problemset5

# problemset5 - 18.034 Solutions to Problemset 5 Spring 2009...

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18.034 Solutions to Problemset 5 Spring 2009 1. (a) Since f n f uniformly on [ a,b ], for > 0 given, there exists N Z + such that | f n ( t ) f ( t ) | < / ( b a ) for t [ ] whenever n N . For n N , ± b ± b ± b f n ( t ) dt f ( t ) dt a | f n ( t ) f ( t ) | dt < a a (b) f n ( t ) is given by ± 1 ± 0 1 0 1 / 2 n 1 /n 1 n f n ( t ) dt = for all n , 1 2 f ( t ) dt = 0. 0 2. In the course of the local existence theorem, | x k ( t ) x k 1 ( t ) | ≤ 1

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± ± ± ± ² ³ ± ± ± ´ ± ³ ± ML k 1 | t k ! t 0 | k for k = 1 , 2 ,... k = n +1 µ ( x k ( t ) x k 1 ( t )) | x ( t ) x n ( t ) | = k = n +1 | x k ( t ) x k 1 ( t ) | µ k ! k = n +1 µ ( LT ) k M L µ L ( n +1)! k ! k =0 = n T n +1 e LT ( n +1)! 3. (a) ( ) Let F ( t ) = f ( t,φ ( t )) and solve x = F ( t ) when x ( t 0 ) = x 0 , x ( t 0 ) = x 1 . ( LT ) n +1 ( LT ) k = M t t d ∂f ( ) Use f ( s,t ) ds = f ( t,t ) + t 0 t 0 ( ) ds . dt ∂t (b) Repeat the proof of the local existence theorem by showing t ( t s ) f ( s,x n 1 ( s ) ds (1) | x n ( t ) x 0 | = | x 1 || t t 0 | + t ± 0 ≤ | x 1 t t 0 | + M t 0 | t s | ds t t 0 2 t t 0 = | x 1 | + M | | 2 B t t 0 t | | t (2) | x n ( t ) x n 1 ( t ) t ± 0 ( t s ) ds L ( t s ) | x n 1 x n 2 | t 0 n 1 | t t 0 | 2 n (2 n )! where L is the Lipshitz constant. f ( n 1 ( s )) f ( n 2 ( s )) | ds | ≤ | t s 2 x 1 1 2 2 e st 4. (a) L dx , where x e x dt = = = st . t t x s 0 0 2 2 e x dx = π/s = s 0 1 e st e st (b) L [ t ] = e st tdt = t dt by parts. 2 t s dt = π/ 2 s s 0 ± 2 s 0 0 0 1 1 3 / 2 e st = 0 + t 2
± ² ³ ³ ³ ³ ± ² t 2 e 0 st dt is indefnite For every real value oF s , no 5. (a) L [ e t 2
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problemset5 - 18.034 Solutions to Problemset 5 Spring 2009...

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