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Algebraic Equation

  1. Oct 12, 2008 #1
    I read in my notes that if you have an algebraic equation of the form:


    And you find an equilibrium solution, [tex]\dot{z}\equiv 0 [/tex]

    Then there exists (J) a function: [tex] z= f(u)[/tex]

    Is this always true of an algebraic equation?

    An algebraic equation being defined as:


    If this is true, could someone show me the proof? (Hopefully its not a big nightmare!)
  2. jcsd
  3. Oct 13, 2008 #2


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    What do you mean by "an equilibrium solution"?
  4. Oct 13, 2008 #3


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    exactly what he said: z'= 0. He is actually talking, at first, about differential equations:
    [itex]f(z,\dot{z}, u, t)= 0[/itex] is a first order differential equation with [itex]\dot{z}[/itex] the derivative of z, t, I assume, the independent variable, and u a parameter. I don't know why Cyrus refers to that as an "algebraic equation" since the link he gives refers only to polynomial equations with many variables. Perhaps he meant f(z, a, u, t)= 0 ignoring the fact that the second variable was the derivative of the first variable. In that case, of course, the "[itex]\dot{z}= 0[/itex]" is irrelevant and the statement is NOT generally true.
  5. Oct 13, 2008 #4
    That's correct, but the order may or may not be first because its really a vector function of variables. What I meant was that suppose the differential equation can be written as an algebraic function [tex]f(\dot{z},z,u,t)[/tex]. Then does that mean the equilibrium solution can be written in the form of u=f(z) ?

    z- state variables
    t- time
    u- control inputs
    \dot{z} -time derivative of state variables
  6. Oct 13, 2008 #5


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    If f is a functio of 4 variables (f(z,z',u,t), then "f(z)" makes no sense.
  7. Oct 13, 2008 #6
    Sure it does. If \dot{z} is equal to zero, then f(z,z',u,t)=f(z,u,t)

    because its an *equilibrium solution, its NOT a function of time, so:

    f(z,u,t) = f(z,u) =0

    So this means:

    f(z,u)=0 or u=f(z)

    Now my question is if it is true that f(z,z',u,t) is algebraic from the start, does that mean we can write u=f(z) as an equilibrium solution? Because this means u is specifically a function of z, i.e. I can move all the z's to the RHS and all the u's to the LHS.

    What is not clear to me is if it could be mixed in such a way that you cannot separate out u as a function of z all by itself on the RHS. (i.e. explicit vs implicit).
  8. Oct 15, 2008 #7
    [tex] f(z,\dot z,u,t) = 0 [/tex] can not be a algebraic equation.
    by definition, algebraic equation can not contains differential terms.
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