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FLAP M6.4 Waves and partial differential equationsCOPYRIGHT © 1998 THE OPEN UNIVERSITY S570 V1.1

Module M6.4 Waves and partial differential equations1 Opening items

1.1 Module introduction

1.2 Fast track questions

1.3 Ready to study?

2 Waves and oscillations

2.1 Oscillations

2.2 Waves

2.3 Functions of two variables

2.4 The general wave

2.5 Partial derivatives

2.6 Higher partial derivatives

2.7 The wave equation

2.8 Deriving the wave equation

2.9 Standing waves

3 The Schrödinger equation

3.1 The time-dependent Schrödinger equation

3.2 The time-independent Schrödinger equation

4 Closing items

4.1 Module summary

4.2 Achievements

4.3 Exit test

Exit module


1 Opening items

1.1 Module introductionThe purpose of this module is to provide an introduction to two topics: waves and partial differentiation.Waves occur in many physical situations, including vibrating strings, ripples on the surface of a lake, soundwaves, electromagnetic radiation and so on. The study of waves will lead us to functions of two variables(in this case position x and time t) and to the derivatives of such functions. Functions of more than one variableare common throughout physics and finding the derivatives of such functions requires a knowledge of partialdifferentiation.


The structure of this module is as follows. In Subsection 2.1 we consider oscillations and the differentialequation of simple harmonic motion (SHM). Subsection 2.2 introduces waves and draws your attention to thedistinction between waves and oscillations. In Subsection 2.3 we consider functions of two variables and suchfunctions are used in the Subsection 2.4 to describe a general wave which propagates without changing its shape.Subsections 2.5 and 2.6 introduce first- and second-order partial derivatives, respectively. (The techniques givenhere have wide applicability throughout physics and not just to the study of waves.) Subsection 2.7 derives thepartial differential equation for a wave propagating in one dimension without changing its shape, andSubsection 2.8 shows how such a wave equation arises in the study of waves on a string. Subsection 2.9 showshow standing waves can occur on a vibrating string with fixed ends. Section 3 is devoted to the time-dependentand time-independent Schrödinger equations that arise in quantum mechanics. It explains the relationshipbetween these two equations, describes the significance of some of their ‘wave-like’ solutions (called wavefunctions), and demonstrates the crucial role of partial derivatives in quantum mechanics.

Study comment

Having read the introduction you may feel that you are already familiar with the material covered by this module and thatyou do not need to study it. If so, try the Fast track questions given in Subsection 1.2. If not, proceed directly to Ready tostudy? in Subsection 1.3.


1.2 Fast track questionsStudy comment Can you answer the following Fast track questions. If you answer the questions successfully you needonly glance through the module before looking at the Module summary (Subsection 4.1) and the Achievements listed inSubsection 4.2. If you are sure that you can meet each of these achievements, try the Exit test in Subsection 4.3. If you havedifficulty with only one or two of the questions, you should follow the guidance given in the answers and read the relevantparts of the module. However, if you have difficulty with more than two of the Exit questions you are strongly advised tostudy the whole module.

Question F1

The functions F1(x, y) and F2(x, y) are defined by

F1 ( x , y) = − ∂φ ( x , y)∂x

and F2 ( x , y) = − ∂φ ( x , y)∂y

where φ x , y( ) = q

4πε0 x2 + y2. ☞

Find F12 + F2

2 .


Question F2

Show that f(x − v1t) is a solution of the wave equation

∂ 2 f

∂x2− 1

v2

∂ 2 f

∂t2= 0

Question F3

The one-dimensional time-dependent Schrödinger equation for a particle of mass m with a potential energyfunction U(x, t) is

− ˙2

2m

∂ 2Ψ x , t( )∂x2

+ U x , t( )Ψ x , t( ) = i˙ ∂Ψ x , t( )∂t

Show that Ψ x , t( ) = exp − mωx2 (2˙)( ) exp − iEt ˙( ) ☞

is a solution of this equation with U ( x , t ) = mω 2 x2

2 where ω is a constant (so that U is a function of the single

variable x), and find an expression for E in terms of ω.


Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal routethrough the module and to proceed directly to Ready to study? in Subsection 1.3.

Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to theClosing items.


1.3 Ready to study?

Study comment To begin the study of this module you need to be familiar with the following topics: the trigonometricfunctions and trigonometric identities (specific identities will be supplied when needed); Cartesian coordinates and graphsketching (including the concept of a tangent to a graph, and the determination of the gradient of a straight line); thedifferentiation of functions of one variable (including the definition of a derivative in terms of a limit and the rules ofdifferentiation, particularly the chain rule); the derivatives of the trigonometric and exponential functions, polynomials andpowers; differential equations (including the role of boundary conditions in determining a solution to such an equation), andfor Section 3, the properties of complex numbers (including the rules for their addition and multiplication, and the meaningof the modulus of a complex number). You need to have some acquaintance with the basic ideas of Newtonian mechanics,including energy, momentum, power and Newton’s laws of motion; it would also be useful to have had some previousexperience of waves (including their wavelength, frequency and speed) since this module only provides a brief review oftheir properties. In a similar spirit it would be helpful, but not essential, to have some knowledge of simple harmonic motion,and Hooke’s law. Finally, you will also need to have some idea of the everyday meaning of the term probability, though youare not expected to be familiar with any of the mathematical properties of probability. The following Ready to studyquestions will help you to establish whether you need to review some of the above topics before embarking on this module.


Question R1

Differentiate the following functions with respect to x:

(a) f0(x) = 3x2 + 4x5

(b) g x( ) = exp(λ x3 ) , where λ is a constant,

(c) h(x) = cos1(x + 3x3)

(d) w x( ) = x + 1x2 + 2

(e) u( x ) = x2 + c2( )2, where c is a constant,

(f) v( x ) = x2 + sin ( yx )[ ]3 , where y is a constant,

(g) p( x ) = sin(ax − bt ) , where a, b and t are constants.


Question R2

Show that:

(a) if k is an arbitrary constant, y x( ) = x3

6+ k

x3 is a solution of the differential equation

dy

dx+ 3y

x= x2 .

(b) if a and b are arbitrary constants, y (x) = a e −x + be−2x is a solution of the differential equationd2 y

dx2+ 3

dy

dx+ 2 y = 0 .


Question R3

(a) Evaluate the limit lim∆x→0

(a + ∆x )2 − a2

∆x

. What does this limit represent?

(b) If y = f0(x) then the derivative dy

dx is sometimes written alternatively as ′f ( x ).

What does the limit lim∆x→0

′f ( x + ∆x ) − ′f ( x )∆x

represent?


y

t

Figure 13Sinusoidal oscillation.

y

t

Figure 23Triangular oscillation.

y

t

Figure 33Square-wave oscillation.

2 Waves and oscillations

2.1 OscillationsAn oscillation is said to occur when some quantity (such as position) repeatedly cycles about an equilibriumvalue.

Although oscillations are often taken to be sinusoidal ☞ (see Figure 1), this is not necessarily the case and other

forms (which arise, for example, in the context of electric circuits) are shown in Figure 2 and Figure 3.


y

displaced position at

time t

weight (mass m)

Figure 43An oscillating system; a masson a spring.

A typical physical example of an oscillating system is a mass on aspring after the mass has been displaced from its equilibrium positionand released (as in Figure 4). After its release, the displacement of sucha mass exhibits an approximately sinusoidal oscillation similar to that inFigure 1. Hooke’s law is often used to provide a mathematical model ofa spring, and using it we can derive an equation which describes themotion of the mass. So-called ‘ideal springs’, which satisfy Hooke’slaw, have the property that the extension of the spring y, is proportionalto the restoring force Fy that causes the mass to return towards itsequilibrium position. Thus, for such a spring

Fy = − k0y (1)

where k is a (positive) constant.

If m is the mass, then Newton’s second law allows us to replace Fy

by m d2 y t( )

dt2, which shows that y(t) satisfies

md2 y t( )

dt2= −ky t( ) (2)


If we introduce the positive constant ω defined by ω = k / m we may rewrite Equation 2

md2 y t( )

dt2= −ky t( ) (Eqn 2)

in the form

d2 y t( )dt2

= −ω 2 y t( ) (3)

Equation 3 is a differential equation. It can be solved for y(t), and the general solution is

y(t) = A1sin1(ω1t) + B1cos1(ω1t) (4)

where A and B are arbitrary constants. This general solution can also be written in the alternative(and often more useful form)

y(t) = C1sin1(ω0t + φ) (5)

where C and φ are arbitrary constants, and we still have ω = k / m .


Equation 5

y(t) = C1sin1(ω0t + φ) (Eqn 5)

is an especially valuable form of the solution because it makes it easy to identify the physical significance ofeach of the terms.

C is the amplitude and represents the maximum value of the displacement from the equilibrium position;

(ω1t + φ) is called the phase and determines the stage that the oscillation has reached in its cycle at time t;

ω = k / m is called the angular frequency and is 2π times the number of oscillations completed per second;

φ is called the phase constant (or the initial phase) since it determines the value of the phase at t = 0, and hencethe displacement from equilibrium at that time (y(0) = Csinφ).

The sort of oscillatory motion described by Equation 5, and characterized by the parameters C, ω and φ, isknown as simple harmonic motion.


✦

Show that y(t) = C1sin1(ω0t +φ) is a solution of the differential equation d2 y t( )

dt2= −ω 2 y t( )

Other examples of physical systems which display oscillatory behaviour include the simple pendulum and somea.c. circuits, but the basic idea remains the same; some physical quantity, for example position, voltage orcurrent, cycles repeatedly about an equilibrium value.


2.2 WavesPerhaps the first thing that comes to mind when you see the word ‘wave’ is the wave that you see at the beach.The undulations of the surface of the sea occur in a regular fashion, and appear to move towards the shore1—1a phenomenon that surfers are able to exploit to the full. We say that the surface ‘appears to move’ because theactual motion of the water is a good deal more complicated than you would at first think. Clearly somethingmoves towards the shore, but it is certainly not a coherent body of water, for otherwise the entire ocean wouldend up on the beach. As further evidence of this you need only note that a swimmer tends to bob up and down asa wave passes, and this would certainly not happen if the velocity of the water at each point in the wave wasdirected towards the shore.


x

y

(a)

t1

t1t2

x

y

(b)

>

There is a well known Tom and Jerry cartoon that illustrates what isreally happening. Tom is seated at a grand piano and runs his thumbrapidly along the keys from left to right, while Jerry is inside the pianoand moves along with the wave of displaced hammers (see Figure 5).The essential point to appreciate is that the hammers are simplymoving up and down, while the wave moves from left to right. Thewave is a travelling disturbance in the position of each hammer, not atravelling set of hammers.

A water wave is a very similar phenomenon. The motion of the wateris predominantly vertical and oscillatory, while the motion of the waveis horizontal and progressive.

Figure 53Snapshots at two different times of the heights of all the hammersin a piano show how vertical movements can produce a wave that travelshorizontally. Notice particularly that the vertical displacement y at any positionx (measured from the left-hand end of the keyboard) is dependent on twofactors: the particular hammer and the time, i.e. the value of x and the value oft. The time t2 in (b) is later than t1 in (a).


Another place where we are used to seeing waves is on strings, ropes and cables. These waves take many forms.For instance, if you hold one end of a long string and give it a single flick, then a solitary wave (often referred toas a pulse) will travel along the string. On the other hand if you jerk the string up and down repeatedly in aregular way, you may produce a repetitive wave that travels along the string. Note that the disturbance does nothave to be repetitive in order to qualify as a wave. Both kinds of disturbance travel along the string, so both areexamples of travelling waves.


y for a fixed value of t

x/m

(a)

ymaxλ

t

(b)

T

y for a fixed value of x

ymax

Although travelling waves do not have to be repetitive, there can beno doubt that the best known and most intensively studied kind oftravelling wave is one which has a sinusoidal shape at eachsuccessive instant of time. Figure 6 attempts to illustrate such asinusoidal wave, such as one travelling along a string. As you cansee, the figure is in two parts. Figure 6a shows the displacement y atall points x at a fixed time t; in effect, this is a ‘snapshot’ of the wave.Figure 6b, on the other hand, shows the way in which thedisplacement y changes with time at a fixed position x.If you compare Figure 6b with Figure 1 you will see that at a fixedvalue of x the disturbance caused by the sinusoidal travelling wave isjust a sinusoidal oscillation. Indeed, one way of picturing asinusoidal travelling wave is as an array of simple harmonicoscillators, where each oscillator is slightly out of phase, i.e. has aslightly different initial phase compared with its neighbours.

Figure 63(a) The ‘shape’ of a sinusoidal travelling wave at a particularinstant of time1—11the wave profile. (b) The oscillation caused by a sinusoidal travelling wave at a particular point1—1the wave form.



x/m

(a)

ymaxλ

t

(b)

T


ymax

A ‘snapshot’ of a wave at a particular instant, such as Figure 6a, isgenerally referred to as a wave profile. A trace showing the changingdisturbance caused by a wave at a fixed point, such as Figure 6b iscalled a wave form. Both of these terms will be used freely in therest of this module. Remembering that both are needed tocharacterize a wave will help you to remember the distinctionbetween a wave and an oscillation. If you were to photograph atravelling wave on a string you would obtain a wave profile.To determine the wave form you would have to record thedisplacement of a particular segment of string at different times.

Figure 63(a) The ‘shape’ of a sinusoidal travelling wave at a particularinstant of time1—11the wave profile. (b) The oscillation caused by a sinusoidaltravelling wave at a particular point1—1the wave form.


The need for both a wave profile and a wave form in order to characterize a wave points to the essentialmathematical difference between a wave and an oscillation. In the case of an oscillating mass, the displacementfrom the equilibrium position y, is only a function of time; so it may be represented by y(t). On the other hand,for a wave on a string the displacement from the equilibrium position is a function of position and time; so itmay be represented by y(x, t). Thus, even when it only travels in one dimension (x), the wave disturbance y is afunction of two variables. We shall consider such functions in more detail in Subsection 2.3, but first we shallfinish this subsection by briefly introducing some of the parameters commonly used to describe waves.



x/m

(a)

ymaxλ

t

(b)

T


ymax

If we stay with our example of a wave on a string, then theamplitude A is the maximum displacement (in the y-direction) of thestring from its equilibrium position. The wavelength λ is thedistance between two successive peaks (or troughs) on the waveprofile, as shown in Figure 6a. ☞ One might determine thewavelength by photographing the string, and then measuring thedistance between the peaks. The period T is the time between twosuccessive peaks (or troughs) on the wave form as shown inFigure 6b, and this might be found by timing the interval betweenpeaks at a fixed position on the string. The frequency f is the numberof troughs (or peaks) which pass a fixed location in one second, andis related to the period by

f = 1T

(6)



The units of frequency will therefore be s−1, which are called hertz (Hz) in SI units. Since f is the rate at whichpeaks pass a fixed point, and since the distance between successive peaks is λ , it follows that thespeed of propagation v at which the waves move along the string is given by

v = f1λ (7)

The angular frequency ω is related to the period by the equation

ω = 2π f = 2πT

(8)

so ω = 2πf, and ω is in units of rad1s−1.


Similarly, the wavenumber σ and the angular wavenumber k are related to the wavelength λ by the equations

σ = 1λ

(9)

and k = 2πλ

(10) ☞

It is probably worthwhile to try to remember Equations 6 to 10,

f = 1T

(Eqn 6)

v = f1λ (Eqn 7)

ω = 2π f = 2πT

(Eqn 8)

since there are many useful relationships that can be derived from them. The basic definitions are summarized inTable 1 on the next page.


Table 13The parameters used to describe a wave.

Amplitude A The maximum displacement from equilibrium,see Figure 6a or 6b.

Wavelength λ The distance that separates adjacent equivalentpoints on the wave profile, see Figure 6a.

Period T The time that separates successive equivalentpoints on the wave form, see Figure 6b.

Frequency f = 1/T

Angular frequency ω = 2πf = 2π/T

Wavenumber σ = 1/λ

Angular wavenumber k = 2π/λ

Speed of propagation v = fλ = ω/k

✦ Use the definitions that have been presented to show that v = ω0/k.



x/m

(a)

ymaxλ

t

(b)

T


ymax

Waves which cause a disturbance perpendicular to their direction ofpropagation are known as transverse waves. Waves on a string,ripples on a pond and electromagnetic waves are all examples oftransverse waves, but not all waves are transverse.

Sound waves in air are caused by compression and rarefaction of theatmosphere as molecules move back and forth along the direction ofpropagation. Such waves are known as longitudinal waves.For a sound wave, we can still represent the displacement from theequilibrium position by y and draw graphs similar to Figure 6, but xand y are not physically perpendicular in that case.



2.3 Functions of two variablesAs we have seen, if y is the displacement from the equilibrium position for a one-dimensional wave, then y is afunction of two variables, the position x, and the time t. In order to emphasize that y is a function of both x andt we may write it as y(x, t). Here is an example of such a function:

y(x, t) = A1sin1(at − bx) (11) ☞

where A = 2.371m, a = 161s−1, and b = 3.231m−1. Note that the units of a and b are such that when we substitutevalues for t and x (in seconds and metres respectively) the argument of the sine function will be dimensionless,as it should be.

✦ Evaluate y(1.91m, 12.01s) from the definition of y(x, t) given in Equation 11. ☞


h

y

x

Figure 73The height h(x, y) abovethe (x, y) plane.

Functions of two variables occur quite often in physical situations, ☞here are some examples.

1 Suppose we want a mathematically precise description of a hill or amountain range. We could obtain such a description by first using atwo-dimensional coordinate system to specify points on a horizontalreference surface (e.g. sea-level), and then using a function of twovariables h(x, y), to represent the height of the Earth’s surface abovesea-level at a point with coordinates (x, y).

A function of this kind is difficult to illustrate on a sheet of paper but auseful picture can often be obtained by means of a three-dimensionalgraph of the sort shown in Figure 7.


φ

y

x

Figure 83The function φ(x, y) ofEquation 12.

2 If a fixed electric charge q (measured in coulombs) is located at theorigin of a Cartesian coordinate system the electrostatic potentialat any point in the (x, y) plane with coordinates (x, y) is given by

φ x, y( ) = q

4πε0 x2 + y2 (12)

where φ is a function of the two variables x and y, and ε0 is aconstant. This function is shown in Figure 8.


(b)

I

V

R

Figure 93The current I as a function of V and R.

(a)

I

V

R

3 The current I, the voltage V, and the resistance R in a circuit suchas that shown in Figure 9a are related by Ohm’s law

I = V

R(13)

In a situation in which V and R can be varied independently, I is afunction of these two variables and can be written as I(V , R).This function is shown in Figure 9b.


It is sometimes convenient to display the arguments of the same function in different ways.

For example, if φ x , y( ) = q

4πε0 x2 + y2, as in Equation 12,

then, although φ is a function of the two variables x and y, it can also be considered as a function of the singlevariable u = x2 + y2, and we can write

φ u( ) = q

4πε0 u(14)

where u = x2 + y2. In such a case we may avoid having to introduce the extra variable u by writing the function φin the form

φ x2 + y2( ) = q

4πε0 x2 + y2 (15) ☞


✦

If y x , t( ) = sin

2πλ

x − vt( )

write y as a function of a single variable.

You will see shortly that the above discussion is relevant to the study of waves.


2.4 The general waveIn this module we are interested in waves for which the displacement from the equilibrium position depends onthe time (t) and a single space variable (x). Such waves are called one-dimensional waves, and a typicalexample is provided by a wave on a string of the kind we considered earlier. In general, a one-dimensional waveconsists of any disturbance which moves along a line, but we are more often interested in waves which maintaintheir shape as they move with a constant speed. Our first example is a disturbance in the form of a parabola.

✦ Sketch the graphs of the following functions

(a) y(x) = (x − vt)2 when t = 01s and v = 11m1s−1

(b) y(x) = (x − vt)2 when t = 11s and v = 11m1s−1

(c) y(x) = (x − vt)2 when t = 21s and v = 11m1s−1

Does the ‘pulse’ move to the left or to the right as t increases? In which direction would the pulse move if

( x − vt )2 was replaced by ( x + vt )2?


y

x0 v T

v T

Figure 113A general wave moving adistance vT to the right.

The continuous curve in Figure 11 shows a snapshot of a solitary wavepulse at some initial time t = 0, when it has a profile given by y = f0(x).The dashed curve shows another snapshot of the same wave at a latertime t = T where T > 0. If the wave moves to the right at constant speedv, without changing its shape, then during the time T that elapsesbetween the two snapshots each part of the wave will travel a distancevT to the right. It follows that the disturbance y at any position x at timet = T will be equal to the disturbance that would have existed at x − vTat t = 0. Hence, at t = T the wave profile is given by y = f0(x − vT),where f is the same function that described the wave profile at t = 0.Of course, there was nothing special about the value of T that we chose,so we can say that at any positive time t (including t = 0) any wave thatmoves in the positive x-direction with unchanging shape and constantspeed v can be represented by

y = f0(x − vt)

Similarly a wave that moves to the left with unchanging shape can be represented by

y = f0(x + vt)

where y = f(x) describes the profile of the wave at t = 0.


Question T1

A function θ

0

(x) is defined by

θ x( ) =0 for x < 0

1 for x ≥ 0

(16)

and a function y (x, t) is defined by y (x , t ) = θ

00

(x − v t). Sketch graphs of y, as a function of x , forvt = 01m, 11m, 21m. Repeat the exercise for y(x, t) = θ1(x + vt).3❏

To a physicist, sinusoidal waves are of paramount interest for several reasons:

1 Many waves that occur in physics are sinusoidal, including monochromatic light (i.e. of a singlewavelength), sound generated by a fixed frequency oscillator and radio waves (when unmodulated byspeech, etc.).

2 Sinusoidal waves are comparatively simple to analyse.

3 In most cases, a complicated wave can be considered as the sum of sinusoidal waves of different frequency.

Mike Tinker


A sinusoidal wave, moving to the right with constant speed v, as t increases, may be written in the general form

y x , t( ) = A sin k x − vt( ) + φ[ ] (17) ☞

where the constants A, k and φ, respectively, represent the amplitude of the wave, its angular wavenumber andits phase constant.

Unless we are comparing two waves, the phase constant is of little interest and we may assume that it is zero; inwhich case Equation 17 may be written in any one of the following equivalent forms (using Equations 6 to 10,which are repeated for your convenience):


y( x , t ) = A sin k ( x − vt )[ ] (18)

y( x , t ) = A sin kx − ω t[ ] (19)

y( x , t ) = A sin 2π(σ x − f t )[ ] (20)

y( x , t ) = A sin 2π x

λ− f t

(21)

y( x , t ) = A sin 2π x

λ− t

T

(22)

f = 1T

4 (Eqn 6)

v = f λ4 (Eqn 7)

ω = 2π T4 (Eqn 8)

σ = 1λ

(Eqn 9)

k = 2πλ

4 (Eqn 10)

Each of these forms has its uses, but Equation 19 is probably the most practical, and you can use Equations 6to 10 to generate the others when you need them. Figure 12 (on the next page) illustrates the function given inEquation 19. Notice that each fixed value of t corresponds to a sinusoidal curve drawn on the surface. Such acurve represents the wave profile at that particular value of t (i.e. a plot of y against x at a fixed value of t).If we were to take a slightly larger fixed value of t then the wave profile would move to the right. A fixed valueof x also produces a sinusoidal curve on the surface, but this time it corresponds to the wave form.


A

−A

0

0

λ

2λ

3λ

y1(x, t)

x 0 t

TFigure 124A three-dimensional graphof the functiony( x , t ) = A sin k x − ω t[ ] .


2.5 Partial derivatives

Given a function of a single variable, f0(x), its rate of change at x = a is equal to the gradient of the graph y = f0(x)at x = a, ☞ and may be determined by evaluating the derivative dy/dx = f00′(x) at x = a. The rate of change of afunction of one variable is an important concept because it often arises in mathematical models of physicalsystems, notably in the context of differential equations.

Rates of change are also important for functions of two variables such as f0(x, t), but in that context the presenceof more than one independent variable introduces additional complications. Functions of two variables may havemany different rates of change at a given point, and care must be taken to distinguish between them. In order todo this we need to extend the notion of an ordinary derivative to that of a partial derivative. The main purpose ofthis subsection is to introduce you to these partial derivatives, but before doing so we need to review the rulesnormally employed in the differentiation of a function of one variable.


The derivative of a function of a single variable

The rules of differentiation for a function of one variable are the following:

The constant multiple rule:4d cf ( x )( )

dx= c

df ( x )dx

The sum rule:4d f ( x ) + g( x )( )

dx= df ( x )

dx+ dg( x )

dx

The product rule:4d f ( x )g( x )( )

dx= f ( x )

dg( x )dx

+ g( x )df ( x )

dx

The quotient rule:4 d

dx

f ( x )g( x )

= g( x ) df dx − f ( x ) dg dx

g( x )[ ]2

The chain rule:4 df (u( x ))dx

= df (u)du

du( x )dx


As an example of the chain rule, suppose that we are given a function f(u) and that we wish to differentiatey = f0(x2 + 3x) with respect to x.

Using the chain rule we can writedy

dx= df

du

du

dx

where u = x2 + 3x, and this gives

dy

dx= du

dx

df

du= 2 x + 3( ) df

du u= x2 +3x

(23)

where the final term on the right means that df

du must be evaluated for u = x2 + 3x.

The notation on the right of Equation 23 is rather cumbersome, and it is often preferable to use the notation

′f (u) in place of df

du, in which case Equation 23 becomes

dy

dx= du

dx

df

du= 2 x + 3( ) ′f ( x2 + 3x ) (24)


Notice that this notation also has the advantage that we can avoid introducing the unnecessary variable u.A more general statement of this form of the chain rule then becomes

d

dxf g( x )( ) = ′g ( x ) ′f g( x )( ) (25)

(In the example on the previous page, g(x) = x2 + 3x so that in this notation ′g ( x ) = 2 x + 3 .)

✦ Using the chain rule, write down the derivative with respect to x of each of the following functions:

(a) y = sin1(3x + 5),

(b) y = sin1(ax + b), where a and b are constants

(c) y = sin1(kx − ωt) where k, ω and t are constants,

(d) y = f0(3x + 5) where f0(u) is an arbitrary function


✦

(a) Given that f0(x) = (x2 + 5x + 2)3 write down an expression for f0′ (x).

(b) Given that φ(t) = cos1(at + b) where a and b are constants, write down an expression for φ0′(t).

✦

(a) Given that y = f(kx − ω0t) where k, ω and t are constants, write dy

dx in terms of f0′.

(b) Given that y = f(kx − ω0t) where k, ω and x are constants, write dy

dt in terms of f0′.


The derivatives of a function of two variables

The last question, in which first x and then t was a constant indicates an important feature of the behaviour offunctions such as F(x, t) in which both x and t are variables. Such functions may generally have several differentderivatives. We can find one derivative by treating t as a constant and differentiating the function with respect tox, and we can find an entirely different derivative by treating x as a constant and differentiating with respect to t.This technique of holding one variable constant and differentiating with respect to another is calledpartial differentiation and the resulting derivatives are called partial derivatives. Such derivatives aredistinguished from ordinary derivatives by using a special kind of ‘curly’ dee (∂0) when writing them. Thus, for agiven function F(x, t) we can introduce the two derivatives:

∂F

∂x at constant t

which may be abbreviated to ∂F

∂x

t

☞ or more briefly ∂F

∂x and

∂F

∂t at constant x

which may be abbreviated to ∂F

∂t

x

☞ or more briefly ∂F

∂t.


If we interpret the function y = F(x, t) as a wave then, at any given point (x, t), the partial derivative ∂F

∂x

represents the gradient of the relevant wave profile at the given value of x, while ∂F

∂t represents the gradient of

the relevant wave form at the given value of t. ☞ In other words, ∂F

∂x and

∂F

∂t represent the slopes of the three-

dimensional graph of y = F(x, t) in the direction of the x-axis and the t-axis respectively, at the given point (x, t).These gradients in the x- and t-directions are indicated in Figure 13 on the next page. (Clearly, they are just twoof an infinite number of gradients that might be found at any given point, since we can identify an infinitenumber of directions between the x- and t-directions.)


1

0.5

0

the gradient of this line is the value of

the gradient of this line is the value of

−1

−0.5

00

1

2

3

4

5

1

2

3

4

∂∂

at x = xA, t = tAyt

∂∂

at x = xA, t = tAyx

y

xA

x t

tA

A

Figure 134The graphicalinterpretation of the partialderivatives with respect tox and t of a functiony = F(x, t) at a given point.


✦

Given that F( x , t ) = a sin k x + b cos ω t( )3 find ∂F

∂x and

∂F

∂t and evaluate them at x = 0, t = 0.

Question T2

Given that y( x , t ) = A sin k x − ω t[ ] find ∂y

∂x and

∂y

∂t.

What is the physical significance of ∂y

∂x and

∂y

∂t if y(x, t) represents the wave of hammers moving inside the

piano in the Tom and Jerry cartoon?4❏


Notice particularly that ∂ is used rather than d in the partial derivative ∂f

∂x

t

, and the t subscript can be used to

emphasize that t is to be held constant. Similarly the x subscript in ∂f

∂t

x

can be used to emphasize that x is

held constant. Usually we shall write the partial derivatives as ∂f

∂x and

∂f

∂t, and drop the suffices. ☞

Finding partial derivatives

When dealing with partial derivatives we are by no means restricted to functions of the variables x and t.We might, for example, choose to use x and y as the independent variables, and write

z = f ( x , y)

No matter what symbols we use to represent the variables, the process of finding partial derivatives involvessimilar techniques to finding ordinary derivatives of functions of one variable, the only difference being that wehave to remember to treat one variable as a constant.

An example should help to clarify this.


Example 1

If f0(x, y) = x2 − y2, find ∂f

∂x

y

and ∂f

∂y

x

Solution ∂f

∂x

y

=∂ x2 − y2( )

∂x=

∂ x2( )∂x

−∂ y2( )

∂x=

d x2( )dx

= 2 x

and∂f

∂y

x

=∂ x2 − y2( )

∂y=

∂ x2( )∂y

−∂ y2( )

∂y= −

d y2( )dy

= −2 y4❏

There are two points to note in this example.

First, when we differentiate a function of y alone with respect to x we get zero, because y is treated as a constant.

In this particular case we have ∂ y2( )

∂x= 0 , and similarly

∂ x2( )∂y

= 0 .

Second, when differentiating a function of x alone with respect to x, the partial derivative and the ordinary

derivative are identical, and for this reason we can replace ∂ by d, so that, for example, ∂ x2( )

∂x =

d x2( )dx

= 2x.


Example 2

Given that f0(x, y) = e2x cos(3y), find ∂f

∂x and

∂f

∂y .

Solution ∂f

∂x=

d e2 x( )dx

cos 3y( ) = 2e2x1cos1(3y)

and111 ∂f

∂y= e2 x

d cos 3y( )( )dy

= −3e2x1sin1(3y)4❏


Question T3

Given that f0(x, y) = x2 − xy + 2y2, find ∂f

∂x and

∂f

∂y.4❏

Question T4

Given that f0(x, y) = x2 1sin1y + y21cos1x, find

∂f

∂x and

∂f

∂y.4❏


It should be clear from these examples and questions that most of the rules of ordinary differentiation aredirectly applicable to partial differentiation. However, the extension of the chain rule is a little morecomplicated. Before using it in the context of partial differentiation let us briefly return to its application in theordinary differentiation of a function of one variable.

✦

Given that f0(x) = sin1(3x2), find df ( x )

dx.

✦

Given that φ1(y) = sin1(c2y4) where c is constant, find dφ (y)

dy.

Now let us apply the same approach to a function of two variables. Naturally, we shall have to modify the chainrule somewhat to apply it in this case.


Example 3

Given that f0(x, y) = sin(x2 0y4), find ∂f

∂x and

∂f

∂y .

Solution

In this case, if we write f0(x, y) = sin1(u), then u = x2 y4 is a function of two variables and the chain rule takes theform ☞

∂f

∂x=

d sin(u)( )du

∂u

∂xUsing the chain rule1 244 344

= cos1(u) × 2xy4 =

2x y4

∂u ∂x123

cos(x2 y4

u{

)

and

∂f

∂y=

d sin(u)( )du

∂u

∂yUsing the chain rule1 244 344

= cos1(u) × 4x21y3 =

4x2 y3

∂u ∂y123

cos(x2 y4

u{

)4❏

In the above example, note that we use d when differentiating a function of one variable with respect to thatvariable, and we use ∂ when differentiating a function of several variables. Apart from that the chain rule isessentially unchanged.


Question T5

Given that f0(x, t) = cosh1(Ax2 t3), ☞ where A is constant, find ∂f

∂x and

∂f

∂t.4❏

Question T6

Given that f x , y( ) = logex

y

find x∂f

∂x+ y

∂f

∂y.4❏


Formal definitions

The ordinary derivative df/dx of a function of a single variable f(x) is formally defined by the following limit,provided that the limit exists:

df

dx= lim

∆x→0

f x + ∆x( ) − f x( )∆x

In a similar way, the partial derivatives of a function of two variables f(x, y) may be defined by ☞

∂f

∂x

y

= lim∆x→0

f x + ∆x , y( ) − f x , y( )∆x

∂f

∂y

x

= lim∆y→0

f x , y + ∆y( ) − f x , y( )∆y

These definitions are mainly of interest to mathematicians, but you should be able to recognize them since theycan easily arise in a physical context.


2.6 Higher partial derivativesIf we are told that f(x, y) = e2x cos(3y) then we may easily find the partial derivatives

∂f

∂x= 2e2 x cos(3y) and

∂f

∂y= −3e2 x sin 3y( )

Now these partial derivatives are themselves functions of the two variables x and y, and so can be differentiatedpartially with respect to x or y.

It is generally the case that a partial derivative of a function of two variables is also a function of these variables,and if we choose we can differentiate a second time. ☞ For example:

If f0(x, y) = e2x1cos1(3y)

then∂f

∂x= 2e2 x cos(3y)

and

∂f

∂y= −3e2 x sin 3y( )

so that ∂∂x

∂f

∂x

= ∂∂x

2e2 x cos(3y)( ) = 4e2x1cos(3y)

while∂∂y

∂f

∂y

= ∂∂y

−3e2 x sin(3y)( ) = −9e2x1cos(3y)


It is common practice to abbreviate these second partial derivatives as follows:

∂∂x

∂f

∂x

= ∂ 2 f

∂x2(26)

and∂∂y

∂f

∂y

= ∂ 2 f

∂y2(27)

These results are straightforward extensions of what happens with a function of one variable; here we simplydifferentiate with respect to x (or y) while keeping y (or x) constant. However, there are two extra possibilitieswhich cannot occur with a function of a single variable.


Firstly, having found (in our earlier example)

∂f

∂x=

∂f x, y( )∂x

y= constant

Writing it in full toremind you of whatit really means.

1 2444 3444

= 2e2 x cos(3y)

we could then differentiate the result with respect to y while keeping x constant. Doing this we obtain

∂∂y

∂f

∂x

= ∂∂y

2e2 x cos(3y)[ ] = −6e2 x sin(3y) (28)

Notice that first we keep y constant and differentiate with respect to x and then we keep x constant anddifferentiate with respect to y. (Notice also that we don’t use x and y subscripts to indicate which variable isbeing held constant since doing so would make the notation unnecessarily complicated.) We normally abbreviatethese second derivatives as follows:

∂∂y

∂f

∂x

= ∂ 2 f

∂y∂x(29)


Secondly however, we could differentiate in reverse order, so that in place of ∂2f/∂y∂x we obtain

∂ 2 f

∂x∂y= ∂

∂x−3e2 x sin 3y( )( ) = −6e2 x sin 3y( ) (30)

It is no coincidence that Equations 28 and 30 give the same result,

∂∂y

∂f

∂x

= ∂∂y

2e2 x cos(3y)[ ] = −6e2 x sin(3y) (Eqn 28)

and in fact, for all the functions that you are likely to meet, it is generally true that

∂ 2 f

∂x∂y= ∂ 2 f

∂y∂x(31)

The derivatives defined in Equations 29 and 31

∂∂y

∂f

∂x

= ∂ 2 f

∂y∂x(Eqn 29)

are called mixed partial derivatives.


Example 4

If V(x, y) is defined by V(x, y) = arctan 1(y/x) ☞ find ∂ 2V

∂x2 and

∂ 2V

∂y2.

Hence verify that V(x, y) satisfies ∂ 2V

∂x2+ ∂ 2V

∂y2= 0 (32)

(An equation (such as Equation 32) which involves partial derivatives is known as a partial differentialequation. Equation 32 is known as Laplace’s equation (in two dimensions) and is crucial to the study of manyareas of theoretical physics, including electrostatics and hydrodynamics.


Solution Using the standard1 derivative (derived elsewhere in FLAP)

d

duarctan u( ) = 1

1 + u24we obtain

∂V

∂x= ∂

∂xarctan

y

x

= 1

1 + y

x

2

∂∂x

y

x

4from the chain rule

33111110 = 1

1 + y

x

2 × − y

x2

= − y

x2 + y2(33)


so∂ 2V

∂x2= 2 xy

x2 + y2( )2

11 ∂V

∂y= 1

x× 1

1 + y

x

2 = x

x2 + y2(34)

so∂ 2V

∂y2= − 2 x y

x2 + y2( )2

It follows that∂ 2V

∂x2+ ∂ 2V

∂y2= 2 x y

x2 + y2( )2 − 2 x y

x2 + y2( )2 = 04❏


Question T7

Given that V(x, y) = arctan1y

x

, verify that

∂ 2V

∂y∂x= ∂ 2V

∂x∂y.4❏

Question T8

If V(r, θ) = (α 0rn + β0r0−n)1cos1(nθ) where α, β and n are constants, find ∂V

∂r,

∂ 2V

∂r2 and

∂ 2V

∂θ 2.

Hence show that

∂ 2V

∂r2+ 1

r

∂V

∂r+ 1

r2

∂ 2V

∂θ 2= 0 4❏ (35)


Question T9

The pressure P, temperature T and volume V of a certain gas are related by van der Waals’ equation ☞

P + a

V 2

V − b( ) = RT (36)

where a, b and R are all constant. Regarding P as a function of V and T, find the values of P, V and T (in terms ofthe constants a, b and R) for which the equations

∂P

∂V= 0 and

∂ 2P

∂V 2= 0

are satisfied simultaneously.4❏


2.7 The wave equation

A general wave

In Subsection 2.4 we showed that a general one-dimensional wave that travels with unchanging shape atconstant speed v has the form y = f0(x ± vt). We now set out to find the equation for which this is a solution. ☞For this restricted class of functions of two variables it is possible to simplify the process of finding the partialderivatives. If we put u = x ± vt, then we can find the partial derivatives of f with respect to x in terms of thederivative with respect to u. ☞

✦

Given that f0(x, t) = sin1(x + vt), find ∂f

∂x, then put u = x + vt and calculate

df

du.

Hence show that ∂f

∂x= df

du in this case.


It is no accident that ∂f

∂x= df

du in the previous exercise. If u = x ± vt we have

∂u

∂x= 1, and we can use the chain

rule to write

∂f

∂x= df

du

∂u

∂xThis is equal to 1

{

= df

du(37) ☞

✦

Given that f0(x, t) = sin(x + vt), find ∂f

∂t, then put u = x + vt and calculate

v

df

du.

Hence show that

∂f

∂t= v

df

du in this case.


Once again, this result is no accident since if u = x ±vt then

∂u

∂t= ±v , and the chain rule gives

∂f

∂t= df

du

∂u

∂tThis is equalto ±v

{

= ±vdf

du(38)

We can now use a similar process to find the second-order partial derivatives. Since df

du can also be regarded as

a function of x and t, we can differentiate once again with respect to x to obtain

∂ 2 f

∂x2 = ∂∂x

∂f

∂x

This is justthe definitionof the secondderivative

1 24 34

= ∂∂x

df

du

From Eqn 37

123

= d

du

df

du

This comes fromEqn 37 withdf du in place of f

1 24 34

= d2 f

du2 (39)


Similar operations ☞ can be performed with respect to t, so that

∂ 2 f

∂t2= ∂

∂t±v

df

du

= ±v

d2 f

du2

∂u

∂t= ±v( )2 ∂ 2 f

∂u2= v2 d2 f

du2(40)

We can now eliminate d2 f

du2 from Equation 39

∂ 2 f

∂x2 = ∂∂x

∂f

∂x

This is justthe definitionof the secondderivative

1 24 34

= ∂∂x

df

du

From Eqn 37

123

= d

du

df

du

This comes fromEqn 37 withdf du in place of f

1 24 34

= d2 f

du2 (Eqn 39)

and Equation 40 to obtain

∂ 2 f

∂x2− 1

v2

∂ 2 f

∂t2= 0 (41)


This is a second-order partial differential equation with the required solution, and is known as the(one-dimensional) wave equation. Many interesting physical quantities can be shown to satisfy this equation,and very often we would like to find its solution subject to various boundary conditions. From our previousdiscussion we know that any function of the form y( x , t ) = f ( x ± vt ) is a solution of Equation 41,

∂ 2 f

∂x2− 1

v2

∂ 2 f

∂t2= 0 (Eqn 41)

but there are two points to appreciate here. First, this may not be the most general solution; and second, weusually need to find some particular solution in order to solve a given problem.

So there are two important questions to consider: what is the most general solution of Equation 41, and how canone select a solution that solves a given problem from this vast array of possible solutions? We shall not be ablefully to answer these questions in this module, but we shall show that certain specific functions are indeedsolutions of Equation 41. ☞ We shall also discuss some general features of solutions to the wave equation.


Question T10

Show that if two functions f1 and f2 are solutions of the wave equation

∂ 2 f

∂x2− 1

v2

∂ 2 f

∂t2= 0

then any function of the form α 0f1 + β0f2, where α and β are arbitrary constants, is also a solution.4❏

Partial differential equations are very common throughout physics and there is a large literature devoted totechniques which in some circumstances lead to explicit solutions. ☞ No matter what method is used to find asolution, it can always be verified by substituting back into the differential equation. We shall be concerned withshowing that we really do have a solution to the wave equation, rather than with general techniques for findingsolutions.


✦ Verify that y(x, t) = A1sin1(ω0t − kx + φ) is a solution of a wave equation of the form

∂ 2 f

∂x2− 1

v2

∂ 2 f

∂t2= 0 (Eqn 41)

provided we identify f with y, and providedv2 = ω02/k2.


2.8 Deriving the wave equationWe have shown that the function y(x, t) = A1sin1(ω0t − kx + φ) is a solution of the wave equation (Equation 41),

∂ 2 f

∂x2− 1

v2

∂ 2 f

∂t2= 0 (Eqn 41)

but why should there be waves on the surface of a lake, on a string or sound waves in air? How do we know thatsuch physical systems are governed, at least approximately, by the wave equation?

As an example of how waves occur in physical systems, we now derive the wave equation for a stretched string.Other physical systems, such as sound waves in air, can be analysed in a similar way.


y

x

(a)

x

l

AB

0(b)

A

B

θ

0

FT

FT1sin1θ

FT

∆l

FT1sin(θ + ∆θ)

θ + ∆θ

x + ∆ x

∆l

x + ∆ x 1

We start by consideringa segment AB of thestring which is oflength ∆l, as shown inFigure 14. The string isassumed to be stretchedtightly along thehorizontal x-direction,and then set in motionin the y -directionperpendicular to itslength. (We ignore theeffects of gravity in thefollowing discussion.)

Figure 144(a) A vibrating string, and(b) an enlargement of thesegment AB.


The angle θ is the angle between the tangent to the curve and the x-direction, and it will be small if theoscillations in the string are small. ☞

The following quantities are needed in our derivation:

x = position of one end of the segment as measured from a fixed point along the line of the string when atrest;

t = time at which the segment is in the position shown;

y(x, t) = displacement from the rest position of the segment ΑΒ;FT = the magnitude of the force acting on the ends of the string due to tension in the string

(assumed to be constant);

µ = the mass per unit length of the string (the linear mass density);

θ = the angle associated with the point at the ‘lower’ end of the segment;

θ + ∆θ = the angle associated with the point at the ‘upper’ end of the segment.


y

x

(a)

x

l

AB

0(b)

A

B

θ

0

FT

FT1sin1θ

FT

∆l

FT1sin(θ + ∆θ)

θ + ∆θ

x + ∆ x

∆l

x + ∆ x 1

The mass of thesegment is µ ∆l whichis approximately µ1∆x ifθ is small (the size of θhas been exaggerated inthe figure). Resolvingthe forces acting on thesegment, as in Figure14b, shows that there isa net transverse forcewhich causes thesegment to accelerate inthe y-direction.

Figure 144(a) A vibrating string, and(b) an enlargement of thesegment AB.


In other words, we treat x as a constant and differentiate y twice partially with respect to t to obtain the

acceleration ∂ 2 y

∂t2 of the segment.

Newton’s second law enables us to relate the acceleration to the force as follows

µ ∆xmassof AB

{

∂ 2 y

∂t2

acceleration of AB in the y-direction

{

= FT sin θ + ∆θ( ) − FT sin θ( )y-component of the force acting on AB1 244444 344444

(42)

However for small angles we have sin(θ ) ≈ tan(θ ) so that Equation 42 can also be written

µ ∆x∂ 2 y

∂t2≈ FT tan θ + ∆θ( ) − FT tan θ( ) (43)


y

x

(a)

x

l

AB

0(b)

A

B

θ

0

FT

FT1sin1θ

FT

∆l

FT1sin(θ + ∆θ)

θ + ∆θ

x + ∆ x

∆l

x + ∆ x 1

Suppose now that wekeep t fixed, andconsider the functionf0(x) defined by

f0(x) = y(x, t)

for constant tthe graph of which isshown in Figure 14a.The gradient of thisgraph at the point A is

f0′ 0(x) = tan1θwhile the gradient ofthe graph at the point B

is ′f ( x + ∆x )

= tan1(θ + ∆θ) Figure 144(a) A vibrating string, and(b) an enlargement of thesegment AB.


This means that we can rewrite the right-hand side of Equation 43

µ ∆x∂ 2 y

∂t2≈ FT tan θ + ∆θ( ) − FT tan θ( ) (Eqn 43)

to obtain

µ ∆x∂ 2 y

∂t2≈ FT ′f ( x + ∆x ) − FT ′f ( x ) (44)

so thatµFT

∂ 2 y

∂t2≈ ′f ( x + ∆x ) − ′f ( x )

∆x

If we now take the limit as ∆x tends to zero, the approximation becomes increasingly accurate, and therefore

(see Question R3) ☞

µFT

∂ 2 y

∂t2= ′′f ( x ) (45)


However, we defined f0(x) to be equal to y(x, t) for constant t, so that ′′f ( x ) = ∂ 2 y

∂x2, and this means that

Equation 45

µFT

∂ 2 y

∂t2= ′′f ( x ) (Eqn 45)

becomes

µFT

∂ 2 y

∂t2= ∂ 2 y

∂x2(46)

Now if we identify µ0/FT with 1/v2 we see that Equation 46 is actually a form of the wave equation.This shows that small amplitude waves on a stretched string propagate with a speed v given by

v = FT

µ(47) ☞


Question T11

The y-component of an electric field Ey(x, t) and the z-component of a magnetic field Bz(x, t) are related by thepartial differential equations ☞

∂Bz

∂t= −

∂Ey

∂x(48)

µ0ε0∂Ey

∂t= − ∂Bz

∂x (49)

where µ0 and ε0 are constants. Show that Ey and Bz both satisfy wave equations, and in each case find the speedof the wave. (Hint: Differentiate one equation with respect to t and the other respect to x.)4❏

Study comment

It turns out that any pair of physical quantities, f(x, t) and g(x, t), which satisfy the pair of differential equations

∂f

∂t= α ∂g

∂x3and3

∂g

∂t= β ∂f

∂x

where α and β are constants, lead to wave equations for f and g.


Energy flow for a wave on a string

If a wave propagates along a string there is an instantaneous flow of energy past a fixed point on the string. ☞As a pulse moves past a particular point there is clearly energy present at that point (since the string is moving)but, once the wave has moved on, the energy returns to zero (since the string is locally at rest). We can see thismathematically by observing that the instantaneous rate of flow of energy past a point x on the string, i.e. thepower P, is equal to the product of the upward force on the point multiplied by the transverse component of thestring’s velocity

∂y

∂t.

(The component of velocity of the segment along the string is assumed to be negligibly small.)

✦ Would it be more sensible to write ‘speed’ rather than ‘the transverse component of velocity’?


From our earlier discussion we know that the component of force in the y-direction at the point A is

− FT sin θ ≈ − FT tan θ = − FT∂y

∂x(50)

and therefore the power is given by

P = − FT∂y

∂x

∂y

∂t (51)

If we consider a wave moving from left to right, then y(x, t) = f(x − v t) and, putting u = x − vt, we obtain(from Equations 37 and 38)

∂f

∂x= df

du

∂u

∂xThis is equal to 1

{

= df

du(Eqn 37)

∂f

∂t= df

du

∂u

∂tThis is equalto ±v

{

= ±vdf

du(Eqn 38)

∂y

∂t= df

du

∂u

∂t= −v

df

duand

∂y

∂x= df

du

∂u

∂x= df

du


Hence, from Equation 51,

P = − FT∂y

∂x

∂y

∂t (Eqn 51)

the power is given by

P = vFT

df

du

2

(52)

The sign of P shows that the energy flows from left to right for a wave that travels from left to right.

✦ Show that P is negative for a wave moving from right to left.


(a)

(b)

(c)

2.9 Standing wavesSo far we have visualized waves as fixed shapes which change position,and we have referred to such waves as travelling waves. However, not allwaves are travelling waves. Figure 15 shows three of the ways in which astretched string with fixed endpoints can be made to move. In each casethe displacement of the string from its equilibrium position is a functionof position and time, and will satisfy the wave equation. Yet in this casethe ‘waves’ are not going anywhere; they do not appear to travel to theleft or to the right. In fact, they are called standing waves. On a standingwave there are some positions, called nodes, at which the displacement isalways zero.

Figure 15 Standing waves for a string fixed at its ends. This is the wave profile,and the horizontal axis is distance.


(a)

(b)

(c)

As we will now demonstrate, a standing wave can be regarded as aparticular combination of progressive waves which are travelling inopposite directions.☞ Consider the sum of two progressive waves whichhave the same angular frequencies and wavenumbers but are moving inopposite directions

y(x, t) = C1sin1(ω0t − kx + φ1) + D1sin1(ω0t + kx + φ2) (53)

and suppose we are looking for standing waves in the stretched stringshown in Figure 15. Let the string be of length L. The parameters in theexpression for y(x, t) are not completely arbitrary since y(x, t) must satisfythe condition that neither end of the string can move.

If x = 0 corresponds to one end of the string, then this implies that

y(0, t) = 0 for all values of t (54)

and therefore

C1sin1(ω0t + φ1) + D1sin1(ω0t + φ2) = 0 for all values of t



Since this equation is true for all values of t the functions C1sin1(ω0t + φ1) and −D1sin1(ω0t + φ2) must be identical,and this implies that

C = − D4and4φ1 = φ2.

If we replace φ1 and φ2 by φ, and D by −C our solution (Equation 53) to the wave equation

y(x, t) = C1sin1(ω0t − kx + φ1) + D1sin1(ω0t + kx + φ2) (Eqn 53)

reduces to

y(x, t) = C1[sin1(ω0t − kx + φ) − sin1(ω0t + kx + φ)]

This expression can be converted into a more convenient form by using the trigonometric identity

sin1α − sin1β = 2 sinα − β

2

cosα + β

2

(55)

to give y(x, t) = −2C1cos1(ω0t + φ)1sin1(kx) (56)


The point (0, L) corresponds to the other end of the string, which is also fixed; so y(L, t) = 0 for all values of t.

So, cos1(ω0t + φ)1sin1(kL) = 0 (57)

Since this is also true for all values of t, we must have

sin1(kL) = 0 (58)

The sine function is zero only when its argument is an integer multiple of π, so it follows that

kL = nπ4where n = 0, ±1, ±2, … (59)

Equation 59 is of profound importance because it tells us that the value of k cannot be chosen arbitrarily; onlycertain values of k are possible if this end of the string is to remain fixed. Substituting the allowed values ofk = nπ/L in Equation 56 we find

y(x, t) = −2C1cos1(ω0t + φ)1sin1(kx) (Eqn 56)

all the allowed forms of sinusoidal standing waves that can exist on a string of length L with fixed end points

y(x, t) = A1cos1(ω0t + φ) sin1nπx

L

4where n = 1, 2, 3,… (60)


Notice that in writing Equation 60

y(x, t) = A1cos1(ω0t + φ) sin1nπx

L

4where n = 1, 2, 3,… (Eqn 60)

we have replaced −2C by A for convenience and we have dropped the non-positive values of n. The negativevalues of n do not lead to expressions for y(x, t) which are independent of the positive values since

sin − nπx

L

= − sin

nπx

L

☞

and all these waves can be reproduced by selecting an appropriate value for the phase constant φ.

Also note that the ‘wave’ corresponding to n = 0 in Equation 60 is the trivial case of the string at rest and istherefore of little interest.


(a)

(b)

(c)

✦

(a) What values of n in the function

y(x, t) = A1cos1(ω0t + φ)1sin1nπx

L

correspond to the three standing waves of Figure 15?

(b) What expressions for y(x, t) correspond to these standing waves inFigure 15?



The values for φ in the above function are indeterminate, since the wave profile tells us nothing about time.Although each expression for y(x, t) given by a particular value of n is a solution of the wave equation

∂ 2 y

∂x2− 1

v2

∂ 2 y

∂t2 = 0

these waves do not appear to propagate with speed v since they are in fact standing waves. As we have justdemonstrated, each of these standing waves is actually the sum of two progressive waves travelling in oppositedirections with speed v. The standing waves cannot be expressed as a single function of f0(x ± vt), since they arealways combinations of two such functions with opposite signs within the bracket.

✦ Are the frequencies of the standing waves that we have found for the stretched string arbitrary?


Study comment Make sure you study the answer to the above question. The answer is important in its own right, but italso leads on to the important finding that:

The standing waves can be written in the form

y x , t( ) = A cos

nπvt

L+ φ

sin

nπx

L

where n = 1, 2, 3… 3(Eqn 61)


Question T12

If the standing waves on a string are described by the function y(x, t) given in Equation 61,

y x , t( ) = A cos

nπvt

L+ φ

sin

nπx

L

where n = 1, 2, 3… 3(Eqn 61)

show that the rate of energy flow past a fixed point x on the string is given by

P = FTv

nπA

2 L

2

sin2nπvt

L+ 2φ

sin

2nπx

L

Also find the values of P at x = 0 and x = L and explain your results.

Hints: The previous section gives an expression for P in terms of partial derivatives, and you may need to usethe identity sin1(2θ) = 21sin1(θ)1cos1(θ).3❏


3 The Schrödinger equationStudy comment The Schrödinger equation is the central equation of quantum mechanics. The purpose of this section is tointroduce the Schrödinger equation as another example of a partial differential equation with wave-like solutions, and toexamine some of its mathematical properties1—1it is not designed to teach you quantum mechanics. A very brief survey ofquantum mechanics is contained in Subsection 3.1, to enable you see something of the physical context in which theSchrödinger equation arises, but for a comprehensive introduction to this very important part of physics you should consultthe appropriate blocks in the physics strand of FLAP.


3.1 The time-dependent Schrödinger equation

In the early years of the 20th century it became increasingly apparent that classical physics1—1the mechanics ofNewton, the electromagnetism of Maxwell, the thermodynamics of Clausius, and so on1—1was incapable ofaccounting for many of the phenomena involving atoms and nuclei that were being discovered at that time.Because of this the period from 1900 to 1925 saw the development of a number of poorly formulated ‘theories’that tried to explain the new phenomena in terms of an ad hoc mix of classical ideas and radical new proposals.The need for a new theoretical framework to replace classical physics, one that would bring old and new ideastogether in a more coherent way, was widely perceived, but the experimental data were so fragmentary, and therequired shift in thinking so great that it was not until 1925/26 that the way forward became clear. It was duringthese years that a relatively small number of European physicists including Werner Heisenberg (1901–1976),Erwin Schrödinger (1887–1961) and Max Born (1882–1970) began to develop the new quantum physics, thatwas destined to displace classical physics from its position as the most comprehensive and coherent descriptionof physical reality. Quantum physics is now a wide-ranging discipline with many subdivisions that influencealmost every other branch of physics. However, the first part of quantum physics to be developed systematicallywas quantum mechanics, the quantum physical counterpart to classical mechanics, and it is this part ofquantum physics that we shall consider in this section. ☞


Quantum mechanics addresses questions such as ‘how do the electrons in atoms behave?’ and hence, ‘how muchenergy will the electrons in a given atom absorb or emit under given circumstances?’ These are questions that a19th century scientist would have tried to answer using classical mechanics and classical electromagnetism, butby 1925 it was clear that when classical physics was applied to individual atoms it was often inadequate,inconsistent or just plain wrong. The ability of quantum mechanics to account for atomic phenomena caused arevolution in physics, not merely because quantum mechanics agreed with experiment where classical physicsfailed, but also because the answers it provided were often of a profoundly different kind and were arrived at in aprofoundly different way from those of classical physics. To clarify this difference let us examine the ways inwhich classical and quantum physicists might have tried to formulate a simple one-dimensional model of anatom, in which a negatively charged electron is bound to a positively charged nucleus by an electrical force.


The classical physicist would have started with a mathematical formula for the electrical force Fx on theelectron, or equivalently, a formula relating the electrical potential energy U(x) of the electron to its position x.☞ The classical physicist would then use that force (or the related potential energy) to write down Newton’ssecond law of motion as a differential equation. In a simple one-dimensional case this might have the generalform

md2 x

dt2= Fx

or, in terms of the potential energy,

md2 x

dt2= −dU

dx

where m is the mass of the electron and x its position at time t.


Given an explicit expression for Fx or U(x), the classical physicist could then solve the relevant differentialequation, subject to various boundary conditions, to find the position of the electron as a function of time x(t).Equipped with a solution of this kind it would be relatively easy to determine all the other attributes of theelectron such as its velocity (vx = dx/dt), its momentum (px = mvx) and its energy ((mvx

2/2) + U(x)), at any time.Of course, having done all this, the classical physicist would probably find that the final results bore littlerelation to the measured properties of electrons even if they were confined to one dimension.Classical mechanics works well for macroscopic (laboratory scale) objects but often fails miserably in themicroscopic (atomic scale) domain.

For the quantum physicist the approach is very different. Force is a concept that sits rather uneasily in quantummechanics, so the quantum physicist would certainly start from the relevant potential energy function. Takinginto account the fact that the potential might change with time as well as position, the quantum physicist wouldthen write down the time-dependent Schrödinger equation. This is as fundamental to quantum mechanics asNewton’s second law is to classical mechanics. In a simple one-dimensional case it would take the form

− ˙2

2m

∂ 2Ψ x , t( )∂x2

+ U x , t( )Ψ x , t( ) = i˙ ∂Ψ x , t( )∂t

(62)

where the symbols have the following meaning.


h =

h

2π = 1.055 × 10−34

1J1s is a physical constant (pronounced ‘h-bar’ and referred to as “Planck’s constant

divided by 2π”).

m represents the mass of the electron, as before.

U(x, t) is the potential energy function particular to the physical system being considered. If we wanted toconsider a different kind of atom then the function U(x, t) would have to be changed to represent the newproblem. (This would have been equally true in classical physics.)

Ψ(x, t), the subject of the equation, is a function of two variables (in this one-dimensional case) and is called

the wave function. ☞ It is the role of the wave function that really distinguishes quantum physics from

classical physics, so we shall have a lot more to say about it shortly. (In what follows we shall alwaysassume that the wave functions that satisfy the Schrödinger equation for a given potential energy functionU(x, t), satisfy an additional technical requirement known as normalization.)

i is the imaginary constant defined by i2 = −1 and often referred to as ‘the square root of −1’. Its presence isanother of the features that distinguishes quantum mechanics from classical mechanics.

x and t are position and time coordinates, as before, but it is no longer true to say that x is the position of theelectron at time t.


Aside Notice that Equation 62 differs from our one-dimensional wave equation (Eqn 41) in two fundamental ways. Firstly, itinvolves the first not the second time derivative of the function. Secondly, it is a complex equation and its solutions are ingeneral complex functions. In a technical sense then, the Schrödinger equation is not strictly a wave equation, but it doeshave wave-like solutions, and for this reason is often referred to as a wave equation.

− ˙2

2m

∂ 2Ψ x , t( )∂x2

+ U x , t( )Ψ x , t( ) = i˙ ∂Ψ x , t( )∂t

(Eqn 62)

∂ 2 f

∂x2− 1

v2

∂ 2 f

∂t2= 0 (Eqn 41)


Knowing the explicit form of U(x, t) the quantum physicist can solve the time-dependent Schrödinger equation,subject to appropriate boundary conditions, to find a wave function Ψ(x, t). ☞ However, unlike the solution toNewton’s second law x(t), a quantum mechanical wave function Ψ(x, t) will involve i (i.e. it will involvecomplex numbers) and will not be a physically measurable quantity. How then is the quantum physicist todetermine the precise location of the particle at any time? The short answer, according to quantum mechanics, isthat the quantum physicist cannot do so, because the electron may not have a precisely defined location at everypossible time. This odd-sounding limitation is sometimes described by saying that ‘quantum mechanics is atheory of behaviour, not a theory of attributes’. This means that rather than providing information about somepresumed attribute of the electron, such as its position at all possible times, what quantum mechanics provides isinformation about the observable behaviour of the electron, such as the likelihood that it will be found in acertain region of space at a particular time. According to quantum mechanics, this behavioural information iscontained in the wave function.

For instance, the probability ☞ that at some particular time t = T, the electron will be found in some smallregion of fixed length ∆X centred on a given point x = X at time t = T, is |1Ψ0(X, T) 1|021∆X, where |1Ψ0(X, T)1|represents the modulus of the complex quantity Ψ0(X, T). Of course, there is nothing special about the particularvalues X, T and ∆X that we have chosen, so we can say that in general the probability of finding the electron in asmall range ∆x, centred on x at time t is |1Ψ0(x, t)1|21∆x.


Unlike the classical trajectory x(t), this information about the probability of finding the electron within a certainrange of x values at a certain time is of little help in determining other important attributes such as energy andmomentum. However, in keeping with its nature as a theory of behaviour rather than attributes, quantummechanics does provide the answer to a behavioural question such as; ‘what are the possible outcomes of anexperiment to measure the energy of the electron?’ Perhaps surprisingly, it is a common feature of quantummechanics that confined particles, such as electrons contained within atoms, are only allowed to have certaindiscrete energy values. These allowed energy values may be determined by examining all possible solutions to agiven time-dependent Schrödinger equation. A particular wave function satisfying that equation will thendetermine the relative likelihood of each of those allowed energy values turning up as the outcome of anexperiment to measure the energy. Similar comments may be made about other observable quantities such asmomentum or position.

Thus a particular wave function may be said to describe a particular quantum state of the system being studied.To a classical physicist a ‘state’ of the system would be defined by the values of attributes such as energy,momentum, position etc., but to a quantum physicist the state is synonymous with a given wave function thatdetermines a particular set of probabilities of particular experimental outcomes. Even when the system is in aparticular state, attributes such as energy and momentum need not have definite values, only possible valueswith definite probabilities of being observed under specified conditions.


Of course, you are not meant to understand quantum theory from this brief discussion. Rather, the points thatyou should take away from this subsection are the following:

o Quantum mechanics involves solving a partial differential equation called the time-dependent Schrödingerequation.

o The precise form of this equation differs from problem to problem, since each problem is characterized by apotential energy function U that appears in the time-dependent Schrödinger equation.

o The solution to the time-dependent Schrödinger equation (subject to specified boundary conditions) is calledthe wave function. In a one-dimensional problem it will be a function of two variables and may be writtenΨ(x, t). This is generally a complex quantity that is not directly measurable in an experiment.

o The answers to behavioural questions about a system, in so far as such answers exist at all, are obtained bycarrying out various mathematical operations that involve the wave function. The wave function maytherefore be regarded as a description of the state of the system under the given boundary conditions.


3.2 The time-independent Schrödinger equationMany quantum mechanical problems involve potential energy functions that do not depend on time. Under thesecircumstances the one-dimensional time-dependent Schrödinger equation takes the form

− ˙2

2m

∂ 2Ψ x, t( )∂x2 + U x( )

Not afunction of t

{

Ψ x, t( ) = i˙ ∂Ψ x, t( )∂t

(63)

It can be shown that because the potential is independent of t in this case, Equation 63 may have separable

solutions, that is solutions Ψ(x, t) that can be written as a product of a function of x and a function of t. ☞When dealing with a separable wave function Ψ(x, t), the following notation is used to indicate its separability

Ψ(x, t) = ψ(x)φ(t) (64)

The function ψ(x) on the right-hand side of Equation 64 is then called the spatial part of the wave function(though this is often contracted to spatial wave function), and the function φ(t) is referred to as thetemporal part of the wave function, (or just the temporal wave function). Note the distinction between theupper case Greek psi (Ψ) used to denote the full time-dependent wave function and the lower case psi (ψ) usedfor its spatial part.


Using Equation 64

Ψ(x, t) = ψ(x)φ(t) (Eqn 64)

we may rewrite Equation 63

− ˙2

2m

∂ 2Ψ x, t( )∂x2 + U x( )

Not afunction of t

{

Ψ x, t( ) = i˙ ∂Ψ x, t( )∂t

(Eqn 63)

in the following form:

− ˙2

2m

∂ 2

∂x2ψ x( )φ t( )[ ] + U x( ) ψ x( )φ t( )[ ] = i˙

∂ ψ ( x )φ (t )( )∂t

Since t (and hence φ(t)) behaves as a constant when we differentiate partially with respect to x, and since x(and hence ψ(x)) behaves as a constant when we differentiate partially with respect to t, we can rewrite this asfollows

φ t( ) − ˙2

2m

∂ 2ψ x( )∂x2

+ U x( )ψ x( )

= i˙ψ x( ) ∂φ∂t


However ψ(x) is a function of the single variable x, so ∂ 2ψ ∂x2 = d2ψ d x2 and φ(t) is a function of thesingle variable t, so ∂φ ∂t = dφ dt .

Hence φ t( ) − ˙2

2m

d2ψ x( )dx2

+ U x( )ψ x( )

= i˙ψ x( ) dφdt

(65)

Notice that by restricting our attention to the separable solutions that can arise when the potential energy isindependent of time we have been able to replace the partial derivatives of Equation 63

− ˙2

2m

∂ 2Ψ x, t( )∂x2 + U x( )

Not afunction of t

{

Ψ x, t( ) = i˙ ∂Ψ x, t( )∂t

(Eqn 63)

by ordinary derivatives.

Dividing both sides of Equation 65 by ψ0(x)φ0(t) we get the following

φ t( )ψ x( )φ t( )

− ˙2

2m

d2ψ x( )dx2

+ U x( )ψ x( )

= i˙ψ x( )ψ x( )φ t( )

dφ t( )dt


i.e.

1ψ x( )

− ˙2

2m

d2ψ x( )dx2 + U x( )ψ x( )

This is a function of x only1 2444444 3444444

= i˙φ t( )

dφ t( )dt

This is a function of t only

1 24 34

(66)

This is a clever step because, as indicated, it produces an ordinary differential equation in which the onlyvariable appearing on the left-hand side is x, while the only variable on the right-hand side is t. Think about thatfor a moment. How is it possible for a function of x (on the left of Equation 66) to be equal to a function of t(on the right of Equation 66) when x and t are independent variables? There is only one way that this canhappen; and that is for each side of Equation 66 to be constant. A constant introduced in this way is called aseparation constant; in this case we shall denote it by C. We can then write Equation 66 as two ordinarydifferential equations


1ψ x( )

− ˙2

2m

d2ψ x( )dx2

+ U x( )ψ x( )

= C (67)

andi˙

φ t( )dφ t( )

dt= C (68)

It is easy to verify that the second equation, Equation 68, has the solution

φ(t) = Κ exp(−i0C0t/˙) (69) ☞

for some constant K.

Moreover, it can be shown that the constant C is of considerable physical significance; it represents the totalenergy E, of the particle.


So a solution to Equation 63

− ˙2

2m

∂ 2Ψ x, t( )∂x2 + U x( )

Not afunction of t

{

Ψ x, t( ) = i˙ ∂Ψ x, t( )∂t

(Eqn 63)

that describes a state in which the particle has energy E is given by

Ψ(x, t) = ψ(x)φ(t)

where φ(t) = K1exp(−iEt/˙) (70)

and − ˙2

2m

d2ψ x( )dx2

+ U x( )ψ x( ) = Eψ x( ) (71)


where Equations 70 and 71 have been obtained from Equations 69 and 67

φ(t) = K1exp(−iEt/˙) (Eqn 70)

− ˙2

2m

d2ψ x( )dx2

+ U x( )ψ x( ) = Eψ x( ) (Eqn 71)

φ(t) = Κ exp(−i0C0t/˙) (Eqn 69)

1ψ x( )

− ˙2

2m

d2ψ x( )dx2

+ U x( )ψ x( )

= C (Eqn 67)

by replacing the separation constant C by the energy E.

Clearly, if we want to find the explicit form of the solution Ψ(x, t), we must now solve Equation 71 to find ψ(x),just as we solved Equation 68

i˙φ t( )

dφ t( )dt

= C (Eqn 68)

to find φ(t).


Equation 71

− ˙2

2m

d2ψ x( )dx2

+ U x( )ψ x( ) = Eψ x( ) (Eqn 71)

appears elsewhere in FLAP and is known as the time-independent Schrödinger equation. Finding andinterpreting solutions to this equation for various explicit forms of U(x) is often a major preoccupation inintroductory courses on quantum mechanics. We shall not pursue that subject here except to note the followingpoints:

o For a given potential energy function U(x) it is often possible to find a number of different solutions to theone-dimensional time-independent Schrödinger equation, each of which corresponds to a different value ofthe energy E. These different solutions may be conveniently labelled ψ1(x), ψ2(x), ψ3(x), ψ4(x), … and thecorresponding values of E may be labelled E1, E2, E3, E4, …

o For many choices of U(x) it will be the case that E1, E2, E3, E4, … are separate and distinct numbers, i.e. thepossible values of the total energy are discrete. In this situation we say that the energy of the system isquantized. However, it should be noted that quantum mechanics does not automatically demand thediscrete quantization of energy in all systems1—1it is perfectly possible for the time-independentSchrödinger equation to have solutions that correspond to a continuous range of values for E, if U(x) has theappropriate form.


o The possible existence of many different solutions to Equation 71,

− ˙2

2m

d2ψ x( )dx2

+ U x( )ψ x( ) = Eψ x( ) (Eqn 71)

with different values of E for a given choice of U(x), implies that there may be many different separablewave functions Ψ(x, t) that satisfy Equation 63.

− ˙2

2m

∂ 2Ψ x, t( )∂x2 + U x( )

Not afunction of t

{

Ψ x, t( ) = i˙ ∂Ψ x, t( )∂t

(Eqn 63)

If so these may be conveniently labelled Ψ1(x, t), Ψ2(x, t), Ψ3(x, t), Ψ4(x, t), …, where

Ψ1(x, t) = exp(− iE1t ˙)ψ1(x), 3Ψ2(x, t) = exp(− iE2t ˙) ψ2(x),

Ψ3(x, t) = exp(− iE3t ˙)ψ3(x), …


o Note that each of these wave functions describes a state in which the system has a fixed amount of energy.

o If Ψi(x, t) and Ψj(x, t) are both solutions to the time-dependent Schrödinger equation for a given potentialenergy function, then

Ψ(x, t) = αΨ i(x, t) + βΨj0(x, t)

where α and β are complex constants, will also be a solution. In the case of Equation 63,

− ˙2

2m

∂ 2Ψ x, t( )∂x2 + U x( )

Not afunction of t

{

Ψ x, t( ) = i˙ ∂Ψ x, t( )∂t

(Eqn 63)

this means that there may be solutions of the form

Ψ(x, t) = α exp(− iEit ˙)ψi(x) + β exp(− iEjt ˙)ψj(x)


o Assuming that i and j represent different numbers, so that Ei ≠ Ej, it is clear that this wave function

Ψ(x, t) = α exp(− iEit ˙)ψi(x) + β exp(− iEjt ˙)ψj(x)

describes a state in which the system has no definite energy. However, according to quantum mechanics, ifthe energy of the particle is measured then the result will be either Ei or Ej and the probability of obtainingeither result in a sequence of identical measurements is predictable. In physical terms this is associated withthe system being in a superposition state. If the quantities α and β are themselves time-dependent, this canoften be associated with the system making a transition between the two states.

Aside Many authors refer to ‘the Schrödinger equation’, without specifying whether they mean the time-dependent equationor the time-independent equation. Generally it is clear from the context which equation is relevant and the explicit labels‘time-independent Schrödinger equation’ and ‘time-dependent Schrödinger equation’ are only used when there is likely to beconfusion.


4 Closing items

4.1 Module summary1 An oscillation occurs when a physical quantity repeatedly cycles about an equilibrium value. The physical

quantity is therefore a periodic function of time. An example is the simple harmonic oscillator which obeysthe ordinary differential equation

d2 y t( )dt2

= −ω 2 y t( ) (Eqn 3)

which, for a given angular frequency ω, has the solution

y(t) = C1sin1(ω0t + φ) (Eqn 5)

where C and φ are constants.

2 A wave is a change in a physical quantity which propagates in space. The physical quantity is therefore afunction of both position and time.


3 A function of two variables x and y, may be written as f0(x, y). The partial derivative of f with respect to x iswritten as

∂f

∂x

y=constant

or ∂f

∂x

y

or ∂f

∂x

which means that f is differentiated with respect to x while holding y constant. Similarly, the partial

derivative of f with respect to y is written as∂f

∂y

x =constant

or ∂f

∂y

x

or ∂f

∂y

which means that f is differentiated with respect to y while holding x constant.

4 To obtain the partial derivatives of a function of two variables, use the rules for differentiating a function ofone variable and treat the other variable as a constant.


5 A wave which travels in the positive x-direction while maintaining its shape corresponds to a function of theform y(x − vt), while a wave which travels in the negative x-direction while maintaining its shapecorresponds to a function of the form y(x + vt), where v is the speed of propagation in either case.Such waves obey the (one-dimensional) wave equation

∂ 2 y

∂x2− 1

v2

∂ 2 y

∂t2= 0

A particularly important solution is the sinusoidal travelling wave

y(x, t) = A1sin1[k(x − vt) + φ] (Eqn 17)

where A, k (= 2π/λ) and φ are constants.

6 A wave equation for a vibrating string can be obtained by considering the forces acting on a short segmentof the string. A partial differential equation for the displacement of the string from its equilibrium positionis obtained in the limit as the length of this segment tends to zero.


7 Any pair of physical quantities, f0(x, t) and g(x, t), which satisfy the pair of differential equations

∂f

∂t= α ∂g

∂x33and33

∂g

∂t= β ∂f

∂xwhere α and β are constants, lead to wave equations for f and g.

8 A standing wave is a particular combination of two travelling (or progressive) waves moving in oppositedirections.

9 The time-dependent Schrödinger equation, for a system consisting of a single particle of mass m moving inone dimension with potential energy U(x, t), is the partial differential equation

− ˙2

2m

∂ 2Ψ x , t( )∂x2

+ U x , t( )Ψ x , t( ) = i˙ ∂Ψ x , t( )∂t

(Eqn 62)

This is the central equation of quantum mechanics. Its solutions Ψ(x, t), subject to appropriate boundaryconditions, are called wave functions and describe the possible states of the system. A wave function isgenerally a complex quantity and is not directly measurable, but it yields (probabilistic) information aboutthe behaviour of the system.


10 If U is a function of x only so that U(x, t) = U(x), and E is the energy of the particle, the time-dependentSchrödinger equation may have separable solutions of the form Ψ(x, t) = ψ0(x) exp(−i0Et0/0˙), where ψ0(x) iscalled the spatial part of the wave function and satisfies the time-independent Schrödinger equation

− ˙2

2m

d2ψ x( )dx2

+ U x( )ψ x( ) = Eψ x( ) (Eqn 71)

For appropriate choices of U(x), this equation may have many different solutions which may correspond todiscrete values of E. When this occurs we say the energy of the system is quantized. This is the crucialfeature of quantum mechanics which is lacking in classical mechanics.


4.2 AchievementsHaving completed this module, you should be able to:

A1 Define the terms that are emboldened and flagged in the margins of this module.

A2 Understand the difference between an oscillation and a wave and give examples of each.

A3 Find the partial derivatives of a function of two variables.

A4 Recognize and derive the wave equation for a disturbance which propagates without changing its shape.

A5 Recognize, use and interpret typical solutions to the wave equation, such as

y x , t( ) = A sin k x − vt( ) + φ( )A6 Derive the differential equation for a wave on a string and relate the speed of propagation to physical

parameters for the string.

A7 Describe the role of partial derivatives in the time-dependent Schrödinger equation, explain its relationshipto the time-independent Schrödinger equation, and carry out simple operations involving these equationsand their solutions. (Note: you are not expected to understand quantum mechanics on the basis of the briefintroduction provided in this module.)


Study comment You may now wish to take the Exit test for this module which tests these Achievements.If you prefer to study the module further before taking this test then return to the Module contents to review some of thetopics.


4.3 Exit test

Study comment Having completed this module, you should be able to answer the following questions, each of which testsone or more of the achievements.

Question E1

(A3)3Given that f x , y( ) = y2

x find

∂f

∂x and

∂f

∂y.

Express ∂f

∂y in terms of a limit, then evaluate this limit to verify your previous answer.


Question E2

(A1 and A3)3Use the standard derivatives of functions of one variable to find ∂f

∂x and

∂f

∂y for

(a) f(x, y) = xy2 + 4xy − 2x,3(b) g x , y( ) = e x2 cos y3( )

(c) h x , y( ) = x

x3 + y3,34(d) w x , y( ) = loge

x2 + y2

x3.

Question E3

(A12and 11A3)3If2V x , y( ) = x

x2 + y22show that2 ∂ 2V

∂x2+ ∂ 2V

∂y2= 0 .


Question E4

(A2)3Explain the difference between an oscillator and a wave, giving an example of each.

Question E5

(A3 and A4)3Derive the second-order partial differential equation obeyed by any one-dimensional wave whichmoves without changing its shape.


Question E6

(A3 and A5)3A fluid which is subject to a small disturbance in the x-direction obeys the equations∂V

∂t= − 1

ρ∂P

∂x(72)

and

∂P

∂t= − K

∂V∂x

(73)

where V(x, t) is the velocity of fluid in the x-direction due to disturbance of fluid, P(x, t) is the excess pressuredue to disturbance of fluid, ρ is the the fluid density (a constant) and K is a constant (known as the bulkmodulus).

Show that both V(x, t) and P(x, t) obey wave equations, and find the propagation speeds in terms of the constantsK and ρ.


Question E7

(A3 and A6)3Show that the one-dimensional time-dependent Schrödinger equation for a freely moving particleof mass m has a solution of the form

Ψ(x, t) = A 1exp[i(kx − Et/˙)]

(where A is a constant). Express k in terms of m, E and ˙. Hint: For a free particle U = 0.

Study comment This is the final Exit test question. When you have completed the Exit test go back to Subsection 1.2 andtry the Fast track questions if you have not already done so.

If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave ithere.

flexible learning approach to physics ÊÊÊ module m6.4 ...€¦ · module m6.4 waves and partial...

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