INTRODUCTION
We will consider a wave to be a signal which propagates in time carrying energy and information. To describe a wave will require functions, u(x,t), which give the displacement from equilibrium of some quantity as a function of position, x, and time, t. The simplest model of a wave would be to assume a function of the form
u(x,t) = f(x−ct), which describes a righttravelling wave. The shape of the signal
Position, x
remains the same but its position moves to the right at speed c. We know from experience that the shape of the pulse can sometimes change (waves on the sea do this as they approach the shore for example), so we will have to be flexible with this definition as we go along.
Systems which allow wave excitations are governed by the same laws of mechanics, and of electricity and magnetism, as particles are. We should therefore expect the equations of motion to be differential equations. But the functions involved are functions of more than one variable and the corresponding equations are called partial differential equations or PDEs. We will start by reviewing the notation for partial derivatives. We will look at the important distinction between linear and nonlinear equations and then concentrate on linear systems, which, at least for small displacements, describe the situation of light (electromagnetic), sound and other mechanical waves. It will be important to become familiar with wavenumbers (k = 2π/λ, here λ is the wavelength), which have a natural generalisation to higher dimension as wavevectors.
PARTIAL DERIVATIVES
The partial derivatives of u = u(x,y) are defined as follows:
∂u
∂x ∂u
∂y
= lim u(x+δx,y)−u(x,y) (1) δx→0
= lim u(x,y +δy)−u(x,y). (2) δy→0
The partial derivative of u(x,y) with respect to x is just the ordinary derivative of u with respect to x with the variable y held constant. Similarly ∂u/∂y is the ordinary derivative with respect to y holding x constant. We can therefore take over all the standard results for differentiating functions, e.g.
(i)


∂xexy = yexy (ii)
∂yynex = nyn−1ex
(y treated as a constant)
(x treated as a constant)

(iii) If f is an arbitrary (differentiable) function of its argument
∂xf(x+ct) =

∂tf(x+ct) =
f′(x+ct) (3)
cf′(x+ct) (4)
Here f′(s) = df/ds is the (ordinary) derivative of f with respect to its argument (in this case x+ct).
HIGHER ORDER DERIVATIVES
Just as there are 2nd and higher order ordinary derivatives for functions of one variable,
we can define higher order partial derivatives:

∂u(x+δx,y) ∂u(x,y)

= lim ∂x ∂x δx→0
∂u(x,y+δy) ∂u(x,y)

∂y ∂y ∂y2 δy→0 δy
Again these are simply higher order ordinary derivatives taken with respect to one variable while holding the other variable constant.
There are also mixed derivatives:


∂x∂y = lim ∂u(x+δx,y) − ∂u∂y y) /δx ∂2u ∂u(x,y +δy) ∂u(x,y)
∂y∂x δy→0 ∂x ∂x
For all the functions, for which the derivatives exist and are continuous, the order in
(5)
(6)
(7)
(8)
which the limits (δx → 0,δy → 0) are taken does not matter, i.e. (Clairaut’s theorem)
∂2u ∂2u ∂x∂y ∂y∂x
EXAMPLE
For u(x,y) = x3y2 +yex
∂u
∂x ∂u
∂y
= 3x2y2 +yex
= 2x3y +ex.
2


By differentiating again we obtain the second order derivatives
uxx =
uyx =
uxy =
uyy =
2

∂x2 = ∂x ∂x = 6xy2 +yex,

2
∂y∂x = ∂y ∂x = 6x2y +ex,



∂x∂y = ∂x ∂y = 6x2y +ex, ∂2u ∂ ∂u 3
∂y2 ∂y ∂y
Note that (as required by Clairaut) the two mixed derivatives, uxy and uyx, are the same.
NOTATION
A number of different notations are used to denote partial derivatives. Three of the
most common are ∂u
∂x
ux and

∂u ∂xy
In the last of these the subscript y is there to emphasise that the variable y is held constant. The second of these (ux for ∂u/∂x) is commonly used when discussing PDEs.
Notation can become cumbersome for partial derivatives, particularly for higher order derivatives, and abbreviated notations are often used. A commonly used notation is
2 uxx for ∂x2,
2 uxy for ∂x∂y,
3

uxxy for ∂x2∂y, etc.
EXERCISE–1
(a) Find ∂u/∂x and ∂u/∂y for the following (marks for each part in brackets):
(i) u(x,y) = x3y2 +2xy. [2] (ii) u(x,y) = sinxy. [2]
(b) Find (take log to mean natural log)
∂2u
∂x2
if u(x,y) = log(x2 +y2). [2]
(c) Find ∂u/∂x, ∂2u/∂x2, ∂u/∂t and ∂2u/∂t2 for
u(x,t) = g(x+ct)
where g is some arbitrary differentiable function. [4]
WAVE EQUATIONS
There are many equations which admit travelling wave solutions. For example, there is the socalled wave equation
utt −c2uxx = 0 or ∂tu −c2∂x2 = 0. (9)
We will not go into its derivation (this is covered in lectures), but it should seem plausible that this comes from applying the laws of mechanics to some matter: utt is an acceleration (2nd derivative with respect to time of the displacement) and this has (presumably) been set equal to a force per unit mass. We can check whether a travelling wave satisfies this equation by inserting u(x,t) = f(x−ct) to find (−c)2f′′(x−ct)−c2f′′(x−ct) = 0.
The advection equation


ut +cux = 0 or ∂t +c∂x = 0 (10)

also has u = f(x−ct) as a solution, as can be seen by direct substitution into the equation. This equation models the advection downstream of the density of some additive in a river flowing at speed c. For example, if the density at x at time t is u(x,t), the density at time t = t+δt will be the density cδt upstream, i.e. the density currently at x−cδt. Provided that the density varies slowly enough with position, u(x−cδt,t) ≈ u(x,t)−cδt ∂u = u(x,t)−cδtux. Hence the change in u at position x in time δt is
δu ≡ u(x,t+δt)−u(x,t) ≈ −cδt ∂x = −cδtux.
Dividing by δt and taking the limit δt → 0 gives ut = −cux, which is the advection equation (10).
The advection equation is not that interesting as it is a mathematical model of something moving at constant speed. This equation
ut +c(u)ux = 0 with c(u) = c0(1 −2u) (11)
looks similar but has interesting solutions. It is used to model trafic flow on motorways. The idea is that u is the density of trafic (normalised to some maximum value), which moves along at an average speed c0 when the density is low. When the density rises, the average speed c0(1 −u) of trafic is reduced and is zero when the maximum capacity is reached. (The factor of c0(1 −2u) in the equation occurs because the effect of the density on wave speed turns out to be different from that on the trafic speed (1 −u).) This equation helps explain many phenomena including why pesky trafic jams can appear, apparently spontaneously, without there having been any accident or other obvious cause.
LINEAR EQUATIONS
We write a partial differential equation for the solution, u(x,t), as
Lu = f(x,t).
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Here all the terms involving u are on the left hand side. For the wave equation the operation, L, is then defined by
Lu = utt −c2uxx
and the function f(x,t) = 0. If we take two function u1(x,t) and u2(x,t) and form u(x,t) = u1(x,t)+ u2(x,t) and
Lu = Lu1 +Lu2 (12)
then we say the equation is linear. There is also the requirement that, for α a constant, Lαu = αLu. These two conditions are the same as those for Ordinary Differential Equations (ODEs) which are treated in lectures.
If we take the wave equation we find
L(u1 +u2) =
=
∂2(u1 +u2) 2∂2(u1 +u2) ∂2u1 ∂2u2 2∂2u1 2∂2u2 ∂t2 ∂x2 ∂t2 ∂t2 ∂x2 ∂x2
Lu1 +Lu2
so the wave equation is linear. On the other hand, for the trafic flow equation
L(u1 +u2) =
=
=
=
∂(u1 +u2) −c0(1 −2(u1 +u2))∂(u∂+u2)
∂u1 −c0(1 −2u1)∂u1 + ∂u2 −c0(1 −2u2)∂u2 −2c0u1∂u2 −2c0u2∂u1
Lu1 +Lu2 −2c0u1 ∂x −2c0u2 ∂x Lu1 +Lu2.
This equation is not linear.
The rough and ready way to see whether an equation is linear is to check whether the power of u is 1 or zero in every term. So u,uxx,ut all have u to the power one. In the trafic flow problem there are terms uux which has u to the power 2 and the equation is not linear.

The important result for linear systems is that if we find some solutions, ui, with
i = 1,…n, any linear combination of these solutions, u = i αiui with the αi constants, will also be a solution. This is sometimes called the principle of superposition.
EXERCISE–2
(a) Verify whether the following equations are linear or not:
(i) Schrodinger’s eqn: iut +uxx = 0. [2] (ii) Burger’s eqn: ut +uux = 0 . [2]
(iii) Transverse beam equation (γ is a positive constant): utt +γuxxxx = 0. [1]
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(iv) KdV eqn (k is a positive constant): ut +uux +kuxxx = 0. [1]
(b) Verify that a superposition of a righttravelling wave and a lefttravelling wave,
u = f(x−ct)+ g(x+ct), with f and g arbitrary differentiable functions, satisfies the wave equation. [2]
(c) Insert a solution u = f(x−ct) into the equation in 2(a)(i) and show that

′′
f′ = ic,
where f′′ is the second derivative of f with respect to its argument. [2]
BOUNDARY CONDITIONS The wave equation (9) has solutions
u(x,t) = f(x−ct)+ g(x+ct)
where f and g are arbitrary (differentiable) functions. This was what you showed in your answer to Question 2(b). However, what we usually want to solve are the wave equation (or some other equation describing waves) subject to some boundary conditions: a wave may be incident on a boundary or a wave may be trapped in some finite region. When the boundary conditions specify some initial configuration we call them initial conditions. Problems involving partial differential equations and associated boundary conditions are called boundary value problems. It is similar to what happens in particle mechanics: we often specify the initial position of a particle and its initial velocity. We then integrate Newto to time.
The b em, the bound xample would
u
x = 0 v T x
Here the boundary condition is at x = 0 and is u(0,t) = 0. However, if the block had finite mass, then the boundary condition would need to reflect the mechanics of the block. This would mean computing the forces on the block from the string (and elsewhere) and setting them equal to the product of its mass and acceleration. The point is that we need to understand the physics at the boundary to deduce what boundary condition to impose on the displacement at x = 0.
Once we have the equation and have specified the boundary conditions, we need to solve for the solution u(x,t). (Proving mathematically that the solution is unique can be involved. We will simply assume that boundary value problems, which model physical systems, have unique solutions.) In linear systems, there are usually natural solutions to
6
look at, which we can then combine in linear combinations to satisfy the boundary conditions. The most familiar solutions are the plane waves.
PLANE WAVES
For linear systems, the natural waves to look at are the socalled plane waves. (In one dimension there are no planes involved but, in three dimensions, there are surfaces of constant phase which are planes.) The standard form for such waves is
u(x,t) = Aeikx−iωt where A = Aeiϕ. (13)
Here A = Aeiϕ is a complexvalued constant. If u is a measurable quantity like a displacement, we have to extract a real function usually by taking the real part, Re[u]. This gives u = Acos[kx−ωt +ϕ].1
The plane wave is a travelling wave solution: ei(kx−ωt) = eik(x−vt) where v = ω/k is a velocity. As any travelling wave is a solution of the wave equation, the plane wave must be a solution. (From now on, will adopt the notation v for a speed and reserve c for the speed of light.) The sign of k determines the direction of travel of the wave—for k > 0 it describes a righttravelling wave and for k < 0 a lefttravelling wave.


It is important to get used to the notation of wavenumbers, k = 2π/λ where λ is the wavelength, and angular frequency ω = 2πν where ν (Greek letter nu) is the frequency. This is the notation normally used in physics. This gives us the familiar result that the speed of the wave
v = k = 2π/λ = νλ,
i.e. the speed is the frequency (ν) multiplied by the wavelength (λ). When we are taking the real part of the complex function, u, the sign of a frequency is not measurable and, as a convention, we will always write the timedependence as e−iωt and take ω > 0. So, our lefttravelling plane waves will be u = e−ikx−iωt with ω > 0. This is the convention in quantum mechanics (QM), where the function u is complexvalued (but is not directly measurable). In QM, ω is related to energy and the sign of the frequency does have a meaning as energies can be negative.
One might think that we need to get involved in the theory of partial differential equations for wave systems. But in cases involving linear equations this is not always necessary. We know that the plane waves are travelling wave solutions. For linear systems, we can form linear combinations of plane waves and obtain a resultant wave which is also a solution of the wave equation (principle of superposition). We look for those linear combinations of plane waves which satisfy the boundary conditions. This approach is used to find and understand a large number of phenomena like diffraction, the formation of standing waves, the idea of a group velocity.
1An advantage of using complex numbers over real numbers is that all the phase information is automatically taken care of (without the need for multiple angle formulae). For linear operations on u (see 12), such as adding, differentiating/integrating with respect to a real variable, multiplying by a real constant, we can take the real part at the end of the calculation to find the physical quantity. However, once any operation involves nonlinear combinations of u, the approach will no longer work and we will need to work explicitly with Re[u].
7

If we take the example of a plane wave incident from the right on the boundary at x = 0 shown above, we assume an incident left travelling wave ui = ei(−kx−ωt). This clearly does not satisfy the boundary condition at x = 0 as Re[ui(0,t)] = cosωt = 0. Yet intuitively we know what happens: the wave is reflected. This suggests looking for a solution which is the linear superposition of the incident (lefttravelling) plane wave, ui, and a reflected (righttravelling) plane wave, ur, at the same frequency:
u(x,t) = ui(x,t)+ ur(x,t) ≡ ei(−kx−ωt) +rei(kx−ωt).
The constant, r, is called the reflection coeficient or reflection amplitude. In general it is a complex number, r = reiϕr. Its value is fixed by enforcing the boundary condition:
u(0,t) = 0 ⇒ 0 = e−iωt(1+ r).
This is satisfied by choosing r = −1. In this case the wave is totally reflected (r = 1) and is inverted with respect to the incident wave (ϕr = π). The full solution is

−ikx ikx
u(x,t) = ei(−kx−ωt) −ei(kx−ωt) = 2ie−iωt 2i = −2ie−iωt sinkx
= −2i(cosωt −isinωt)sinkx = −2sinkx(sinωt −icosωt). (14)
When we take the real part, we obtain Re[u(x,t)] = −2sinkxsinωt. Note that this is not a travelling wave. At each point the displacement oscillates with frequency ω and amplitude 2sinkx. The incident and reflected waves have interfered to give a standing wave, which is not propagating anywhere.
Perfect(!) Join
T
Mass per unit length ρ1 Mass per unit length ρ2
u(x,t)
0 T x

Now let us look at a more general boundary condition. The figure shows two strings with different values for the mass per unit length, ρ1 and ρ2. There is supposed to be a perfect join between them. The mass per unit length affects the wave speed along the string as this is given by T/ρ. The displacement, u, therefore satisfies
utt −v2uxx = 0

utt −v2uxx = 0


with v1 = ρ1

with v2 = ρ2
for x < 0,
for x > 0.
However neither equation is valid at x = 0. Instead there are boundary conditions. One is
lim u(x,t) = lim u(x,t), x→0− x→0+
8

i.e. the two strings need to be joined. Here limx→0± means the limit as x tends to zero from below () or above (+). Less obvious is the requirement that
lim ∂u(x,t) = lim ∂u(x,t), x→0− x→0+
which relates to the fact that the string is under tension. Newton’s third law tells us that at each point the tension pulling one way is exactly balanced by the tension pulling the other way. If ∂u/∂x is not continuous there would be a net force pulling in some direction and no mass to accelerate, which would contradict Newton’s laws.
How do we solve this problem? Well, we work with plane waves. We imagine sending in an incident righttravelling plane wave from the far left, ei(k1x−ωt), and ask what happens. It can be reflected back into the lefttravelling wave, ei(−k1x−ωt), or transmitted into a righttravelling wave beyond the boundary, ei(k2x−ωt). (Once we have picked a frequency, ω, the wavenumbers k1 = ω/v1 and k2 = ω/v2 are fixed by the respective wave equations in the two strings.) We write
x < 0 : u(x,t) =
x > 0 : u(x,t) =
eik1x−iωt +re−ik1x−iωt
seik2x−iωt
Here r and s are reflection and transmission amplitudes (or coeficients) which we need to find.
The two boundary conditions give
1+ r = s (continuity of u), ik1(1 −r) = ik2s (continuity of gradient),
and so k1 −k2 k1 +k2
and s = k1 +k2. (15)
The wavenumbers are related to ω by the wavespeeds, ki = ω/vi, and hence we could also write
r = v1 +v2 and s = v1 +v2. (16)
Are r and s all we need to know? The answer is yes, because the system is linear. If we w
b t t
E
x = 0 u
x
m T
9
The displacement from the xaxis, u(x,t), of a tight string satisfies the wave equation. The string extends to infinity for x > 0 and is attached to a ring, of mass m, at x = 0. The ring is free to move up and down a frictionless pole. The wave velocity in the string is v. In the limit, ∂u/∂x ≪ 1, the boundary condition on u is

m∂tu(0,t) = T ∂x(0,t). Gravity is ignored.
(a) Explain the origin of the boundary condition. You should note that for small values of ∂u/∂x,
∂u/∂x = tanθ ≈ sinθ,
where θ is the angle the string makes to line parallel to u = 0 at position x. [4]
(b) Assume a solution
u(x,t) = e−ikx−iωt +reikx−iωt
and find an expression for r. [4]
(c) ∗C∗ Your answer for r will be a complex number. This question is about its amplitude and its argument.
(If you have not already done so, you should use v = ω/k to eliminate k from your answer to part (b).)
(i) The modulus of r is equal to one. What physical reason do you think is behind this? [3]
(ii) Write r = eiϕ(ω). Give the value of ϕ(ω) in the limits ω → 0 and ω → ∞. (You should be able to find these answers without finding an explicit expression for ϕ(ω).) [4]
ENERGY in WAVES


We stated earlier that waves carry energy. What is the energy flux? This is not obvious and the derivation needs consideration of the resolved forces and velocities. We will not derive it here. The answer for the tight string problem is that the energy flux in the positive x direction is:
P(x,t) = −T ∂t ∂x. (17) This clearly has the right dimensions (force × velocity). However, because this involves the product of two derivatives of u, we need to work directly with the real part of u.
We can check that (17) gives sensible results for the energy flux for the travelling waves, u ∼ eikx−iωt, and standing waves, u ∼ eiωt sinkx. In the two cases we find:
(v = ω/k = T/ρ)
Ptrav(x,t) = Tωksin2(kx−ωt) = pTρω2 sin2(kx −ωt)
Pstan(x,t) = Tωksin(kx)cos(kx)sin(ωt)cos(ωt) = Tρω2 sin(2kx)sin(2ωt)
10


When averaged over a period, Ptrav(x) = √Tρω2/2, while for the standing wave the corresponding average, Pstan(x) = 0, as we should expect. To average over a period, T = 2π/ω, we compute the average value of the energy flux

Z T
T 0 P(x,t)dt.
EXERCISE–4
Consider the case of a wave incident on a join between two strings under tension, T.
(a) We want to compute the energy fluxes of the component waves and their superposition.
(i) Compute the quantities ux and ut for the incident and reflected waves, ui = eik1x−iωt, ur = re−ik1x−iωt and for their superposition u = ui +ur.
(Treat only the case that r is real and leave your answers as complexvalued quantities); [4]
(ii) Find the real parts of the quantities ux and ut for both incident and reflected waves and their superposition in part (i);[4]
(iii) Insert your answers to part (ii) into the expression for the energy flux, Eq 17, and compute the energy flux in the incident and reflected waves and their superposition. [2]
(iv) Compute the energy flux in the transmitted wave (x > 0), u(x,t) = seik2x−iωt (again treat only the case that s is real). [5]
(b) ∗C∗ Insert the values for r and s given in Equation (15). Is energy conserved? [5]
WAVE GROUPS
A general case of a superposition (remember this means a linear superposition of waves with different wavenumbers and frequencies) might be:

∞
u(x,t) = akj ei(kjx−ω(kj)t) (18) j=1
which is a sum over a set of wavenumbers of travelling waves with amplitude akj . A simple case would have 2 different wavenumbers, k1 = k +∆k and k2 = k −∆k. Here k and ∆k characterise the average wavenumber k = (k1 +k2)/2 and their difference k1 −k2 = 2∆k. The corresponding frequencies are ω(k1) = ω +∆ω and
ω(k2) = ω −∆ω, where we have defined the average frequency ω = (ω(k1)+ ω(k2))/2 and their difference ω(k1)−ω(k2) = 2∆ω. This gives

i∆kx−i∆ωt −i∆kx+i∆ωt u(x,t) = eik1x−iω(k1)t +eik2x−iω(k2)t = 2eikx−iωt 2
= 2eikx−iωt cos(∆kx −∆ωt). (19)
11
One case to consider could be to take k2 = −k1. We would expect ω(k1) = ω(−k1), as in most systems waves with the same wavelength have the same frequency whether they are travelling to the right or to the left. The linear superposition of the two waves gives u(x,t) = 2e−iωt cos∆kx, which is a standing wave. This is similar to the case of the wave on a string incident on the boundary at x = 0, which we treated before—see Eq. (14). In that case we superposed two waves with equal and opposite wavenumbers but the reflected wave was inverted. We obtained a standing wave which varied as sinkx instead of the cos∆kx we have here.
Suppose we look at the case with ∆k ≪ k, i.e. with two wavenumbers, which are close on the scale of their absolute value. We can then write that
∆ω/∆k ≈ dω/dk ≡ vg(k). We will call the quantity vg(k) the group velocity. We obtain
u(x,t) = 2eikx−iωt cos(∆k(x−vgt)), (20)
or, if we take the real parts explicitly u(x,t) = 2cos(k(x−vt))cos(∆k(x−vgt)), where v = ω/k. We have the product of two travelling waves. One travels at what we have been calling the wavespeed (also called the phase velocity) and one travels at vg(k). For waves on a tight string and for light waves, ω = vk with v a constant ( T/ρ for the tight string and c for case of light). In these cases, vg = v.
There are many cases where ω(k) is not a linear function of k. In these cases the waves are said to be dispersive. For dispersive waves:


vg(k) = dk = k.
Dispersion is the reason we see rainbows or the splitting of white light into its constituent colours. The amount light is refracted, when passing from air into a prism or raindrop, depends on the speed of the light in the droplet or prism, which is different for different wavelengths. As a result different colours are refracted by different amounts.
For dispersive waves, the two waves in the product (20) do not propagate at the same speed. Examples you will meet include waves on deep water, where ω(k) = gk with g the acceleration due to gravity, and quantum mechanics where ω ∝ k2.
Phenomena associated with dispersion are common and not just found for the case of the superposition of two waves. They are observed for any superposition of waves, where the wavenumbers are grouped around some average value. For example,

N
u(x,t) = akj eikjx−iω(kj)t. (21) j=1
If all the kj are close to some average value, k0, then we write ∆kj = kj −k0 and approximate
ω(kj) ≈ ω(k0)+(kj −k0)dω(k) = ω(k0)+∆kjvg.
12
This gives

N
u(x,t) ≈ eik0x−iω(k0)t akj ei∆kj(x−vgt) ≡ eik0x−iω(k0)tf(x−vgt), (22) j=1

where the function, f(s) = Pj=1 akj ei∆kjs. The function u can be thought of as a wave, ∼ ei(k0x−ω(k0)t), with a time and spacedependent amplitude, f(x−vgt). We would need to know the values of the akj to compute the function f. However, provided that the values kj are close enough to some value k0, we have shown that the function f is a right travelling waveform with speed vg (see figure). Close enough in this context means that the errors involved in assuming that ω(kj) = ω(k)+∆kjvg are not significant.
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� �ω/��
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�(���) ω/� �
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The wavepattern (oscillating continuous line), u(x,t) , and “amplitude” functions (dashed lines), ±f(x−vgt), are shown for a possible wave group at a fixed time t. The wave group amplitudes, a(k) centred on k0, are shown in the insert. The carrier wave travels at speed
v = ω/k, while the amplitude modulation pattern, described by f, moves at speed vg = dω/dk (shown for this example to be v/2). A continuous distribution, a(k) ∼ e−(k−k0)2/4, has been
assumed for this example, instead of a distribution, a(kj), defined for discrete kj (see Eq. 21).
GROUP VELOCITY
The quantity vg (see Eq. 22) is called the group velocity. The name is used because it is associated with the velocity of the pattern formed by a group of waves with wavenumbers distributed close to some value. We will not go through the arguments here, but it is the group velocity, vg, not the phase velocity, v, at which energy and/or information usually propagate in waveforms.
EXERCISE–5
Assume a plane wave solution, eikx−iω(k)t, to the following equations and find ω(k) and vg/v where the phase velocity v = ω/k. State whether the waves are dispersive or not.
(a) utt −uxx = 0 [2]
(b) iut +uxx = 0. [4]
(c) utt +γuxxxx = 0, (γ is a positive constant) [4]
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EXERCISE–6


The displacement of a tight string is given by
u(x,t) = Acos 2π(x−vt) .
(a) Find an expression for the velocity ∂u/∂t at which a piece of the string moves. [2]
(b) What is the maximum value of ∂u/∂t? [2]
(c) For what values of A is this maximum value greater than the wave propagation speed v? [2]
(d) ∗C∗ Would you expect the behaviour of the string to be well modeled by the wave equation under the conditions in part (c)? [4]

(Show that an element of string of length dx becomes (1+ u2)1/2 dx. The derivation of the wave equation assumes that the extension can be neglected. Is this true when ut is comparable to the wave speed, v?)
EXERCISE–7
In exercise 2, we wrote that iut +uxx = 0 is called Schrodinger’s equation. But this is “mathstalk”. When we meet this equation in physics it is usually written for a free particle (free means no electrostatic potential applied) of mass m as follows:


2 2
iℏ∂t = −2m∂x2. (23)
Here ℏ is Planck’s constant divided by 2π. (We should expect any equation concerned with quantum mechanics to include Planck’s constant after all.) However, the two equations are the same once a scaling of the variables has been applied.
Write t = t′ℏ and x = ℏx′/√2m. Here t′ and x′ are the original variables scaled by constants. The chain rule gives

∂u ∂t ∂u ∂x∂u ∂u ∂t′ ∂t′ ∂t ∂t′ ∂x ∂t
(a) By using the chain rule, find
∂u
∂x′
and
∂2u
∂x′2
in terms of ∂x and ∂xu and write the equation (23) using the variables t′ and x′. [7] (b) Insert u = f(x−vt), where v is a constant, into (23) and show that
f′′ = i2mv.
Here f′′ is the second derivative of f with respect to its argument s, s = (x−vt). You should recognise that the left hand side is a perfect derivative. Integrate the equation, find f′ and integrate again to find f. You should ignore any solution of the type f =constant. [8]
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EXERCISE–8
The form u = eikx−iωt, or sometimes just u(x) = eikx (with the e−iωt factor assumed but not written explicitly), is often referred to as a plane wave. This is because it is the 1D case of
u(r,t) = eik.r−iωt,
where the vectors r = (x,y,z) and k = (kx,ky,kz) in 3D and r = (x,y) and
k = (kx,ky) in 2D. The exponent, k.r−ωt, is the phase of the wave. The troughs of the wave (if we take the real part of u) occur when this phase is (2n+1)π and the peaks occur when the phase is 2nπ.
(a) The points (x,y), where the phase of the wave in 2D is constant in space at a particular time t, satisfy kxx+kyy −ωt = ϕ0, with ϕ0 a constant. Why is this the equation of a straight line? [2]
(b) Find the perpendicular distances between the lines of constant phase for the same value of ϕ0 at increasing values of t. [2]
(c) ∗C∗ Repeat the analysis for the 3D case. You should find that the surfaces of constant phase are planes perpendicular to the direction k. [4]
(d) ∗C∗ If, in the 2D case, ϕ0 is increased by 2π, how far does the line of constant phase move in the perpendicular direction as a function of {kx,ky}? [2]
EXTENSION WORK–9 (not for credit)
A boundary value problem with the nonlinear trafic flow model
ut +(1 −2u)ux = 0 with u(x,0) = f(x),
has a relation involving the solution which we can write down but it is not an explicit solution. Here, we are measuring speeds in units of the maximum vehicle speed, c0. The density is assumed normalised to its maximum value, so that 0 ≤ u ≤ 1. One of the points of this question is to see why the apparent wave speed, (1 −2u) is different from the assumed trafic speed, (1 −u).
The equation looks like the advection equation, Eq. (10). We can guess a righttravelling wave solution, f(x−vt), and take v = (1 −2u):
u(x,t) = f(x−(1 −2u)t).
When a solution, u, appears on both sides of an equation, it is called an implicit equation for a solution. With s = (x−(1 −2u)t), we find
ux = f′(s)[1+2tux]
f′(s)

x 1−2tf′(s)
Here we have used the chain rule to differentiate u with respect to x and noted that s depends on x through the factor x and through the xdependence of u.
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(a) Find ∂u. (Do this in the same way ux is computed above, i.e. do not assume that u satisfies the equation.)
(b) Verify by direct substitution that the assumed form for u does satisfy the equation.
(c) Is the solution a valid model for trafic flow when 1−2tf′(s) = 0?
(d) ∗C∗ Discuss the nature of the solution for the two different cases, f′(s) > 0 for all s, and f′(s) < 0 for all s.
(e) ∗C∗ Derivation of Model
Let the trafic density as a function of time be u(x,t), and assume that the trafic flows at speed c(u) > 0. Consider the road at position x. Convince yourself that the trafic flow into this point in time interval δt is
u(x,t)×c(u(x,t)) ×δt.
Now consider an element of road of length dx between x and x+dx.
(i) Explain why the net flow into the road element in a time interval, δt, is
δt(u(x,t)c(u(x,t)) −u(x+dx,t)c(u(x+dx),t)) ≈ −δt ∂(c(u)u)dx.
Set this equal to the net increase in vehicles between t and t+δt, so
δu = dx(u(x,t+δt)−u(x,t)).
(ii) Make the Ansatz that c(u) = (1 −u), divide by δt to obtain δu/δt on the left hand side, let δt → 0 and derive the equation given at the beginning of the question.