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?
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.