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I have the set below of three nonlinear equations: \begin{align} Y_1=&\;\frac{X_1+GX_2X_3}{1+X_2X_3} \tag1\\ Y_2=&\;\frac{X_1+GX_2X_3+GX_2(1-X_3)^2}{1+X_2X_3+X_2(1-X_3)^2} \tag2\\ Y_3=&\;\frac{X_1+GX_2X_3+FGX_2(1-X_3)^2}{1+X_2X_3+FX_2(1-X_3)^2} \tag3 \end{align} Note that all variables and constants are complex numbers.

I need to find expressions for $X_1$, $X_2$ and $X_3$ each as a function of $Y_1$, $Y_2$, $Y_3$, $F$ and $G$.

What I did so far is:

I solved $(1)$ for $X_1$ $$ X_1=Y_1+X_2(Y_1X_3-GX_3) \tag4 $$

I then substituted $(4)$ in $(3)$ and solved for $X_2$ $$ X_2=\frac{X_1-Y_3}{X_3(Y_3-G)+(1-X_3)^2(Y_3F-FG)} \tag5 $$ let $$ R=Y_1 -G \\ W=Y_3-G\\ Q=Y_3F-FG\\ H=Y_1-Y_3\\ M=W-R $$ Hence $$ X_2=\frac{H}{MX_3+Q(1-X_3)^2} \tag6 $$ then substituted $(6)$ in $(4)$ $$ X_1= Y_1+\frac{HRX_3}{MX_3+Q(1-X_3)^2}\tag7 $$ Now $(6)$ and $(7)$ are $X_2$ and $X_1$ as functions of $X_3$, $Y1$, $Y_3$, $F$ and $G$. When substituting $(6)$ and $(7)$ back in $(2)$ and solve for $X_3$, and simplify it to a conventional quadratic form $aX_3^2+bX_3+c=0$, I end up with almost zero values for $a$, $b$ and $c$. And hence cannot find $X_3$.

Thanks a lot in advance.

mzp
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Rayan Jordan
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1 Answers1

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Define \begin{eqnarray*} Z_{1} &=&X_{1}+GX_{2}X_{3}, \\ Z_{2} &=&1+X_{2}X_{3}, \\ Z_{3} &=&X_{2}\left( 1-X_{3}\right) ^{2}, \end{eqnarray*} then the system becomes \begin{eqnarray*} Y_{1} &=&\frac{Z_{1}}{Z_{2}}, \\ Y_{2} &=&\frac{Z_{1}+GZ_{3}}{Z_{2}+Z_{3}}, \\ Y_{3} &=&\frac{Z_{1}+FGZ_{3}}{Z_{2}+FZ_{3}}. \end{eqnarray*} Solving for $Z_1$ and $Z_2$ from the first two equations yields \begin{eqnarray*} Z_{1} &=&Y_{1}Z_{2},\\ Z_{2} &=&\frac{Y_{2}-G}{Y_{1}-Y_{2}}Z_{3}. \\ \end{eqnarray*} Substituting this into the third equation implies $$Y_{3} =\frac{Y_{1}\left( G-Y_{2}\right) -FG\left( Y_{1}-Y_{2}\right) }{% \left( G-Y_{2}\right) -F\left( Y_{1}-Y_{2}\right) }.$$ Hence, if the parameters are such that this equation does not hold, the system does not have a solution. If this condition is satisfied, then there is an infinity of solutions, since you can drop equation $(3)$, and for any $X_3\neq0$,

\begin{eqnarray*} X_{1} &=&G-\frac{\left( G-Y_{1}\right) \left( G-Y_{2}\right) \left( 1-X_{3}\right) ^{2}}{\left( Y_{1}-Y_{2}\right) X_{3}+\left( G-Y_{2}\right) \left( 1-X_{3}\right) ^{2}}, \\ X_{2} &=&\frac{Y_{2}-Y_{1}}{\left( Y_{1}-Y_{2}\right) X_{3}+\left( G-Y_{2}\right) \left( 1-X_{3}\right) ^{2}}. \end{eqnarray*}

while for $X_3=0$,

\begin{eqnarray*} X_{1} &=&Y_{1}, \\ X_{2} &=&\frac{Y_{2}-Y_{1}}{G-Y_{2}}. \end{eqnarray*}

mzp
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  • Thank you for your answer. I did check the rank of the coefficient matrix and based on the parameters $G$ and $F$, the rank changes and hence the system may have infinite solutions. – Rayan Jordan Apr 29 '18 at 16:08