Condition for Vector Field to satisfy Poisson's Equation

Theorem

Let $\mathbf V$ be a vector field over a region of space $R$.

Then:

$\mathbf V$ is conservative but not solenoidal

if and only if:

$\mathbf V$ is the gradient of a scalar field $F$ over $R$ which satisfies Poisson's equation over $R$:

$\nabla^2 F = \phi$

where $\phi$ is a function which is not identically zero.

Proof

Sufficient Condition

Let $\mathbf V$ be conservative but not solenoidal.

From Vector Field is Expressible as Gradient of Scalar Field iff Conservative:

$\mathbf V = \grad F$

for some scalar field $F$ over $R$.

Because $\mathbf V$ is not solenoidal, we have:

$\exists \mathbf v \in R: \operatorname {div} \mathbf v \ne 0$

that is:

$\operatorname {div} \grad F \ne 0$

for at least some $\mathbf v \in R$.

Hence by Laplacian on Scalar Field is Divergence of Gradient:

$\nabla^2 F = \phi$

where:

$\phi$ is not identically zero

$\nabla^2$ is the Laplacian on $F$.

Hence $F$ satisfies Poisson's equation.

$\Box$

Necessary Condition

Let $\mathbf V$ be the gradient of a scalar field $F$ over $R$ which satisfies Poisson's equation:

$\nabla^2 F = \phi$

where $\phi$ is not identically zero.

Thus $F$ is such that:

$\mathbf V = \grad F$

and from Vector Field is Expressible as Gradient of Scalar Field iff Conservative it follows that $\mathbf V$ is conservative.

Then by Laplacian on Scalar Field is Divergence of Gradient:

$\exists \mathbf v \in R: \operatorname {div} \grad F \ne 0$

That is:

$\exists \mathbf v \in R: \operatorname {div} \mathbf V \ne 0$

and so by definition $\mathbf V$ is specifically not solenoidal.

$\blacksquare$

Sources

• 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $7$. The Classification of Vector Fields: $\text {(ii)}$