2 4 GBNS (Good and Bad Numerical Schemes.) 15 Introduction GBNS 01 Scheme A. Forward differences for

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1 Problem List for for , MIT. Rodolfo R. Rosales Λ. May 4, 2001 New problems will/may be added to this list at random times. Check (always) the list date to make sure you have the latest version. Contents 1 MeVi (Mechanical Vibrations) problems Problem MeVi 01. Experiments with slinkies Problem MeVi 02. Dimensional analysis Problem MeVi 03. Pendulum with torsion Problem MeVi 04. Two torsion coupled pendulums Problem MeVi 05. N torsion coupled pendulums. Continuum limit as N! ExPD (Exercises in Partial Differentiation) problems Problem ExPD 01. General solution of the 1D wave equation Problem ExPD 02. Initial value problem for the 1D wave equation Problem ExPD 03. One directional wave equation Problem ExPD 04. Laplace's equation in 2D Problem ExPD 05. Kinematic transport equation in 1D Problem ExPD 06. Nonlinear plane wave profile behavior Problem ExPD 07. Check 1D wave equation solution Problem ExPD 08. Check one directional wave equation solution SEPARATION OF VARIABLES Problem ExPD 20. Separation of Variables for the 1D Wave Equation ExID (Exercises in Implicit Differentiation) problems ExID 01. Simple single variable y = y(x) dependence Λ MIT, Department of Mathematics, room 2-337, Cambridge, MA

2 2 4 GBNS (Good and Bad Numerical Schemes.) 15 Introduction GBNS 01 Scheme A. Forward differences for u t + u x = GBNS 02 Scheme B. Backward differences for u t + u x = GBNS 03 Scheme C. Centered differences for u t + u x = GBNS 04 Scheme D. Centered differences for u t = u xx GBNS 05 Scheme E. Friedrichs centered differences scheme for u t + u x = AENS (Associated Equation to a Numerical Scheme.) 18 Introduction AENS 01 Scheme A. Forward differences for u t + u x = AENS 02 Scheme B. Backward differences for u t + u x = AENS 03 Scheme C. Centered differences for u t + u x = Answer to: AENS 03 Scheme C. Centered differences for u t + u x = AENS 04 Scheme D. Centered differences for u t = u xx Answer to: AENS 04 Scheme D. Centered differences for u t = u xx AENS 05 Scheme E. Friedrichs centered differences scheme for u t + u x = vnsa (von Neumann Stability Analysis.) 26 Introduction vnsa 01 Scheme A. Forward differences for u t + u x = vnsa 02 Scheme B. Backward differences for u t + u x = vnsa 03 Scheme C. Centered differences for u t + u x = Answer to: vnsa 03 Scheme C. Centered differences for u t + u x = vnsa 04 Scheme D. Centered differences for u t = u xx Answer to: vnsa 04 Scheme D. Centered differences for u t = u xx vnsa 05 Scheme E. Friedrichs centered differences scheme for u t + u x = FouS (Computer Exercise in Fourier Series.) Introduction The Problem

3 3 8 CHRO (Chromatography.) Introduction The problem DWMo (Dispersive Waves and Modulation.) Introduction The problem EIKO (Eikonal Equation.) Introduction The problem. Part I The problem. Part II TFPa (Traffic Flow Problems: Part A.) Problem TFPa 01. Traffic data and processing Problem TFPa 02. Near constant initial data: Linearize Problem TFPa 03. Near constant initial data: Nonlinear solution Problem TFPa 04. Translations of a function Problem TFPa 05. Wave number and frequency units Problem TFPa 06. Solve a first order linear P.D.E Problem TFPa 07. Solve a first order linear P.D.E Problem TFPa 08. The wave equation Problem TFPa 09. Traffic flow signaling problem Problem TFPa 10. Traffic light red to green Problem TFPa 11. Last car to make green light Problem TFPa 12. Find shock locus Problem TFPa 13. Shock formation time Problem TFPa 14. Solve using dimensionless variables Problem TFPa 15. Differential equation for shock Problem TFPa 16. Equation for an envelope Problem TFPa 17. Shock interaction with a traffic light

4 Problem TFPa 18. Envelopes for families of curves Problem TFPa 19. Quasilinear equation and characteristics Problem TFPa 20. Quasilinear equation and characteristics Problem TFPa 21. Characteristics in 3 D Problem TFPa 22. Two space D and time characteristics TFPb (Traffic Flow Problems: Part B.) Problem TFPb 01. Flood waves Problem TFPb 02. Wave equation Problem TFPb 03. Kinematic equation solution Problem TFPb 04. Kinematic equation solution (example.) Problem TFPb 05. Traffic flow linearization failure Problem TFPb 06. Damped kinematic equation Problem TFPb 07. Shocks in damped kinematic equation Problem TFPb 08. Envelope of a family of straight lines Problem TFPb 09. Quasi-linear equation solution Problem TFPb 10. Quasi-linear equation solution Problem TFPb 11. Check solution by implicit differentiation Problem TFPb 12. Check red light turns green solution directly Problem TFPb 13. Expansion fans for quadratic flows Problem TFPb 14. Check if discontinuities are allowed shocks Problem TFPb 15. Traffic lights at the ends of a tunnel Problem TFPb 16. Linear Traffic Flow Problem TFPb 17. "Regulated entrance exit" tunnel Problem TFPb 18. Breaking times Problem TFPb 19. Envelope of Characteristics Problem TFPb 20. An initial value problem Problem TFPb 21. Damped equations

5 MIT. (R. R. Rosales). MeVi (Mechanical Vibrations) problems. 5 List of Figures 1 Problem TFPa 10: Typical density flow relationship for traffic flow Problem TFPa 17: Shock in traffic density Problem TFPb 01: Typical river bed Problem TFPb 05: Typical density flow relationship for traffic flow MeVi (Mechanical Vibrations) problems. 1.1 Problem MeVi 01. Experiments with slinkies. Consider a homogeneous cylindrical rod (made of an elastic material), subject to (small amplitude) longitudinal deformations. Let x be the length coordinate (measured along the axis of the cylinder) when the cylinder is in its relaxed position. Use x as a label for the mass elements in the cylinder. 1 For every mass element x, let u = u(x; t) be its position at time t, measured along the axis of the cylinder (note that u = x corresponds to the cylinder at rest.) Then u describes the state of the cylinder at any time and obeys the wave equation: u tt c 2 u xx =0; where c = q k=ρ : (1.1.1) Here ρ is the density (mass per unit length) of the rod, and k characterizes the elastic response of the material: if we stretch the cylinder by an amount L, then the elastic force is k L L, where L is the length of the rod (note that k has the dimensions of a force, thus c is a speed.) Remark The basic assumption here is that the cylinder remains at all times within the regime of applicability of Hooke's law. This means that the deformation (given by u x ) is small enough everywhere. In particular, this also implies that variations in the cross section of of the cylinder can be ignored (e.g.: if volume is preserved, the cross section will be larger in regions under compression than in those under tension.) 1 Since we are considering only longitudinal motions, points in a cross section move inunison and we need not label them separately.

6 MIT. (R. R. Rosales). MeVi (Mechanical Vibrations) problems. 6 In the derivation of equation (1.1.1) it is assumed that the elastic forces are dominant, so that other forces (e.g.: gravity) can be ignored. For a rod with a vertical orientation, such that the elastic forces are not dominant over gravity, equation (1.1.1) must be modified to: u tt c 2 u xx = g; (1.1.2) where g is the acceleration of gravity, and we assume that the vertical coordinate x increases upwards. In particular, consider the case of a rod hanging vertically without any motion (i.e.: u = u(x), with no time dependence), and measure x from the bottom of the rod. Then: u =0 and u x =1 at x =0; (1.1.3) where the second equation follows because there is no force at the lower end (no section of the rod below that must be supported.) Then the equation for u = u(x), namely: c 2 u xx = g; can be integrated to yield: u = g 2c 2 x2 + x: (1.1.4) A particular example where this should apply to is that of a slinky. The object of this problem is for you to check how well a slinky obeys equation (1.1.4). Proceed as follows: 1. Get a slinky in good condition and draw a straight line along its edge, parallel to the slinky's axis. Draw the line so that, when it reaches one end (the bottom" end), it does so at the end of the coil that makes the slinky. 2. Starting from the bottom" end of the slinky, name the points at which each coil is marked by the line as n =0;1;::: Then (if w is the width of a coil) when the slinky is at rest, the position of the n th point isgiven by x n = nw : 3. Measure w by measuring the total length and dividing the result by the number of coils. You also can easily measure the density" ρ of the coil by weighting it and dividing the result by its length.

7 MIT. (R. R. Rosales). MeVi (Mechanical Vibrations) problems Hang the slinky in a vertical position (with the bottom end down) and wait till it is at rest. Then measure the distance u n of the n th point from the point n = 0 at the bottom. One way to do this is to tape a measuring tape to a wall, and then hang the slinky by the wall. 5. Equation (1.1.4) predicts that u n = g 2c 2 x2 n + x n = gρ 2k x2 n + x n : (1.1.5) 6. The question is now: How well does equation (1.1.5) match your measurements? Of course, you do not have k, but you will have several values of n. If (1.1.5) applies, then u n+1 u n = u n = gρ k w2 n + gρ 2k w2 + w: Thus a plot of u n+1 u n versus n should give a straight line of slope gρw 2 =k (note that you can measure u n directly.) From this you can get k, which is the hardest quantity to measure directly in this context. Next suspend the slinky from one end and set it to vibrate (longitudinally.) In this case, if we set the origin for the coordinate x at the top (where the slinky does not move), the governing equation will still be (1.1.2), but the boundary conditions are now: u(0; t)=0 and u x ( L; t) =1; (1.1.6) where L is the length of the slinky in its relaxed state. The first condition here follows because x = 0 corresponds to the clamped end at the top, while the second simply states that there are no elastic forces at the bottom end (same reason used when deriving the second condition in (1.1.3).) It is easy to see that these conditions (and the equation) are satisfied by the function u = a sin ß 2L x sin cß 2L t + g 2c 2 x gl c 2 x; (1.1.7) which oscillates with a period T = 4L c : Here a is an arbitrary constant. Now, continue with the experiment: 7. In the situation described in the paragraph above, measure the period of the slinky do not try to measure a single period, time several and then divide by the number of periods timed.

8 MIT. (R. R. Rosales). MeVi (Mechanical Vibrations) problems Compare the result of your measurement of the period with the one given by the formula for T above from the prior steps you should, by now, have avalue for c. 9. Discuss the results of your experiment. 1.2 Problem MeVi 02. Dimensional analysis. For the wave equation u tt c 2 u xx =0; (1.2.1) c>0 is the velocity along which disturbances propagate (say, as in the case of waves along a string under tension.) Later on, we will encounter the equation u tt + d 2 u xxxx =0; (1.2.2) where d is some constant. This equation also describes the propagation of waves, albeit not the same kind as (1.2.1). These are the questions now: 1. What are the dimensions of the constant d>0? 2. Consider now a wave of wavelength whose propagation is described by (1.2.2). Without actually solving the equation, in fact, without doing almost any calculation: give a formula for the speed c at which this wave propagates. The question here is: How does the speed c depend on? You should be able to answer this question with at most one constant you will not be able to determine. Hint For the second part, think in terms of dimensions: Whatever the speed is, the only dimensional quantities available to construct it are: d and. 1.3 Problem MeVi 03. Pendulum with torsion. Consider an horizontal axle A, of length `, with a rod R of length L attached (perpendicularly) at its midpoint. Assume that a mass M is attached to the other end of the rod R. Assume now that:

9 MIT. (R. R. Rosales). MeVi (Mechanical Vibrations) problems The axle A is free to rotate, and we can ignore any frictional forces (i.e.: they are small.) 2. Any deformations to the axle and rod shapes are small enough that we can ignore them. 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 that the rod R makes with a vertical line pointing down. This angle satisfies the pendulum equation where g is the acceleration of gravity and t is time. d 2 dt = g sin( ) (1.3.1) 2 L 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 is small enough that the torsional forces are proportional to the angle of twist (Hooke's law), derive the equation of motion for the angle in both cases (one end of the axle fixed, or both ends fixed.) Hint 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 ` of the axle has been twisted by an amount (angle) Ψ. Then, if T is the torque generated by this twist, one can write T =» Ψ ` ; where» is a constant that depends on the axle material and the area of its cross-section (assuming the axle is an homogeneous cylinder.) What are the dimensions of»? This torque then translates on a tangential force of magnitude F = T=L on the mass M at the end of the rod R of length L (be careful with the signs, remember that the torque will act so as to try to undo the twist.) Finally, note that in this problem, when the rod is at an angle from it's vertical reference position, a twist of magnitude will be generated on any side where the axle ends are tied.

10 MIT. (R. R. Rosales). MeVi (Mechanical Vibrations) problems Problem MeVi 04. Two torsion coupled pendulums. A variation on problem 1.3 is as follows: 1. The axle has two equal rods attached to it (equally spaced along the axle, so they are a distance `=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. 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. 1.5 Problem MeVi 05. N torsion coupled pendulums. Continuum limit as N! 1. Generalize the result of problem 1.4 to the case when there are N equal rods attached (equally spaced) to the axle, each with a mass 1 M at its end. Notice then that any two neighboring rods N ` will be separated by a distance along the axle. N +1 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)bythe integers n =1;:::;N. Then the n th rod corresponds to the position x = x n = n N +1`along the axis, and it is characterized by the angle n = n (t). Consider now the limit N!1, in which you look at solutions for which you can write n (t) = (x n ;t), where = (x; t) is a nice" function (i.e.: has as many derivatives as you need, so that Taylor expansions are valid.) Derive a PDE 2 for the function = (x; t). 2 The equation you will obtain is known as the Sine-Gordon equation.

11 MIT. (R. R. Rosales). ExPD (Exercises in Partial Differentiation.) 11 Hint The continuum (N!1) limit here proceeds very much in the same way as the derivation of the wave equation (done in the lectures) for the continuum limit of a set of masses connected by springs. Notice that the masses at the ends of the rods scale in such a way (as N varies) that the mass per unit length (density) ρ is a constant in fact ρ = M=`: Similarly, hint gives the appropriate way in which the torsional force generated byeach segment of axle (between any two neighboring rods) scales as N becomes large. 2 ExPD (Exercises in Partial Differentiation) problems. 2.1 Problem ExPD 01. General solution of the 1D wave equation. Consider the wave equation for u = u(x; t), where c>0 is a constant: u tt c 2 u xx =0: (2.1.1) Introduce the new independent variables = x ct and ο = x + ct: Change variables to write the equation for u as a function of these new variables: u = u( ; ο). Using this transformed form of the equation, integrate it twice to show that it must be u = f( )+g(ο); (2.1.2) for some arbitrary functions f and g. This shows that any solution of the wave equation (2.1.1) must have the form u = f(x ct)+g(x+ct). 2.2 Problem ExPD 02. Initial value problem for the 1D wave equation. Using the result of problem 2.1, show that if initial values u(x; 0) = U(x) and u t (x; 0) = V (x) (2.2.1) are given for the wave equation, then the solution is u = 1 2 (U(x ct)+u(x+ct)) + 1 2c Z x+ct x ct V(s)ds : (2.2.2)

12 MIT. (R. R. Rosales). ExPD (Exercises in Partial Differentiation.) 12 In particular, show that if U and V are periodic functions of x (of period P ) then: u is periodic in x of period P and periodic in t of period T = P=c. Use the GBNS GScheme MatLab script in the Athena Toolkit and check how closely the numerical calculation reproduces the periodicity in time (for initial conditions where, say, U is periodic of period 2 and V 0.) Note that in GBNS GScheme c = Problem ExPD 03. One directional wave equation. Consider the one directional wave equation for u = u(x; t), where c is a constant: u t + cu x =0: (2.3.1) Introduce the new independent variables = x ct and ο = x + ct: Change variables to write the equation for u as a function of these new variables: u = u( ; ο). Using this transformed form of the equation, integrate it to show that it must be u = f( ) ; (2.3.2) for some arbitrary function f. This shows that any solution of equation (2.3.1) must have the form u = f(x ct). 2.4 Problem ExPD 04. Laplace's equation in 2D. By direct substitution, show that u = f(x iy)+g(x+iy) ; (2.4.1) (where f and g are arbitrary" functions) is a solution of Laplace's equation: u xx + u yy =0: (2.4.2)

13 MIT. (R. R. Rosales). ExPD (Exercises in Partial Differentiation.) Problem ExPD 05. Kinematic transport equation in 1D. By direct substitution, show that u = f(x V (u) t) ; (2.5.1) (where f is an arbitrary function and V = V (u) isagiven function) is a solution of the equation: u t + V (u) u x =0: (2.5.2) Notice that: 1. Equation (2.5.1) defines u implicitly, so you will have to find u t and u x by implicit differentiation. 2. Equation (2.5.2) is nonlinear, except for the trivial case when V is a constant function in which case this problem reduces to problem Problem ExPD 06. Nonlinear plane wave profile behavior. The solution for equation (2.3.1) given by equation (2.3.2) as u = f(x ct) can be interpreted as representing a wave of fixed shape f, propagating at speed c. Can you give a similar interpretation for the solution of the kinematic transport equation in (2.5.1)? Note: this is rather non-trivial. 2.7 Problem ExPD 07. Check 1D wave equation solution. By direct substitution, show that u = f(x ct)+g(x+ct) (where f and g are arbitrary functions and c is a constant) is a solution of the 1D wave u = u: 2

14 MIT. (R. R. Rosales). ExID (Exercises in Implicit Differentiation.) Problem ExPD 08. Check one directional wave equation solution. By direct substitution, show that u = f(x ct) (where f is an arbitrary function and c is a constant) is a solution of u u =0: Separation of variables problems. 2.9 Problem ExPD 20. Separation of Variables for the 1D Wave Equation. Consider the wave equation u tt u xx =0: (2.9.1) Elsewhere it was shown that any solution of this equation can be written in the form: u = f(x t)+g(x+t); (2.9.2) where f and g are arbitrary functions. Find now ALL the separation of variables solutions to the wave equation (2.9.1), for the cases below (Note: the separation of variable solutions are those that have the form u = T (t)x(x), for some functions T = T (t) and X = X(x).) ffl Case A. The solution u = u(x; t) =T(t)X(x) is periodic in x, of period 2ß. ffl Case B. The solution u = u(x; t) =T(t)X(x)vanishes at x = 0, and at x = ß. This second case corresponds to the problem of a string of nondimensional length ß, tied at both ends. The solutions you will find are the standing wave modes for the wave equation. In each case, show that the solutions you have found have the form given in equation (2.9.2) above. In particular, this will show you how the standing wave solutions can be written as a combination of a left and right moving waves.

15 MIT. (R. R. Rosales). GBNS (Good and Bad Numerical Schemes.) 15 3 ExID (Exercises in Implicit Differentiation) problems. 3.1 ExID 01. Simple single variable y = y(x) dependence. In each case compute dy dx 1. x 3 + xy +2=0. 2. y = sin(y + x). 3. ln(y) =x. 4. cos 2 (y) =x, for x>0. as a function of y and x, given that y = y(x) satisfies: 5. y = f(c yx), where f is an arbitrary function and c is a constant. 6. y = f(x cy), where f is an arbitrary function and c is a constant. 4 GBNS (Good and Bad Numerical Schemes.) Experimenting with Numerical Schemes. Consider the numerical schemes that follow after this introduction, for the specified equations. Your task here is to experiment (numerically) with them so as to answer the question: Are they sensible? Specifically: ffl Which schemes give rise to the type of behavior illustrated by the bad" scheme in the GBNS lecture MatLab script ( Athena Toolkit.) ffl Which ones behave properly as x and t vanish? Show that they all arise from some approximation of the derivatives in the equations, similar to the approximations used to derive the good" and bad" schemes used in the GBNS lecture MatLab script ( Athena Toolkit.)

16 MIT. (R. R. Rosales). GBNS (Good and Bad Numerical Schemes.) 16 Remark Some of these schemes are good" and some are not. For the good" schemes you will find that as x gets small restrictions are needed on tto avoid bad behavior (fast growth of grid scale oscillations.) Specifically, note that in all the schemes a parameter appears, the parameter being = t= x in some cases and ν = t=( x) 2 in others. Then you will need to keep this parameter smaller than some constant (to get the "good" schemes to behave). That is: < c ; or ν <ν c : For the bad" schemes, it will not matter how small (or ν) is. Figuring out these constants is also part of the problem. Remark In order to do this problems you may have to write your own programs. If you choose to use MatLab for this purpose, there are several scripts in the Athena Toolkit that can easily be adapted for this. Copy them from the MenuMatlab subdirectory of the locker, and then modify them to be used in these problems. The relevant scripts are: ffl The schemes used by the GBNS lecture MatLab script in the Athena Toolkit are implemented in the script InitGBNS. ffl The series of scripts PS311 Scheme A, PS311 Scheme B, etc., has several examples of schemes already setup in an easy to use format. Some of the algorithms below are already implemented in these scripts. These scripts are written so that modifying them to use a different scheme involves editing only a few (clearly indicated) lines of the code. ffl Note that the MatLab scripts in the Athena Toolkit are written avoiding the use of for" loops and making use of the vector forms MatLab allows things run a lot faster this way. Do your programs this way too, it is good practice. Remark Do not put lots of graphs and numerical output in your answers. Explain what you did and how you arrived at your conclusions, and illustrate your points with an extremely few selected graphs. Remark In all cases we use the notation: x n = x0 + n x; t k = t0+k t; and u k n = u(x n;t k ):

17 MIT (R. R. Rosales.) AENS (Associated Equation to a Numerical Scheme.) GBNS 01 Scheme A. Forward differences for u t + u x = 0. Equation: u t + u x =0... Scheme: u k+1 n where = t x : 4.2 GBNS 02 Scheme B. Backward differences for u t + u x = 0. Equation: u t + u x =0... Scheme: u k+1 n where = t x : = u k n (u k n+1 u k n), = u k n (u k n u k n 1), 4.3 GBNS 03 Scheme C. Centered differences for u t + u x = 0. Equation: u t + u x =0... Scheme: u k+1 n where = t x : = u k n 1 2 (uk n+1 u k n 1). 4.4 GBNS 04 Scheme D. Centered differences for u t = u xx. Equation: u t = u xx... Scheme: u k+1 n where ν = t ( x) : GBNS 05 Scheme E. = u k n + ν (u k n+1 2u k n + u k n 1), Friedrichs centered differences scheme for u t + u x = 0. Equation: u t + u x =0... Scheme: u k+1 n = u k n 1 2 (uk n+1 u k n 1)+ 1 2 ffi 2 (u k n+1 2u k n + u k n 1). where = t x and ffi>0 is a parameter. This scheme is unstable for ffi<ffi c, for some value ffi c > 0. When ffi>ffi c, if < c (for some c = c (ffi)) the scheme is stable. Find ffi c and (for some value ffi>ffi c ), find c.

18 MIT (R. R. Rosales.) AENS (Associated Equation to a Numerical Scheme.) 18 5 AENS (Associated Equation to a Numerical Scheme.) The Associated Equation to a Numerical Scheme. Consider the numerical schemes that follow after this introduction, for the specified equations. Your task here is to find their associated equations. Then answer the questions: ffl Does the associated equation indicate stable or unstable behavior? ffl Are there any conditions that will suggest instability? ffl The associated equation to a numerical scheme does not provide an absolute" test of numerical stability, because it only describes the behavior of the long" waves in the numerical solution. It says nothing about the behavior of short, grid scale oscillations (more about this below), which can make a scheme behave very badly indeed. ffl On the other hand, the associated equation often tells us a lot about what is wrong with a numerical scheme, and provides good hints as to how a scheme can be fixed". ffl The main advantages of the associated equation approach over that of a von Neumann stability analysis are: A. It is much simpler to do. B. It is not restricted to linear constant coefficients equations. Of these, the second is by far the most important advantage. Next we carry out an example of finding the associated numerical equation, to illustrate the idea. We will pick the example of the bad" scheme used in the GBNS lecture MatLab script ( Athena Toolkit.) This scheme is supposed to solve the wave equation, written as the system of equations: u t = v; and v t = u xx ; (5.0.1) where the subscripts indicate partial derivatives and the equations have been nondimensionalized (so the wave speed satisfies c = 1.) A description of the scheme follows. Consider a uniform grid in space and time f(x n ; t k )g, with grid spacings x and t (assumed small".) That is x n+1 = x n + xand t k+1 = x k + t. On this grid we assume that the solution is

19 MIT (R. R. Rosales.) AENS (Associated Equation to a Numerical Scheme.) 19 approximated by the grid functions u k n and v k n. That is: u(x n ; t k ) ß u k n ; and v(x n; t k ) ß v k n ; (5.0.2) where the grid functions satisfy the following discretized version of equations (5.0.1) (the bad" numerical scheme): u k+1 n v k+1 n u k n t v k n t = v k n ; (5.0.3) = uk n+1 2u k n + u k n 1 ( x) 2 : (5.0.4) Remark The idea behind this is that, if u k n = U(x n ; t k ) and v k n = V (x n ; t k ); for some smooth" functions U = U(x; t) and V = V (x; t), then equations ( ) indicate 3 that U and V will (approximately) satisfy the wave equation system (5.0.1) with errors that vanish as t and x vanish. At least, this is the idea, but it does not always work. The fact that, upon substituting U and V as above (and expanding in Taylor series in the small quantities x and t), we obtain the equations we want to solve (plus small errors that vanish as x and t vanish) is only a necessary condition for the scheme to work (known as CONSISTENCY.) But it is not sufficient. To get an idea of what can go wrong, let us next calculate the leading order error terms in the Taylor expansions mentioned above in remark These will show that U and V above, if they existed, would have to satisfy a wave equation system (i.e.: (5.0.1)) with some small" correction terms, whose effect happens to be rather dramatic. This corrected system of equations below in ( ) constitutes the Associated Equations to the Numerical Scheme given by ( ). Remark Note that we say in the paragraph above if they existed," when referring to U and V. This is the key fact: the argument in remark to show that equations ( ) provide an approximation to the wave equation system (5.0.1), makes sense only if we have such smooth U and V. However, as we will show below using the associated equation approach, this is not true! Assuming that there are such smooth U and V leads to a contradiction. 3 This follows upon expanding u k+1 n = U(x n ;t k + t); u k n+1 = U(x n + x; t k ); etc., in a Taylor series centered at (x n ;t k ).

20 MIT (R. R. Rosales.) AENS (Associated Equation to a Numerical Scheme.) 20 Let us now proceed with the construction of the associated equations for the scheme ( ). As explained earlier, this begins by assuming that we can write u k n = U(x n ; t k ) and v k n = V (x n ; t k ); for some smooth functions U = U(x; t) and V = V (x; t). Then we have: u k n = U; (5.0.5) u k+1 n = U +( t)u t + 1 2! ( t)2 U tt + ::: ; (5.0.6) u k n+1 = U +( x)u x + 1 2! ( x)2 U xx + 1 3! ( x)3 U xxx + 1 4! ( x)4 U xxxx + ::: ; (5.0.7) u k n 1 = U ( x) U x + 1 2! ( x)2 U xx 1 3! ( x)3 U xxx + 1 4! ( x)4 U xxxx + ::: ; (5.0.8) with similar expansions for the v's here U and its derivatives are all evaluated at (x n ; t k ). Substituting these expansions into equations ( ), we obtain: U t + 1 2! ( t) U tt + O(( t) 2 ) = V ; (5.0.9) V t + 1 2! ( t) V tt + O(( t) 2 ) = U xx + 2 4! ( x)2 U xxxx + O(( x) 4 ) : (5.0.10) Now we use that U t = V + O( t) and V t = U xx + O( t; ( x) 2 ), as given by these last equations, to write U tt = U xx + O( t; ( x) 2 ) and V tt = V xx + O( t; ( x) 2 ). Substituting these equalities into the equations, we obtain the Associated Equations to the Numerical Scheme given by ( ): U t = V 1 2! tu xx + O(( t) 2 ; ( t)( x) 2 ) ; (5.0.11) V t = U xx 1 2! tv xx + O(( t) 2 ; ( x) 2 ) : (5.0.12) Now, here comes the key point: notice that the perturbation to the wave equation system (5.0.1) in these equations is, at leading order, a NEGATIVE DIFFUSION. That is to say, wehave the terms U xx and V xx multiplied by a negative coefficient. This means that these terms will induce a behavior that is equivalent to the situation that would arise in a universe where heat flowed from cold to hot, instead of from hot to cold: any small non-uniformity in the temperature distribution would grow immediately without bounds, leading to the formation of singularities. Thus we see that, generally, we cannot have smooth functions U and V satisfying the assumptions we made in remark see remark If we did, they would have to

21 MIT (R. R. Rosales.) AENS (Associated Equation to a Numerical Scheme.) 21 satisfy ( ) which generally does not have smooth solutions. 4 In conclusion: the analysis above shows that the scheme given by ( ) cannot possibly work. It also suggests a way to fix the problem: add to the scheme terms that will neutralize" the negative diffusion terms at the level of the associated system of equations ( ). For example, modify the scheme so it becomes: u k+1 n v k+1 n u k n t v k n t where ffi>1is a constant. 5 = v k n ffi t uk n+1 2u k n + u k n 1 ( x) 2 ; (5.0.13) = uk n+1 2u k n + u k n 1 ( x) ffi t vk n+1 2v k n + v k n 1 ( x) 2 ; (5.0.14) The jargon here is that we say that we have added numerical dissipation to stabilize the scheme. This actually works: provided = t= x is kept below some critical value c (which depends on ffi), this last scheme converges to the solution of the wave equation as t and x vanish. Remark Earlier we mentioned that the associated equation approach only provides a partial answer to the issue of numerical stability, because it only looks at the behavior of long" waves in the numerical solution. By long waves here we mean: disturbances whose wavelength and period are much larger than the grid spacings x and t. That this is so should be obvious, for once we assume that u k n = U(x n ; t k ) and v k n = V (x n ; t k ) (for some smooth" functions U = U(x; t) and V = V (x; t) as done in remark 5.0.1) we exclude from consideration any possibility of high" frequency oscillations in the numerical solution, with wavelengths and periods close to x and t. It is possible to have an unstable numerical scheme, having unbounded growth of high frequency oscillations, with the associated equation completely oblivious" to the fact. An example is provided by the scheme above in ( ). The associated equations for this scheme, given by U t = V (ffi 1) tu xx + O(( t) 2 ; ( t)( x) 2 ) ; (5.0.15) V t = U xx (ffi 1) tv xx + O(( t) 2 ; ( x) 2 ) : (5.0.16) 4 The mathematical jargon to describe systems such as this, is that they are ill-posed" problems. We will talk more about these type of problems later in the course. The main point is that, whenever you get one, it means something went very wrong with your reasoning. 5 For ffi = 2 this new scheme is the good" scheme used in the GBNS lecture MatLab script ( Athena Toolkit.)

22 MIT (R. R. Rosales.) AENS (Associated Equation to a Numerical Scheme.) 22 show no difficulties as long as ffi>1. However, unless one takes = t= x < c for some c = c (ffi) > 0 the scheme is unstable! How can we see that the instability mentioned in the prior paragraph exists, in a fast and easy way? The answer is simple: just as the associated equation approach looks at the behavior of a scheme in the long wave" limit, we can look directly at the other end (short waves.) It is clear that the shortest waves a numerical grid can support are those where the solution alternates (in space) between two values. Thus we look for special solutions for the scheme ( ) of the form: u k n = UG k ( 1) n and v k n = V G k ( 1) n ; (5.0.17) where U, V, and G are constants to be determined. Substituting into the scheme equations, we get the eigenvalue problem: GU = (1 2 ffi 2 )U + tv GV = 4νU + (1 2 ffi 2 )V 9 >= >; ; (5.0.18) where = t= x and ν = t=( x) 2. It follows that we must have G =1 2ffi 2 ±2i, with jg 2 j =1 4(ffi 1) 2 +4ffi 2 4 <1 as long as ffi 2 2 <ffi 1. This turns out to be, precisely, the stability condition for the scheme ( ). That is: < c = p ffi 1 ; (5.0.19) ffi where ffi>1. Remark In all cases we use the notation: x n = x0 + n x; t k = t0+k t; u k n = U(x n; t k ); and calculate the associated equation up to the leading order terms beyond the equation the numerical scheme is intended to solve. For most of the schemes below you will find out that the associated equation has a negative diffusivity when the scheme is unstable. In most cases just making sure that the parameters in the scheme are selected so that the numerical diffusion is positive, will be enough to keep the scheme stable. Check this out numerically.

23 MIT (R. R. Rosales.) AENS (Associated Equation to a Numerical Scheme.) AENS 01 Scheme A. Forward differences for u t + u x = 0. Equation: u t + u x =0... Scheme: u k+1 n where = t x : 5.2 AENS 02 Scheme B. Backward differences for u t + u x = 0. Equation: u t + u x =0... Scheme: u k+1 n where = t x : = u k n (u k n+1 u k n), = u k n (u k n u k n 1), 5.3 AENS 03 Scheme C. Centered differences for u t + u x = 0. Equation: u t + u x =0... Scheme: u k+1 n where = t x : = u k n 1 2 (uk n+1 u k n 1). 5.4 Answer to: AENS 03 Scheme C. Centered differences for u t + u x = 0. This is the scheme implemented by the PS311 Scheme C MatLab script in the Athena Toolkit. This scheme is unstable: there is no value c > 0 such that (for < c )the numerical solution does not develop exponentially large oscillations as x! 0. Let us see now what its associated equation tells us. We write the scheme as u k+1 n u k n t + uk n+1 u k n 1 2 x =0; and substitute Taylor expansions for u k n = U(x n ; t k ). This yields: U t tu tt + O(( t) 2 )+U x ( x)2 U xxx + O(( x) 4 )=0:

24 MIT (R. R. Rosales.) AENS (Associated Equation to a Numerical Scheme.) 24 We use now that U t = U x + O( t; ( x) 2 )toeliminate U tt, and get the associated equation: U t + U x = 1 2 tu xx 1 6 ( x)2 U xxx + O(( t) 2 ; ( t)( x) 2 ; ( x) 4 ) : Again, this always has negative diffusion, and no value of can fix this. Remark You may wonder here what is the role of the term U xxx in the associated equation above. After all, if we take t ß ( x) 2 this term can be just as significant as the negative diffusion term. Thus: can the term U xxx in the associated equation, somehow, counterbalance the bad behavior produced by the negative diffusion? The answer is no. This term corresponds to a mechanism called Dispersion, which we will encounter later in the course. What it does is to produce a wavelength dependent speed of propagation, nothing else. It should be fairly obvious that this is not enough to prevent the sort of catastrophe that heat moving from cold to hot induces. As a simple example with dispersive effects, consider the equation u t + u x + ffiu xxx =0, where ffi is some constant. It is easy to see that this equation has solutions of the form u = A sin (k(x ct)), where k =2ß= is the wave number ( is the wavelength), A is an arbitrary constant (the wave amplitude) and c =1 ffik 2 is the wave speed. It is clear that, for ffi =0the wave speed is independent of the wavelength, but ffi 6= 0introduces wavelength dependence on the wave speed. A physical example of dispersion is provided by the propagation of light in matter: in vacuum the speed of light is a constant, independent of wavelength. But this is no longer true when light propagates through matter (rainbows would not occur otherwise.) One more example of dispersion is given by the equation u tt u xx + sin(u) =0,which is derived in the context of torsion coupled pendulums in another problem. Assume that u is small, linearize the equation to u tt u xx + u =0,and look for solutions of the form u = A sin (k(x ct)). You will see that the wave speed c depends on k. The last example should show you that terms like U xxx are not the only ones capable of producing dispersion (though it is a fairly common type occurring in applications.) 5.5 AENS 04 Scheme D. Centered differences for u t = u xx. Equation: u t = u xx... Scheme: u k+1 n = u k n + ν (u k n+1 2u k n + u k n 1),

25 MIT (R. R. Rosales.) vnsa (von Neumann Stability Analysis.) 25 where ν = t ( x) 2 : 5.6 Answer to: AENS 04 Scheme D. Centered differences for u t = u xx. This is the scheme implemented by the PS311 Scheme D MatLab script in the Athena Toolkit. This scheme is stable for ν <ν c =0:5. Let us see now what its associated equation tells us. We write the scheme as u k+1 n u k n t uk n+1 2u k n + u k n 1 ( x) 2 =0; and substitute Taylor expansions for u k n = U(x n ; t k ). This yields: U t tu tt + O(( t) 2 ) U xx 1 12 ( x)2 U xxxx + O(( x) 4 )=0: Weuse now that U t = U xx + O( t; ( x) 2 ) to eliminate U tt, and get the associated equation: U t = U xx (( x)2 6 t)u xxxx + O(( t) 2 ; ( t)( x) 2 ; ( x) 4 ) : This is an example where the associated equation tells us nothing about the behavior of the scheme. This is because the equation we want to solve already has O(1) dissipation, so that any scheme instabilities in the long wave regime are completely irrelevant. The stability of this scheme is completely dominated by the behavior of the high frequency grid size oscillations. This is easy to see by substituting into the scheme the form for the solution 6 given by: u k n =( 1) n G k ; where G is a constant to be determined. We obtain G =1 4ν. Thus, it is clear that to avoid unbounded growth of these oscillations we must have ν < 0:5 so that 1 <G< AENS 05 Scheme E. Friedrichs centered differences scheme for u t + u x = 0. Equation: u t + u x =0... Scheme: u k+1 n = u k n 1 2 (uk n+1 u k n 1)+ 1 2 ffi 2 (u k n+1 2u k n + u k n 1). where = t x and ffi>0 is a parameter. 6 Note that this corresponds to the shortest possible wavelength that the numerical grid can support.

26 MIT (R. R. Rosales.) vnsa (von Neumann Stability Analysis.) 26 6 vnsa (von Neumann Stability Analysis.) von Neumann Stability Analysis for a Numerical Scheme Consider the numerical schemes that follow after this introduction, for the specified equations. Your task here is to do a von Neumann stability analysis for each scheme, to answer the question: Can the scheme be made stable, and if so, under which conditions? A von Neumann stability analysis can be carried out only in the case of constant coefficients linear difference schemes. It is based on the fact that (for this class of schemes) the general solution to the scheme equations can be found as a linear combination of special solutions found by separation of variables. This works because: ffl For linear schemes, linear combination of solutions are solutions. ffl If the scheme coefficients are constant (independent of the indexes) exponential and power dependencies on the indexes factor out, leaving algebraic equations for the parameters. Basically, the same reasons that allow the solution of linear constant coefficients differential equations using exponentials. Note that the limitation to linear, constant coefficients equations is not as bad as it might seem. One can often get information on the behavior of schemes where these conditions do not apply (e.g: either non-linear or non-constant coefficients situations) by doing a von Neumann stability analysis on a frozen" coefficients version of the schemes (this can be tricky, sowe will not go into this now.) Next we carry out an example of doing a von Neumann stability analysis for a particular scheme, to illustrate the idea. We will pick the example of the good" scheme used in the GBNS lecture MatLab script ( Athena Toolkit.) This scheme is supposed to solve the wave equation, written as the system of equations: u t = v; and v t = u xx ; (6.0.1) where the subscripts indicate partial derivatives and the equations have been nondimensionalized (so the wave speed satisfies c = 1.) A description of the scheme follows.

27 MIT (R. R. Rosales.) vnsa (von Neumann Stability Analysis.) 27 Consider a uniform grid in space and time f(x n ; t k )g, with grid spacings x and t (assumed small".) That is x n+1 = x n + xand t k+1 = x k + t. On this grid we assume that the solution is approximated by the grid functions u k n and vn. k That is: u(x n ; t k ) ß u k n ; and v(x n; t k ) ß v k n ; (6.0.2) where the grid functions satisfy the following discretized version of equations (6.0.1) (the good" numerical scheme): u k+1 n v k+1 n u k n t v k n t = v k n ffi t uk n+1 2u k n + u k n 1 ( x) 2 ; (6.0.3) = uk n+1 2uk n + uk n 1 ( x) ffi t vk n+1 2v k n + v k n 1 ( x) 2 ; (6.0.4) where ffi>1 is a constant. 7 This scheme is obtained by adding a small amountofnumerical viscosity 8 to the naive (and intuitive) discretization used in the bad" scheme of the GBNS lecture script in the MatLab Athena Toolkit. Namely: u k+1 n v k+1 n u k n t v k n t = v k n ; (6.0.5) = uk n+1 2u k n + u k n 1 ( x) 2 : (6.0.6) We now look for solutions of the following form u k n = UG k e i»n and v k n = V G k e i»n (6.0.7) for the scheme equations in ( ). Here U and V are constants, G is the Growth Factor and ß»»» ß is the grid wave number. Remark We restrict ß»»» ß because the terms e i»n above, in (6.0.7) are periodic in», of period 2ß. Thus there is no point in using values of» outside a 2ß range. Note that, for» = p=q a rational number, the solution above is periodic, of period q in n. 7 For ffi = 2 this is the good" scheme used in the GBNS lecture MatLab script ( Athena Toolkit.) 8 See: Introduction, in the AENS (Associated Equation to a Numerical Scheme) problems course Problem List.

28 MIT (R. R. Rosales.) vnsa (von Neumann Stability Analysis.) 28 Remark The idea is that the general solution to the scheme equations ( ), will be a linear combination of the special solutions given by the fundamental modes in (6.0.7). In order for the scheme to behave appropriately, it is important that none of these modes grow i.e.: we need jgj»1 since otherwise the numerical errors will be amplified. Thus, the stability conditions for the scheme will be the conditions that guarantee jgj»1. If no choice of the parameters ffi and = t= x can be made that yields jgj»1,then the scheme is useless. Typically G is a function of», x, t, and any parameters in the scheme (such as ffi in the current case.) In order to prevent errors in the calculation from growing, it is necessary that jgj»1. A scheme will be useful if we can do this while simultaneously letting x and t vanish. In this situation we say that the scheme is stable. Typically this imposes some restriction on the parameters involved; for example we may need = t to be less than some constant. For some schemes, x such as the "bad" scheme of the GBNS lecture script in the MatLab Athena Toolkit, this is impossible and G is always outside the unit circle. What you are being asked to do in these problems is to calculate the growth factor G for each one of the given schemes, and then discuss conditions for G to be inside the unit circle (if possible) so that the scheme is stable. Substituting (6.0.7) into ( ), we obtain the equations: GU = 1 2 ffi 2 sin 2 (»=2) U + tv ; GV = 4 2 t sin2 (»=2) U ffi 2 sin 2 (»=2) V ; 9 >= >; (6.0.8) where we have used that e i» 2+e i» = 4 sin(»=2). Clearly, (6.0.8) is an eigenvalue problem, with eigenvalue G. Solving this problem, we obtain: G =1 2ffi 2 sin 2 (»=2) ± 2 i sin(»=2) : (6.0.9) Thus jg 2 j = 1 2 ffi 2 sin 2 (»=2) sin 2 (»=2) = 1 4(ffi 1) sin 2 (»=2) 2 +4ffi 2 sin 4 (»=2) 4 : (6.0.10)

29 MIT (R. R. Rosales.) vnsa (von Neumann Stability Analysis.) 29 For any fixed» this is a quadratic polynomial in 2. It is easy to see that jg 2 j < 1 will be satisfied provided that the following conditions hold ffi>1 and 2 < ffi 1 ffi 2 sin 2 (»=2) : (6.0.11) Thus the stability conditions for the scheme in ( ), obtained from requiring that (6.0.11) hold for all values of», are given by: ffi>1 and < p ffi 1 : (6.0.12) ffi These guarantee that jgj < 1 for» 6= 0 of course, G(0)=1(which is always true for consistent schemes.) Remark On consistency: Generally speaking, for a constant coefficients linear P.D.E., a Fourier mode in space (i.e.: space dependence via an exponential e ikx ) has an associated time dependence of the form e i!t, where! =!(k) is some function of the wave number k! is called the wave frequency with a Taylor expansion (for small k) of the form:! = ck+c 2 k 2 +::: (6.0.13) When these Fourier mode solutions are evaluated on the numerical grid as in equation(6.0.2) they give rise to a dependence on the grid indexes of the form 9 u` n / exp(ikx n i!t`)/exp(ik xn i! t`)=(g e )`e i»n ; (6.0.14) where» = k x and G e = exp( i! t). This has the same form as the dependence in equation (6.0.7), with the numerical G = G replaced by the exact G e. If the numerical solution is to approximate the actual solution of the P.D.E., it had better be that G e and G are close to each other when t and x are small this is what CONSISTENCY means. To be more precise, generally we will have that G is a function of», x, t, and whatever parameters the scheme equations have. That is G = G(»; t; x; parameters). Then, what is needed is: G(»; t; x; parameters) =G e +ffl= exp ( i!(k) t)+ffl; (6.0.15) 9 Here we use ` for the time index (instead of k) toavoid confusion with the wave number k.