# sol4 - solve ˙ x = p x,t x τ = x ψ solve ˙ z = p z,t z...

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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 ( τ ) = x , ψ solve ˙ z = p ( z,t ) , z ( τ ) = x + δx . Then φ ( t ) = x + Z t τ p ( φ ( σ ) ,σ ) dσ, ψ ( t ) = x + δx + Z t τ p ( ψ ( σ ) ,σ ) + f ( σ ) dσ, therefore | φ ( t )-ψ ( t ) | = µ µ µ µ δx + Z t τ p ( φ ( σ ) ,σ )-p ( ψ ( σ ) ,σ )-f ( σ ) dσ µ µ µ µ ≤ | δx | + Z t τ | f ( σ ) | dσ + Z t τ | p ( φ ( σ ) ,σ )-p ( ψ ( σ ) ,σ ) | dσ ≤ ε + ε 1 ( t-τ ) + Z t τ K ( σ ) | φ ( σ )-φ ( σ ) | dσ. Therefore by Bellman-Gronwall, | φ ( t )-ψ ( t ) | ≤ ( ε + ε 1 ( t-τ )) exp ¶ R t τ K ( σ ) dσ · ....
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