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MIT OpenCourseWarehttp://ocw.mit.edu 18.306 Advanced Partial Differential Equations with Applications Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Problem Set Number 05 18.306 — MIT (Fall 2009) Rodolfo R. Rosales MIT, Math. Dept., Cambridge, MA 02139 November 22, 2009 Due: Monday November 30, Contents 1.1 Statement: Wave equations (problem 01) . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Statement: Wave equations (problem 02) . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3 Statement: Homogenization (problem 01). Slow compression. . . . . . . . . . . . . 3 Slow compression in 1-D Gas Dynamics. . . . . . . . . . . . . . . . . . . . . . . 3 Part 1. Simple Linearization. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Part 2. Homogenization. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.1 Statement: Wave equations (problem 01). Consider an elastic (homogeneous) string under tension, tied at one end, initially at rest, and forced by a (small amplitude) harmonic shaking of the other end. To simplify the situation, assume that all the motion is restricted to happen in a plane. After a proper adimensionalization, the situation is modeled by the mathematical problem below for the wave equation in 1-D — where u = u(x, t) is the displacement from equilibrium of the string. utt − uxx = 0, for 0 < x < 1, and t > 0, (1.1) with initial data u(x, 0) = ut(x, 0) = 0, and boundary conditions u(0, t) = 1 − cos(ω t) and u(1, t) = 0. (1.2) 1FIND the solution to this problem, for the times 0 < t ≤ 4. Furthermore: note that the solution, while making sense in the classical sense (no need to invoke generalized function derivatives), is not infinitely differentiable. There are certain lines along which “singularities” occur. FIND these lines of singularity, and describe what the situation is along them (nature of the singularities) — the lines are, of course, characteristics. FINALLY: does anything special happen if ω = π? 1.2 Statement: Wave equations (problem 02). Consider an elastic (homogeneous) string under tension, undergoing small amplitude oscillations, and assume that all the motion is restricted to happen in a plane. Under these conditions, and after a proper adimensionalization, the displacements u = u(x, t) from equilibrium of the string can be shown to satisfy the 1-D wave equation utt − uxx = 0, (1.3) to which appropriate initial data and boundary conditions must be applied to determine a unique solution. In the lectures we showed that the second order (in space and time) equation in (1.3) is equivalent to a system of two first order equations. We did this by introducing the variables v = ut and w = ux, for which it can be seen that vt − wx = 0 and wt − vx = 0 (1.4) apply — the first equation is (1.3) and the second follows from equality of cross-derivatives. On the other hand, the second equation in (1.4) guarantees that there is a u such that v = ut and w = ux, and then the first equation yields (1.3). Consider now the situation where the string is attached to an (homogeneous) “elastic bed”, instead of being free in space.1 In this case, in addition to the forces caused by the tension in the string, forces are generated by the bed — which are functions of the displacements u only. Thus the 1For example: imagine a ribbon made of some elastic material, with one edge attached to a rigid surface, the other edge attached to the string, and thin enough that we can ignore its mass. 2governing equation above in (1.3) must be modified to utt − uxx + g(u) = 0, (1.5) where g characterizes the elastic response by the bed. If Hooke’s law applies, then g = κ u — for some elastic constant κ > 0. By introducing appropriate variables, SHOW THAT the second order equation in (1.5) is equivalent to a first order system of two equations in two unknowns functions. Hint: because the function u appears in equation (1.5), the trick that we used for (1.3) does not work for (1.5). If you introduce v = ut and w = ux as new variables, you will also have to keep u, and then you will end up with three variables (not two). Instead, try introducing as a new variable an appropriate combination of ut and ux. Note that the approach that you develop here should work for any g = g(u). In particular, for g ≡ 0 it will give you a different (from the one used in the lectures) way to show that (1.3) is equivalent to a system of two first order equations. 1.3 Statement: Homogenization (problem 01). Slow compression. Consider the slow compression/decompression by a piston of a gas in a closed container. On the one hand, intuition tells us that gas properties (such the density) should be uniform across the container, varying in time (only) to adjust for the volume changes. On the other hand, the behavior should be governed by the equations of gas dynamics, which tells us that changes in the gas state happen only through waves propagating at the speed of sound. Since these waves should start at the piston, the solution cannot be just a function of time. Here we show, with a simple model problem, that these two, seemingly contradictory scenarios, coexist without trouble: they are BOTH correct! Consider a gas in a closed cylindrical pipe, where one of the ends is closed by a piston, that can be moved to change the pipe length. Let us neglect motion of the gas in the directions perpendicular to the pipe axis, neglect viscous forces and thermal conductivity, and approximate the evolution as 3� � � being at constant entropy (adiabatic).2 Thus we arrive at the equations 1 ρt + (ρ u)x = 0 and ut + u ux + px = 0, for 0 < x < a and t > 0, (1.6)ρ where ρ is the density, u is the flow velocity, p = p(ρ) is the pressure, c = c(ρ) = dp/dρ > 0 is the sound speed, and a is the varying length of the pipe (piston position). We assume non-dimensional variables, related to the dimensional variables (denoted with tildes) via L x˜ = L x, t˜= t, a˜ = L a, ρ˜ = ρ0 ρ, u˜ = c0 u, c˜ = c0 c, and p˜ = ρ0 c 20 p, (1.7) c0 where L is the “average” pipe length, ρ0 is a typical density, and c0 is the sound speed …


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MIT 18 306 - Problem Set Number 05

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