Laplacian vector field

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:
 * $$\begin{align}

\nabla \times \mathbf{v} &= \mathbf{0}, \\ \nabla \cdot \mathbf{v} &= 0. \end{align}$$

From the vector calculus identity $$\nabla^2 \mathbf{v} \equiv \nabla (\nabla\cdot \mathbf{v}) - \nabla\times (\nabla\times \mathbf{v})$$ it follows that
 * $$\nabla^2 \mathbf{v} = \mathbf{0}$$

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 $${\bf v} = (xy, yz, zx)$$ 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 :
 * $$ \mathbf{v} = \nabla \phi. \qquad \qquad (1) $$

Then, since the divergence of v is also zero, it follows from equation (1) that
 * $$ \nabla \cdot \nabla \phi = 0 $$

which is equivalent to
 * $$ \nabla^2 \phi = 0.$$

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