If $$w=z\tan^{-1}\Big(\frac{x}{y}\Big),$$

find$$\frac{\partial^2 {w}}{\partial{x^2}}+\frac{\partial^2 {w}}{\partial{y^2}}+\frac{\partial^2 {w}}{\partial{z^2}}=?$$

I calculated this and the answer is zero. But It was long calculations ( find $\frac{\partial {w}}{\partial{x}}$ then calculate $\frac{\partial^2 {w}}{\partial{x^2}}$ and do this again for $y$ and $z$ ): \begin{align} \frac{\partial {w}}{\partial{x}}&=\frac{yz}{x^2+y^2}\\ \frac{\partial^2 {w}}{\partial{x^2}}&=\frac{2xyz}{(x^2+y^2)^2}\\ \frac{\partial {w}}{\partial{y}}&=\frac{-zx}{x^2+y^2}\\ \frac{\partial^2 {w}}{\partial{y^2}}&=\frac{-2xyz}{(x^2+y^2)^2}\\ \frac{\partial^2 {w}}{\partial{z^2}}&=0. \end{align}

I wonder if there is a shorter answer or another approach to calculate this? (Because the answer is zero I guess maybe there is different approach too.)

  • $\begingroup$ it is obvious that $\frac{\partial^2 {w}}{\partial{z^2}}$ is zero, maybe the other two are easier to see after the first derivative, write out the details in the post $\endgroup$ – Arjang Feb 28 at 18:55
  • $\begingroup$ Ok, I edited this $\endgroup$ – Soheil Feb 28 at 19:06
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    $\begingroup$ Maybe recast the Laplacian in cylindrical coordinates? $\endgroup$ – Daniel Schepler Feb 28 at 19:23

The second derivative with respect to $z$ is clearly zero. Furthermore, $$\arctan(y/x) = \operatorname{Im}(\operatorname{Log}(x + \mathrm iy))$$ and we know that the imaginary part of an analytic function is harmonic (satisfies Laplace's equation). The result for $\arctan(x/y)$ follows by symmetry.

  • $\begingroup$ How we know: $\arctan(y/x) = \operatorname{Im}(\operatorname{Log}(z))$? $\endgroup$ – Soheil Feb 28 at 19:23
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    $\begingroup$ Maybe better write $\operatorname{Log}(x+iy)$ since the variable name $z$ is already in use. $\endgroup$ – David K Feb 28 at 19:24
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    $\begingroup$ Thanks, I edited my answer. @Soheil See the definition of the complex logarithm, especially the principal value: en.wikipedia.org/wiki/Complex_logarithm $\endgroup$ – Jan Feb 28 at 19:48
  • $\begingroup$ @Jan Ok I saw the proof here: math.stackexchange.com/questions/2260028/… But still have a question: How we should recognize a function is imaginary part of an analytic function like this? should Memorize them? $\endgroup$ – Soheil Feb 28 at 19:54
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    $\begingroup$ See for example here: math.stackexchange.com/questions/48980/… It is of course also helpful if you know some definitions or memorize some common functions. $\endgroup$ – Jan Feb 28 at 19:57

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