Question

Problem 2.Given the following dynamic system Given the Lyapunov (energy) function: V = 1. What is the definiteness (positive definite PD, negative definite ND, PSD, NSD) of? 2. What is the definiteness of V - dl 3. Based on Lyapunov Stability theorem, is the system stable? 4. Using the eigenvalues technique, is the system stable? dt

Answer 1

(1)

Positive Definite

A scalar function
V(r)
is said to be positive definite if
V(r)
is such that
V(r)
is positive at all points in the state space except at the origin, where
V(r)
is equal to zero.

V (z) > OY.r,メ0.1 = 1, 2, 3, n

V(z) 0Y.r, = 0.1 = 1, 2, 3, : n

Negative Definite

A scalar function
V(r)
is said to negative definite if
V(r)
is such that
V(r)
is negative at all points in the state space except at the origin, where
V(r)
is equal to zero.

$V(x) < 0 \forall x_{i} \neq 0,i=1,2,3,...,n$

V(z) 0Y.r, = 0.1 = 1, 2, 3, : n

Positive Semi Definite

A scalar function
V(r)
is said to be positive semidefinite if
V(r)
is such that
V(r)
is positive at all points in the state space except at one or more points in the state space including the origin where
V(r)
it is equal to zero.

Positive definite function will be zero only at the origin.
The positive semi-definite function can be zero at several points in the state space with function remaining positive at all other places.

$V(x) >0 \forall x_{i} = 0,i=1,2,3,...,n$

$V(x) >0 \forall x_{i}=0$and some $x_{i}\neq 0, i = 1,2,....,n$

Negative Semi Definite

A scalar function
V(r)
is said to be negative semi-definite if
V(r)
is such that
V(r)
is negative at all points in the state space except at one or more points in the state space including the origin where it is equal to zero.

$V(x) < 0 \forall x_{i} \neq 0,i=1,2,3,...,n$

$V(x) < 0 \forall x_{i}=0$ and some $x_{i}\neq 0,i= 1,2,...,n$

(2)

V = 22.12: 1 2x2x2

$\dot{V}= 2x_{1}(-x_{1}-x_{2})+2x_{2}(x_{1}-x_{2})=-2x_{1}^{2}-2x_{2}^{2}$

which is a negative definite.

(3)

Equilibrium points are :
0 T1
and $x_{2}=0$

is positive definite.

V(r)

is negative definite

V(z)

so equilibrium state at origin is uniformly asymptotically stable in Large.

(4)

2 72

where  $A=\begin{bmatrix} -1 & -1\\ 1 & -1 \end{bmatrix}$

eigenvalues calculates as $\begin{vmatrix} A-\lambda I \end{vmatrix} = 0$

eigen values are : -1+i and -1-i both lies on negative axis so system is stable.