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 Monday March 12, Turn it in (by 3PM) at the Math. Problem Set Boxes, right outside room There is a box/slot there for 306. Be careful to use the right box (there are many slots). Contents 1 Nonlinear solvable ODEs Statement: A nonlinear solvable ode #02 Equation: 0 = x + (1 y)/y + (x y/y ) Linear 1st order PDE (problem 04) 2 3 Linear 1st order PDE (problem 08) 2 Solve (x y) u x + (x + y) u y = x 2 + y 2 using characteristics Semi-Linear 1st order PDE (problem 04) 2 5 Fundamental Diagram of Traffic Flow #01 3 The simplest velocity-density relationship Fundamental Diagram of Traffic Flow #03 3 Traffic flow as prescribed by state laws Envelope of the characteristics and cusp 4 Show that the envelope of the characteristics has a cusp at the breaking time/location Small vibrations of a 2D string under tension (sv2ds) Introduction: Small vibrations of a 2D string under tension Statement: sv2ds01 (no other forces/effects) Statement: sv2ds02 (vibrations in a viscous fluid) Nonlinear solvable ODEs 1.1 Statement: A nonlinear solvable ode #02 Generally nonlinear ode, even simple scalar and first order ones, do not admit explicit exact solutions. There are, however, some exceptions. For example, consider the ode 0 = x + 1 y ( y + x y ) 2 y, (1.1) where y = dy. This ode is nonlinear with variable coefficients, and not separable. Yet it can be solved exactly, even dx though the standard methods do not apply. Your task here is to do this. Hint. Introduce the variables, u = 1 y and v = x y = x y u, (1.2) y 1

2 2 and investigate what equation (1.1) implies for them. 2 Linear 1st order PDE (problem 04) Statement: Linear 1st order PDE (problem 04) Discuss the two problems u x + 2 x u y = y, with { (a) u(x, x 2 ) = 1 for 1 < x < 1, (b) u(x, x 2 ) = 1 3 x3 + π for 1 < x < 1. (2.1) How many solutions exist in each case? Note: the data in these problems is prescribed along a characteristic! 3 Linear 1st order PDE (problem 08) Statement: Linear 1st order PDE (problem 08) Solve the problem below, using the method of characteristics: (a) Compute the characteristics, as done in the lectures, starting from each point in the data set. (b) Next solve for the solution u along each characteristic. (c) Finally, eliminate the characteristic variables ζ and s from the expression 1 for u obtained in step (b) using the result in step (a) to obtain the solution as a function of x and y. with data u(x, 0) = (1/2) x 2 for 1 x < exp(2π). (x y) u x + (x + y) u y = x 2 + y 2, (3.1) Answer this question: Where in the (x, y) plane does the problem above define the solution u? That is: what is the region of the plane characterized by the property that: through each point in this region there is exactly one characteristic connecting it with the curve where the data is given? 4 Semi-Linear 1st order PDE (problem 04) Statement: Semi-Linear 1st order PDE (problem 04) Consider the following problem, in the region x, t > 0, u t x 1 + t u x = u 2, with data u(x, 0) = x and u(0, t) = t. (4.1) 1. Use the method of characteristics to solve the problem. Write explicit formula(s) for u for all the values x, t > 0 for which the solution is defined. Describe the domain of definition Ω. 1 Here ζ is the label for each characteristic, and s is a parameter along the characteristics.

3 3 2. Use the result in 1 to show that the solution is smooth (infinite partial derivatives) in the region Ω, except along a special curve Γ describe Γ. 3. Is the solution continuous for points on Γ? What happens with u t and u x for points on Γ? 4. For the solution u = u(x, t) found in item 1, does the pde 2 in (4.1) make sense for points on Γ? In other words, does u satisfy the pde everywhere in x, t > 0? 5 Fundamental Diagram of Traffic Flow #01 Statement: Fundamental Diagram of Traffic Flow #01 The desired car velocity u = U(ρ) has its maximum, u m, at ρ = 0, and vanishes at the jamming density, ρ J. Assuming that U is a linear function of ρ, write a formula for the flow rate q = Q(ρ). What is the road capacity q m? What is the wave velocity c = c(ρ)? Note: The road capacity is the maximum flow rate that a road allows. 6 Fundamental Diagram of Traffic Flow #03 Statement: Fundamental Diagram of Traffic Flow #03 Many state laws state that: for each 10 mph (16 kph) of speed you should stay at least one car length behind the car in front. Assuming that people obey this law literally (i.e. they use exactly one car length), determine the density of cars as a function of speed (assume that the average length of a car is 16 ft (5 m)). There is another law that gives a maximum speed limit (assume that this is 50 mph (80 kph)). Find the flow of cars as a function of density, q = q(ρ), that results from these two laws. The state laws on following distances stated in the prior paragraph were developed in order to prescribe a spacing between cars such that rear-end collisions could be avoided, as follows: a. Assume that a car stops instantaneously. How far would the car following it travel if moving at u mph and a1. The driver s reaction time is τ, and a2. After a delay τ, the car slows down at a constant maximum deceleration α. b. The calculation in part a may seem somewhat conservative, since cars rarely stop instantaneously. Instead, assume that the first car also decelerates at the same maximum rate α, but the driver in the following car still takes a time τ to react. How far back does a car have to be, traveling at u mph, in order to prevent a rear-end collision? c. Show that the law described in the first paragraph of this problem corresponds to part b, if the human reaction time is about 1 sec. and the length of a car is about 16 ft (5 m). Note: What part c is asking you to do is to justify/derive the state law prescription, using the calculations in part b to arrive at the minimum car-to-car separation needed to avoid a collision when the cars are forced to brake. 2 That is: u t + 1+x 1+t ux = u2.

4 4 7 Envelope of the characteristics and cusp Statement: Envelope of the characteristics and cusp Consider the problem c t + c c x = 0, where c(x, 0) = C(x), for < x <. (7.1) e1. Write the characteristic curves for this problem, each one parameterized by time, and labeled by the value of x = s at time t = 0. That is, write formulas for the characteristics of the form x = X(s, t). e2. Write the equation for the envelope of the characteristics, and express the envelope in parametric form x = X e (s) and t = T e (s). e3. Assume that C has an inflection point at x = 0. In fact, assume that where the primes denote derivatives. C(0) = 0, C (0) = a < 0, C (0) = 0, and C (0) = 2 a b > 0, (7.2) Let x c = X e (0) and t c = T e (0). Show that T e has a local minimum at s = 0, and that the envelope has a cusp at (x c, t c ). HINT: Expand the equations for s small. e4. Replace (7.1) by c t + c c x = c, where c(x, 0) = C(x), for < x <. (7.3) Repeat steps e1 and e2 for this problem. Question: what condition is needed on C so that the characteristics of (7.3) actually have an envelope in the upper half space-time plane? That is, so that T e (s) > 0 somewhere. Note: Assume a one parameter set of curves in the plane, defined by an equation of the form F (x, t, s) = 0, where s is the parameter. Then the envelope is the set of points in the plane which are the intersections of two infinitesimally close curves. That is, points that satisfy both F (x, t, s) = 0 and F (x, t, s + ds) = 0. Expanding the second equation we see that the envelope is defined by F (x, t, s) = 0 and F s(x, t, s) = 0. [A] Given the definition above of the envelope, it should be clear that the envelope of characteristics is precisely the place where the derivatives of the solution by characteristics of (7.1) become infinity: at t = 0, dx = ds, and there is some dc given by the initial conditions that result in a finite dc. However, at the envelope dc is still the same, while dx = 0; hence cx = ±. dx An alternative definition of the envelope is: it is a curve x = X(s) and y = Y (s) such that the point (X(s), Y (s)) in the curve is also in the curve with label s belonging to the family; and both curves are tangent there. 3 neighboring characteristics that intersect at the envelope, are also tangent to the envelope there. What this means is that the 8 Small vibrations of a 2D string under tension (sv2ds) 8.1 Introduction: Small vibrations of a 2D string under tension Here you are asked to derive equations for the transversal vibrations of a thin elastic string under tension, under several scenarios. The following hypotheses are common to all of them (8.1) 1. The string is homogeneous, with mass density (mass per unit length) ρ = constant. 2. The motion is restricted to the x-y plane. At equilibrium the string is described by y = 0 and 0 x L = string length. The tension T > 0 is then constant. See remark 8.1, # You can check that this also leads to the equations in [A].

5 5 3. The string has no bending strength. 4. The amplitude of the vibrations is very small compared with their wavelength, and any longitudinal motion can be neglected. Thus: The string can be described in terms of a deformation function u = u(x, t), such that the equation for the curve describing the string is 4 y = u(x, t). The tension remains constant throughout see remark 8.1, #4. Remark 8.1 Some details: #1 We idealize the string as a curve. This is justified as long as λ d, where λ = scale over which motion occurs, and d = string diameter. This condition justifies item 3 as well. #2 The tension is generated by the elastic forces (assume that the string is stretched). At a point along the string, the tension is the force with which each side (to the right or left of the point) pulls on the other side, 5 and it is directed along the direction tangent to the string (there are no normal forces, see item 3). #3 At equilibrium the tension must be constant. For imagine that the tension is different at two points a < b along the string. Then the segment a x b would receive a net horizontal force (the difference in the tension values), and thus could not be at equilibrium. #4 The hypotheses imply that string length changes can be neglected. Thus the stretching generating the tension does not change significantly, and the tension remains equal to the equilibrium tension T. 8.2 Statement: sv2ds01 (no other forces/effects) Here we consider the simplest situation, where there is no other force beside the tension T on the string, nor any dissipation. Under these conditions, plus those in 8.1: Use the conservation of the transversal momentum to derive an equation for u. Thus: i) Obtain a formula for the transversal momentum density along the string (momentum per unit length). ii) Obtain a formula for the transversal momentum flux along the string (momentum per unit time). iii) Use the differential form of the conservation of transversal momentum to write a pde for u. Hint. Remember that internal string forces 6 are momentum flux! Since there is no longitudinal motion, momentum can flow only via forces. Be careful with the sign here! Remember also that u x is very small. The pde for u derived above applies for 0 < x < L. For a solution to this equation to be determined, extra conditions on the solution u at each end are needed. These are called boundary conditions, and must be derived/modeled as well. Later on we will see that exactly one boundary condition (a restriction on u and its derivatives) is needed at each end, x = 0 and x = L. Answer the following extra questions: iv) What boundary condition applies at an end where the string is tied? (e.g., as in a guitar). v) What boundary condition applies at an end where the string is free? For example, imagine a no-mass cart that can slide up and down (without friction) a vertical rod at x = 0, and tie the left end of the string to this cart. This keeps the left end of the string at x = 0, with the rod canceling the horizontal component of the tension force, but u(0, t) can change. Note that a no-mass-no-friction cart is an idealization, approximating the set-up in (vi) in the limit where the mass and the friction are small. vi) What boundary condition applies at an end where there is a mass (M) attached to the string, with a force f = b f v m opposing the mass motion (here b f = constant and v m = mass velocity). An example of this would be the cart in (v), but neglecting neither the cart mass, nor the friction as it moves up or down along the vertical rod. 4 In this approximation the x-coordinate of any mass point on the string does not change in time. 5 If you were to cut the string, this would be the force needed to keep the lips of the cut from separating. 6 Internal forces = forces by one part of the string on another.

6 6 Warning: the form of boundary condition in this case depends on which end of the string the bead is on. This is because the force by the string tension on the bead has a sign change (why?) vii) Let λ be a typical wavelength for the string vibrations. Then τ = λ/c, where c = T/ρ, is the typical time scale (why?). Under which conditions on ρ, T, M, b f, and λ is the free end boundary condition in (v) a good approximation to the boundary condition in (vi)? Hint. Let a denote a typical amplitude for the string vibrations. Then we expect that u t = O(a/τ), u x = O(a/λ), and so on. Use this to compare the size of the terms in the boundary condition derived in (vi). From this you should be able to obtain two a-dimensional numbers that must be small for the boundary condition in (vi) to reduce to the one in (v). One of them measures the ratio of the bead inertial forces to the string tension, and the other the ratio of the friction forces to the string tension. 8.3 Statement: sv2ds02 (vibrations in a viscous fluid) Here we consider the simple situation, where (on the string) there is no other force beside the tension T, nor any dissipation. However, the string is vibrating in a viscous fluid, which produces a force (per unit length) opposing to the direction of motion. Assume that this force per unit length has the simple form ν f u t, (8.2) where ν f > 0 is a constant and u t is the string velocity. Under these conditions, plus those in 8.1: Task #1. Use the conservation of the transversal momentum to derive an equation for u. Thus: i) Obtain a formula for the transversal momentum density along the string (momentum per unit length). ii) Obtain a formula for the transversal momentum flux along the string (momentum per unit time). iii) Use the differential form of the conservation of transversal momentum to write a pde for u. Note that there is a momentum source as well. Hint. Remember that internal string forces 7 are momentum flux! Since there is no longitudinal motion, momentum can flow only via forces. Be careful with the sign here! Remember also that u x is very small. Task #2. Use the conservation of energy to derive another equation for u. Furthermore, show that the solutions to the equation derived in task #1 satisfy the conservation of energy equation. Hint. The energy density is the sum of the kinetic energy per unit length (e k ), and the elastic energy per unit length stored in the string deformation (e d ). You will need an expression for e d. In fact, you only need to compute the difference between the elastic energy, and some constant energy specifically: the elastic energy of the string at equilibrium. But the tension is constant (see 8.1) over the stretching process taking the string from equilibrium y = 0 to y = u(x, t). Thus the elastic energy stored is the product of T times the change in length of each segment dx. All you need now is the formula for the change in length of any interval dx (actually, just the leading order approximation to this, in the u x small limit). This yields e d = T ( 1 + u 2 x 1) = 1 2 T u2 x, since the arc-length is ds = 1 + u 2 x dx, and u x is small. Task #3. Assuming that the string is tied at both ends u(0, t) = u(l, t) = 0, write an equation for the evolution of the total energy. That is: de dt =?? THE END 7 Internal forces = forces by one part of the string on another.