Assignment 10
Transcription
Assignment 10
2014 Fall Mathematical Analysis III 1 Assignment 10 Problems 1–3 are for systems of differential equations. 1. Consider the Cauchy problem for the third order equation: y 000 = f (x, y, y 0 ), y(x0 ) = y0 , y 0 (x0 ) = y1 , y 00 (x0 ) = y3 . Express it as the Cauchy problem for a system of three first order equations. Solution. Let u1 (x) = y(x), u2 (x) = y 0 (x). Then u(x) = (u1 (x), u2 (x)) solves the system of first order equations u01 = u2 , u02 = f (x, u1 , u2 ), (u1 , u2 )(x0 ) = (y0 , y1 ). Similarly one can express the Cauchy problem for a third order equation as the Cauchy problem for a system of three first order equations. 2. Optimal. Consider (2.3) for a system of differential equations. Let f ∈ C(G) satisfies the Lipschitz condition on every compact subset of the open set G. Show that there exists a solution y ∗ of (2.3) defined on some (α, β) with the properties: (a) whenever y is a solution of (2.3) on some I, I ⊂ (α, β); and (b) when β is finite, for each compact K in G, there exists some ρ > 0 such that (x, y ∗ (x)) ∈ G \ K for x ∈ [β − ρ, β). Solution. Omitted. It is similar to the scalar case. 3. Consider the Cauchy problem (2.3) for a system of differential equations. Assuming that f (x, y) ∈ C(R2 ) satisfying the Lipschitz condition and the growth condition f (x, y) ≤ C(1 + |y|), ∀(x, y) ∈ R2 . Show that (2.3) admits a global solution, that is, aPsolution defined on (−∞, ∞). Hint: Study the differential inequality satisfied by |y|2 = nj=1 yj2 . 1 Solution. Let E(x) = |y(x)|2 . Then 2 |E 0 (x)| = |y(x) · y 0 (x)| = |y(x) · f (x, y)| ≤ C|y|(1 + |y|) ≤ C0 (1 + |y|2 ) = C0 (1 + |E(x)|). Hence it follows from the comparison principle that E(x) ≤ (E(0) + 1)eC0 x − 1. Hence (2.3) admits a global solution. 4. Let In = [−n, n] and kf kn = sup{|f (x)| : x ∈ [−n, n]}. For f, g ∈ C(R), define ∞ X 1 kf − gkn d(f, g) = . 2n 1 + kf − gkn n=1 (a) Show that this is a complete metric on C(R). Hint: Use Problem 8, Ex 4. (b) Show that {fj } converges to f in this metric implies {fj } converges to f uniformly on every bounded interval. Solution. (a). kf −Pgk/(1 + kf − gk) is a metric by a previous exercise. As the infinite series is bounded by n 2n ≤ 1, it is convergent so d(f, g) is a well-defined metric on C(R). (b). Well, when {fj } is a Cauchy sequence in this metric, it is also a Cauchy sequence in the metric kf − gkn /(1 + kf − gkn ) for every n. But this metric is equivalent to kf − gkn . 2014 Fall Mathematical Analysis III 2 By the completeness of kf − gkn there exists some f n such that {fj } converges to f n uniformly on In . From In ⊂ In+1 and the uniqueness of limit we know that f n+1 = f n on In , so the function f (x) = f n (x), x ∈ In is a continuous function P on R.−nWe claim that d(fj , f ) → 0 as j → ∞. For, given ε > 0, fix a large N such that ∞ < ε/2. Then n=N 2 ∞ ∞ X X 1 kfj − f kn ε 1 ≤ < . 2n 1 + kfj − f kn 2n 2 n=N n=N On the other hand, we can find a large j0 such that kfj − f k1 , · · · , kfj − f kN −1 < ε , 2N ∀j ≥ j0 . It follows that d(fj , f ) < ε, for all j ≥ j0 . (c). From (b) we see that {fj } converges to f on every In . As every bounded set is contained in some In for large n, our desired conclusion is drawn. 5. Optional. Inspired by the previous problem, introduce a metric on C ∞ (R), the vector space of all smooth functions on the real line, so that {fn } converges to f in this metric implies {dk fn /dxk } converges to dk f /dxk uniformly on each bounded interval for each k ≥ 0. Solution. Try the metric ∞ X 1 kf − gk∞ + · · · + kf (n) − g (n) kn . ρ(f, g) = 2n 1 + kf − gkn + · · · + kf (n) − g (n) kn n=1 6. Let E be a bounded, convex set in Rn . Show that a family of equicontinuous functions is bounded in E if it is bounded at a single point, that is, if there is some constant M such that |f (x0 )| ≤ M for all f in this family. Solution. By equicontinuity, for ε = 1, there is some δ0 such that |f (x) − f (y)| ≤ 1 whenever |x − y| ≤ δ0 . Let BR (x0 ) a ball containing E. Then |x − x0 | ≤ R for all x ∈ E. We can find x0 , · · · , xn = x where nδ0 ≤ R ≤ (n + 1)δ0 so that |xn+1 − xn | ≤ δ0 . It follows that n−1 X R |f (x) − f (x0 )| ≤ |f (xj+1 − f (xj )| ≤ n ≤ . δ0 j=0 Therefore, |f (x)| ≤ |f (x0 )| + n + 1 ≤ M + R δ0 ∀x ∈ E, ∀f ∈ F. 7. Let {fn } be a sequence in C(G) where G is open in Rn . Suppose that on every compact subset of G, it is equicontinuous and bounded. Show that there is a subsequence {fnj } converging to some f ∈ C(G) uniformly on each compact subset of G. Solution. Let Kn be an ascending family of compact sets in G satisfying G = ∪n Kn . Applying Ascoli-Arezela theorem to {fn } on each Kn step by step and then take a Cantor’s diagonal sequence. 8. Let {fn } be a sequence of bounded functions in [0, 1] and let Fn be Z x Fn (x) = fn (t)dt. 0 2014 Fall Mathematical Analysis III 3 (a) Show that the sequence {Fn } has the Bolzano-Weierstrass property provided there is some M such that kfn k∞ ≤ M, for all n. (b) Show that the conclusion in (a) holds when boundedness is replaced by the weaker condition: There is some K such that Z 1 |fn |2 ≤ K, ∀n. 0 Solution. Rx Rx (a) Since |Fn | ≤ 0 |fn (t)|dt ≤ M , and |Fn (x) − Fn (y)| ≤ y |fn (t)|dt ≤ |x − y|M , {Fn } is uniformly bounded and equicontinuous. Then it follows from Arzela-Ascoli theorem that {Fn } has the Bolzano-Weierstrass property. (b) It follows from the Cauchy-Schwarz inequality that Z x x Z |fn (t)|dt ≤ |Fn (x) − Fn (y)| ≤ y 1/2 Z 1 dt 2 y x 1/2 √ p |fn (t)| dt ≤ K |x − y|. 2 y Similarly one can show that {Fn } is uniformly bounded. Then apply Arzela-Ascoli theorem. 9. Let K ∈ C([a, b] × [a, b]) and define the operator T by Z (T f )(x) = b K(x, y)f (y)dy. a (a) Show that T maps C[a, b] to itself. (b) Show that whenever {fn } is a bounded sequence in C[a, b], {T fn } contains a convergent subsequence. Solution. (a) Since K ∈ C([a, b] × [a, b]), given ε > 0, there exists δ > 0 such that |K(x, y) − K(x0 , y)| < ε, whenever |x − x0 | < δ. Then for x, x0 ∈ [a, b], |x − x0 | < δ, one has 0 Z |(T f )(x) − (T f )(x )| ≤ b |K(x, y) − K(x0 , y)||f (y)|dy ≤ |a − b|kf k∞ ε. a Hence T f ∈ C[a, b]. (b) Suppose supn kfn k∞ ≤ M < ∞. It follows from the proof of (a) that δ can be taken independent of n. Hence {fn } is equicontinuous. Furthermore, since |(T fn )(x)| ≤ Rb a |K(x, y)||fn (y)|dy ≤ M (b − a)kKk∞ , {fn } is uniformly bounded. Then it follows from Arzela-Ascoli theorem that {T fn } contains a convergent subsequence. 10. Let f be a bounded, uniformly continuous function on R. Let fa (x) = f (x + a). Show that for each l > 0, there exists a sequence of intervals In = [an , an + l], an → ∞, such that {fan } converges uniformly on [0, l]. Solution. Let {an } be a sequence with an → ∞. Since f is bounded and uniformly continuous on R, it follows that {fan } is uniformly bounded and equivcontinuous on [0, l]. Applying Ascoli-Arezela theorem to obtain a subsequence converging uniformly on [0, l]. 2014 Fall Mathematical Analysis III 4 11. Optional. Let {hn } be a sequence of analytic functions in the unit disc satisfying |hn (z)| ≤ M, ∀z, |z| < 1. Show that there exist an analytic function h in the unit disc and a subsequence {hnj } which converges to h uniformly on each smaller disc {z : |z| ≤ r}, r ∈ (0, 1). Suggestion: Use a suitable Cauchy integral formula. Solution. Let D = {|z| < 1} be the unit disc. Let rn % 1 be a strictly-increasing positive sequence and let Dn = {|z| < rn }. For each n ∈ N, since hj is analytic in D, it follows from the Cauchy integral formula that Z hj (ζ) 1 hj (z) = dζ, for |z| < rn . 2πi |ζ|=rn+1 ζ − z Hence |h0j (z)| 1 =| 2πi Z |ζ|=rn+1 f (ζ) M dζ| ≤ , (ζ − z)2 2π|rn+1 − rn |2 for |z| < rn . Since |h0j (z)| is uniformly bounded on Dn , it follows that {hj } is equicontinuous on each Dn . Applying Ascoli-Arezela theorem to {hj } on each Dn step by step and then taking a Cantor’s diagonal sequence, one obtains a {hnj } which converges to h uniformly on each smaller disc {z : |z| ≤ r}, r ∈ (0, 1). It follows from uniform convergence that Z h(ζ) 1 dζ, for |z| < rn . h(z) = 2πi |ζ|=rn+1 ζ − z Hence h is analytic in D.