Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
1  
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! :)

share|cite|improve this answer
    
Cylindrical coordinates $(r, \phi, z)$ are also probably better suited to this computation. mathworld.wolfram.com/CylindricalCoordinates.html – 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.