proving the laplacian of a vector in cylindrical coordnates I am proving the following identity for the laplacian of a vector $\vec{v}=<v_r,v_\theta,v_z>$ in cylindrical coordinates:
$$\nabla^2 \vec{v}=\left( \frac{\partial^2 v_r}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2 v_r}{\partial \theta^2}+\frac{\partial^2 v_r}{\partial z^2}+\frac{1}{r}\frac{\partial v_r}{\partial r}-\frac{2}{r^2}\frac{\partial v_\theta}{\partial \theta} -\frac{v_r}{r^2}\right )\vec{e_r} \\ + \left (\frac{\partial^2 v_\theta}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2 v_\theta}{\partial \theta^2}+\frac{\partial^2 v_\theta}{\partial z^2}+\frac{1}{r}\frac{\partial v_\theta}{\partial r}+\frac{2}{r^2}\frac{\partial v_r}{\partial \theta}-\frac{v_\theta}{r^2} \right )\vec{e_\theta} \\ \left( \frac{\partial^2 v_z}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2 v_z}{\partial \theta^2}+\frac{1}{r}\frac{\partial v_z}{\partial r}+\frac{\partial^2 v_z}{\partial z^2} \right)\vec{e_z}$$ I am able to derive the following identity for the Laplacian operator in cylindrical coordinates $$\nabla^2=\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}+\frac{\partial^2}{z^2} $$. So to prove the desired identity, $$\nabla^2 \vec{v}=\left(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}+\frac{\partial^2}{z^2}  \right)(v_r\vec{e_r}+v_\theta \vec{e_\theta}+v_z\vec{e_z}) \\
= \left(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}+\frac{\partial^2}{z^2}  \right)(v_r\vec{e_r})+ \left(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}+\frac{\partial^2}{z^2}  \right)(v_\theta\vec{e_\theta})+ \left(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}+\frac{\partial^2}{z^2}  \right)(v_z\vec{e_z})$$. And upon distributing the vector components to the operator I finally get $$\nabla^2 \vec{v}=\left( \frac{\partial^2 v_r}{\partial r^2}+\frac{1}{r}\frac{\partial v_r}{\partial r}+\frac{1}{r^2}\frac{\partial^2 v_r}{\partial \theta^2}+\frac{\partial^2 v_r}{\partial z^2}-\frac{v_r}{r^2} \right)\vec{e_r} \\
+\left( \frac{\partial^2 v_\theta}{\partial r^2}+\frac{1}{r}\frac{\partial v_\theta}{\partial r}+\frac{1}{r^2}\frac{\partial^2 v_\theta}{\partial \theta^2}-\frac{v_\theta}{r^2}+\frac{\partial^2 v_\theta}{\partial z^2} \right)\vec{e_\theta} \\
\left( \frac{\partial^2 v_z}{\partial r^2}+\frac{1}{r}\frac{\partial v_z}{\partial r}+\frac{1}{r^2}\frac{\partial^2 v_z}{\partial \theta^2}+\frac{\partial^2 v_z}{\partial z^2} \right)\vec{e_z}$$ which is not the same with the identity. I am confused how the $-\frac{2}{r^2}\frac{\partial v_\theta}{\partial \theta}$ and $\frac{2}{r^2}\frac{\partial v_r}{\partial \theta}$ appeared in the $\vec{e_r}$ and $\vec{e_\theta}$ components, respectively. Where did I go wrong? Need help...thanks
 A: The product rule for second order differentiation is $(fg)'' = f''g + 2f'g'+ fg''$. You simply omitted the middle value.
A: In Cartesian coordinates, the Laplacian of a vector can be found by simply finding the Laplacian of each component, $\nabla^{2} \mathbf{v}=\left(\nabla^{2} v_{x}, \nabla^{2} v_{y}, \nabla^{2} v_{z}\right)$. However, as noted above, in curvilinear coordinates the basis vectors are in general no longer constant but vary from point to point. You will need either to derive these vectors or use the general definition of the Laplacian of a vector,
$$\nabla^{2} \mathbf{v}=\nabla(\nabla \cdot \mathbf{v})-\nabla \times(\nabla \times \mathbf{v})$$
A: Acheson = Acheson, D.J.: Elementary Fluid Dynamics, Oxford: Clarendon Press, 2005.
Wangsness = Wangsness, R. K.: Electromagnetic Fields, New York: John Wiley & Sons, 1986.
$\nabla^2\pmb{u}=(\nabla^2u_r-\frac{u_r}{r^2}-\frac{2}{r^2}\frac{\partial u_\theta}{\partial \theta})\hat{\pmb{r}} 
+(\nabla^2u_\theta+\frac{2}{r^2}\frac{\partial u_r}{\partial \theta}-\frac{u_\theta}{r^2})\hat{\pmb{\theta}}
+(\nabla^2u_z)\hat{\pmb{z}}$.
Proof
A. Apply $\nabla^2=\frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial}{\partial r})+\frac{1}{r}\frac{\partial^2}{\partial\theta^2}+\frac{\partial^2}{\partial z^2}$ to $u_\theta\hat{\pmb{\theta}}$. Except for the usual $\hat{\pmb{\theta}}$-component term $\nabla^2u_\theta\hat{\pmb{\theta}}$, we obtain  two extra terms [Wangsness, p.29, (1-79)]{Wangsness}]: they are the last two terms of $\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}(u_\theta\hat{\pmb{\theta}})=\frac{1}{r^2}(\frac{\partial^2u_\theta}{\partial \theta^2}\hat{\pmb{\theta}}-2\frac{\partial u_\theta}{\partial \theta}\hat{\pmb{r}}-u_\theta\hat{\pmb{\theta}})$.
B. Apply $\nabla^2=\frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial}{\partial r})+\frac{1}{r}\frac{\partial^2}{\partial\theta^2}+\frac{\partial^2}{\partial z^2}$ to $u_r\hat{\pmb{r}}$. Except for the usual $\hat{\pmb{r}}$-component term $\nabla^2u_r\hat{\pmb{r}}$, we obtain  two extra terms [Wangsness, p.29, (1-79)]{Wangsness}]: they are the last two terms of $\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}(u_r\hat{\pmb{r}})=\frac{1}{r^2}(\frac{\partial^2u_r}{\partial \theta^2}\hat{\pmb{r}}-2\frac{\partial u_r}{\partial \theta}\hat{\pmb{\theta}}-u_r\hat{\pmb{r}})$.
Remark 1. There is a Chinese saying, "Dismembering an ox should be as skillful as a butcher." When encountering a mammoth expansion, we should cut it properly and make its form compact and easy to operate. $\nabla^2\pmb{u}$ is used in the Navier--Stokes equation [Acheson, p.42, l.3]]. The above expansion is a professional way for fluid mechanists to write the Laplacian of a vector field in cylindrical coordinates.
Remark 2. The advantages of the above expansion are that it is more readable, easier to handle and the arrangement of various terms is more in line with the need for proof reference.
Remark 3. Only when we know the physical meaning of the concept of Laplacian of a vector field in fluid mechanics, where it is used and what role it plays can we understand the Laplacian of a vector field at a high level.
Remark 4. [Acheson, p.43, l.1-l.9] provides another proof [You may cross out [Acheson, p.43, l.3]]. That is, we want to prove the formula for $\nabla^2\pmb{u}$ in cylindrical coordinates using  [Acheson, p.43, (2.25)]. The formulas to be used here are [Wangsness, p.30, (1-85); p.31, p.31, (1-87), (1-88) & (1-89)]. This proof is more routine and straightforward because it is unnecessary to use the troublesome trick given in [Wangsness, [p.29, (1-79)].
This answer is excerpted from §1.12.(A), Remark 7, Proof. Part II in https://sites.google.com/view/lcwangpress/%E9%A6%96%E9%A0%81/papers/quantum-mechanics.
