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Assume $$(x^{2}+y^{2})\cos^{2}\psi+z^{2}\cot^{2}\psi=A^2$$ which $A$ is constant. How we can show $\psi(x,y,z)$ satisfies the Laplacian equation $\psi_{xx}+\psi_{yy}+\psi_{zz}=0$ ($\operatorname{div}\nabla\psi=0$) without calculating $\psi(x,y,z)$? I calculate $\psi(x,y,z)$ itself and differentiate, but I'm looking for easier methods, It's not important to use what, only the time that it takes is important.

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Hmmm...WolframAlpha can't simplify the expression, but you could try implicit differentiation and Laplacian in the cylindrical coordinates. May I know the origin of this question? – Shuhao Cao Jul 28 '13 at 3:17
@ShuhaoCao, about the origin of the question, I don't know. A physics student asked it from me and I had no idea so I put it here. – AmirHosein SadeghiManesh Jul 28 '13 at 7:32
Are you really sure this is true? Where does this come from? The simplest harmonic functions ($\psi(x,y,z)=x, \psi(r, \theta, z)=\theta$, etc...) don't seem to satisfy that equation. – Giuseppe Negro Dec 16 '15 at 16:14
@GiuseppeNegro really? So you checked and this equation failed to satisfy Laplacian? I'll try later when I found time, as I answered to Shushao Cao, this question was asked from me by a Physics student. – AmirHosein SadeghiManesh Dec 17 '15 at 9:04
@GiuseppeNegro: $\psi$ is not just any harmonic function! In fact, this implicit equation tells us what $\psi$ is! Hence, it is not $x$ or $\theta$! We just want to find the Laplacian of this given $\psi$ and observe that it is zero. :) – H. R. Dec 17 '15 at 9:46

I just want to introduce the proper implicit differentiation formula for this problem. If you don't want to go by directly finding $\psi$ then you should use implicit differentiation and I just give you the tools. For this purpose, you need a product rule and also a chain rule for the Laplacian operator and also a chain-rule for the gradient operator. First, define the following functions

$$\eqalign{ & f = f(t) \cr & g = g(x,y,z) \cr & h = h(x,y,z) \cr} $$

Then the rules you need will be

$$\begin{align} {\nabla ^2}\left( {h g} \right) &= g{\nabla ^2}h + h{\nabla ^2}g + 2\nabla h \cdot \nabla g \cr {\nabla ^2}\left( {f \circ g} \right) &= \left( {f'' \circ g} \right){\left\| {\nabla g} \right\|^2} + \left( {f' \circ g} \right){\nabla ^2}g \cr \nabla \left( {f \circ g} \right) &= \left( {f' \circ g} \right)\nabla g \cr\end{align}$$

You can prove them easily by using index notation or some vector identities. Now, combining the above relations by replacing $g$ with $f \circ g$ in the first formula leads to

$$\eqalign{ {\nabla ^2}\left[ {h\left( {f \circ g} \right)} \right] &= \left( {f \circ g} \right){\nabla ^2}h + h{\nabla ^2}\left( {f \circ g} \right) + 2\nabla h \cdot \nabla \left( {f \circ g} \right) \cr &= \left( {f \circ g} \right){\nabla ^2}h + h\left[ {\left( {f'' \circ g} \right){{\left\| {\nabla g} \right\|}^2} + \left( {f' \circ g} \right){\nabla ^2}g} \right] + 2\nabla h \cdot \left[ {\left( {f' \circ g} \right)\nabla g} \right] \cr } $$

and finally

$$\boxed{{\nabla ^2}\left[ {h\left( {f \circ g} \right)} \right] = \left( {f \circ g} \right){\nabla ^2}h + h\left( {f' \circ g} \right){\nabla ^2}g + 2\left( {f' \circ g} \right)\nabla h \cdot \nabla g + h\left( {f'' \circ g} \right){\left\| {\nabla g} \right\|^2}}$$

Then, just apply the Laplacian operator to the both side of

$$(x^{2}+y^{2})\cos^{2}\psi+z^{2}\cot^{2}\psi = A^{2}$$

to obtain

$${\nabla ^2}\left[ {({x^2} + {y^2}){{\cos }^2}\psi } \right] + {\nabla ^2}\left[ {{z^2}{{\cot }^2}\psi } \right] = 0$$

and do the computations for proper choices of $h$, $f$, and $g$. The arithmetic seems to be lengthy so it takes some time! :)

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Cylindrical coordinates $(r, \phi, z)$ are also probably better suited to this computation. – Giuseppe Negro Dec 16 '15 at 14:23
@GiuseppeNegro: You are right, cylindrical is a good choice. However, any coordinates we use, we need to use the three rules I mentioned. :) – H. R. Dec 16 '15 at 14:36

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