# Offset-Assisted Factored Solution of Nonlinear Systems
## 3 Limitations of the Factored Method When Real Negative or Complex Solutions Are Involved
### Negative Real Solutions  
Let us assume that, as usually happens in engineering and science, one is mostly interested in finding the real solutions of nonlinear systems. Typically, for a particular application, it is known in advance whether the solutions will be positive or negative, according to the nature of the involved variables (e.g., in power engineering, voltage and current magnitudes are positive, by definition). As stated above, when those real solutions are positive, the factored method performs well. However, in the presence of negative solutions, some components of the auxiliary vector \(y\) will eventually become negative throughout the iterative process, driving the respective components of vector \(u\) to the complex domain (more specifically, to be of the form \(u_{k}=\ln|y_{k}|+\pi i\)). In turn, when solving Equation (12) to obtain the _log_ variables \(\alpha\) (linear combinations of \(u\) variables), it may happen that one or several components \(\alpha_{k}\) become also complex, with an angle other than \(\pi\), which means that the original variable \(x_{k}\) would be complex, rather than negative real. In those cases, the iterative process may remain oscillating or be diverted away from the basin of attraction of the intended real solution, eventually reaching one of the unsought complex solutions. This potential risk is easily illustrated with the help of the following simple example. **example 1**. : Consider the solution of the system,  
\[p_{1} = x_{1}x_{2}+x_{2} \tag{13}\] \[p_{2} = x_{2}^{2}+2x_{1}\]  
For \(p_{1}=-10\) and \(p_{2}=19\), there are two real solutions, \(x=[3,-5]\) and \(x=[9,-1]\), both of them containing a negative component. The two constant matrices involved in the factored procedure are in this case:  
\[E=\left( \begin{array}{cccc}0&1&1&0\\ 2&0&0&1\end{array} \right)\hskip 14.226378 pt;\hskip 14.226378 ptC=\left( \begin{array}{cccc}1&0\\ 0&1\\ 1&1\\ 0&2\end{array} \right)\]  
and the nonlinear transform, \(u=f(y)\),  
\[u=\ln\left( \begin{array}{c}y_{1}\\ y_{2}\\ y_{3}\\ y_{4}\end{array} \right)\hskip 14.226378 pt\mbox{with}\hskip 14.226378 pty= \left( \begin{array}{c}x_{1}\\ x_{2}\\ x_{1}x_{2}\\ x_{2}^{2}\end{array} \right)\]  
Starting from \(x^{(0)}=[0,0]\), the first iteration provides the following results:  
\[y^{(1)}=\left( \begin{array}{c}7.6\\ -5\\ -5\\ 3.8\end{array} \right)\hskip 14.226378 pt;\hskip 14.226378 ptu^{(1)}=\left( \begin{array}{c}2.0281\\ 1.6094+\pi i\\ 1.6094+\pi i\\ 1.3350\end{array} \right)\hskip 14.226378 pt;\hskip 14.226378 ptx^{(1)}=\left( \begin{array}{c}-3.9872-6.9061i\\ -0.8853-1.5334i\end{array} \right)\]  
As is clearly seen, the two negative components of \(y\) translate into complex components in the solution \(x\), which are obtained as the exponential of linear combinations of \(u\) components. These complex components of \(x\) propagate through the iterative process until a steady-state cyclic pattern settles down or a complex solution is eventually reached (it is very unlikely that vector \(x\) becomes again real during the transient period). The simplest way of preventing this risk (_i.e._, the factored method being unable to smoothly converge to negative real solutions) is by adding a sufficiently large positive offset, \(m\), to all variables in \(x\), or at least to those components in \(x\) that are clear candidates to become negative, so that the resulting transformed problem has only positive real solutions. This idea is applied below to the previous example. **example 2**. : Introducing new variables, \(x_{o1}=x_{1}+m\) and \(x_{o2}=x_{2}+m\), the original Equation system (13) becomes:  
\[p_{1}-m^{2}+m = x_{o1}x_{o2}-mx_{o1}+(1-m)x_{o2} \tag{14}\] \[p_{2}-m^{2}+2m = x_{o2}^{2}+2x_{o1}-2 mx_{o2}\]  
and the new matrix \(E\),  
\[E=\left( \begin{array}{rrrr}-m&1-m&1&0\\ 2&-2m&0&1\end{array} \right)\]  
Starting again from \(x^{(0)}=[0,0]\), the first iteration provides this time the following results:  
\[y^{(1)}=\left( \begin{array}{rrrr}7.1429\\ 0.0476\\ 2.3333\\ 4.9048\end{array}