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Extra resources for Topics in Engineering Mathematics: Modeling and Methods
Example text
Consequently, writing f = co~t, the function t --+ z( t) satisfies the equation z(t) = -co~DV 1(z(t)) = -coDV H(z(t)). (57) Consequently, t --+ z(t) satisfies the required equation, and is at the same time merely a translation. This completes the proof of the proposition! Remark. All steps above are equally true for the partial differential equation itself, and they are in fact a consequence of that. Since (39) has constant coefficients (independent of x), there is translation symmetry. Related to this, the functional J(u) = J is a constant of the motion.
01. 94646 for the second mode). 005. Comparison of the pictures at the left and at the right indicates that the variation in mode amplitude is only caused by numel·ical errors. = A KALEIDOSCOPIC EXCURSION INTO DIFFERENTIAL EQUATIONS 31 Figure 21: The wave profile obtained with the N-mode Fourier truncation, using as coeeficients the data obtained from solving the constrained optimisation problem. In the picture at the left, calculated with N 3, the low number of modes produce a wavy profile in the valley, which is not expected in reality.
50) There exist solutions of this minimization problem, (one of) which we will call z (there exist also other critical points). This minimizer satisfies the equation found from applying Lagranges multiplier rule for constrained optimisation problems: for some "multiplier" ~ E R, z satisfies "V H(z) = (51) ~"V I(z). This minimization problem can be solved numerically (if necessary, by hand for N = 3). For N = 3 one solution is given by the data (approximated to a few digits) in (48), where ~ is the multiplier and the Fourier coefficients form the vector z, when for the value of the constrained is taken (rather arbitrarily) "f = 211'.
