What is the Laplacian of $k\hat{r}$ where $r=\sqrt{x^2+y^2}$ and $k$ is a constant? [closed]

Question

Using 3D cylindrical coordinates, I get 0 as the answer.

$$ \nabla^{2} (k \hat{r}) = \hat{r} (\frac{1}{r} \frac{\partial }{\partial r}\left(r \frac{\partial (k)}{\partial r}\right) + 0 + 0) + \hat{\phi} [\nabla^{2}(0)] + \hat{z} [\nabla^{2}(0)]= 0$$

Am I correct? (Edit: I'm not)

The solution to this question in the book that I'm using has ignored the unit vector and has given the Laplacian of $kr$ instead which is $k/r$.

Edit: Directly applying $\nabla^2\mathbf A=\nabla(\nabla\cdot\mathbf A)-\nabla\times\nabla\times\mathbf A$ gave me the answer $\frac{-k}{r^2} \hat{r}$ which is correct according to the expression for vector Laplacian in cylindrical coordinates given on page 60 of this pdf: https://ws.engr.illinois.edu/sitemanager/getfile.asp?id=135 Thanks everyone for the replies.

Answer 1

Your problem is complicated by the fact that you operate with the Laplacian on a vector. To quote Wikipedia:

The vector Laplace operator, also denoted by $\nabla^2$, is a differential operator defined over a vector field.[6] The vector Laplacian is similar to the scalar Laplacian; whereas the scalar Laplacian applies to a scalar field and returns a scalar quantity, the vector Laplacian applies to a vector field, returning a vector quantity. When computed in orthonormal Cartesian coordinates, the returned vector field is equal to the vector field of the scalar Laplacian applied to each vector component.

The vector Laplacian of a vector field $\mathbf A$ is defined as $$\nabla^2\mathbf A=\nabla(\nabla\cdot\mathbf A)-\nabla\times\nabla\times\mathbf A$$ In Cartesian coordinates, this reduces to the much simpler form as $$\nabla^2\mathbf A=(\nabla^2 A_x,\nabla^2 A_y,\nabla^2 A_z)^T$$

So, because you act on a vector you can't use the usual formula for the Laplacian in polar coordinates. But the Cartesian form is still straightforward (although cumbersome). For the x-component you will get $$-\frac{x}{\left(x^2+y^2\right)^{3/2}}$$ I checked this with Mathematica. What do you get when you write is as $$c\frac{\hat r}{r^n}?$$ With $c$ a constant and $n$ a natural number.

Answer 2

In Cartesian coordinates, where each vector component may be thought of as an independent function, $$ \hat \rho = \begin{pmatrix} \cos\phi \\ \sin \phi \\ 0\end{pmatrix}, $$ strictly a function of $\phi$; since the (scalar!) Laplacian on such functions is merely ${1\over \rho^2} \partial_\phi^2$, it simply follows that $$ \nabla^2 \hat \rho= -{1\over \rho^2} \hat \rho ~. $$