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If I proceed to compute the chain rule of a second-order partial derivative of a continuous function $\psi(x,y)$ in terms of new coordinates $\xi=\xi(x,y)$ and $\eta=\eta(x,y)$ like $$\frac{\partial^2 \psi}{\partial x^2}=\left(\frac{\partial \xi }{\partial x}\frac{% \partial }{\partial \xi }+\frac{\partial \eta }{\partial x}\frac{% \partial }{\partial \eta }\right)\left(\frac{\partial \xi }{\partial x}\frac{% \partial \psi}{\partial \xi }+\frac{\partial \eta }{\partial x}\frac{% \partial \psi}{\partial \eta }\right),$$ I obtain the following result $$\frac{\partial^2 \psi}{\partial x^2}=\left(\frac{\partial\xi}{\partial x}\right)^2\frac{\partial^2\psi}{\partial \xi^2}+\left(\frac{\partial\eta}{\partial x}\right)^2\frac{\partial^2\psi}{\partial \eta^2}+2\frac{\partial \eta}{\partial x}\frac{\partial \xi}{\partial x}\frac{\partial^2\psi}{\partial\eta\partial\xi}.$$ I know this result is incorrect because I miss two terms with second-order derivatives of $\xi$ and $\eta$, but I do not identify in which step I made the mistake.

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These two terms are missing $$\frac {\partial \psi}{\partial \xi}\frac {\partial^2 \xi}{\partial x^2} \text{ , and }\frac {\partial \psi}{\partial \eta}\frac {\partial^2 \eta}{\partial x^2}$$ You don't apply correctly the derivative of a product of function it seems

$$\frac{\partial \xi }{\partial x}\frac{% \partial }{\partial \xi }\left(\frac{\partial \xi }{\partial x}\frac{% \partial \psi}{\partial \xi }\right)$$

This last one should give you two terms and one missing term and another term that you already have $$\frac{\partial \xi }{\partial x}\frac{% \partial }{\partial \xi }\left(\frac{\partial \xi }{\partial x}\frac{% \partial \psi}{\partial \xi }\right)=\frac {\partial \psi}{\partial \xi}\frac {\partial^2 \xi}{\partial x^2}+\left(\frac{\partial\xi}{\partial x}\right)^2\frac{\partial^2\psi}{\partial \xi^2}$$

You did the same mistake for the other derivative where a term is also missing $$\frac{\partial \eta }{\partial x}\frac{% \partial }{\partial \eta }\left(\frac{\partial \eta }{\partial x}\frac{% \partial \psi}{\partial \eta }\right)=\frac {\partial \psi}{\partial \eta}\frac {\partial^2 \eta}{\partial x^2}+\left(\frac{\partial\eta}{\partial x}\right)^2\frac{\partial^2\psi}{\partial \eta^2}$$

And you missed 4 terms not two. It's juste these 4 terms are 2 $\times$ the same two terms...

Edit

$$\frac{\partial\xi}{\partial x}\frac{\partial}{\partial \xi}\left(\frac{\partial \xi}{\partial x}\right)\frac{\partial\psi}{\partial \xi}=\frac{\partial \psi}{\partial \xi}\frac{\partial^2\xi}{\partial x^2},$$ Simplifiy the dervative you take ...we need x So $$\frac{\partial\xi}{\partial x}\frac{\partial}{\partial \xi}=\frac{\partial}{\partial x}$$

You see we simply simplify by $$\partial \xi$$

Therefore

$$\frac{\partial\xi}{\partial x}\frac{\partial}{\partial \xi}\left(\frac{\partial \xi}{\partial x}\right)\frac{\partial\psi}{\partial \xi}=\frac{\partial}{\partial x}\left(\frac{\partial \xi}{\partial x}\right)\frac{\partial\psi}{\partial \xi}=\frac{\partial \psi}{\partial \xi}\frac{\partial^2\xi}{\partial x^2},$$

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  • $\begingroup$ I gave an answer to ask in detail. If you do, I delete it. Thanks. $\endgroup$ Commented Mar 23, 2018 at 16:45
  • $\begingroup$ I will add some lines @Janstillerion $\endgroup$ Commented Mar 23, 2018 at 16:47
  • $\begingroup$ Great! Finally I get it. Thanks! $\endgroup$ Commented Mar 23, 2018 at 16:53
  • $\begingroup$ yw @Janstillerion ........... $\endgroup$ Commented Mar 23, 2018 at 16:54

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