18.300-MIT. Principles of Applied Mathematics. Spring 2022. Tu-Th 9:30--11:00. Room 2-131. Rodolfo R. Rosales. Brief points/topics to be covered in the lectures. The Lecture Summaries file refers to these #nnn points. [PSQ] means Problem Set Question. These are brief summaries used by the instructor to prepare the lectures. They ARE *NOT* "lecture notes" which you can use to study and/or replace attending the lectures and taking notes [or reading books, etc]. They are provided for your convenience, so that you have a brief summary of each topic. These summaries will make sense to you only *after* you understood the material that they cover. You can use them as a quick "refresher" for what are the main points in each topic. They are *not* intended for you to use as "study" material. % ========================================================================== % -------------------------------------------------------------------- #000 Outline of the mathematical topics to be covered: --- Modeling in continuum physics. --- Examples from traffic flow, river and channel flows, etc. --- Mathematical properties of the pde involved. Method of characteristics, shocks, etc. --- Numerical solution of pde by Finite Differences. Basic theory: stability analysis and convergence. --- From stability analysis to the Discrete Fourier Transforms to Fourier Series. --- FFT and spectral methods. --- Fourier and Laplace Transforms. --- Other topics, maybe. See syllabus. Physical phenomena that motivate the material: --- Hydraulic Jumps [kitchen; river floods; flash floods; dams; etc]. --- Shock waves [sonic boom, explosions, super-novas and crab nebula]. --- Traffic flow waves. Others: --- Solitary waves [say, in lakes]. --- Diffusion. During the first lecture videos and pictures that <========= illustrate these physical phenomena will be shown. <========= % % ========================================================================== % ----------------------------------------------------------------------- % % --------- NOTE: for the material we start covering now, see ----------- % % --------- the conservation law notes on the WEB page. ----------- % % ------ The books by Haberman and Whitham cover these topics too. ------ % % ----------------------------------------------------------------------- % % ========================================================================== % % -------------------------------------------------------------------- #001 EXAMPLE: Lighthill-Whitham-Richards model for Traffic Flow. % Derive the model equation rho_t + q_x = 0, where q is a function of rho. Brief discussion of the continuum limit involved: --- What is rho and under what assumptions one can define it. --- What is q and under what assumptions one can define it. --- Point out analogy with continuum hypothesis in fluids and solid matter. The difference between 10^20 and 10^1.5. --- Conservation of cars: Integral and differential forms. % % -------------------------------------------------------------------- #002 CLOSURE PROBLEM: more unknows than equations. % -1- Quasi-equilibrium approximation leading to q = Q(rho). Assumption: there is a stable equilibrium to which system adjusts on a time scale faster than the time scales of interest in problem. Explain how drivers, at steady state, and given a velocity, adjust to an optimal separation between cars. When changes slow enough, they have time to adjust to the optimal separation, so one can use q = Q(rho) to fill-in for the "missing" equation in the conservation of cars [one equation for two unknowns: rho and q]. Typical time scale here is seconds [dominated not by human reaction time, but but car inertia!]. Compare this with the adjustment times in thermodynamics. Quasi-equilibrium is a very important and non-trivial component. This is where the details of the physics behind the equations is built in. For Gas-Dynamics quasi-equilibrium is Thermodynamics. For Traffic Flow this is a step for which a satisfactory theory is not known. There are many topics where this is the case, with ``equations of state'' which are incomplete and problematic. Note: ``equations of state'' is the generic name given to the relationships that complete the conservation laws, so that in the end the number of equations matches the number of unknowns. -2- Discuss generic form of Q. Existence of a maximum throughput q_m = Q(rho_m). % % -------------------------------------------------------------------- #003 Lighthill-Whithams model for Traffic Flow. % The flux function q = Q(rho): --- Discuss practical significance of the existence of q_m = Q(rho_m). Leads to entrance lights at highway access at peak time. Explain purpose, and limitations (no dynamics and not enough space to have cars wait till optimal rho_m is reached). % % -------------------------------------------------------------------- #004 GENERAL FRAMEWORK. Discrete to continuum limits. % --- Densities and fluxes. --- Issues relating to the meaning of the "continuum" limit. EXAMPLES: Car densities and flux. Interstellar media (where do super-nova shocks travel on?). River density and flux. (see #005) Pressure in a gas. Other fluid properties. (see #006) Other examples: Grade "curve" (idealization versus reality: histograms). % % @#@ SKIP: Forestry, molds, bacteria, ... --- examples later. % % -------------------------------------------------------------------- #005 EXAMPLE: River flow equations. A_t + Q_x = 0. % Describe quasi-equilibrium function Q = Q(A). dQ/dA increases with A. Justify: at equilibrium forces (friction and gravity) must balance. --- Note that dq/drho (traffic flow) and dQ/dA (river floods) have the dimensions of a velocity. But, the velocity of what? We will answer this question soon. --- Compare q = Q(rho) and q = Q(A) for traffic flow and river flow. Concave versus convex. In one case c = dq/drho decreases and in the other increases. % % -------------------------------------------------------------------- #006 % -------------------------------------------------------------- #006a EXAMPLE: Euler Equations of Gas Dynamics in 1-D. rho_t + (rho*u)_x = 0 .... conservation of mass (rho*u)_t + (rho*u^2 + p)_x = 0 .... conservation of momentum (rho*E)_t + (rho*u*E + p*u)_x = 0 .... conservation of energy where E = (1/2)*u^2 + e is the total energy per unit mass and e is the internal energy per unit mass % -------------------------------------------------------------- #006b Closure: Quasi-equilibrium = thermodynamics: e = e(p, rho) [Tg] Polytropic gas ..................... e = p/((gamma-1)*rho) [Tp] For an ideal gas p = R*T*rho. T is the temperature, and R is the gas constant. In addition v = 1/rho is the specific volume. A polytropic gas is an ideal gas with constant specific heats: c_v = specific heat at constant volume. c_p = specific heat at constant pressure. Then e = c_v*T and c_p = c_v + R. Eliminating T from p = R*T*rho and e = c_v*T yields equation [Tp] above, where ...................... gamma = c_p/c_v. % ------------------------------------------------------------ % % IMPORTANT POINT regarding gas dynamics versus traffic flow. % % There is a massive difference in the adjustment time scales, % % as well as the number of discrete elements used to compute % % the averages. Because of this the gas dynamic equations are % % far more accurate in their predictions than the equations % % used for traffic flow. % % ------------------------------------------------------------ % % -------------------------------------------------------------- #006c Isentropic Gas Dynamics If one assumes adiabatic changes with entropy constant, then .............................. p = p(rho), and the equations simplify to the p-system: rho_t + (rho*u)_x = 0 .... conservation of mass (rho*u)_t + (rho*u^2 + p)_x = 0 .... conservation of momentum There is no need to use conservation of energy to get a closed system. More accurately: the equation for the conservation of energy is subsumed by the adiabatic changes with constant entropy hypothesis. % ----------------------------------------------------------- In general we use the equation T*dS = de + pdv, v=1/rho [a] where T = temperature S = entropy Then dS = 0 provides an extra equation that can be used to get, from e = e(p, rho), p as a function of rho only. Where does come from? Thermodynamics! Thermodynamics says: If an isolated system does an infinitesimal amount of work, \delta W, then it should be ... \delta W = T*dS. This is always true (see [b] below), but a gas can do work in only two ways: via use of internal energy, or mechanically via work by the pressure. Hence [a] above. Example [polytropic gas]. Then e = c_v T and p = R*rho*T = R*T/v Thus T*dS = c_v*dT - R*T*(drho/rho). Dividing by T and integrating S-S_0 = c_v*ln(T) - R*ln(rho) or (S-S_0)/c_v = ln(T) - (gama-1)*ln(rho) or T = rho^{gamma-1} * exp{(S-S_0)/c_v} or p = rho^gamma * R * exp{(S-S_0)/c_v} ==> p = kappa rho^gamma if S = constant. [b] Note: in general, for an amount of work \delta W, it will be \delta W = \sum_1^N lambda_n dpsi_n, where the psi_n correspond to the quantities capable of doing work, such as internal energy, specific volume, changes in chemical composition (chemical reactions), phase transitions, etc., and the lambda_n represents how much work dpsi_n does --- Note that each \lambda_n is a function of the psi_j. For N > 2, \delta W = T*dS is a very strong restriction on the lambda_n, since it means that they can all be written in terms of just two functions, T and S. That is \lambda_n = T S_{psi_n}. % -------------------------------------------------------------- #006d SOURCES for conservation laws. General 1-D formulation changed to: rho_t + q_x = S, where rho = density of conserved stuff (stuff per unit length) q = flux of conserved stuff (stuff per unit time) S = sources of conserved stuff (stuff per length, per unit time) Examples: --- body forces in gas dynamics. (#006a) S = F ..... in conservation of momentum S = F*u ... in conservation of energy where F = body force per unit mass --- feeder roads to main artery in traffic flow. (#001) --- feeder streams in river flows. (#005) % % @#@ SKIP: 2-D and 3-D. Compressible and incompressible cases. % % -------------------------------------------------------------------- #007 GENERAL FRAMEWORK. How conservation principles lead to pde's: --- Integral and differential form of a conservation law. --- Closure issues and "constitutive equations." Quasi-equillibrium approximations and time-scale limitations. The case of traffic flow and the traffic flow curve. SEE/READ: conservation law notes on the WEB page. Further examples will clarify these ideas: Small transverse vibrations of a string ............ #008, #066. Small longitudinal vibrations of an elastic rod .... #009. #059 Heat flow along an insulated wire .................. #010. Heat flow in 2-D or 3-D ............................ #012. Diffusion equation (e.g.: Salt in water) ........... #013. Viscosity in fluids ................................ #016. Slow granular flow in a silo ....................... #017. Gas Dynamics ....................................... #050. p-system ........................................... #059. String on an elastic bed ........................... #067. % % -------------------------------------------------------------------- #008 % @#@ SKIP. Will return to this later [see #066]. % EXAMPLE: Small transverse vibrations of a string under constant tension T, with motion restricted to a plane. See problem set series "Conservation Laws", "Problem CoLa07. Small vibrations of a 2D string under tension." The general conservation law machinery can be used to obtain the wave equation u_{tt} - c^2 u_{xx} = 0, where c > 0 is a constant (wave speed). Here x = coordinate along string. u = u(x, t) transversal string displacement from equilibrium. Thus the string is described by y = u(x, t). Assumptions: the string is homogeneous, under tension, and the deviations from equilibrium (a straight line) are small. The derivation should be compared with that in elementary physics books where the equation is derived using force [F = m*a] balance on string differential elements. Conservation is a more powerful method, generalizable to many other contexts. Key assumption: If T is the tension along the string, and theta is the angle of the string with the horizontal, the projection of the force in the transversal direction is T*sin(theta), and the projection along the longitudinal direction is T*cos(theta). In the small deviation limit theta is small, so that u_x = tan(theta) ~ theta ~ sin(theta). On the other hand cos(theta) = 1 + O(theta^2), so longitudinal force is near constant [hence no longitudinal motion]. ALTERNATIVE EXAMPLE: taut string over an elastic bed. Get Klein-Gordon equation. % % -------------------------------------------------------------------- #009 % @#@ SKIP. May return later #059, or assign in a problem set ....... [PSQ] % EXAMPLE: Small longitudinal vibrations of an elastic rod. x = Lagrangian coordinate (particle position at equillibrium). rho = density (mass per unit length) of rod u = u(x, t) displacement. Thus rho*u_t = Longitudinal linear momentum density. T = T(x, t) tension: T(x0, t) = force by x > x0 on x < x0. Thus -T = momentum flux Elasticity: T is a function of the strain = f(u_x - 1) Explain why u_x-1 is strain. Derive equation and consider Hooke's law case when T = \kappa*(u_x-1). Alternative derivation using F = m*a on rod differential elements. % % -------------------------------------------------------------------- #010 EXAMPLE: Heat flow along an insulated wire. % Describe closure, Fick's law, and do intuitive justification using stat. mech. interpretation of heat and temperature. % % -------------------------------------------------------------------- #011 More than 1-D. Conservation laws in 2-D or 3-D. Flux q is now a vector. Use Gauss theorem to obtain general form rho_t + div(q) = S (S equal sources & sinks). Examples of S: Cars flowing in/out of highway through commuter township. Water flowing into river from small affluents. Heating by electromagnetic waves [microwave oven]. % % -------------------------------------------------------------------- #012 EXAMPLE: Heat flow in 2-D or 3-D. % Then rho = r c_v T = conserved stuff (heat) per unit mass where c_v = specific heat of material, r = mass density T = temperature, Fick's law applies q = - kappa gradient(T), and kappa = heat conductivity. Thus: Heat equation: T_t = nu Laplacian T, where nu = kappa/(r c_v) = coeff. thermal diffusion. What are the dimensions of kappa? nu? c_v? % % -------------------------------------------------------------------- #013 EXAMPLE: Diffusion equation (Salt in water, sugar in coffee, ink in water, ...) Same as heat equation C_t = nu*Laplacian C where C = concentration (salt, ink, sugar, ...) nu = difussion coefficient. % % -------------------------------------------------------------------- #014 DIMENSIONS and DIMENSIONAL ANALYSIS. % % -------------------------------------------------------------- #014a Example with ode, say harmonic oscillator \ddot{x} + omega^2 x = 0, showing how the fact that the solution has to be a ``mathematical function'' yields x = f(omega*t). We can say this without knowing anything about ode, or the actual solution to the equations. The examples in #014b and #014d give more striking examples of the reasoning involved. Point out that this type of analysis explains why in special situations where the behavior is controlled by a few parameters, power law behaviors are observed. Vice-versa, an observed power law behavior is an indication that the phenomena that obeys it is controlled by a few parameters only. % % -------------------------------------------------------------- #014b How long does it take sugar to sweeten a coffee cup without stirring? Idealized problem: start with a very small blob of ink, and ask: What is the radius of the blob of ink, R = R(t), as the blob expands due to diffusion? Dimensional analysis says R(t) \propto sqrt{nu*t} In particular, let L be the size of the coffee cup. Sugar will reach whole cup when R = O(L) ===> time = O(L^2/nu). Also the relevant time needed to cool/heat a size L vessel. These times are very long when measured in human-relevant scales. Hence stirring needed. Boiling and convection speed up heating. Questions: why does stirring help? why does convection occur? what would happen when heating something with a flame in the absence of gravity? At room temperature, in cm^2/sec Thermal diffusivity: water ~ 0.0014 mercury ~ 0.042 Diffusion in water: NaCl ~ 10^{-5} Example: How long does it take to reach R = 5 cm [salt in water]? % % -------------------------------------------------------------- #014c Blast wave from a point release of a large amount of energy E. If the energy is very large, the ambient fluid velocity and pressure become irrelevant. Only the density matters. Then what happens is controlled by just two parameters: E, with units M*L^2/T^2; and rho with units M/L^3 From these two, the only way to eliminate mass is via the ratio E/rho, with units L^5/T^2; Hence the radius of the blast wave should satisfy: R = k*(E*t^2/rho)^{1/5}, where k is a numerical constant [e.g.: 1.5, pi, whatever]. Note that we can state this without solving for ANY of the details of the physics involved! Knowing R at one value of time, t1, allows you to estimate E --- if you assume that k = O(1). Knowing R at two values, t = t1 and t = t2, is enough to determine E. Is there an Astophysical context where you could use this? % % -------------------------------------------------------------------- #015 Conservation laws and pde: Higher order (TRANSPORT) effects beyond quasi-equillibrium. Note difference between heat/diffusion and the other examples. In one case the fluxes are functions of the densities only, in the other derivatives of the densities appear. --- Important when gradients are not so small. --- Connection with randomness at microscopic level. EXAMPLES: Viscosity in fluids. See #016. Slow granular flow in a silo. See #017. Preventive driving in traffic flow. See #044 % % -------------------------------------------------------------------- #016 EXAMPLE: Viscosity in fluids. Cover lightly. May return to this later. % Another example of a higher order transport effect is viscosity: Forces proportional to the flow velocity gradient. REASON: the flow velocity is a macroscopic "average" variable. The same phenomena that causes heat conduction for thermal energy, produces momentum transfer (forces), as well as kinetic energy transfer. A small peek at the Navier-Stokes equations: to the conservation-law derivation of the Euler equations (#006), add heat conductivity and viscosity. Derive compressible Navier Stokes in 1-D. rho_t + (rho*u)_x = 0 (rho*u)_t + (rho*u^2 + p - mu*u_x )_x = 0 (rho*E)_t + (rho*u*E + p*u - mu*u*u_x - kappa*T_x)_x = 0 where E = (1/2)*u^2 + e is the total energy per unit mass and e = e(p, rho) is the internal energy per unit mass The new terms, relative to Euler, are: - mu*u_x ......... momentum flow due to viscosity - mu*u*u_x ....... work by the viscous forces - kappa*T_x ...... heat flux by thermal conductivity We will/may come back to this later. % % -------------------------------------------------------------------- #017 % EXAMPLE: Slow granular flow in a silo. % @#@ SKIP: Students should read this in the WEB page notes. % % -------------------------------------------------------------------- #018 % Conservation laws in more than 1-D. @#@ SKIP. % Meaning of the flow vector. Why is it a vector? ................... [PSQ] For some conserved quantity, define the flow q as follows: q is the amount of rho per unit time and per unit length in 2D (area in 3-D) crossing a curve in 2D (surface in 3-D) from one side to the other. With this definition q is a (scalar) function that depends on the position and the orientation (normal) of the differentials of -- length for the curves across which the flux occurs (2D). -- area for the surfaces across which the flux ocurrs (3D). Then show that q must have the form q = \hat{n} \cdot \vec{q}, where \vec{q} is a vector, and \hat{n} is the unit normal to the curve/surface. [then q positive means that the net flow is in the direction of \hat{n}]. Argue using the standard rectangular simplex of "infinitesimal" size (limit is involved), and conservation on an area (volume) that vanishes much faster than the fluxes through the sides --- which, hence, must balance. % ------------------------------------------------------------------------- % ----------------------------------------------------------------------- % % --------- NOTE: for the material we start covering now, see ----------- % % --------- the books by Wan, Haberman, Whitham. ----------- % % ----------------------------------------------------------------------- % TOPIC: Solution of 1-st order (scalar) quasilinear equations by characteristics. Examples: traffic flow and river waves. Roughly 6 lectures. % ======================================================================= % % Review of chain rule in several variables. % % It is important that the students be familiar with differentiation % % with several variables, chain rule, and implicit differentiation. % % Please review your 18.02 notes. % % ======================================================================= % % % -------------------------------------------------------------------- #019 Traffic Flow equations. Traffic density, flow and car velocity. Linearize equation near a constant density and solve u_t+c_0*u_x = 0. Do it first by separation of variables/Fourier. Show solution has form u = f(x - c_0*t). This works because equation is linear and constant coefficients. Need approach that will generalize to u_t + c(u)*u_x = 0. But first, note that: for traffic flow u > c and for river flows u < c. Density waves reach cars from ahead. River waves move faster than the flow. Interpret equation as a statement about a directional derivative of the solution in space-time being zero. Leads to the solution being constant along the integral curves for the directional derivatives. Definition of characteristics and solution by characteristics. Applies to any equation that can be translated into statements about the directional derivatives of the solutions. Such equations are called hyperbolic. Scalar hyperbolic equations can be reduced to o.d.e.'s along the curves integrating the corresponding directional derivatives. % % -------------------------------------------------------------------- #020 Examples: Linear problems with constant or simple variable coefficients, where all the calculations can be done exactly. In each example: --- Write characteristics in parametric form. --- Solve and draw the characteristics. --- Eliminate the characteristic variables and find the solution. --- Show where the solution is defined. Example 1: u_t+c_0*u_x = a*u. IVP problem on -inf < x < inf, t > 0. Example 2: x*u_x + y*u_y = y, u(x, 1) = g(x) for -inf < x < inf, y > 0. Example 3: u_x + x^2*u_y = y, with u(x, 0) = g(x) for x > 0. Show that this defines the solution to the right of y = x^3/3. % % -------------------------------------------------------------------- #021 Describe general method, for a problem of the form a(x, y)*u_x + b(x, y)*u_y = c(x, y), u given along some curve. That is: u = U(z) on some curve x = X(z) and y = Y(z). Equations: dx/ds = a, dy/ds = b, and du/ds = c, to be solved with the conditions x = X(z), y = Y(z) and u = U(z) for s = 0. Leads to solution expressed in the form u = u(s, z), with x = x(s, z) and y = y(s, z). where: z = parameter/label for the characteristic curves. s = parameter that results from solving o.d.e.'s above = parameter along each characteristic curve. (s, z) is a curvilinear, coordinate system. Characteristic coordinates. To get the solution must solve for s and z as functions of x and y. % -------------------------------------------------------------- #021b Illustrate the coordinates (s, z) graphically. % -------------------------------------------------------------- #021c *** CAUSALITY: *** When solving problems that involve time, the information from the data to the solution must always be propagated FORWARD in time. Data cannot determine the solution in the past. Physical principle. For continuum approximation models it can be justified because: the neglected terms generally include effects that are non-reversible --- e.g.: dissipation [one cannot solve the heat equation backwards in time. Explain (ill-posed)]. % -------------------------------------------------------------- #021d Another example: x^2*u_x + x*y*u_y = y, with conditions on circle u = U(zeta) for x = cos(zeta), y = sin(zeta). dx/ds = x^2 and dy/ds = x*y and du/ds = y x = cos(zeta) y = sin(zeta) u = U(zeta) for s=0. x = cos(zeta)/(1-s*cos(zeta)) = r*cos(zeta); r = 1/(1-s*cos(zeta)); y = sin(zeta)/(1-s*cos(zeta)) = r*cos(zeta); u = - tan(zeta)*ln(1-s*cos(zeta)) + U(zeta); % % -------------------------------------------------------------------- #022 In all the examples in #020, the characteristic curves are independent of the solution. This follows from the equations being LINEAR. That is a, b, and c do NOT depend on u. % ----------------------------------------------------- % % ----------------------------------------------------- % % ----- % Next we move up to NONLINEAR PROBLEMS % ----- % % ----------------------------------------------------- % % ----------------------------------------------------- % Define: linear, semilinear, and quasilinear. POINT OUT: method of characteristics works for 1-D linear/semilinear/quasilinear scalar eqns., where it provides the full solution. It can be extended to both systems and more dimensions, where it provides very useful information about the solutions of hyperbolic equations [to be defined later], but there it does not quite provide the full solution --- though it underlies many of the numerical methods used for these equations. % % -------------------------------------------------------------- #022b EXAMPLE: a semilinear equation u_t + u_x = u^2, u(x, 0) = f(x). Show that the solution may cease to exist along characteristics. Thus the sln. may not be defined everywhere the characteristics reach. This generally does not happen for linear equations, but it is not un-common for nonlinear ones. Note: equation breaks down when a singularity is reached. Whatever the physical assumptions made deriving the model, they fail there and the solution cannot be continued further. Better model is needed. An example of a semi-linear equation that yields a singularity of this type occurs in nonlinear optics: laser self-focusing and the nonlinear Schodringer equation. % % -------------------------------------------------------------------- #023 Example 4: General kinematic wave equation u_t + c(u)*u_x = 0, with u(x, 0) = f(x). -1- Characteristic form and characteristic speed. -2- Solution generally cannot be written explicitly. -3- Geometrical interpretation of the solution. Leads to a clear picture of how conservation is achieved: "SLIDING SLABS" image. -4- Wave distortion/steepening and wave breaking. -5- Smooth solution do not exist for all time. -6- Show that the characteristics can/do cross in space time. -7- Derivatives become infinity at time of first crossing. % -------------------------------------------------------------- #023a NOTE: dc/du < 0 for traffic flow, and dc/du > 0 for river flows. Consequences for wave steepening: waves steepen backwards (TF) or forwards (RF). Matches observations. % % -------------------------------------------------------------------- #024 Back to nonlinear kinematic waves and wave-breaking: After wave breaking, need more physics to fix the problem. Explain: why does the model fail there? (quasi-equilibrium). Fix to the problem is NOT in the mathematics. Need NEW physics! Will study traffic flow to learn how to fix it. % % -------------------------------------------------------------------- #025 Back to traffic flow, and example problems there. % -------------------------------------------------------------- #025a Red light turns green. Gap in characteristic field. Argue nice dependence on perturbations to data. Fill gap by taking limit of smeared discontinuity. Obtain expansion fan solution c = x/t. Explicitly solve for rho in the case: q = q(rho) is quadratic. Alternative: use the "sliding-slab evolution" picture (#023) to get the solution Physical meaning of the various waves observed: --- Path of the first car in line behind the red light. A characteristic, and also a car path, since u(0) = c(0). --- Location where cars start moving behind the light. Light wave: locus where brake lights turn off. % -------------------------------------------------------------- #025b Green light turns red. Part 1. Split into two initial-boundary value problems (ahead and behind the light). We will do both problems, and clarify the physical meaning of the various waves observed. "Light" traffic first. Ahead of light: characteristics cross. Resolve issue by introduction of the last car to make the light. Discontinuity in solution where characteristics are chopped. Speed of this discontinuity is obvious (path of last car through the light). Introduce notion of curves along which characteristics die as a solution to the problem of multiple values. That is: paths in space time that are "characteristic cemeteries" --- Along these curves the solution is discontinuous. --- These curves are called shocks. --- Key point: *** CHARACTERISTICS END/DIE AT SHOCKS *** This works in this case, at least. We'll see it is generic. **** Notice that this is new math. into the model. **** Note that, since the discontinuity here is at the last car through the light, conservation follows trivially. Below we will see what is the general rule for the speed of the places where characteristics die [discontinuities/shocks]. We will also see that the general rule below reduces to: Shock speed = car speed when state behind is rho = 0. % -------------------------------------------------------------- #025c Green light turns red. Behind the light. Again, characteristics cross. Argue that in real life drivers wait till the last moment to break. Another discontinuity needed: **** Location of thin layer where cars break. **** **** Wave moving backwards from the light. **** How does one compute it's velocity? Get law: speed = [Q]/[\rho] by 1) Arguing from the integral form of the conservation law. Split integral \int_a^b = \int_a^shock + \int_shock^b. 2) Move into the shock frame and argue conservation "flow of cars into discont. = flow of cars out" Note flow = rho*(u-s). ** SHOCK LAW is CONSERVATION! ** Note: limit of shock speed as [rho] vanishes = characteristic speed. Again: CHARACTERISTICS CHOPPED AT DISCONTINUITY AND CROSSING AVOIDED. % -------------------------------------------------------------- #025d Graphical interpretation of shock jump condition: ***** slope of secant line in rho-q=q(rho) diagram. ***** Again: as shock jump vanishes, this becomes characteristic speed. It is ALL in the Fundamental Diagram: --- characteristic velocity = slope of tangents. --- car velocity = slope of secant through origin. --- shock speed = slope of secant line. % -------------------------------------------------------------- #025e % --------------------- Recap of points needed for problem sets. % Domain of dependence for % a(x,y)*u_x + b(x, y)*u_y = c(x, y)*u + d(x, y) % with u given along some curve Gamma. % % Region where the solution is defined by the characteristics that go % through Gamma. % Example: x*u_x + y*u_y = u, with u=1/(1+x^2) on y = 1. % Sln. determined for y > 0 only, even though % the formula for u that one gets by solving % the equation, u = y^3/(x^2+y^2), % has a value for any x, y, not both zero. % % Point out: can "extend" this solution to the lower half plane y < 0 in % many ways. % Example: u = C*y^3/(x^2+y^2), C any constant, solves equation for y < 0, % and matches above solution with continuous derivatives! % % Note, general solution is u = r*f(theta) in polar coordinates, because % equation is r*u_r = u. Any f that vanishes and has a derivative that % vanishes, at theta = 0 and pi, can be used to extend solution below! % % Another point: A student asked during lecture about shocks and shock % crossings: % What if the characteristics hit each other "head on"? % An alternative way to put the question is: % Imagine a characteristic that "turns around in time", % what do you do? % Example: a curve like t = t0 - x^2, -sqrt{t0} <= x <= sqrt{t0}. % Time must proceed forward, so this is 2 characteristics. % 1) x = + sqrt{t0-t}, for 0 <= t <= t0 [x decreasing!] % 2) x = - sqrt{t0-t}, for 0 <= t <= t0 [x increasing!] % % At t = t0 these two characteristics collide, head on, and kill % each other. % --- This provided that a shock did not cut them off % earlier. % --- At x = 0, the solution will, generally, have some % singular behavior, as it is getting info from two % different characteristics. % % An example of this type occurs in the problem % Linear 1st order PDE # 09 (surface evolution). % There the IC are special, so no singularity occurs. % % % NOTE ABOUT SOLVING ODE's! % How do you solve dx/ds = x-y and dy/ds = x+y, % x(0) = z and y(0) = 0 .... ? % CANNOT DO SEPARATION OF VARIABLES! % Review. % ------------------------------------------------------------------------- % % -------------------------------------------------------------------- #026 Recap extended theory: -1- smooth parts ........................ use pde and characteristics. -2- when characteristics cross .......... introduce shocks. shock speed = [q]/[rho] -3- IMPORTANT: use shocks only to avoid characteristic crossings. Characteristics must end at the shock. Mathematical formulation of this: ENTROPY CONDITION. Explain how this: -4- Allows construction of solution for all times. But need to identify where characteristics first cross, to insert a shock there. See #027. -5- Introduces the arrow of time: solutions with shocks are *** NOT TIME REVERSIBLE *** Example of a place where NO SHOCK is needed: rarefaction. NOT ALL DISCONTINUITIES THAT SATISFY speed = [Q]/[rho] ARE ALLOWED! Example 1: Compare behavior, with initial smooth "bump" profile rho = f(x), or A = f(x) for TRAFFIC FLOW and FLOOD WAVES. List resulting differences in behavior. Example 2: A_t + (0.5*A^2)_x = 0 [Flood waves with Q = 0.5*A^2] with IC: A = 0 for x > 1, and A = A_0 > 0 for x < 0. A linear between x=0 and x=1. Show show condition is s = average(A). Compute characteristics. Draw, and find crossing. Introduce shock. Example 3: A_t + (0.5*A^2)_x = 0, with a > 0 a constant, and A = a for x < 0 % a > 1, t --> infty get shock. A = x for 0 < x < 1 % a < 1, t --> infty get rarefaction. A = 1 for 1 < x % Note that there is always a shock starting at x=t=0. But for a < 1 this shock never reaches the A = 1 region, and becomes smaller and smaller as t grows. Draw solution [A as a function of x] for typical values of t. % % -------------------------------------------------------------------- #027 Characteristics almost always cross. WRITE THE PRECISE CONDITIONS NEEDED FOR THIS TO HAPPEN. WHEN/WHERE DO CHARACTERISTICS CROSS. FIRST CROSSING. Simple problem: rho_t + q_x = rho_t + c(rho)*rho_x = 0, rho(x, 0) = R(x), c(rho) = dq/drho. Solution by characteristics: x = X(s, t) = C(s)*t + s, (#1) rho = R(s), where C(x) = c(R(x)) = wave speed along initial data. Characteristics do not cross if and only if can solve for s as a function of x and t --- s = S(x, t) --- from (#1) if and only if map s --> x is monotone: X_s not 0. That is: inspect X_s = C'(s)*t + 1. So, if C'(s) < 0 somewhere, there will be a time when X_s = 0. Thus, the condition for crossing is: dc/dx < 0 somewhere in the initial data. Graphics: show how x = X(s, t) = C(s)*t+s looks like as a function of s, for t fixed, as t grows, if C(s) is a localized hump. t = 0 straight line x = s. t >0 moderate straight line develops a wiggle. t >0 large wiggle large enough to produce a local max. and a local min. Hence a range where map is not 1-to-1. Formula for critical time t_c and location x_c where the characteristics cross first, assumming C'(s) < 0 somewhere (no crossings otherwise) Let s_c be the value of s at which C'(s) reaches its largest negative value (i.e.: absolute minimum). Then t_c = - 1/C'(s_c) ... so X_s vanishes. x_c = C(s_c)*t_c + s_c. % % -------------------------------------------------------------- #027b Extras: To (graphically) visualize infinities of rho_x and rho_t, characterized by the solutions to 1+t*C'(s), plot y = C'(s) versus y = - 1/t (horizontal line). % % -------------------------------------------------------------------- #028 % RECAP POINT. Cover briefly. % Evolution of wave profile, as given by the characteristic solution. Graphical interpretation: -- Move each point on graph at velocity c(rho). Evolution as sliding of horizontal slabs at different velocities (guarantees conservation). -- Causes wave steepening, wave breakdown, and multiple values. -- Mathematical model breakdown. Quasi-equillibrium assumption fails and PDE model breaks down. -- Back to "physics". Need to augment model with new physics, namely: shocks in the case of traffic flow or flood waves. % % -------------------------------------------------------------------- #029 % RECAP POINT. Cover briefly. Do examples from #031. % First look at EXPANSION FANS: A discontinuity in the initial, or boundary, conditions gives rise to an expansion fan if the edge characteristics do not cross and leave a gap. That is: c is increasing across the discontinuity: EXAMPLES: Red-light turns green. Red-green-red-green ... Will see more later. DEFINITION: An expansion fan is the solution produced by a collection of characteristics, all starting at one single point, but with a range of values for the solution there. % % -------------------------------------------------------------------- #030 % RECAP POINT. Cover briefly. See examples in #032. % First look at SHOCKS: A discontinuity in the initial, or boundary, conditions gives rise to shock if the edge characteristics cross. That is: c is increasing across the discontinuity: EXAMPLE: green-light turns red problem. One shock on each side of light. Shock on the right: last car through light. Shock on the left: locus where cars break behind the light. Shock conditions: Rankine-Hugoniot shock speed S = [q]/[rho]. Derive from conservation: In shock frame: flow from the left = rho_*(u_ - S) = F_ In shock frame: flow from the right = rho+*(u+ - S) = F+ Then conservation is F_ = F+, which gives S = [q]/[rho]. Graphical interpretation of Rankine-Hugoniot: slope of secant line in rho-q diagram. Note: limit for infinitesimal strength shocks is characteristic speed. % % -------------------------------------------------------------------- #031 EXAMPLES with EXPANSION FANS. Do (at least) one, say: 1. % -1- u_t + u*u_x = 0, with u(x, 0) = 0 for x < 0, and u(x, 0) = 1 for x > 0. -2- u_t + u*u_x = -u, with u(x, 0) = 0 for x < 0, and u(x, 0) = 1 for x > 0. -3- u_t + u*u_x = 0, with u(x, 0) = 1 for x > 0, u(0, t) = 1 for 0 < t < 1, and u(0, t) = 0 for 1 < t. -4- u_t + u*u_x = 0, with u(x, 0) = 0 for x > 0, u(0, t) = 1 for 0 < t < 1, and u(0, t) = 0 for 1 < t. % % -------------------------------------------------------------------- #032 IMPORTANT point: When dealing with shocks, it is CRUCIAL TO KNOW WHAT IS CONSERVED. Note that the equation u_t + u*u_x = 0 can arise from various conservation laws: (a) Conserved density rho = u, flux q = (1/2)*u^2, (b) Conserved density rho = u^2, flux q = (2/3)*u^3 (c) Conserved density rho = u^3, flux q = (3/4)*u^4 .... Each of this gives rise to the same solution as long as there are no shocks, but give rise to DIFFERENT shock speeds, hence different solutions once shocks arise. Example: u_t + u*u_x, with IC: u = 1 x < 1, and u = 0 for x > 0. Compute solution for the three cases above. % % -------------------------------------------------------------------- #033 RECAP: IMPORTANT point concerning the ENTROPY condition: % We need shocks to stop the crossing of the characteristics: characteristics end at shock, and do not continue \ #A on the other side. Hence crossings are avoided. / This requires the condition c_ > S > c+ ``Entropy condition''. Why name? Note that #A implies that INFORMATION IS LOST AT SHOCKS, and problem becomes IRREVERSIBLE once SHOCKS FORM. Hence #A is determining the arrow of time. In physics, information contents is measured by the entropy, that must be non-decreasing (second law of thermodynamics). Hence the name. Example: look at red light turns green problem. In addition to the solution with the rarefaction fan, in principle the problem admits a solution with a ``shock'', but this shock does not satisfy entropy, and in fact generates information (characteristics). This solution is NOT ``physical'' and should be considered ``in-admissible'' in the augmented model. % % -------------------------------------------------------------------- #034 RECAP POINT. Back to entropy condition and R-H jump condition: c_ > S > c+ and S = [q]/[rho]. Inspect graphical meaning of this in rho-q diagram. Consistent for traffic flow and river flows (q concave and convex). Leads to: For traffic flow rho increases across shock. For river flow A decreases across shock. % ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ % % IMPORTANT NOTE ABOUT THE ENTROPY CONDITION: % % It applies if one assumes standard drivers and standard driving % % conditions. Using "special" drivers one can arrange to have % % discontinuities that do not satisfy the entropy condition. % % 1-Example (traffic flow): Situation at the start of a car race, % % with all the racing cars neatly organized in a pack behind a % % lead car. This gives rise a "square" wave density shape. The % % entropy condition is violated by the front discontinuity. % % But this is "car ballet", not traffic flow. % % 2-Example (traffic flow): driver that goes slower than the road % % conditions allow, creating long line of cars behind [common % % in mountain roads]. Again, it requires a special driver to % % maintain the discontinuity at the front of the pack. But the % % discontinuity at the end of the pack is a standard shock. % % 3-Example (river flow): push water from behind with a paddle % % This is the equivalent of example 2 above in traffic flow. % % ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ % % % -------------------------------------------------------------------- #035 SUMMARY: SOLUTION PROCESS FOR KINEMATIC CONSERVATION LAWS THAT SUPPORT SHOCKS (A) Solve for the characteristics starting at EVERY point along the curves with data (initial, boundary, whatever). If the data is given with different formulas for various segments, solve each segment separately. (B) Check for GAPS in the characteristics, caused by jumps (discontinuity in the data), and fill-in the corresponding fans of characteristics. (C) Inspect the set of characteristics thus obtained and check for crossings. Use shocks to eliminate the crossings (with the techniques to be described below). -- Shocks used ONLY to prevent characteristics from crossing. Characteristics converge into them and STOP there! (D) After inserting the shocks, and solving for their paths in space time, check again characteristics. Make sure that all crossings have been resolved, and that (for each shock) you do have the characteristics on each side ending there. Shocks MUST satisfy both the Rankine-Hugoniot jump conditions and the entropy condition. (E) Note that a shock may pass through different "regions" where the solution on each side is given by changing formulas (characteristics starting at different parts of data). Keep this in mind when solving for their paths. (F) Check for possible shock "collisions" and resolve them. % % -------------------------------------------------------------------- #036 % Cover briefly. Return to this later after more hyperbolic stuff? % EXAMPLE: RIEMANN PROBLEM. % Define Riemann problem, and do it for u_t + ((1/2)*u^2)_x = 0, with u the conserved density, and q = (1/2)*u^2 the flux. Why is Riemann problem important? Talk a bit about Godunov. Possible Topics: [A] Describe solution to Riemann Problem for Traffic Flow, Flood Waves, etc. Two cases: shock wave or rarefaction wave. [B] Describe Godunov's method and how the Riemann Problem is critical. Important: explain role of finite speed of propagation. Preview of domain of dependence and domain of influence ... to be seen in more detail later. [C] Describe how to get higher order Godunov, using piece-wise linear and higher order approximants. Important: keep conservation! Generalized Riemann Problems [briefly mention]. [D] Explain need for switches, to prevent undesirable oscillations. [E] Method easy to extend to systems; e.g.: Gas Dynamics. Sketch how the Riemann problem works for Gas Dynamics: In isentropic Gas Dynamics [or Shallow Water] there are two sets of characteristics, thus two types of shocks and two types of rarefactions. This leads to 4 cases, instead of two: shock-shock; shock-rarefaction; rarefaction-shock, and rarefaction-rarefaction. Hopefully this will all become clearer later in the semester. [F] Derive the characteristic equations for Isentropic Gas Dynamics. [G] Note: the solution to the Riemann problem is always a function of x/t, which basically allows to reduce its solution to an algebraic problem (albeit maybe a hard one). [H] Extension to multi-D. Can be reduced to 1-D Riemann problems, upon: In 2-D neglect corners and consider fluxes along sides only. In 3-D neglect corners and sides and consider fluxes along faces only. % % -------------------------------------------------------------------- #037 % Optional exercise for students. Strongly recommended. % Example: u_t + ((1/2)*u^2)_x = 0, with u the conserved density, and q = (1/2)*u^2 the flux. Initial Value problem: u(x, 0) = 1 for x > 0 and x < -1 u(x, 0) = -x for -1 < x < 0. % % -------------------------------------------------------------------- #038 % Optional exercise for students. Strongly recommended. QUADRATIC FLOW. Next we will do a few Traffic Flow examples, where we consider the quadratic flow special case: q = (4*q_m/rho_j^2)*rho*(rho_j-rho). -1- Non-dimensionalize the equations and show that they can be reduced to rho_t + q_x = 0, with q = 4*rho*(1-rho). Example: Use interval between light changes to non-dim time. -2- Note then c = 4*(1-2*rho) is linear in c. Hence, since rho is conserved, in this case c is also conserved. Equation can thus be reduced to the simple form c_t + ((1/2)*c^2)_x = 0. In particular: -3- In this case the shock speed s satisfies: s = average of the characteristic speeds across the shock. % % -------------------------------------------------------------------- #039 % Optional exercise for students. Strongly recommended. Example, for #038 (quadratic flow). % a) Red light (at x = 0) turns green at t = 0. Thus rho(x, 0) = 1 for x < 0, ..... c(x, 0) = - 4, and rho(x, 0) = 0 for x > 0, ..... c(x, 0) = 4. b) Green light turns red at t = 1. c) Red light turns green at t = 2. Write full solution, for all times. Track down the shocks that form at x = 0 and t = 1 [last car through light and stopping locus on x < 0] as they interact first with the rarefaction fan from x = t = 0, and later with the one from x = 0 and t = 2. Do in detail the setting of the ode's for the shocks (from the Rankine Hugoniot conditions) and solving them. Check that the entropy conditions are satisfied, and that the shocks do not interact with any other waves that may change their path (for example: with the first car through the light at t = 0). % % -------------------------------------------------------------------- #040 % Optional exercise for students. Strongly recommended. Example, for #038 (quadratic flow). % a) Red light turns green, with a finite line of cars stopped behind the light: rho = 0 for 0 < x; rho = 1 for -2 < x < 0; rho = 1/2 for x < -2; Solve using characteristics. Then plug in expansion fans and shocks. b) Let the light turn back to red at t = T > 1. Continue the solution. Identify shocks that correspond to car paths (last car through light) and check shock speed = u for them. c) Let the light turn back to green at t = 2*T. Continue solution. % % -------------------------------------------------------------------- #041 % Optional exercise for students. Strongly recommended. EXAMPLE of a signaling problem (quadratic flow). Solve: Part I. c_t + ((1/2)*c^2)_x = 0 on x > 0 with c(0, t) = 1 for t < 0; c(0, t) = 1/2 for t > 0; Part II. Same as above, but now let c return to 1 after t=1. That is c(0, t) = 1/2 for 0 < t < 1; c(0, t) = 1 for 1 < t; Find shock, and calculate path all the way to t = infinity. Note that shock starts with constant velocity, but then gets into a rarefaction fan, where velocity is variable. Note that, as t --> infinity, the shock approaches the forward edge of the rarefaction fan in (1/2)*t < x < t (t > 0), where c = x/t. In both cases, display the solution in space-time, as well as do plots of c as a function of x for various typical times. Part III. Qualitative: What happens if c(0, t) = D for 1 < t; where 1/2 < D constant? Case D > 1. Shock is able to cross through the rarefaction fan, and returns to a constant velocity, separating c = D behind from c = 1 ahead. Case D < 1. As t --> infty, the shock approaches the line x = D*t (inside rarefaction fan), becomes very weak, and approaches a characteristic. % % -------------------------------------------------------------------- #042 INFORMATION LOSS AT SHOCKS. Cover lightly now, and specially do (3) below. May come back to this after #044--#049 % % For more details see the two problems: % Information loss and traffic flow shocks. % Preventive driving and dissipation. % in the problem series: KiNe (Kinematic Waves) % Example: Can one calculate how much "information" is lost at a shock? Answer depends on how one measures the amount of information that one has. However, a "partial" answer to this question goes as follows 1) As long as the solution has derivatives, rho_t + q(rho)_x = 0 implies the conservation of any f(rho), with flux F(rho) given by the equation F' = c*f'. 2) When shocks appear, this is no longer true. Hence one can measure the "loss" caused by a shock by picking an "appropriate" f(rho). Notice that, if one knows ``int f(rho) dx'' for all possible f's, one can recover rho (provided rho is, say, continuous). 3) Example: A_t + ((1/2)*A^2)_x = 0, with A conserved [baby example of a river floods equation]. Show then d/dt (\int_a^b (1/2)*A^2 dx) = (1/3)*A^3(a, t) - (1/3)*A^3(b, t) - (1/12) [A]^3 \-------------------------------/ \-------------/ Conservative part (a flux) Stuff lost at shock. where [A] = A_ - A+ > 0 is the jump across the shock in A. Note that \int A^2 is a rough measure of the amount of information, in the following sense: for fixed total area \int A in some interval, and A > 0, the minimum value for \int A^2 occurs when A is constant. The more structure (info) the function A has, the larger \int A^2 will be. % % -------------------------------------------------------------------- #043 Yet another recap on the laws governing shocks/hydraulic jumps. IMPORTANT: Shocks are not always the answer to wave breakdown and multiple values. Shocks introduce new physics. Without shocks, system is reversible. With shocks, system is not reversible. Physics needed to get shocks: flow against gradient and locality. --- Rankine Hugoniot jump conditions for shocks and integral form of the conservation laws. s = [q]/[rho]. Recall simple derivation using conservation in shock frame, where flux is (u-s)*rho. Hence ((u-s)*rho)- = ((u-s)*rho)+. --- Shocks are curves in space time along which characteristics end, so crossing does no occur. They are NOT NEEDED OTHERWISE! Shocks MUST satisfy the ENTROPY condition: c- > s > c+. At a discontinuity where c+ < c-, insert an EXPANSION FAN, not a shock. --- Arrow of time and causality: characteristics die at shock, hence INFORMATION IS LOST. Irreversible evolution, FORWARD IN TIME ONLY. Entropy involves death (of characteristics, at shocks). --- The entropy condition allows causality, with the shock (and the solution on each side) determined by the data (see next). --- Assume piecewise smooth solution with shocks. At a shock, the problem splits into three subproblems: - Solution ahead of the shock. Smooth and satisfying characteristic form. Determined at each point by a single characteristic connecting to data. - Solution behind the shock: same deal as solution ahead of shock. - Shock position determined by solving o.d.e. given by jump cond. --- The shock conditions in the Q-rho Flow-Density plane. a) Need consistency between entropy condition and jump condition. Jump condition must yield shock velocity satisfying c- > s > c+. Works for for Q concave or convex. More complicated otherwise. May get back to this later. FOR NOW WE ONLY CONSIDER SITUATIONS WHERE THE FLUX IS CONVEX OR CONCAVE. b) Traffic flow: backward facing shocks, moving slower than cars. Car enter shock region from behind. (Q concave). c) River flow: forward facing shocks, moving faster than water flow. Water enters hydraulic jumps from ahead. % % ------------------------------------------------------------------ #043.5 Dimensional analysis, similarity solutions, and Group invariance. FOLLOW UP ON #014. In various problems we have found information about the solutions using dimensional arguments [diffusion without stirring #014b, blast wave from single point #014c], while in others we have found special ``similarity'' solutions where the pde can be reduced to an ode [Ex1 through Ex4 below]. Mathematically, all these situations result because the pde is invariant under the action of a continuous group of transformations (a ``Lie'' group). Formally, for every element a in the group there is a transformation u --> G_a(u) such that if u is a solution, so is G_a. Further: G_b * G_a = G_{b*a}. The similarity solutions are the ones such that u = G_a(u) for all a. For EXAMPLE, consider Ex1 with initial conditions u(x, 0) = g(x) that satisfy g(a*x) = g(x) [Riemann problem]. Then if u is a sln., G_a(u) is a sln. as well. So, by uniquenes u = F_a(u). This means: u(x, t) = u(a*x, a*t) for any a > 0. Evaluate this at t = 1/a, to get u(x, t) = u(x/t, 1). But a [i.e. t] is arbitrary, thus u(x, t) = f(x/t), where f(s) = u(s, 1). In general, solutions that satisfy u = G_a(u) involve fewer independent variables, and reduce the complexity of the equations to be solved. Examples % --------------------------------------------------------- Ex1 Rarefaction Fans ....................... u = f(x/t); u_t + q(u)_x = 0. G_a(u) = u (a*x, a*t); Group: Positive real numbers under multiplication. Ex2 Traveling waves ........................ u = f(x-c*t); Constant coefficients pde. G_a(u) = u(x-c*a, t-a); c = constant; Group: real numbers under addition. Ex3 Problems KINE03 and KINE04 ............. u = t*f(x/t*2); u_t + (0.5*u^2)_x = 0. G_a(u) = u(a^2 x, a*t)/a; Group: Positive real numbers under multiplication. Ex4 Diffusion .............................. u = t^(-n) f(x/sqrt(t)); u_t = Laplacian(u) G_a(u) = a^n u(sqrt(a)*x, a*t); Group: Positive real numbers under multiplication. % ------------------------------------------------------------------ Invariance under a group of transformations is very important in physics. For example: classical mech. is invariant under Galilean Transformations (this is ``why'' all inertial frames are equivalent), while relativistic mechanics (and EM) are invariant under Lorentz Transformations. This is often turned around in High Energy Physics, with the equations designed constrained by some desired Group invariances. You can use invariances to not only get information about the solutions, but also as a SANITY CHECK. Example: if you are dealing with a classical system, and your conclusions are not Galilean invariant, then something is ``fishy'', and you'd better figure out what is going on. Note: Traffic flow is not Galilean invariant because it is done relative to a road, which is fixed. Thus there is a preferred frame. Dimensional analysis is related to this because the solutions should not depend on the units used. Example; time unit change is: t \to a*(t-t_0). % % ------------------------------------------------------------------------- % ======================================================================= % % NEW TOPIC: % % ************* Shock structure due to higher order effects ************* % % ======================================================================= % % % -------------------------------------------------------------------- #044 Diffusion effects in traffic flow: "look ahead" by drivers. When the gradients are not small, one must consider the fact that reasonable people drive "preventively" --- i.e.: they attempt to predict "future" traffic conditions, and proceed accordingly. A rough model for this is: If density is larger ahead, rho_x > 0, slow down below what local density indicates, and conversely. A simple mathematical model incorporating this is Take: q = Q(rho) - nu*rho_x in the conservation of cars law, where nu > 0 is a constant (nu is a diffusivity). This leads to: rho_t + (Q(rho))_x = nu*rho_xx Here "nu" is small away from shocks. What this means is that, if L = typical length scale for the traffic flow away from shocks, and T = typical time scale for the traffic flow away from shocks, then nu*T/L^2 is small. Of course, for the time and space scales near shocks this fails, as we will see next. % % -------------------------------------------------------------------- #045 The role diffusion plays in stopping steepening and wave breaking. Note that new term adds a contribution to the car flux which is of the same type as the one diffusion (as in heat) causes. As a wave steepens, this term gets larger and larger, till it eventually can stop the steepening^\# # Notice that it produces a flow against the gradient, thus it "fights" the wave steepening produced by the nonlinear terms. % % -------------------------------------------------------------------- #046 --- Shock as thin layer where diffusion and nonlinearity balance. --- Shock structure argument: the shock zone in space time is very thin, thus shock look ``locally'' like a plane wave. That is: a traveling wave. % % -------------------------------------------------------------------- #047 Traveling wave for augmented equation. Substitute rho = R((x-s*t)/nu) into the equation. Then the equation reduces to ode for R. Study this ode and: --- Analyze conditions under which a wave connecting two given states as (x-s*t)/nu \to \pm \infty exists. This gives shock as nu vanishes. --- Recover jump and entropy conditions. ODE is: R^\prime = Q(R) - [s*R + const.]. Use graphical analysis: curve Q(R) vs straight line s*R+const. Do recap of "graphical theory" for \dot{y} = f(y). Phase line, and qualitative plot of solutions y = y(t). Behavior near critical points by linearization. Inflection points. Etc. Exact solution by separation is useless, unless f(y) is pretty simple [e.g.: 1st/2nd order polynomial]. % % -------------------------------------------------------------------- #048 % @#@ SKIP EXAMPLE: Quadratic flow case. Wave speed c is linear in rho ......... [PSQ] Show that then rho_t + (Q(rho))_x = nu*rho_xx is the same as c_t + c*c_x = nu*c_xx Burgers's Equation. Get explicit solution for traveling waves. % % -------------------------------------------------------------------- #049 Compute time derivative of \int (1/2)*rho^2 dx for viscous problem rho_t + (Q(rho))_x = nu*rho_xx Show decreasing (dissipation). Show time derivative does not go to zero (as nu vanishes) if there are shocks [plugin form rho = R((x-s*t)/nu)]. RECALL/RECAP: #042, Information loss at shocks. This is intimately related to the topic here: When diffusion is small, we can neglect it EXCEPT at at shocks. This diffusion causes the dissipation/loss of info. there --- as shown in this point. % ======================================================================= % % NEW TOPIC: % % **************** Gas Dynamics, Acoustics, and Strings. **************** % % ======================================================================= % % % -------------------------------------------------------------------- #050 Gas Dynamics: Derive equations in 1-D (using conservation of % Recall. mass and momentum), plus the quasi-equilibrium isentropic % Done assumption p = p(rho). Example: p = kappa*rho^gamma. [#] % before. [#] polytropic gas (ideal gas with constant specific heats). --- Boundary conditions at the end of the pipe: Closed pipe u = 0; and Open pipe p = p_0; --- For smooth solutions, manipulate equations into the form rho_t + u*rho_x + rho*u_x = 0 u_t + (a^2/rho)*rho_x + u*u_x = 0 where a^2 = dp/drho > 0 [note that dp/drho > 0]. Calculate a for ideal gas case: a^2 = gamma*p/rho. Check dimensions: a is a velocity (sound speed, as we will see). Derive also Shallow Water equations on a flat bottom. Simplify derivation by neglecting air pressure [makes no difference, since adding a constant to the pressure does not change the forces]. Note that these equations are THE SAME AS GAS DYNAMICS WITH GAMMA = 2. % % -------------------------------------------------------------------- #051 Review topic from thermodynamics for ideal gases: Show p = rho^gamma for ideal gas with constant specific heat at constant entropy. MORE DETAILS ASSIGNED IN A PROBLEM SET ...... [PSQ] % % -------------------------------------------------------------------- #052 For gas dynamics, at constant entropy, so that p = p(rho): Property of p(rho): a^2 = dp/d rho > 0. Show a has dimensions of speed. Compute what a is for Shallow Water. Calculate a for tidal wave in the deep ocean (4.000 m). Some rough numbers: p = 1 atmosphere ~ 10^5 kg/(m*s^2) [about 10m of water depth]. rho air ~ 1 kg/m^3 [about 1/1000 that of water]. gamma air ~ 1.4 This yields a ~ 370 m/s [actual is ~ 340 m/s] For shallow water: p = (1/2)*g*h^2, a = sqrt(g*h). 4000 m depth yields: 200 m/s % % -------------------------------------------------------------------- #053 Gas Dynamics: Notice form Y_t + A(Y)*Y_x, similar to the scalar case, but with the wave velocity replaced by a matrix. Look for o.d.e. forms [i.e. characteristics] by doing linear combinations of the equations. Need to find combinations that produce only one directional derivative in space-time. ** HYPERBOLIC IN 1-D ** Equations can be reduced to statements about directional derivatives of the solution. Equivalent: A is real-diagonalizable. Show it works if using eigenvalues/eigenvectors L*A = c*L: Characteristic form: L*(Y_t + c*Y_x) = 0, or L*dY/dt = 0 along dx/dt = c. Then, along the curves dx/dt = c, solution behaves (sort of) like an o.d.e. Hyperbolic: have enough linearly independent (real) eigenvectors (in this case, 2) so that equation is equivalent to stuff above. This happens if and only if A is real diagonalizable. Before applying these ideas to the full Gas Dynamics problem, LINEARIZE near the equilibrium solution u = 0, and rho = rho_0, and analyze the resulting problem (this is ACOUSTICS - NEXT ITEM/TOPIC below). For a linear, constant coefficients, hyperbolic system: Y_t + A*Y_x = 0. Start with an orthonormal base of left and right eigenvectors: Ln*A = cn*Ln \ Linear Algebra A*Rn = cn*Rn | Explain and Ln.Rm = delta_{n, m} / justify. #053b Then show general solution is Y = sum_n y_n(x-cn*t)*Rn. % % -------------------------------------------------------------------- #053b Needed linear algebra facts (sketched proofs for most of them). Note: here we are only interested in real matrices, though the results have easy generalizations to complex matrices. a) A symmetric matrix has real eigenvalues, and the eigenvectors can be taken to be an orthonormal basis. b) If {n_j}, 1 <= j <= N, is an orthonormal basis of R^N, then any vector Y can be written in the form Y = sum_1^N y_j n_j, where y_j = , where < , > = scalar product. c) Let A be a real-diagonalizable square matrix. Let R_j, 1 <= j <= N, be a set of N linearly independent right (column) eigenvectors of A A R_j = lamda_j R_j, where the lambda's are the eigenvalues [they need not be distinct]. Then a set of N linearly left (row) eigenvectors of A L_j A = lambda_j L_j can be selected such that L_n^T R_m = delta_{n, m}, where delta_{n, m} is the Kronecker delta. Then any (column) vector Y can be written in the form Y = sum_1^N y_j R_j, where y_j = L_j^T Y. This formula generalizes the one in (a-b), which applies for symmetric matrices. % ------------------------------------------------------------------------- % =============================================================== % % The material below can be (partly) found in the book by Wan. % % =============================================================== % % % -------------------------------------------------------------------- #054 % % -------------- Linear Gas Dynamics - Acoustic in a pipe -------------- % % % -------------------------------------------------------------- #054a Write equations: R_t + rho_0*u_x = 0 and u_t + (a_0^2/rho_0)*R_x = 0, where R is the density perturbation to rho_0 [rho = rho_0 + R] Boundary conditions for closed (u = 0) and open (R = 0) pipe ends. % % -------------------------------------------------------------- #054b Find eigenvalues c = +/- a_0 and and eigenvectors L = (+/- a_0, rho_0) Write in characteristic form and solve. Solution superposition of two waves: left and right going sound waves. Get form (u + (a_0/rho_0)*R)_t + a_0*(u + (a_0/rho_0)*R)_x = 0, (u - (a_0/rho_0)*R)_t - a_0*(u - (a_0/rho_0)*R)_x = 0. Thus: u + (a_0/rho_0)*R = f(x-a_0*t), u - (a_0/rho_0)*R = g(x+a_0*t). So: u = (1/2)*[f(x-a_0*t) + g(x+a_0*t)], R = (rho_0/(2*a_0))*[f(x-a_0*t) - g(x+a_0*t)], P = (a_0/2)*[f(x-a_0*t) - g(x+a_0*t)], where P = (a_0^2/rho_0)*R is the pressure perturbation. % % -------------------------------------------------------------- #054c Show equivalent to wave equation: Eliminate either R or u. Better yet, introduce velocity potential: u = phi_x and (a_0^2/rho_0)*R = - phi_t \-------------/ Note this is the pressure perturbation. Hence, second equation is satisfied. Then first equation gives phi_tt - a_0^2*phi_xx = 0. Wave equation. Note boundary conditions: Closed pipe: phi_x = 0. Open pipe: phi = 0. Give other examples where these boundary conditions occur: --- Shallow water: Closed and open channel. --- String equation: Free end and clamped. Using the solution above for gas-dynamics, we see that phi = F(x-a_0*t) + G(x+a_0*t)], where F' = (1/2)*f and G' = (1/2)*g. % % -------------------------------------------------------------- #054d Use (2-3) to write general solution for the wave equation. % ====================================================== % % *** This is, cover: 064, 065, 066, and 067 below. *** % % ====================================================== % % ------------------------------------------------------------------------- % ================================================= % % The material in #055, #056, #057 is (partially) % % From: Various lecture notes for 18311 % % Section: Isentropic Gas Dynamics. % % ================================================= % % -------------------------------------------------------------------- #055 Derive characteristics for the full 1-D isentropic Gas Dynamics. rho_t + u*rho_x + rho*u_x = 0 u_t + (a^2/rho)*rho_x + u*u_x = 0 where a^2 = dp/drho. Write in the form Y_t + A(Y)*Y_x = 0, and find the eigenvalues and left eigenvectors for A (namely: solve L*A = c*L). Then c = u +/- a, and the characteristic form is: +/- (a/rho)*(drho/dt) + du/dt = 0 along dx/dt = u +/- a. Two sets of characteristics, which interact and couple. Situation similar to u_tt - u_xx + V(u) [see #065], but more complicated: Now the characteristic speed is no longer constant. Characteristics in the same family may cross ... leading to shocks. Introduce h = h(rho) by property dh/drho = a/rho. Show for ideal gas h = 2*a/(gamma-1). Then d/dt (u +/- h) = 0 along dx/dt = u +/- a. i.e (u +/- h) is constant along characteristics. Show how this, in principle, determines the solution. At each point in space time two characteristics [C+ and C-], each carrying information from a different part of the initial data, which combined gives the solution at the point. But now the characteristics are neither straight, nor can we solve for them explicitly, because they interact with each other. % % -------------------------------------------------------------------- #056 Example: simple waves in Gas Dynamics. Assume u-h = constant for initial data. Then characteristics yield u-h = L = constant, as long as characteristic form applies. Hence u = h + L = U(rho) is a function of rho only. The equations then reduce to d/dt (u+h) = 0 along dx/dt = u+a. That is (U+h)_t + (u+a)*(U+h)_x = 0, which is a first order equation of the same type as Traffic Flow and River Flows. Hence characteristics can cross and once this happens we need to re-examine the physics to see what to do beyond breakdown. In this case, again, shocks are the appropriate answer. Note: the basic idea behind a simple wave was introduced by S. Earnshaw On the mathematical theory of sound. Phil. Trans. vol. 150, pp. 133-148, 1858 He looked for solutions of the isentropic Gas Dynamics equations were u = u(rho). This leads to the same answer as above. The Riemann invariants were introduced by Riemann, at about the same time: B.Riemann. Uber die Fortpflanzung ebener Luftwellen von endlicher Schwingungsweite. Gottingen Abhandlungen, Vol. viii, p. 43, 1858. While the Riemann invariants provide an elegant and powerful tool to solve the isentropic Gas Dynamics, they are a property of only some hyperbolic systems. On the other hand, all hyperbolic systems have simple wave solutions. As follows: Consider Y_t + A(Y) Y_x = 0, and let lambda(Y) be an eigenvalue of A, with eigenvector R(Y). That is A(Y)*R(Y) = lambda(Y) R(Y). Let U(s) be a solution to the ode dU/ds = R(U). [there are always solutions!] Then Y = U(r), where r satisfies the scalar equation r_t + \lambda(U(r)) r_x = 0 solves Y_t + A(Y) Y_x = 0. [proof left to the students]. This solution is called a ``simple wave'', and generalizes Earnshaw's idea of looking for a solution to isentropic Gas Dynamics which depends of a single variable [in his case, rho]. % % -------------------------------------------------------------------- #056.5 Preview of domain of dependence and domain of influence. Show how to solve an hyperbolic equation by infinitesimal time steps using characteristics, and how this yields the generic result: finite speed of propagation [e.g.: in relativity no signals beyond light cone]. Consider only the simpler case where Riemann variables exist so equation is d R_n/dt = F(\vec{R}) along dx/dt = lambda_n(\vec{R_n}). See picture in https://math.mit.edu/classes/18.306/index.html % % -------------------------------------------------------------------- #057 % Shocks in Gas Dynamics: Rankine-Hugoniot jump conditions. Two types of shocks, needed to suppress either C+ or C- characteristics from crossing Lax Entropy conditions. Show how the "equation counting" is correct: two unknowns, shock speed and the "surviving" characteristic, and two eqns. Graphical interpretation of the Rankine-Hugoniot jump conditions. % % -------------------------------------------------------------------- #058 % SKIP?? Optional material. Piston problem. Use Riemann invariant form of equations in simple problem: -- Gas in a pipe, initially at rest: rho = rho_0 and u = 0, x > 0. -- Some signal at x = 0, t > 0 (sound) u = sigma(t) -- Argue left Riemann invariant constant through-out flow, provided no shocks. Reduce problem to finding right R. I. -- Point out problem is now very similar to Traffic flow. -- Do case sigma(t) = V < 0 (constant). Show need for expansion fan in C+ characteristics -- Do case sigma(t) = V > 0 (constant). Show C+ characteristics cross. Need shock, and left Riemann invariant assumption has to be re-examined. % % -------------------------------------------------------------------- #059 % SKIP?? Optional material. EXAMPLE: The p-system. % --- Characteristic form. Riemann invariants and SIMPLE waves. Shocks. --- Isentropic Gas Dynamics in Lagrangian Coordinates is the p-system (change of variables). --- p-system same as equation for a nonlinear vibrating elastic rod --- see #009. % % -------------------------------------------------------------------- #060 % SKIP?? Optional material. Weak shocks in Kinematic Waves. --- Weak shock velocity (average of characteristic speeds). --- Weak shocks and quadratic approximation for flux. % % -------------------------------------------------------------------- #061 % SKIP?? Optional material. General problem of shock formation in Kinematic Waves. -- Shock formation time and envelope of characteristics. -- Envelopes. Examples of envelopes and shocks. % % -------------------------------------------------------------------- #062 % SKIP?? Optional material. Burgers' equation. -- Cole-Hopf transformation: exact solution. -- Solution of heat equation by Green's function and application to Burgers. -- The small viscosity limit in Burgers'. Laplace's method. % % -------------------------------------------------------------------- #063 % SKIP?? Optional material. Sonic booms (toy model). Full problem requires 3-D and motion of finite body through air. -- Very messy. Simplify dramatically, to bare bone effects. Use axe if needed: -- Not interested in details near the plane, so think of the plane as a point source of momentum. -- Knock the problem to 1-D. But then loose the fact that in 3-D the waves, as they move away from the plane, decay [geometrical effect due to expansion]. Replace this by adding some ``damping'' into the system. -- Replace Gas Dynamics by the simplest (nonlinear) model with wave velocity dependence on the solution, e.g. use a simple wave model. Thus, the TOY MODEL is p_t + ((1/2)*p^2)_x = delta(x-V*t) - a*(p-p_0), a > 0. where p_0 = "ambient" state into which plane moves. Now -- Use a Galilean transformation to set p = p_0 -- Explain meaning of delta function (point source of p) and derive jump conditions across source path x = V*t. -- Write characteristic equations for problem away from x = V*t. Assume initial conditions p = p_0 = 0, and solve problem: Case V < 0. -1- Solution behind plane trivial. -2- Show characteristics starting behind the plane reach it, go through it, and determine solution ahead of plane. These do not cross, and set-up a steady state ahead of plane (show solution by characteristics yields p = p(x-V*t) ahead of plane. -3- There is a shock separating -2- from the air at rest far ahead of the plane. But this shock has strength that vanishes as it moves further and further away from plane. -4- Solution is smooth steady state in t --> infinity limit. Case V > 0 (but not too large) -5- Show need expansion fan behind the plane. -6- Steady state solution analog to 2 has characteristic crossing and needs a shock. -7- Solve for steady state and show solution has a square root singularity. -8- Steady state, t ---> infinity solution, now has shock riding ahead of plane. Case V > 0 "large" (plane catches up to shock ahead). Still to write. % % -------------------------------------------------------------------- #064 % NOTE: This was/may have been covered with #054d. Characteristic form for wave equation: u_tt - c^2*u_xx = 0. WAY #1: use relationship with acoustics and its characteristics. [Done earlier]. Equivalently: Transform equation into 1st order system introducing a potential. -- Equation is curl(V) = 0 for vector field V = [u_t, c^2 u_x]. -- Remind/prove theorem: Irrotational vector fields in simply connected domains are gradients. -- Remind students to review the Green/Gauss/Stokes theorems in calculus. -- HENCE u_t + \phi_x = 0 and \phi_t + c^2 u_x = 0, which is acoustics with phi = delta p = (a_0^2/rho_0)*R and c = a_0. WAY #2 More symmetric: Look for linear combinations of u_t and u_x that yield equations with directional derivatives only. This gives: (u_t + c*u_x)_t - c*(u_t + c*u_x)_x = 0 (u_t - c*u_x)_t + c*(u_t - c*u_x)_x = 0 Solve this and get again solution. % EXTRA DETAILS (review of prior material) Linearized gas dynamics: rho_t + rho_0*u_x = 0 and u_t + (c_0^2/rho_0)*rho_x = 0 Show equivalent to wave equation (eliminate either rho or u). Write in characteristic form and solve: 1) u_t + c_0*u_x + (c_0/rho_0)*(rho_t + c_0*rho_x) = 0 ===> rho_0*u + c_0*rho = constant on characteristic x = c_0*t + x_0 2) u_t - c_0*u_x - (c_0/rho_0)*(rho_t - c_0*rho_x) = 0 ===> rho_0*u - c_0*rho = constant on characteristic x = -c_0*t + x_0 Solution superposition of two waves. WAY #3 The wave equation u_tt - u_xx = 0 (set speed c = 1). Convert to characteristic form by introducing v = u_t - u_x. Solve and find general solution to equation. Solution is two waves, moving right and left at speed c = 1. u_t - u_x = v and v_t + v_x = 0. % % ================================================================ % % Left in the ink-pot: % % % % Transform wave equation u_{tt} - c^2 u_{xx} = 0 into a 1st order % % system by introducing a potential --- % % interpret equation as a curl(V) = 0 % % for vector field V = [u_t, c^2 u_x]. % % --- Remind/prove theorem: % % irrotational vector fields in simply connected domains % % are gradients. % % --- Remind students to review the Green/Gauss/Stokes theorems % % in calculus. % % % % Show: boundary conditions for a tied string of length l lead to % % a solution of space period L = 2*l --- extend solution by % % "reflecting" across ends]. % % ================================================================ % % % -------------------------------------------------------------------- #065 % NOTE: This was/may have been covered with #054d. Another example: u_tt - c^2*u_xx + V(u) = 0, which yields (u_t + c*u_x)_t - c*(u_t + c*u_x)_x = - V(u), (u_t - c*u_x)_t + c*(u_t - c*u_x)_x = - V(u). Two families of characteristics, that interact with each other. Can no longer produce an exact solution (i.e.: cannot solve explicitly for u_t + u_x, or u_t - u_x, along their corresponding characteristics. However: 1. Examine how the solution advances in space time. 2. Notions of domain of dependence and domain of influence. 3. Light/sonic cone. No information can propagate faster than c. If solution vanishes outside an interval, it stays zero outside light cone later. Better way to write the equation in characteristic form: u_t + c*u_x = v v = new variable, defined by this eqn. v_t - c*v_x = -V(u) % % -------------------------------------------------------------------- #066 % NOTE: This was/may have been covered with #054d. % EXAMPLE: Waves on strings [See #008]. % Recall that the wave equation u_tt - c^2*u_xx = 0, can be used to describe the small transversal motion of an homogeneous string under tension. Do the derivation if problem CoLa07 has not been assigned, including the boundary conditions: If string is tied at one end: u vanishes there. If string is free at one end: u_x vanishes there. Show, explicitly, that the units for c (as given in terms of the tension and the string density) are those of a velocity. % % -------------------------------------------------------------------- #067 % NOTE: This was/may have been covered with #054d. % EXAMPLE: String on an elastic bed. % This adds a restoring force at each point (source term). The equation becomes u_tt - c^2*u_xx + b*u = 0, where b = k/rho, k is the elastic restoring force constant, and rho is the string density. A more general restoring force yields u_tt - u_xx + V(u) = 0. % ====================================================================== % % NEW TOPIC: % % **************** Separation of variables/normal modes **************** % % ====================================================================== % % % -------------------------------------------------------------------- #068 Separation of variables, review. Another way to solve the wave equation (works for other equations too). Explain idea of separation of variables, write solution as product, etc. Example: do heat equation in 1D with T=0 at ends. do wave equation in 1D with u_x = 0 at ends. Note: for the wave equation the solutions obtained in this way should be compatible with the form u = f(x-c*t) + g(x+c*t). Exercise: Typical separation of variables solution has the form: u = cos(n*pi*t/L)*sin(n*pi*x/L) which yields u = (1/2)*cos((n*pi*/L)*(x+t)) + (1/2)*cos((n*pi*/L)*(x-t)) using trigonometric equalities. % % -------------------------------------------------------------------- #069 Normal modes. Equations of the form u_t = Lu Relationship with separation of variables: equation invariant under time shift allows separation u = exp(lambda*t) U(x) Example: write wave equation as u_t = v and v_t = u_xx. heat equation u_t = u_xx --- Note analogy with linear o.d.e. dY/dt = A*Y, A NxN matrix, solved by finding eigenvalues and eigenvectors of A. --- Hence look for solutions of the form u = e^{\lambda*t} v(x) Solve and find normal modes (eigenvalues and eigenfunctions). General solution: Superposition ... leads to Fourier Series, etc. Example: heat equation with various B.C. 1. In a ring: periodic. 2. Zero T at ends. 3. Zero flux at ends. Various types of Fourier series. Explain this works, for example, as long as the assocated eigenvalue problem is self-adjoint: 1) Interpretation of matrices as linear operators. 2) Interpretation of self-adjoint for matrices in terms of the scalar product. 3) Definition of scalar product and Hilbert space. Will not go into details of this: Sturm-Liouville problems. Series and Transforms. When do normal modes work. Etc. % % -------------------------------------------------------------------- #070 % SKIP?? May assign in a problem set ............................... [PSQ] Consider a string tied at the ends. Use a-dimensional variables. Then: u_tt - u_xx = 0 and u = 0 at x = 0 and x = L Find normal modes (or separate variables), and find connection with characteristics: Normal modes as superpositions of a right and a left traveling wave. % % -------------------------------------------------------------------- #071 % SKIP?? May assign in a problem set ............................... [PSQ] Wave equation. Show that: Boundary conditions for a tied string of length L lead to a solution of space period P = 2*L --- extend solution "reflecting" across ends. % % -------------------------------------------------------------------- #072 Example of normal modes. T_t = T_xx for 0 < x < 1, with: T = 0 for x = 0, \ [E] and T_x + T = 0 for x = 1. / Physical meaning of the boundary conditions (heat and elasticity) --- Dirichlet: ice bath or rigid clamped end. --- Neumann: flux prescribed or free end (no stress). --- Robin: fluid cooling or elastically clamped end. Show space operator in [E] is self-adjoint and negative ==> eigenvalues real and negative. Calculate eigenvalues and use to write solution in terms of the initial value T(x, 0) = f(x). Graphical solution of the equation for the eigenvalues: lambda = - k^2, where k*cos(k) + sin(k) = 0, and k > 0. Plot k versus tan(k) and show solutions k_n, k_n ~ pi*(n-1/2), n = 1, 2, ... Explain how to solve using Newton's method. % % ====================================================================== % % NEW TOPIC: % % ***************** Finite differences schemes for pde ***************** % % ====================================================================== % % ====================================================================== % % The material below is (partly) contained in the WEB page notes: % % Stability of Numerical Schemes for PDE's. % % Further stuff can be found in the introductions to the problem % % suites (also to be posted in course WEB page) % % --- GBNS Good and Bad Numerical Schemes. % % --- vNSA von Neumann Stability Analysis. % % --- AENS Long wave equation associated to a numerical scheme. % % ====================================================================== % % % -------------------------------------------------------------------- #073 Finite difference schemes for PDE's. Follow notes in WEB page: --- Stability of Numerical Schemes for PDE's. 1 Naive Scheme for the Wave Equation. USE the MatLab script in the 18311 toolkit: GBNS_lecture. See #074b. 2 von Neumann stability analysis for PDE's. 3 Numerical Viscosity and Stabilized Scheme. 4 Model equation. About 2.5 weeks of lectures. Reminder of details to include here: 1) von Neumann stability analysis. % ---------------------------- #073b Recall that solutions to time-evolution linear PDE's can be found by separation of the time variable --- recall analogy with ODE approach --- leading to an eigenvalue problem. Extend idea to constant coefficients linear FD schemes --- key to the von-Neuman stability analysis. 2) Examine instabilities using associated equation: % ----------- #073c Explain behavior via forward & backward heat equations. 3) Introduce stabilization by artificial viscosity % ------------ #073d (general idea). Relationship with the solutions of the heat equation by separation of variables exp(-k^2*t + i*k*x}. Inspect what happens when a term u_xx is added to an equation like u_t+u_x=0 . What does it do to the normal modes? Other examples [0 < D << 1] --- u_t + u_xxx = D*u_xx --- u_t = v + D*u_xx and v_t = u_xx + D*v_xx 4) Define CONSISTENCY. % ---------------------------------------- #073e von Neumann and consistency: Numerical and exact growth rates; comparison in the small k limit. 5) Define STABILITY. % ------------------------------------------ #073f 6) Lax Theorem: % ----------------------------------------------- #073g for linear schemes, Consistency+Stability ==> Convergence. % % -------------------------------------------------------------------- #074 Further details: 6) Show that the general solution to a finite differences linear scheme can be written as a linear combination of the solutions G^j*e^(i*k*n) obtained by the von Neuman stability analysis. In other words, show that the matrix {w^{n*m}} has an inverse, where w = e^{i*2*pi/N}. This leads to the Discrete Fourier Transform (DFT), which will be the next course topic. 7) Associated equation. More examples of von Neuman stability analysis; associated equation, and stabilization by artificial viscosity: Lax-Friedrich sheme. % % -------------------------------------------------------------------- #074b Computer illustrations Numerical instabilities ............... GBNS_lecture scheme.m Wave breaking and steepening .......... demoWBRch_v02.m Convergence of Fourier Series ......... fourierSC.m Talk about Gibb's phenomena [related to new topic below]. % ======================================================================= % % * NEW TOPIC: Fourier Series, Discrete Fourier Transform (DFT) * % % * and Fast Fourier Transform (FFT), Spectral Methods, Laplace * % % * Transform and Fourier Transform. * % % ======================================================================= % % % -------------------------------------------------------------------- #DFTa von Neumann Stability Analysis and DFT: Show that any periodic sequence {u_n} of period N can be written as a linear combination of the "modes" u_n = exp(i*k_m*n), where 0 <= m < N and k_m = 2*pi*m/N. That is: let w = e^{i*2*pi/N}, then show that the NxN matrices A with entries A_{n, m} = w^{ n*m} = exp(i*k_m*n) and B_{n, m} = (1/N)*w^{-n*m} are inverses of each other. This leads to the DFT (Discrete Fourier Transform) IDFT (Inverse DFT) pair f_n = sum_{m=1:N} a_m exp(i*k_m*n) = sum_{m=1:N} a_m w^{n*m}; [#1] a_m = sum_{n=1:N} (1/N)*w^{-n*m}*f_n; [#2] This is implemented by the fft/ifft MatLab scripts. FFT = Fast Fourier Transform IFFT = Inverse FFT. NOTE: point out w is the fundamental N-th root of unity. This plays a crucial role in the proof that B is the inverse of A. % % -------------------------------------------------------------------- #f01 Extend DFT/IDFT formulas for periodic sequences of period N. Show that [#1-#2] in #DFTa can be extended to f_n = sum_{m, *} a_m exp(i*k_m*n) = sum_{m, *} a_m w^{n*m}; [1] a_m = sum_{n, *} (1/N)*w^{-n*m}*f_n; [2] where the sums are over any set m=a+1:a+N or n=a+1:a:N; Argument: the summands are periodic of period N, thus the sum over any N consecutive terms is the same. Show that DFT/IDFT formulas follow from the Fourier series formulas, as follows [do the case N = 2*N+1, which is simpler] #1 Discretize the integral for the Fourier coefficients using the trapezoidal rule over an equispaced grid with N intervals in the period 2*pi [other periods work the same]. #2 Take the complex Fourier series f = sum f_n e^{i*n*x} and discretize it by #2.1 Evaluate f on the same discrete grid used in 1. #2.2 Chop the Fourier sum to go only from -M to M. % % -------------------------------------------------------------------- #f02 Show, using integration by parts, that the Fourier coefficients decay at least as n^{-p}, where p is the number of derivatives that the function has [need f^(p) to exist and be integrable]. Use this to show that #1 The trapezoidal rule, as in #1 of #f01 is as accurate as the number of derivatives for integral of periodic functions over a period. Note: in general, for functions with one bounded derivative, trapezoidal yields an error O(1/N^2). Proof: apply the trapezoidal rule to the Fourier series. #2 Derivatives can be computed very accurately via DFT/IDFT using the fact that, on the Fourier "side", taking derivatives can be done by multiplication: If f(x) = sum f_n exp(i*n*x), then f'(x) = sum i*n*f_n exp(i*n*x). Again: note that decay of f_n connected with derivatives. The faster the decay, the more derivatives one can take. % % ------------------------------------------------------------------- #f03 Brief description of spectral methods, building on the results of #f01. % % ------------------------------------------------------------------- #FFT Describe key idea for FFT. How to reduce the N^2 operations the DFT involves, to N*log(N). Use hand-written notes ............................................. [PTS] % % ------------------------------------------------------------------- #f04 Start with a function f = f(x) defined for -infty < x < infty that decays "very fast" as |x| --> infty, and construct F(x) = sum f(x+n*L) ........ periodic of period L. Note "F --> f" as L --> infty. Now take the Fourier series (and inverse) formulas, apply them to F, and formally get the Fourier Transform and Inverse Fourier Transform formulas. Again: point out that smoothness and decay are duals of each other. % % ------------------------------------------------------------------- #f05 Compute Fourier Transform of Gaussian, and compute the product of their standard deviations. As Gaussian gets narrower, its transform gets wider (and viceversa). Take limit where delta arises. Relationship with uncertainty principle in QM. % % ------------------------------------------------------------------- #f06 Look at Fourier Transform for a one-sided function. Defined on Real(z) geq 0, not just the real axis. Rotate coordinates and get Laplace transform. Use Fourier Transform Inverse formula to get inverse of Laplace. Check how much they know of complex variables. If enough, get formulas for inverse Laplace using poles, etc. % % ------------------------------------------------------------------- #f07 Do proof of Fourier series formula for C^2 functions. From the "Various lecture notes for 18311". % % ========================================> MORE POINTS WILL BE ADDED LATER. % ======================================================================== % EOF