# Factorization for frozen state methods

Considering the inverted pendulum equation as follows:

$$\begin{split} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= \sin(x_1) - 0.5x_2 \end{split}$$

Now if I want to apply frozen state methods, I can use factorization to get a pseudo linear system

$$\dot{\mathbf{x}} = \mathbf{A}(\mathbf{x})\mathbf{x}$$

for example by factorizing out $x_1$ like

$$\begin{split} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= \sin(x_1)\frac{1}{x_1}x_1 - 0.5x_2 \end{split}$$

which gives the state dependent $\mathbf{A}$-matrix as

$$\mathbf{A}(\mathbf{x}) = \begin{bmatrix} 0 & 1\\ \frac{\sin(x_1)}{x_1} & -0.5 \end{bmatrix}\,.$$

Remark: The singularity can be removed here using L'Hopitals rule.

However: To me it is not clear how to choose such a factorization. In especially I could also have chosen $x_2$ instead of $x_1$ for the factorization - which one should be prefered then? In this special case it seems to be $x_1$ because the term $\sin(x_1)$ depends only on $x_1$ and it allows nicely to remove the singularity.

But: How to do this in general? Assuming a system like

$$\begin{split} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= f(x_1, x_2) - 0.5x_2 \end{split}$$

with $f$ nonlinear. Should I go for $f(x_1, x_2)\frac{1}{x_1}x_1$, for $f(x_1, x_2)\frac{1}{x_2}x_2$ or maybe even a linear combination of these?

Is there any theory/best practise available how to do this?

• Never heard of this method, but according to Hadamard's lemma if your function $f(x_1, x_2)$ is zero at the origin and smooth enough, it has unique representation as $f(x_1, x_2) = x_1 \cdot g_1 (x_1, x_2) + x_2 \cdot g_2(x_1, x_2)$, where $g_1$ and $g_2$ are one-degree-less-smooth-than-$f$. – Evgeny May 13 '17 at 19:17
• Unfortunatly this is only the case for $n = 1$ (with $n$ being the number of states). For $n > 1$ the factorization is not unique anymore. – SampleTime May 13 '17 at 19:47
• Maybe I'm overlooking something, the uniqueness part was the only one about which I had doubts. Can you show a counterexample? – Evgeny May 13 '17 at 19:49
• Consider for example $x_1^2x_2^2$ which can be written as $f_1(\mathbf{x}) = (x_1^2x_2)x_2$ or as $f_2(\mathbf{x}) = (x_1x_2^2)x_1$ so there exist at least two different factorizations such that $f_1(\mathbf{x}) = f_2(\mathbf{x}) \, \forall \mathbf{x}$. Therefore $f_{\alpha}(\mathbf{x}) = \alpha f_1(\mathbf{x}) + (1 - \alpha)f_2(\mathbf{x})$ is parameterizing with $\alpha \in [0, 1]$ a family of factorizations. Hence there are infinitly many factorizations. – SampleTime May 13 '17 at 19:59