SYSTEM OF EQUATIONS, ALGEBRA, DIFFERENTIAL EQUATIONS
Thanks for taking the time to check this. I have been struggling to
solve the following system of equations:
(x^?)*((1+x)^(-2))*(((1+y^(1-?))/(1-?))-1-y)=k
(y^?)*((1+y)^(-2))*(((1+x^(1-?))/(1-?))-1-x)=(1/k)

where 0<x<1, 0<y<1 are the variables, 0<?<1 and k are the parameters.
It comes up as the equilibrium condition to a game theoretic model,
and i tried a lot of things to find an explicit solution but could not
get anything. Do you have any suggestions? Thank you for your time.

I doubt that there is an explicit solution. So, you will have to
settle for solving numerically. See

http://www.math.gatech.edu/~carlen/2507/notes/NewtonMethod.html

for solving two simultaneous implicit equations numerically using the
Newton Raphson method. Evidently, you must start with a good guess.

Some observations that may be useful for formulating an approximate
solution and thus a numerical strategy:

As I'm sure baroqqque-ga has already observed, the equations can be restated:

f(x) * g(y) = k

f(y) * g(x) = 1/k

where the functions on [0,1] are defined:

f(z) = (z^?)/(1+z)^2

g(z) = ((1+z^(1-?))/(1-?) - (1+z)

Bearing in mind that 0 < ? < 1 as well, we find taking the derivative:

g'(z) = 1/(z^?) - 1

Hence g(z) is strictly increasing on [0,1], with:

g(0) = ?/(1-?)  and  g(1) = 2?/(1-?).

Also g"(z) = -?/(z^(1+?)) < 0 on (0,1), ie. the curve is concave down.

The picture is less simple for f(z), which is not monotonic:

f'(z) = ?*z^(?-1)/(1+z)^2  -  2*(z^?)/(1+z)^3

      = [?/z + ? - 2] * (z^?)/(1+z)^3

Since 0 < ? < 1, we have:

 f increasing for 0 < z < ?/(2-?),

 f decreasing for ?/(2-?) < z < 1.

In other words f(z) reaches its maximum at z = ?/(2-?), while:

f(0) = 0  and  f(1) = 1/4

f(?/(2-?)) = (?/(2-?))^?/(1 + ?/(2-?))^2

           = (?/(2-?))^?/(2/(2-?))^2

           = (?^?)*((2-?)^(2-?))/4

Given these points it would be helpful to know more about the specific
values of ? (and k) for which a solution is sought.

In principle the invertibility of g(z) can be exploited to eliminate
one of the two variables.  Let G(g(z)) = z on [0,1] be such an
inverse:

y = G(k/f(x))

x = G(1/(k*f(y)))

  = G(1/(k*f(G(k/f(x)))))

Thus one obtains x as the fixed point of the function given last in
terms of G,f, and k.  Conceivably a simple fixed point iteration will
produce a satisfactory starting approximation for a more sophisticated
Newton iteration.

regards, mathtalk-ga

Here are some more observations that constrain k and ?.

Combining the bounds on f and g:

  0 < f(x)g(y) <  1/2 * ?/(1-?) * ?^? * (2-?)^(2-?)  = U(?)

We must have
 
   0  <  k  <  U(?)

and similarly

   0  <  1/k  <  U(?)

giving

   1 / U(?)  <  k  <  U(?).

This requires that

   U(?) > 1

Solving numerically,  ? > 0.6358.

Therefore, there are no solutions if  ? < 0.6358. If  ? > 0.6358 
there are no solutions if  k  is outside the bounds  1/U(?) and U(?).
Further investigation is required to determine if there are 0, 1 or
more solutions otherwise.

The following web site is useful for graphing some of these functions:
  http://wims.unice.fr/wims/en_home.html

Number of solutions:

The graph of G(1/(k*f(G(k/f(x))))) generally looks like a "W" shaped
curve. Some parts may be missing when the inverse of g is not defined.
Also, the "W" may be deformed, looking more like the normal
(statistical) curve.

My experiments show that the number of solutions is 0, 1, 2, 3 or 4.
Here are some numerical results:

0 solutions:  ? = 0.7  k = 0.8

1 solution:  ? = 0.8 k = 1  
  (x=0.1123,y=0.1123)

2 solutions: ? = 0.7 k = 0.95  
  (0.1559,0.8957)
  (0.1850,0.2925)

3 solutions: ? = 0.68  k = 1.03
  (0.2271,0.9666)
  (0.3441,0.2278)
  (0.6863,0.1956)

4 solutions:  ? = 0.66  k = 1.005
  (0.2802,0.6015)
  (0.3948,0.3138)
  (0.5045,0.2832)
  (0.8927,0.9222)

Graphs

Make obvious changes to the website name: 
cs*senecac*on*ca/~lew*baxter to see graphs of f(x)g(y)=k  and 
f(y)g(x) = 1/k  where ? and k can be set.