A Particular Lyapunov Function

In document lectures.pdf - Nptel (Page 151-157)

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 relationx=P y, whereP is a non-singular constant matrix. The system (5.60) then transforms to

y0= (P1AP)y.

Now choose the matrixP such that

P1AP

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 matrixA is such that its main diagonal consists of eigenvalues ofA and fori < j, aij = 0. In other words the matrixA is of the following form:

A=







λ1 0 0 · · · 0 a21 λ2 0 · · · 0 a31 a32 λ3 · · · 0 ... ... ... . .. ...

an1 an2 an3 · · · λn





 .

The equation (5.62) is





λ1 a21 a31 · · · an1 0 λ2 a32 · · · an2 ... ... ... . .. ... 0 0 0 · · · λn









r11 r12 r13 · · · r1n r21 r22 r23 · · · r2n ... ... ... . .. ... rn1 rn2 rn3 · · · rnn





+





r11 r12 r13 · · · r1n r21 r22 r23 · · · r2n ... ... ... . .. ... rn1 rn2 rn3 · · · rnn









λ1 0 0 · · · 0 a21 λ2 0 · · · 0 ... ... ... . .. ...

an1 an2 an3 · · · λn





=





q11 q12 q13 · · · q1n q21 q22 q23 · · · q2n ... ... ... . .. ... qn1 qn2 qn3 · · · qnn



.

Equating the corresponding terms on both sides results in the following system of equations (λj+λk)rjk =−qjk+δjk(· · · , rhk,· · ·),

whereδjk is a linear form inrhk, h+k > j+k, with coefficients inars. Hopefully the above system determinesrjk. 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

λj+λk.

In such a case the matrix R is uniquely determined if none of the characteristic roots λi is zero and further the sum of any two different roots is not zero. The following example illustrates the procedure for the determination ofR.

Example 5.8.1. Let us construct a Lyapunov function for the system x01 =3x1+kx2, x02 =2x14x2

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

·3 k

2 4

¸

. LetQ be an arbitrary positive definite matrix, say Q=

·2 0 0 2

¸ . Now Eq. (5.62) is

·3 2

k 4

¸ ·r11 r12 r21 r22

¸ +

·r11 r12 r21 r22

¸ ·3 k

2 4

¸

=

·2 0 0 2

¸ .

Consequently (equating the terms on both sides solving the system of equations) we have r11= 16 +k

7(k+ 6), r12=r21= 3 + 2k

7(k+ 6), r22= 21 + 2k+k2 14(k+ 6) or else

R= 1

14(k+ 6)

·32 + 2k 6 + 4k

6 + 4k 21 + 2k+k2

¸ . NowR is positive definite if

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

(ii) (32 + 2k)(21 + 2k+k2)(4k−6)2

14(k+ 6) >0.

Consequently, it is true ifk >−6 ork <−16. Thus for anykbetween (16,−6) the matrix Ris positive definite and therefore, the zero solution of the system is asymptotically stable.

Lecture 39

Letg:SρRn be a smooth function . Let us consider the following system of equations ( in a vector form)

x0=g(x), (5.63)

whereg(0) = 0. Let us denote ∂x∂gij by aij. Then, equation (5.63) may be written as

x0 =Ax+f(x), (5.64)

wheref contains terms of order two or more in (x)andA= [aij]. Now we study the stability of the zero solution of the system (5.64). The system (5.60) namely,

x0 =Ax, x∈Rn,

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 whenAis 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) =xTRx,

whereR is the unique solution of the equation (5.62). We have already discussed a method for the determination of a matrixR.

For the asymptotic stability of the zero solution system (5.64), the function f naturally has a role to play. We expect that iff 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) =x0TRx+xTRx0 = (xTAT +fT)Rx+xTR(Ax+f)

=xT(ATR+RA)x+fTRx+xTRf =−xTQx+ 2xTRf, (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 onf. 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 pointsx0 such that

t→∞lim x(t, t0, x0) = 0.

If the stability region is the whole of Rn 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

·x1 x2

¸0

=

·1 3

3 1

¸ ·x1 x2

¸ +

·0 x22

¸ . LetV(x) =xTRx, 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(x1, x2) = 2(x21+x22)

V˙(x1, x2) = 4(x1x01+x2x02) = 4[−x21−x22(1−x2)]

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

(x1, x2) : ˙V(x1, x2) = 4[−x21−x22(1−x2)] = 0.

When

x2 <1, V˙(x1, x2)<0 for allx1. Hence,

(x1, x2) :V(x) = 2(x21+x22)1 is the region which lies in the region

V˙(x1, x2)<0.

The size of the stability region thus obtained depends on the choice of a matrixQ.

EXERCISES

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

2. IfR is a solution of the equation (5.62) then, prove that so isRT and hence, RT =R.

3. The matricesAandQare given below. Find a matrixR 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

x1 x2 x3

0

=

 0 p 0

0 2 1

1 1 1

x1 x2 x3

.

Choose

Q=

 2 0 0 0 0 0 0 0 0

.

Determine the value/values ofp for which the matrixR is positive definite.

5. For the system

x01=−x1+ 2x2, x02=2x1+x2+x22 find the region of the asymptotic stability.

6. Prove that the zero solution of the system

(x1, x2)0 = (−x1+ 3x2,−3x1−x2−x32) is asymptotically stable.

In document lectures.pdf - Nptel (Page 151-157)

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