# Chain Rule

#### Julio

##### Member
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 $v=x-y-1$?

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?

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#### Chris L T521

##### Well-known member
Staff member
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 $v=x-y-1$?

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?

#### Julio

##### Member
Is this what you were after?

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?

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#### Julio

##### Member
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?