### Lecture 38

A consequence :

In the light of the above proposition we again repeat that that the positive definiteness of
the matrices *R* and *Q* is a necessary and sufficient condition for the asymptotic stability of
the zero solution of the linear system (5.60).

Remark: The stability properties of zero solution of the equation (5.62)is unaffected if
the system (5.60) is transformed by the relation*x*=*P y*, where*P* is a non-singular constant
matrix. The system (5.60) then transforms to

*y*^{0}= (*P*^{−1}*AP*)*y.*

Now choose the matrix*P* such that

*P*^{−1}*AP*

is a triangular matrix. Such a transformation is always possible by Jordan normal form. So
there is no loss of generality by assuming in (5.60) that, the matrix*A* is such that its main
diagonal consists of eigenvalues of*A* and for*i < j, a*_{ij} = 0. In other words the matrix*A* is
of the following form:

*A*=

*λ*_{1} 0 0 *· · ·* 0
*a*_{21} *λ*_{2} 0 *· · ·* 0
*a*_{31} *a*_{32} *λ*_{3} *· · ·* 0
... ... ... . .. ...

*a*_{n1} *a*_{n2} *a*_{n3} *· · ·* *λ*_{n}

.

The equation (5.62) is

*λ*_{1} *a*_{21} *a*_{31} *· · ·* *a*_{n1}
0 *λ*_{2} *a*_{32} *· · ·* *a*_{n2}
... ... ... . .. ...
0 0 0 *· · ·* *λ*_{n}

*r*_{11} *r*_{12} *r*_{13} *· · ·* *r*_{1n}
*r*_{21} *r*_{22} *r*_{23} *· · ·* *r*_{2n}
... ... ... . .. ...
*r*_{n1} *r*_{n2} *r*_{n3} *· · ·* *r*_{nn}

+

*r*_{11} *r*_{12} *r*_{13} *· · ·* *r*_{1n}
*r*_{21} *r*_{22} *r*_{23} *· · ·* *r*_{2n}
... ... ... . .. ...
*r*_{n1} *r*_{n2} *r*_{n3} *· · ·* *r*_{nn}

*λ*_{1} 0 0 *· · ·* 0
*a*_{21} *λ*_{2} 0 *· · ·* 0
... ... ... . .. ...

*a*_{n1} *a*_{n2} *a*_{n3} *· · ·* *λ*_{n}

=*−*

*q*_{11} *q*_{12} *q*_{13} *· · ·* *q*_{1n}
*q*_{21} *q*_{22} *q*_{23} *· · ·* *q*_{2n}
... ... ... . .. ...
*q*_{n1} *q*_{n2} *q*_{n3} *· · ·* *q*_{nn}

.

Equating the corresponding terms on both sides results in the following system of equations
(*λ*_{j}+*λ*_{k})*r*_{jk} =*−q*_{jk}+*δ*_{jk}(*· · ·* *, r*_{hk}*,· · ·*)*,*

where*δ*_{jk} is a linear form in*r*_{hk}*, h*+*k > j*+*k*, with coefficients in*a*_{rs}. Hopefully the above
system determines*r*_{jk}. The solution of the linear system is unique if the determinant of the

coefficients is non-zero. Obviously the determinant is the product of the coefficients of the form

*λ*_{j}+*λ*_{k}*.*

In such a case the matrix *R* is uniquely determined if none of the characteristic roots *λ*_{i}
is zero and further the sum of any two different roots is not zero. The following example
illustrates the procedure for the determination of*R*.

Example 5.8.1. Let us construct a Lyapunov function for the system
*x*^{0}_{1} =*−*3*x*_{1}+*kx*_{2}*, x*^{0}_{2} =*−*2*x*_{1}*−*4*x*_{2}

to find values of*k* which ensures the asymptotic stability of the zero solution. In this case
*A*=

·*−*3 *k*

*−*2 4

¸

. Let*Q* be an arbitrary positive definite matrix, say
*Q*=

·2 0 0 2

¸ . Now Eq. (5.62) is

·*−*3 *−*2

*k* *−*4

¸ ·*r*_{11} *r*_{12}
*r*_{21} *r*_{22}

¸ +

·*r*_{11} *r*_{12}
*r*_{21} *r*_{22}

¸ ·*−*3 *k*

*−*2 4

¸

=

·*−*2 0
0 *−*2

¸ .

Consequently (equating the terms on both sides solving the system of equations) we have
*r*_{11}= 16 +*k*

7(*k*+ 6)*, r*_{12}=*r*_{21}= *−*3 + 2*k*

7(*k*+ 6)*, r*_{22}= 21 + 2*k*+*k*^{2}
14(*k*+ 6)
or else

*R*= 1

14(*k*+ 6)

·32 + 2*k* *−*6 + 4*k*

*−*6 + 4*k* 21 + 2*k*+*k*^{2}

¸
*.*
Now*R* is positive definite if

(i) 32 + 2*k*
14(*k*+ 6) *>*0*,*

(ii) (32 + 2*k*)(21 + 2*k*+*k*^{2})*−*(4*k−*6)^{2}

14(*k*+ 6) *>*0.

Consequently, it is true if*k >−*6 or*k <−*16. Thus for any*k*between (*−*16*,−*6) the matrix
*R*is positive definite and therefore, the zero solution of the system is asymptotically stable.

### Lecture 39

Let*g*:*S*_{ρ}*→*R^{n} be a smooth function . Let us consider the following system of equations (
in a vector form)

*x*^{0}=*g*(*x*)*,* (5.63)

where*g*(0) = 0. Let us denote _{∂x}^{∂g}^{i}_{j} by *a*_{ij}. Then, equation (5.63) may be written as

*x*^{0} =*Ax*+*f*(*x*)*,* (5.64)

where*f* contains terms of order two or more in (*x*)and*A*= [*a*_{ij}]. Now we study the stability
of the zero solution of the system (5.64). The system (5.60) namely,

*x*^{0} =*Ax, x∈*R^{n}*,*

is called the homogeneous part of the system (5.63) ( which sometimes is also called the
linearized part of the system (5.64).We know that the zero solution of the system (5.60) is
asymptotically stable when*A*is a stable matrix. We now make use of the Lyapunov function
given by (5.61) to study the stability behavior of certain nonlinear systems which are related
to the linearized system (5.60). Let the Lyapunov function be

*V*(*x*) =*x*^{T}*Rx,*

where*R* is the unique solution of the equation (5.62). We have already discussed a method
for the determination of a matrix*R*.

For the asymptotic stability of the zero solution system (5.64), the function *f* naturally
has a role to play. We expect that if*f* is small then, the zero solution of the system (5.64)
may be asymptotically stable. With this short introduction let us employ the same Lyapunov
function (5.61) to determine the stability of the origin of (5.64). Now the time derivative of
*V* along a solution of (5.64) is

*V*˙(*x*) =*x*^{0T}*Rx*+*x*^{T}*Rx*^{0} = (*x*^{T}*A*^{T} +*f*^{T})*Rx*+*x*^{T}*R*(*Ax*+*f*)

=*x*^{T}(*A*^{T}*R*+*RA*)*x*+*f*^{T}*Rx*+*x*^{T}*Rf* =*−x*^{T}*Qx*+ 2*x*^{T}*Rf,* (5.65)
because of (5.62) and (5.64). The second term on the right side of (5.65) contains terms
of degree three or higher in *x*. The first one contains a term of degree two in *x*. The first
term is negative whereas the sign of the second term depends on*f*. Whatever the second
term is, at least a small region containing the origin can definitely be found such that the
first term predominates the second term and thus, in this small region the sign of ˙*V* remains
negative. This implies that the zero solution of nonlinear equation (5.64) is asymptotically
stable. Obviously the negative definiteness of ˙*V* is only in a small region around origin.

Definition 5.8.2. The region of stability for a differential equation (5.64) is the set of all
initial points*x*_{0} such that

*t→∞*lim *x*(*t, t*_{0}*, x*_{0}) = 0*.*

If the stability region is the whole of R^{n} then the we say the zero solution is asymp-
totic stability in the large or globally asymptotically stable. We give below a method of
determining the stability region for the system (5.64).

Consider 3 a surface *{x* : *V*(*x*) = *k}* (where *k* is a constant to be determined) lying
entirely inside the region *{x* : ˙*V*(*x*) *≤* 0*}*. Now find *k* such that *V*(*x*) = *k* is tangential to
the surface ˙*V*(*x*) = 0. Then, stability region for the system (5.64) is the set*{x*:*V*(*x*)*≤k}*.

Example 5.8.3 given below illustrates a procedure for finding the region of stability.

Example 5.8.3. Consider a nonlinear system

·*x*_{1}
*x*_{2}

¸_{0}

=

·*−*1 3

*−*3 *−*1

¸ ·*x*_{1}
*x*_{2}

¸ +

·0
*x*^{2}_{2}

¸
.
Let*V*(*x*) =*x*^{T}*Rx*, where *R* is the solution of the equation

·*−*1 *−*3
3 *−*1

¸
*R*+*R*

·*−*1 3

*−*3 *−*1

¸

=*Q*
Choose *Q*=

·4 0 0 4

¸

, so that *R*=

·2 0 0 2

¸
. Thus
*V*(*x*_{1}*, x*_{2}) = 2(*x*^{2}_{1}+*x*^{2}_{2})

*V*˙(*x*_{1}*, x*_{2}) = 4(*x*_{1}*x*^{0}_{1}+*x*_{2}*x*^{0}_{2}) = 4[*−x*^{2}_{1}*−x*^{2}_{2}(1*−x*_{2})]

with respect to the given system. To find the region of asymptotic stability consider the surface

(*x*_{1}*, x*_{2}) : ˙*V*(*x*_{1}*, x*_{2}) = 4[*−x*^{2}_{1}*−x*^{2}_{2}(1*−x*_{2})] = 0*.*

When

*x*_{2} *<*1*,* *V*˙(*x*_{1}*, x*_{2})*<*0 for all*x*_{1}*.*
Hence,

(*x*_{1}*, x*_{2}) :*V*(*x*) = 2(*x*^{2}_{1}+*x*^{2}_{2})*≤*1
is the region which lies in the region

*V*˙(*x*_{1}*, x*_{2})*<*0*.*

The size of the stability region thus obtained depends on the choice of a matrix*Q*.

EXERCISES

1. Prove that the stability of solutions the equation (5.62) remain unaffected by a trans-
formation*x*=*P y*, where *P* is a non-singular matrix.

2. If*R* is a solution of the equation (5.62) then, prove that so is*R*^{T} and hence, *R*^{T} =*R*.

3. The matrices*A*and*Q*are given below. Find a matrix*R* satisfying the equation (5.62)
for each of the following cases.

(i)*A*=

· 0 1

*−*2 *−*3

¸

, *Q*=

·2 0 0 2

¸

;

(ii)*A*=

·*−*1 3

*−*3 *−*1

¸

, *Q*=

·4 0 0 4

¸

; and
(iii)*A*=

·*−*3 *−*5

*−*2 *−*4

¸

, *Q*=

·2 0 0 2

¸ . 4. For the system

*x*_{1}
*x*_{2}
*x*_{3}

*0*

=

0 *p* 0

0 *−*2 1

*−*1 *−*1 *−*1

*x*_{1}
*x*_{2}
*x*_{3}

.

Choose

*Q*=

2 0 0 0 0 0 0 0 0

.

Determine the value/values of*p* for which the matrix*R* is positive definite.

5. For the system

*x*^{0}_{1}=*−x*_{1}+ 2*x*_{2}*, x*^{0}_{2}=*−*2*x*_{1}+*x*_{2}+*x*^{2}_{2}
find the region of the asymptotic stability.

6. Prove that the zero solution of the system

(*x*_{1}*, x*_{2})^{0} = (*−x*_{1}+ 3*x*_{2}*,−*3*x*_{1}*−x*_{2}*−x*^{3}_{2})
is asymptotically stable.