18.034, Honors Differential Equations Lecture 4 Prof. Jason Starr Page 1 of 4 18.034, Honors Differential Equations Prof. Jason Starr Lecture 4: Existence & Uniqueness, Part II Feb. 11, 2004 1. Cauchy sequences and complete metric spaces. A metric space is complete if it “has no holes”. What does this mean? A metric space “has a hole” if there is a sequence that “should converse”, but such that there is no limit. The sequences which “should converse” are Cauchy sequences. Def’n: Let (X,d) be a metric space. A sequence of elements in X, (Pi)i=0,1,…, is a Cauchy sequences if for every 0>ε, there is an integer 0M ≥ such that for all pairs of integers Mnm, ≥ , the distance ε<)P,d(Pnm. Thm [≈Heine-Borel thm] For the Euclidean metric space (IR n, Eucl∂ ), sequence (Pi) is convergent if and only if it is Cauchy. This property of (IR n, Eucl∂) has a name. Def’n: A metric space )(X, ∂ is complete if for every sequence of elements in X, (Pi), the sequence is convergent if and only if it is Cauchy. Exercise: Prove that for any metric space, if (Pi) is convergent, then (Pi) is Cauchy. 2 Cauchy test Let [a,b] be a bounded interval in IR , and let C([a,b], IR ) be the metric space of continuous real valued functions on [a,b] with the metric =∂ )y,(y12 maximum value of )()(12tyty − on [a,b]. The main theorem about this metric space is the following. Thm [Cauchy test = Thm A.21]: Let (yi(t)) be a sequence in C([a,b], IR ). If (yi) is Cauchy, then it converses, i.e. ∂), b],C([a, is a complete metric space. Pf: Suppose (yi) is Cauchy. Then for every t in [a,b], the sequence of real numbers (yi,(t)) i=0,1,… , is a Cauchy sequence. Because IR is complete, (yi,(t)) converges to some real number. Call this number y(t). The claim is that, for every 0>ε, there is an integer 0N ≥ such that for all Nn ≥ , ε y(t)}-(t)max{(yn< . To prove this, observe by the Cauchy condition that there is 0N ≥ such that for all Nnm, ≥, max{}2)()(maxε<− tytynm. Now, for each t, because (yi,t)) converges to y(t), there exists 0M(t)M ≥= such that for all Mm ≥ , 2)()(ε<− tytym. But then, for every Nn ≥ (N doesn’t depend on t), )()( tytyn− ))()(())()(( tytytytymmn−+−= ))()(())()(( tytytytymmn−+− 2ε< 2ε+ ε=. (Cauchy condition) (ym(t) converses to y(t)) This proves the claim. IR18.034, Honors Differential Equations Lecture 4 Prof. Jason Starr Page 2 of 4 If we knew that y(t) is continuous, it would follow that (yi) converses to y. So let’s prove y(t) is continuous. Let t be any element in [a,b]. By the claim, there exists N≥ 0 such that for all n ≥N, 3y(t)}-)(max{(ynεδ< . Because yn(t) is continuous, there exists z >0 such that 3)()(ε<− tysynn whenever z <− st , Then, whenever z <− st )()()()()()()()( tytytysysysytysynnnn−+−+−≤− (by the triangle inequality) ≤ 3ε + 3ε + 3ε {}( )3maxε<− yycbn ( )ctstycbn)( {}( )3maxε<− yycbn ε = So y(t) is continuous. Therefore y(t) is an element of C([a,b], IR ), and the sequence (yi(t)) converges to y(t). Variation: The metric space b),(yB0• is complete (defined as in lecture 3). Proof: The only new step is to prove that the limit y(t) has graph contained in b]yb,-[y x C]t,[t0000++ . But for each t, y(t) is the limit of (yi(t)). Because yi(t) is a sequence in the closed interval b]yb,-[y00+ , its limit is also in this interval. So y(t)b-y0≤ by0+≤ , i.e. y(t) is in b),(yB0•. (Contd. on next page… )18.034, Honors Differential Equations Lecture 4 Prof. Jason Starr Page 3 of 4 3. Thm [Contraction mapping fixed point theorem, Part II]: Let ) (X, ∂ be a complete metric space. Let ε be a number with 10 <≤ε. Let T be an ε-contraction mapping on ) (X, ∂ . Then there exists a unique fixed point of T. Proof: We already proved there is at most one point. The real content is that there exists a fixed point. Let p0 be any point of (X, ∂ ). Define )T(pp01= , define )T(pp12= , etc. In other words, define a sequence .0,1,i i)(p…= of elements in X inductively by 1iip)T(p+= . Denote by D the distance )p,(pD01∂= . The claim is that for all i=0,1,…, ∂ D)p,(pii1i•≤+ε. This is proved by induction on i, the base case being done. If )p,(pi1i+∂ Di•≤ε, then ∂ ∂=++)p,(p1i2i(T(pi+1),T(pi)) ≤+))T(p),(T(pi1i∂•ε)p,(pi1i+ (b/c T is ε-contracting) D••≤iεε (by hypothesis) D21i•=+ε This proves the induction step, thus )p,(pi1i+∂ D•≤iε for all i. The claim is that (pi) is a Cauchy sequence. Let 0' >ε be any number. Let N be an integer such that '1.εεε<−DN, i.e. ()()εεεlog'1logDN−>. Then for all N n m ≥≥ , )p,(pnm∂ ≤ )p,(p + ..… + )p,(p + )p,(p1-mm1-n2+nn1+n∂∂∂ D .. D2 D1 D1-m2n1nn•+…+•++•++•≤++εεεε D .. D2 D1 Dn2n1nn•=…+•++•++•≤++εεεεε−•11. Since '1.εεε<−DN for Nn ≥ . ' )p,(pnmε<∂ . So (pi) is a Cauchy sequence. Because (X,∂ ) is a complete metric space, the X Cauchy sequence (pi) converses to an element p of X. Let 0' >ε be a number. There exists N such that for all N n m ≥≥ , 3 )p,(pnmε<∂ . Also, there is Nn ≥ such that <∂ )p(p,n3ε. Thus, T(p))(p,∂≤ +∂ )p(p,n+∂+ )p,(p1nnT(p)),(p1n+∂ . But ∂ T(p)),(p1n+∂= T(p))),(T(pn∂< p),(pn. So ∂ T(p)),(p1n+≤∂•2+ )p(p,n∂ •<++2 )p (p1nn3εεε 3=+ . Since ∂T(p))(p, (p,T(p)) is less than ε for all 0>ε, ∂0T(p))(p, = . Therefore T(p)p = , i.e. p is a fixed point of T. 4. Existence of a solution to the IVP. Let R, D, f, M, L, a, b and c be as in Lecture 3. Thm: There exists a differentiable function y(t) defined on [t0,t0+c] such that (1) 00y )y(t =18.034, Honors Differential Equations Lecture 4 Prof. Jason Starr Page 4 of 4 (2) y(t))f(t, (t)y' = (3) b)(0≤− yty Proof: As proved in Lecture 3, on the metric space b),(yB0• of continuous functions y(t) on [t0,t0+c] such that b)(0≤− yty, the mapping z T(y)= , ∫+=ttdssysf0))(,( y : z(t)0. is a 1/2 – contraction mapping. By the Cauchy test, b),(yB0• is a complete metric space. By the contraction mapping fixed point theorem, part II, there exists a continuous function y(t) in b),(yB0• such that T(y(t))y(t)=, i.e. ∫+=ttdssysf0))(,( y y(t)0. By the fundamental theorem of calculus, y(t) is differentiable and y(s))f(s,(t)y'=.
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