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 0 M ≥ such that for all pairs of integers M n m, ≥ , the distance ε < ) P , d(P n m . 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 , (y 1 2 maximum value of ) ( ) ( 1 2 t y t y − 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 0 N ≥ such that for all N n ≥ , ε y(t)} -(t) max{(yn < . To prove this, observe by the Cauchy condition that there is 0 N ≥ such that for all N n m, ≥ , max { } 2 ) ( ) ( max ε < − t y t y n m . Now, for each t, because (yi,t)) converges to y(t), there exists 0 M(t) M ≥ = such that for all M m ≥ , 2 ) ( ) ( ε < − t y t ym . But then, for every N n ≥ (N doesn’t depend on t), ) ( ) ( t y t yn − )) ( ) ( ( )) ( ) ( ( t y t y t y t y m m n − + − = )) ( ) ( ( )) ( ) ( ( t y t y t y t y m m n − + − 2 ε < 2 ε + ε = . (Cauchy condition) (ym(t) converses to y(t)) This proves the claim. IR 18.034, 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, 3 y(t)} -) ( max{(yn ε δ < . Because yn(t) is continuous, there exists z >0 such that 3 ) ( ) ( ε < − t y s y n n whenever z < − s t , Then, whenever z < − s t ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( t y t y t y s y s y s y t y s y n n n n − + − + − ≤ − (by the triangle inequality) ≤ 3 ε + 3 ε + 3 ε { } ( ) 3 max ε < − y y c b n ( ) cts t y c b n ) ( { } ( ) 3 max ε < − y y c b n 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) , (y B 0 • 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] y b, -[y x C] t , [t 0 0 0 0 + + . But for each t, y(t) is the limit of (yi(t)). Because yi(t) is a sequence in the closed interval b] y b, -[y 0 0 + , its limit is also in this interval. So y(t) b -y0 ≤ b y0 + ≤ , i.e. y(t) is in b) , (y B 0 • . (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 1 0 < ≤ 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(p p 0 1 = , define ) T(p p 1 2 = , etc. In other words, define a sequence . 0,1, i i) (p … = of elements in X inductively by 1 i i p ) T(p + = . Denote by D the distance ) p , (p D 0 1 ∂ = . The claim is that for all i=0,1,…, ∂ D ) p , (p i i 1 i • ≤ + ε . This is proved by induction on i, the base case being done. If ) p , (p i 1 i+ ∂ D i • ≤ ε , then ∂ ∂ = + + ) p , (p 1 i 2 i (T(pi+1),T(pi)) ≤ + )) T(p ), (T(p i 1 i ∂ • ε ) p , (p i 1 i+ (b/c T is ε-contracting) D • • ≤ i ε ε (by hypothesis) D 21 i • = + ε This proves the induction step, thus ) p , (p i 1 i+ ∂ D • ≤ i ε 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 . ε ε ε < − D N , i.e. ( ) ( ) ε ε ε log ' 1 log D N − > . Then for all N n m ≥ ≥ , ) p , (p n m ∂ ≤ ) p , (p + .. … + ) p , (p + ) p , (p 1 -m m 1 -n 2 + n n 1 + n ∂ ∂ ∂ D .. D 2 D 1 D 1 -m 2 n 1 n n • + … + • + + • + + • ≤ + + ε ε ε ε D .. D 2 D 1 D n 2 n 1 n n • = … + • + + • + + • ≤ + + ε ε ε ε ε − • 1 1 . Since ' 1 . ε ε ε < −D N for N n ≥ . ' ) p , (p n m ε < ∂ . 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 , (p n m ε < ∂ . Also, there is N n ≥ such that < ∂ ) p (p, n 3 ε . Thus, T(p)) (p, ∂ ≤ + ∂ ) p (p, n + ∂ + ) p , (p 1 n n T(p)) , (p 1 n+ ∂ . But ∂ T(p)) , (p 1 n+ ∂ = T(p)) ), (T(pn ∂ < p) , (pn . So ∂ T(p)) , (p 1 n+ ≤ ∂ • 2 + ) p (p, n ∂ • < + + 2 ) p (p 1 n n 3 ε ε ε 3 = + . Since ∂ T(p)) (p, (p,T(p)) is less than ε for all 0 > ε , ∂ 0 T(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) 0 0 y ) 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 ≤ − y t y Proof: As proved in Lecture 3, on the metric space b) , (y B 0 • of continuous functions y(t) on [t0,t0+c] such that b ) ( 0 ≤ − y t y , the mapping z T(y) = , ∫ + = t t ds s y s f 0 )) ( , ( y : z(t) 0 . is a 1/2 – contraction mapping. By the Cauchy test, b) , (y B 0 • is a complete metric space. By the contraction mapping fixed point theorem, part II, there exists a continuous function y(t) in b) , (y B 0 • such that T(y(t)) y(t) = , i.e. ∫ + = t t ds s y s f 0 )) ( , ( y y(t) 0 . By the fundamental theorem of calculus, y(t) is differentiable and y(s)) f(s, (t) y' = . Moreover, 0 0 0 0 y 0 y )) ( , ( y ) y(t 00 = + = + = ∫tt ds s y s f . Finally since y(t) is in b) , (y B 0 • , ≤ − 0 ) ( y t y b for all t in [t0,t0+c].