Math425_Practice_Homework_Solutions

Exercise 5 suppose that w 0 r solves the ode 1 aw 00

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Exercise 5. Suppose that w : [0 , + ) R solves the ODE: (1) aw 00 + bw 0 + cw = 0 for some constants a,b,c . Furthermore, we assume that b 0 . a) Let us define the Energy to be: E ( t ) := 1 2 ² a ( w 0 ( t )) 2 + c ( w ( t )) 2 ³ . Without solving the ODE (1) , show that E 0 ( t ) 0 . b) Under the additional assumption that a > 0 and c > 0 , show that w (0) = 0 and w 0 (0) = 0 implies that w ( t ) = 0 for all t 0 . c) Assume again that a > 0 and c > 0 . Show that if w 1 and w 2 solve the ODE (1) and if w 1 (0) =
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PRACTICE HOMEWORK FOR MATH 425, SOLUTIONS 3 w 2 (0) ,w 0 1 (0) = w 0 2 (0) , then one can deduce that w 1 ( t ) = w 2 ( t ) for all t 0 . In this way, we obtain uniqueness of solutions to (1) . Solution: a) We use the product rule to calculate: E 0 ( t ) = aw 0 w 00 + cww 0 . We can now use the ODE to deduce that w 00 = - bw 0 - cw . Hence: E 0 ( t ) = w 0 ( - bw 0 - cw ) + cww 0 = - b ( w 0 ) 2 0 since b 0. In other words, E ( t ) is a decreasing function of t on [0 , + ). b) By assumption E (0) = 1 2 ± a ( w 0 (0)) 2 + c ( w (0)) 2 ² = 0. Since a,c > 0, it follows that E ( t ) is non-negative. Finally, from part a), it follows that E ( t ) is a decreasing function on [0 , + ), hence E ( t ) is identically zero on [0 , + ). In particular, since both a and c are positive, it follows that w ( t ) = 0 for all t 0. c) If w 1 and w 2 solve the ODE, then so does w := w 1 - w 2 . The function w then satisfies the conditions of part b) and the claim follows. ±
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Exercise 5 Suppose that w 0 R solves the ODE 1 aw 00 bw cw...

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