# Normal form of a vector field in $\mathbb {R}^4$.

EDIT (In responce to xpaul's answer): I'm looking for the exact normal form, not the one up to $O(|x^5|)$

Those are two analogous problems, the first one of which I have already accounted for.

Find the normal form of the vector fields:

a) Solved.

b) $$\dot x_1=x_2$$ $$\dot x_2=-x_1$$ $$\dot x_3=\sqrt 2 x_4+x_1^3$$ $$\dot x_4=-\sqrt 2 x_3+x_3x_4^2$$

Solution:

The matrix of the linearised system is $\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & \sqrt 2 \\ 0 & 0 & -\sqrt 2 & 0 \\ \end{array} \right)$.

The eigenvalues are $\{\pm i\sqrt 2, \pm i\}$. Therefore, we have infinitely many resonances $\lambda_i=\lambda_i+k(\lambda_1+\lambda_2)+p(\lambda_3+\lambda_4), (i=\overline{1,4}), (k+p\geq1)$. There are no other resonances.

Furthermore, by Poincare-Dulac's theorem, I can map the vector field to its linear part up to resonant monomes, but how should I proceed?

EDIT:

The resonant monomes are then $$x_i\dot{}x_1^k\dot{}x_2^k\dot{}x_3^p\dot{}x_4^p\dot{}e_i$$ where $e_i$ is the unit vector along the i-th coordinate.

Then by Poincare-Dulac's theorem, the vector field can be "reduced" to: $$\dot x_1=x_2 + \sum_{k+p\geq 1} c_{1kp}x_1^{k+1}\dot{}x_2^k\dot{}x_3^p\dot{}x_4^p$$

$$\dot x_2=-x_1 + \sum_{k+p\geq 1} c_{2kp}x_1^k\dot{}x_2^{k+1}\dot{}x_3^p\dot{}x_4^p$$

$$\dot x_3=\sqrt 2 x_4 + \sum_{k+p\geq 1} c_{3kp}x_1^k\dot{}x_2^k\dot{}x_3^{p+1}\dot{}x_4^p$$

$$\dot x_4=-\sqrt 2 x_3+ \sum_{k+p\geq 1} c_{4kp}x_1^k\dot{}x_2^k\dot{}x_3^p\dot{}x_4^{p+1}$$

Is that the correct answer? Is there a complexification of the system that leads to this answer, or gives it in a more pleasing form?

EDIT 2: Is it unreasonable to consider the normal form in order to study the dynamics of this system. What would be a good approach to find the stationary points and examine their stability?

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In your case, $\mathcal{O}(|x|^5)$ will not be in the normal form and there is no parameter. In the link, you can find formulas to compute the exact normal form. You have to get $q_1,q_2,p_1,p_2$ and $B(x, y), C(x,y,z)$ first, then use them to compute $h_{ijkl}$ and finally $G_{ijkl},H_{ijkl}$. The best way to perform complicated computation is to Mathematica or Maple. –  xpaul May 30 '13 at 12:53