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To obtain d'Alembert's soltion to the wave equation: $$\frac{\partial^2 u}{\partial t^2}=\alpha^2\frac{\partial^2 u}{\partial x^2} $$ Change of variables is used: $\psi=x+at$, $\eta=x-at$. Then chain rule can be applied to find the first order partial derivatave of $u$: $$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial \psi}+\frac{\partial u}{\partial \eta}$$ But I don't understand how the second order partial derivative can be expressed into this form:$$\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial \psi^2}+2\frac{\partial^2 u}{\partial \psi\partial \eta}+\frac{\partial^2 u}{\partial \eta^2}$$ In particular, $$ \frac{\partial^2 u}{\partial x\partial \psi}=\frac{\partial^2 u}{\partial \psi^2}+\frac{\partial^2 u}{\partial \psi\partial \eta}$$ Can someone please explain explain how the calculation is carried out? I refer to Thomas Calculus but there is no where inside the book can explain the second order partial differentiation.

P/S: I found this on physics forum: $$\frac{\partial}{\partial x}=\left(\frac{\partial}{\partial r}+\frac{\partial }{\partial s}\right)$$ $$\frac{\partial^{2}}{\partial x^{2}}=\left(\frac{\partial}{\partial r}+\frac{\partial }{\partial s}\right)\left(\frac{\partial}{\partial r}+\frac{\partial }{\partial s}\right)$$ Why does the partial derivative act somehow like polynomial?

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$$ \frac{\partial^2 u}{\partial x\partial \psi}=\frac{\partial}{\partial\psi}\left[\frac{\partial u}{\partial x}\right]=\frac{\partial}{\partial\psi}\left[\frac{\partial u}{\partial \psi}+\frac{\partial u}{\partial \eta}\right]=\frac{\partial^2 u}{\partial \psi^2}+\frac{\partial^2 u}{\partial \psi\partial \eta}\ , $$ where one uses the first derivative you had computed earlier $$ \frac{\partial u}{\partial x}=\frac{\partial u}{\partial \psi}+\frac{\partial u}{\partial \eta}\ . $$

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    $\begingroup$ Is it always true that $\frac{\partial^2 u}{\partial x\partial \psi}=\frac{\partial}{\partial\psi}\left[\frac{\partial u}{\partial x}\right]$? $\endgroup$ – Dave Clifford Feb 28 '16 at 10:25
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    $\begingroup$ @Dave Clifford What is always true is $\frac{\partial^2 u}{\partial x\partial \psi}=\frac{\partial}{\partial x}\left[\frac{\partial u}{\partial \psi}\right]$ [as this is just the definition of a mixed second derivative]. If then this (mixed) second derivative is everywhere continuous in $\mathbb{R}^2$, then this is also equal to $\frac{\partial}{\partial\psi}\left[\frac{\partial u}{\partial x}\right]$ by Schwarz' theorem en.wikipedia.org/wiki/Symmetry_of_second_derivatives $\endgroup$ – Pierpaolo Vivo Feb 28 '16 at 10:31

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