. 2' 22 2xsint. v = xx =- x + --< -2v + 1 l+x2. u= -2u + 1, u(o) = 2. Ix(t)1 = vv(t) $ 2. l~txtxl $ 2I/xIl2I1f(t,x)1I2 $ 2Lllxll~

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1 52 CHAPTER 3. o 3.16 Let v = x2 let u(t) be the solution ofthe differential equation. 2' 22 2xsint v = xx =- x + --< -2v + 1 l+x2 u= -2u + 1, u(o) = 2 Then Thus ft 1 + 3e-2t vet) $ u(t) = 2e- 2t + 10 e- 2 (t-r) dr = 2 VI +3e- 2t Ix(t)1 = vv(t) $ 2 <it 3.17 (a) ~XTX =2xTx =2xT f(t,x) dt l~txtxl $ 2I/xIl2I1f(t,x)1I2 $ 2Lllxll~ (b) Let Vet) = xt(t)x(t) and Vo = x6'xo, then from part (a) we have -2LV(t) $ i'(t) $ 2LV(t) Divide through by Vet), multiply by dt, and integrate to obtain Taking the square root yields., 3.18 Let z(t) = y(t)eo'(t-to). Then _ ft 2L dt $ fv dv v $ ft 2L dt lto lvo lto. (V(t») -2L(t - to) $ In Vo $ 2L(t - to) Voexp[-2L(t - to)] $ Vet) $ Voexp[2L(t - to)] IIxoll2 exp[-l(t - to)] $ IIx(t)1I2 $ IIxol/2 exp[l(t - to)] z(t) $ kl + ( eo'(r-to) [k 2 y(r) + k:rj dr = kl + ft [k 2 z(r) + k 3eO'(r-to)] dr J~ ho - kl + kal: z(r) dr + (ka(o:) [expq(t-to) -1] From Gronwall-Bellman inequality, z(t) :5 kl + (k3/a.) [expq(t-~) -1] +. ~t {kl + (k3/a) [expq(6-to) -I]} k 2 ek2(t-s) ds ltd By evaluating the integtal, it can be shown that z(t) $ k3 eck(t-to) + (kl _ k3) e.l:2(t-to) + k2 k3 react-to) _ e.l:2(t-to>] 0: a a(a - k2)

2 53 Hence yet) 3.19 Choose the covering of S as described in the hint. Within each neighborhood we have If x, Y E SnN(ai, Ti) for some i, then the Lipschitz condition holds with L = Li. Otherwise, IIx-yll ;::: mini Ti. Since f(x) is uniformly bounded on S, we have Therefore, whenever IIx - yll ;::: mini T.j, we have Take IIf(x) - f(y)1i ~ c, 'r/ x, yes, C > 0 C IIf(x) - f(y)1i ~ -.-lix - yll rmni ri 3.20 We have IIf(x) - fey)!! ~ Lllx - yll, 'r/ X,y E W Given e > 0 1 let fj = ell. For all IIx - yll < fj, we have IIf(x) - f(y)1i < LfJ = e, which implies uniform continuity The vector y is defined only for xi: O. For x = 0, we can take y as any vector with IIYllq = 1. Now, for x 1= 0 we have TX_~.x._~xfSign(xf) _ 1 ~ x. p IIxll~ - x y - ~y. - ~ IIxll~ 1 - IIxll:-1 ~I.1 - IIxll~-l -II 111' For p = 00, take { I, if i =argmax IXil Yi = 0 otherwise Then, ytx = IIxli oo and liylll = If 11:~(t'X)11 ~ L, 'r/ (t,x) E [a,b] x R n then, from Lemma 3.1, IIi(t, y) - f(t, x)1i $ Lllx - yll, 'r/ (t, x) E [a, b] x Rn

3 54 CHAPTER 3. Alternatively, suppose I(t, x) is globally Lipschitz. By the mean value theorem where z == ax + (1- a)y and 0 < a < 1. Then ali h(t, y) -h(t, x) == ax (t, z) (y - x) II ~(t,z) (y - X)II = IIh(t,y) -h(t,x)1i :5 LillY - xii, V x,y ERn, V t E [a, b) Hence 1I~(t,z)(y - x)11 n By _ xii $ L i, V x, Y E R,V t E [a, b) Taking y == {3x with IJ > 1, we have z == [a + (1 - a){3)x 1t -yx, and Thus 11~(t,-yx) x I IIxll $ Li, Vx ERn, Vt E [a, b) By letting fj approach 1, we conclude that which shows that 1I~(t'X)1I :5 Li, V x ERn, V t E [a,b] Since this inequality holds for every 1 :5 i :5 n, we conclude that the Jacobian matrix [af/ax] is globally bounded Set g(o-) = I(o-x) for 0 $ 0- $ 1. Since D is convex, o-x E D for 0 $ 0- $ (a) (b) Since af ao'x af g'(o') == -(o-x)- = -Co-x) x ax ao- ax f(x) == f(x) - f(o) = g(l) - g(o) = 10 1 g'(o') do- = 10 1 :~ (o-x) del x V(t, x) = 10 1 ~: (t,o-x) del x :510111~: (t,o-x)iii1xii do- $10 1 C40- do- IIxll2 :5 ~c411x1l2 we must have Cl $!C4. (c) Consider two points Xl and X2 such that axl + (1 - a)x2 =I- 0 for all 0 $ a $ 1; that is, the origin does

4 55 not lie on the line connecting Xl and X2. The Jacobian raw/ax] is defined for every X = axi + (1 - a)x2 and given by aw t 1 av t ax (,x) = 2JV(t,x) ax (,x) By the mean value theorem, there is a* E (0,1) such that, with z = a*xi + (1- a*)x2' Hence 1 C4 IW(t,X2) - W(t,xI)1 $ 2VCIlizil c411zllllx2 - xiii $ 2VCIllx2 - XIII Consider now the case when the origin lies on the line connecting Xl and X2i that is, = aoxi + (1 - ao)x2 for some ao E [.0,1]. We have IW(t,X2) - W(t,O)1 = IW(t,X2)i = JV(t,X2) $ ~IIX21I IW(t,xd- W(t,c~)1 = IW(t,XI)l = JV(t,Xl) $ ~IIXdl IW(t,X2) - W(t,xdl = IW(t,X2) - W(t, 0) + W(t,O) - W(t,xl)1 $ ~(IIX211 + \lxlll) Since the origin lies on the line connecting Xl and X2, we have IIX211 + IIXIII = IIX2-1 :::; JC4/2cl. Therefore, C4 IW(t,X2) - W(t,xI)1 $ 2VCIllx2 - XIII XIII. We also have Thus, the preceding inequality is satisfied for all Xl, X2 ED (a) X(t) = x(a) + it f(r,x(r» dr, V [a,t) C [to,t) Since f(t, x) is piecewise continuous in t and continuous in x, there exists a constant M > such that IIf(t,x(t»1I $ M for all t E [to,t). Therefore IIx(t) - x(a)1i = li/a t f(r,x(r» drll $ /at M dr =M(t - a) which shows that x(t) is uniformly continuous on [to, T). (b) since x(t) is uniformly continuous. Thus x(t) = x(to) + lim t f(r,x(r» dr = x(to) + rt f(r,x(r» dr t-+t lto lto x(t) = x(to) + r' f(r,x(r» dr, V t E [to, T] lto is a solution on [to,t]. Since W is closed, x(t) E W. (c) Apply the local existence and uniqueness theorem at (T,x(T».

5 56 CHAPTER Suppose there is no such t. Then, yet) E W for all t E [to, T). From the previous exercise we can extend the solution beyond T, which contradicts the claim that [to, T) is the maximal interval of existence Set yet) = Xl (t) - X2(t) and I-" = 1-"1 + 1-"2 liy(t)1i = IIxl (t) - X2(t)1I = IIxl(t) - h(t,xl(t» - X2(t) + h{t,x2(t» + h(t,xl(t» - h(t,x2(t»1i < 1-"1 + 1-"2 + Lllxl (t) - x2{t)1i = I-" + IlY(t)1I liy(t) II = liy(to) + it yes) dsll :5 I' + it liy(s)1i ds to to < I' + I-"(t - a) + 1t Llly(s)1I ds Application of Gronwall-Bellman inequality yields t liy{t)1i :5 "Y + I-"{t - a) + 10 h + I-"(S - a)]lej r t L d T ds After integrating the right-hand side by parts, we obtain liy(t}1i :5l'eL(t-o) + i [el(t-o) -1] 3.28 Let where to 2: to. Then x{t} =Xo + (t I(s,x(s» ds, ito yet) =Xo + t I{s,y(s» ds it' o to it, x(t) - yet) =. l I(s,x(s» ds + [J(s,y(s» -/(s,x(s»] ds ~ ~.. 0 We have III(s, x(s» II :5 M for all t E [to, tl]' and IIf(s, y) - f(s, x} II :5 Lllx - yll. Therefore By Gronwall-Bellman inequality Hence, over any compact interval of time, we have IIx(t) - y(t)1i :5 M(tri - to) + t Lllx(s) - y(s)1i ds it' o IIx(t) - y(t)ii":5 M(t~ - to}el(t-to).3.29 "x(t) - y(t)1i :5 K(t~ - to) Setting y = X -f}, we obtain x=i(t,x), x(to) =11 y = I(t, Y + 1]), y(to} = 0 Since I is continuously differentiable in x, the solution will be continuously differentiable in 11. Let Y (t...) - (}y(t,11) _ ox(t,1]) I -... (t...) I '1 '",. - 8'f] - ar, - - "''I '"'

6 57 From (3.4) where a 8t y,,(t,.,,) =A(t,.,,)y,,(t,.,,) + B(t,'I]), y,,(to,.,,) =0 af af A(t,.,,) = 8y (t, y(t,.,,) +,,) = ax (t, x(t,,,».3.30 B(t,.,,) = aa f (t, yet, '1]) + y) = aa f (t, x(t,.,,» = A(t, '1]) ". x Thus, x~(t,,,) satisfies the variational equation a atx,,(t,.,,) = A(t,,,)x,,(t, '1]), x,,(to,,,) = 1 Therefore Xo{t) + x,,(t)f(a,.,,) = it.{ ~~ (s,x(8,a'''))[xo(8) + x,,(8)f(a,'i])]} ds Differentiating with respect to t, we see that Xo (t) + x"(t)f (a, "1) satisfies the differential equation with initial condition Thus Put a af at [xo(t) + x,,(t)f(a,,,)] = ax (t, x(t, a, ",))[xo (t) + x,,(t)f(a,,,)] xo(a) + x,,(a)f(a,,,) = - f(a,,,) + f(a,.,,) =0 xo(t) + x,,(t)f(a, "1) == 0, V t E [a, tt] z(t) =x(a) +1t f(s,y(8» ds so that z(a) = x(a) and z(t) =5 yet) for a =5 t =5 b. z=f(t, y(t» =5 f(t, z(t» from the comparison lemma, we conclude that z(t) ~ x(t) => yet) $; x(t), Va$; t $; b