1

Can the Poisson equation be solved exactly when we know that the source term is a divergence?

$\nabla^2 \psi = \nabla \cdot \mathbf{F}$

user250673
  • 11

1 Answers1

1

If $F$ is a conservative vector field, there exists $\phi$ such that $\nabla\phi=F$. Even if $F$ is not conservative we can in many cases decompose it into $F=\nabla\phi+v$ where $\nabla\cdot v=0$ by Helmholtz's Theorem. Then $\phi$ is a "particular solution," to borrow a term from O.D.E.'s. If $\mu$ is a solution to Laplace's equation, $\psi=\phi+\mu$ is a solution to $$\nabla^2 \psi=\nabla\cdot F.$$

Plutoro
  • 22,730
  • 3
  • 41
  • 71
  • @user250673 There are versions of Hodge/Helmholtz decomposition with boundary conditions, too. Look them up. –  Jun 26 '15 at 01:47
  • Ok, all these approaches just express the same thing in different terms, and there appears to be no easy way to solve the Poisson equation. – user250673 Jun 26 '15 at 03:34
  • Do you know what $F$ is? – Plutoro Jun 26 '15 at 03:52
  • I was able to write it as $\mathbf{F}=(\nabla u \cdot \nabla u) \nabla u$ where we also know $\nabla^2 u = 0$. – user250673 Jun 26 '15 at 04:12
  • 1
    If that is the case $$\nabla\cdot F=\nabla\cdot\left[(\nabla u\cdot\nabla u)\nabla u\right]\=\nabla(\nabla u\cdot\nabla u)\cdot\nabla u+(\nabla u\cdot\nabla u)\nabla^2u\=(2\nabla u\nabla^2u)\cdot\nabla u\=0,$$ and you are left with Laplace's equation which is probably either very good for you, or very bad for you. – Plutoro Jun 26 '15 at 04:22
  • Thanks for your reply. I believe $\nabla (\nabla u \cdot \nabla u) = 2 (\nabla u \cdot \nabla)\nabla u$ and that is not zero. – user250673 Jun 26 '15 at 04:30
  • Ah, you are right. I'm misremembering my vector identities. – Plutoro Jun 26 '15 at 04:33