How to prove this derivative problem

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Homework Help Overview

The discussion revolves around proving a relationship between second derivatives in terms of new variables, specifically \(\xi\) and \(\eta\), defined as \(\xi = x + y\) and \(\eta = x - y\). Participants are exploring how to express the second derivatives with respect to these new variables and relate them back to the original variables \(x\) and \(y\).

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the chain rule to relate derivatives in terms of \(\xi\) and \(\eta\) to those in terms of \(x\) and \(y\). Some express uncertainty about applying the chain rule correctly for second derivatives. Others question the validity of the relationship being proven, noting discrepancies in calculated second derivatives.

Discussion Status

There is an ongoing exploration of how to apply the chain rule for the derivatives involved. Some participants have provided partial insights and examples, while others are seeking further clarification and resources. Multiple interpretations of the problem are being considered, and no consensus has been reached yet.

Contextual Notes

Participants mention the challenge of handling derivatives and express concerns about the correctness of their calculations. There is a sense of confusion regarding the application of the chain rule in this context.

Spartakhus
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Given that

[itex]\xi = x + y[/itex] and [itex]\eta = x - y[/itex]



How do I show that:

[itex]\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2} = \frac{\partial^2}{\partial \xi^2}+\frac{\partial^2}{\partial \eta^2}[/itex]


I know that

[itex]\xi^2 + \eta^2 = 2(x^2 + y^2)[/itex] and [itex]\xi^2 - \eta^2 = 4xy[/itex]

But I do not know how to handle these derivatives :(

Sorry about this newbie question.
 
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Spartakhus said:
Given that

[itex]\xi = x + y[/itex] and [itex]\eta = x - y[/itex]



How do I show that:

[itex]\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2} = \frac{\partial^2}{\partial \xi^2}+\frac{\partial^2}{\partial \eta^2}[/itex]


I know that

[itex]\xi^2 + \eta^2 = 2(x^2 + y^2)[/itex] and [itex]\xi^2 - \eta^2 = 4xy[/itex]

But I do not know how to handle these derivatives :(

Sorry about this newbie question.

Use the chain rule to relate [itex]\partial /\partial x,[/itex] etc, to [itex]\partial/ \partial \xi[/itex] and [itex]\partial / \partial \eta .[/itex]

RGV
 
There's a problem.
Suppose you have [itex]f(\xi, \eta) = \xi ^2 + \eta^2[/itex]
 
Thanks for the reply. Could you write me an example or tell me where I can read about this?

I know that dy/dx = dy/du*du/dx
But I don't know how to apply this in case of d/dx.
 
Sorry, I had to finish, but it seems it's not possible to edit your own messages.
Anyway:

[itex]f(\xi,\eta)=\xi^2+\eta^2[/itex]

This is equal to
[itex]f(x,y)=2x^2+2y^2[/itex]

Notice that [itex]\frac{\partial^2 f}{\partial\xi^2}+\frac{\partial^2 f}{\partial\eta^2}=4[/itex]
while [itex]\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=8[/itex]

so the thesis doesn't hold, there's something to fix, if I'm not wrong.
 
[itex]\frac{\partial}{\partial x} = \frac{\partial}{\partial \zeta}\frac{\partial \zeta}{\partial x} + \frac{\partial}{\partial \eta}\frac{\partial \eta}{\partial x}[/itex]

you following (I'm using zeta because I can't remember what that other squiggle is called)

[itex]\frac{\partial \zeta}{\partial x} = 1[/itex]
[itex]\frac{\partial \eta}{\partial x} = 1[/itex]

therefore

[itex]\frac{\partial}{\partial x} = \frac{\partial}{\partial \zeta} + \frac{\partial}{\partial \eta}[/itex]

do that again, and do the same for y, you should pick up some negatives on the y side that will cancel out everything and leave you with what you're trying to get
 
Thanks for the replies...
 

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