Proof of Equilibrium Solution for Algebraic Equation

[Cyrus]

I read in my notes that if you have an algebraic equation of the form:

0=f(\dot{z},z,u,t)

And you find an equilibrium solution, \dot{z}\equiv 0

Then there exists (J) a function: z= f(u)

Is this always true of an algebraic equation?

An algebraic equation being defined as:

http://mathworld.wolfram.com/AlgebraicEquation.html

If this is true, could someone show me the proof? (Hopefully its not a big nightmare!)

 

[morphism]

What do you mean by "an equilibrium solution"?

 

[HallsofIvy]

What do you mean by "an equilibrium solution"?

exactly what he said: z'= 0. He is actually talking, at first, about differential equations:
f(z,\dot{z}, u, t)= 0 is a first order differential equation with \dot{z} 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 "\dot{z}= 0" is irrelevant and the statement is NOT generally true.

 

[Cyrus]

exactly what he said: z'= 0. He is actually talking, at first, about differential equations:
f(z,\dot{z}, u, t)= 0 is a first order differential equation with \dot{z} 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 "\dot{z}= 0" is irrelevant and the statement is NOT generally true.

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 f(\dot{z},z,u,t). 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

 

[HallsofIvy]

If f is a functio of 4 variables (f(z,z',u,t), then "f(z)" makes no sense.

 

[Cyrus]

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).

 

[Jang Jin Hong]

f(z,\dot z,u,t) = 0 can not be a algebraic equation.
by definition, algebraic equation can not contains differential terms.