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I want to derive the laplacian for cylindrical polar coordinates, directly, not using the explicit formula for the laplacian for curvilinear coordinates.

Now, the laplacian is defined as $\Delta = \nabla \cdot (\nabla u)$

In cylindrical coordinates, the gradient function, $\nabla$ is defined as: $$\frac{\partial }{\partial r}\boldsymbol{e_r} + \frac{1}{r}\frac{\partial }{\partial \phi}\boldsymbol{e_{\phi}} + \frac{\partial}{\partial Z}\boldsymbol{e_Z}$$

So the laplacian would be $$(\frac{\partial }{\partial r}\boldsymbol{e_r} + \frac{1}{r}\frac{\partial }{\partial \phi}\boldsymbol{e_{\phi}} + \frac{\partial}{\partial Z}\boldsymbol{e_Z})\cdot(\frac{\partial u }{\partial r}\boldsymbol{e_r} + \frac{1}{r}\frac{\partial u }{\partial \phi}\boldsymbol{e_{\phi}} + \frac{\partial u}{\partial Z}\boldsymbol{e_Z})$$

Now, due to orthogonality, the only terms that would remain are $(\frac{\partial }{\partial r}\boldsymbol{e_r})\cdot (\frac{\partial u }{\partial r}\boldsymbol{e_r}), (\frac{1}{r}\frac{\partial }{\partial \phi}\boldsymbol{e_{\phi}})\cdot (\frac{1}{r}\frac{\partial u }{\partial \phi}\boldsymbol{e_{\phi}}), (\frac{\partial}{\partial Z}\boldsymbol{e_Z})\cdot(\frac{\partial u}{\partial Z}\boldsymbol{e_Z}).$

I know we have to use the product rule here as the basis vectors are not constant with respect to eachother.

So by the product rule, the first term becomes $\frac{\partial^2 u}{\partial r^2}$ and the third term becomes $\frac{\partial^2 u}{\partial Z^2}$, but I seem to be going wrong on the second term.

Now, I thought the second term would be evaluated like this; $(\frac{1}{r^2}\boldsymbol{e_{\phi}})\cdot(\frac{\partial^2 u}{\partial \phi^2}\boldsymbol{e_{\phi}} + \frac{\partial \boldsymbol{e_{\phi}}}{\partial \phi}\frac{\partial u}{\partial \phi})$, which i thought would be equal to $\frac{1}{r^2}(\frac{\partial^2 u}{\partial \phi^2})$ as $\frac{\partial \boldsymbol{e_{\phi}}}{\partial \phi} = -\boldsymbol{-e_r}$ so by orthogonality the term should be zero.

But I get the wrong expression, so where is my mistake?

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  • $\begingroup$ @Bye_World Is that not what i've done? $\endgroup$
    – the man
    Commented Oct 16, 2016 at 18:57
  • $\begingroup$ @Bye_World That is what I'm trying to do, as that is the definition of divergence. I know I can do it the method in which you're stating, but I'm trying to derive it this way, directly. $\endgroup$
    – the man
    Commented Oct 16, 2016 at 19:02
  • $\begingroup$ @Bye_World How did you arrive at that first term? $\endgroup$
    – the man
    Commented Oct 16, 2016 at 19:10

1 Answer 1

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The differentiation operations must be applied before the scalar products, and not the inverse way. With the present notations, the "gradient operator" in cylindrical coordinates writes \begin{equation} \nabla = \boldsymbol{e}_r \frac{\partial}{\partial r} + \boldsymbol{e}_\phi \frac{1}{r} \frac{\partial}{\partial \phi} + \boldsymbol{e}_z\frac{\partial}{\partial z} \, , \end{equation} where \begin{equation} \boldsymbol{e}_r = \cos\phi\, \boldsymbol{e}_x + \sin\phi\, \boldsymbol{e}_y \, ,\\ \boldsymbol{e}_\phi = \cos\phi\, \boldsymbol{e}_y - \sin\phi\, \boldsymbol{e}_x \, , \end{equation} and $(\boldsymbol{e}_x, \boldsymbol{e}_y, \boldsymbol{e}_z)$ is an orthonormal basis of a Cartesian coordinate system such that $\boldsymbol{e}_z = \boldsymbol{e}_x\times \boldsymbol{e}_y$. Some basis vectors depend on the coordinates, according to the rule \begin{equation} \frac{\partial \boldsymbol{e}_r}{\partial \phi} = \boldsymbol{e}_\phi \qquad\text{and}\qquad \frac{\partial \boldsymbol{e}_\phi}{\partial \phi} = -\boldsymbol{e}_r \, . \end{equation} When expanding $\nabla\cdot (\nabla u)$ and using the product rule of differentiation, \begin{aligned} &\nabla\cdot (\nabla u) = \left(\boldsymbol{e}_r \frac{\partial}{\partial r} + \boldsymbol{e}_\phi \frac{1}{r} \frac{\partial}{\partial \phi} + \boldsymbol{e}_z\frac{\partial}{\partial z}\right) \cdot \left(\frac{\partial u}{\partial r} \boldsymbol{e}_r + \frac{1}{r} \frac{\partial u}{\partial \phi} \boldsymbol{e}_\phi + \frac{\partial u}{\partial z} \boldsymbol{e}_z\right)\\ &\phantom{\nabla\cdot (\nabla u)} = \boldsymbol{e}_r \cdot \frac{\partial}{\partial r} \left(\frac{\partial u}{\partial r} \boldsymbol{e}_r + \frac{1}{r} \frac{\partial u}{\partial \phi} \boldsymbol{e}_\phi + \frac{\partial u}{\partial z} \boldsymbol{e}_z\right) \\ &\phantom{\nabla\cdot (\nabla u) =}+ \boldsymbol{e}_\phi \cdot \frac{1}{r} \frac{\partial}{\partial \phi} \left(\frac{\partial u}{\partial r} \boldsymbol{e}_r + \frac{1}{r} \frac{\partial u}{\partial \phi} \boldsymbol{e}_\phi + \frac{\partial u}{\partial z} \boldsymbol{e}_z\right)\\ &\phantom{\nabla\cdot (\nabla u) =}+ \boldsymbol{e}_z \cdot \frac{\partial}{\partial z} \left(\frac{\partial u}{\partial r} \boldsymbol{e}_r + \frac{1}{r} \frac{\partial u}{\partial \phi} \boldsymbol{e}_\phi + \frac{\partial u}{\partial z} \boldsymbol{e}_z\right) \\ &\phantom{\nabla\cdot (\nabla u)} = \frac{1}{r}\frac{\partial}{\partial r}\left(r \frac{\partial u}{\partial r}\right) + \frac{1}{r^2}\frac{\partial^2 u}{\partial \phi^2} + \frac{\partial^2 u}{\partial z^2} \, , \end{aligned} the correct Laplacian is obtained.

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  • $\begingroup$ I see that you got the Laplacian at the end but that derivation does not seem right to me. Del u is a dyadic product which is a Tensor. But it looks like you took the dot product of Del and u $\endgroup$ Commented Oct 2, 2021 at 4:02
  • $\begingroup$ @user1721803 In the end the correct Laplace operator $\Delta^2 u = \nabla\cdot(\nabla u)$ is obtained. In fact the divergence operator is often viewed as a dot product. If you know a more rigorous derivation, you are welcome to propose it as an answer. $\endgroup$
    – EditPiAf
    Commented Oct 4, 2021 at 10:47

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