8.6 Illustrative example

This example is to display the fact that the initial conditions of βNH(t) and u(t) must satisfy some conditions in order that the system is impulsive free and the fact that the proposed approach is efficient in analyzing the impulsive property of the system.

Consider the following LNHMDE:

ρ+1ρ201β1(NH)(t)β2(NH)(t)=ρ2+1ρ10ρ+11u1(t)u2(t)u3(t),t0.

si128_e  (8.32)

with the initial values

βNH(0):=β1(NH)(0)β2(NH)(0),βNH(1)(0):=β1(NH)(1)(0)β2(NH)(1)(0)

si129_e

and

u0=u10u20u30,u10=u110u210u310

we have

A2=0100,A1=1000,A0=1001,

B

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