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MIT 18 034 - Lecture Notes

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18.034, Honors Differential Equations Page 1 of 6 Prof. Jason Starr 18.034, Honors Differential Equations Prof. Jason Starr Notes from Lecture 36, 5/5/04 I. The winding number Let R ⊂ IR 2 be an open region, let ( )yxFx ,' = ( )yxGy ,' = be an autonomous differential system on R , and let C⊂ R be an oriented, simpler closed curve in R . In other words, C is the image of the circle under a 1-to-1 map whose derivative vector is always nonzero, say [ ]( )()01,1,0: hhRh =→ . If C contains no equilibrium point of the system, the following function is well-defined and continuous: :f C → δ1⊂ IR 2 , ()() ()( )()⎥⎥⎦⎤⎢⎢⎣⎡+=qGqFqGqFqf221. The composition [ ]1,0:foh →δ1 is (essentially) a cts. map from the circle to the circle. To such a map there is an associated integer n, the degree of the map. This integer counts the number of times ( )tfoh rotates counterclockwise around the circle as t rotates once counterclockwise around the circle. If h , F and G are all continuously differentiable function, ⎥⎦⎤⎢⎣⎡=21ggg , and then the degree is simply () () () ()[]∫−102'1'2121dttgtgtgtgπ This integer turns out to be independent of ( )th (although it does depend on the orientation of C ). it is called the winding number of ( )GF, about C . Let Rp ∈ be an equilibrium point. It is isolated if there exists 0>ε such that p is the only equilibrium point in the ε- ball about p . For any ε<< q0 , consider the circle qC of radius q centered at p. The winding number of ( )6,F about qC is independent of q and is called the index of ( )6,F at p (or sometimes the Poincare index). Examples: (1) Let λ, 0>µ and let yGxFµλ== , . Then ( )0,0=p is an isolated equilibrium point. Consider ()ttqtqthq<⎥⎥⎦⎤⎢⎢⎣⎡= 0,)(sin.)(cos. Then ()() ()( )()⎥⎥⎦⎤⎢⎢⎣⎡+=tttttgsincossincos12222µλµλ.18.034, Honors Differential Equations Page 2 of 6 Prof. Jason Starr And () () () ()tgtgtgtg'2'1'21− = () ()tt2222sincosµλλµ+. This is closely related to the Poisson kernel. It is nontrivial, but the integral () ()dttt∫+πµλλµ202222sincos can be computed by elementary methods, and it equals π2 (consider the case that λ=µ). So the index is +1. (2) λ, 0<µ. This is the same as above when λ→-λ, µ→-µ. Notice the integral does not change. So the index is +1. (3) λ<0, µ>0. now the integral is -( )( ) () ()dttt∫+−−πµλµλ202222sincos. This is -1 times the integral from (1). So the index is -1. Theorem: Let C be a simple closed curve that contains no equ. pts. in R oriented so the interior is always on the left. If the interior is contained in R , and if the interior contains only finitely many equilibrium points, npp ,.....,1, then the winding number about C is index ()++ ....1p index( )np . (and 0 if there are no eq. pts). Proof: This is proved, for instance in Theorem 3, § 11.9 on p.442 of Wilfred Kaplan, Ordinary Differential Equations, Addison-Wesley, 1958. Corollary:If R is simply-connected, then every cycle C contained in R contains an equilibrium point in its interior. Rmk: A region R in IR2 is simply-connected if for every simple closed curve C in R , the interior of C is contained in R . A cycle is a periodic orbit (that is necessarily a simple closed curve). Pf: By construction, ()6,F is parallel to the tangent vector of C. Therefore the winding number is +1. So, by the theorem, there is an equilibrium point in the interior of C . II. Lyapunov functions Let R⊂IRn be an open region. Let x ’= F( )x be an autonomous system on R . Let Rp ∈ be a point. Definition: A function RV : → IR is positive definite (resp. negative definite) if (1) ()0≥qV (resp. ()0≤qV ) for all Rq ∈ (2) ()0=qV iff pq = . Let p be an equilibrium point.18.034, Honors Differential Equations Page 3 of 6 Prof. Jason Starr Definition: A strong Lyapunov function is a continuously differentiable function RV : →IR such that (1) V is positive definite (2) the function ()()xFdxdVViniix∑==1': is negative definite. Remark : It is often the case that there is no strong Lyapunov function on R , yet there is an open subregion R ’⊂R containing p and a strong Lyapunov function on R ’. In this case, simply replace R by R ’ in what follows. Hypothesis: Suppose a strong Lyapunov function exists. There is a minor issue that your book does not deal with: long-time existence of solution curves. Let K ⊂ IRn be a bounded closed region whose interior contains p and such that K ⊂ R . Define 0r = minimum of V on the bounded closed set K∂(a continuous function on a bounded closed subset of IRn always attains a minimum). Because ∈p interior of K , 00>r . Define R ’ to be R ’= []( )(){ }001/,0 rVKqrKnVq<∈=−. Observe this is an open region in R that contains p and is contained in the interior of K . Theorem:(1) For every ∈0xR’, the solution curve ( )tx is defined for all 0>t . (2) Moreover, ()ptxt=→∝lim . Therefore p is an attractor and R ’ is in the basin of attraction of p . Proof: For any ∈0xR, if ()tx is defined on the interval [)1,0 t , consider ()()txV defined on [)1,0 t . By the Chain Rule, ( )( )txV is differentiable and ()() ()() ()tdtdxtxxVtxVdtdinii.1∑=∂∂=. By hypothesis, () ()()txFtxii='. Thus ()() ()()txVtxVdtd'= . By hypothesis, this is nonpositive. Therefore ( )( )txV is a non-increasing function. In particular, if ∈0xR’, then ()tx is in R ’ for all [)1,0 tt ∈ . (1) Let ∈0xR’. By way of contradiction, suppose that ( )tx is defined only on [)1,0 t where 1t is finite. By the theorem on maximally extended solutions, 1limtt →()tx exists and is in K∂ . Therefore V (1limtt →( )tx ) ≥ 0r . Since V is continuous V (1limtt →()tx) =1limtt →V (( )tx). For all 0≥t , V (( )tx) ≤( )00rxV < . So 1limtt →V (()tx ) ≤ ()00rxV < . This contradiction proves ( )tx is defined for all t>0.18.034, Honors Differential Equations Page 4 of 6 Prof. Jason Starr (2) Let ε>0 and let ()pBε denote the open ball of radius ε centered at p . The set difference ()()pKnBKε\ is closed and bounded. Therefore V attains a mimimum value 1r on this set. Since p is not in this set 01>r . Also, [)()∞−,11rKnV is a closed set contained in K . So it


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MIT 18 034 - Lecture Notes

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