Is there a common way to find an inverse vector field to a known one?
For example I have a real vector field defined by the following two equations: $$ x' = x + Δ_x $$ $$ y' = y + Δ_y $$
Where in turn:
$$ Δ_x = p_1xy+p_2(3x^2 + y^2) $$ $$ Δ_y = p_2xy+p_1(3y^2 + x^2) $$
Where $p_1,p_2$ - are real parameters
Where $<x',y'>$ is the resulting field vector at point $<x, y>$
I'm looking for a way to find an inverse vector field defined analytically by a function $f$ such that $f(x',y')=<x,y>$.
The vector field above represents tangential image distortion model, the actual distortion is represented by $Δ_x$ and $Δ_y$ parts.
Here's a visual example of real world example of such distortion $<Δ_x,Δ_y>$:
Here's it's stream plot:
How to approach this problem?
P.S.
Short version: There are also several properties that can be noted for above vector field. All of the field vectors that lay on a line crossing the XY origin are oriented in same direction, while magnitude is equal to magnitude of $<x,y>$ squared multiplied by a constant.
Long version Since a line can be represented by the following system of equations:
$$ x = ρ * cos(φ) $$ $$ y = ρ * cos(φ) $$
Then $<Δ_x,Δ_y>$ can be represented this way:
$$ Δ_x = 2 p_1 ρ^2 cos(φ) sin(φ) + p_2 (3 ρ^2 cos(φ)^2 + ρ^2 sin(φ)^2) $$ $$ Δ_y = 2 p_2 ρ^2 cos(φ) sin(φ) + p_1 (ρ^2 cos(φ)^2 + 3 ρ^2 sin(φ)^2) $$
Which can be represented by the following two equations in polar coordinates:
$$ Δ_Ρ = ρ^2 \sqrt{-8 \cos(φ)^2 p_1^2 + 8 \cos(φ)^2 p_2^2 + 16 \cos(φ) sin(φ) p_1 p_2 + 9 p_1^2 + p_2^2} $$ $$ Δ_Φ = -\arctan\frac{2 \cos(φ)^2 p_1 - 2 \cos(φ) sin(φ) p_2 - 3 p_1}{2 \cos(φ)^2 p_2 + 2 \cos(φ) sin(φ) p_1 + p_2} $$
which basically means that in polar coordinates vector field has a simpler representation:
$$ Δ_Ρ = ρ^2 constant_1 $$ $$ Δ_Φ = constant_2 $$