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I am trying to find $(x_1,x_2)$ solving the system of equations

$$ p + a_1 x_1 = \dfrac{k_1}{2} a_1 \left[ p + a_1 x_1 + \left(1 + \dfrac{x_1}{s_1} \right) b_2 x_2 \right]^2 \\\\ p + a_2 x_2 = \dfrac{k_2}{2} a_2 \left[ p + a_2 x_2 + \left(1 + \dfrac{x_2}{s_2} \right) b_1 x_1 \right]^2 $$

which looks solvable, but none of the approaches I have tried so far (including https://math.stackexchange.com/a/2148580/449277) has worked.

Any solution or guidance would be most appreciated!

user449277
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    Those aren't quadratic equations. If you were to multiply out the square brackets ${[...]}^2$, you'd get fourth order terms proportional to $x_1^2 x_2^2$. – John Barber Sep 20 '24 at 02:01
  • Good point! Is there any hope for an analytical solution? – user449277 Sep 20 '24 at 15:32
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    First step would be to simplify your constants and variables. For example $y_i = a_ix_i, A_i = \frac{k_1a_1}2, t_i = a_is_i, c_i =\frac{b_i}{a_i}$ yields $$p+y_1 = A_1\left[p + y_1+\left(1 + \frac{y_1}{t_1}\right)c_2y_2\right]^2\p+y_2 = A_2\left[p + y_2+\left(1 + \frac{y_2}{t_2}\right)c_1y_1\right]^2$$ is at least a bit easier to manipulate. Some additional simplifications may be better yet. – Paul Sinclair Sep 20 '24 at 17:55
  • Thank you! Mine already was a simplification of a much more complicated parametrization, but I like yours better still! But I'm afraid it doesn't seem to get me closer to a solution. – user449277 Sep 20 '24 at 20:23
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    Possibly yet cleaner in the form $p + a_1 x_1 = d_1 ( p + a_1 x_1 + b_1 x_2 + c_1x_1x_2 )^2$ and symmetrically (mind a change in indexing). The squared expression is a bilinear function, but I don't think that knowing this is much useful. –  Sep 24 '24 at 08:30
  • It is no surprise that a system of two quartic equations leads to an octic univariate one. –  Sep 24 '24 at 08:34

1 Answers1

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I tried to find a rather simple analytical approach, but, unfortunately, it still looks cumbersome and should lead to an equation of eighth degree, see below.

Let us proceed from Paul Sinclair's simplification. If $A_1=0$ then $y_1=-p$ and we have a quadratic equation for $y_2$, if $A_2=0$ then the situation is similar. So we assume that both $A_1$ and $A_2$ are nonzero.

Put $$z_1=p+y_1+\left(1+\frac {y_1}{t_1}\right)c_2y_2 \mbox{ and } z_2=p+y_2+\left(1+\frac {y_2}{t_2}\right)c_1y_1.$$

Then $y_1=A_1z_1^2-p$ and $y_2=A_2z_2^2-p$, so we have the system

$$\begin{cases} z_1=A_1z_1^2+c_2(A_2z_2^2-p)+\frac{c_2}{t_1}(A_1z_1^2-p)(A_2z_2^2-p)\\ z_2=A_2z_2^2+c_1(A_1z_1^2-p)+\frac{c_1}{t_2}(A_1z_1^2-p)(A_2z_2^2-p). \end{cases} $$

If $A_1z_1^2+t_1=p$ or $c_2=0$ then we have at most two possible values for $z_1$ and for each of them two (maybe incompatible) quadratic equations for $z_2$.

So we suppose that $A_1z_1^2+t_1\ne p$ and $c_2\ne 0$. Then

$$\frac{c_1}{t_2}(A_1z_1^2-p)(A_2z_2^2-p)=$$ $$z_2-A_2z_2^2-c_1(A_1z_1^2-p)= \tag{1}$$ $$(z_1-A_1z_1^2-c_2(A_2z_2^2-p))\frac{c_1t_1}{c_2t_2},$$

If $c_1t_1=t_2$ then $$z_2=c_1(A_1z_1^2-p)+\frac{z_1-A_1z_1^2}{c_2}+p.$$

Plugging this value of $z_2$ into the first equation of the system, we obtain a quartic equation for $z_1$.

So we suppose that $c_1t_1\ne t_2$.

Then $$A_2z_2^2-p=\frac{A_1c_1(c_2t_2-t_1)z_1^2+c_1t_1z_1-c_2t_2z_2-c_1c_2t_2p}{c_2(c_1t_1-t_2)}.$$

Plugging this value of $A_2z_2^2-p$ into the first equation of the system, we obtain

$$(z_1-A_1z_1^2)t_1(c_1t_1-t_2)=$$ $$(t_1+A_1z_1^2-p)(A_1c_1(c_2t_2-t_1)z_1^2+c_1t_1z_1-c_2t_2z_2-c_1c_2t_2p),$$

that is a linear equation with respect to $z_2$. Plugging the obtained expression for $z_2$ into the second equality from (1), we should obtain an equation of eighth degree for $z_1$.

Alex Ravsky
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