拉薩爾不變集理論和利亞季諾夫穩(wěn)定性證明

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 龍(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): 說71 說 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"一說 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 龍=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 =+ 彷鄉(xiāng)一4) 帀2 4龍彳覺2 + x2xl + 化$ 4)and consider the equilibrium point 龍 = 0.Given the function:V(a?) =+ a?2(37)its derivative V along any system trajectory is:V = 2(爲(wèi) + 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 覺(覺扌 + 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)=(說 + 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鷗)(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覺鄉(xiāng) = 10 simply defines the limit cycle. The second equation + 3覺? = 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處2 鄉(xiāng) (46) is negative semidefinite in D. It is not negative definite since Vx) = 0 for a?2 0 irrespective the value of 龍iTherefore, using Lyapunov's

20、 stability theory it is only possible to conclude that 龍=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(龍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 龍=0 is asymptotically stable.5123 拉薩爾不變原理(LaSalle9s invaria

22、nt principle)一個沒有外力的彈簧質(zhì)量系統(tǒng)狀態(tài)方程為x2 = -bx2 - icxl其中/；0鼻>0,總能最可以表示為/. V = x2x2 + kxixl = -hx < 0V為負(fù)半定的，但除x: = 0外可知U(x)<0。系統(tǒng)如果耍保持V(x) = 0，則系統(tǒng)軌線必須限定在x2=0o而實際上不可能，除非可=0。因為x2 (/) = 0 => (Q = 0 => x、(/)三 const于是x2(t) = 0=> X(f) = 0如果推導(dǎo)出V(x) < 0而且還知道對T- v(x) = 0除原點(diǎn)外，沒有任何系統(tǒng)軌線 能永遠(yuǎn)保留在V(x) =

23、 o中。那么我們可以知道"0是漸近穩(wěn)定的。這種思想就是 拉薩爾不變原理(LaSalle's invariance pimciple)。.為了敘述并證明拉薩爾不變原理，下而介紹兒個定義。正極限點(diǎn)：(血一極限點(diǎn))P稱為X。)正極限點(diǎn),如果存在時間序列時,使得P 當(dāng)"I00 X(t)的所有正極限點(diǎn)的集介稱為X(t)的正極限集。 集合M稱為x=f(x)的不變集，如果x(0)e M =>x(f)eR即：如果一個解任某-時刻屬丁M,則在所有過公和未來時間均®T-MO 集介M稱為正不變集(positively invariant set)如果。x(0) g M =

24、> x(t) e M Vr > 0稱當(dāng)X(/)趨向 TA/，是指若 mPwM ,和-TOO (/2T8 )使IK，x0)-P|->0(或：V>0,3T>0,使得對于 Vr>T,有 dist (x(t)M)vw,這甲.dist(P, M)表示點(diǎn)P到集合M的距離，即P到M任一點(diǎn)的繪小距離。 dist(P, M)= inf | P - M |)x«A/例：漸近穩(wěn)定平衡點(diǎn)：對J：起始J：充分靠近平衡點(diǎn)的解是正極限點(diǎn)。例：穩(wěn)定極限環(huán)對丁起始丁充分靠近極吸環(huán)的任何解是正極限集。IS 時，x(/)趨向丁極限環(huán)，注意到：x(/)并不趨向丁極限環(huán)上任何一個 特定的點(diǎn)

25、，也就是說并不惠味著極限存在。例：平衡點(diǎn)與極限環(huán)是不變集，因為任何解起始于它們，將對VteR保 持在其中。例：如果V(x)>0, V(x)<0, Qc=xeRnV(x)<c,對TxgQ.是正的 不變集。因為一個解x()起始于將保持在Q對于r>0o引理2.3.1：如果上=幾丫)的解2)對丁0是有界的，則它的任極限集厶是:(1) 非空的(nonempty) (2)緊的(compact) (3)不變集(invariant set),而且(4) fTS 有證明：因為x(f)有界，則由Weierstrass定理可知，f ts時，x(/)有聚點(diǎn)， 因此止極限集芒非空。(當(dāng)“T8,對

26、丁任意則X(/,A0，/0)是有界序列, 由Weierstrass定理，有界序列必有收斂子列)。對T Vy e L* ,存在時間序列too,假設(shè)，->qo時，x(/() y ,因為對于i是一致有界的，所以，極限y是有界的，因此芒是有界的。為證Z7是閉的，即證對于序列牙廠丿8時，有yel：.3對于每一個i，由T X G C ,存在序列”,當(dāng)J->00時，陽T8,而且使 得雙4) -> y(?，F(xiàn)在構(gòu)造一個特殊序列。對于陽，選取r2 > tl2，使得卜(八)_ y j V丄； 選取f3 > ZB使得卜(4)-兒|<?依此選取乙込 ，顯然,當(dāng)iTS時，：；T8, 并

27、且|心)-山以對于S。IV> 0,37V, N2正格數(shù)，使得|心)-則弓gN、h->il<|， Vi>N,第一個不等式由|.)-);|<|得出，笫二個不等式由極限x-y得出。丁是|卜(巧)一 y I < S, v，> N 二 maxNl, NJ即當(dāng)/T8時，有x(y)Ty,從而y也是極限點(diǎn)，yeT,所以Z；是閉的。 乂因為廠是有界的，所以廠為緊的。為證ZT是不變集,設(shè)ye U , 0(f,y)記為方程f = /(x)在/= 0時經(jīng)過y的解, 即炎0,刃=八只需證對J'/r67?* (p(t,y)gL+由T* yd 存在序列匕,當(dāng)i ->o

28、o時>oo使得x(tf) -> y o記 x(ti)=如,x0), £為x(f)右汀=0時初值。由于解的唯一性，處/ +厶,兀)=如g、x$ =刃也)對于充分大的i,有/ + r,>0,由連續(xù)性知lim 0(/ + t.9x0) = lim 0(/、也)=0(/, y)l->xdTR于是(p(t> y) e r o最后證/T8時XL+O用反證法。假設(shè)不成立，則3fo>0,存在序列 f,”T00時f,->oo,使得dist(x(tt)X)> o因為x(f,)有界，存在一個收斂子 列當(dāng) i->oo,x(/：)->F。F 為極限點(diǎn)

29、，所以 x* g U ,同時 dist(xU)>sQ 盾，4證畢。定理231：設(shè)G為緊集，從。出發(fā)的方程x = f(x)的解對Tf>0均停留在Q 內(nèi)，如果存在函數(shù)是連續(xù)可微的，A Q 'I* V(x)<0 , 乂設(shè) E=,v| V(x) = 0,xe Q, M u E，為E中的最大不變集，則對于0兀w Q,當(dāng)/ ->oo 時，有o證明：設(shè)x(f)為起始于G的方程x=/(x)的解。在G中V(x)<0,所以V(.v(r) 是/的非增函數(shù)。由于V在緊集G中連續(xù)，因此有下界,且V(x(r)在/T8時 有極限，設(shè)為g即limV(x(/) = «fT*注意到

30、G為有界閉的，根據(jù)引理，心在正的極限集廠，而且ZTuQ,對丁 Vp G Z；,存在序歹|J /,/»-> 00 時而且 x(tn) -> p o 由 T*V(x)連續(xù)，于是可 得V(p) = hniV(x(tn) = a即函數(shù)V(x)在集合Z?中任何點(diǎn)x±,有V(x) = a o又由于是正的不變集(引理),因此中包含系統(tǒng)的整條正半軌線,在軌 線的任何時間點(diǎn)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 時 x(/)->Z7 (引理)，因此 t ->00 時，x(/)tM。證畢說明：(1)

31、拉薩爾不變性定理放松了對于李雅普諾夫泄理中卩負(fù)定的耍求， 而且也沒有耍求函數(shù)叫小是心定的。(2) 拉薩爾定理可以應(yīng)用到系統(tǒng)有平衡點(diǎn)集而不僅僅是平衡點(diǎn)的情形。如 F面所耍提到的簡單自適應(yīng)控制例中所示。(3) 對丁緊的正不變集G,并不一定依賴y-V(x)的構(gòu)造。但在很多應(yīng)用中,U(x)的構(gòu)造可以I動保證集介。的心在。比如取G< =xg/?h |V(x)<c假設(shè)0°是有界的，并且在Q. ±V(.v)<0,可以選取Q=Q( o它也可以作為 吸引域的一個估計。(4) 對TV(x)而言，如果V(x)是正定的，對丁充分小的c>0, 是有 界的。但如果V(x)不是正

32、定的,則以上結(jié)論未必成立。例如取V(x) = (x1-x2)2, 對丁無論多么小的c>0, G,都不是有界的。如果V(x)是徑向無界的，即Mts時則無論V(x) iE定與否, 對地任意的c, Q.都是有界的。檢驗徑向無界對丁 正定函數(shù)容易進(jìn)行，因為只需檢證沿各個坐標(biāo)軸XTS 時，是否趨向s就足夠了，但對于不是正定函數(shù)，這樣驗證不是充分的。例如對 于 V(x) = (x1-x2)2,沿x1 = 0,x2 = 0 坐標(biāo)軸，|x|ts 時均有，V(x) ->oo ,但當(dāng)沿 X, = X2 f |x| -> 8 時 v(X)三 0 o如果我們考察原點(diǎn)的漸近穩(wěn)定性，我們需耍指出E中的最

33、大不變集是原點(diǎn), 即需指出在E中除原點(diǎn)外，沒有方程的解可以永遠(yuǎn)停留在E中。假設(shè)"(X)是正 定的，可以紂到如卜定理231的推論。這兩個推論被稱為Baibasluii-Kiasovskii 定理，在拉廬爾不變性定理給出之前就己經(jīng)由Barbashin和Kiasovskii給出了證 明。推論231設(shè)x = 0為x=f(x)的平衡點(diǎn)，是連續(xù)可微的正定函 數(shù),其中。為x = 0鄰域，在G中，V(x)<0,令M =xeQ|V(x) = 0,假設(shè)沒 有非零解包含在M中，則原點(diǎn)是漸近穩(wěn)定的。推論232推論2.3.1屮V(x)如果還具有徑向無界的性質(zhì)，則x = 0是全丿漸近穩(wěn)定的。利用拉薩爾不變

34、集理論也可以得出關(guān)丁零平衡點(diǎn)不穩(wěn)定的一個結(jié)論，如推論233,該結(jié)果被稱為克拉索夫斯某不穩(wěn)定性定理。推論233若存在可微兩數(shù)V(x):ClR, V(0) = 0,在原點(diǎn)的任意小的鄰域內(nèi)，存在兀，使”(兀)>0, 乂V>0但集介M=x|V = O,xwQ中除x二0夕卜，不禽系統(tǒng)非平凡解的幣條正半軌線，則 系統(tǒng)的平凡解不穩(wěn)定。例：考察系統(tǒng)呂=兀2勺=一8(不)一心)這里g()與h(.)是局部李普希斯的,并滿足g(0) = 0, yg(y) > 0, Vy 工 0, y w (-a衛(wèi))/i(0) = 0、)7/( y) > 0, Vy m 0, y w (-a,a)原點(diǎn)是孤立平

35、衡點(diǎn)，V換數(shù)可取為令Q = xeR2-a<xi <aj = l,2. V(x)在Q中正定，V(x)沿系統(tǒng)軌線的導(dǎo)數(shù)為V(x) = gMx2 + x2-g(xx)-/!(x2) = -x2h(x2) < 0"(X)負(fù)半定。由于V(x) = 0=> x2h(x2) = 0 => x, = 0所以E = xeDx2=0f假設(shè)x(t)E的系統(tǒng)軌線，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中的最大不變集為M=x = O,所以 x = 0是漸近穩(wěn)定的。例：考察一階系統(tǒng)y = ay + u有如卜的H適應(yīng)控制規(guī)律Il = -kyk=/y />0令x1 = y,x2=,則閉環(huán)系統(tǒng)可以表示為若=-(x2 - 0)兀A =卩;舌=0是系統(tǒng)的平衡點(diǎn)集。下面說明系統(tǒng)軌線在/>8時漸近丁平衡點(diǎn)集人=0,這樣在IH適應(yīng)控制作用下)0)漸近丁()，選取v函數(shù)V(X)= *X；-b)其中>g, V(x)沿系統(tǒng)軌線的導(dǎo)數(shù)為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、徑向無界，集合Qt =xeRnV(.x)<c是緊的正不變集。 取Q=Q(,則f = xgQ< |x1 = 0o因為E中任一點(diǎn)均為平衡點(diǎn)，E為不變集， 這樣定理2.3.1中由定理2.3.1可知任何起始于雪的軌線，當(dāng)/too時均 漸近TE,即：r ->oo 時兀(/)->0乂因為#(兀)是徑向無界的，該結(jié)論乂是全局的，這是因為對丁任何x(o),常數(shù)C可以取得非常大，使得x(0)6 Q. o注：上例中李雅普諾夫函數(shù)依賴J：常數(shù)b,并要求這是因為在H適應(yīng) 控制中常數(shù)d未知，但我們可以根據(jù)d的界來選取債例：考察系統(tǒng)若=耳+為(0一彳一丘)X, = _兀 + .V, （0 _ 兀

38、_ X；）顯然原點(diǎn)為平衡點(diǎn)，圓彳+疋=0是一個不變集，因為4(x： + 疋 一 0：) = 2(x; + x;)(/32-x;-x;)dt在圓彳+卅=0上恒為零。這就意味著起始丁圓上的任何軌線，對Tt>tQ永 保留在圓上。在這個不變集上，軌線町以由以卜方程描述這樣彳+工=0，實際上是一個順時針運(yùn)動的極限環(huán)。下而根據(jù)拉薩爾定理考察該極限環(huán)的穩(wěn)定性，選取V(x) = l(x； + x：-/72)24且U(x)沿系統(tǒng)軌線的導(dǎo)數(shù)為V (x) = -(x； +x； -pzy <0而且，V(x) = 0當(dāng)且僅當(dāng)以下條件滿足(Cl)x； + X； = 0(b)f + 丘 _ 0? = 0即

39、63; = x|V(x) = 0,xg Q是原點(diǎn)與圓f + X； = 0，的并集。 下而應(yīng)用拉薩爾定理(1)對丁o丄伊，定義G如下：4Q = xe/?n |V(x)<cG是閉的有界的，因此是緊的，乂由丁在Q中V(x)<0,則任何起始丁-G中 的軌線將保留在G中，因此G是不變集。(2) 尋找£： = x|V(x) = O,xgQ,這里E = (O,O)Uxw/?'|f + W=0' 即£為原點(diǎn)與極限環(huán)的并集。(3) 尋找M,即E中的最大不變集。由丁 £本身是不變集，顯然根據(jù)拉薩爾定理，任何起始丁S的軌線有這電任何起始TG的軌線趨向丁原點(diǎn)或

40、極限環(huán)。觀察V(x)表達(dá)式可以看出，V(x)實際上是任何一點(diǎn)到極限壞距離的變帚:。V(x) = -(xf+x；-2)2V(.v) = O當(dāng)彳+疋=0當(dāng)x=09#選取Ovcv丄04,則集合Q = xeR2V(x)<c包括了極限環(huán)，但不包括原對于選取e<c<-p £任意小構(gòu)造的G，再次應(yīng)用拉薩爾定理可以看到, 任何起始丁的軌線收斂到極限環(huán)，因此極限環(huán)是穩(wěn)定的。而乂由丁*任意小,可以看出原點(diǎn)是不穩(wěn)定的。例討論方程d2x dx f# c f 小+a += 0 (a > 0,/?> 0)d廠dt的平凡解的穩(wěn)定性.先化為等價方程組y = -bx-ay-x2取函數(shù)叫3

41、)=扣+撲+撲#令a = G0>O,作如下有界閉區(qū)域10V <ap2:=x>-/3 ； W2 = y-ax> -ap.對任意x0 e Q ,當(dāng) m。,今證x（r,/0,x0）恒停留在G中。11若不然，設(shè)解x（Ho，x。）離開G，必與曲線ABCD,或H線4E, DE相交而 穿出。因為卩=-ay2 < 0,當(dāng)）v0 ,故x（/,/0,x0）不能由ABCD由里向外穿出。在DE段，因為y >0,故dtdxdt軌線的走向是從左到右，即從外到里。在AE上，因為兀<0,且-x = 0 +丄<0,當(dāng) ”0 a故當(dāng)Ov0vb時，有b> p> -x, b

42、 + x>0,當(dāng)工0從而Jivj = _x（b + x）>0.當(dāng)”0dt因此 軌線走向也是由外向電的。于是，解“兒人）一氏停留在。中。乂 V = -ay2 ,故E = y = 0 = x軸，M = 0,0當(dāng)0<“ <b時，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| < 幾 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)護(hù) y 6x8.If a 齊 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 讓 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