Math425_Practice_Homework_Solutions

Exercise 4 a for what c r does there exist a non zero

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Exercise 4. a) For what c R does there exist a non-zero function w : [0 , 2 π ] C such that: w 00 - c 2 w = 0 and such that w (0) = w (2 π ) = 0 ? b) What if w instead solves w 00 + c 2 w = 0 (again with the assumption that w (0) = w (2 π ) = 0 )? Solution: Let us first suppose that c 6 = 0. From ODE theory, we know that w = a 1 e ct + a 2 e - ct for some (complex numbers) a 1 , a 2 . The condition w (0) = w (2 π ) = 0 then implies that: ( a 1 + a 2 = 0 a 1 e 2 πc + a 2 e - 2 πc = 0 From the above two equations, it follows that a 1 = a 2 = 0 and so w is identically zero. If c = 0, then w = a 1 + a 2 t . In this case, w (0) = 0 implies that a 1 = 0 and w (2 π ) = 0 implies that a 2 = 0, and so w is again identically zero. Hence, in a), it is not possible to find such a function w . b) We now consider what happens when w 00 + c 2 w = 0. Based on part a ), we need to assume that c 6 = 0. In this case, we recall that w ( t ) = a 1 cos( ct ) + a 2 sin( ct ). Since w (0) = a 1 = 0, it follows that w ( t ) = a 2 sin( ct ) . We then obtain that w (2 π ) = a 2 sin(2 πc ). Since we want a 2 6 = 0 (since otherwise, w is identically zero), it follows that we need to have sin(2 πc ) = 0, and hence 2 πc = for some k Z . Consequently, c = k 2 for some k Z \ { 0 } .
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