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Lipschitz condition

# Lipschitz condition - Math 128A Spring 2002 Sergey Fomel...

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Math 128A Spring 2002 Handout # 26 Sergey Fomel April 30, 2002 Answers to Homework 10: Numerical Solution of ODE: One-Step Methods 1. (a) Which of the following functions satisfy the Lipschitz condition on y ? For those that do, find the Lipschitz constant. i. f ( x , y ) = x 2 + y 2 for x [ - 1,1] ii. f ( x , y ) = | y | iii. f ( x , y ) = | y | iv. f ( x , y ) = | y | / x for x [ - 1,1] Answer: i. f ( x , y ) = x 2 + y 2 for x [ - 1,1] satisfies the Lipschitz condition. It is proved by the following chain of equalities and inequalities: | f ( x , y 1 ) - f ( x , y 2 ) | = x 2 + y 2 1 - x 2 + y 2 2 = | y 2 1 - y 2 2 | x 2 + y 2 1 + x 2 + y 2 2 = | y 1 + y 2 | x 2 + y 2 1 + x 2 + y 2 2 | y 1 - y 2 | | y 1 |+| y 2 | x 2 + y 2 1 + x 2 + y 2 2 | y 1 - y 2 | ≤ | y 1 - y 2 | . The Lipschitz constant is 1. ii. f ( x , y ) = | y | satisfies the Lipschitz condition. We have f ( x , y 1 ) - f ( x , y 2 ) = | y 1 |-| y 2 | . From the equality y 1 = y 2 + y 1 - y 2 , it follows that | y 1 | ≤ | y 2 |+| y 1 - y 2 | and | y 1 |-| y 2 | ≤ | y 1 - y 2 | . From the equality y 2 = y 1 + y 2 - y 1 , it follows that | y 2 | ≤ | y 1 |+| y 1 - y 2 | and -| y 1 - y 2 | ≤ | y 1 |-| y 2 | . 1

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Putting it together, -| y 1 - y 2 | ≤ | y 1 |-| y 2 | ≤ | y 1 - y 2 | or || y 1 |-| y 2 || ≤ | y 1 - y 2 | . This proves the Lipschitz condition. The Lipschitz constant is 1. iii. f ( x , y ) = | y | does not satisfy the Lipschitz condition. It is sufficient to consider a particular case y 2 = 0. Then f ( x , y 1 ) - f ( x , y 2 ) = | y 1 | and | f ( x , y 1 ) - f ( x , y 2 ) | | y 1 - y 2 | = 1 | y 1 | The right-hand side is unbounded for y 1 approaching zero, which shows that the Lipschitz condition cannot be satisfied. iv. f ( x , y ) = | y | / x for x [ - 1,1] does not satisfy the Lipschitz condition. Again let us consider a particular case y 2 = 0. Then | f ( x , y 1 ) - f ( x , y 2 ) | | y 1 - y 2 | = 1 | x | The right-hand side is unbounded for x = 0, which shows that the Lipschitz condition cannot be satisfied. (b) Prove that the function f ( x , y ) = - | 1 - y 2 | does not satisfy the Lipschitz condition and find two different solutions of the initial-value problem y ( x ) = - | 1 - y 2 ( x ) | y (0) = 1 (1) on the interval x [0, π ]. Answer: To disprove the Lipschitz condition, let us consider the special case y 2 = 1. Then f ( x , y 1 ) - f ( x , y 2 ) = - | 1 - y 2 1 | and | f ( x , y 1 ) - f ( x , y 2 ) | | y 1 - y 2 | = | 1 - y 2 1 | | 1 - y 1 | = | 1 + y 1 | | 1 - y 1 | The right-hand side is unbounded for y 1
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