GGR203 - INTRODUCTION TO CLIMATOLOGY 2.0 Radiation The radiation referred to here is the energy emitted by all matter warmer than 0 K in the form of electromagnetic (EM) waves. 2.1 What are EM waves? They are coupled, oscillating, electric and magnetic fields that move out (at the speed of light) in all directions from an oscillating electric charge. Speed of wave, c = speed of individual crests and troughs Frequency of wave, ν = number of crests or troughs that pass a given point per second Wavelength, λ = distance between corresponding points on successive waves Clearly, c = λν (because if, for example, we see 10 wave crests passing a point in one second and the wavelength is 10 cm, the furthest crest was 100 cm away at the start of the second) For EM waves, c is fixed, so λ α 1/ν (wavelength varies inversely with frequency) Frequency of an electric scillator = number of up & down cycles it does per second Frequency of an EM wave = frequency of the oscillator that generated it. THUS: faster frequency oscillator ↔ shorter wavelength of EM radiation See Figure 2.1, illustrating the electromagnetic spectrum - the division of radiation of all possible wavelengths into different categories. Visible light is one very small set of wavelengths within the spectrum, with different colours corresponding to different wavelengths (and hence, different frequencies) (Note that violet is next to ultraviolet, and red next to infrared, with violet being of shorter wavelength and red longer wavelength). When an EM wave encounters another charge, it tends to set that charge in motion – because the oscillating electric field causes an oscillatory force on the charge [force on charge = electric field x magnitude of charge] Interaction of EM wave with a charge tends to set it oscillating at the same frequency as the EM radiation (or not at all) In so doing – the charge acquires energy - the EM radiation is absorbed To conserve energy, it must be that - the EM radiation itself carries energy - an oscillator loses energy when it emits EM radiation. 2.2 Emission of EM radiation All matter contains electric charges (protons, electrons). Although non-ionized atoms and molecules are electrically neutral (# protons = # electrons), the “centre of mass” of the positive and negative charges might not coincide. This gives an electric dipole. Electric dipoles in atoms and molecules can arise in 3 ways: 1. From asymmetry of a molecule. Example: H2O: -electrons are pulled toward the oxygen atom 2.From the presence of an electric field, which pushes electron and protons in opposite directions 3.From asymmetric vibrations. Example: CO2, a symmetric linear molecule O=C=O CO2 can vibrate symmetrically (both bonds being stretch at the sametime), asymmetrically in two ways: bending, or asymmetric bond stretching, where one bond shortens while the other lengthens, and viceversa Heat is the random motion of molecules and atoms. As all matter above 0 K is in thermal motion, and as all matter contains electric dipoles – all matter radiates EM radiation. Consider a solidbody: - At a given temperature, the atoms or molecules will oscillate at a given set of frequencies with a given set of amplitudes – to there will a range of wavelengths at which radiation is emitted, corresponding to the frequencies of the oscillators As T increases, - more energy goes into existing oscillation frequencies by increasing the amplitudes of the oscillations - oscillations at higher frequencies (↔ shorter wavelengths) can occur Hence, as T increases, more EM energy is radiated (emitted), and the emitted radiation shifts to shorter wavelengths. The relationship between temperature and the maximum amount of radiation that can be emitted in a given wavelength interval is given by the Planck Function: where c = speed of light = 2.998 x 108 m/s h = Planck’s constant k = Boltzman constant B(λ,T) is the energy emitted perm2 of surface area per μm of wavelength interval. The energy emitted in a λ interval of width Δλ and centred at λ is B(λ,T)Δλ W m-2 For example, to estimate the amount of radiation between wavelengths of 1.00 μm and 1.02 μm, calculate B(λ,T) using λ in the centre of the interval (1.01 μm) and multiply by Δλ=0.02 μm. Sum this approach over many small intervals: We will designate the different intervals and the wavelength at the centre of each interval by the subscript i, where i=1 for the first interval, i=2 for the second interval and so on up to the last interval, n. The energy emitted in each interval i is B(λi,T)Δλi, and the total energy emitted is the sum of the radiation emitted in all the intervals. That is Total energy emitted = ∑1 B(λ i , T)Δ λ i ~ “Area” under that curve, where the “area” has units = units of height x units of width of each rectangle. (in the above, ∑1 xi is the standard summation notation and means add all the values of xi (whatever xi is) from x1 to xn) As stated above, the Planck function gives the maximum amount of radiation that can be emitted in a given wavelength interval. An object that emits this amount at all wavelengths is called blackbody. Layout calculations for Q2 of PS1 in columns: 2.3 Radiation Laws Derived From the Planck Function (or: properties of blackbodies) 1. The emission at all wavelengths increases as T increases 2. As Δλ i → 0, ∑Ni=1B(λi, T)Δλi → ∫0 ∞ B(λ, T)dλ = σT4 Total radiation emitted by a blackbody, B(T) = σT4, where σ = 5.673 x 10-8 W m-2 K-4 is the Stefan-Boltzman constant, and the equation is called the Stefan-Boltzman Law 3. Wavelength at which maximum emission occurs decreases as T increases. λmax = 2898/T Wien’s Displacement Law, T isin K and λ is in μm At higher T, higher frequency oscillations – corresponding to shorter λ radiation – can be excited. 4. Almost all the radiation emitted by the Sun is at λ 4.0 μm See Figure 2-2. To recap – the Planck function gives the blackbody emission - which is the maximum amount permitted at a given temperature. The ratio of actual emission (F(T)) to BB emission is called the emissivity, ε . That is, ε = F(T)/σT4, so 11 F(T) = εσT4. ε ≤= 1 , ε = 1 for a BB Emissivity at a particular wavelength is called the spectral emissivity, ελ . The emission is F(λ,T) = ελB(λ,T) 2.4 Absorption of electromagnetic radiation The fraction of total incident radiation absorbed by an object (or gas) is called the absorptivity, a. The fraction of incident radiation at a specific wavelength absorbed is called the spectral absorptivity, aλ . An oscillator will absorb a photon of a given frequency (wavelength) only if it can oscillate at that frequency. Likewise, it will emit at that frequency if it has enough energy (that is, if it is warm enough). These ideas and others lead to Kirchoff’s Law: a = ε, absorptivity = emissivity Also true at every wavelength: aλ . = ελ . If something is a good absorber at a given wavelength, it is potentially a good emitter too. For a BB, ε = 1 so a = 1. A blackbody absorbs all the radiation falling on it. Although emissivity = absorptivity, this does not mean that emission = absorption Absorbed energy = a I = ε I … .. I depends on the temperature of the surroundings Emitted energy = εσT4 … . depends on temperature of itself 2.5 Types of possible electric oscillators As we have seen, emission and absorption of electromagnetic radiation involve oscillating electric charges or dipoles. These can arise in 3 ways: -oscillation of electrons in their orbitals – v. high frequency, short λ radiation: 0.1-0.7 μm -vibration of dipole molecules – medium frequency, medium λ radiation: 0.8- 18 μm -rotation of dipole molecules – low frequency, long λ rad’n mostly > 50 μm, some effects down to 12 μm The part of the electromagnetic spectrum of importance to climate is: As electrons oscillate, or dipole molecules vibrate or rotate, they give offEM radiation but at the same time drop to lower energy levels. Electron transitions/oscillations - correspond to UV, Visible radiation Vibrational transitions - correspond to NIR, IR Rotational transitions - correspond to IR, far IR Recall: a vibrating or rotating molecule will emit EM radiation only if it has a dipole (centre of positive charge distribution ≠ centre of negative distribution) 2.6 Temperature Temperature is a measure of the kinetic energy of atoms or molecules. This kinetic energy occurs as - Vibrations - Rotations - Translation = uniform. from one point to another Only vibrational and rotational KE involves emission/absorption of EM radiation (and then, only if there is a dipole). Conversely, if amolecule has a dipole but was simply moving at a uniform velocity (having translational kinetic energy only), it would not emit radiation. Temperature is directly related to the translational KE but not the vibrational and rotational KE.
Table of Contents PROJECT GOAL………………………………………………………………………………………………………………… 2 Part 1: Setup………………………………………………………………………………………………………………… 2 Part 2: Files Layout……………………………………………………………………………………………………….. 2 Part 3: TODOs………………………………………………………………………………………………………………. 3 Part 4: Testing and Debugging……………………………………………………………………………………….. 5 Part 5: Assumptions and Clarifications……………………………………………………………………………. 6 What to Turn In………………………………………………………………………………………………………………. 7 What you can and cannot share………………………………………………………………………………………… 7 Rubric…………………………………………………………………………………………………………………………….. 8 In the lectures, you learned about Spanning Trees which are used to prevent forwarding loops in a network. In this project, you will add to those concepts by developing a distributed algorithm that can run on an arbitrary layer 2 topology. We will simulate the communications between the switches with Messages. This project is NOT an implementation of the standard Spanning Tree Protocol, so be sure not to reference any outside material or concepts that could lead you astray. Referencing any outside material[1] is also a violation of the Honor Code. If you have questions about the project, post them on Ed Stem.Download the project files from Canvas. You can do this project on your host system if it has Python 3.11.x. The project does not have any dependencies outside of Python. You must be sure that your submission runs properly in Gradescope. Gradescope is the environment where your project will be graded. Gradescope and the VM are the only valid environments for this course. There are many files in the SpanningTree directory, but you should only modify Switch.py.The files in the project skeleton are described below. DO NOT modify these files. All of your work must be in Switch.py ONLY. You should study the other files to understand the project framework.msg = Message(claimedRoot, distanceToRoot, originID, destinationID, pathThrough, timeToLive)and assigning the correct value to each input. Message format may NOT be changed. See the comments in Message.py for more information about the variables. This is an outline of the code you must implement in Switch.py with suggestions for implementation. Keep in mind that certain update rules will take precedence over others. To run your code on a specific topology (SimpleLoopTopo.py in this case) and output the results to a text file (out.txt in this case), execute the following command:python run.py SimpleLoopTopo “SimpleLoopTopo” is not a typo in the example command – don’t include the .py extension. We have included several topologies with correct solutions for you to test your code against. You can (and are encouraged to) create more topologies and test suites with output files and share them on Ed Discussion. There will be a designated post where students can share these files.You will only be submitting Switch.py – your implementation must be confined to modifications of that file. We recommend testing your submission against a clean copy of the rest of the project files prior to submission.You may add print statements to facilitate debugging during your development process, but they should be removed or commented out prior to submission.You may assume the following: Submit ONLY your Switch.py file to Gradescope as a single file. Do not modify the name of Switch.py. You may make an unlimited number of submissions to Gradescope before the deadline. Your last submission will be your grade unless you activate a different submission.Before submission:After submission:Honor Code/Academic Integrity: Do NOT share any code from Switch.py with your fellow students, on Ed Discussion, or publicly in any form, even after the course ends. You may share log files for any topology, and you may share any code you write that will not be turned in, such as new topologies or testing suites.All work must be your own, and consulting Spanning Tree Protocol solutions, even in another programming language or just for reference, are considered violations of the honor code. Do not reference solutions on Github! Do not use IDE extensions (like Github Copilot or any AI) that write or recommend blocks of code to you (autocomplete for function names is fine). For more information see the Syllabus Definition of Plagiarism. We provide you with all the materials you need to complete this project without help from Google/Stack Overflow (Searching basic Python syntax is fine). Do not risk an honor code violation for a very doable project.Start early, ask questions in Ed Discussion, and attend TA chat sessions if needed. [1] Searching general Python syntax is fine but do not search any material related to the Spanning Tree algorithm
1. The KdV equation for ion-acoustic waves in plasmas. Ion-acoustic waves are low-frequency electrostatic waves in a plasma consisting of electrons and ions. We consider the case with a single ion species. Consider the following system of one-dimensional equations ∂n ∂t + ∂ ∂z (nv) = 0 ∂v ∂t + v ∂v ∂z = − e m ∂ϕ ∂z ∂ 2ϕ ∂z2 = e ε0 N0 exp eϕ κTe − n Here n denotes the ion density, v is the ion velocity, e is the electron charge, m is the mass of an ion, ϕ is the electrostatic potential, ε0 is the vacuum permittivity, N0 is the equilibrium density of the ions, κ is Boltzmann’s constant, and Te is the electron temperature. (a) Verify that cs = q κTe m , λDe = qε0κTe N0e 2 , and ωpi = qN0e 2 ε0m have dimensions of velocity, length and frequency, respectively. These quantities are known as the ion acoustic speed, the Debye wavelength for the electrons, and the ion plasma frequency.(b) Nondimensionalize the above system, using n = N0n ∗ , v = csv ∗ , z = λDez ∗ , t = t ∗ ωpi , ϕ = κTe e ϕ ∗ .(c) You have obtained the system ∂n ∂t + ∂ ∂z (nv) = 0 ∂v ∂t + v ∂v ∂z = − ∂ϕ ∂z ∂ 2ϕ ∂z2 = e ϕ − n for the dimensionless variables. Note that we have dropped the ∗’s, to ease the notation. Find the linear dispersion relation for this system, linearized around the trivial solution n = 1, v = 0, and ϕ = 0.(d) Rewrite the system using the “stretched variables” ξ = ϵ 1/2 (z − t), τ = ϵ 3/2 t Given that we are looking for low-frequency waves, explain how these variables are inspired by the dispersion relation.(e) Expand the dependent variables as n = 1 + ϵn1 + ϵ 2n2 + . . . v = ϵv1 + ϵ 2 v2 + . . . ϕ = ϵϕ1 + ϵ 2ϕ2 + . . . Using that all disturbances return to their equilibrium values as ξ → ±∞, τ → ∞, find a governing equation which determines how ϕ1 depends on ξ and τ .2. Obtaining the KdV equation from the NLS equation. We have shown that the NLS equation may be used to describe the slow modulation of periodic wave trains of the KdV equation. In this problem we show that the KdV equation describes the dynamics of long-wave solutions of the NLS equation. Consider the defocusing NLS equation iat = −axx + |a| 2 a. (a) Let a(x, t) = e i R V dxρ 1/2 . Derive a system of equations for the phase function V (x, t) and for the amplitude function ρ(x, t), by substituting this form of a(x, t) in the NLS equation, dividing out the exponential, and separating real and imaginary parts. Write your equations in the form ρt = . . ., and Vt = . . .. Due to their similarity with the equations of hydrodynamics, this new form of the NLS equation is referred to as its hydrodynamic form.(b) Find the linear dispersion relation for the hydrodynamic form of the defocusing NLS equation, linearized around the trivial solution V = 0, ρ = 1. In other words, we are examining perturbations of the so-called Stokes wave solution of the NLS equation, which is given by a signal of constant amplitude.(c) Rewrite the system using the “stretched variables” ξ = ϵ(x − βt), τ = ϵ 3 t Given that we are looking for long waves, explain how these variables are inspired by the dispersion relation. What should the value of β be?(d) Expand the dependent variables as V = ϵ 2V1 + ϵ 4V2 + . . . ρ = 1 + ϵ 2 ρ1 + ϵ 4 ρ2 + . . . Using that all disturbances return to their equilibrium values as ξ → ±∞, τ → ∞, find a governing equation which determines how V1 depends on ξ and τ . This equation should be equivalent to the KdV equation.3. Consider the previous problem, but with the focusing NLS equation iat = −axx − |a| 2 a. The method presented in the previous problem does not allow one to describe the dynamics of long-wave solutions of the focusing NLS equation using the KdV equation. How does this show up in the calculations?4. The mKdV equation considered in the text is known as the focusing mKdV equation, because of the behavior of its soliton solutions. This behavior is similar to that of the focusing NLS equation. In this problem, we study the defocusing mKdV equation 4ut = −6u 2ux + uxxx. You have already seen that you can scale the coefficients of this equation to your favorite values, except for the ratio of the signs of the two terms on the right-hand side. (a) Examine, using the potential energy method and phase plane analysis, the traveling-wave solutions.(b) If you have found any homoclinic or heteroclinic connections, find the explicit form of the profiles corresponding to these connections.5. Consider the so-called Derivative NLS equation (DNLS) bt + αb|b| 2 x − ibxx = 0. This equation arises in the description of quasi-parallel waves in space plasmas. Here b(x, t) is a complex-valued function. (a) Using a polar decomposition b(x, t) = B(x, t)e iθ(x,t) , with B and θ real-valued functions, and separating real and imaginary parts (after dividing by the exponential), show that you obtain the system Bt + 3αB2Bx + 1 B (B 2 θx)x = 0, θt + αB2 θx + θ 2 x − 1 B Bxx = 0(b) Assuming a traveling-wave envelope, B(x, t) = R(z), with z = x − vt and constant v, show that θ(x, t) = Φ(z) − Ωt, with constant Ω, is consistent with these equations. You can show (but you don’t have to) that assuming a traveling-wave amplitude results in only this possibility for θ(x, t). At this point, we have reduced the problem of finding solutions with traveling envelope to that of finding two one-variable functions R(z) and Φ(x). The problem also depends on two parameters v (envelope speed) and Ω (frequency like).(c) Substituting these ansatz in the first equation of the above system, show that Φ ′ = C + vs − 3s 2 2s , where C is a constant and s = αR2/2.(d) Lastly, by substituting your results in the second equation of the system, show that s(z) satisfies 1 2 s ′2 + V (s) = E, the equation for the motion of a particle with potential V (s). Find the expression for V (s) and for E.6. Consider example 5.2 in the notes. Check that y = x 2/t and t 1/2 q are both scaling invariant. Find the ordinary differential equation satisfied by G(y), for similarity solutions of the form q(x, t) = t −1/2G(y). Show that this results in the same similarity solutions as in the example.7. One way to write the Toda Lattice is dan dt = an(bn+1 − bn), dbn dt = 2(a 2 n − a 2 n−1 ), where an, bn, n ∈ Z, are functions of t. (a) Find a scaling symmetry of this form of the Toda lattice, i.e., let2 an = αAn, bn = βBn, t = γτ , and determine relations between α, β and γ so that the equations for the Toda lattice in the (An, Bn, t) variables are identical to those using the (an, bn, τ ) variables.(b) Using this scaling symmetry, find a two-parameter family of similarity solutions of the Toda lattice. If necessary, find relations among the parameters that guarantee the solutions you found are real for all n and for t > 0.8. Consider the equation ut = 30u 2ux + 20uxuxx + 10uuxxx + u5x, which we will encounter more in later chapters, due to its relation to the KdV equation. Show that it has a scaling symmetry. When we look for the scaling symmetry of the KdV equation, we have two equations for three unknowns: we have three quantities (x, t, u) to scale, and after normalizing one coefficient to 1, two remaining terms that need to remain invariant. Thus it is no surprise that we find a one-parameter family of scaling symmetries. The above equation has two more terms, and it should be clear that some “luck” is needed in order for there to be a scaling symmetry.9. Consider a Modified KdV equation ut − 6u 2ux + uxxx = 0. (a) Find its scaling symmetry(b) Using the scaling symmetry, write down an ansatz for any similarity solutions of the equation.(c) Show that your ansatz is compatible with u = (3t) −1/3w(z), with z = x/(3t) 1/3 .(d) Use the above form of u to find an ordinary differential equation for w(z). This equation will be of third order. It can be integrated once (do this) to obtain a second-order equation. The second-order equation you obtain this way is known as the second of the Painlev´e equations. We will see more about these later.
1. Consider the Modified Vector Derivative NLS equation Bt + (∥B∥ 2B)x + γ(e1 × B0) (e1 · (Bx × B0)) + e1 × Bxx = 0. This equation describes the transverse propagation of nonlinear Alfv´en waves in magnetized plasmas. Here B = (0, u, v), e1 = (1, 0, 0), B0 = (0, B0, 0), and γ is a constant. The boundary conditions are B → B0, Bx → 0 as |x| → ∞. By looking for stationary solutions B = B(x − W t), one obtains a system of ordinary differential equations. Integrating once, one obtains a first-order system of differential equations for u and v. a) Show that this system is Hamiltonian with canonical Poisson structure, by constructing its Hamiltonian H(u, v).b) Find the value of the Hamiltonian such that the boundary conditions are satisfied. Then H(u, v) equated to this constant value defines a curve in the (u, v)-plane on which the solution lives. In the equation of this curve, let U = u/B0, V = v/B0, and W0 = W/B2 0 . Now there are only two parameters in the equation of the curve: W0 and γ.c) With γ = 1/10, plot the curve for W0 = 3, W0 = 2, W0 = 1.1, W0 = 1, W0 = 0.95, W0 = 0.9. All of these curves have a singular point at (1, 0). This point is an equilibrium point for the Hamiltonian system, corresponding to the constant solution which satisfies the boundary condition. The curves beginning and ending at this equilibrium point correspond to soliton solutions of the Modified Vector Derivative NLS equation. How many soliton solutions are there for the different velocity values you considered? Draw a qualitatively correct picture of the solitons for all these cases.2. Show that the canonical Poisson bracket {f, g} = X N j=1 ∂f ∂qj ∂g ∂pj − ∂f ∂pj ∂g ∂qj satisfies the Jacobi identity {{f, g}, h} + {{g, h}, f} + {{h, f}, g} = 0.3. Show that the Sine-Gordon equation utt − uxx + sin(u) = 0 is Hamiltonian with canonical Poisson structure and Hamiltonian H = Z 1 2 p 2 + 1 2 q 2 x + 1 − cos(q) dx, where q = u, and p = ut .4. Check explicitly that the conserved quantities F−1 = R udx, F0 = R 1 2 u 2dx, F1 = R 1 6 u 3 − 1 2 u 2 x dx, F2 = R 1 24u 4 − 1 2 uu2 x + 3 10u 2 xx dx are mutually in involution with respect to the Poisson bracket defined by the Poisson structure given by ∂5. Find the fourth conserved quantity for the KdV equation ut = uux + uxxx, i.e., the conserved quantity which contains 1 24 R u 4dx6. Recursion operator For a Bi-Hamiltonian system with two Poisson structures given by B0, B1, one defines a recursion operator R = B1B −1 0 , which takes one element of the hierarchy of equations to the next element. For the KdV equation with B0 = ∂x and B1 = ∂xxx + 1 3 (u∂x + ∂xu), we get B −1 0 = ∂ −1 x , integration with respect to x. Write down the recursion operator. Apply it to ux (the zero-th KdV flow) to obtain the first KdV flow. Now apply it to uux+uxxx to get (up to rescaling of t2) the second KdV equation. What is the third KdV equation?7. Consider the function U(x) = 2∂ 2 x ln 1 + e kx+α . Show that for a suitable k, U(x) is a solution of the first member of the stationary KdV hierarchy (as you’ve already seen, it is the one-soliton solution): 6uux + uxxx + c0ux = 0. (Note: it may be easier to define c0 in terms of k, instead of the other way around) Having accomplished this, let u(x, t1, t2, t3, . . .) = U(x)|α=α(t1,t2,t3,…) . Determine the dependence of α on t1, t2 and t3 such that u(x, t1, t2, t3, . . .) is simultaneously a solution of the first, second and third KdV equations: ut1 = 6uux + uxxx, ut2 = 30u 2ux + 20uxuxx + 10uuxxx + u5x, ut3 = 140u 3ux + 70u 3 x + 280uuxuxx + 70uxxuxxx + 70u 2uxxx + 42uxuxxxx + 14uu5x + u7x Based on this, write down a guess for the one-soliton solution that solves the entire KdV hierarchy+1
1. Show that N¯(x, k) is analytic in the open lower-half plane, Imk < 0, by showing N¯(x, k) and ∂N/∂k ¯ are bounded there. What are the conditions you need for this to be true?2. Recall the second member of the stationary KdV hierarchy from HW4: 30u 2ux + 20uxuxx + 10uuxxx + u5x + c1(6uux + uxxx) + c0ux = 0. Integrating once, it can be rewritten as (10u 3 + 10u 2 x + 10uuxx − 5u 2 x + uxxxx) + c1(3u 2 + uxx) + c0u + c−1 = 0. (1) This is an ordinary differential equation for u as a function of x. You already know that the two soliton is a solution of this. You know this equation can be written as δT2 δu = 0 ⇐⇒ δ δu (F2 + c1F1 + c0F0 + c−1F−1) = 0, (2) where Fk, k = −1, 0, . . . are the conserved quantities of the KdV equation. These conserved quantities are in involution, {Fj , Fk} = 0 ⇒ {Tj , Fk} = 0, for j, k = −1, 0, . . .. From HW4, you know that this implies (using j = 2) δT2 δu d dx δFk δu = dHk dx , k = 0, 1. (3) For some functions H0 and H1. Thus, H0 and H1 are conserved quantities of (1)1 . (a) Find H0 and H1 explicitly.(b) Check explicitly, by taking an x derivative, that H0 and H1 are conserved along solutions of (1).3. The Ostrovsky equation is used to model weakly nonlinear long waves in a rotating frame. It is given by (ηt + ηηx + ηxxx)x = γη, with γ ̸= 0. In what follows, we assume that as |x| → ∞, η and its derivatives approach zero as fast as we need them to. • Show that R ∞ −∞ ηdx = 0. In other words, not only is R ∞ −∞ ηdx conserved, but its value is fixed at zero.• Using this result, show that R ∞ −∞ η 2dx is a conserved quantity. Do this by rewriting the equation in evolution form, with an indefinite integral on the right-hand side.• Use the definition of the variational derivative to verify that the Ostrovsky equation is Hamiltonian with Poisson operator ∂x and Hamiltonian H = 1 2 Z ∞ −∞ η 2 x − 1 3 η 3 − γϕ2 dx, where ϕx = η.4. Using the Painlev´e test, discuss the integrability of ut = u pux + uxxx.
1. Show that X = −iζ q ±q ∗ iζ , T = −iζ2 ∓ i 2 |q| 2 qζ + i 2 qx ±q ∗ ζ ∓ i 2 q ∗ x iζ2 ± i 2 |q| 2 are Lax Pairs for the Nonlinear Schr¨odinger equations iqt = − 1 2 qxx ± |q| 2 q. Here the top (bottom) signs of one matrix correspond to the top (bottom) signs of the other. In other words, show that the X, T with the top (bottom) sign are a Lax pair for the Nonlinear Schr¨odinger equation with the top (bottom) sign.2. Let ψn = ψn(t), n ∈ Z. Consider the difference equation ψn+1 = Xnψn, and the differential equation ∂ψn ∂t = Tnψn. What is the compatibility condition of these two equations? Using this result, show that Xn = z qn q ∗ n 1/z , Tn = iqnq ∗ n−1 − i 2 (1/z − z) 2 i z qn−1 − izqn −izq∗ n−1 + i z q ∗ n −iq∗ n qn−1 + i 2 (1/z − z) 2 is a Lax Pair for the semi-discrete equation i ∂qn ∂t = qn+1 − 2qn + qn−1 − |qn| 2 (qn+1 + qn−1) Note that this is a discretization of the NLS equation. It is known as the AblowitzLadik lattice. It is an integrable discretization of NLS. For numerical purposes, it is far superior in many ways to the “standard” discretization of NLS: i ∂qn ∂t = qn+1 − 2qn + qn−1 − 2|qn| 2 qn.3. For the KdV equation ut + 6uux + uxxx = 0 with initial condition u(x, 0) = 0 for x ∈ (−∞, −L) ∪ (L, ∞), and u(x, 0) = d for x ∈ (−L, L), with L and d both positive, consider the forward scattering problem. • Find a(k), for all time t. • Knowing that the number of solitons emanating from the initial condition is the number of zeros of a(k) on the positive imaginary axis (i.e., k = iκ, with κ > 0), discuss how many solitons correspond to the given initial condition, depending on the value of 2L 2d. You might want to use Maple, Mathematica or Matlab for this. • What happens for d < 0? • In the limit L → 0, but 2dL = α, u(x, 0) → αδ(x). What happens to a(k) when you take this limit? Discuss.5. The Liouville equation. Consider the horribly nonlinear1 PDE uxy = e u , known as Liouville’s equation. Consider the transformation vx = −ux + √ 2e (u−v)/2 , vy = uy − √ 2e (u+v)/2 , where u(x, y) satisfies Liouville’s equation above. (a) Find an equation satisfied by v(x, y): vxy = . . .. Your right-hand side cannot have any u’s. Those should all be eliminated.(b) Write down the general solution for v(x, y) from the equation you obtained.(c) Use this solution for v in your B¨acklund transformation and solve for u, obtaining the general solution of the Liouville equation!6. The sine-Gordon equation. Consider the sine-Gordon equation uxt = sin u, also horribly nonlinear. (a) Show that the transformation vx = ux + 2 sin u + v 2 , vt = −ut − 2 sin u − v 2 , is an auto-B¨acklund transformation for the sine-Gordon equation. In other words, v satisfies the same equation as u.(b) Let u(x, t) be the simplest2 solution of the sine-Gordon equation. With this u(x, y) solve the auto-B¨acklund transformation for v(x, t), to find a more complicated solution of the sine-Gordon equation. Congratulations! You just found the one-soliton solution of the sine-Gordon equation
1. Problem 1: The Benjamin-Ono equation ut + uux + Huxx = 0 is used to describe internal water waves in deep water. Here Hf(x, t) is the spatial Hilbert transform of f(x, t): Hf(x, t) = 1 π − Z ∞ −∞ f(z, t) z − x dz and − R denotes the Cauchy principle value integral. Write down the linear dispersion relationship for this equation linearized about the zero solution.2. Problem 2: Derive the linear dispersion relationship for the one-dimensional surface water wave problem by linearizing around the trivial solution ζ(x, t) = 0, ϕ(x, z, t) = 0 : (a) ∇2ϕ = 0 at −h < z < ζ(x, t) (b) ϕz = 0 at z = −h (c) ζt + ϕxζx = ϕz at z = ζ(x, t) (d) ϕt + gζ + 1 2 (ϕ 2 x + ϕ 2 z ) = T ζxx (1+ζ 2 x ) 3/2 at z = ζ(x, t) Here z = ζ(x, t) is the surface of the water wave, ϕ(x, z, t) is the velocity potential so that v = ∇ϕ is the velocity of the water, g is the acceleration of gravity, and T > 0 is the coefficient of surface tension3. Problem 3: Having found that for the surface water wave problem without surface tension the linear dispersion relationship is ω 2 = gk tanh kh, find the group velocities for the case of long waves in shallow water (kh small), and for the case of deep water (kh big).4. Problem 4: Whitham wrote down what is now known as the Whitham equation to incorporate the full effect of water-wave dispersion for waves in shallow water by modifying the KdV equation ut+vux+uux+γuxxx = 0 (where we have included the transport term) to ut + uux + Z ∞ −∞ K(x − y)uy(y, t)dy = 0 where K(x) = 1 2π − Z ∞ −∞ c(k)e ikxdk p and c(k) is the postive phase speed for the water-wave problem: c(k) = g tanh(kh)/k. (a) What is the linear dispersion relation of the Whitham equation? (b) Show that the dispersion relation of the KdV equation is an approximation to that of the Whitham equation for long waves, i.e., for k → 0. What are v and γ? Note that using this process of ”Whithamization”, one could construct a KdV-like equation (i.e., an equation with the KdV nonlinearity) that has any desired dispersion relation. Similar procedures can be followed for other equations, like the NLS equation, etc.5. Problem 5: Consider the linear free Schr¨odinger (”free”, because there’s no potential) equation iψt + ψxx = 0 where −∞ < x < ∞, t > 0, ψ → 0 as |x| → ∞. With ψ(x, 0) = ψ0(x) such that R ∞ −∞ |ψ0| 2dx < ∞. (a) Using the Fourier transform, write down the solution of this problem. (b) Using the Method of Stationary Phase, find the dominant behavior as t → ∞ of the solution, along lines of constant x/t. (c) With ψ0(x) = e −x 2 , the integral can be worked out exactly. Compare (graphically or otherwise) this exact answer with the answer you get from the Method of Stationary Phase. Use the lines x/t = 1 and x/t = 2 to compare. (d) Use your favorite numerical integrator (write your own, or use maple, mathematica or matlab) to compare (graphically or other) with the exact answer and the answer you get from the Method of Stationary Phase.6. Problem 6: Everything that we have done for continuous space equations also works for equations with a discrete space variable. Consider the discrete linear Schr¨odinger equation: i dψn dt + 1 h 2 (ψn+1 − 2ψn + ψn−1) = 0 where h is a real constant, n is any integer, t > 0, ψn0 as |n| → ∞, and ψn(0) = ψn,0 is given. (a) The discrete analogue of the Fourier transform is given by ψn(t) = 1 2πi I |z|=1 ψˆ(z, t)z n−1 dz and its inverse ψˆ(z, t) = X∞ m=−∞ ψm(t)z −m Show that these two transformations are indeed inverses of each other. (b) The dispersion relation of a semi-discrete problem is obtained by looking for solutions of the form ψn = z ne −iωt. Show that for the semi-discrete Schr¨odinger eqaution ω(z) = − (z − 1)2 zh2 How does this compare to the dispersion relation of the continuous space problem? Specifically, demonstrate that you recover the dispersion relationship for the continuous problem as h → 0.
1. Problem 1.2: The KdV equation ut = uux + uxxx is often written with different coefficients. By using a scaling transformation on all variables (dependent and independent), show that the choice of the coefficients is irrelevant: by choosing a suitable scaling, we can use any coefficients we please. Can you say the same for the modified KdV (mKdV) equation ut = u 2ux + uxxx?2. Problem 1.4: (Use a symbolic computing software for this problem.) Consider the KdV equation ut + uux + uxxx = 0. Show that u = 12∂ 2 x ln 1 + e k1x−k 3 1 t+α is a one-soliton solution of the equation (i.e., rewrite it in sech2 form). Now check that u = 12∂ 2 x ln 1 + e k1x−k 3 1 t+α + e k2x−k 3 2 t+β + k1 − k2 k1 + k2 2 e k1x−k 3 1 t+α+k2x−k 3 2 t+β ! is also a solution of the equation. It is a two-soliton solution of the equation, as we will verify later. By changing t, we can see how the two solitons interact. With α = 0 and β = 1, examine the following 3 regions of parameter space: (a) k1 k2 > √ 3 (b) √ 3 > k1 k2 > q (3 + √ 5)/2 (c) k1 k2
1. Consider the sound waves governed by in a circular cylinder of radius a and length L. Assume that the sound produced in this tube is symmetric, i.e. no dependence. Find the lowest three frequencies. Take2. Consider the wave function for an electron of mass in a sphere surrounded by an infinite potential at a radius a from the nucleus, which just mean that Find the energy levels for the symmetric case, where does not depend on . Your answer should be exact and in terms of parameters given.3. Consider the Legendre’s equation: Compute the first four coefficients in the Legendre expansion (similar to Fourier sine or cosine series expansion): for Plot the approximation of the sum consisting of one, two, three and four terms along with the original function ∂2 ∂t 2ψ = c 2 ∇2 ψ, ∇2 = 1 r ∂ ∂r (r ∂ ∂r )+ 1 r 2 ∂2 ∂θ2 + ∂2 ∂z 2 ψ = 0 at r = a; ψ = 0 at z = 0,L. θ c = 300m /s, a = 1cm, L = 0.5m. ψ µ ψ = 0 at r = a. i! ∂ ∂t ψ = − !2 2µ ∇2 ψ. ψ θ and φ d dx (1− x 2 ) d dx y(x) ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ + n(n +1)y(x) = 0, −1 ≤ x ≤1, with the condition that y(±1) are bounded. The solutions are the Legendre polynomials, Pn (x), which are given by the Rodrigue’s formula: Pn (x) = 1 2n n! dn dxn (x 2 −1) n . For example P2 (x) = 1 2 (3x 2 −1). f (x) = an n=0 ∞ ∑ Pn (x), where an = 2n +1 2 f (x)Pn (x)dx −1 1 ∫ f (x) = 0 for −1 < x < 0 x for 0 < x
1. For the 1-dimensional heat equation for conduction in a copper rod: (a) solve using separation of variables. Show that the time-dependence for the nth standing mode is . (b) For , find the mode that dominates the solution, and thus write down the approximate solution (in time and in space). Describe in words how an initial condition, which may not look like a sine wave, becomes the sine shape with the largest wavelength fitting in between the boundaries.2. Sound waves in a box satisfies PDE: BC: Use separation of variables to solve this problem for the following configurations and find the quantized frequency of oscillation w, where w appears in the time dependence of the solution in the form of a sine or cosine of wt. (a) V is a one-dimensional box: 0
1. Green’s function of the 1-D heat equation in a semi-infinite domain, , is defined by: subject to zero initial condition: The boundary condition is either (a): or (b): The solution in a semi-infinite domain can be constructed from the solution in the infinite domain by adding or subtracting another source located at , so that the contributions cancel at for (a), or the contributions are symmetric about Find the Green’s function defined above for boundary condition (a). Then repeat the problem for boundary condition (b).2. Find the Greens function for the wave equation in two-dimensions governed by The solution is (a) Derive this solution using Fourier transform in and . Hint: In the inverse transform, use polar coordinates to get(b) Derive this solution using Laplace transform in . Hint: First show that the Laplace transform of G is Gxt (,; , ) x t 2 2 ( ) ( ) ( ), 0 , , 0, 0. DGx t x t t x d xd t x t ¶ ¶ – = – – < < ¥ > > ¶ ¶ G t = = 0 at 0. Gx x = = 0 at 0 and , ® ¥ Gx x 0 at 0 and . x ¶ = = ® ¥ ¶ x = -x x = 0 x = 0. 2 22 2 22 222 ( ) ( ) ( ) ( ). 0 as , where . 0 for 0. G G txy t xy G r rxy G t dd d ¶ ¶¶ – + = ¶ ¶¶ ® ®¥ = + º
1. (a) Solve using Fourier transform in x and Laplace transform in t: PDE: BC: IC:(b) Same problem as in (a), except that you do not use Laplace transform in t. You need to figure out the matching condition for your ODE across t=t. 2 2 uD u x t x t ( ) ( ), , 0, , 0 t x d xd t x t ¶ ¶ – = – – – ¥ < < ¥ > -¥ < < ¥ > ¶ ¶ uxt x t ( , ) 0 as , 0 ® ® ±¥ > ux x ( ,0) 0, . = – ¥ < < ¥
1. Consider the wave equation: , which is a special case of the general quasi-linear equation: , with . Find the slope of each of the two characteristics:Find the expression in terms of and for and , so that the wave equation simplifies to2. Use the Fourier transform method to solve the 2-D Laplace equation in the upper plane for the bounded solution: Assume is of compact support;3. Solve the following problem in two ways: (a) by the method of similarity transformation. Look for the value of such that the PDE reduces to an ode in (b) by an integral transform in t, in this case a Laplace transform (You can use a table of Laplace transform to do the inverse transform). 2 0 in 0, . ( ,0) ( ), . uy x ux f x x Ñ = > – ¥ < < ¥ = – ¥ < < ¥ f(x) u(x, y)→0 as x → ±∞. ∂ ∂t u = ∂2 ∂x 2 u , 0 < t;0 < x < ∞ u(x,0) = 0, u(x,t) bounded as x → ∞. u(0,t) = T0 , a constant, t > 0. a , / ; x ta h h =
1. Solve the PDE: , subject to the initial condition:2. Consider the initial value problem in infinite domain: (a) Find where and when a shock first forms. (b) Solve the problem and sketch or plot the solution before when a shock first forms. (c) Find the shock speed using the Rankine-Hugoniot condition. (d) Solve the problem and sketch or plot the solution after the shock has formed.3. Solve the PDE: , subject to the initial condition: Where in the x-t plane is the solution valid?
1. Consider the inverted pendulum dynamics: y 00 + (δ + cos ωt) sin y = 0 (a) Perform a Floquet analysis (computationally) of the pendulum with continous forceing cos ωt. (b) Evaluate for what values of δ, and ω the pendulum is stabilized.
1. Consider the Optical Parametric Oscillator as given in Lecture 23 of the notes (Pages 99-102) (a) Assuming slow time τ = 2 t and slow space ξ = x, derive the Fisher-Kolmogorov equation for the slow evolution of the instability (the expression after Eq. (518)) (b) Derive the Swift-Hohenberg type expression which is governed by Eq. (519) with the scalings detailed in the notes.
1. Consider the singular equation: y00 + (1 + x) 2 y 0 + y = 0 with y(0) = y(1) = 1 and with 0 < 1. (a) Obtain the leading order uniform solution using the WKB method. (b) Plot the uniform solution for = 0.01, 0.05, 0.1, 0.2.
1. Consider the singular equation ϵy′′ + (1 + x) 2 y ′ + y = 0 with y(0) = y(1) = 1 and with 0 < ϵ ≪ 1. (a) Obtain a uniform approximation which is valid to leading order.(b) Show that assuming the boundary layer to be at x = 1 is inconsistent. (Hint: use the stretched inner variable ξ = (1 − x)/ϵ).(c) Plot the uniform solution for ϵ = 0.01, 0.05, 0.1, 0.2.2. Consider the singular equation: ϵy′′ − x 2 y ′ − y = 0 with y(0) = y(1) = 1 and with 0 < ϵ ≪ 1. (a) With the method of dominant balance, show that there are three distinguished limits: δ = ϵ 1/2 , δ = ϵ, and δ = 1 (the outer problem). Write down each of the problems in the various distinguished limits.(b) Obtain the leading order uniform approximation. (Hint: there are boundary layers at x = 0 and x = 1).(c) Plot the uniform solution for ϵ = 0.01, 0.05, 0.1, 0.2.
