(2 {\eta}_x+\frac{2y}{x} \eta_y)\partial_{\xi \eta}+(({\eta}_x)^2+ \frac{ 2y }{ x } {\eta}_x {\eta}_y+\frac{y^2}{x^2}(1+y^2)({\eta}_y)^2) \partial_{\eta \eta}+\frac{2y^3}{x^2} {\eta}_y \partial_{\eta})u=0$$

It should be:
$$2 {\eta}_x+\frac{2y}{x} {\eta}_y=0 \Rightarrow x n_x=-y n_y \Rightarrow \eta_y = - \frac x y \eta_x$$
and since the coefficient of $\partial_{\eta \eta}$ should be 1:
$$({\eta}_x)^2+ \frac{ 2y }{ x } {\eta}_x {\eta}_y+\frac{y^2}{x^2}(1+y^2)({\eta}_y)^2 = 1$$

So we have the following:

$$({\eta}_x)^2+ \frac{ 2y }{ x } {\eta}_x (- \frac x y \eta_x)+\frac{y^2}{x^2}(1+y^2)(- \frac x y \eta_x)^2 = 1 \Rightarrow ({\eta}_x)^2 -2 {\eta}_x^2 +{\eta}_x^2+y^2 {\eta}_x^2 = 1 \Rightarrow y^2 {\eta}_x^2=1 \Rightarrow y \eta_x= \pm 1 \Rightarrow \eta_x=\pm \frac{1}{y} \Rightarrow \eta= \pm \frac{x}{y}+g(y) \Rightarrow \eta_y= \mp \frac{x}{y^2}+g'(y)=\mp \frac{x}{y^2} \Rightarrow g'(y)=0 \Rightarrow g(y)=c \text{ let } c=0$$

So $\eta=\pm \frac{x}{y}$. So let $\eta = \frac{x}{y}$.

So the differential equation becomes:

$$(\partial_{\xi \xi}+ \partial_{\eta \eta}-\frac{2y^3}{x^2} \frac{x}{y^2}\partial_{\eta})u=0 \Rightarrow (\partial_{\xi \xi}+ \partial_{\eta \eta}-\frac{2y}{x} \partial_{\eta})u=0 \Rightarrow (\partial_{\xi \xi}+ \partial_{\eta \eta}-\frac{2}{n} \partial_{\eta})u=0 \Rightarrow $$ $$u_{\xi \xi}+ u_{\eta \eta}-\frac{2}{n} u_{\eta}=0$$

Is it now correct??
And is also the way I formulated it correct??

 

[I like Serena]

Is it now correct??

There is a smiley missing. (Crying)

And is also the way I formulated it correct??

It looks good!

Still, there seem to be a couple of lower order terms missing.

$$\partial_{y}={\eta}_y \partial_{\eta}$$
$$\partial_{yy}=({\eta}_y)^2 \partial_{\eta \eta}$$

What happened to the $\eta_{yy}$ term?
As long as you're only analyzing the 2nd order derivatives of $u$, I guess you can choose to leave them out, but they should be there in the final expression.

For instance, it seems to me it should be:
$$\partial_{yy}=\partial_{\eta \eta}\eta_y^2+\partial_\eta \eta_{yy}$$

 

[mathmari]

There is a smiley missing. (Crying)

(Worried)

What happened to the $\eta_{yy}$ term?
As long as you're only analyzing the 2nd order derivatives of $u$, I guess you can choose to leave them out, but they should be there in the final expression.

For instance, it seems to me it should be:
$$\partial_{yy}=\partial_{\eta \eta}\eta_y^2+\partial_\eta \eta_{yy}$$

I calculated this as followed:
$\frac{\partial}{\partial{y}}=\frac{\partial{\xi}}{\partial{y}} \frac{\partial}{\partial{\xi}}+\frac{\partial{\eta}}{\partial{y}} \frac{\partial}{\partial{\eta}}=\frac{ \partial{\eta}}{\partial{y}} \frac{\partial}{\partial{\eta}} \Rightarrow \partial_y=\eta_y \partial_{\eta}$

$\partial_{yy}=\partial_y \partial_y=(\eta_y)^2 \partial_{\eta \eta}$You're right! (Yes)
At an other exercise I could calculate this in the above way because $\eta_y$ was a constant, but in this case it depends from $y$.

So the differential becomes:
$$(\partial_{xx}+\frac{2y}{x}\partial_{xy}+\frac{y^2}{x^2}(1+y^2)\partial_{yy}+\frac{2y^3}{x^2} \partial_y)u=0 \Rightarrow

(\partial_{\xi \xi}+2 {\eta}_x\partial_{\xi \eta}+({\eta}_x)^2 \partial_{\eta \eta}+\frac{2y}{x}({\eta}_y \partial_{\xi \eta}+{\eta}_x {\eta}_y\partial_{\eta \eta})+\frac{y^2}{x^2}(1+y^2)(\partial_{\eta \eta}\eta_y^2+\partial_\eta \eta_{yy})+\frac{2y^3}{x^2}{\eta}_y \partial_{\eta})u=0 \Rightarrow

(\partial_{\xi \xi} +(2 {\eta}_x+\frac{2y}{x}{\eta}_y)\partial_{\xi \eta}+(\frac{2y}{x}{\eta}_x {\eta}_y+\frac{y^2}{x^2}(1+y^2)\eta_y^2+\eta_x^2) \partial_{\eta \eta}+(\frac{y^2}{x^2}(1+y^2)\eta_{yy}+\frac{2y^3}{x^2}{\eta}_y)\partial_{\eta})u=0$$It should be:
$2 {\eta}_x+\frac{2y}{x}{\eta}_y=0$
and
$\frac{2y}{x}{\eta}_x {\eta}_y+\frac{y^2}{x^2}(1+y^2)\eta_y^2+\eta_x^2=1$

So
$$(\partial_{\xi \xi} + \partial_{\eta \eta}+(\frac{y^2}{x^2}(1+y^2)\eta_{yy}+\frac{2y^3}{x^2}{\eta}_y)\partial_{\eta})u=0 \Rightarrow (\partial_{\xi \xi} + \partial_{\eta \eta}+\frac{2}{xy}\partial_{\eta})u=0 \Rightarrow (\partial_{\xi \xi} + \partial_{\eta \eta}+\frac{2}{\frac{\xi^2}{\eta}}\partial_{\eta})u=0 \Rightarrow (\partial_{\xi \xi} + \partial_{\eta \eta}+\frac{2 \eta}{{\xi^2}}\partial_{\eta})u=0 \Rightarrow u_{\xi \xi} + u_{\eta \eta}+\frac{2 \eta}{{\xi^2}}u_{\eta}=0$$

How is it now?? (Thinking)

 

[I like Serena]

So the differential becomes:
$$(\partial_{xx}+\frac{2y}{x}\partial_{xy}+\frac{y^2}{x^2}(1+y^2)\partial_{yy}+\frac{2y^3}{x^2} \partial_y)u=0 $$

$$\Rightarrow (\partial_{\xi \xi}+2 {\eta}_x\partial_{\xi \eta}+({\eta}_x)^2 \partial_{\eta \eta}+\frac{2y}{x}({\eta}_y \partial_{\xi \eta}+{\eta}_x {\eta}_y\partial_{\eta \eta})+\frac{y^2}{x^2}(1+y^2)(\partial_{\eta \eta}\eta_y^2+\partial_\eta \eta_{yy})+\frac{2y^3}{x^2}{\eta}_y \partial_{\eta})u=0 $$

I think that in $\partial_{xx}$ you are missing a term $\eta_{xx}\partial_\eta$, although admittedly the factor $\eta_{xx}$ turns out to be 0.

More importantly, I think that in $\partial_{xy}$ you are missing a term $\eta_{xy} \partial_\eta$, which turns out not to be zero!

 

[mathmari]

I think that in $\partial_{xx}$ you are missing a term $\eta_{xx}\partial_\eta$, although admittedly the factor $\eta_{xx}$ turns out to be 0.

More importantly, I think that in $\partial_{xy}$ you are missing a term $\eta_{xy} \partial_\eta$, which turns out not to be zero!

$$\partial_x=\partial_{\xi}+\eta_x \partial_{\eta}$$
$$\partial_y= \eta_y \partial_{\eta}$$
$$\partial_{xx}=\partial_{\xi \xi}+2 \eta_x \partial_{\xi \eta}+ \eta_{xx} \partial_{\eta}+\eta_x^2 \partial_{\eta \eta}$$
$$\partial_{yy}=\eta_y^2 \partial_{\eta \eta}+\eta_{yy} \partial_{\eta}$$
$$\partial_{xy}=\eta_y \partial_{\xi \eta}+\eta_x \eta_y \partial_{\eta \eta}+\eta_{xy}
\partial_{\eta}$$

Could you tell me if these partial derivatives are correct? (Thinking)

 

[I like Serena]

$$\partial_x=\partial_{\xi}+\eta_x \partial_{\eta}$$
$$\partial_y= \eta_y \partial_{\eta}$$
$$\partial_{xx}=\partial_{\xi \xi}+2 \eta_x \partial_{\xi \eta}+ \eta_{xx} \partial_{\eta}+\eta_x^2 \partial_{\eta \eta}$$
$$\partial_{yy}=\eta_y^2 \partial_{\eta \eta}+\eta_{yy} \partial_{\eta}$$
$$\partial_{xy}=\eta_y \partial_{\xi \eta}+\eta_x \eta_y \partial_{\eta \eta}+\eta_{xy}
\partial_{\eta}$$

Could you tell me if these partial derivatives are correct? (Thinking)

Yep!
All correct! (Star)

 

[mathmari]

Yep!
All correct! (Star)

Nice! (flower)

So replacing these at the differential equation we have the following:

$$(\partial_{\xi \xi}+(2 \eta_x +\frac{2y}{x} \eta_y) \partial_{\xi \eta}+(\eta_x^2+\frac{2y}{x}\eta_x \eta_y+\frac{y^2}{x^2}(1+y^2) \eta_y^2)\partial_{\eta \eta}+(\eta_{xx}+\frac{2y}{x} \eta_{xy}+\frac{y^2(1+y^2)}{x^2} \eta_{yy}+\frac{2y^3}{x^2} \eta_y ) \partial_{\eta})u=0 (*)$$

By the constraints:
$2 \eta_x +\frac{2y}{x} \eta_y=0$ and $\eta_x^2+\frac{2y}{x}\eta_x \eta_y+\frac{y^2}{x^2}(1+y^2) \eta_y^2=1$ we get that $ \eta =\frac{x}{y}$

So $$(*) \Rightarrow (\partial_{\xi \xi}+\partial_{\eta \eta})u=0$$
$$u_{\xi \xi}+u_{\eta \eta}=0$$

Is this correct? (Wondering)

 

[I like Serena]

Nice! (flower)

So replacing these at the differential equation we have the following:

$$(\partial_{\xi \xi}+(2 \eta_x +\frac{2y}{x} \eta_y) \partial_{\xi \eta}+(\eta_x^2+\frac{2y}{x}\eta_x \eta_y+\frac{y^2}{x^2}(1+y^2) \eta_y^2)\partial_{\eta \eta}+(\eta_{xx}+\frac{2y}{x} \eta_{xy}+\frac{y^2(1+y^2)}{x^2} \eta_{yy}+\frac{2y^3}{x^2} \eta_y ) \partial_{\eta})u=0 (*)$$

By the constraints:
$2 \eta_x +\frac{2y}{x} \eta_y=0$ and $\eta_x^2+\frac{2y}{x}\eta_x \eta_y+\frac{y^2}{x^2}(1+y^2) \eta_y^2=1$ we get that $ \eta =\frac{x}{y}$

So $$(*) \Rightarrow (\partial_{\xi \xi}+\partial_{\eta \eta})u=0$$
$$u_{\xi \xi}+u_{\eta \eta}=0$$

Is this correct? (Wondering)

I believe so. (Handshake)

That's a nice equation. (Inlove)

 

[mathmari]

I believe so. (Handshake)

That's a nice equation. (Inlove)

Good! Thank you very much! (Dance)

 

[mathmari]

I have the following exercise:
Write in the normal form the differential equation
$$u_{xx}+\frac{2y}{x}u_{xy}+\frac{y^2}{x^2}[(1+y^2)u_{yy}+2yu_y]=0$$
Hint: You can suppose that the one new variable is given by $\xi=x$

I have done the following:

$a=1, b=\frac{y}{x}, c=\frac{y^2}{x^2}(1+y^2)$

$b^2-ac=\frac{y^2}{x^2}-\frac{y^2}{x^2}(1+y^2)=-\frac{y^4}{x^2}=(\frac{y^2}{x}i)^2<0$

So the differential equation is elliptic.

Do we not have to take also the case when $y=0$, so $b^2-ac=0$, so the differential equation is parabolic?? (Wondering)

Or can we not take this case, because then:

$$u_{xx}=0$$

$$a=1, b=c=0$$
$$b^2-ac=0$$

$$\xi=x$$

To find the $\eta$, the Jacobian should be $\neq 0$:
$$\begin{vmatrix}
\xi_x & \xi_y\\
\eta_x & \eta_y
\end{vmatrix}= \xi_x \eta_y-\xi_y \eta_x =\eta_y \neq 0$$

But since $y=0$, $\eta_y$ cannot be $\neq 0$.

Is this correct?? (Wondering)

 

[I like Serena]

Do we not have to take also the case when $y=0$, so $b^2-ac=0$, so the differential equation is parabolic?? (Wondering)

Or can we not take this case, because then:

$$u_{xx}=0$$

$$a=1, b=c=0$$
$$b^2-ac=0$$

$$\xi=x$$

To find the $\eta$, the Jacobian should be $\neq 0$:
$$\begin{vmatrix}
\xi_x & \xi_y\\
\eta_x & \eta_y
\end{vmatrix}= \xi_x \eta_y-\xi_y \eta_x =\eta_y \neq 0$$

But since $y=0$, $\eta_y$ cannot be $\neq 0$.

Is this correct?? (Wondering)

Hmm...
I'd say that generally $b^2-ac<0$.
The only exception is the x-axis where $y=0$.
But that doesn't really characterize the ODE!

What it literally means, is for $y>0$ the ODE is elliptic, for $y<0$ the ODE is elliptic, and for $y=0$ I guess we might consider the ODE parabolic. (Thinking)

 

[mathmari]

Hmm...
I'd say that generally $b^2-ac<0$.
The only exception is the x-axis where $y=0$.
But that doesn't really characterize the ODE!

What it literally means, is for $y>0$ the ODE is elliptic, for $y<0$ the ODE is elliptic, and for $y=0$ I guess we might consider the ODE parabolic. (Thinking)

At this exercise we have that $b^2-ac=-\frac{y^4}{x^2}$ and we took at the beginning that it is $<0$.
Do we also have to take the case $b^2-ac=-\frac{y^4}{x^2}=0$, or are we sure that $y \neq 0$?? (Wondering)

 

[I like Serena]

At this exercise we have that $b^2-ac=-\frac{y^4}{x^2}$ and we took at the beginning that it is $<0$.
Do we also have to take the case $b^2-ac=-\frac{y^4}{x^2}=0$, or are we sure that $y \neq 0$?? (Wondering)

Let's make a comparison.

Suppose we have a family of regular hyperboles that may be the solution of some ODE.
Let's say that this family is given by $xy=C$.
Then the lines $x=0$ respectively $y=0$ are also likely to be solutions of the ODE.

Would you then say that lines are characteristic of the ODE?
Or would you simply say it is a hyperbolic ODE? (Thinking)

 

[mathmari]

Let's make a comparison.

Suppose we have a family of regular hyperboles that may be the solution of some ODE.
Let's say that this family is given by $xy=C$.
Then the lines $x=0$ respectively $y=0$ are also likely to be solutions of the ODE.

Would you then say that lines are characteristic of the ODE?
Or would you simply say it is a hyperbolic ODE? (Thinking)

I got stuck right now.. (Worried)
Could you explain it further to me?? (Wondering)

 

[I like Serena]

I got stuck right now.. (Worried)
Could you explain it further to me?? (Wondering)

The solution of your ODE is of the form $u(x,y)$, which is supposedly defined for all $x$ and for all $y$.
The solution $u(x,y)$ that you're interested in will be hyperbolic everywhere, except where $y=0$, but this is an infinitely small portion of the domain of the solution. (Nerd)