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Numbertheory

# Numbertheory - Research Sudarshan Balakrishnan Linearly...

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Research Sudarshan Balakrishnan June 28, 2011 Linearly dispersive wave equation The purpose of this write-up is to examine the solution to the linearly dis- persive wave equation at different times and observe any number-theoretic results that might arise. Using the program qd.m, we notice the following results: Solution profile t x u ( t, x ) t = π 0 < x < π 1 π < x < 2 π 0 t = π 2 0 < x < π 2 1 π 2 < x < π 0 3 π 2 < x < 2 π 1 t = π 3 0 < x < π 3 1 π 3 < x < 2 π 3 0 2 π 3 < x < π 0 π < x < 4 π 3 0 4 π 3 < x < 5 π 3 1 5 π 3 < x < 2 π 1 t = 2 π 3 0 < x < π 3 1 π 3 < x < 2 π 3 1 2 π 3 < x < π 0 π < x < 4 π 3 0 4 π 3 < x < 5 π 3 0 5 π 3 < x < 2 π 1 An immediate observation here is for every t = q , the solution u ( t, x ) is constant for every subinterval π q < x < π + π q . In other words, for the interval of length π from π q to π + π q the solution u ( t, x ) is constant. The 1

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next step was to observe if this holds good for further intervals, we proceed similarly: Solution profile t x u ( t, x ) t = π 4 0 < x < π 4 1 π 4 < x < π 2 0 π 2 < x < 3 π 4 0 3 π 4 < x < π 0 π < x < 5 π 4 0 5 π 4 < x < 6 π 4 1 6 π 4 < x < 7 π 4 1 7 π 4 < x < 2 π 1 As we notice, the pattern exists for π 4 . For 3 π 4 , we notice the same : Solution profile t x u ( t, x ) t = 3 π 4 0 < x < π 4 1 π 4 < x < π 2 1 π 2 < x < 3 π 4 1 3 π 4 < x < π 0 π < x < 5 π 4 0 5 π 4 < x < 6 π 4 0 6 π 4 < x < 7 π 4 0 7 π 4 < x < 2 π 1 However, we don’t notice the same pattern for t = 5 which only takes identical values for intervals 4 π 5 - 6 π 5 . There might be an interesting result for 5 , but I can’t seem to notice it. But the earlier pattern remains for t = π 6 , as we observe the results : 2
Solution profile t x u ( t, x ) t = π 6 0 < x < π 6 1 π 6 < x < 2 π 6 0 2 π 6 < x < 3 π 6 0 3 π 6 < x < 4 π 6 0 4 π 6 < x < 5 π 6 0 5 π 6 < x < π 0 π < x < 7 π 6 0 7 π 6 < x < 8 π 6 1 8 π 6 < x < 9 π 6 1 9 π 6 < x < 10 π 6 1 10 π 6 < x < 11 π 6 1 11 π 6 < x < 2 π 1 We notice the same at t = 2 π 6 = π 3 , t = 3 π 6 = π 2 and t = 4 π 6 = 2 π 3 . Now, at t = 5 π 6 we get: Solution profile t x u ( t, x ) t = π 6 0 < x < π 6 1 π 6 < x < 2 π 6 1 2 π 6 < x < 3

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Numbertheory - Research Sudarshan Balakrishnan Linearly...

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