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Re: anything faster than Solve[] for solving sets of polynomial equations symbolically?

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The code runs here, exactly as you see it below.

eqs = {-k5 x1 x3 + k6 x4 == 0,
    k1 x1 - k4 x2 - 2 k2 x2^2 + 2 k3 x3 + k7 x4 == 0,
    k2 x2^2 - k3 x3 - k5 x1 x3 + k6 x4 == 0, x1 + x4 - Xtot == 0};
nonNeg = Thread[{k1, k2, k3, k4, k5, k6, k7, Xtot} >= 0];
real = Element[vars = {x1, x2, x3, x4}, Reals];
problem = Flatten@{eqs, nonNeg};
Solve[problem, Reals] // FullSimplify

$Version

"8.0 for Mac OS X x86 (64-bit) (February 23, 2011)"

Bobby

On Wed, 04 May 2011 18:47:36 -0500, dantimatter <google@dantimatter.com>  
wrote:

>
> Hi Bobby,
>
>> eqs = {-k5 x1 x3 + k6 x4 == 0,
>>     k1 x1 - k4 x2 - 2 k2 x2^2 + 2 k3 x3 + k7 x4 == 0,
>>     k2 x2^2 - k3 x3 - k5 x1 x3 + k6 x4 == 0, x1 + x4 - Xtot == 0};
>> nonNeg = Thread[{k1, k2, k3, k4, k5, k6, k7, Xtot} >= 0];
>> real = Element[vars = {x1, x2, x3, x4}, Reals];
>> problem = Flatten@{eqs, nonNeg};
>> Solve[problem, Reals] // FullSimplify
>
> I'm getting an error message that 'Xtot >= 0 is not a well-formed  
> equation'.  Is this something that you encountered as well?  Also, is  
> your Solve[] function call missing a reference to the variables, or am I  
> just confused?
>
> Thanks!
> Dan
>
>


-- 
DrMajorBob@yahoo.com

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Re: anything faster than Solve[] for solving sets of polynomial equations symbolically? DrMajorBob <btreat1@austin.rr.com> - 2011-05-07 10:22 +0000

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