1
$\begingroup$

I have to solve a nonlinear system (for some reactions) using newton's method to get molar fractions (positive values), sometimes I get negative values depending on the initial vector, to prevent this, I want to replace variables in the system by their absolute values, but I have no idea if that will change the problem or not.

$\begin{cases} 2x_1+x_5+2x_6+x_8-8\alpha = 0 \\ x_1+2x_2+2x_3+x_4+x_5+x_7-4 = 0\\ 2x_3+x_4+x_9-4\alpha-2 = 0\\ x_4^2-K_{P7}x_3x_2 = 0\\ x_5^2x_6P_T-K_{PT}x_1^2x_{10} = 0 \\ x_6^2x_2P_T-K_{P3}x_1^2x_{10} = 0 \\ x_7^2P_T-K_{P5}x_2x_{10} = 0 \\ x_8^2P_t-K_{P4}x_6x_{10}= 0 \\ x_9^2P_t-K_{P6}x_3x_{10}= 0 \\ \left(\sum\limits_{i=1}^{9}{x_i}\right)-x_{10} = 0 \\ \end{cases}$

where $\alpha, K_{Pi}, PT$ are parameters

I think it is an optimization problem, do that change my initial problem?

Thank you

$\endgroup$
1
$\begingroup$

Replacing variables by their absolute values is likely to give you spurious solutions or trap you in a cycle (I don't know if it will in your particular case). I suspect it's better to try a different initial point if you encounter solutions with negative values. If you are lucky enough to start your Newton's method with an initial point close to a positive solution, you shouldn't encounter solutions with negative values.

$\endgroup$
1
  • $\begingroup$ Thank you sir, So mathematically, that will change the solutions of the real problem? $\endgroup$ – user265759 Jun 17 '16 at 0:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy