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Let $x,y$ be two variables. Consider the following system:

$$ \begin{cases} x=a_0+b (1+y^2/x^2)^{-1/2} \\ y=a_1+b(1+x^2/y^2)^{-1/2}, \end{cases}$$

where $a_0,a_1,b$ are parameters.

I can solve this using numerical solvers, but it's very inefficient. The two equations appear very symmetrical and somewhat 'simple' in formulation, and I wonder if there might be a way to solve this (maybe even obtain an approximation) in some analytic manner. Any hints or directions will be appreciated.

Thanks.

yoki
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1 Answers1

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$$\begin{cases} x=a_0+b (1+y^2/x^2)^{-1/2} \\ y=a_1+b(1+x^2/y^2)^{-1/2} \end{cases} \implies \begin{cases} (x-a_0)^2=\dfrac{b^2x^2}{x^2+y^2} \\ (y-a_1)^2=\dfrac{b^2y^2}{x^2+y^2} \end{cases}\implies \dfrac{(x-a_0)^2}{x^2}=\dfrac{(y-a_1)^2}{y^2}$$

$$\implies \left|\dfrac{x-a_0}{x}\right|= \left|\dfrac{y-a_1}{y}\right|\implies \dfrac{y-a_1}{y}=\pm \dfrac{x-a_0}{x}.$$ So, we get that $$y=\dfrac{a_1x}{a_0} \quad \mathrm{or}\quad y=\dfrac{a_1x}{2x-a_0}.$$

If $y=\dfrac{a_1x}{a_0}$ one gets $x:$ $$x=a_0+b (1+a_0^2/a_1^2)^{-1/2}$$ and $y=\dfrac{a_1}{a_0}(a_0+b (1+a_0^2/a_1^2)^{-1/2}).$

If $y=\dfrac{a_1x}{2x-a_0}$ then $\dfrac{(x-a_0)^2}{x^2}=\dfrac{(y-a_1)^2}{y^2}$ that $$a_1^2(x-a_0)^2=(a_1x+a_0)^2$$ is an identity. So, we consider

$$\dfrac{a_1x}{2x-a_0}=a_1+b\left(1+\dfrac{x^2(2x-a_0)^2}{a_1^2x^2}\right)^{-1/2}=a_1+b\left(1+\dfrac{(2x-a_0)^2}{a_1^2}\right)^{-1/2}.$$ Thus

$$\left(\dfrac{a_1x}{2x-a_0}-a_1\right)^2=\dfrac{b^2}{1+\dfrac{(2x-a_0)^2}{a_1^2}}=\dfrac{a_1^2b^2}{a_1^2+(2x-a_0)^2}.$$ It follows that

$$\left(\dfrac{a_0-x}{2x-a_0}\right)^2=\dfrac{b^2}{a_1^2+(2x-a_0)^2}.$$ From this we get that

$$a_1^2(a_0-x)^2=(2x-a_0)^2(b^2-(a_0-x)^2).$$

Since $a_1^2(a_0-x)^2\ge 0$ it must be $b^2\ge (a_0-x)^2$ (unless $2x=a_0$) But, from $x=a_0+b (1+y^2/x^2)^{-1/2} $ we have

$$(x-a_0)^2=\dfrac{b^2}{1+\dfrac{y^2}{x^2}}< b^2.$$ This is not possible. Now, if $2x=a_0$ then it must be $a_1^2(a_0-x)^2=0.$ That is, $a_1=0,$ from where $y=0.$ But, since we are taking division by $y$ in the original equation this is not possible.

In conclusion, the solutions of the system (when $a_0a_1\ne 0$) are given by

$$x=a_0+b (1+a_0^2/a_1^2)^{-1/2}, y=\dfrac{a_1}{a_0}(a_0+b (1+a_0^2/a_1^2)^{-1/2}).$$

Note that from $$\dfrac{(x-a_0)^2}{x^2}=\dfrac{(y-a_1)^2}{y^2}$$ we get that $a_0=0\iff a_1=0.$ In such a case it is $x=y=\sqrt{b^2-1}.$

mfl
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  • Thanks, great! One question - when you take square roots after for each case (after getting $y$ as a function of $x$), why don't you consider the plus/minus solutions there as well? – yoki Feb 03 '16 at 08:32
  • Also, I didn't get why from $y=\frac{a_1x}{2x-a_0}$ and the ratio equation (the third line to the end) you got the second line to the end. I just get a 'tautology' and no new information... – yoki Feb 03 '16 at 08:44
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    You're right. Sorry for my mistake. I have edited partially but I don't have time at this moment to complete the answer. I'll come back to try to complete the answer this night. – mfl Feb 03 '16 at 13:39
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    I have edited the answer. I think it is correct but, of course, it can contain some mistake. Please, check it an tell me. I am bit tired to revise it today. – mfl Feb 03 '16 at 21:41
  • Thank you, you've been very helpful. – yoki Feb 04 '16 at 00:33