I have just read (without further explanation) that any vector field $(v_x(x,y),v_y(x,y))$ from $\mathbb{R}^2$ to $\mathbb{R}^2$, which has a continuous derivative, can be uniquely written as the sum of a gradient and a hamiltonian vector field, i.e. $$ v_x(x,y) = \partial_xP(x,y) + \partial_y S(x,y) $$ $$ v_y(x,y) = \partial_yP(x,y) -\partial_x S(x,y)$$for $P$,$S$ : $\mathbb{R}^2 \to \mathbb{R}$.
If this is true, how can I show it? It seems non-obvious to me.
Thanks!
Observe that if you replace $P$ by $\bar{P} = P + ax + by + c$ and $S$ by $\bar{S} = S + bx - ay + d$, we have that
$$ \partial_x \bar{P} + \partial_y \bar{S} = \partial_x P + a + \partial_y S - a = \partial_x P + \partial_y S $$
and
$$ \partial_y \bar{P} - \partial_x \bar{S} = \partial_y P + b - \partial_x S - b = \partial_y P - \partial_x S $$
so strictly speaking, your source was wrong about uniqueness of the decomposition.
In more generality, the decomposition you mentioned is a manifestation of the Hodge decomposition, and in the case of three dimensions instead of two, it often goes by the name of Helmholtz decomposition.
Roughly speaking, we can compute $P$ by solving
$$ \triangle P = \operatorname{div} v = \partial_x v_x + \partial_y v_y $$
and
$$ \triangle S = \operatorname{curl} v = -\partial_x v_y + \partial_y v_x $$
That the decomposition is true hinges on
This is the case, for example, if we assume that the derivatives of the coefficients $v_x$ and $v_y$ decays rapidly (faster than any polynomial). In that case using the Newton potential we can solve the Poisson equations for $P$ and $S$, and verify that they are up to constants the unique bounded potentials for the decomposition (notice that $ax + by + c$ is not bounded unless $a = b = 0$).
