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I have to show the identity I wrote in the title: it should be $\partial_z\partial_{\bar z}=\frac14\left(\partial_r^2+\frac1r\partial_r+\frac1{r^2}\partial_{\theta}^2\right)$ but some computation doesn't match with my book with these operators.

We know that $\partial_z\partial_{\bar z}=\frac14(\partial_x^2+\partial_y^2)$. Then we pass to polar coordinates writing $x=r\cos\theta$ and $y=r\sin\theta$, from which, obviously we have $r=\sqrt{x^2+y^2}$ and $\theta=\arctan\frac yx$. We use than the chain rule to write $$ \partial_x=\frac{\partial r}{\partial x}\partial_r+\frac{\partial\theta}{\partial x}\partial_{\theta}=\cos\theta\partial_r-\frac{\sin\theta}{r}\partial_{\theta} $$ and $$ \partial_y=\frac{\partial r}{\partial y}\partial_r+\frac{\partial\theta}{\partial y}\partial_{\theta}=\sin\theta\partial_r+\frac{\cos\theta}{r}\partial_{\theta} $$

and till here all matches with my book. My problem comes now: I used the above cited $\partial_z\partial_{\bar z}=\frac14(\partial_x^2+\partial_y^2)$ but this gives me $$ \partial_z\partial_{\bar z}=\frac14\left(\partial_r^2+\frac1{r^2}\partial_{\theta}^2\right) $$ because the "crossed" terms are equal and opposite: squaring $\partial_x$ the "crossed" term is $-\frac2r\cos\theta\sin\theta\partial_r\partial_{\theta}$, squaring $\partial_y$ the "crossed" term is $\frac2r\cos\theta\sin\theta\partial_r\partial_{\theta}$ hence in the sum they should vanish.

Where is the error?

EDIT: following Andrea's suggest, I wrote \begin{align*} \partial_x^2+\partial_y^2 &=\left(\cos\theta\partial_r-\frac{\sin\theta}{r}\partial_{\theta}\right)\left(\cos\theta\partial_r-\frac{\sin\theta}{r}\partial_{\theta}\right)+\\ &\;\;\;\;\;\,\left(\sin\theta\partial_r+\frac{\cos\theta}{r}\partial_{\theta}\right) \left(\sin\theta\partial_r+\frac{\cos\theta}{r}\partial_{\theta}\right)\\ &=\cos^2\theta\partial_r^2+\frac{\sin\theta}{r}\cos\theta\partial_{\theta} +\frac{\sin^2}{r}\partial_r+\cos\theta\frac{\sin\theta}{r^2}\partial_\theta+\\ &\;\;\;\;\;\;\sin^2\theta\partial_r^2-\frac{\cos\theta}{r}\sin\theta\partial_{\theta} +\frac{\cos^2}{r}\partial_r-\sin\theta\frac{\cos\theta}{r^2}\partial_\theta\\ =&\partial_r^2+\frac1r\partial_r \end{align*}

which is false again! I checked several times but I cannot find where my mistake is!

Many thanks!

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  • $\begingroup$ When trying to derive it, it can be useful to consider the identity acting on a test-function, i.e. try to calculate $(\partial_x^2+\partial_y^2)f$. This can help you avoid missing some terms in the derivation . $\endgroup$ – Winther Mar 3 '15 at 23:45
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You have correctly derived $$ \frac{\partial}{\partial x} = \cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial\theta} \\ \frac{\partial}{\partial y} = \sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial\theta} $$ Therefore, $$ \begin{align} \frac{\partial^{2}}{\partial x^{2}} & =\left(\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\right)\left(\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\right) \\ & = \cos\theta\frac{\partial}{\partial r}\cos\theta\frac{\partial}{\partial r} \\ & -\cos\theta\frac{\partial}{\partial r}\frac{\sin\theta}{r}\frac{\partial}{\partial\theta} \\ & -\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\cos\theta\frac{\partial}{\partial r} \\ & + \frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\frac{\sin\theta}{r}\frac{\partial}{\partial\theta} \end{align} $$ It is true that $$ \cos\theta\frac{\partial}{\partial r}\cos\theta\frac{\partial}{\partial r} = \cos^{2}\theta\frac{\partial^{2}}{\partial r^{2}} $$ However, the product rule gives $$ \begin{align} \cos\theta\frac{\partial}{\partial r}\frac{\sin\theta}{r}\frac{\partial}{\partial\theta} & = \cos\theta\sin\theta\frac{\partial}{\partial r}\frac{1}{r}\frac{\partial}{\partial\theta} \\ & = \cos\theta\sin\theta\frac{1}{r}\frac{\partial^{2}}{\partial r\partial\theta} - \cos\theta\sin\theta\frac{1}{r^{2}}\frac{\partial}{\partial\theta} \end{align} $$ Therefore, keeping things on separate lines for you to inspect, $$ \begin{align} \frac{\partial^{2}}{\partial x^{2}} & = \cos^{2}\frac{\partial^{2}}{\partial r^{2}} \\ & -\frac{\cos\theta\sin\theta}{r}\frac{\partial^{2}}{\partial r\partial \theta} +\frac{\cos\theta\sin\theta}{r^{2}}\frac{\partial}{\partial \theta} \\ & -\frac{\sin\theta\cos\theta}{r}\frac{\partial^{2}}{\partial\theta\partial r} +\frac{\sin^{2}\theta}{r}\frac{\partial}{\partial r} \\ & + \frac{\sin^{2}\theta}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}} +\frac{\sin\theta\cos\theta}{r^{2}}\frac{\partial}{\partial\theta} \end{align} $$ Every first order derivative term came from a product rule. Similarly, $$ \begin{align} \frac{\partial^{2}}{\partial y^{2}} & = \sin^{2}\frac{\partial^{2}}{\partial r^{2}} \\ & +\frac{\sin\theta\cos\theta}{r}\frac{\partial^{2}}{\partial r\partial\theta}-\frac{\sin\theta\cos\theta}{r^{2}}\frac{\partial}{\partial\theta} \\ & +\frac{\cos\theta\sin\theta}{r}\frac{\partial^{2}}{\partial\theta\partial r} + \frac{\cos^{2}\theta}{r}\frac{\partial}{\partial r} \\ & +\frac{\cos^{2}\theta}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}} -\frac{\sin\theta\cos\theta}{r^{2}}\frac{\partial}{\partial\theta} \end{align} $$ Now when you add these together, you get $$ \frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}} = \frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}} $$

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    $\begingroup$ MANY MANY MANY THANKS! I'm not used to handle these objects, I made a lot of confusion! Thanks again $\endgroup$ – Joe Mar 4 '15 at 8:25
  • $\begingroup$ @Joe : You're welcome. However well-intentioned, the other answers weren't helping you. $\endgroup$ – Disintegrating By Parts Mar 4 '15 at 9:14
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When you "multiply" two differential operators, like $\cos\theta\,\partial_r$ and $\frac{\sin\theta}r\,\partial_\theta$, the derivative operator in the first factor acts on the coefficient in the second factor.

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  • $\begingroup$ All clear, thank you so much! $\endgroup$ – Joe Mar 3 '15 at 16:09
  • $\begingroup$ Andrea, I compute with the rule you said, but something's still wrong. Where did I get wrong? $\endgroup$ – Joe Mar 3 '15 at 17:05
  • $\begingroup$ Christian Blatter's answer explains the remaining errors. $\endgroup$ – Andreas Blass Mar 3 '15 at 20:46
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One has $\theta=\arctan{y\over x}$ (not $=\arctan{x\over y}$) when $x>0$, but $$\theta_x={1\over1+ {y^2\over x^2}}\left(-{y\over x^2}\right)={-y\over x^2+y^2}=-{\sin\theta\over r}$$ and $$\theta_y={1\over1+ {y^2\over x^2}}{1\over x}={x\over x^2+y^2}={\cos\theta\over r}$$are valid in all of $\dot{\Bbb R}^2$.

You have the sign wrong in your displayed formula for $\partial_y$.

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  • $\begingroup$ Many thanks! But it was a typo and in my computations the sing seems to be correct! $\endgroup$ – Joe Mar 3 '15 at 22:58

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