2.1.3 Well-posedness of standard OdE system models As it was mentioned before, not all ODE models are adequate for design and analy purposes. The notion of well-posedness introduces some typical constraints aimed at insuring their applicability. Definition A standard ODE model ODE(, g) is called well posed if for every signal v(t)E V and for every solution T1: 0, ti- X of(2.4)with 1(0)E Xo there exists a solution R++X of (2. 4)such that c(t)=a1(t)for all t E 0, t1 The OdE from Example 2.1.1 can be used to define a standard autonomous Ode tem model i(t)=-sgn(ar(t)), w(t)=r(t), here V=X=Xo=R, f(a, v, t)=-sgn(a)and g(a, v, t)=r. It can be verified that this autonomous system is well-posed. However, introducing an input into the model destroys well-posedness, as shown in the following example Example 2.2 Consider the standard ODE model i(t)=sgn(ar(t))+o(t),w(t)=r(t) where u(t)is an unconstrained scalar input. Here V=X= XO=R, f(, u, t)=-sgn(r)+v, g(a, v, t) While this model appears to describe a physically plausible situation(velocity dynamics subject to dry friction and external force input u), the model is not well-posed To prove this, consider the input v(t)=0.5= const. It is sufficient to show that no solution of the ODe i(t)=0.5-sgn(x(1) satisfying c(0)=0 exists on a time interval [0, t,] for tf>0. Indeed, let a=r(t) be such solution. As an integral of a bounded function, a =a(t) witll be a continuous function of time. A continuous function over a compact interval always achieves a maximum. Let tm E0, t,] be an argument of the maximum over tE 0, te If c(tm)>0 then tm >0 and, by continuity, r(t)>0 in a neighborhood of tm, hence there exists e>0 such that a(t)>0 for all tE tm-E, tm. According to the differential equation, this means that r(tm-e)=r(tm)+0.5e>a(tm), which contradicts the selection of tm as an argument of maximum. Hence max c(t)=0. Similarly, min c(t)=0. Hence c(t)=0 for all t. But the constant zero function does not satisfy the differentlial equation It can be shown that the absense of solutions in Example 2.1.3 is caused by lack of continuity of function f=f(a, v, t) with respect to r(discontinuity with respect to v and t would not cause as much trouble3 2.1.3 Well-posedness of standard ODE system models As it was mentioned before, not all ODE models are adequate for design and analysis purposes. The notion of well-posedness introduces some typical constraints aimed at insuring their applicability. Definition A standard ODE model ODE(f, g) is called well posed if for every signal v(t) ⊂ V and for every solution x1 : [0, t1] ∈� X of (2.4) with x1(0) ⊂ X0 there exists a solution x : R+ ∈� X of (2.4) such that x(t) = x1(t) for all t ⊂ [0, t1]. The ODE from Example 2.1.1 can be used to define a standard autonomous ODE system model x˙ (t) = −sgn(x(t)), w(t) = x(t), where V = X = X0 = R, f(x, v, t) = −sgn(x) and g(x, v, t) = x. It can be verified that this autonomous system is well-posed. However, introducing an input into the model destroys well-posedness, as shown in the following example. Example 2.2 Consider the standard ODE model x˙ (t) = −sgn(x(t)) + v(t), w(t) = x(t), (2.6) where v(t) is an unconstrained scalar input. Here V = X = X0 = R, f(x, v, t) = −sgn(x) + v, g(x, v, t) = x. While this model appears to describe a physically plausible situation (velocity dynamics subject to dry friction and external force input v), the model is not well-posed. To prove this, consider the input v(t) = 0.5 = const. It is sufficient to show that no solution of the ODE x˙ (t) = 0.5 − sgn(x(t)) satisfying x(0) = 0 exists on a time interval [0, tf ] for tf > 0. Indeed, let x = x(t) be such solution. As an integral of a bounded function, x = x(t) witll be a continuous function of time. A continuous function over a compact interval always achieves a maximum. Let tm ⊂ [0, tf ] be an argument of the maximum over t ⊂ [0, tf ]. If x(tm) > 0 then tm > 0 and, by continuity, x(t) > 0 in a neighborhood of tm, hence there exists π > 0 such that x(t) > 0 for all t ⊂ [tm − π, tm]. According to the differential equation, this means that x(tm−π) = x(tm)+0.5π > x(tm), which contradicts the selection of tm as an argument of maximum. Hence max x(t) = 0. Similarly, min x(t) = 0. Hence x(t) = 0 for all t. But the constant zero function does not satisfy the differentlial equation. Hence, no solution exists. It can be shown that the absense of solutions in Example 2.1.3 is caused by lack of continuity of function f = f(x, v, t) with respect to x (discontinuity with respect to v and t would not cause as much trouble)