1
$\begingroup$

Vector Calc has been a while, and I can't seem to find anything about this online. How can I solve for D in an equation like so: $$\nabla\cdot D = \rho_v$$ How do I "move" the gradient to the other side?

  • 0
    Hi, could you elaborate on what are $D$ and $\rho$ ? Can you solve the equation in dimension 1 (where you probably get an ODE) ? – 2017-02-28
  • 1
    It appears to be the divergence operator of a vector field, from [Gauss's law](https://en.wikipedia.org/wiki/Gauss%27s_law). – 2017-02-28
  • 0
    @ThibautDumont It is gauss's law, but you can just treat them as arbitrary variables. – 2017-02-28

1 Answers 1

1

Depending on smoothness and boundary conditions there may or may not be a solution and it is not uniquely determined since you can add any divergence-free field and still satisfy the PDE.

Helmholtz's theorem states that any sufficiently smooth vector field $\mathbf{D}$ that decays sufficiently fast can be decomposed into curl-free and divergence-free components of the form

$$\mathbf{D} = \nabla \phi + \nabla \times\mathbf{A}.$$

Hence,

$$\nabla \cdot \mathbf{D} = \nabla \cdot \nabla \phi + \nabla \cdot \nabla \times \mathbf{A} = \nabla^2\phi,$$

and $\phi$ must be a solution of the Poisson equation

$$\nabla^2 \phi = \rho_v.$$

Given suitable boundary conditions a unique solution for $\phi$ can be obtained leaving $\mathbf{D}$ determined up to the addition of an arbitrary divergence free field.

On an unbounded domain for well-behaved $\rho_v$ we have the solution

$$\phi(\mathbb{x}) = \frac{-1}{4 \pi}\int_{\mathbb{R}^3} \frac{\rho_v(\mathbb{x}')}{|\mathbb{x} - \mathbb{x}'|}d\, \mathbb{x}'.$$

Ignoring the divergence-free part, this effectively "moves" $\nabla$ to the other side:

$$\mathbf{D} = \frac{-1}{4 \pi} \nabla \int_{\mathbb{R}^3} \frac{\rho_v(\mathbb{x}')}{|\mathbb{x} - \mathbb{x}'|}d\, \mathbb{x}'.$$

In physical terms,if $\rho_v$ is the electric charge distribution, then the integral is the electric potential and the electric field $\mathbf{D}$ is the gradient of the potential.