- Source: 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:
∇
×
v
=
0
,
∇
⋅
v
=
0.
{\displaystyle {\begin{aligned}\nabla \times \mathbf {v} &=\mathbf {0} ,\\\nabla \cdot \mathbf {v} &=0.\end{aligned}}}
From the vector calculus identity
∇
2
v
≡
∇
(
∇
⋅
v
)
−
∇
×
(
∇
×
v
)
{\displaystyle \nabla ^{2}\mathbf {v} \equiv \nabla (\nabla \cdot \mathbf {v} )-\nabla \times (\nabla \times \mathbf {v} )}
it follows that
∇
2
v
=
0
{\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
v
=
(
x
y
,
y
z
,
z
x
)
{\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 :
v
=
∇
ϕ
.
(
1
)
{\displaystyle \mathbf {v} =\nabla \phi .\qquad \qquad (1)}
Then, since the divergence of v is also zero, it follows from equation (1) that
∇
⋅
∇
ϕ
=
0
{\displaystyle \nabla \cdot \nabla \phi =0}
which is equivalent to
∇
2
ϕ
=
0.
{\displaystyle \nabla ^{2}\phi =0.}
Therefore, the potential of a Laplacian field satisfies Laplace's equation.
See also
Potential flow
Harmonic function
References
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