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

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

My question:

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

I am aware that if $$d$$ is a differential operator, then $$d$$ $$(\triangledownF(x,y,z))=0$$, 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.

• I think you mean $F(x,y,z)$ is the potential associated with a conservative force. Oct 22 '18 at 3:09
• @spaceisdarkgreen No, $F(x,y,z)$ is the force associated with a conservative vector field. That is, $\exists$$U(x,y,z) such that \triangledown$$U(x,y,z)=-F(x,y,z)$ Oct 22 '18 at 3:16
• Well, you describe it as a function, not a vector field. And in the case of a conservative field/potential, both the potential and the vector field have zero laplacian. Oct 22 '18 at 3:22
• @spaceisdarkgreen Sorry for the confusion, I have seen vector fields referred to as function as well. It is generally done for ease of notation. In fact, a vector field is a function of $x, y$, and $z$. Thanks for the input that potential and the vector field has zero Laplacian. Oct 22 '18 at 3:33
• Yeah, I think that a vector field is a function, in the same sense that a scalar field is also a function (both with $>1$ variables). Oct 22 '18 at 3:52

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 $$C$$.
• The part where I ask about the relation between differential of a gradient, and the Laplacian. It starts from: "I am aware if $d$ is a differential operator..." Regard. Mar 29 '19 at 4:25
• its from differential geometry, wedge product. en.m.wikipedia.org/wiki/One-form. The problem is I know that differential of a gradient is $0$ if the gradient is a conservative field, but I don't know why that is. Mar 29 '19 at 4:57