Chain Rule in 2D Function Transformation

[Julio1]

Let $z:\mathbb{R}^2\to \mathbb{R}$ an function of kind $C^2(\mathbb{R}^2)$. What transforms the equation $2\dfrac{\partial^2 z}{\partial x^2}+\dfrac{\partial^2 z}{\partial x\partial y}-\dfrac{\partial^2 z}{\partial y^2}+\dfrac{\partial z}{\partial x}+\dfrac{\partial z}{\partial y}=0$ under the change of variable $u=x+2y+2$ and $v=x-y-1$?

Spoiler

Hi, I have this problem I don't understand how to solve it. I have calculated the following:

$\dfrac{\partial z}{\partial x}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial x}=\dfrac{\partial z}{\partial u}+\dfrac{\partial z}{\partial v}.$

$\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial y}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial y}=2\dfrac{\partial z}{\partial u}-\dfrac{\partial z}{\partial v}.$

However, the case $\dfrac{\partial^2 z}{\partial x^2}$ I don't understand how solve, anyone can help me?

 

[Chris L T521]

Let $z:\mathbb{R}^2\to \mathbb{R}$ an function of kind $C^2(\mathbb{R}^2)$. What transforms the equation $2\dfrac{\partial^2 z}{\partial x^2}+\dfrac{\partial^2 z}{\partial x\partial y}-\dfrac{\partial^2 z}{\partial y^2}+\dfrac{\partial z}{\partial x}+\dfrac{\partial z}{\partial y}=0$ under the change of variable $u=x+2y+2$ and $v=x-y-1$?

Spoiler

Hi, I have this problem I don't understand how to solve it. I have calculated the following:

$\dfrac{\partial z}{\partial x}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial x}=\dfrac{\partial z}{\partial u}+\dfrac{\partial z}{\partial v}.$

$\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial y}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial y}=2\dfrac{\partial z}{\partial u}-\dfrac{\partial z}{\partial v}.$

However, the case $\dfrac{\partial^2 z}{\partial x^2}$ I don't understand how solve, anyone can help me?

By the chain rule, note that

\[\frac{\partial}{\partial x} = \frac{\partial u}{\partial x}\frac{\partial}{\partial u} + \frac{\partial v}{\partial x} \frac{\partial}{\partial v}\]

(and similarly with respect to $y$). Therefore, for $u=x+2y+2$ and $v=x-y-1$, you should have that

\[\begin{aligned} \frac{\partial^2 z}{\partial x^2} &= \frac{\partial}{\partial x}\left[\frac{\partial z}{\partial x}\right] \\ &= \left(\frac{\partial u}{\partial x} \frac{\partial}{\partial u} + \frac{\partial v}{\partial x} \frac{\partial}{\partial v}\right) \left(\frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial x}\right)\\ &= \frac{\partial u}{\partial x}\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial x}\right) + \frac{\partial v}{\partial x}\frac{\partial}{\partial v} \left(\frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial x}\right)\\ &= \frac{\partial}{\partial u}\left( \frac{\partial z}{\partial u} + \frac{\partial z}{\partial v}\right) + \frac{\partial}{\partial v}\left(\frac{\partial z}{\partial u} + \frac{\partial z}{\partial v}\right) \\ &= \frac{\partial^2 z}{\partial u^2} + 2\frac{\partial^2 z}{\partial u\partial v} + \frac{\partial^2 z}{\partial v^2}\end{aligned}\]

In a similar manner (verify),

\[\frac{\partial^2 z}{\partial y^2} = 4\frac{\partial^2 z}{\partial u^2} - 4\frac{\partial^2 z}{\partial u\partial v} + \frac{\partial^2 z}{\partial v^2}\]

and (verify)

\[\frac{\partial^2 z}{\partial x \partial y} = 2\frac{\partial^2 z}{\partial u^2} + \frac{\partial^2 z}{\partial u\partial v} - \frac{\partial^2 z}{\partial v^2}\]

Therefore,

\[2\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial x\partial y} - \frac{\partial^2 z}{\partial y^2} + \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = 0 \implies 3\frac{\partial^2 z}{\partial u\partial v} + \frac{\partial z}{\partial u} = 0\]

Is this what you were after?

I hope this made sense! (Smile)

 

[Julio1]

Is this what you were after?

I hope this made sense! (Smile)

Hello Chris L T521 :), yes, that's what should do. Sorry, I wanted to express it better in English, but I could not.

By the chain rule, note that
\[\frac{\partial}{\partial x} = \frac{\partial u}{\partial x}\frac{\partial}{\partial u} + \frac{\partial v}{\partial x} \frac{\partial}{\partial v}\]

Thanks for helping, but know I do not quite understand this. What I understand from this is that $\dfrac{\partial}{\partial x}=D^{0},$ namely $\dfrac{\partial}{\partial x}$ is an operator, but is possible do $D^{0}=\dfrac{\partial u}{\partial x}\cdot D^{0}+\dfrac{\partial v}{\partial x}\cdot D^0.$ In other words, is only one notation?, because $\dfrac{\partial u}{\partial x}$ it should not be understood as a fraction, or as a product fraction, but is an differential.

\begin{aligned} &= \dfrac{\partial u}{\partial x}\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial x}\right) + \frac{\partial v}{\partial x}\frac{\partial}{\partial v} \left(\frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial x}\right)\\ &= \frac{\partial}{\partial u}\left( \frac{\partial z}{\partial u} + \frac{\partial z}{\partial v}\right) + \frac{\partial}{\partial v}\left(\frac{\partial z}{\partial u} + \frac{\partial z}{\partial v}\right) \\
\end{aligned}

Thanks, but know you that I don't understand how pass from the third to the fourth line?. Is there something that simplify?

 

[Julio1]

Thanks, I solved so:

$\begin{eqnarray*}
\dfrac{\partial^2 z}{\partial x^2}&=&\dfrac{\partial}{\partial x}\left(\dfrac{\partial z}{\partial x}\right)\\
&=&\left(\dfrac{\partial }{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial}{\partial v}\dfrac{\partial v}{\partial x}\right)\left(\dfrac{\partial z}{\partial u}+\dfrac{\partial z}{\partial v}\right)\\
&=&\dfrac{\partial}{\partial u}\dfrac{\partial u}{\partial x}\dfrac{\partial z}{\partial u}+\dfrac{\partial}{\partial u}\dfrac{\partial u}{\partial x}\dfrac{\partial z}{\partial v}+\dfrac{\partial}{\partial v}\dfrac{\partial v}{\partial x}\dfrac{\partial z}{\partial u}+\dfrac{\partial}{\partial v}\dfrac{\partial v}{\partial x}\dfrac{\partial z}{\partial v}\\
&=&\dfrac{\partial u}{\partial x}\dfrac{\partial^2 z}{\partial u^2}+\dfrac{\partial u}{\partial x}\dfrac{\partial^2 z}{\partial u\partial v}+\dfrac{\partial v}{\partial x}\dfrac{\partial^2 z}{\partial v\partial u}+\dfrac{\partial v}{\partial x}\dfrac{\partial ^2 z}{\partial v^2}\\
&=&\dfrac{\partial^2 z}{\partial u^2}+\dfrac{\partial^2 z}{\partial u\partial v}+\dfrac{\partial^2 z}{\partial v\partial u}+\dfrac{\partial^2 z}{\partial v^2}\\
&\overbrace{=}^{\text{Schwarz}}&\dfrac{\partial^2 z}{\partial u^2}+2\dfrac{\partial^2 z}{\partial u\partial v}+\dfrac{\partial^2 z}{\partial v^2}.
\end{eqnarray*}
$

In analogy with the others, okay?