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The Wave Equation and Energy Conservation

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The Wave Equation and Energy Conservation The wave equation and energy conservation Peter Haggstrom www.gotohaggstrom.com [email protected] May 21, 2017 1 Problem 10, Chapter 3 of "Fourier Analysis: An Introduc- tion" by Elias Stein and Rami Shakarchi Problem 10 in Chapter 3, page 90, of Elias Stein and Rami Shakarchi's textbook "Fourier Series: An Introduction" [1] contains the following problem which exhibits some fundamental techniques for physics and analysis students. Here is the problem. Consider a vibrating string whose displacement u(x; t) at time t satisfies the wave equation: 1 @2u @2u τ = ; c2 = (1) c2 @t2 @x2 ρ ρ is the constant density and τ is the coefficient of tension of the string. The string is subject to the initial conditions: @u u(x; 0) = f(x) and (x; 0) = g(x) (2) @t where it is assumed that f 2 C1 and g is continuous. The total energy for the string is defined as: ρ Z L @u τ Z L @u E(t) = 2 dx + 2 dx (3) 2 0 @t 2 0 @x 1 (there is a typo in the textbook where T appears instead of τ). The length of the string is L. The first term in (3) corresponds to the "kinetic energy" of the string (in analogy 1 2 with 2 mv , the kinetic energy of a particle of mass m and velocity v), and the second term corresponds to the "potential energy". Show that the total energy of the string is conserved, in the sense that E(t) is constant. Therefore: L L ρ Z τ Z 0 E(t) = E(0) = g(x)2 dx + f (x)2 dx (4) 2 0 2 0 2 Detailed solution The crux of this problem is the definition of total energy in (3) which is merely given, not derived. The kinetic energy component is simply the differential velocity- mass integrated over the length of the string. If the string is displaced a small q @u @u 2 amount so that @x is small then the differential arc length ds = 1 + @x dx (this is just Pythagoras' theorem ds2 = dx2 + du2). Because @u is small we can q @x @u 2 approximate ds = 1 + @x dx by dx and so the mass is simply ρ dx and velocity @u ρ R L @u 2 @t so that the kinetic energy is 2 0 @t dx. The potential energy is a bit trickier. If we assume that the potential energy is proportional to a differential increase in the length of the string compared with its rest length of L we can get an expression for the potential energy as follows. The factor of proportionality is the tension τ. The change in length is (using Taylor's theorem): r Z L @u Z L 1 @u 1 Z L @u 1 + 2dx − L ≈ 1 + 2 dx − L = 2 dx (5) 0 @x 0 2 @x 2 0 @x τ R L @u 2 Accordingly the potential energy is 2 0 @x dx. There is a more indirect way of arriving at the expression for total energy E(t). Because this involves differentiating under the integral sign, a general theorem is as follows (see [2] pages 268-9): 2 2.1 Theorem @g(x;t) Let g : R × R ! R be a continuous function such that @t exists and is con- tinuous. Suppose R 1 jg(x; t)j dx and R 1 j @g(x;t) j dx and exist for each t and that −∞ −∞ @t R @g(x;t) dx ! 0 as R ! 1 uniformly in t on every interval [a; b] . Then jxj>R @t R 1 −∞ g(x; t) dx is differentiable with: d R 1 R 1 @g(x;t) dt −∞ g(x; t) dx = −∞ @t dx Let us leave for the moment whether we can actually do the differentiation inside the integral (I will come to the subtleties of that shortly) and simply formally differentiate (4) as follows: L 2 L 2 L 2 L 2 0 ρ Z @u @ u τ Z @u @ u Z @u @ u Z @u @ u E (t) = 2 2 dx+ 2 dx = ρ 2 dx+τ dx 2 0 @t @t 2 0 @x @t @x 0 @t @t 0 @x @x @t Z L @u @2u h@u @u L Z L @u @2u Z L @u @2u Z L @u @2u = ρ dx+τ − dx = ρ dx−τ dx 2 2 2 2 0 @t @t @x @t 0 0 @t @x 0 @t @t 0 @t @x Z L 2 Z L 2 2 @u @ u @u @ u = ρ c 2 dx − τ 2 dx = 0 (6) 0 @t @x 0 @t @x @2u @2u Note that in equating @x @t dx with @t @x dx we are implicitly assuming continuity of the second derivative. In the step involving integration by parts we are assuming @u @u initial conditions on the derivatives ie @t (0; t) = @t (L; t) = 0 for all t. In other words the ends are fixed so that there is no movement and hence no velocity. ρ R L 2 Equation (6) shows that E(t) is a constant so that E(t) = E(0) = 2 0 g(x) dx + τ R L 0 2 2 0 f (x) dx where (2) has been used. We can derive equation (3) in a more general context by starting with the kinetic energy ie: ρ Z 1 @u KE = 2dx (7) 2 −∞ @t To get convergence of the integral we have to assume that the integrand vanishes outside of some large interval jxj ≤ R. To see whether the KE is conserved in time we differentiate with respect to time and we get: 3 Z 1 2 Z 1 2 Z 1 2 Z 1 2 d ρ @u @ u @u @ u @u 2 @ u @u @ u KE = 2 2 dx = ρ 2 dx = ρ c 2 dx = τ 2 dx dt 2 −∞ @t @t −∞ @t @t −∞ @t @x −∞ @t @x (8) The final integral in (8) does not look like it would be zero but if we integrate by parts we come up with something useful: Z 1 @u @2u @u @u 1 Z 1 @u @2u τ dx = τ − dx (9) 2 −∞ @t @x @t @x −∞ −∞ @x @x @t 1 @u @u Now @t @x should vanish because the speed of propogation is finite and the −∞ derivatives ought to approach zero in the limit. So we are left with: d Z 1 @u @2u d 1 Z 1 @u KE = − dx = − τ 2 dx (10) dt −∞ @x @x @t dt 2 −∞ @x If we define the potential energy to be: 1 Z 1 @u PE = τ 2 dx (11) 2 −∞ @x Then what we have is this: d d d KE = − PE or (KE + PE) = 0 (12) dt dt dt So (12) says that the total energy defned by the sum of (7) and (11) ( which is just (3) ) is conserved. Now let's go right back to basics. If we start with the classic plucked string set up (see [3] pages 39-40) we get the following situation. The string is stretched between (0; 0) and (2; 0) and the mid-point (1; 0) is raised to a height h above the x axis. Thus the initial position of the string is defined by: n hx when 0 ≤ x ≤ 1 f(x) = (13) −hx + 2h when 1 ≤ x ≤ 2 8h nπ The Fourier sine series of (13) has coefficients bn = π2n2 sin 2 . The formal solution @2u 2 @2u to the wave equation (obtained by separation of variables) @t2 = a @x2 for 0 < x < c and t > 0 is (see [3] pages 39-40) : 4 1 X nπx nπat u(x; t) = b sin cos (14) n c c n=1 Thus we get the formal solution: 1 8h X 1 nπ nπx nπat u(x; t) = sin sin cos (15) π2 n2 2 2 2 n=1 which can be shown to be represented as: 1 8h X (−1)m+1 (2m − 1)πx (2m − 1)πat u(x; t) = sin cos (16) π2 (2m − 1)2 2 2 m=1 Neither (15) or (16) is actually twice differentiable with repect to x or t. To see 8h 1 nπ this look at (15) for instance and note that the damping factor π2 n2 sin 2 will nπx nπat be "neutralised" by two differentiations of either the sin 2 factor or the cos 2 factor so you will get a pure non-damped oscillatory infinite sum ie something that will not converge. More detailed inforrmation concerning differentiating under the integral sign can be found in [5]. Clearly, then, further conditions are needed to ensure existence and uniqueness of a solution to the wave equation. Churchill provides the proof in Chapter 10 of [3]. He generalizes the problem as follows: @2u(x; t) @2u(x; t) = a2 + φ(x; t); 0 < x < c; t > 0 (17) @t2 @x2 u(0; t) = p(t); u(c; t) = q(t) t ≥ 0 (18) @u u(x; 0) = f(x); (x; 0) = g(x) 0 ≤ x ≤ c (19) @t It is now assumed that u is of class C2 in the region R defined by 0 ≤ x ≤ c; t ≥ 0. Thus u and its first and second derivatives including the mixed second partial derivatives are assumed to be continuous in this region R . Now suppose there are 2 two C solutions u1(x; t) and u2(x; t) in the region. Hence the difference z = u1 −u2 is of class C2 in the region and it will satisfy the homogeneous problem: 5 @2z(x; t) @2z(x; t) = a2 ; 0 < x < c; t > 0 (20) @t2 @x2 z(0; t) = 0; z(c; t) = 0; t ≥ 0 (21) @z z(x; 0) = 0; (x; 0) = 0; 0 ≤ x ≤ c (22) @t We have to show that z = 0 throughout the region so that u1 = u2.
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