In vector calculus, a Laplacian vector field is a vector field which is both irrotational and incompressible. If the field is denoted as v, then it is described by the following differential equations:

${\displaystyle {\begin{aligned}\nabla \times \mathbf {v} &=\mathbf {0} ,\\\nabla \cdot \mathbf {v} &=0.\end{aligned))}$

From the vector calculus identity ${\displaystyle \nabla ^{2}\mathbf {v} \equiv \nabla (\nabla \cdot \mathbf {v} )-\nabla \times (\nabla \times \mathbf {v} )}$ it follows that

${\displaystyle \nabla ^{2}\mathbf {v} =\mathbf {0} }$

that is, that the field v satisfies Laplace's equation.

However, the converse is not true; not every vector field that satisfies Laplace's equation is a Laplacian vector field, which can be a point of confusion. For example, the vector field ${\displaystyle {\bf {v))=(xy,yz,zx)}$ satisfies Laplace's equation, but it has both nonzero divergence and nonzero curl and is not a Laplacian vector field.

A Laplacian vector field in the plane satisfies the Cauchy–Riemann equations: it is holomorphic.

Since the curl of v is zero, it follows that (when the domain of definition is simply connected) v can be expressed as the gradient of a scalar potential (see irrotational field) φ :

${\displaystyle \mathbf {v} =\nabla \phi .\qquad \qquad (1)}$

Then, since the divergence of v is also zero, it follows from equation (1) that

${\displaystyle \nabla \cdot \nabla \phi =0}$

which is equivalent to

${\displaystyle \nabla ^{2}\phi =0.}$

Therefore, the potential of a Laplacian field satisfies Laplace's equation.