What are the necessary and sufficient condions for a laplacian to be zero?

Question

Let $F$ be a function of $x,y,z$, namely $F(x,y,z)$.

My question:

What are the necessary and sufficient conditions for $\triangledown$$^2$$F(x,y,z)$=$0$, what does it signify?

I am aware that if $d$ is a differential operator, then $d$ $(\triangledown$$F(x,y,z))$$=0$, where $i, j, k$ are $dx,dy,dz$ respectively. That is, $d(\frac{\partial F}{\partial x}dx,\frac{\partial F}{\partial y}dy,\frac{\partial F}{\partial z}dz)=0$.

I know this is true if $F(x,y,z)$ is the force associated with a conservative potential function or in other words conservative vector field. However, I can not seem to relate it with the Laplacian. Are the two related, or they just seem related to me?

Thank you.

Answer 1

Remember that for any vector field $\mathbf{F}$, $\nabla^2 \mathbf{F} = \boldsymbol\nabla(\boldsymbol\nabla\cdot \mathbf{F}) - \boldsymbol\nabla\times(\boldsymbol\nabla\times \mathbf{F}$). So $\nabla^2 \mathbf{F} = 0$ is equivalent to $\boldsymbol\nabla (\boldsymbol\nabla\cdot \mathbf{F}) = \boldsymbol\nabla\times(\boldsymbol\nabla\times \mathbf{F})$.

If $\mathbf{F}$ is conservative, then $\boldsymbol\nabla\times\mathbf{F} = 0$, and so we must have $\boldsymbol\nabla(\boldsymbol\nabla\cdot\mathbf{F}) = 0$. A scalar function with vanishing gradient is a constant.

Thus, if $\mathbf{F}$ is a conservative vector field, $\nabla^2 \mathbf{F} = 0$ if and only if $\boldsymbol\nabla\cdot \mathbf{F} = C$ for some constant $C$.