MIT OpenCourseWare http://ocw.mit.edu 18.034 Honors Differential Equations Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.034 Solutions to Problemset 5 Spring 2009 1. (a) Since fn → f uniformly on [a, b], for � > 0 given, there exists N ∈ Z+ such that |fn (t) − f (t)| < �/(b − a) for t ∈ [a, b] whenever n ≥ N . For n ≥ N , �� b � � b � b � � � �≤ f (t) dt − f (t) dt |fn (t) − f (t)| dt < � n � � a a a (b) fn (t) is given by n 0 � 1 fn (t) dt = �0 1 1 1/2n 1/n 1 for all n, 2 f (t) dt = 0. 0 2. In the course of the local existence theorem, 1 |xk (t) − xk−1 (t)| ≤ M Lk−1 |t − t0 |k for k = 1, 2, . . . k! � � � � � ∞ � � | x(t) − xn (t)| = � (xk (t) − xk−1 (t))� � k=n+1 � ∞ � ≤ |xk (t) − xk−1 (t)| ≤ = = k=n+1 ∞ M � (LT )k L k! k=n+1 ∞ M (LT )n+1 � (LT )k L (n+1)! k! k=0 T n+1 LT n M L (n+1)! e 3. (a) (⇒) Let F (t) = f (t, φ(t)) and solve x�� = F (t) when x(t0 ) = x0 , x� (t0 ) = x1 . � � t d t ∂f (⇐ ) Use f (s, t) ds = f (t, t) + (s, t) ds. dt t0 t0 ∂t (b) Repeat the proof of the local existence theorem by showing �� t � � � � (1) |xn (t) − x0 | = |x1 | |t − t0 | + � (t − s)f (s, xn−1 (s) ds�� t0� t ≤ |x1 | |t − t0 | + M |t − s| ds t0 2 = |x1 | |t − t0 | + M 2 |t − t0 | ≤ B | t − t0 | � t (2) |xn (t) − xn−1 (t)| ≤ (t − s)| f (s, xn−1 (s)) − f (s, xn−2 (s))| ds t0� t ≤ L (t − s)|xn−1 − xn−2 | ds t0 |t − t0 |2n (2n)! where L is the Lipshitz constant. � � � ∞ √ � ∞ 1 −st 1 −x2 s 2x 4. (a) L √ = e √ dt = e dx, where x 2 = st. x s t � t 0 0 � 2 ∞ −x2 =√ e dx = π/s s 0 � � ∞ √ √ √ e−st ∞ � ∞ e−st 1 −st √ dt by parts. (b) L[ t] = e t dt = t − −s 0 −s 2 t 0 �0 ∞ √ 1 1 = 0 + e−st √ dt = π/2s3/2 2s 0 t ≤ M Ln−1 2 5. (a) 2 L[e t ] � ∞ = et 2 −st dt is indefinite for every real value of s, no 0 matter how large, since t2 − st > 0 for t > s and so � ∞ � ∞ � ∞ t2 −st t2 −st e dt > e dt > e 0 dt = ∞ 0 s s � � � ∞ 1 1 (b) L k = e−st k dt, (s > 0). The trouble here is when t = 0. t t 0 � τ � τ dt −st 1 −st Near t = 0, e ≈ 1 and therefore e dt � = k k t 0 0 t ⎧ ⎪ ⎪ 1−k �τ � � ⎨ � t 1 � k = � 1 . Therefore, L k exists for k < 1. � ⎪ 1 − k t ⎪ 0 ⎩ log t|τ0 k = 1 6. (a) 5 cos 2t − 3 sin 2t + 2. √ (b) e−t/3 cos 2t. 3