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?