

Subject:
SYSTEM OF EQUATIONS
Category: Science > Math Asked by: baroqqquega List Price: $50.00 
Posted:
10 Sep 2006 23:40 PDT
Expires: 19 Oct 2006 05:56 PDT Question ID: 764063 
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?))1y)=k (y^?)*((1+y)^(2))*(((1+x^(1?))/(1?))1x)=(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. 

There is no answer at this time. 

Subject:
Re: SYSTEM OF EQUATIONS
From: maplekiwiga on 13 Sep 2006 18:53 PDT 
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. 
Subject:
Re: SYSTEM OF EQUATIONS
From: mathtalkga on 19 Sep 2006 21:27 PDT 
Some observations that may be useful for formulating an approximate solution and thus a numerical strategy: As I'm sure baroqqquega 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, mathtalkga 
Subject:
Re: SYSTEM OF EQUATIONS
From: maplekiwiga on 21 Sep 2006 10:31 PDT 
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 
Subject:
Re: SYSTEM OF EQUATIONS
From: maplekiwiga on 24 Sep 2006 16:05 PDT 
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) 
Subject:
Re: SYSTEM OF EQUATIONS
From: maplekiwiga on 30 Sep 2006 14:02 PDT 
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. 
If you feel that you have found inappropriate content, please let us know by emailing us at answerssupport@google.com with the question ID listed above. Thank you. 
Search Google Answers for 
Google Home  Answers FAQ  Terms of Service  Privacy Policy 