Problem Set Number 01, MIT (Winter-Spring 2018)

Transcription

1 Problem Set Number 01, MIT (Winter-Spring 2018) Rodolfo R. Rosales (MIT, Math. Dept., room 2-337, Cambridge, MA 02139) February 28, 2018 Due Thursday, March 8, Turn it in (by 3PM) at the Math. Problem Set Boxes, right outside room There is a box/slot there for 377. Be careful to use the right box (there are many slots). Contents 1 The flux for a conserved quantity must be a vector 1 2 Forced-damped nonlinear oscillator response to a sinusoidal force 2 3 Pendulum with torsion 3 4 Two torsion coupled pendulums 4 5 N torsion coupled pendulums, and continuum limit as N 5 6 Relaxation oscillations while pulling a heavy load 5 List of Figures 3.1 Problem: Pendulum with torsion Problem: Relaxation oscillations while pulling a heavy load (the friction force function) The flux for a conserved quantity must be a vector Statement: The flux for a conserved quantity must be a vector Consider some conserved quantity, with density ρ = ρ( x, t) and flux vector q = q( x, t) in 2D here we present the arguments, and questions, in 2D, but they apply just as well in 3D (or any number of dimensions). Then, in the absence of sources or sinks, we made the argument that conservation leads to the (integral) equation ρ t dx 1 dx 2 = q ˆn ds, (1.1) Ω for any region Ω in the domain where the conserved stuff resides, where: Ω is the boundary of Ω, s is the arc-length along Ω, and ˆn is the outside unit normal to Ω. There are two implicit assumptions used above 1. The flux of stuff is local: stuff does not vanish somewhere and re-appears elsewhere (this would not violate conservation). For most types of physical stuff this is reasonable. But one can think of situations where this is not true e.g.: when you wire money, it disappears in your local bank account, and reappears elsewhere, and in various bank accounts (fee s). 2. The flux is given by a vector. But, why should this be so? Ω 1

2 2 Given item 1, the flux should be a scalar function of position, time, and direction q = q( x, t, ˆn), (1.2) where q is the amount of stuff, per unit time and per unit length, crossing a curve 1 with unit normal ˆn, from one side to the other (with direction 2 given by ˆn). Then (1.1) takes the form ρ t dx 1 dx 2 = q( x, t, ˆn) ds. (1.3) Ω However, in this form we cannot use Gauss theorem to transform the integral on the right over Ω, into one over Ω. This is a serious problem, for this is the crucial step in reducing (1.1) to a pde. Your task: Show that, provided that ρ and q are smooth enough (e.g.: continuous partial derivatives), equation (1.3) can be used to show that q must have the form q = ˆn q, (1.4) for some vector valued function q( x, t). Hints. A. It should be obvious that the flux going across any curve (surface in 3D) from one side to the other should be equal and of opposite sign to the flux in the opposite direction. That is, q in (1.2) satisfies Ω q( x, t, ˆn) = q( x, t, ˆn). (1.5) Violation of this would result in the conserved stuff accumulating (or being depleted) at a finite rate from a region with zero area (zero volume in 3D), which is not compatible with the assumption that ρ t is continuous and equation (1.3). Note that there are situations where it is reasonable to make models where conserved stuff can have a finite density on curves or surfaces (e.g.: surfactants at the interface between two liquids, surface electric charge, etc.). Dealing with situations like this requires a slightly generalized version of the ideas behind (1.5). B. Given an arbitrary small curve segment 3 of length h > 0 and unit normal ˆn, realize it as the hypotenuse of a right triangle where the other sides are parallel to the coordinate axes. Then write (1.3) for the triangle, divide the result by h, and take the limit h 0. Note that, if the segment of length h is parallel to one of the coordinate axis, then one of the sides of the triangle has zero length, and the triangle has zero area but the argument still works, albeit trivially (it reduces to the argument in A). 2 Forced-damped nonlinear oscillator response to a sinusoidal force Statement: Forced-damped nonlinear oscillator response to a sinusoidal force Consider a damped mass-spring system, with a nonlinear spring, to which we apply a sinusoidal force. If we assume that the forcing is weak, so that the amplitude of the oscillations is small, the governing equations can be written in the following a-dimensional form ẍ + d ẋ + x + ɛ ν x 3 = α cos(ω t), (2.1) 1 In 3D:... per unit area, crossing a surface... 2 A positive q means that the net flow is in the direction of ˆn. 3 Which you can assume is straight, since a limit h 0 will occur.

3 3 where d > 0, α, and ω are constants; ν = ±1; and 0 < ɛ 1. Note: Here we have assumed that the spring force is symmetric (same for extension and compression), and kept the first nonlinear correction only. Note that ν = 1 corresponds to a hard spring, while ν = 1 is a soft spring. Of course, for the approximation above to make sense, the forcing has to be weak. However, how weak depends on the dissipation coefficient d and the forcing frequency ω. Here we will consider the specific case where d = ɛ, α = 2 ɛ, and the frequency is near resonance, 4 that is: ω = 1 + µ ɛ, where µ is an O(1) constant. Thus the equation becomes ẍ + ɛ ẋ + x + ɛ ν x 3 = 2 ɛ cos(ω t), where ω = 1 + ɛ µ. (2.2) This equation has periodic solutions, with period 2 π/ω. The question is: what is their amplitude, versus µ? Your task: Use a perturbation expansion to (approximately) calculate the periodic solutions to (2.2) a leading order approximation is enough. Then plot their amplitude as a function of µ. Hint 2.1 A variation of the Poincaré-Lindstedt-Stokes method should work, since you are looking for solutions which are 2π-periodic in the variable θ = ω t. However, here you already know ω, and the amplitude is the only unknown: the leading order in the expansion x x 0 + ɛ x will yield x 0 = A e i θ + c.c., and you need to find A. The equation for A is a nasty cubic equation involving both A and κ = A 2. Do not try to solve it analytically. Instead: (1) Extract from it an equation for κ alone, and note that once you know κ, there is an explicit expression for A. (2) Plot the solutions in the κ-ν plane by noticing that you can introduce a parameter ζ in terms of which simple, explicit, expressions giving κ = κ(ζ) and µ = µ(ζ) exist. 3 Pendulum with torsion Statement: Pendulum with torsion Consider an horizontal axle A, of length l, with a rod R of length L attached (perpendicularly) at its midpoint. Let a mass M be attached to the other end of the rod R see figure 3.1. Assume that: Figure 3.1: Problem: Pendulum with torsion. Left: side view of the axle and rod with a mass device in this problem. Right: view along the axle A. The rod is constrained to move in the plane perpendicular to the axle, which can only rotate. Hence the only dynamical variable is the angle θ. 1. The axle A is free to rotate, and we can ignore any frictional forces (i.e.: they are small). 4 Away from resonance the behavior is not particularly interesting.

4 4 2. Any deformations to the axle and rod shapes are small enough that we can ignore them. They are both straight cylinders, and they remain so. 3. The masses of both the rod and the axle are small compared to M, so we ignore them (this assumption is not strictly needed, but we make it to make matters simple). In this case the mass M is restricted to move on a circle in a vertical plane, perpendicular to the axle. We can describe the state of the system at any time by the angle θ that the rod R makes with a vertical line pointing down. This angle satisfies the pendulum equation d 2 θ dt 2 = g sin θ, (3.1) L where g is the acceleration of gravity and t is time. Imagine now that the axle A is not free to rotate, but in fact it is fixed at one (or both) ends, so that it opposes the motion of the rod by elastic torsional forces. Assume that, when the rod is pointing down, these forces vanish (i.e.: the axle has no twist). Then, in the case where the angle θ is small enough that the torsional forces are proportional to the angle of twist (Hooke s law), derive the equation that θ must satisfy in the two cases: (1) one end of the axle is fixed, and the other is free to rotate (without friction), or (2) both ends are fixed. Hint 3.2 In the Hooke s law regime, for a given fixed bar, the torque generated is directly proportional to the angle of twist, and inversely proportional to the distance over which the twist occurs. To be specific: in the problem here, imagine that a section of length l of the axle has been twisted by an amount (angle) Ψ. Then, if T is the torque generated by this twist, we have T = κ Ψ l, where κ is a constant that depends on the axle material and the area of its cross-section (assuming that the axle is an homogeneous cylinder). What are the dimensions of κ? This torque translates on a tangential force of magnitude F = T /L on the mass M at the end of the rod R of length L. The sign of the force is such that it acts to undo the twist. Finally, note that in this problem, when the rod is at an angle θ from it s vertical reference position, a twist of magnitude θ will be generated on any side where the axle ends are tied. 4 Two torsion coupled pendulums Statement: Two torsion coupled pendulums A variation on the problem Pendulum with torsion is as follows: 1. The axle has two equal rods attached to it (equally spaced along the axle, so that they are a distance l/3 from each other and from the ends). 2. Each rod has a mass 1 M at its end The axle is free to rotate without friction. 4. The axle has no twist when both rods are pointing down.

5 5 Write the equations of motion for this system. Note that the state of the system is described by two angles, and that when writing the equations of motion you have to consider the torsional forces that are generated when the two angles are not equal. Assume that the angle differences are small, so Hooke s law applies and the torsional forces are proportional to the angle difference. 5 N torsion coupled pendulums, and continuum limit as N Statement: N torsion coupled pendulums, and continuum limit as N Generalize the result of the problem Two torsion coupled pendulums to the case where there are N equal rods attached to the axle (equally spaced along it, so that they are a distance l/(n + 1) from each other and from the axle s ends), each with a mass M/N at its end. Let x be the length coordinate along the axle. Label the rods (starting from one end of the axle, at which we set x = 0) by the integers n = 1,..., N. Then the n th rod corresponds to the position x = x n = n l along the axis, N + 1 and it is characterized by the angle θ n = θ n (t). Consider now the limit N. Look at solutions for which you can write θ n (t) = θ(x n, t), where θ = θ(x, t) is a nice function (i.e.: θ has as many derivatives as you need, so that Taylor expansions are valid). Derive a P.D.E. 5 for the function θ = θ(x, t). Hint 5.3 The continuum (N ) limit here follows the same process as the derivation of the wave equation for the continuum limit of a set of masses connected by springs. The masses at the rods ends scale with N in such a way that the mass per unit length (density) ρ = M/l is a constant. See the hint in the problem Pendulum with torsion for the appropriate way in which the torsional force between any two neighboring rods scales with N. 6 Relaxation oscillations while pulling a heavy load Statement: Relaxation oscillations while pulling a heavy load Consider the situation where a heavy load is being pulled at constant speed U, over a not-very-smooth surface, using an elastic band e.g.: see the left panel 6 in figure 6.1. Relative to a coordinate frame moving with the puller, this situation can be (roughly) modeled by the equation: M ẍ F (ẋ U) + k x = 0, (6.1) where M is the mass of the load, x is the position of the load, F is the friction force, and k is the elastic constant for the traction band the coordinate system origin located so that x = 0 corresponds to the band/spring at equilibrium. We assume that the friction force is given by F (u) = a M f(u/v ), (6.2) where u is the load velocity relative to the surface, a is a constant [with units of acceleration], V is another constant [a velocity], and f is an a-dimensional function with a graph like the one shown on the right panel of figure The equation you will obtain is known as the Sine-Gordon equation. 6 The constant speed U assumption is not very realistic when the pulling is being done by a human being. On the other hand, when machines (e.g.: tractors) are used to pull loads (e.g.: logs) over rough surfaces, typically chains are used. This problem is heavily idealized.

6 6 Friction motivated force function 1 f v Figure 6.1: Problem: Relaxation oscillations while pulling a heavy load. Left: a picture from the web illustrating the set-up for this problem. Right: friction-like force. Notes: 1. Equation (6.2) incorporates the fact that the force depends on how strongly the surfaces in contact are pushed together. 2. Typically, at zero velocity, frictional forces balance the load applied up to some threshold value [given here by am]. Beyond the threshold motion starts, and the force drops as the velocity increases, till a minimum is reached at a critical velocity [given here by V ]. Beyond this minimum, the force grows again. 3. Normally the regime where friction can balance any load applied below some threshold is modeled via a discontinuity. Here, to simplify the mathematics, we have replaced the discontinuity by a very steep region for very small velocities. Your task: Under appropriate conditions the nonlinear oscillator above gives rise to relaxation oscillations corresponding to the stick-and-shift behavior that is often observed when heavy loads are pushed/pulled across rough surfaces. Find this regime and describe the resulting relaxation oscillations. Hint 6.4 First: note that the equations have a unique equilibrium solution, given by ẋ = 0 and k x 0 = F ( U). Note also that two qualitatively different cases are possible: U > V and U < V (the phenomena occurs for only one of them). 7 Next: (1) Change coordinates so the load position is measured relative to x 0, not the spring equilibrium point used in (6.1). (2) Write the equations in non-dimensional form. Use V for velocity and L for distance where L is the length scale that characterizes the ratio of the friction force to the elastic force. Of course, the time scale is then given by T = L/V. At this point the situation will look remarkably similar to that of the van der Pol oscillator (after an appropriate time scale is introduced, and the transformation x = ż is performed). That is, to the equation: z µ 2 (ż 1 3 ż3 z) = 0. Now you should know how to proceed forward, and identify what is needed for relaxation oscillations to arise. Further question: Suppose that you are in the parameter regime where oscillations can arise. What happens as U crosses from U > V to U < V? The situation here is a bit tricky, and getting all the details right is hard, but you should be able to identify the main features. In particular, (i) What does linearized theory say about the behavior near the equilibrium point as U crosses V? (ii) What happens when you analyze the behavior near the equilibrium point, when U = V, using the leading order nonlinear correction? THE END 7 At equilibrium the load moves at constant velocity U relative to the surface. This is the ideal situation to have when pulling a load.