Chapter 3. f(x) == [. k 1 =? - == k

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1 Chapter (1) The term Ixl is not continuously differentiable at x == 0, but it is globally Lipschitz. The term x2 is continuously differentiable, but its partial derivative is not globally bounded. Thus f == x 2 + Ixl is not continuously differentiable at x == O. It is continuously differentiable on a domain that does not include x == O. It is locally Lipschitz, hence continuous, but not globally Lipschitz. (2) The term sgn(x) is discontinuous at x == O. Thus, f(x) == x + sgn(x) does not have any of the four properties in a domain that contains x == O. (3) f(x) == sin(x) sgn(x) is globally Lipschitz. This can be seen as follows..if both x and yare nonnegative, we have If(x) - f(y)1 == I sin(x) - sin(y)1 :5 Ix - yl If x 2: 0 and y :5 0, we have If(x) - f(y)1 == Isin(x}+.sin(y)I == 12 sin(!(x + V»~ cos(!(x - y»1 :5 Ix - yl Other cases can be dealt with similarly to conclude that If(x) - f(y)1 :5 Ix - yl for all x,y. It follows that f is both locally Lipschitz and continuous. It is not continuously differentiable at x == 0 because lim~-to+ f'(x) == +1 while lim~-to- f'(x) == -1. (4) f(x) == -x + asinx is continuously differentiable. Hence, it is locally Lipschitz and continuous. *== -1 + a cos x is globally bounded. Hence, it is globally Lipschitz. (5) f(x) == -x + 21xl is not continuously differentiable. It is globally Lipschitz because both x and Ixl are so. Hence, it is locally Lipschitz. (6) f(x) == tan(x) is continuously differentiable in the open interval -7r/2 < x < 7r/2. Hence, it is locally Lipschitz and continuous in the same interval. Its derivative sec2(x) is not globally bounded; hence, it is not globally Lipschitz. (7) The function tanh(y) is continuously differentiable and its derivative 1/cosh2(y) is globally bounded; hence it is globally Lipschitz. Clearly, the linear function y is both continuously differentiable and globally Lipschitz. Hence, f has all four properties. (8) f is not continuously differentiable due to the term IX21 in fl. Check the Lipschitz property component by component. fl is globally Lipschitz as can be easily checked. h is continuously differentiable, but its partial derivatives are not globally bounded. Hence h is locally Lipschitz but not globally so. Since both h and h are locally Lipschitz, so is f Since f is locally Lipschitz, it is continuous. f is not globally Lipschitz since h is not so. 3.2 (1) X2 ] af [0 1] f(x) == [. k 1 =? - == k - f smxl - ;nx2 + mf':t ax - f cos Xl -;n [oflax] is globally bounded. Hence f is globally Lipschitz, which implies that it is locally Lipschitz on Dr for any r > O. (2) 45

2 46 CHAPTER 3. [of/ax] is continuous everywhere; hence it is bounded on the bounded set Dr Thus f is locally Lipschitz on Dr for any finite r > O. It is not globally Lipschitz since [of/ax] is not globally bounded. (3) TJ(Xl, X2) is discontinuous at X2 = O. Hence it is not locally Lipschitz at the origin. This means it is not locally Lipschitz on Dr for any r > O. (4) f( ) - [ X2 ] 8f _ [0 1] x - -Xl - eel - X~)X2 :::} 8x eXIX2 -eel - xn [oflax] is continuous on Dr; hence f is locally Lipschitz on Dr for any r > O. bounded; hence f is not globally Lipschitz. (5) Let x = [eo, t/>b </>2]T. amx! + kpx2r(t) + kpxa(xl + Ym(t}) 1 f(t,x) = [ -7Xl r(t) -7Xl(Xl + Ym(t)).. [oflax] is not globally [af/ox] is continuous and bounded on Dr for bounded r(t} and Ym(t}. Hence f is locally Lipschitz. It is not globally Lipschitz since [oflax] is not globally bounded. (6) f(x) = Ax - B'I/J(Cx) where 'I/J(.) is a dead-zone nonlinearity. The dead-zone nonlinearity is globally Lipschitz. Hence f is globally Lipschitz, which implies that it is locally Lipschitz on Dr for any r > O. 3.3 For each Xo E R, there exist positive constants r, L 1, ~, ki, and k2 such that!ft(x) - ft(y)1 $ L1lx - YI, Ih(x) - h(y)1 $ L21x - YI, Ift(x)1 $ k}, Ih(x)1 $ k2 for all X,Y E {x E R Ilx - xo/ < r}. For f = ft + 12, we have For f = hh, we have If(x) - f(y)1 = 1ft (x)h(x) - h(y)h(y)1 = Ift(x)h(x) - ft(x)h(y) + ft(x)h(y) - h(y)h(y)i $ Ih(x)1!h(x) - hey)! + Ih,(y)llh(x) - h(y)1 $ k1l 21x - yl + k2l 1 lx - yl $ (kll2 + ~Ll)ix - y! For f =h 0 h, we have If(x) - f(y)l = Ih(fl(X» - h(h(y»1 $ ~Ih(x) - h(y)1 $ L2L lix - yl 3.4 The function f can be written as f(x) = g(x)kxh('l/j(x» where l. t/1, if'i/j?; 1-'> 0 { h('i/j) = 1. p' if tp < J.' and 'l/j(x) = g(x)iikxll The norm function 11KxII is Lipschitz since I IIKxll-IlKyll I $ IIKx - Kyll $IIKllllx - yll

3 Using the previous exercise, we see that 'I/;(x) is Lipschitz on any compact set. Furthermore, g(x)kx is also Lipschitz. Thus, f(x) will be Lipschitz on any compact set if we can show that h('i/;) is Lipschitz in 'ljj over any compact interval [0, b]. Now if '1/;1 ~ IJ and '1/;2 ~ IJ, we have 47 IT '1/;2 ~ Il and '1/;1 < Il, we have IT '1/;1 < Il and '1/;2 < Il, we have Ih('I/;2) - h('i/;dl = I.!. -.!.I = 0 ~.!.1'I/;2 - '1/;11 Il Il 1J2 Thus h('i/;) is Lipschitz with a Lipschitz constant I/1J There are positive constants C1 and C2 such that Suppose III(y) - f(x)lia ~ LallY - xlla IIf(y) - f(x)lit3 ~ c2i1f(y) - f(x)lia ~ c2laily - xlla ::; c2la IIY - xlli3 C1 Similarly, it can be shown that if f is Lipschitz in the {1-norm, it will be Lipschitz in the a-norm. 3.6 (a) By Gronwall-Bellman inequality Integrating by parts, we obtain x(t) = Xo + rt f(r,x(r» dr IIx(t) II ~ IIxoll + f IIf(r, x'(r» II dr t ~ IIxoll+ r[k1 +k2 I1x(r)lIldr = IIxoll + k1(t - to) + k2 rt IIx(r) II dr Hx(t)1I ~ IIxoll + kl (t - to) + rt [IIxoll + kl (s - to)]k2ek2 (t-s) ds kl IIx(t)1I ~ IIxoll exp[k2(t - to)] + k2 {exp[k2(t - to)] - I}, V t, ~ to (b) The upper bound on IIx(t)1I is finite for every finite t. It tends to 00 as t Hence the solution of the system cannot have a finite escape time.

4 48 CHAPTER It can be easily verified that f(x) is continuously differentiable. Hence, local existence and uniqueness follows from Theorem 3.1. lourthermore, IIg(x) IIf(x)1I2 = 1 + IIg(x)lI~ :s "2 Hence 1 IIx(t)1I2 $ IIxol12 + "2(t - to) which shows that the solution is defined for all t 2: to. " 3.8 It can be easily seen that f(x) is continuously differentiable and for some positive constants kl and k2 Apply Exercise Due to uniqueness of solution, trajectories in the plane cannot intersect. Therefore, all trajectories starting in the region enclosed by the limit cycle must remain in that region. The closure of this region is a compact set. Therefore, the solution must stay in a compact set. Apply Theorem ,10 where R = 1.5, u = 1.2, and the nominal values of e and L are 2 and 5, respectively. Let>. = [e, L]T. The Jacobian matrices [atlax] and [afloa], are given by of = l r - b\(xd b R ] a~ = [ - b[-h(xt} +X2] 1 0 ] ax 1. - T. - L ' v>. 0 - p(-xl - &2 + u) Evaluate these Jacobian matrices at the nominal values e =2 and L =5. Let 8 = oxi = [X3 X5] a>. nominal X4 X6 s= afl 8+ afl ' 8(0)=0 ax nominal' {)>. nominal Xl = 0.5[-h(XI) + X2] X2 = 0.2(-Xl - 1.5X ) X3 = 0.5[-h'(XI)X3 + X4] - O.25[-h(Xl) + X2] X4 = O.2(-X3-1.5X4) XS = 0.5[-h'(Xl)X5 + X6] X6 = O.2(-X5-1.5X6) - O.04(-Xl - 1.5X )

5 Xl =X2, X2 =-Xl + E(l - X~)X2 Denote the nominal values of Eby EO. The Jacobian matrices [8/18x] and [8/18E], are given by Let 8/ [0 1] 8/ [ 0 ] 8x = -1-2EXIX2 e(l- xn ' Be = (1- XnX2 Xl X2 8 = 8x 1 1! = [ X3 ] 8e nomi~al X4 8=8/ /1 ' 8(0) = 0 = X2 8x nominal = -Xl + eo(l - X~)X2 3;3 = X4 X4 8E nominal = - [1 + 2eOXIX2] X3 + Eo(1- XnX4 + (1 - X~)X Denote the nominal values of e by eo. The Jacobian matrices [8/18x] and [8/18e], are given by Let 8/ = [ 0 ~.] 8/ o[ - ttx2. ] 8x -e E(l ~ x~), 8E = - (Xl - X2 + ix~) 8=axl ae nominal =[X3] X4 8= a/i 8+ 8/1 ' 8(0)=0 1 Xl = -X2 eo ax nominal X2 = -eo (Xl - X2 + ~X~ ) 1 1 X3 = -X4 - -X2 eo eg ae nominal X4 = -eox3 + eo(l - X~)X4 - (Xl - X2 + ~X~)

6 50 CHAPTER Let A= [a, b, ct. The nominal values are ao = 1, bo = 0, and Co = 1. The Jacobian matrices [oflox] and [of loa], are given by Let o of = [ H:2z~ - X2 -Xl 1 of = [ H:~zi ox 2b ' oa 0 2 Xl -c Xl s= ofl 8 + ofl,8(0) =0 ox nominal OA nominal Xl = tan-1(xi) - XIX2 X2 = -X2 (1) Xl X3 = ---X2 X3- XI X4+- l+x~ l+x~ X4 = -X4 XS = (1 : x~.x2) Xs - XIX6 2 X6 = -X6 +XI = (_1 X7 X2) X7 - XIXS l+x~ Xs = -XS -X (a) Let p = [ ~ ] be the vector of parameteis. The sensitivity equation is given by xd cosh2(axd - X2/ cosh2(ax2) ] xii cosh2(axl) + X2/ cosh2p.x2) s= Ao8 + Bo, 8(0) = 0 where Ao and Bo are evaluated at the nominal parameters. This equation should be solved simultaneously with the nominal state equation.

7 51 (b) rr = Xl%1 + X2%2 = -(1/T)r2 + Xl [tanh('xx1) - tanh(ax2)] + X2[tanh('xX1) + tanh('xxdj = -(1/T)r 2 + r cos (6) [tanh ('xx1 ) - tanh('xx2)] + rsin(6)[tanh('xx1) + tanh('xxdj < -(1/T)r2 + 2r(I cos(6)i + Isin(6)!) < -(1/T)r 2 + 2V2r :t (c) By the comparison lemma, ret) $ u(t) where '1 satisfies the scalar differential equation u = -(1/T)U + 2V2, '1 (0) =reo) = IIx(0)1I2 The solution of this differential equation is u(t) =. exp(-t/t)lix(o)112 + lot exp[-(t - q)/t]2v2 do 3.15 Let V ::: IIxll~ = xr + x~. = exp(-t/t)iix(o) V2T[1 - exp(-tit)] x1x2 4X1X2 V = 2X1%1 + 2X2%2 = - Xl - x IX21 IX11 :5-2V + 41x x x2 +x 1 Taking W =v'v = IIxll2' we see that for V =F 0, At V =0, we have l+x~ 1+4 :5-2V + 21x x21 (Since 1!~2 :5 ~) :5-2V +2v'2rv (since IIxlh S vnllxll2). 11. r.:: W= -- < -W+v2 2v'V IW(t + h) - W(t)1 = IW(t + h)1 =.!.lix(t + h)1i2 h h h Similar to Example 3.9 of the textbook, it can be shown that 1 [Hh lim -h III(x(1"» 112 dt = 0 h-+o+ t Thus D+W(t) :5 -Wet) + v'2 for all t ~ O. Let u(t) be the solution ofthe differential equation U = -'1 + V2, '1 (0) = IIx(0)1I2 By the comparison lemma,