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Unformatted text preview: Math 124A – October 19, 2010 Viktor Grigoryan 7 The energy method 7.1 Energy for the wave equation Let us consider an infinite string with constant linear density ρ and tension magnitude T . The wave equation describing the vibrations of the string is then ρu tt = Tu xx ,∞ < x < ∞ . (1) Since this equation describes the mechanical motion of a vibrating string, we can compute the kinetic energy associated with the motion of the string. Recall that the kinetic energy is 1 2 mv 2 . In this case the string is infinite, and the speed differs for different points on the string. However, we can still compute the energy of small pieces of the string, add them together, and pass to a limit in which the lengths of the pieces go to zero. This will result in the following integral KE = 1 2 Z ∞∞ ρu 2 t dx. We will assume that the initial data vanishes outside of a large interval  x  ≤ R , so that the above integral is convergent due to the finite speed of propagation. We would like to see if the kinetic energy KE is conserved in time. For this, we differentiate the above integral with respect to time to see whether it is zero, as is expected for a constant function, or is different from zero. d dt KE = 1 2 ρ Z ∞∞ 2 u t u tt dx = Z ∞∞ ρu t u tt dx. Using the wave equation (1), we can replace the ρu tt by Tu xx , obtaining d dt KE = T Z ∞∞ u t u xx dx. The last quantity does not seem to be zero in general, thus the next best thing we can hope for, is to convert the last integral into a full derivative in time. In that case the difference of the kinetic energy and some other quantity will be conserved. To see this, we perform an integration by parts in the last integral d dt KE = Tu t u x ∞∞ Z ∞∞ Tu xt u x dx....
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 Fall '08
 Ponce
 Math, Differential Equations, Equations, Partial Differential Equations, Energy, Potential Energy, Vibrating string, Boundary value problem, wave equation, T uxx

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