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:

$\begin{array}{rl}\mathrm{\nabla }\times \mathbf{𝐯}& =\mathbf{𝟎},\\ \mathrm{\nabla }\cdot \mathbf{𝐯}& =0.\end{array}$

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

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

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 $\mathbf{𝐯}=(xy,yz,zx)$ 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 :

$\mathbf{𝐯}=\mathrm{\nabla }\varphi .(1)$

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

$\mathrm{\nabla }\cdot \mathrm{\nabla }\varphi =0$

which is equivalent to

${\mathrm{\nabla }}^2\varphi =0.$

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

See also

• Potential flow
• Harmonic function