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I have a system of equations as below: $$\frac{dx_1}{dz}=A+B\cdot\frac{x_2}{1-x_1}$$ $$\frac{dx_2}{dz}=C+D\cdot\frac{x_1}{1-x_2}$$

I got derivative from first equation and substitute the value for $x_2$ and $\frac{dx_2}{dz}$, but the resulting equation seems too complicated. Any ideas?

Bita
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  • One possible of set of solutions are of the form $x_i = a_iz+b_i$. – player100 Aug 22 '18 at 21:42
  • Thanks for your comment. I am wondering if I should substitute these answer in the system and look for $a_i$ $b_i$?where the boundary conditions involved? – Bita Aug 23 '18 at 01:25
  • I tried briefly, but I think you will end up with a coupled set of 4 equations with 4 unknowns, resulting in a quartic polynomial. If you have boundary conditions, and the solutions are not merely the lines connecting the two boundaries, then we have to keep looking... – player100 Aug 23 '18 at 11:23
  • See https://math.stackexchange.com/questions/1764919/how-to-solve-the-following-system-frac-textdx-textd-t-ax-fracb/1768953#1768953 – player100 Aug 23 '18 at 15:53
  • Thank you so much. Your solution helped me a lot. However, I m not sure the equation in the link you provided is similar to my problem. I really appreciate the solution you provided as it is exactly what I wanted. – Bita Aug 24 '18 at 06:06
  • Be careful accepting them as the only solution... Nonlinear equations are tricky. If you have access to a nonlinear solver I would recommend playing around with it to see what other behavior this system portrays – player100 Aug 24 '18 at 06:35
  • I have results from numerical simulator and they look linear. Thanks for your advice. I keep it in mind. What kind of nonlinear solver I can use for these system? – Bita Aug 25 '18 at 02:08
  • There are many... Runge-Kutta based solvers with adaptive step sizing is one. – player100 Aug 25 '18 at 02:48

1 Answers1

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Let $y_i = 1-x_i$. Then,

$$\begin{align} -y_1'&=A+B\frac{1-y_2}{y_1} \\-y_2'&=C+D\frac{1-y_1}{y_2} \end{align}$$

Rearranging, $$\begin{align} -y_1y_1'&=-\frac{1}{2}\frac{d}{dz}y_1^2=Ay_1+B(1-y_2) \\-y_2y_2'&=-\frac{1}{2}\frac{d}{dz}y_2^2=Cy_2+D(1-y_1) \end{align}$$

By inspection, one possible solution is $y_i = a_iz+b_i$. Substituting this into the equations:

$$\begin{align}-a_1(a_1z+b_1)&=A(a_1z+b_1)+B(1-a_2z-b_2)\\ -a_2(a_2z+b_2)&=C(a_2z+b_2)+D(1-a_1z-b_1) \end{align}$$

Matching coefficients of $z$ results in a system of 4 equations: $$\begin{align}Aa_1-Ba_2+a_1^2&=0\\ Ca_2-Da_1+a_2^2&=0\\ Ab_1+B(1-b_2)+a_1b_1&=0\\ Cb_2+D(1-b_1)+a_2b_2&=0 \end{align}$$

It is possible to obtain a quartic polynomial equation in terms of $a_i$, and then linear equations for $b_i$ with coefficients in terms of $a_i$. Or, you can employ a nonlinear solver.

player100
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