Finding Partial Derivatives with Transformations

Click For Summary

Discussion Overview

The discussion revolves around finding partial derivatives using transformations between variables \( \xi \) and \( \eta \) as functions of \( x \) and \( y \). Participants explore the application of chain and product rules in the context of second derivatives, addressing both theoretical and mathematical aspects of the problem.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a method for finding the first and second partial derivatives using transformations, but expresses confusion on how to proceed further.
  • Another participant suggests a correction to the initial approach, providing a detailed breakdown of the second derivative using transformations and applying the product rule.
  • A later reply references a textbook definition of the second derivative, questioning the differences between their derivation and the textbook's approach.
  • Further elaboration includes applying both the chain rule and product rule to derive the second derivative of a function \( f(\xi, \eta) \), detailing each step of the process.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to finding the second partial derivative, with multiple competing views and methods presented. The discussion remains unresolved regarding the best method to apply.

Contextual Notes

Some participants note potential discrepancies between their calculations and textbook definitions, indicating that assumptions or interpretations of the rules may vary. The discussion highlights the complexity of applying transformation rules in partial differentiation.

evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
Hello! :)

Having the transformations:
$$\xi=\xi(x,y), \eta=\eta(x,y)$$

I want to find the following partial derivatives:
$$\frac{\partial}{\partial{x}}= \frac{\partial}{ \partial{\xi}} \frac{\partial{\xi}}{\partial{x}}+\frac{\partial}{\partial{\eta}} \frac{\partial{\eta}}{\partial{x}}=\partial_{\xi} \xi_x+\partial_{\eta} \eta_x$$

$$\frac{\partial}{\partial{y}}= \frac{\partial}{ \partial{\xi}} \frac{\partial{\xi}}{\partial{y}}+\frac{\partial}{\partial{\eta}} \frac{\partial{\eta}}{\partial{y}}=\partial_{\xi} \xi_y+\partial_{\eta} \eta_y$$

$$\frac{\partial^2}{\partial{x^2}}=\frac{\partial{(\partial_{\xi} \xi_x+\partial_{\eta} \eta_x})}{\partial{x}}=\frac{\partial{(\partial_{\xi} \xi_x+\partial_{\eta} \eta_x})}{\partial{x}} \frac{\partial{\xi}}{\partial{x}}+\frac{\partial{(\partial_{\xi} \xi_x+\partial_{\eta} \eta_x})}{\partial{\eta}} \frac{\partial{\eta}}{\partial{x}}$$
I got stuck...How can I continue??
 
Physics news on Phys.org
evinda said:
Hello! :)

Having the transformations:
$$\xi=\xi(x,y), \eta=\eta(x,y)$$

I want to find the following partial derivatives:
$$\frac{\partial}{\partial{x}}= \frac{\partial}{ \partial{\xi}} \frac{\partial{\xi}}{\partial{x}}+\frac{\partial}{\partial{\eta}} \frac{\partial{\eta}}{\partial{x}}=\partial_{\xi} \xi_x+\partial_{\eta} \eta_x$$

$$\frac{\partial}{\partial{y}}= \frac{\partial}{ \partial{\xi}} \frac{\partial{\xi}}{\partial{y}}+\frac{\partial}{\partial{\eta}} \frac{\partial{\eta}}{\partial{y}}=\partial_{\xi} \xi_y+\partial_{\eta} \eta_y$$

$$\frac{\partial^2}{\partial{x^2}}=\frac{\partial{(\partial_{\xi} \xi_x+\partial_{\eta} \eta_x})}{\partial{x}}=\frac{\partial{(\partial_{\xi} \xi_x+\partial_{\eta} \eta_x})}{\partial{x}} \frac{\partial{\xi}}{\partial{x}}+\frac{\partial{(\partial_{\xi} \xi_x+\partial_{\eta} \eta_x})}{\partial{\eta}} \frac{\partial{\eta}}{\partial{x}}$$
I got stuck...How can I continue??

Hey! (Mmm)

How about this (correcting a small mistake):
\begin{aligned}
\frac{\partial^2}{\partial{x^2}}

&=\frac{\partial{(\partial_{\xi} \xi_x+\partial_{\eta} \eta_x})}{\partial{x}} \\

&=\frac{\partial{(\partial_{\xi} \xi_x+\partial_{\eta} \eta_x})}{\partial{\xi}}
\frac{\partial{\xi}}{\partial{x}}
+ \frac{\partial{(\partial_{\xi} \xi_x+\partial_{\eta} \eta_x})}{\partial{\eta}}
\frac{\partial{\eta}}{\partial{x}} \\

&=(\partial_{\xi\xi} \xi_x + \partial_{\xi\eta} \eta_x) \xi_x + \partial_\xi \xi_{xx}
+ (\partial_{\eta\xi} \xi_x+\partial_{\eta\eta} \eta_x) \eta_x + \partial_\eta \eta_{xx}\\

\end{aligned}
 
I like Serena said:
Hey! (Mmm)

How about this (correcting a small mistake):
\begin{aligned}
\frac{\partial^2}{\partial{x^2}}

&=\frac{\partial{(\partial_{\xi} \xi_x+\partial_{\eta} \eta_x})}{\partial{x}} \\

&=\frac{\partial{(\partial_{\xi} \xi_x+\partial_{\eta} \eta_x})}{\partial{\xi}}
\frac{\partial{\xi}}{\partial{x}}
+ \frac{\partial{(\partial_{\xi} \xi_x+\partial_{\eta} \eta_x})}{\partial{\eta}}
\frac{\partial{\eta}}{\partial{x}} \\

&=(\partial_{\xi\xi} \xi_x+\partial_{\xi\eta} \eta_x) \xi_x
+ (\partial_{\eta\xi} \xi_x+\partial_{\eta\eta} \eta_x) \eta_x \\

\end{aligned}

According to my textbook it is like that:
$$\frac{\partial^2}{\partial{x^2}}=\frac{\partial{(\xi_x \partial_{\xi}+\eta_x \partial_{\eta})}}{\partial{x}}=\xi_{xx} \partial_{\xi}+\xi_x \partial_x \partial_{\xi}+\eta_{xx} \partial_{\eta}+\eta_x \partial_x \partial_{\eta}$$
Why is it so? :confused:
 
evinda said:
According to my textbook it is like that:
$$\frac{\partial^2}{\partial{x^2}}=\frac{\partial{(\xi_x \partial_{\xi}+\eta_x \partial_{\eta})}}{\partial{x}}=\xi_{xx} \partial_{\xi}+\xi_x \partial_x \partial_{\xi}+\eta_{xx} \partial_{\eta}+\eta_x \partial_x \partial_{\eta}$$
Why is it so? :confused:

Let's apply the sum rule and the product rule:
\begin{aligned}
\frac{\partial{(\xi_x \partial_{\xi}+\eta_x \partial_{\eta})}}{\partial{x}}

&= \frac{\partial(\xi_x \partial_{\xi})}{\partial{x}} + \frac{\partial(\eta_x \partial_{\eta})}{\partial{x}} \\

&= \xi_{xx} \partial_{\xi} + \xi_x \partial_x\partial_\xi
+ \eta_{xx} \partial_{\eta} + \eta_x \partial_x\partial_\eta\\

\end{aligned}
I worked it out completely, but apparently that is not what was asked! :eek:
Ah well, since I have already written it, I'll leave it here.

So let's suppose we have a function $f(\xi, \eta)$.

Then:
$$\frac{\partial f}{\partial x}
= \frac{\partial f}{\partial \xi} \frac{\partial \xi}{\partial x}
+ \frac{\partial f}{\partial \eta} \frac{\partial \eta}{\partial x}$$

Applying both the chain rule and the product rule:
\begin{aligned}
\frac{\partial^2 f}{\partial x^2}

&= \frac{\partial}{\partial x}
\left( \frac{\partial f}{\partial \xi} \frac{\partial \xi}{\partial x}
+ \frac{\partial f}{\partial \eta} \frac{\partial \eta}{\partial x} \right) \\

&= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial \xi} \right)
\frac{\partial \xi}{\partial x}
+ \frac{\partial f}{\partial \xi}
\frac{\partial}{\partial x}\left( \frac{\partial \xi}{\partial x} \right )

+ \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial \eta} \right)
\frac{\partial \eta}{\partial x}
+ \frac{\partial f}{\partial \eta}
\frac{\partial}{\partial x}\left( \frac{\partial \eta}{\partial x} \right) \\

&= \left( \frac{\partial^2 f}{\partial \xi^2}\frac{\partial \xi}{\partial x}
+ \frac{\partial^2 f}{\partial\eta\partial\xi}\frac{\partial \eta}{\partial x} \right)
\frac{\partial \xi}{\partial x}

+ \frac{\partial f}{\partial \xi}
\frac{\partial^2 \xi}{\partial x^2}

+ \left( \frac{\partial^2 f}{\partial \xi\partial\eta} \frac{\partial \xi}{\partial x}
+ \frac{\partial^2 f}{\partial \eta^2} \frac{\partial \eta}{\partial x} \right)
\frac{\partial \eta}{\partial x}

+ \frac{\partial f}{\partial \eta}
\frac{\partial^2 \eta}{\partial x^2} \\

&= f_{\xi\xi}\xi_x^2 + f_{\eta\xi}\eta_x\xi_x
+ f_\xi \xi_{xx}
+ f_{\xi\eta} \xi_x\eta_x + f_{\eta\eta} \eta_x^2
+ f_\eta \eta_{xx} \\

&= f_{\xi\xi}\xi_x^2 + 2 f_{\eta\xi}\eta_x\xi_x + f_{\eta\eta} \eta_x^2
+ f_\xi \xi_{xx} + f_\eta \eta_{xx} \\

&= \left(\ \xi_x^2\partial_\xi\partial_\xi + 2 \eta_x\xi_x\partial_\eta\partial_\xi
+ \eta_x^2 \partial_\eta\partial_\eta
+ \xi_{xx}\partial_\xi + \eta_{xx}\partial_\eta \ \right)\ f \\

\end{aligned}
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 17 ·
Replies
17
Views
3K