Re: no one has a real way to help you

From: Roger Bagula (rlbagulatftn_at_yahoo.com)
Date: 03/25/05

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    Date: Fri, 25 Mar 2005 16:21:36 GMT

    If you use b(i,j) as a matrix M with a(i,i) as it's diagonal
    then you have a Markov like 4by4 matrix already.
    Make that determinant<=1
    The d(i)'s are decay rates: you have just four of them:
    you can start at d(i)=1/2 as the first integer decay not equal to one.
    If b(i,j), a(i,i) are weights , then you can try a(i,j) self weights
    equal zero. And symmetrical matrix as
    b(i,j)=b(j,i)
    To get the symmetrical matrix:
    M={{ 0,b(1,2),b(1,3),b(1,4)}
         {b(1,2),0,b(2,3},b(2,4)},
          {b(1,3),b(2,3},0,{b(3,4)}
          {b(1,4},b(2,4),b(3,4),0}}
    Take the determinant of that as one and you can eliminate
    one more as dependent on the other five make it b(1,4)
    You have now five constants to worry about.
    If you let
    b(1,2)=b(3,4)
    b(1,3)=b(2,4)
    b(1,4)=b(2,3)
    you should be down to two constants over all.
    That may be "too symmetrical", but it is a starting point.
    surya wrote:

    > Its 'disheartening' to know that 'no one has a real way to help
    > me'!!But its impossible to be more specific.I know the equations seem
    > very general.But I doubt that we can reduce the number of constants at
    > this stage.The equations were written from a state diagram and most of
    > the constants like a11 or b12 are the path weightages.And the variables
    > x1,x2,x3,x4 are emotional states.
    > I don't know anything about Markov matrix .Any information about
    > them would be helpful.
    >
    > Surya
    >
    > Roger Bagula wrote:
    >
    >>Dear surya,
    >>Try making it up in a Markov matrix or affine form:
    >>{dx1/dt,dx2/dt,dx3/dt,dx4/dt}=M*{x1,x2,x3,x4}
    >>where M is a 4by4 matrix.
    >>Then take the determinant of the matrix M.
    >>Det[M]<=1 is a good indication that the result
    >>will have a convex hull.
    >>Doing Lyapunov constant analysis on 4d system is extreme
    >>fun and I doubt that you can do it with as general a system as this.
    >>You have to get more specific... it would help to know what kind of
    >>system you are modeling, maybe. Simplifying using group theory
    >>or some other symmetry approach can and get you to fewer constants.
    >>As it stands no one has a real way to help you:
    >>it's too general as you have stated the problem.
    >>surya wrote:
    >>
    >>
    >>>Hi,
    >>>
    >>>
    >>> Can anyone help me? I am a student of Electrical and Computer
    >>>Engineering.I am doing a project with our Artificial Intelligence
    >>>Lab. and I have stumbled on a part that requires solving a set of
    >>>Differential equations to get chaos.I am entirely
    >>>new to chaos theory.Help!
    >>>
    >>>Surjya Sarathi Ray
    >>>
    >>>
    >>>
    >>>PROBLEM: To find the values/range of values of the parameters in
    >>>such a way so that the plot of the variables against time is
    >
    > chaotic.
    >
    >>>
    >>>
    >>>
    >>>The differential Equations are as follows.
    >>>
    >>>Where x1(t), x2(t), x3(t), x4(t) are the respectively the variables
    >>>
    >>>dx1(t)/dt = a11*x1(t)*(1-x1(t)/k1) + b31*x1(t)*(1-exp(-d1*x3(t))) -
    >>>b12*x1(t)*(1-exp(-d2*x2(t)))
    >>>
    >>>dx2(t)/dt = a22*x2(t)*(1-x2(t)/k2) + b12*x2(t)*(1-exp(-d3*x1(t))) +
    >>>b32*x2(t)*(1-exp(-d4*x3(t))) + b42*x2(t)*(1-exp(-d5*x4(t))) -
    >>>b23*x2(t)*(1-exp(-d6*x3(t)))
    >>>
    >>>dx3(t)/dt = a33*x3(t)*(1-x3(t)/k3) - b34*x3(t)*(1-exp(-d7*x4(t))) -
    >>>b31*x3(t)*(1-exp(-d8*x1(t))) - b32*x3(t)*(1-exp(-d9*x2(t))) +
    >>>b23*x3(t)*(1-exp(-d10*x2(t)))
    >>>
    >>>dx4(t)/dt = a44*x4(t)*(1-x3(t)/k3) + b34*x4(t)*(1-exp(-d11*x3(t)))
    >
    > -
    >
    >>>b42*x4(t)*(1-exp(-d12*x2(t)))
    >>>
    >>>
    >>>
    >>>Say x1(t),x2(t),x3(t) and x4(t) lie in the range of 0-1000. Then
    >>>
    >>>the RANGE of the parameters are:
    >>>
    >>>1) k1,k2,k3,k4 : they are POSITIVE numbers.Can lie in any range
    >
    > say
    >
    >>>0 to 1000.
    >>>2)a11,a22,a33,a44,b31,b12,b32,b42,b23,b34-The are also POSITIVE
    >>>numbers;any range say 0-10 may be taken.
    >>>3)d1,d2,d3,d4,d5,d6,d7,d8,d9,d10,d11,d12 are all POSITIVE-the lie
    >
    > in
    >
    >>>the range of 0.1 to 0.0001;
    >>>
    >>>Can anyone suggest the values of the parameters?Will the system
    >
    > give
    >
    >>>chaos at all???
    >>>
    >>
    >>
    >>--
    >>Roger L. Bagula email: rlbagula@sbcglobal.net or
    >>rlbagulatftn@yahoo.com
    >>11759 Waterhill Road,
    >>Lakeside, Ca. 92040 telephone: 619-561-0814}
    >
    > 

    -- 
Roger L. Bagula       email: rlbagula@sbcglobal.net  or 
rlbagulatftn@yahoo.com
11759 Waterhill Road,
Lakeside, Ca. 92040    telephone: 619-561-0814}
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