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Re: Finding a function that makes it possible to Solve[] a system of equations

• To: mathgroup at smc.vnet.net
• Subject: [mg114054] Re: Finding a function that makes it possible to Solve[] a system of equations
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Mon, 22 Nov 2010 07:37:23 -0500 (EST)

----- Original Message -----
> From: "Mauricio Esteban Cuak" <cuak2000 at gmail.com>
> To: mathgroup at smc.vnet.net
> Sent: Saturday, November 20, 2010 5:27:53 PM
> Subject: [mg114029] Finding a function that makes it possible to Solve[ ] a system of equations
> Hey MathGroup,
> 
> I apologise beforehand since this might turn to be more a mathematical
> query
> than a Mathematica one.
> However, I appreciate any suggestion on the matter.
> 
> In[1]:= $Version
> Out[1]= "7.0 for Mac OS X x86 (64-bit) (February 19, 2009)"
> 
> 
> This is my setup:
> 
> 
> f = (g*x + y)^h + d* i[x, y];
> u1 = a*f - x^2;
> u2 = (1 - a)*f - y^2;
> 
> 
> Where x, y are the variables and {a,h,g,d} are just parameters, with 0
> < a <
> 1, g > 0, d bigger or equal to 0
> 
> i[x,y] is the function that I need to specify.
> 
> Further, it must be true that
> 
> 
> i[ x, 0] = 0 & i[ 0, y ] = 0
> 
> 
> For simplicity I've set
> 
> 
> h=1;
> 
> 
> Though if the solution you suggest requires it, h could belong to
> (0,1]
> 
> What I need is the analytic solution for {x, y}, hopefully a unique
> real
> solution (I don't mind if there are complex ones) to:
> 
> 
> sol = Solve[{D[u1, x] == 0, D[u2, y] == 0}, {x, y}];
> 
> 
> Obviously, I need to specify i[x,y] to do that. I need the function
> i[x,y]
> to grow slower that x^2 or y^2.
> 
> 
> i[ x, y] = x^(1/2)*y^(1/2) or
> 
> i[x,y] = Log[1+x] * Log[1+y ]
> 
> 
> do the trick but Mathematica can't find an analytic solution.
> 
> The only thing that works is i[x,y] = x*y but the problem is that it
> grows
> as fast
> 
> as x^2.
> 
> 
> Thanks for your time,
> 
> 
> ME


This much seems to work. Use the product of square roots for i[x,y], and precompute a Groebner basis.

f = (g*x + y)^h + d*i[x, y];
u1 = a*f - x^2;
u2 = (1 - a)*f - y^2;

i[x_, y_] := x^(1/2)*y^(1/2);
h = 1;

In[37]:= Timing[
 sol = Solve[
    GroebnerBasis[{D[u1, x], D[u2, y]}, {x, y}] == 0, {x, y}];]

Out[37]= {26.364, Null}

This solution is a large object.

In[40]:= LeafCount[sol]
Out[40]= 186365

In[41]:= Length[sol]
Out[41]= 4

Possibly version 8 will allow you to forego the precomputation, but I have not explicitly tried this. It is also possible that it might be able to represent a solution set using the sum of logs variant you propose for i[x,y].

As for determining forms of i[x,y] that meet your requirements in general, that is indeed a mathematics question. Offhand I'm not sure whether Mathematica can be useful for that purpose.

Daniel Lichtblau
Wolfram Research

• Follow-Ups:
  • Re: Finding a function that makes it possible to
    • From: Mauricio Esteban Cuak <cuak2000@gmail.com>