sol4 - solve x = p ( x,t ) , x ( ) = x , solve z = p ( z,t...

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EE 221 : Linear Systems HW #4 Solutions — Sam Burden — Oct 8, 2010 1 Note these solutions are overly terse and leave out some details; in your solutions, you should strive for greater clarity and rigor. Exercise 1. Lemma If f,g are PC, so are f + g and fg . Proof. Let D ( f ) ,D ( g ) be the set of discontinuity points for f and g . Then clearly D ( f + g ) ,D ( fg ) D ( f ) D ( g ), so f + g and fg are PC. Let f ( x,t ) := A ( t ) x + B ( t ) u ( t ). Since A,B , and u are PC, f ( x, · ) is PC by the Lemma. Lemma Norms are continuous. Proof. || x | - | y || ≤ | x - y | by the reverse triangle inequality. From Problem 3 on HW3, | f ( x,t ) - f ( y,t ) | = | A ( t ) x - A ( t ) y | ≤ | A ( t ) || x - y | . Now K ( t ) := | A ( t ) | is PC by the lemma, whence f ( · ,t ) is Lipschitz. Exercise 2. We have f ( x,y ) = ± y - g sin x - k m y + T m‘ 2 ² = Df = ± 0 1 - g cos x - k m ² . Thus | Df | ≤ max ³ 1 , g + k m ´ , whence f is globally Lipshitz by the fact from Recitation 6. Exercise 3. Let φ
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Unformatted text preview: solve x = p ( x,t ) , x ( ) = x , solve z = p ( z,t ) , z ( ) = x + x . Then ( t ) = x + Z t p ( ( ) , ) d, ( t ) = x + x + Z t p ( ( ) , ) + f ( ) d, therefore | ( t )- ( t ) | = x + Z t p ( ( ) , )-p ( ( ) , )-f ( ) d | x | + Z t | f ( ) | d + Z t | p ( ( ) , )-p ( ( ) , ) | d + 1 ( t- ) + Z t K ( ) | ( )- ( ) | d. Therefore by Bellman-Gronwall, | ( t )- ( t ) | ( + 1 ( t- )) exp R t K ( ) d ....
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