If $G$ is a function with three components. $G$ is actually a Green's function in my case, like $G( \textbf{x} , \xi)$ with $\textbf{x} = (x,y)$ and $\xi = (\xi _x, \xi _y)$ so I am guessing it is three components...

Is this a vector or a scalar? Could someone give an example. Is it just simply each component differentiated twice (first component wrt $x$, then second is $y$ and third is $z$) and then summed?

  • $\begingroup$ It is exactly what you think it is. $\endgroup$ Commented Mar 9, 2016 at 21:33
  • $\begingroup$ @Omnomnomnom I am just so confused because we are given $\nabla^2 G= \delta (\textbf x - \xi)$ for $x,y>0$. What exactly is this? Are we meant to reverse the differentiation to get an ODE or something. And does $G$ have $2$, $3$ or $4$ components? $\endgroup$
    – snowman
    Commented Mar 9, 2016 at 21:36
  • $\begingroup$ @snowman Here $G$ is a scalar function of two 2D vectors, and the same is true of $\nabla^2 G$. Mathematicians often prefer $\Delta$ for what you are using as $\nabla^2$ (the Laplacian), with a different meaning being used for $\nabla^2$ itself. $\endgroup$
    – Ian
    Commented Mar 9, 2016 at 21:40

2 Answers 2


$\nabla$ is defined by the vector $(\frac{d}{dx},\frac{d}{dy},\frac{d}{dz})$, in Cartesian coordinates.

$G$ can be either a scalar function $G(x,y,z)$ or a vector function $\mathbf G (x,y,z)$.

From here on, it's just the normal vector arithmetic, where the differential operators of $\nabla$ act on either a vector component of $\mathbf G$ --- to end up with the scalar function $$\partial_xG_x(x,y,y)+\partial_yG_y(x,y,z)+\partial_zG_z(x,y,z)$$ or make a vector out of $G(x,y,z)$ $$ \Big(\partial_x G(x,y,z),\ \partial_y G(x,y,z),\ \partial_z G(x,y,z) \Big).$$

For the $\nabla^2$, you just repeat the first $\nabla$ operation to see what happens.

  • $\begingroup$ By convention, $\nabla^2$ means $\nabla \cdot \nabla$ $\endgroup$ Commented Mar 9, 2016 at 21:47

$\nabla^2=\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$ is a scalar. Hence: if $G$ is a scalar, then $\nabla^2 G$ is scalar too; if $G$ is a vector, then so is $\nabla^2 G$.

What $G$ is, must be seen from context. According to your comment, you know it is a scalar.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .