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Given, say $4$ non linear equations with $4$ positive parameters,

$$f_1(x,y,z,t)=a,\quad f_2(x,y,z,t)=b,\quad f_3(x,y,z,t)=c,\quad f_4(x,y,z,t)=d$$

for given $a,b,c,d$, If I am able to show that when the other $3$ variables are fixed, if $f_1$ is increasing with $x$ and $f_2$ is increasing with $y$ and $f_3$ is increasing with $z$ and $f_4$ is increasing with $t$ and all functions have at least a positive point.

Can I claim that this equation system has a unique solution for positive $x,y,z,t$?

Many Thanks.

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up vote 2 down vote accepted

If I understand the problem correctly, this is a counterexample.

Take $f_1(x,y,z,t) = x-y+1$, $f_2(x,y,z,t) = y-z+1$, $f_3(x,y,z,t) = z-t+1$, $f_4(x,y,z,t) = t-x+1$.

Then with $x=y=z=t=1$, we have all functions equal to $1$, but this is also true for $x=y=z=t=2$.

Actually my example is affine, replace the terms $x-y$ for example by $x^2-y^2$ to make it 'more nonlinear'.

Additional elaboration:

Take $f_1(x,y,z,t) = x-y$, $f_2(x,y,z,t) = y-z$, $f_3(x,y,z,t) = z-t$, $f_4(x,y,z,t) = t-x$. $f_1$ is increasing in $x$, $f_2$ is increasing in $y$, $f_3$ is increasing in $z$ and $f_4$ in increasing in $t$.

Then for all $\lambda$ we have $f_1(\lambda,\lambda,\lambda,\lambda) = 0$, $f_2(\lambda,\lambda,\lambda,\lambda) = 0$, $f_3(\lambda,\lambda,\lambda,\lambda) = 0$ and $f_4(\lambda,\lambda,\lambda,\lambda) = 0$, so the solution is not unique by any means.

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thanks for your post. Perhaps I couldnt explain very well. In your example we have such equations for example $(x-1)^2=1$, $(y-1)^2=1$, $(z-1)^2=1$, $(t-1)^2=1$. For example for the first it is first decreasing $x\in[0,1]$ and then it is increasing $x>1$. I must know that for positive $x$ it is only increasing and lets assume that the beginning point of increase is $0$, namely $f1(x=0,...)=0$. –  Seyhmus Güngören Nov 7 '12 at 19:10
@SeyhmusGüngören: I modified my answer slightly so it is increasing for positive $x$, but in any case the affine example will satisfy this condition. Subtract 1 from all functions above to satisfy $f_k(...) = 0$. Perhaps I am missing your point? –  copper.hat Nov 7 '12 at 19:20
Ok. If the function is starting from $0$ and increasing, then it can pass through a positive number only once. Therefore there will be a unique solutions for any of the $f$ functions by the chosen parameters such as $x$ for $f_1$ and $y$ for $f_2$ etc.. Now my question is; does this guarrantee that the whole system has a unique solution for some $x,y,z,t$ or is it still possible that there exist other $x,y,z,t$ that can be another solution. –  Seyhmus Güngören Nov 7 '12 at 19:33
Hi @SeyhmusGüngören, I think I am missing your point. I added some elaboration above, which seems to meet all the conditions in the question and comments? –  copper.hat Nov 7 '12 at 19:46
I think there is a mismatch in the understanding)) ok so for the first one $f_1(x,y,z,t)=x-\lambda$ and $f_1$ is increasing in $x$. Then $x=\lambda$ is the unique solution for $f_1$. –  Seyhmus Güngören Nov 7 '12 at 19:51
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