拉萨尔不变集理论和利亚季诺夫稳定性证明

1、Lecture 3CY4A2: Advaneed Nonlinear ControlLa Salle's Invariant Set TheoryAsymptotic stability of a control system is often an important property to be determined.Lyapunovas stability theorems studied above are often difficult to apply to establish this property, as it often happens that V (the d

2、erivative of the Lyapunov function candidate) is only negative semi-definite.In this kind of situation, it is still possible to draw conclusions on asymptotic stability, with the help of the invariant set theorems, which are attributed to La Salle.40Definition 14 (Invariant set). A set S is an invar

3、iant set for a dynamic system x = /(a?) if every trajectory 龙(t) which starts from a point in S remains in S for all time.For example, any equilibrium point is an invariant set.The domain of attract!on of an equilibrium point is also an invariant set.The invariant set theorems reflect the intuition

4、that the decrease of a Lyapunov function V has to gradually vanish (In other words V has to converge to zero) because V is lower bounded.Theorem 5 (Local Invariant set theorem). Consider an autonomous system of the form x = fx)f with f continuous and let Vx): 说71 说 be a scalar function with continuo

5、us first partial derivatives. Assume that for some I > 0, the set Q/ defined by V(a?) < I is bounded. V(a?) < 0 for all x in Q/.Let R be the set of all points within Q/ where p(爼)=0 and M be the largest invariant set in R. Then, every solution a?(t) originating in Q； tends to M as t oo Proo

6、f. See Slotine and Li (1991).In Theorem 5, the word largest means that M is the union of all invariant sets within R. The geometrical interpretation of the theorem is illustrated in Figure 1.Convergence to the largest invaria nt set MNotice that R is not necessarily connected, nor is the set M. Now

7、we are ready to state the following important theorem.Theorem 6 (La Salle's principle to establish asymptotic stability). Let V(a?):況"一说 be such that on 筠=a? w 況"：卩3) < / we have V(a?) < 0. Define R= xV(a?)=0. Then, if R contains no other trajectories other than x = 0, then the o

8、rigin 0 is asymptotically stable.Proof. It follows directly from Theorem 5.If Q/ in Theorem 6 extends to the whole space 況匕 then global asymptotic stability can be established.La Salle's theory is a useful extension to Lyapunov theory. In summary, If V(a?) is a negative semi-definite in a region

9、 where V(x) < Z, then a solution starting in the interior of Q/ remains there. If, in addition, no solutions (except the equilibrium 龙=0) remain in R (the subset of where V(a?) = 0), then all solutions starting in the interior of Q/ will converge to the equilibrium.Example 10. Consider the system

10、(36)x =+ 彷乡一4) 帀2 4龙彳觉2 + x2xl + 化$ 4)and consider the equilibrium point 龙 = 0.Given the function:V(a?) =+ a?2(37)its derivative V along any system trajectory is:V = 2(爲 + 2)(1 + 於 一 4)(38)Notice that V(a?) < 0 within a circle of radius 2. Hence, using Lyapunov's stability theorem (Theorem 1)

11、 we infer that the origin is locally asymptotically stable. For I = 2, the region Q2 defined by V(a?)=好 + < 4 is bounded. The set R is simply the origin 0, which is an invariant set. All the conditions of Theorem 6 are satisfied, hence any trajectory starting within the circle of radius 2 converg

12、e to the origin and this region is called the domain of attraction.To run from Matlab >> exlO.mExample 11. Consider the system:(39)x± = X2 觉(觉扌 + 2a?2 10) X2 = 33?2(1 + 2a?2 10)Notice that the set defined by += 10is invariant, since:-(+2-10) = -4(10+126)(+2-10) dt(40)which is zero on the

13、set. The motion on this set is described by either of these equations:(41)We can infer that the invariant set represents a limit cycle, along which the state vector moves clockwise.To check if the limit cycle is attractive, define the Lyapunov function candidate:V(ir)=(说 + 22 一 IO)? (42) This functi

14、on represents the distance to the limit cycle. For an arbitrary positive number lt the region Q/, which surrounds the limit cycle, is bounded. The time derivative of V along trajectories of the system is given by:V(x) = 8(屁0 + 3鸥)(迸 + 2於-10)2(43) Then V(a?) is strictly negative except ifa?扌 + 2a?2 =

15、 10 or= 0, in whichcase Vx) = 0 (so that Vx) is in fact negative semi-definite). The first equation re# + 2觉乡 = 10 simply defines the limit cycle. The second equation + 3觉? = 0 is satisfied only at the origin. Since both the limit cycle and the origin are invariant sets, the set M consists of their

16、union. Then, all system trajectories starting in Q/ converge either to the limit cycle, or to the origin.Consider for example, the case when I = 100 e, for some small g0.Notice that V(0) = 100 > 100 e, such that the origin does not belong to Q/.Then, while the express!on for V is the same as the

17、already given, now the set M is just the limit cycle.Hence, since e is arbitrary, applying once again the invariant set theorem shows that any trajectory starting from the region within the limit cycle, excluding the origin, actually converges to the limit cycle.This implies that the origin is unsta

18、ble!.-505Con vergence to an attractive limit cycle with an unstable equilibrium point at the originTo run from Matlab >> exllExample 12. Consider the autonomous pendulum with friction:(44)xi = a?2X2 = a Sin xi bx2where a，b0 and the candidate Lyapunov function (the energy function in this case)

19、:卩&) = a(l - COS(a?i) +(45)Let D = x E 5R2| tv < xi < 7r Then V(a?) is positive definite in D. The derivative Vx)V(a?) = axi sin(a?i) + 化2处2 乡 (46) is negative semidefinite in D. It is not negative definite since Vx) = 0 for a?2 0 irrespective the value of 龙iTherefore, using Lyapunov's

20、 stability theory it is only possible to conclude that 龙=0 is stable. However, we know that 彷 = 0 is asymptotically stable. In this case, we can use La Salle's Theorem 2 to prove asymptotic stability. Note that R = x e Dx2 = 0, so that V(a?) vanishes at x e R Let be a solution that belongs to R,

21、 so that %2(t) = 0. This implies by looking at the equation for x2 that sin(龙i(t) = 0, and hence that 力i(t) = 0. So that the only solution that can stay identically in R is the trivial solution 爼(t)= 0. Using Theorem 6 we can con elude that 龙=0 is asymptotically stable.5123 拉萨尔不变原理(LaSalle9s invaria

22、nt principle)一个没有外力的弹簧质量系统状态方程为x2 = -bx2 - icxl其中/；0鼻>0,总能最可以表示为/. V = x2x2 + kxixl = -hx < 0V为负半定的，但除x: = 0外可知U(x)<0。系统如果耍保持V(x) = 0，则系统轨线必须限定在x2=0o而实际上不可能，除非可=0。因为x2 (/) = 0 => (Q = 0 => x、(/)三 const于是x2(t) = 0=> X(f) = 0如果推导出V(x) < 0而且还知道对T- v(x) = 0除原点外，没有任何系统轨线 能永远保留在V(x) =

23、 o中。那么我们可以知道"0是渐近稳定的。这种思想就是 拉萨尔不变原理(LaSalle's invariance pimciple)。.为了叙述并证明拉萨尔不变原理，下而介绍儿个定义。正极限点：(血一极限点)P称为X。)正极限点,如果存在时间序列时,使得P 当"I00 X(t)的所有正极限点的集介称为X(t)的正极限集。 集合M称为x=f(x)的不变集，如果x(0)e M =>x(f)eR即：如果一个解任某-时刻属丁M,则在所有过公和未來时间均®T-MO 集介M称为正不变集(positively invariant set)如果。x(0) g M =

24、> x(t) e M Vr > 0称当X(/)趋向 TA/，是指若 mPwM ,和-TOO (/2T8 )使IK，x0)-P|->0(或：V>0,3T>0,使得对于 Vr>T,有 dist (x(t)M)vw,这甲.dist(P, M)表示点P到集合M的距离，即P到M任一点的绘小距离。 dist(P, M)= inf | P - M |)x«A/例：渐近稳定平衡点：对J：起始J：充分靠近平衡点的解是正极限点。例：稳定极限环对丁起始丁充分靠近极吸环的任何解是正极限集。IS 时，x(/)趋向丁极限环，注意到：x(/)并不趋向丁极限环上任何一个 特定的点

25、，也就是说并不惠味着极限存在。例：平衡点与极限环是不变集，因为任何解起始于它们，将对VteR保 持在其中。例：如果V(x)>0, V(x)<0, Qc=xeRnV(x)<c,对TxgQ.是正的 不变集。因为一个解x()起始于将保持在Q对于r>0o引理2.3.1：如果上=几丫)的解2)对丁0是有界的，则它的任极限集厶是:(1) 非空的(nonempty) (2)紧的(compact) (3)不变集(invariant set),而且(4) fTS 有证明：因为x(f)有界，则由Weierstrass定理可知，f ts时，x(/)有聚点， 因此止极限集芒非空。(当“T8,对

26、丁任意则X(/,A0，/0)是有界序列, 由Weierstrass定理，有界序列必有收敛子列)。对T Vy e L* ,存在时间序列too,假设，->qo时，x(/() y ,因为对于i是一致有界的，所以，极限y是有界的，因此芒是有界的。为证Z7是闭的，即证对于序列牙厂丿8时，有yel：.3对于每一个i，由T X G C ,存在序列”,当J->00时，阳T8,而且使 得双4) -> y(。现在构造一个特殊序列。对于阳，选取r2 > tl2，使得卜(八)_ y j V丄； 选取f3 > ZB使得卜(4)-儿|<?依此选取乙込 ，显然,当iTS时，：；T8, 并

27、且|心)-山以对于S。IV> 0,37V, N2正格数，使得|心)-则弓gN、h->il<|， Vi>N,第一个不等式由|.)-);|<|得出，笫二个不等式由极限x-y得出。丁是|卜(巧)一 y I < S, v，> N 二 maxNl, NJ即当/T8时，有x(y)Ty,从而y也是极限点，yeT,所以Z；是闭的。 乂因为厂是有界的，所以厂为紧的。为证ZT是不变集,设ye U , 0(f,y)记为方程f = /(x)在/= 0时经过y的解, 即炎0,刃=八只需证对J'/r67?* (p(t,y)gL+由T* yd 存在序列匕,当i ->o

28、o时>oo使得x(tf) -> y o记 x(ti)=如,x0), £为x(f)右汀=0时初值。由于解的唯一性，处/ +厶,兀)=如g、x$ =刃也)对于充分大的i,有/ + r,>0,由连续性知lim 0(/ + t.9x0) = lim 0(/、也)=0(/, y)l->xdTR于是(p(t> y) e r o最后证/T8时XL+O用反证法。假设不成立，则3fo>0,存在序列 f,”T00时f,->oo,使得dist(x(tt)X)> o因为x(f,)有界，存在一个收敛子 列当 i->oo,x(/：)->F。F 为极限点

29、，所以 x* g U ,同时 dist(xU)>sQ 盾，4证毕。定理231：设G为紧集，从。出发的方程x = f(x)的解对Tf>0均停留在Q 内，如果存在函数是连续可微的，A Q 'I* V(x)<0 , 乂设 E=,v| V(x) = 0,xe Q, M u E，为E中的最大不变集，则对于0兀w Q,当/ ->oo 时，有o证明：设x(f)为起始于G的方程x=/(x)的解。在G中V(x)<0,所以V(.v(r) 是/的非增函数。由于V在紧集G中连续，因此有下界,且V(x(r)在/T8时 有极限，设为g即limV(x(/) = «fT*注意到

30、G为有界闭的，根据引理，心在正的极限集厂，而且ZTuQ,对丁 Vp G Z；,存在序歹|J /,/»-> 00 时而且 x(tn) -> p o 由 T*V(x)连续，于是可 得V(p) = hniV(x(tn) = a即函数V(x)在集合Z?中任何点x±,有V(x) = a o又由于是正的不变集(引理),因此中包含系统的整条正半轨线,在轨 线的任何时间点上有V(x(t) = at因此V(x) = O于是有Z? u A/ u E u G由 T x(t)有界，/T8 时 x(/)->Z7 (引理)，因此 t ->00 时，x(/)tM。证毕说明：(1)

31、拉萨尔不变性定理放松了对于李雅普诺夫泄理中卩负定的耍求， 而且也没有耍求函数叫小是心定的。(2) 拉萨尔定理可以应用到系统有平衡点集而不仅仅是平衡点的情形。如 F面所耍提到的简单自适应控制例中所示。(3) 对丁紧的正不变集G,并不一定依赖y-V(x)的构造。但在很多应用中,U(x)的构造可以I动保证集介。的心在。比如取G< =xg/?h |V(x)<c假设0°是有界的，并且在Q. ±V(.v)<0,可以选取Q=Q( o它也可以作为 吸引域的一个估计。(4) 对TV(x)而言，如果V(x)是正定的，对丁充分小的c>0, 是有 界的。但如果V(x)不是正

32、定的,则以上结论未必成立。例如取V(x) = (x1-x2)2, 对丁无论多么小的c>0, G,都不是有界的。如果V(x)是径向无界的，即Mts时则无论V(x) iE定与否, 对地任意的c, Q.都是有界的。检验径向无界对丁 正定函数容易进行，因为只需检证沿各个坐标轴XTS 时，是否趋向s就足够了，但对于不是正定函数，这样验证不是充分的。例如对 于 V(x) = (x1-x2)2,沿x1 = 0,x2 = 0 坐标轴，|x|ts 时均有，V(x) ->oo ,但当沿 X, = X2 f |x| -> 8 时 v(X)三 0 o如果我们考察原点的渐近稳定性，我们需耍指出E中的最

33、大不变集是原点, 即需指出在E中除原点外，没有方程的解可以永远停留在E中。假设"(X)是正 定的，可以紂到如卜定理231的推论。这两个推论被称为Baibasluii-Kiasovskii 定理，在拉庐尔不变性定理给出之前就己经由Barbashin和Kiasovskii给出了证 明。推论231设x = 0为x=f(x)的平衡点，是连续可微的正定函 数,其中。为x = 0邻域，在G中，V(x)<0,令M =xeQ|V(x) = 0,假设没 有非零解包含在M中，则原点是渐近稳定的。推论232推论2.3.1屮V(x)如果还具有径向无界的性质，则x = 0是全丿渐近稳定的。利用拉萨尔不变

34、集理论也可以得出关丁零平衡点不稳定的一个结论，如推论233,该结果被称为克拉索夫斯某不稳定性定理。推论233若存在可微两数V(x):ClR, V(0) = 0,在原点的任意小的邻域内，存在兀，使”(兀)>0, 乂V>0但集介M=x|V = O,xwQ中除x二0夕卜，不禽系统非平凡解的幣条正半轨线，则 系统的平凡解不稳定。例：考察系统呂=兀2勺=一8(不)一心)这里g()与h(.)是局部李普希斯的,并满足g(0) = 0, yg(y) > 0, Vy 工 0, y w (-a卫)/i(0) = 0、)7/( y) > 0, Vy m 0, y w (-a,a)原点是孤立平

35、衡点，V换数可取为令Q = xeR2-a<xi <aj = l,2. V(x)在Q中正定，V(x)沿系统轨线的导数为V(x) = gMx2 + x2-g(xx)-/!(x2) = -x2h(x2) < 0"(X)负半定。由于V(x) = 0=> x2h(x2) = 0 => x, = 0所以E = xeDx2=0f假设x(t)E的系统轨线，有x2(O = 0=> 大(/) = 0=> 兀(/)三 c.c g (-a, a)而且x2(t) = 0=>x2(r)= 0=> g(c) = 0 => c = 0I大1此，只有x =

36、0可以停留在E中，E中的最大不变集为M=x = O,所以 x = 0是渐近稳定的。例：考察一阶系统y = ay + u有如卜的H适应控制规律Il = -kyk=/y />0令x1 = y,x2=,则闭环系统可以表示为若=-(x2 - 0)兀A =卩;舌=0是系统的平衡点集。下面说明系统轨线在/>8时渐近丁平衡点集人=0,这样在IH适应控制作用下)0)渐近丁()，选取v函数V(X)= *X；-b)其中>g, V(x)沿系统轨线的导数为V(X)=心勺 +y(x：- b)X2=-彳(D - a) + X；(兀 - b)= -x(b-a)<0所以V(x)<0, 乂U(x)

37、径向无界，集合Qt =xeRnV(.x)<c是紧的正不变集。 取Q=Q(,则f = xgQ< |x1 = 0o因为E中任一点均为平衡点，E为不变集， 这样定理2.3.1中由定理2.3.1可知任何起始于雪的轨线，当/too时均 渐近TE,即：r ->oo 时兀(/)->0乂因为#(兀)是径向无界的，该结论乂是全局的，这是因为对丁任何x(o),常数C可以取得非常大，使得x(0)6 Q. o注：上例中李雅普诺夫函数依赖J：常数b,并要求这是因为在H适应 控制中常数d未知，但我们可以根据d的界來选取債例：考察系统若=耳+為(0一彳一丘)X, = _兀 + .V, （0 _ 兀

38、_ X；）显然原点为平衡点，圆彳+疋=0是一个不变集，因为4(x： + 疋 一 0：) = 2(x; + x;)(/32-x;-x;)dt在圆彳+卅=0上恒为零。这就意味着起始丁圆上的任何轨线，对Tt>tQ永 保留在圆上。在这个不变集上，轨线町以由以卜方程描述这样彳+工=0，实际上是一个顺时针运动的极限环。下而根据拉萨尔定理考察该极限环的稳定性，选取V(x) = l(x； + x：-/72)24且U(x)沿系统轨线的导数为V (x) = -(x； +x； -pzy <0而且，V(x) = 0当且仅当以下条件满足(Cl)x； + X； = 0(b)f + 丘 _ 0? = 0即

39、63; = x|V(x) = 0,xg Q是原点与圆f + X； = 0，的并集。 下而应用拉萨尔定理(1)对丁o丄伊，定义G如下：4Q = xe/?n |V(x)<cG是闭的有界的，因此是紧的，乂由丁在Q中V(x)<0,则任何起始丁-G中 的轨线将保留在G中，因此G是不变集。(2) 寻找£： = x|V(x) = O,xgQ,这里E = (O,O)Uxw/?'|f + W=0' 即£为原点与极限环的并集。(3) 寻找M,即E中的最大不变集。由丁 £本身是不变集，显然根据拉萨尔定理，任何起始丁S的轨线有这电任何起始TG的轨线趋向丁原点或

40、极限环。观察V(x)表达式可以看出，V(x)实际上是任何一点到极限坏距离的变帚:。V(x) = -(xf+x；-2)2V(.v) = O当彳+疋=0当x=09#选取Ovcv丄04,则集合Q = xeR2V(x)<c包括了极限环，但不包括原对于选取e<c<-p £任意小构造的G，再次应用拉萨尔定理可以看到, 任何起始丁的轨线收敛到极限环，因此极限环是稳定的。而乂由丁*任意小,可以看出原点是不稳定的。例讨论方程d2x dx f# c f 小+a += 0 (a > 0,/?> 0)d厂dt的平凡解的稳定性.先化为等价方程组y = -bx-ay-x2取函数叫3

41、)=扣+扑+扑#令a = G0>O,作如下有界闭区域10V <ap2:=x>-/3 ； W2 = y-ax> -ap.对任意x0 e Q ,当 m。,今证x（r,/0,x0）恒停留在G中。11若不然，设解x（Ho，x。）离开G，必与曲线ABCD,或H线4E, DE相交而 穿出。因为卩=-ay2 < 0,当）v0 ,故x（/,/0,x0）不能由ABCD由里向外穿出。在DE段，因为y >0,故dtdxdt轨线的走向是从左到右，即从外到里。在AE上，因为兀<0,且-x = 0 +丄<0,当 ”0 a故当Ov0vb时，有b> p> -x, b

42、 + x>0,当工0从而Jivj = _x（b + x）>0.当”0dt因此 轨线走向也是由外向电的。于是，解“儿人）一氏停留在。中。乂 V = -ay2 ,故E = y = 0 = x轴，M = 0,0当0<“ <b时，j（r,r0,.v0）-> 0。#LaSalle's Invariance PrincipleLecture 23Math 63410/22/99Linearization versus Lyapunov FunctionsIn the previous two lectures, we have talked about two dif

43、ferent tools thatcan be used to prove tlnit an equilibrium point of ail ciutoiioinous systemis asymptotically stable: linearization and Lyapunov*s direct method One might ask which of these methods is better. Certainly, linearization seems easier to apply because of its straightforward liatiue: Comp

44、ute the eigenvalues of Df(xo). The direct method requires you to find an appropriate Lyajiuiiov function, whicli doesift seem so straightforward But, in fact, anytime linearization works, a simple Lyapunov function works, as well.To be more precise, suppose = 0 and all the eigenvalues of A := D/(0)

45、have negative real part. Pick ail inner product, and induced norm | -1 such that, for some c > 0，for all rr C Rn. Pick r > 0 small enough that |/(x) Ax| < (c/2)|ar| whenever |x| < 几 letV = xe | x < r,and define V : R x P R by the formula x) = |x|2. Since | | is a uoriih V is positive

46、definite. Also卩仏 w) = 2(些血+ x, f(x) - Ax)< 2(-ch|2 + h|/(x) - Ar|) < -c|x|2,so V is negative definite.On the other luuid there are very simple examples to illustrate that the direct method works in some cases where linearization doesn't. For example, consider x = x3 on R. Tlie equilibrium

47、point at the origin is not hyperbolic, so linearization fails to dctcrniinc stability, but it is casv to check that x2 is positive definite and has a negative definite orbital derivative, thus ensuring the asymptotic stability of 0.A More Complicated ExampleThe previous example is so simple that it

48、might make one question whether the direct method is of any use on problems where stability cannot be determined by linearization or by inspection. Thus, let's consider something more complicated Consider the pkmar system(x = -y Yy = x5.The origin is a non hyperbolic equilibrium point, with 0 be

49、ing the only eigenvalue, so the principle of linearized stability is of no use. A sketch of the phase portrait indicates that orbits circle the origin in the counterclockwise direction, but it is not obvious whether tliey spiral in, spiral out. or move on closed curves.The simplest potential Lyapuno

50、v function that often turns out to be useful is the square of the standard Euclidean norm, vvliich in this case isV := x2 y2. The orbital derivative isV = 2xx + 2yy = 2x5y 2xy 2x4.(2)For some points (x, y) near the origin (e.g.、(d.d) V < 0, while for other points near the origin (e.g.、(d, d) V &g

51、t; 0, so this function doesn't seem to be of much use.Sometimes when the square of the standard Euclidean norm doesift work some other homogeneous quadratic function does. Suppose we tryV := x2 + axy + 0y2, with a and 0 to be deterinincd. ThenV = (2x + ay)x + (ax + 23y)y = -(2t + oy)(y + x3) + (

52、ax + 2/3y 疋 =2x4 + axG 2ry ax:iy + 2(3x5y a/Setting (x, y) = (d, -<52) for J positive and small, vve see that V is not going to be negative semidefinite, no matter what we pick a and (i to he.If these quetdratic functions doift work, maybe something customized for the particular equation might. N

53、ote that the right-hand side of the first equation in (2) sort of suggests that x3 and y should be treated as quantities of the same order of magnitude. Let's try V := x6 + ay2 for some a > 0 to be determined Clearly, V is positive definite, andV = 6x5i + 2ayy = (2a 6)护 y 6x8.If a 齐 3. then V

54、 is of opposite signs for (些 y) = (d. 6) and for (t, y) = (d. 6) when 6 is small. Hence, we should set a = 3, yielding V = -6.r8 < 0. Thus U is positive definite andis negative seniidefinite, implying that the origin is Lyapunov stable.Is the origin asymptotically stable? Perhaps we can make a mi

55、nor modification to the preceding formula for V so as to make V strictly negative in a deleted neighborhood of the origin without destroying the positive definiteness of U If we added a small quantity wliosc orbital derivative was strictly negative when t = 0 and y is small and positive, this might

56、work. Experimentation suggests that a positive multiple of xy3 might work, since this quantity changes from positive to negative as we cross the ?/-axis in the counterclockwise direction. Also, it is at least of liigher order than 3y2 near the origin, so it has the potential of preserving the positi

57、ve definiteness of V.In fact, we claim that V := t6 + xys 3y2 is positive definite with negative definite orbital derivative near 0. A handy inequality, sometimes called Young's inequality, that can be used in verifying this claim (and in other circumstances, as well) is given in tlie following

58、lemma.(3)Lemma (Young s Inequality) If a. b > 0, then for every pair of numbers p.q e (1< oc) satisfying=1.Proof. Assume that (4) holds. Clearly (3) holds 让 b = 0, so assume that b > 0, and fix it Define g : 0. oc) by the formula(、xp bq .q(x) := + 也p qNote that g is continuous, and gf(x) = 一 b for every x E (0, oo). Since ()= -6 < 0? linixioc = oo, and gf is increasing on (0. oo), we know that g has a unique minimizer at r0 =Thus, for everyx