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:
- [math]\displaystyle{ \begin{align} \nabla \times \mathbf{v} &= \mathbf{0}, \\ \nabla \cdot \mathbf{v} &= 0. \end{align} }[/math]
From the vector calculus identity [math]\displaystyle{ \nabla^2 \mathbf{v} \equiv \nabla (\nabla\cdot \mathbf{v}) - \nabla\times (\nabla\times \mathbf{v}) }[/math] it follows that
- [math]\displaystyle{ \nabla^2 \mathbf{v} = \mathbf{0} }[/math]
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 [math]\displaystyle{ {\bf v} = (xy, yz, zx) }[/math] 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 :
- [math]\displaystyle{ \mathbf{v} = \nabla \phi. \qquad \qquad (1) }[/math]
Then, since the divergence of v is also zero, it follows from equation (1) that
- [math]\displaystyle{ \nabla \cdot \nabla \phi = 0 }[/math]
which is equivalent to
- [math]\displaystyle{ \nabla^2 \phi = 0. }[/math]
Therefore, the potential of a Laplacian field satisfies Laplace's equation.
See also
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