near identity change of coordinates

Question

Problem: Consider the scalar differential equation

$$x' = \frac{4x – 24x^2 – 16x^3}{1 – 12x – 12x^2}.$$

which has a fixed point at $x^* = 0 $. For $x$ close to $x^* = 0 $ find a near identity change of coordinates $y = \phi(x) = x + b_2x_2 + b_3x_3 $ (i.e., find $b_2, b_2 ∈ R$) such that close to $y = 0 $ we have $y' = Df(x^*)y$.

I found $ Df(x^*)$, I substituted in $y' = Df(x^*)y$, and I compared the coefficients in both sides, but the fraction made me confused. So could you help me guys to do this problem?

thanks,

Answer 1

You can go all the way and compute the coefficients, but I'll try to give a quicker argument.

First of all, I assume that $y=\phi(x)$ should read $y=\phi(x)=x+b_2x^2+b_3x^3$.

Next I assume that you already know that $Df(x^*)=4$. Which means that you are interested in solving:

\begin{eqnarray} \frac{\partial\phi}{\partial x}x' &=4(x+b_2x^2+b_3x^3)\\ (1+2b_2x+3b_3x^2)\frac{4x–24x^2–16x^3}{1–12x–12x^2} &= 4(x+b_2x^2+b_3x^3). \end{eqnarray}

We can then rewrite as follows: \begin{eqnarray} (1+2b_2x+3b_3x^2)\frac{4(x–6x^2–4x^3)}{1–12x–12x^2} &= 4(x+b_2x^2+b_3x^3)\\ (1+2b_2x+3b_3x^2)\frac{x–6x^2–4x^3}{1–12x–12x^2} &= (x+b_2x^2+b_3x^3). \end{eqnarray}

In such format it is evident that the coefficients are: $b_2=-6$ and $b_3=-4$.