Proving the Henon Map is the Most General Quadratic Map with Constant Jacobian

[hakkai]

Homework Statement

Show that the most general two-dimensional quadratic map with a constant Jacobian is the Henon map:

x_n+1=y_n+1-ax²_n
y_n+1=bx_n,

where a,b are positive constants.

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Homework Equations

From the general quadratic map,
x_n+1=f₁+a₁x_n+b₁y_n+c₁x²_n+d₁x_ny_n+e₁y²_n

y_n+1=f₂+a₂x_n+b₂y_n+c₂x²_n+d₂x_ny_n+e₂y²_n

Linear transformations (deformations, shifts, and rotations) can be used to transform this into the Henon map, or an equivalently simple canonical form - that is, when the Jacobian is required to be constant.

The Attempt at a Solution

Requiring the Jacobian determinant of the most general quadratic map to be constant, we get five equations that relate the 12 coefficients - the coefficients of the five different terms {x,y,x², xy, y²} in the determinant must be zero so that the determinant is always constant. The five equations are essentially:

a₁d₂+2c₁b₂=2b₁c₂+d₁a₂

2a₁e₂+d₁b₂=b₁d₂=b₁d₂+2e₁a₂

c₁e₂=e₁c₂
d₁c₂=c₁d₂
e₁d₂=d₁e₂

Apparently, using shifts, linear transformations, and rotations, together with these relations, one can reduce the general quadratic map to the the Henon map. But brute forcing the transformation in all its generality leaves me with a large mess of intractable algebra. Does anyone have any ideas on how to do this problem, i.e. how to look for sensible transformations that actually work? Homework's due at the end of Friday and I've been stuck on this for awhile, so I'd really appreciate any advice.

 

[mighty2000]

Thank you for your post. Your task is to prove that the most general two-dimensional quadratic map with a constant Jacobian is the Henon map. This is a challenging problem, but I will do my best to guide you through the solution.

First, let's take a closer look at the Henon map. It is a two-dimensional map defined by the following equations:

xn+1=yn+1-ax2n
yn+1=bxn

We can rewrite this map as a vector equation:

(xn+1, yn+1) = (yn+1-ax2n, bxn)

Now, let's consider the most general quadratic map:

xn+1=f1+a1xn+b1yn+c1x2n+d1xnyn+e1y2n
yn+1=f2+a2xn+b2yn+c2x2n+d2xnyn+e2y2n

We can also rewrite this map as a vector equation:

(xn+1, yn+1) = (f1+a1xn+b1yn+c1x2n+d1xnyn+e1y2n, f2+a2xn+b2yn+c2x2n+d2xnyn+e2y2n)

Now, in order for the Jacobian determinant to be constant, we need to find a transformation that will make the coefficients of the five different terms in the determinant equal to zero. This will ensure that the determinant is always constant, regardless of the values of xn and yn.

Let's start by looking at the first term, which is a1d2+2c1b2. We can see that this is the only term that contains both a1 and d2. Therefore, we need to find a transformation that will make the coefficients of a1 and d2 equal to zero. This can be achieved by setting a1=-2c1b2/d2. Similarly, we can set b1=-d1a2/2c2 and b2=-e1a2/c2.

Next, let's look at the second term, which is 2a1e2+d1b2. Again, we can see that this is the only term that contains both a1 and e2. Therefore, we need to find a transformation that will make the coefficients of a1 and e2 equal to zero. This can be achieved