18.376-MIT. Wave propagation. Spring 2023. Tu-Th 9:30--11:00. Room E25-117. 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. % ========================================================================== % --------------------------------------------------------------------- #001 What is a wave? Will not "define" the concept. Possible definitions are too restrictive (and eliminate wave examples). We will work by examples. When you see a wave, you will know! Note: there is much more to the concept than what "the wave equation" provides. Examples: Mechanical waves: need a restoring force (e.g.: gravity, elasticity) in an extended media. But there are many other, non-mechanical, types of waves; examples: Traffic flow; Bio-waves (on neurons, the heart, etc.); Chemical reactions (flames, detonations, and others); Population waves (in growing spores, slime molds; etc.); Spiral waves in galaxies; etc. Applications: seismology, weather prediction, water waves (surface, internal, tides, tsunamis, etc.), E.M. waves, optics, sound/acoustics, combustion, chemical waves, neural waves, etc. Mathematically waves are (mostly) described by pde. In particular, small amplitude (linear) waves arise as perturbations to neutrally stable (or only weakly stable: small dissipation) media at equilibrium. But there are also large amplitude, nonlinear, waves. A few examples (to be justified later in #003 below). % % --------------------------------------------------------------------- #002 Exponential notation, part 1: The simplest type of wave: linear waves in homogeneous media. [I am sure you all know this, but this serves to set-up notation, which is not 100% uniform (factors of 2*pi show up sometimes)]. Small amplitude waves in space-time homogeneous media (invariant under translation). Governed by linear and constant coefficients pde, mostly. --- Solutions can be written as sums of exponentials: exp(i*k*x + lambda*t), [#002a] where k is real and lambda depends on what k is. (Explain). There may be restrictions on k, imposed by (say) boundary conditions [Example: we may only be interested in periodic solutions of some period]. But we will ignore them for now. However, not everything of the form [#002a] is a wave. Wave conditions: lambda has the form lambda = - i*omega, \ Neither decay, nor where omega = omega(k) is real valued. / amplification. Hence the solution is proportional to exp{i*(k*x - omega*t)} = exp{i*theta} In more than 1-D, k is a vector, and k*x becomes a dot product. ......... Plane, monochromatic, waves. Note: Generally, for a physically valid solution, we only take the real part of the exponential. However, there are exceptions: Schrodinger equation in QM. Important concepts: Wave-number, wave-frequency, wave-length, period, phase, phase velocity, amplitude, and phase shift. Explain phase shift: must take real part of complex solution, so the phase of the complex amplitude comes into theta. For these type of waves, the Fourier Transform is very useful. Notation: F(k) = (1/(2*pi)) int f(x) e^{-i*k*x} dx f(x) = int f(k) e^{i*k*x} dk where the integrals are from -infinity to infinity. % % --------------------------------------------------------------------- #003 Exponential notation, part 2: Examples (derivations later). Wave equation. Klein-Gordon [string over an elastic bed]. Linear KdV equation. Schrodinger [both QM and nonlinear optics] Wave equation: General solution by Fourier Transform. Dispersion-less pulses. Klein-Gordon: General solution by Fourier transform. Explain dispersion: phase of Fourier components has to be just right to get the cancelation needed to obtain a localized solution. Definition of dispersive: d^2\omega/dk^2 \neq 0. To be explained in detail later. KdV equation, to be used in some examples: u_t + c0*u_x = epsilon*u_xxx - delta*u*u_x. Derivation later, but context/rough idea (for water waves): Take unidirectional plane wave in the same limit where linear shallow water applies, and compute the leading order corrections due to finite depth (dispersion) and finite amplitude (nonlinearity). The Airy paradox: linear versus nonlinear waves. Soliton observed by J. Scott Russell (1834). Boussinesq (1871), Lord Rayleigh (1876), Korteweg and de Vries (1895) KdV and solitary wave. Note that the KdV equation applies for many more examples than just water waves. It is the "canonical" equation modeling the balance of weak dispersion and a "hyperbolic" weak nonlinearity. % % --------------------------------------------------------------------- #004 Derivation of equations [examples for class material]. Conservation law to pde in continuum physics. Example: Traffic Flow Lighthill-Whithams model. Discuss mechanics of conservation laws to p.d.e. Discuss quasi-equilibrium assumption to obtain equation of state. Fundamental equation of Traffic Flow. Linearize near constant state and obtain "simplest wave equation". [A] Do intuitive argument for wave steepening, based on [A]. Shocks. If we have time we will come back to this, later, in more detail. Example: Euler equations of Gas Dynamics in 1-D. Use conservation laws, and thermodynamics. What quasi-equilibrium is in this case? Mention isentropic equations, when the initial conditions are near constant entropy and the flow is adiabatic. Then p = p(rho) and the energy equation can be dropped. More details in #009 % % --------------------------------------------------------------------- #005 Example: Traffic Flow Payne-Whitham model. What if quasi-equilibrium assumption does not quite apply because the time scales are not fast enough? Simple modeling for the relaxation time to equilibrium: Instead of assuming q = Q(rho) = rho*U(rho), where U is the "desired" velocity, we take q = rho*u, and assume that the car/flow velocity relaxes towards U, on some time scale tau. Hence: rho_t + (rho*u)_x = 0 and u_t + u*u_x = (1/tau)(U - u). [A] tau: determined by vehicle inertia, not response time; tau ~ seconds. This model is not good: leads to wild oscillations! (see note#4) [A] corresponds to new drivers. Experienced drivers anticipate traffic conditions [preventive driving]. They do not accelerate towards U for the current state, but to the expected U after the time lag tau. A way to model preventive driving is (Payne-Whitham) by a "traffic pressure" p = p(rho) [see note#1] rho_t + (rho*u)_x = 0 and u_t + u*u_x + p_x = (1/tau)(U - u). [B] Justify p: Posit acceleration towards a corrected desired velocity given by U(rho) - nu(rho)*rho_x; nu > 0. [C] Then u_t + u*u_x = (1/tau)(U-nu*rho_x-u) is [B] with p_x = (nu/tau)*rho_x. That is: dp/d rho = nu/tau, for some nu(rho) > 0. NOTE#1. The equations in [B] are analog to the isentropic equations of Gas Dynamics in 1-D ............. with a force (1/tau)(U(rho)-u). Hence the name "traffic pressure". NOTE#2. [C] is just one way of modeling ``preventing driving'', only a rough approximation to what actual human drivers do. Others have been proposed. In particular, in the ARZ (Aw-Rascle-Zhou) model it is argued that drivers do not have access to rho_x, so the above is incorrect. Thus they propose a preventive driving formulation in terms of rho_t + u rho_x, the time derivative of rho in the driver's reference frame. NOTE#3. The problem with *all* these models is: they involve unknown functions, such as U(rho) and p(rho), which are very difficult to pin down given the imprecise nature of most traffic data. In fact, this is an issue with all traffic models, which (unfortunately) thus have very limited predictive capabilities beyond small changes from the data used to ``determine'' their parameters. NOTE#4. To see that [A] is bad, examine the ``high frequency'' limit, where rho_t + (rho*u)_x = 0 and u_t + u*u_x = 0. The first equation here says the u is the flow velocity for rho, while the second shows that space-time lines along which u = u_0 = constant are straight with velocity u_0. Hence any region where u decreases with x, leads to the lines crossing ... at which point a delta function forms in rho. This model predict that the cars ``clump'' into very tight bundles! Alternative: linearize [A] near a steady state rho=rho_0, u=U(rho_0). Then show that the growth rate for these equations is not bounded for high frequencies: they have solutions proportional to e^{i*k*x + lambda*t}, with Re(lambda) \to \infty as k \to infty. What physicists call an ultraviolet catastrophe. ..................... [PSQ] % % --------------------------------------------------------------------- #006 Exponential notation, part 3: Group velocity by conservation laws. Dispersion relation: omega = omega(k) or G(k, omega) = 0. Interpretation of k = 2*pi/lambda as wave density, omega = 2*pi/T as wave flux. For "slowly varying" wave fields, use conservation of waves to derive k_t + omega_x = 0, and omega = omega(k). [B] Group velocity: cg = domega/dk, speed at which wave properties propagate. [*] [*] Including energy. We will see this later: when we fully justify [B] using asymptotics and/or Fourier, and obtain equations for the wave amplitude. % % --------------------------------------------------------------------- #007 Example: small transverse vibrations of a string under constant tension T, with the motion restricted to a plane. x = coordinate along string u = u(x, t) transversal displacement of string from equillibrium. rho = density of string (assume constant). Thus rho*u_t = transversal linear momentum density. T*u_x = Transversal component of tension = transversal momentum flux. p = p(x, t) applied load per unit length. Note T = E*S*(\Delta L)/L, where L is the string length, S is the string cross section, E is the Young's modulus of elasticity, and \Delta L is the change in length relative to the unstretched length. Argue: for small transversal deviations, the extra change in length is higher order. Thus we can ignore variations in T. Use general conservation law machinery: --- Obtain wave equation: u_{tt} - c^2 u_{xx} = p/rho; c^2 = T/rho. --- Note c is a velocity (future lectures will show of what). Example where the load is (linear) elastic: p proportional to u; i.e.: String on an elastic bed. Get Klein Gordon equation. Note: this derivation ignores gravity. What happens if one adds gravity to the ``string on an elastic bed'' [heavy string]? ................... [PSQ] % % --------------------------------------------------------------------- #008 Conservation laws in more than 1-D. Discuss the flux. Why is it a vector? ................................ [PSQ] Applications to group speed concepts. Group speed in n-D [#006 has 1d]. Derive equation for conservation of waves in n-D. Argue that in each direction [n-1 variables frozen] bumps/waves should be conserved for a slowly varying wave field, and derive n-D group speed. This yields k_t + grad(omega) = 0, with omega = Omega(k) [A] Furthermore: wave crest do not end abruptly in space either. Thus, freezing all variables, including time, but 2 (say x_i and x_j) leads to [conservation like argument] to (k_i)_j = (k_j)_i. That is curl(k) = 0 [B] Justification: Think of the x_i direction as ``space'', with density of waves given by rho = (k_i)/(2\pi), and the x_j direction as the ``time'', with flux q = - (k_j)/(2\pi). The reason for the negative sign in the q is: if there are (k_j)/(2\pi) waves along x_j, for some fixed x_i, then if k_j < 0, the wave crests are ``moving to the right'' as x_j grows [so the flux should be positive], while k_j > 0 leads to the opposite conclusion. Hint: draw the wave-crests in some (x_i, x_j) region (they should be nearly straight on the scale of their separation because of the slow variation assumption), and think of them as the ``paths'' of the stuff being conserved. The flow velocity is then u = -k_j/k_i, which then gives q = rho u as above. ALTERNATIVE DERIVATION OF [B]: k has to be the gradient of the local phase, because it is normal to the waves and inversely proportional to the wave length [define phase as constant on wave crests, with variations normal to the waves, increasing linearly in such a way that it changes by 2*pi over one wave length]. Use then [A] and [B] to show that k_t + (c_g.grad) k = 0, [C] where c_g = grad Omega ............................................ [PSQ] % % --------------------------------------------------------------------- #009 Example: Isentropic equations of Gas Dynamics and acoustics. Show how full Euler equations of Gas Dynamics lead to the isentropic equations when the initial conditions are constant entropy. Derive equation for evolution of entropy along particle paths, using the Second law of Thermodynamics T*dS = de + p*dv, with v = 1/rho. Polytropic gases: Derivation of the equation of state Linearize Euler equations. Get wave equation. Sound in pipes. Derive appropriate B.C. for open/closed ends. % % --------------------------------------------------------------------- #010 Example: Longitudinal vibrations of an elastic rod. Skip ............. [PSQ] 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. Note that u = x + U, where U = U(x, t) is the longitudinal displacement from equillibrium. T = T(x, t) tension (T(x0, t) is the force by x > x0 on x < x0). Thus -T = momentum flux. f = f(x, t) applied longitudinal force per unit length. Elasticity: T is a function of the strain = F(u_x - 1) = F(U_x) Explain why U_x is the strain. Derive equation and consider Hooke's law case: T = kappa*(u_x-1). In fact: kappa = E*S, where E = Young's modulus and S = cross-section. % % --------------------------------------------------------------------- #011 Shallow Water Wave equations; 1d and 2d. Analogy with polytropic gas with gamma = 2. Linearization. Wave equation. Boundary conditions. Analogy of wall condition [u=0] with closed pipe in acoustics. Waves moving from a channel into a big open, quiescent, lake provide analogy with open pipe in acoustics [h = constant at outlet]. % % --------------------------------------------------------------------- #012 Sketch derivations: --- Fluid flow on an elastic, flexible pipe [notes on web page]. Point out derivation similarity to the one used for shallow water: "long wave assumption". Equations "the same" as shallow water or isentropic gas dynamics, but for the expression of the pressure as given from the balance with the pipe wall tension. Discuss the effects of non-thin pipe walls; dissipation and/or dispersion due to bending resistance. --- Vibrations of a taut membrane under constant, isotropic, tension (e.g.: a drum). % % --------------------------------------------------------------------- #013 Wave equation, homogeneous media: % --------------------------------------------------------------- #013a I.V. problem solution for the infinite line/string. D'Alembert's formula, derived from u = f(x-c*t) + g(x+c*t). Define: **domain of influence** and **domain of dependence.** Concept applies to many other equations, e.g.: Klein-Gordon. % --------------------------------------------------------------- #013b Signaling problem in semi-infinite string: u(0, t) = sigma(t) for t > 0; some initial condition at t = 0. Start from general solution u = f(t-x/c) + g(t+x/c). Do space-time diagram of characteristics + zones of dependence/influence. If IV only affect a small portion of the string, after a while they are gone, and only the signal from the origin remains u=sigma(t-x/c). % --------------------------------------------------------------- #013c String in a finite interval, with signaling on both ends: picture of the solution in space-time, with the various regions. Sketch how to solve problem by following the wave reflections [from both ends] as they go back and forth in the interval. % -------------------------------------------------------------- #013d Reflection/transmission at a junction (two strings tied). Same tension but different density: cj^2 = T/rho. rho1*u_tt - T*u_xx = 0 for x < 0, \ u continuous at x = 0. rho2*u_tt - T*u_xx = 0 for x > 0, / T*u_x continuous at x = 0. u = I(t-x/c1) + R(t+x/c1) x < 0 \ R = (c2-c1)/(c2+c1)*I u = J(t-x/c2) x > 0 / J = 2*c2/(c2+c1)*I Waves: I = incident; R = reflected; J = transmitted. Normally one uses T for the transmitted wave, but here T is the tension! Note: here I, R, J are functions. For generic wave systems this calculation has to be done Fourier mode by Fourier mode; but for the wave equation all modes behave the same. % % --------------------------------------------------------------------- #014 Impedance calculation for string: rho*u_tt - T*u_xx = 0. % -------------------------------------------------------------- #014a Apply force F = a*e^{i*omega*t} at end [x = 0] and compute response u = A*e^{i*omega*(t-x/c)}. % ----------- % The boundary condition at x = 0 ............. F + T*u_x = 0 [total force on zero mass must vanish] implies that a and A are related by ......... a = i*omega*(T/c)*A Hence, the impedance (defined below) is Z = i*omega*(T/c) = i*omega*sqrt{rho*T} % -------------------------------------------------------------- #014b IMPEDANCE = coefficient relating applied forcing amplitude to system response amplitude: a_f = Z*a_r. Applies to oscillatory situations (time dependence e^{i*omega*t}). In electric circuits Z relates applied voltage to intensity: V = Z*I. Generalizes Ohm's law to AC: R ---> I. % -------------------------------------------------------------- #014c IMPEDANCE MATCHING (for strings example). At a junction between two different strings, the forces by each string on the other are equal and opposite. If the two impedances are the same, their responses [for a wave traveling along the string] are equal; and no reflected wave is generated. On the other hand: IMPEDANCE MISSMATCH leads to a reflected wave. % -------------------------------------------------------------- #014d Example of impedance matching for strings: At a ``free'' junction between two different strings [densities rho1 \neq rho2], the two tensions must be equal. Hence impedance matching cannot be done. HOWEVER: add a "ring and rod" device at the junction, so as to be able to have distinct tensions on each side, to achieve impedance matching. rho1*u_tt - T1*u_xx = 0 for x < 0, rho2*u_tt - T2*u_xx = 0 for x > 0, u ............... continuous across junction. force = T*u_x ... continuous across junction. u = I(t-x/c_1) + R(t+x/c_1) x < 0 \ R = (q1-q2)/(q1+q2)*I u = J(t-x/c_2) x > 0 / J = 2*q1/(q1+q2)*I where qj = sqrt(rhoj*Tj) ---> Compare with #014a; Zj = i*omega*qj Impedance matching yields q1 = q2, so that R = 0. % -------------------------------------------------------------- #014e SKIP ................................................................. [PSQ] Another example of impedance matching: Sound waves through a constriction in tube cross-section + a diaphragm [two gases]. Very similar to the example in #014d. % -------------------------------------------------------------- #014f Generally Z is complex and has a nontrivial frequency dependence. The dependence shown above for the wave equation is special, and allows impedance matching for ALL frequencies simultaneously, as in #014d. This is very much not the general case. In general: **** reflection can be avoided for only a few frequencies. **** % -------------------------------------------------------------- #014g Examples of practical applications of impedance matching: --- Joints at transmissions lines (e.g.: wave-guides). --- Coating of complex lenses in optics, to avoid loses. See #015 [this is not exactly impedance matching]. To get around the issue in #014f, complicated junctions with many parameters are used. This allows the suppression of reflections for many frequencies, with ``small'' reflections for intermediate frequencies [effectively, getting no reflections for a range of frequencies]. In high quality optics this is achieved by multiple layers of coating on the lenses. Add to this composite lenses to decrease aberrations, and you can see why these things are rather expensive. % % --------------------------------------------------------------------- #015 Simple example, 1-D analog of the situation in optics [see #014g]. Two strings with different properties, but (instead of joining them directly) insert a third string of length L in between. This can be used to eliminate reflections at certain frequencies. Sketch how/why this works: cancelation of reflections by being out of phase. Details of the calculation will go in a ....................... [PSQ] % % --------------------------------------------------------------------- #016 Energy budget across a junction between two strings. rhoj*u_tt - T*u_xx = 0, % j = 1 for x < 0 and cj = sqrt(T/rhoj) = wave speed. % j = 2 for x > 0. Energy equation: E_t + F_x = 0. E = 0.5*(rho*u_t^2 + T*u_x^2) energy density and F = -T*u_x*u_t energy flux. Why is this the energy flux? Note: force (T*ux) times velocity. so F is the work done by the left side (on the right side) at any point along the string. Reflected and transmitted wave: u and u_x continuous across zero, and u = I(t-x/c1) + R(t+x/c1) for x < 0 u = J(t-x/c2) for x > 0 Then R(s) = (c2-c1)/(c2+c1)*I(s) and J(s) = 2*c2/(c2+c1)*I(s). This was done in #013d. Now calculate the energy fluxes -T*u_x*u_t for each of the waves Energy flux for I ... Fi = sqrt(rho1*T) (dI/ds)^1 Energy flux for R ... Fr = ((c2-c1)/(c2+c1))^2*Fi Energy flux for J ... Fj = (2*c2/(c2+c1))^2*(c1/c2)*Fi Thus Fi = Fr + Fj [energy in = energy out]. % % --------------------------------------------------------------------- #017 String on an elastic bed. Klein Gordon equation. % % --------------------------------------------------------------- #017a a1 - Dispersion relation, allowed and forbidden frequencies. Show that c_phase > c > c_group, and point out that: c = maximum speed of propagation. See picture in the web page [home]. "Prove" this by writing equation in characteristic form u_t + c*u_x = v v_t - c*v_x = - m^2*u and use it to show how the solution outside the domain of influence of the initial data vanishes. Sketch: Advance small time steps tau = Delta t, and solve the charac. eqns. approximately. Can show then that in 1-time step solution changes by O(tau^2). Hence, at a finite T, the change is T*O(tau), because the number of steps is T/tau. Now let tau --> 0. It follows that phase speed is not physically very significant. It is the velocity at which the phase moves, but it does not actually carry any information, or energy. This fact is not very intuitive, as the phase speed is the speed at which ``the wave bumps'' move, it is the stuff you see when you look at (say) waves on the surface of a pond. However, if you look at the edges of a wave package, you will see that the bumps either die or are born there. The package itself moves at the group speed, which is highly significant, as it is the speed at which actual information (and energy) propagates. % % --------------------------------------------------------------- #017b (Steady state) signaling problem on a semi-infinite string: b1 - Evanescent waves: Subcritical forcing. No propagation. Penetration length [compute] for wave coming "from the left" in string with parameters that allow propagation, into a string on the right where the frequency is subcritical. SKETCH details ......................................... [PSQ]. b2 - Supercritical waves: Waves propagate. First look at radiation condition: compute average work done on string by forcing. Select solution where this is positive (else energy must be arriving from infinity at steady state). Note relationship with group speed. rho*u_tt - T*u_xx + b*u = 0, c^2 = T/rho and m^2 = b/rho. Apply signal at x = 0: u(0, t) = a*cos(omega*t), with omega supercritical omega^2 > m^2 and compute resulting wave: u = a*cos(k*x-omega*t). Pick k out of the two possible ones by computing power [energy flow] at x = 0: P = -T*u_x*u_t = T*k*omega*sin^2(omega*t) Must be P \geq 0, hence k*omega > 0. Picks sln. with c_g > 0. b3 - Notes about b2. For energy/power computations MUST pick real solution. Complex notation not so good here! Use energy equation to get the same answer. The sign of the energy flux is the same as that of c_g. b4 - Finally: compute average energy and energy flux densities for a sinusoidal wave. Show: Fave = c_g*Eave. Thus: c_g = wave energy flow velocity. Let u = a*cos(k*x-omega*t), with omega^2 = m^2 + c^2*k^2. Then [use the fact that average(cos^2) = average(sin^2)=1/2] E = (1/2)*rho*[u_t^2 + c^2*u_x^2 + m^2*u^2] ==> Eave = (1/4)*rho*[omega^2 + c^2*k^2 + m^2]*a^2 = (1/2)*rho*omega^2*a^2. F = - T*u_x*u_t = - rho*c^2*u_x*u_t ==> Fave = (1/2)*rho*c^2*omega*k*a^2. Hence Fave = (k/omega)*c^2*Eave = c_g*Eave, since c_g = c^2*k/omega. More examples of the point made in b4 above ............. [PSQ] % % --------------------------------------------------------------- #017c Do other examples as well; like KdV or Schrodinger [even wave equation]. 0 = u_t + u_x + u_xxx yields 0 = E_t + F_x, with E = (1/2)*u^2; and F = (1/2)*u^2 + u*u_xx - (1/2)*u_x^2; % % --------------------------------------------------------------------- #018 Radiation Damping. Many applications --- e.g.: wave drag on ships. See, in Lecture Topics for 18376, the Radiation Damping Section. -- Semi-infinite string with mass-spring at end. -- Semi-infinite string over elastic bed with mass-spring at end. Free trapped modes. Harmonic forcing. Laplace Transform. See also #019 below. Below: Radiation damping simple examples. % % --------------------------------------------------------------- #018a Example: semi-infinite string under tension, with a mass-spring system at one end: rho*u_tt - T*u_xx = 0 for x > 0, M*u_tt = -k*u + T*u_x at x = 0. Write general solution [with radiation condition] for x > 0 u = f(t - x/c), where c = sqrt{T/rho} Then plug into BC at x = 0 and find an ode for f. Solve ode for f. Note then that using the solutions of the ode for all times leads to an u=f(t-x/c) that blows up as x --> infty. -- Contradiction is apparent only. The ode solution cannot have been there for arbitrarily large negative times [infinite energy needed]. Hence solution above must be cut-off beyond some point; i.e.: for x > c*(t-t0), where t0 = start time. -- For the general IVP we need both waves. The wave going left forces the oscillator at x = 0, till it is gone. Then decay starts. Write equation with both inbound and outbound waves. % % --------------------------------------------------------------- #018b Example: linear shallow-water with a moving paddle (attached to a spring) at one end. SKETCH (Math. same as Example 1). Problem: Radiation damping # 02. ............................. [PSQ] % % --------------------------------------------------------------- #018c Example: semi-infinite string under tension, on an elastic bed, with a mass-spring system at one end. The two prior examples are simple because there is no dispersion, hence a simple exact solution, frequency independent, is available. When a simple exact solution is not available, and the calculations are made mode-by-mode, the fact that there is no finite energy steady state complicates things. How do we then compute the damping? In this example we consider ways to deal with this issues. Details in Lecture Topics for 18376, Radiation Damping Section. % % --------------------------------------------------------------------- #019 Review: Laplace Transform Laplace transform as function in the complex plane, and formula for the inverse. Relationship with normal modes via residues, but more general (allows singularities other than poles). Application to #018c [See Lecture Topics for 18376, Radiation Damping]. % % --------------------------------------------------------------------- #020 Moving sources problems and wave generation. Examine first a 1-D problem. [hand written notes] ............... [PTS] For steady state solution must have c_p(k) = V, where V = source speed. Then c_g decides if the waves are ahead (c_g > V) or behind (c_g < V) the source. If c_p(k) = V has no solution, then there may not be a steady state, or there may be a "trapped" steady state (evanescent waves near the body). % % --------------------------------------------------------------- #020a Example: string on an elastic bed with a source. See ``Lecture Topics'' notes as well. u_tt - c^2*u_xx + omega_0^2*u = f(x - V*t), 0 < V. Because of linearity, this can be reduced to u_tt - c^2*u_xx + omega_0^2*u = delta(x - V*t), 0 < V. Analyze cases V > c and 0 < V < c. Construct solution by looking at ode steady state must satisfy. In this case, when V < c, get evanescent waves near source for a steady state. % % --------------------------------------------------------------------- #021 First look at dispersive amplitude modulation equations: [PTS] % % --------------------------------------------------------------- #021a Amplitude modulation with frozen carrier frequency. u = A(chi, tau) e^{i*theta_0}, with theta_0 = k_0*x-omega_0*t, chi = epsilon*x, tau = epsilon*t, and epsilon is small. Show: to leading order, A satisfies: A_tau + c_g^0*A_chi = 0. Do it by "direct" calculation for a few examples; e.g.: u_t + u_xxx = 0; .......... Linear KdV. u_tt - u_xx + u = 0; ...... Klein Gordon [note c_p > c_g]. u_tt + u_xxxx = 0; ........ Beam equation [note c_p < c_g]. Do it for the "generic" equation u_t + i*Omega(-i*\partial_x) u = 0; where omega = Omega(k) is a polynomial. This requires showing: i*Omega(-i*\partial_x) u = i*Omega(k_0)*u + c_g(k_0)*A_chi*e^{i*theta} which can be done term by term. Note: ........................................................... [PSQ] In the n-D case the equation is: A_tau + (c_g^0.grad) A = 0. % % --------------------------------------------------------------- #021b Fourier Transform Approach [narrow band spectrum]. Note that u = A(chi, tau) e^{i*theta_0} corresponds to initial conditions where all the Fourier modes are near k_0. Thus, consider solutions of the form u = \int a((k-k_0)/\epsilon) e^{i(k*x-omega*t)} dk where a(zeta) decays for |zeta| large. Then change variables: k = k_0 + epsilon*kappa, and expand omega = Omega(k), to obtain the same result as in #021a, but now for a generic Omega. The "expand" is what is behind the group speed being the derivative of omega. % % --------------------------------------------------------------- #021c Note that the approach in #021b is more "powerful" than the one in #021a, as it treats all possible dispersion relations at once, including ones where Omega might not be something simple like a polynomial. On the other hand, the approach in #021a can be generalized to weakly nonlinear problems, or variable coefficients, ....................................... [PSQ] while the one in #021b cannot. % % --------------------------------------------------------------- #021d As an example where Omega is a transcendental function, consider the equations for linear water waves. In 2-D these are: phi_xx + phi_yy = 0 for 0 < y < h. phi_y = 0 on y = 0. eta_t - phi_y = 0 on y = h. phi_t + g*eta = 0 on y = h. Here: phi = velocity potential, x = horizontal coordinate, y = vertical coordinate, y = 0 is the bottom, y = h is the equilibrium level of the water, eta = deviation from equilibrium of the surface, and g = acceleration of gravity. These equations neglect viscosity, surface tension, and assume that the flow is irrotational. "Derive" the equations and point out that they give rise to a transcendental dispersion relation [involving a hyperbolic tangent]. Task for the students: compute the dispersion relation ......... [PSQ] Separate the horizontal dependence as e^{i*(k*x-omega*t)} and solve for the vertical dependence. % % --------------------------------------------------------------- #021e Generic modulation: u = A(chi, tau)*e^{i*theta} where chi = epsilon*x, tau = epsilon*t, epsilon is small, and theta = Theta(chi, tau)/epsilon. Then k = theta_x and omega = -theta_t. Justify the form above as corresponding to a wave whose amplitude, wave-number and wave-frequency are slowly variable. Explain how a local Taylor expansion at each point provides a "plane" wave solution near any point. Obtain the modulation equations, by direct calculation, for a few examples [same as #021a], as well as for the "generic" equation u_t + i*Omega(-i*\partial_x) u = 0; where omega = Omega(k) is a polynomial. For the calculations note that: Omega(-i*\partial_x) A*e^{i*theta} = Omega(k-i*epsilon*\partial_chi) A*e^{i*theta}. Then, by expanding Omega(zeta) = sum a_n zeta^n show that: Omega(k-i*epsilon*\partial_chi) = Omega(k) - i*epsilon*0.5*Omega''(k)*k_chi - i*epsilon*Omega'(k)*\partial_chi + O(epsilon^2) Thus, omega = Omega(k) ---> Note pde: Theta_tau + Omega(Theta_chi) = 0, and A_tau + c_g*A_\chi + 0.5*(c_g)_\chi*A = 0; where c_g = Omega'(k). Show this second equation implies rho_tau + (c_g*rho)_chi = 0 for rho = |A|^2. Then show that, for any E = e(k)*rho [in particular, the energy] E_tau + (c_g*E)_chi = 0. Hence: energy flux = c_g*(energy density) NOTE: if phi is the argument of A [i.e.: A = sqrt(rho)*e^{i*phi} then phi_tau + c_g*phi_chi = 0 ...................... [PSQ] This analysis is valid for Omega = Omega(k) a polynomial, and it can extended to the case Omega a rational function. But it does not cover "arbitrary" Omega. This can be done using Fourier Transforms, but it will have to wait till after we do Stationary Phase .......................................................... [PSQ] % % ========================================> MORE POINTS WILL BE ADDED LATER. % ======================================================================== % EOF