Let F,G be vector �fields on the plane such that both com-positions F dot� G, G dot� F : R^2 -> R^2 -> R^2 are the identity (x, y)->(x, y).
Prove that the fi�eld F is incompressible if and only if G is incompressible.
(A vector fi�eld is called incompressible if its divergence is zero everywhere.)
To prove that the field F is incompressible if and only if G is incompressible, we need to show two implications:
1. If F is incompressible, then G is incompressible.
2. If G is incompressible, then F is incompressible.
Let's start with the first implication:
1. If F is incompressible, then G is incompressible:
To prove this, we need to show that if the divergence of F is zero everywhere, then the divergence of G is also zero everywhere.
The divergence of a vector field F = (F1, F2) in two dimensions is given by the following formula:
div(F) = ∂F1/∂x + ∂F2/∂y
We know that F dot G = (x, y). Taking the dot product of F with G, we have:
F dot G = F1G1 + F2G2 = x
Now, let's compute the divergence of G using the given information:
div(G) = ∂G1/∂x + ∂G2/∂y
Taking the dot product of both sides with F, we get:
F dot div(G) = F dot (∂G1/∂x + ∂G2/∂y)
Using the properties of the dot product, we can distribute F dot over each term:
F dot div(G) = F1(∂G1/∂x) + F1(∂G2/∂y) + F2(∂G1/∂x) + F2(∂G2/∂y)
Notice that this expression is symmetric because F dot G = G dot F = (x, y). So we can rewrite it as:
F dot div(G) = (∂G1/∂x)(F1 + F2) + (∂G2/∂y)(F1 + F2)
Since F dot G = x, we can substitute (F1 + F2) with x in the expression:
F dot div(G) = (∂G1/∂x)x + (∂G2/∂y)x
Now, let's recall that F dot G = (x, y) implies F1G1 + F2G2 = x. We can rearrange this equation to solve for F1 in terms of F2:
F1 = (x - F2G2)/G1
Substituting this expression for F1 in the previous equation, we have:
F dot div(G) = (∂G1/∂x)(x - F2G2)/G1 + (∂G2/∂y)x
Simplifying this expression, we get:
F dot div(G) = (∂G1/∂x)(x - F2G2) + (∂G2/∂y)x
Now, notice that both F dot div(G) and x are equal to F dot G. Therefore, we have:
F dot G = (∂G1/∂x)(x - F2G2) + (∂G2/∂y)F dot G
Since F dot G = x, we can substitute it in the equation:
x = (∂G1/∂x)(x - F2G2) + (∂G2/∂y)x
Expanding the terms and rearranging, we get:
0 = (∂G1/∂x)(F2G2) + (∂G2/∂y)(1 - x)
Now, recall that G dot F = (x, y). We can rearrange this equation to solve for G2 in terms of G1:
G2 = (y - F1G1)/F2
Substituting this expression for G2 in the previous equation, we have:
0 = (∂G1/∂x)[F2(y - F1G1)/F2] + (∂G2/∂y)(1 - x)
Simplifying this expression, we get:
0 = (∂G1/∂x)(y - F1G1) + (∂G2/∂y)(1 - x)
Expanding the terms, we have:
0 = (∂G1/∂x)y - (∂G1/∂x)F1G1 + (∂G2/∂y) - (∂G2/∂y)x
Rearranging, we get:
0 = (∂G1/∂x)y + (∂G2/∂y) - [(∂G1/∂x)F1 + (∂G2/∂y)x]G1
Comparing this equation with the divergence formula, we can conclude that div(G) = 0. Therefore, if F is incompressible, then G is also incompressible.
Now, let's move on to the second implication:
2. If G is incompressible, then F is incompressible:
To prove this, we need to show that if the divergence of G is zero everywhere, then the divergence of F is also zero everywhere.
Using similar steps as in the previous implication, we can derive an equation:
div(F) = (∂F1/∂x) + (∂F2/∂y) = (∂F2/∂y)(F1 + F2) + (∂F1/∂x)
Now, using the fact that G dot F = (x, y), we can substitute (F1 + F2) with y:
div(F) = (∂F2/∂y)(y) + (∂F1/∂x)
Notice that this expression is symmetric to the equation obtained in the previous implication for div(G). Therefore, we can conclude that if G is incompressible (i.e., div(G) = 0), then F is also incompressible (i.e., div(F) = 0).
Hence, we have proven that F is incompressible if and only if G is incompressible.