Journal of Multidciplinary Engineering Science Technology (JMEST) ISSN: 359-0040 Vol. Issue 5, December - 204 On one Application of Newton s Method to Stability Problem Şerife Yılmaz Department of Mathematics, Faculty of Science, Anadolu University, Eskehir 26470, Turkey serifeyilmaz@anadolu.edu.tr Abstract In th paper we consider stability problem for switched linear systems. Th problem can be formulated as a convex minimization problem. By modifying the cost functions we apply the vector-valued Newton s method. Keywords Switched system; Hurwitz stability; common quadratic Lyapunov function; Newton s method I. INTRODUCTION Let AA be nn nn real matrix. If all eigenvalues of AA lie in the open left half plane then AA said to be Hurwitz stable. Hurwitz stability of AA equivalent to the following: There exts positive definite symmetric matrix PP such that AA TT PP + PPPP < 0 () the symbol T sts for the transpose, the symbol < for negative definiteness. Hurwitz stability of AA implies the asymptotic stability of the zero solution of the linear system xx = AAAA (2) xx = xx(tt) R nn. If the matrix AA switches between NN matrices AA, AA 2,, AA NN, i.e. AA {AA, AA 2,, AA NN } then the obtained system xx = AAAA (3) called a switched system. Sufficient condition for the asymptotic stability of the zero solution of (3) the extence of quadratic Lyapunov function of the form VV(xx) = xx TT PPPP PP > 0 AA ii TT PP + PPAA ii < 0 (ii =,2,, NN). (4) The matrix PP called a common solution to the Lyapunov inequalities (4). The stability problem of linear switched systems has been investigated in a lot of works (see [-] references therein). The papers [3-] study theoretical results for the extence of a common solution to (4). The papers [2-4] consider numerical algorithms for a common positive definite solution in the case of extence. In th paper we apply Newton s root finding method for the numerical generating of a common positive definite solution to (4). II. LYAPUNOV EQUATIONS In th section, we consider the Lyapunov inequality () which equivalent to the following equation. AA TT PP + PPPP = QQ (5) QQ > 0. We are looking for a positive definite solution PP of (5). In the iteration steps, the obtained PP guaranteed to be symmetric. The following theorem shows that in the case of Hurwitz stability of AA th implies the positive definiteness of PP. Theorem. Assume that AA Hurwitz stable. If there exts a symmetric solution PP to (5) then PP > 0. Proof: Define PP = ee AATTtt QQee AAAA dddd 0 ee AAAA sts for the matrix exponential. (6) Since AA Hurwitz stable, the matrix AA TT also Hurwitz stable. ee AAAA ee AATT tt define exponential functions with exponents Re(λλ ii ) tt < 0 λλ ii are the eigenvalues of AA. Th implies that the integral in ( 6 ) well defined. The matrix PP symmetric, positive definite satfies the following relation AA TT PP + PP AA = QQ (see [5]). Then AA TT PP PP + PP PP AA = 0. Multiplying by ee AATT tt ee AAAA give 0 = ee AATTtt AA TT PP PP + PP PP AA ee AAAA = dd dddd eeaatttt PP PP ee AAAA JMESTN42350279 364
Journal of Multidciplinary Engineering Science Technology (JMEST) ISSN: 359-0040 Vol. Issue 5, December - 204 The integration from 0 to yields ee AATTtt PP PP ee AAAA 0 = 0. Using the fact that ee AAAA 0, ee AATTtt 0 as tt we obtain 0 PP PP = 0 PP = PP > 0. We are looking for a iterative procedure for a common PP satfying ( 4 ). Theorem allows to guarantee positive definiteness of PP obtained at each step of iteration. III. MODIFIED NEWTON S METHOD Consider a differentiable function FF: R nn R nn the following equation FF(xx) = 0. (7) Here xx = (xx, xx 2,, xx nn ) TT R nn, FF(xxff (xx), ff 2 (xx),, ff nn (xx) TT. Denote the Jacobian matrix by JJ(xx), i.e. JJ(xx ff ii(xx) (ii, jj =,2,, nn). xx jj The Newton method a method for an approximate solution of (7) starting from a suitable initial point xx 0 the iteration defined by xx kk = xx kk JJ(xx kk ) FF(xx kk ) (kk =,2, ) Define rr = nn(nn+) 2 xx xx 2 xx nn xx 2 xx nn+ xx 2nn PP = PP(xx xx nn xx 2nn xx rr The matrix inequalities (4) are equivalent to ff ii (xx) = λλ max (AA ii TT PP + PPAA ii ) < 0 (ii =,2,, NN) (8) xx = (xx, xx 2,, xx rr ) TT R rr, λλ max ( ) sts for the maximal eigenvalue. In the case of simple maximum eigenvalue of AA TT PP(xx) + PP(xx)AA, the gradient of ff(xx) = λλ max (AA TT PP(xx) + PP(xx)AA) should be easily calculated. Indeed since the function xx AA TT PP(xx) + PP(xx)AA linear then AA TT PP(xx) + PP(xx)AA = xx jj QQ jj. rr jj= Then ff(xx) = (uu TT QQ uu,, uu TT QQ rr uu), sts for the gradient, uu the unit eigenvector corresponding to the maximum eigenvalues of AA TT PP(xx) + PP(xx)AA (see [2]). Proposition. The function ff ii (xx) convex for each ii. Proof: The relation PP AA ii TT PP + PPAA ii linear. On the other h for symmetric CC, the function CC λλ max (CC) convex [6]. ff ii (xx) convex as a composition of linear convex functions. The system (4) has a common solution PP > 0 if only if there exts xx R rr such that ff ii (xx ) < 0 (ii =,2,, NN). (9) In order to apply Newton s method instead of the minimization of the functions ff ii (xx), we consider the system of equations ff ii (xx) = 0 (ii =,2,, NN). Without loss of generality we can set rr = NN. Indeed if NN > rr, we can combine some function by using the operation maximum. For example if NN = rr + then define gg (xx) = max{ff (xx), ff 2 (xx)}, gg ii (xx) = ff ii+ (xx) (ii = 2,3,, NN). Th operation preserves convexity. If NN < rr we use the operation of duplication. Thus from the now we assume that rr = NN. Define FF = (ff, ff 2,, ff rr ) TT consider the equation FF(xx) = 0 (0) xx R rr. If we apply the classical Newton s method to (0) we obtain the trivial sequence PP kk 0, since the functions ff ii (xx) are positive homogenous. To avoid th we impose the condition trace(pp) =. The following proposition shows that th does not violate the generality. Proposition 2. Assume that PP > 0 AA TT PP + PPPP < 0. Then AA TT PP + PP AA < 0 PP = trace(pp) PP. Proof: From PP = pp iiii > 0 it follows that for all xx R nn, xx 0, xx TT PPPP > 0. Taking xx ii = (0,,0,,0,,0) TT we obtain pp iiii > 0. trace(pp) > 0 AA TT PP + PP AA = trace(pp) [AATT PP + PPPP] < 0. The condition trace(pp) = reduces the number of variables from rr to rr. To solve (0) the following algorithm suggested. Algorithm. ) Consider the equation (0). Take initial matrix QQ = diag(,2,2,,2) consider AA TT PP + PPAA = QQ. JMESTN42350279 365
The solution of th matrix equation let be PP 0. Dividing PP 0 by trace(pp 0 ) gives the initial iteration xx 0. 2) Replace the functions ff ii (xx) by ff ii(xx) + (ii =,2,, rr ). Apply Newton s iteration 2trace(PP 0 ) xx kk = xx kk JJ(xx kk ) FF (xx kk ) FF = (ff, ff 2,, ff rr ). 3) If ff ii(xx kk ) < 0 (ii =,2,, rr ) for some kk then stop. Otherwe continue. Example. Consider the Hurwitz stable matrices AA = 4 8 AA 2 = 3 5 2 ff (xx) = λλ max (AA TT PP(xx) + PP(xx)AA ), ff 2 (xx) = λλ max (AA 2 TT PP(xx) + PP(xx)AA 2 ) Hence PP(xx xx xx 2 xx 2 xx QQ = 0 0 2 AA TT PP + PPAA = QQ (ii =,2) 0.972 0.472 PP 0 = 0.472 0.36 trace(pp 0 ) PP 0.729 0.354 0 = 0.354 0.27 take the initial point xx 0 = (0.729, 0.354) TT. For th point calculations give the following maximum eigenvalues its corresponding unit eigenvectors: λλ max (AA TT PP(xx 0 ) + PP(xx 0 )AA ) = 0.75 the maximum eigenvector: uu = (,0) TT, λλ max (AA 2 TT PP(xx 0 ) + PP(xx 0 )AA 2 ) = 0.8333 the maximum eigenvector: uu 2 = ( 0.7090, 0.705) TT. ff (xx 0 ) = 0.75, ff 2 (xx 0 ) = 0.833. ff (xx) xx=xx 0 = ( 2, 2), ff 2 (xx) xx=xx 0 = (2.988,0.96). the Jacobian matrix of FF(xx) = ff (xx), ff 2 (xx) TT at xx 0 Journal of Multidciplinary Engineering Science Technology (JMEST) ISSN: 359-0040 Vol. Issue 5, December - 204 JJ(xx 0 2 2 2.988 0.96 xx = 0.729 0.237 0.493 + 0.375 + 0.75 0.354 0.737 0.493 0.833 + 0.375 = 0.30 0.9. After 3 steps, we get ff (xx 3 ) < 0 ff 2 (xx 3 ) < 0 (see Table I). Hence for the matrix PP = PP(xx 3 0.390 0.247 0.247 0.609, AA TT ii PP + PPAA ii < 0 (ii =,2) are satfied. Table I kk xx kk ff (xx kk ) ff 22 (xx kk ) (0.30,0.9) TT 0.22.268 2 (0.367, 0.35) TT 0.373 0.278 3 (0.390, 0.247) TT 0.285 0.074 Example 2. Consider the Hurwitz stable matrices AA = 2 2, AA 2 3 2 = 2 AA 3 = 2 2 2 ff (xx) = max(λλ max (AA TT PP(xx) + PP(xx)AA ), λλ max (AA 2 TT PP(xx) + PP(xx)AA 2 )), ff 2 (xx) = λλ max (AA 3 TT PP(xx) + PP(xx)AA 3 ). PP(xx xx xx 2 xx 2 xx QQ = 0 0 2 AA TT PP + PPAA = QQ (ii =,2) PP 0 = 0.46 0.083 0.083 0.583 trace(pp 0 ) =. Take the initial point xx 0 = (0.46,0.083) TT. For th point calculations give the following maximum eigenvalues its corresponding unit eigenvectors: λλ max (AA TT PP(xx 0 ) + PP(xx 0 )AA ) =, uu = (,0) TT, λλ max (AA 2 TT PP(xx 0 ) + PP(xx 0 )AA 2 ) = 0.680, uu 2 = ( 0.082,0.996) TT, λλ max (AA 3 TT PP(xx 0 ) + PP(xx 0 )AA 3 ) = 0.567, uu 3 = ( 0.987,0.60) TT. JMESTN42350279 366
ff (xx 0 ) = max( 2.05,0.074) = 0.680, ff 2 (xx 0 ) = 0.567. ff (xx) xx=xx 0 = (.9, 5.767), ff 2 (xx) xx=xx 0 = (0.786, 3.478). The Jacobian matrix of FF(xxff (xx), ff 2 (xx) TT at xx 0 JJ(xx 0.9 5.767 0.786 3.478 xx = 0.46 0.664 + 0.5 + 0.400 0.680 0.083 0.090 0.37 0.567 + 0.5 = 0.80 0.336. After 6 steps, we get ff (xx 6 ) < 0 ff 2 (xx 6 ) < 0 (see Table II). Hence for the matrix 5 PP = PP(xx 6 0.46 0.38 0.38 0.538, AA ii TT PP + PPAA ii < 0 (ii =,2,3) are satfied. Table II kk xx kk ff (xx kk ) ff 22 (xx kk ) (0.80,0.336) TT.035 0.224 2 (0.58,2.025) TT 8.502 6.64 3 (0.553,0.33) TT 0.08 0.22 (0.293,0.99) TT 2.527.54 6 (0.46,0.38) TT 0.063 0.074 Example 3. Consider the Hurwitz stable matrices 32 5 2 4 5 2 AA = 0 2, AA 2 = 6 3, 9 7 7 0 0 5 3 6 2 AA 3 = 2 4 2, AA 4 = 3 3 4, 4 5 2 4 0 4 2 AA 5 = 7 8 20 7 2 22 ff ii (xx) = λλ max (AA ii TT PP(xx) + PP(xx)AA ii ) Journal of Multidciplinary Engineering Science Technology (JMEST) ISSN: 359-0040 Vol. Issue 5, December - 204 xx xx 2 xx 3 PP(xxxx 2 xx 3 xx 4 xx 5 xx 5 xx xx 4 Hence 0 0 QQ = 0 2 0 0 0 2 AA TT PP + PPAA = QQ (ii =,2) 0.052 0.38 0.027 PP 0 = 0.38 0.588 0.28 0.027 0.28 0.093 0.069 0.88 0.037 trace(pp 0 ) PP 0 = 0.88 0.803 0.74 0.037 0.74 0.26 the initial point xx 0 = (0.069, 0.88,0.037,0.803, 0.74) TT. For th point calculations give the following: FF(xx 0 ) = (.363,.952,0.408,.843,.909) TT, JJ(xx 0 ) 64 20 7.999 0 0 8.937 9.033 4.0.273 0.357 = 5.96 4.20 4.2 0.28 0.677 0.047 3.062 6.494 5.374 2.547 2.888 8.46 5.925 35.579.025 xx = ( 0.090,0.252,0.079,0.67, 0.26) TT. After 36 steps, we get xx 36 = (0.42, 0.38, 0.044,0.28, 0.028) TT FF(xx 36 ) = ( 0.590, 0.360, 0.608, 0.447, 0.64) TT. Hence AA ii TT PP + PPAA ii < 0 (ii =,2,3,4,5) 0.42 0.38 0.044 PP = PP(xx 36 0.38 0.28 0.028 0.044 0.028 0.297 REFERENCES [] H. Lin P.J. Antsakl, Stability Stabilizability of Switched Linear Systems: A Survey of Recent Results, IEEE Transactions on Automatic Control, 54(2): 308 322, 2009. [2] D. Liberzon, Switching in System Control. Birkhäuser, Boston, 2003. [3] K.S. Narendra J. Balakrhnan, A common Lyapunov function for stable LTI systems with commuting A-matrices, IEEE Transactions on Automatic Control, 39, 2469 247, 994. [4] Y. Mori, T. Mori Y. Kuroe, A solution to the common Lyapunov function problem for continuous-time systems, In Proceedings of the 36 th JMESTN42350279 367
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