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453.701 Linear Systems, S.M. Tan, The University of Auckland 1-1

Chapter 1 Dirac delta functions and distributions

1.1 Motivation

In many applications in physics, we would like to be able to handle quantities such as

1. The density ½(r) of a point mass at r0,

2. The probability density function of a discrete distribution,

3. The force F (t) associated with an �instantaneous� collision,

4. The derivative of a function with a step discontinuity.

We may regard each of these as the result of a limiting process in which a quantity gets more andmore concentrated at a point in space or time. Consider the sequence of functions

fn(t) =

½n for jtj < 1

2n0 otherwise

: (1.1)

These functions get narrower and taller as n ! 1. The area in any interval enclosing the origintends to one as n becomes large.

These functions do not tend pointwise to any limit as n ! 1 but physically it is useful to think of�point� sources such as point charges, point masses etc. We would like to invent a �limit� for thissequence of functions which should have certain properties. It is called the (Dirac) delta functionor the unit impulse and is denoted by ±(t). We would like ± (t) to satisfyZ b

a±(t) dt =

½1 if 0 2 (a; b)0 if 0 =2 [a; b]

: (1.2)

Given any function Á(t) continuous at the origin, we have that

limn→∞

Z ∞

−∞fn(t)Á(t) dt = Á(0); (1.3)

and so we would like ± (t) to have the property thatZ ∞

−∞±(t)Á(t) dt = Á(0): (1.4)

Many other sequences of functions may be found which also satisfy the limit (1.3) for a suitableclass of functions Á. These provide alternative representations of the delta function.

1.2 Distributions

We wish to develop a theory which will allow us to deal with objects like delta functions in a correctway. These are not ordinary functions but are rather examples of distributions or generalizedfunctions. The extra conceptual power and elegance of distribution theory will (hopefully) becomeobvious as we proceed. In particular, we shall Þnd that any distribution can be differentiated (togive another distribution) any number of times and that distributions justify many operations whichwe often carry out in Physics calculations which are incorrect for �ordinary� functions.


1.2.1 Some concepts from mathematical analysis

The following is a rather brief summary of a few deÞnitions and results which will be useful later.They should also serve to emphasize that we do need to be rather careful when manipulatingordinary functions.

Suppose that fxng for n = 1; 2; ::: is a sequence of real or complex numbers. We say that thesequence converges to x as n ! 1 if jxn ¡ xj can be made to be as small as we like whenevern is sufficiently large. Formally, this is written 8² > 0, 9N 2 N, such that n ¸ N ) jxn ¡ xj < ²where the quantiÞers 8 and 9 represent �for all� and �there exist� respectively.A function f(t) is said to have limit b as x ! a and we write

limt→a f(t) = b; (1.5)

provided that jf(t) ¡ bj can be made as small as we like whenever t is sufficiently close (but notequal) to a. Formally, 8² > 0, 9± > 0 such that 0 < jt¡ aj < ± ) jf(t)¡ bj < ².

A function f(t) is said to be continuous at a if

limt→a f(t) = f(a); (1.6)

i.e., 8² > 0, 9± > 0 such that jt¡ aj < ± ) jf(t)¡ f(a)j < ².

Note that continuity is a property that a function may have at a point. There are functions whichare continuous only at a single point and are discontinuous everywhere else. Consider for example

f (t) =

½t if t is a rational number0 if t is irrational.

(1.7)

You should be able to show (using the deÞnition of continuity) that this function is continuous att = 0 and nowhere else.

It should be familiar result from analysis that a function f(t) which is continuous on a closed andbounded set K is bounded, i.e., 9M 2 R such that jf(t)j · M for all t 2 K.

We now wish to consider sequences of functions. Suppose ffng is a sequence of functions alldeÞned on some set S. The functions are said to converge pointwise to a function f as n ! 1if for each t 2 S,

f(t) = limn→∞ fn(t): (1.8)

Notice that for a given t, the limit is simply the limit of a sequence of numbers.

Examples of sequences of functions

1. DeÞne

fn(t) =

8<:0 for t < 0tn for 0 · t · 11 for t > 1

:

This sequence of functions (see Þgure 1.1) converges pointwise to the limit function

f(t) =

½0 for t < 11 for t ¸ 1

: (1.9)

Note that the limiting function is discontinuous at t = 1; even though every function fn (t)in the sequence is continuous at t = 1:


0 0.5 1 1.50

0.5

1

1.5

t

f n(t)

n=1n=2n=3n=4

Figure 1.1 An example of a sequence of functions which converges pointwise but not uniformly to alimiting function as n ! 1:

2. The sequence of functions

fn(t) =

½n jtj < 1

2n0 otherwise

; (1.10)

does not converge pointwise to any function since the sequence of numbers ffn(0)g does notconverge.


fn(t) =

8<:n2t for 0 · t · 1

nn2

¡2n ¡ t

¢for 1n < t · 2

n0 otherwise

(1.11)

converges pointwise to the zero function. i.e., for every t; the sequence of numbers fn (t)converges to zero (see Figure 1.2). This is because given any t > 0; there is an integer N;such that for all n > N; fn (t) = 0: For all other values of t; the value of fn (t) is zero forevery n.


fn(t) =1

nsin(n2t) (1.12)

converges pointwise to the zero function.

Example 1 shows that even when all the functions in a sequence are continuous, the pointwiselimit can fail to be continuous (in this case at t = 1).

In example 3, each function fn(t) satisÞesZ ∞

−∞fn(t) dt = 1: (1.13)

Thus the limiting value of the integral as n ! 1 is one. However, the limiting function is zero andso the integral of the limit function is zero. We see in general that

limn→∞

Z ∞

−∞fn(t) dt 6=

Z ∞

−∞f(t) dt even if fn(t) ! f(t) pointwise. (1.14)


0 0.5 1 1.5 20

0.5

1

1.5

2

2.5

3

3.5

4

t

f n(t)

n=1n=2n=3n=4

Figure 1.2 An example of a sequence of functions which converges pointwise but not uniformly to thefunction which is identically zero.

In example 4, the derivative sequence

f 0n(t) = n cos(n2t) (1.15)

does not converge pointwise to any function, and certainly not to f 0(t), the derivative of the limitingfunction. Thus in general

limn→∞

d

dtfn(t) 6= d

dtf(t) even if fn(t) ! f(t) pointwise. (1.16)

One way of trying to improve matters is to strengthen what we mean by convergence of functions.We thus deÞne the concept of uniform convergence. A sequence of functions all deÞned on someset S is said to converge uniformly on S to a function f as n ! 1 if 8² > 0, 9N 2 N such that8t 2 S, jfn(t)¡ f(t)j < ² whenever n ¸ N .

It is important to know how this is different from the concept of pointwise convergence. Translatingthe deÞnition of pointwise convergence given previously into symbols reads 8t 2 S, 8² > 0, 9N 2 Nsuch that jfn(t)¡ f(t)j < ² whenever n ¸ N . We see that for pointwise convergence, the value ofN depends both on ² and on t whereas for uniform convergence, the same N must work for allt 2 S. Notice that if a sequence of functions converges uniformly to some limit, the sequence mustalso converge pointwise to that limit, but the converse does not necessarily hold.

Exercise: Show that the convergence in examples 1 and 3 is not uniform on R but that convergenceis uniform in example 4.

Another way of checking whether a pointwise-convergent sequence is uniformly convergent is tolook at the difference sequence dn(t) = fn(t)¡f(t); which is another sequence of functions. Theconvergence of fn(t) to f(t) is uniform on S provided that for all t 2 S; jdn(t)j may be boundedabove by a sequence which converges to zero. i.e., if we can Þnd a sequence Mn such that Mn ! 0as n ! 1 and jdn(t)j · Mn for all t 2 S.

If fn converges to f uniformly on S, then

1. If each fn is continuous at a 2 S, f is also continuous at a.


2. Provided that fn and f are integrable, the limit of the integral is equal to the integral of thelimit, i.e.,

limn→∞

Z x

afn(t) dt =

Z x

af(t) dt; (1.17)

where [a; x] is contained in S. The convergence of the integral is uniform.

Exercise: Prove the above assertions using the deÞnitions given.

However, example 4 shows that we still cannot interchange the limiting process and a derivative,even given uniform convergence. This whole situation is rather unsatisfactory, especially sinceuniform convergence is quite a strong condition which may be difficult to establish in practicalsituations. By contrast, by using the theory of distributions, we shall see that it is always possibleto differentiate distributions and to interchange the limiting process with derivatives of any order.

1.2.2 Test functions, functionals and distributions

In equation (1.4) we found it convenient to think about the delta function in terms of its actionon another function Á which is called a test function. The delta function may be regarded as amachine which operates on the test function Á to produce the number Á (0) as the result. We shallconsider such machines (Figure 1.3) which take test functions as inputs and produce numbers asoutputs in more detail.

Figure 1.3 A Dirac delta function acts on a test function to return its value at zero. More generally,a functional is a machine which acts on a test function to produce a number.

We must Þrst deÞne the class of test functions on which our machines work. This class is denotedby D, which we shall require to be a vector space. For technical reasons, we shall initially assumethat test functions are taken from the class of inÞnitely differentiable functions which vanishoutside a Þnite interval. It is important Þrst to check that test functions exist.

Exercise: Show that if

h(t) =

½exp(¡1=t) for t > 00 otherwise

: (1.18)

Then Á(t) = h(t)h(1¡ t) is a test function. (i.e., it is inÞnitely differentiable and vanishes outsidea Þnite interval). In Figure 1.4, graphs of h (t) and Á (t) are shown.

A functional T is a mapping from test functions to numbers. Given a test function Á 2 D, wedenote the number obtained by applying T to Á by the notation hT; Ái. We shall sometimes writethis as hT (t); Á(t)i to indicate explicitly the independent variable although T is not to be thoughtof as a function of t. Similarly we shall often refer to �the functional T (t)� by which we mean


−1 0 1 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

t

h(t)

−1 0 1 20

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0.018

0.02

t

h(t)

h(1−

t)Figure 1.4 The function h (t) is inÞnitely differentiable everywhere, and in particular at t = 0; althoughit is constant for t < 0: The function h (t)h (1¡ t) is inÞnitely differentiable everywhere and vanishesidentically outside the interval [0; 1] and is an example of a test function.

that T acts on a test function which is a function of t. In particular, it is not meaningful to speakof �the value of T at t�.

Two functionals S and T are said to be equal (with respect to the class of test functions D) if forevery test function Á 2 D;

hS; Ái = hT; Ái: (1.19)

A distribution is a functional which is both linear and continuous. A functional T is said tobe linear if for any Á1; Á2 2 D and for any numbers c1 and c2, we have that

hT; c1Á1 + c2Á2i = c1hT; Á1i+ c2hT; Á2i: (1.20)

By analogy with the deÞnition for an ordinary function to be continuous, the functional T is saidto be continuous if for any sequence of test functions fÁng which converges to Á in D,

limn→∞hT; Áni = hT; Ái: (1.21)

In order for this deÞnition to make sense, we need to know what is meant by �a sequence of testfunctions converging in D�. It turns out to be convenient to deÞne:A sequence of test functions fÁn(t)g in D converges to zero and we write Án ! 0 if

1. for every k 2 N[f0g, the sequence of k�th derivatives Á(k)1 (t); Á(k)2 (t); ::: converges uniformly

to zero.

2. there is a common interval [a; b] independent of n such that every Án(t) vanishes outside of[a; b].

We then say that fÁng converges to Á in D iff the sequence fÁn ¡ Ág converges to zero in D.


Note: For a linear functional T , the condition for continuity of T may be written in termsof sequences of test functions fÁn(t)g which converge to zero in D. The linear functional T iscontinuous iff for any such sequence of test functions Án ! 0,

limn→∞hT; Áni = 0: (1.22)

Example: The delta distribution ± is deÞned by

h±; Ái = Á(0): (1.23)

Check that this is indeed a distribution by showing that it is both linear and continuous.

Exercise: With the deÞnition of h (t) given by equation (1.18), determine which of the followingsequences of test functions converge to zero in D as n ! 1:

1. ®n (t) = 1nh (t)h (1¡ t)

2. ¯n (t) = 1nh (nt)h (1¡ nt)

3. °n (t) = 1nh (n¡ t)h (n+ 1¡ t) :

1.2.3 Regular distributions

Suppose that f(t) is a locally integrable function (i.e., the integral of jf(t)j exists and is Þnite forevery Þnite interval [a; b]). Then associated with the function f we may deÞne a functional Tf viathe rule

hTf ; Ái =Z ∞

−∞f(t)Á(t) dt: (1.24)

Since Á (t) is continuous on a closed and bounded interval, it is bounded, so the integral existsand is Þnite for any test function Á 2 D. We now wish to prove that the functional is in fact adistribution.

Linearity follows from the linearity of integration. Continuity holds for if Án(t) ! 0 in D, andall the Án vanish outside [a; b],¯̄̄̄Z ∞

−∞f(t)Án(t) dt

¯̄̄̄·

Z b

ajf(t)jjÁn(t)jdt · max

t∈RjÁn(t)j

Z b

ajf(t)j dt:

As n ! 1, the uniform convergence of Án (t) to the zero function means that the maximum ofjÁn (t)j converges to zero. Since f is locally integrable, the integral in the last term is Þnite, and sothe entire right-hand side converges to zero.

Distributions which are induced by some locally integrable function are said to be regular. Otherdistributions (such as the delta distribution) are said to be singular. (As an exercise, prove thatthe delta distribution is not induced by any locally integrable function).

As mentioned above, two distributions S and T are said to be equal iff hS; Ái = hT; Ái for allÁ 2 D. Notice that it is possible for two different locally integrable functions f and g to inducethe same distribution, i.e., for Z ∞

−∞f(t)Á(t) dt =

Z ∞

−∞g(t)Á(t) dt (1.25)

for all Á 2 D. This will happen, for example, if f(t) and g(t) differ only on a set of measure zero,such as on a Þnite or countable set of points.


1.2.4 Some notational matters, generalized functions

Every locally integrable function f(t) induces a regular distribution Tf . We delibrately wish to blurthe distinction between the function f(t) and the distribution it induces and so we shall often talkabout �the distribution f(t)� to mean Tf and write hf(t); Á(t)i instead of hTf ; Ái. Thus for locallyintegrable functions we shall write

hf(t); Á(t)i ´ hTf ; Ái =Z ∞

−∞f(t)Á(t) dt: (1.26)

Since two different locally integrable functions f(t) and g(t) can induce the same distribution ifthey differ on a set of measure zero, we shall say that �f(t) = g(t) distributionally� to mean thatthey induce the same distribution, i.e., that for all Á 2 D,

hf(t); Á(t)i = hg(t); Á(t)i: (1.27)

This is to be understood as being quite distinct from the usual concept of equality of functions.Normally, we would write f (t) = g (t) only if f and g agree at every point t in their commondomain.

For singular distributions, such as the delta distribution ± there is no locally integrable function±(t) satisfying Z ∞

−∞±(t)Á(t) dt = h±; Ái = Á(0): (1.28)

Nevertheless, we still use the notation above with the understanding that the integral form issimply an alternative way of writing the distribution. Unlike an ordinary function, we cannotevaluate ±(t) at an arbitrary value of t.

We shall always feel free to use the integral notation to denote the action of a distribution on a testfunction, whether the distribution is regular or singluar. We call the entity in the integrand whichmultiplies the test function a �generalized function�.

1.3 Operations on distributions

In this section, we shall begin to describe how all this mathematical structure works. We shalluse the regular distributions to tell us how to deÞne operations on general distributions. ThesedeÞnitions will be chosen so as to be consistent with the usual operations on ordinary functions.The role of the test functions and the reasons for choosing them from such a restricted class (i.e.,the inÞnitely differentiable functions which vanish outside a bounded set) will also become clearer.

1.3.1 Differentiation

Suppose that f(t) is a function which is both locally integrable and continuously differentiable.It induces a distribution hf; Ái. The derivative f 0(t) of the function f(t) is also locally integrableand induces another distribution hf 0; Ái. For consistency, we want the distribution f 0 to be thederivative of the distribution f whenever the function f 0 is the derivative of the function f .


For any test function Á 2 D,

hf 0(t); Á(t)i =Z ∞

−∞f 0(t)Á(t) dt (1.29)

= [f (t)Á (t)]∞−∞ ¡Z ∞

−∞f(t)Á0(t) dt

= ¡hf(t); Á0(t)i; (1.30)

where the boundary terms from the integration by parts have been set to zero since test functionsvanish outside a bounded set. The fact that Á is inÞnitely differentiable means that Á0 is alsoinÞnitely differentiable and is also in D.The above shows how we must deÞne the derivative of a regular distribution. The notation suggeststhat we should extend this to deÞne the derivative of an arbitrary distribution T (t) to be thefunctional T 0 where

hT 0(t); Á(t)i ´ ¡hT (t); Á0(t)i: (1.31)

We need to check that T 0 is in fact linear and continuous so that it is actually a distribution. Bothof these facts follow readily from our deÞnitions of D and convergence in this space.

Examples:

1. What is u0(t)? (Note: u(t) is the unit step function deÞned by

u(t) =

½1 for t > 00 for t < 0

: (1.32)

The value of u(0) is usually unspeciÞed and can take on any value).

The function u(t) is locally integrable and induces the regular distribution

hu(t); Á(t)i =Z ∞

−∞u(t)Á(t) dt =

Z ∞

0Á(t) dt: (1.33)

The derivative of u is found by applying the deÞnition

hu0(t); Á(t)i = ¡hu(t); Á0(t)i= ¡

Z ∞

0Á0(t) dt = ¡Á(1) + Á(0)

= Á(0) = h±(t); Á(t)i: (1.34)

Since this holds for all Á 2 D, we see that u0(t) = ±(t) in a distributional sense. This holdsdespite the fact that u(t) is not differentiable in the ordinary sense at the point t = 0.

2. What is the derivative of ±(t)? Again we simply use the deÞnition. Given any test functionÁ 2 D, the action of ±0 on Á is given by

h±0(t); Á(t)i = ¡h±(t); Á0(t)i = ¡Á0(0): (1.35)

Formally, we write Z ∞

−∞±0(t)Ádt = ¡Á0(0): (1.36)

This is a distribution called the dipole or doublet. Just as the charge density for a pointcharge can be written as a delta function, the charge density for an electric dipole can bewritten as a dipole function.

Similarly it is easy to show that

h±(k)(t); Á(t)i = (¡1)kh±(t); Á(k)(t)i = (¡1)kÁ(k)(0): (1.37)


These examples and the deÞnition of the derivative of distributions show that every generalizedfunction can be differentiated an inÞnite number of times. Thus we have not only extended ourclass of locally integrable functions (which can be discontinuous) to the generalized functions butin doing so have also made them all inÞnitely differentiable (in the distributional sense).

1.3.2 Shifting

Suppose that T (t) is a distribution. How should we deÞne the shifted distribution T (t¡ a)? Againwe proceed via consistency with the regular distributions. Suppose that f(t) is a locally integrablefunction. We want to see how the distribution induced by f(t ¡ a) is related to that induced byf(t). Using the deÞnition of a regular distribution in terms of the integral,

hf(t¡ a); Á(t)i =Z ∞

−∞f(t¡ a)Á(t) dt =

Z ∞

−∞f(u)Á(u+ a) du = hf(t); Á(t+ a)i: (1.38)

We thus make the deÞnition:

Given any distribution T (t), the functional T (t¡ a) is deÞned by

hT (t¡ a); Á(t)i = hT (t); Á(t+ a)i; (1.39)

for any Á 2 D. It is easy to check that the shifted distribution is also a distribution.

1.3.3 Scaling

Suppose that T (t) is a distribution. We wish to assign a meaning to T (at) for a real number a.Start once again with a locally integrable function f(t). The distribution induced by f(at) isZ ∞

−∞f(at)Á(t) dt =

1

a

Z a∞

−a∞f(u)Á

³ua

´du: (1.40)

The limits on the right-hand side depend on the sign of a. If a > 0 this is

1

a

Z ∞

−∞f(u)Á

³ua

´du =

1

a

¿f(t); Á

µt

a

¶À: (1.41)

On the other hand if a < 0, the lower limit is 1 and the upper limit is ¡1. To swap these limitsaround, we need to introduce an additional minus sign and so the right-hand side of (1.40) becomes

1

a

Z −∞

∞f(u)Á

³ua

´du = ¡1

a

¿f(t); Á

µt

a

¶À: (1.42)

These results may be summarized by the rule

hf(at); Á(t)i = 1

jaj¿f(t); Á

µt

a

¶À: (1.43)

This holds for real, nonzero values of a. We use this to deÞne the scaling law for an arbitrarydistribution T (t)

hT (at); Á(t)i = 1

jaj¿T (t); Á

µt

a

¶À: (1.44)

Examples:


1. The delta function is even:

h±(¡t); Á(t)i = 1

j ¡ 1j¿±(t); Á

µt

a

¶À= Á(0) = h±(t); Á(t)i (1.45)

2. The shifted delta function:

h±(t¡ ¿); Á(¿)i = h±(¿ ¡ t); Á(¿)i = h±(¿); Á(¿ + t)i = Á(t) (1.46)

Thus we may write Z ∞

−∞Á(¿)±(t¡ ¿) d¿ = Á(t) (1.47)

This equation holds at each point t for any test function Á 2 D.

1.3.4 Product of a C∞ function and a distribution

If T (t) is a distribution and g(t) is inÞnitely differentiable, g(t)T (t) is deÞned by

hg(t)T (t); Á(t)i = hT (t); g(t)Á(t)i (1.48)

It is easy to show that this is a distribution and that it is consistent with the result for regulardistributions. We need g(t) to be C∞ to ensure that g(t)Á(t) 2 D whenever Á(t) 2 D.In general, it is not possible to deÞne the product of two arbitrary distributions.

1.3.5 Integral of a delta function

Consider

f(t) =

Z t

−∞±(¿) d¿ (1.49)

This simply means that we require that f 0(t) = ±(t) and that f(¡1) = 0. Both of these aresatisÞed by the unit step u(t) which is also called the Heaviside function (sometimes written asH(t)).

Since u(t) induces a regular distribution (as it is locally integrable), the successive integrals of u(t)are ordinary functions such as the ramp r(t) = t u(t), the parabola p(t) = 1

2t2 u(t) etc. These all

induce regular distributions.

1.4 Sequences and series of distributions

In the previous section, we have seen how various mathematical operations deÞned on ordinaryfunctions may be extended to apply to distributions. In physics, it is useful to develop an intuitive,pictorial understanding as well as a mathematical understanding of the theory. In this section wewish to see in what senses the intuitive picture of a delta function as an inÞnitely high spike of zerowidth but with unit area is a useful representation. For example, although there is no functionwhich is inÞnitely tall and of zero width, we feel intuitively that a sequence of functions fn (t) ; suchas

fn(t) =

½n jtj < 1

2n0 otherwise

; (1.50)


in some sense �tends to� a delta function, even though the sequence does not have a pointwiselimit. As we shall see, the key idea is not to focus on the above as a sequence of functions, butrather on the sequence of regular distributions that the functions induce. To this end, we need toconsider the properties of sequences of distributions and their convergence.

Suppose that fTn(t)g is a sequence of distributions and fT (t)g is a distribution. We say that Tn(t)converges to T (t) and write Tn(t) ! T (t) distributionally if

limn→∞hTn(t); Á(t)i = hT (t); Á(t)i; (1.51)

for all test functions Á 2 D.Note: It is in fact unnecessary to assume that the limit T is a distribution. If for all Á 2 D, hTn; Áiis a convergent series and we deÞne the action of T in terms of this limit, T can be shown (withdifficulty!) to be a distribution.

Theorem: If fSng and fTng are sequences of distributions which converge to S and T respectively,then

1. aSn + bTn ! aS + bT for any constants a and b,

2. T 0n ! T 0,

3. Tn(at¡ b) ! T (at¡ b) for any real a and b,

4. g(t)Tn(t) ! g(t)T (t) for any C∞ function g(t).

Perhaps the most surprising aspect of these relationships is the ease of the proofs. Consider forexample the proof of the derivative result. For any test function Á 2 D,

hT 0n; Ái = ¡hTn; Á0i ! ¡hT; Á0i = hT 0; Ái (1.52)

These results are nonetheless remarkable, since they tell us that all the common distributionaloperations commute with the limiting process.

Let us again consider the sequence of functions (1.50), which we saw does not converge pointwise.For each n, fn(t) is locally integrable and so induces the regular distribution

hfn(t); Á(t)i = n

Z 1/2n

−1/(2n)Á(t) dt: (1.53)

We wish to consider the sequence of distributions ffng : For any test function Á 2 D; as n ! 1, thesequence hfn(t); Á(t)i tends to Á(0) since Á is continuous at t = 0. By deÞnition of the convergenceof a sequence of distributions, we see that fn ! ± distributionally which conÞrms our originalintuition. This is an example of a sequence of functions which does not converge pointwise, butwhich does converge distributionally.

Let us next consider the sequence of functions

gn(t) =

8<:n2t for 0 · t · 1

nn2

¡2n ¡ t

¢for 1

n < t · 2n

0 otherwise(1.54)

which, as we discovered (see Figure 1.2), converges pointwise to zero. Given a test function Á 2 D;we see that Z ∞

−∞gn(t)Á (t) dt = n2

Z 1/n

0tÁ (t) dt+ n2

Z 2/n

1/n

µ2

n¡ t

¶Á (t) dt (1.55)


We now wish to show that each of the terms on the right hand side is equal to 12Á (0) : Since

n2Z 1/n

0tdt =

1

2; (1.56)

we see that ¯̄̄̄¯n2

Z 1/n

0tÁ (t) dt¡ 1

2Á (0)

¯̄̄̄¯ =

¯̄̄̄¯n2

Z 1/n

0t (Á (t)¡ Á (0)) dt

¯̄̄̄¯

· n2Z 1/n

0jtj jÁ (t)¡ Á (0)jdt: (1.57)

Since Á is a test function, it is continuous at t = 0: By deÞnition of continuity, given any " > 0; wecan Þnd ± > 0 such that whenever jtj < ±; jÁ (t)¡ Á (0)j < ": So if we choose any n > ±−1;¯̄̄̄

¯n2Z 1/n

0tÁ (t) dt¡ 1

2Á (0)

¯̄̄̄¯ · n2

Z 1/n

0jtj "dt = "

2: (1.58)

This establishes that

limn→∞n2

Z 1/n

0tÁ (t) dt =

1

2Á (0) : (1.59)

Similarly, you should check that you can show

limn→∞n2

Z 2/n

1/n

µ2

n¡ t

¶Á (t) dt =

1

2Á (0) ; (1.60)

so that

limn→∞

Z ∞

−∞gn(t)Á (t) dt = Á (0) : (1.61)

This establishes that the sequence of distributions fgng induced by gn (t) also tends distributionallyto the Dirac ±:distribution. Notice that the distributional limit of a sequence can be very differentfrom the pointwise limit. It should also be clear that there are an inÞnite number of sequences ofregular distributions which converge distributionally to the delta distribution. When contemplatingthe Dirac delta function, one should try to think of it as the common limit of all such sequences ofdistributions.

Exercise: Show that the sequence of functions (see Þgure 1.5)

hn (t) =1

¼

µn

n2t2 + 1

¶(1.62)

converges distributionally to ± (t) as n ! 1:

Let us now see how the theorem helps us to visualize what we mean by the derivative of a deltafunction, which we denoted by ±0 (t) : According to the theorem if we can Þnd any sequence ofdistributions fTng with the property that Tn ! ± distributionally, we are guaranteed that T 0n willalso converge to the distribution ±0: We have found three sequences ffng ; fgng and fhng ; all ofwhich tend distributionally to ±: By differentiating each of these, we will get three sequences ofdistributions which tend to ±0 (t) : It is straightforward to Þnd h0n (t) and g0n (t) ; which are

h0n (t) = ¡ 2

¼

n3t

(n2t2 + 1)2(1.63)


−5 0 50

0.2

0.4

0.6

0.8

1

1.2

1.4

t

h n(t)

n=1n=2n=3n=4

Figure 1.5 A sequence hn (t) which tends distributionally to ± (t) :

and

g0n (t) =

8<:n2 for 0 · t · 1

n¡n2 for 1n < t · 2

n0 otherwise

A graph of h0n (t) for n = 1; :::; 4 is shown in Figure 1.6.

The common feature of g0n and h0n is that as n ! 1; they both have a large positive peak followedby a large negative peak. The areas of each of the two peaks increases proportionally to n. Wecan also Þnd the sequence of distributions f 0n (t) ; each of which turns out to be singular, since thefunctions fn are not differentiable. We may write fn (t) as the difference of two unit step functions,

fn (t) =

½n jtj < 1

2n0 otherwise

= n

·u

µt+

1

2n

¶¡ u

µt¡ 1

2n

¶¸(1.64)

From our previous result (1.34) that u0 (t) = ± (t) distributionally, we see that

f 0n(t) = n

·±

µt+

1

2n

¶¡ ±

µt¡ 1

2n

¶¸; (1.65)

which consists of two delta functions of strengths §n located at distance 1=n apart. As n ! 1;the two spikes get closer to each other and their areas become larger. From this we can see why ±0

is referred to as a dipole.

Given a test function Á (t) 2 D; it is clear that−f 0n (t) ; Á (t)

®=

¿n

·±

µt+

1

2n

¶¡ ±

µt¡ 1

2n

¶¸; Á (t)

À= n

¿±

µt+

1

2n

¶; Á (t)

À¡ n

¿±

µt¡ 1

2n

¶; Á (t)

À= n

·Á

µ¡ 1

2n

¶¡ Á

µ1

2n

¶¸: (1.66)


−5 0 5−4

−3

−2

−1

0

1

2

3

4

t

h’n(t

)

n=1n=2n=3n=4

Figure 1.6 A sequence h0n (t) which tends distributionally to ±0 (t) ; the derivative of the Dirac ±function.

Taking the limit as n ! 1; we see that

limn→∞

−f 0n (t) ; Á (t)

®= ¡Á0 (0) ; (1.67)

which is consistent with what we found (1.35) from the mathematical deÞnition of the derivativeof a distribution.

These results often allow us to visualize arbitrary (singular) distributions since we can alwaysexpress such a distribution as the limit of a sequence of regular distributions associated with asequence of locally integrable functions. It is possible to calculate with the singular distribution bytaking limits over the results obtained with the sequence of regular distributions.

Exercise: Show that for any test function Á(t),

limn→∞n

Z ∞

−∞Á(t) cos(n2t) dt = 0 (1.68)

We can similarly treat a series of distributionsP∞k=1 Tk which is said to converge to the distribution

S if the sequence of numbers hPnk=1 Tk(t); Á(t)i converges as n ! 1 for every Á 2 D. We may

treat a series in terms of the sequence of its partial sums and apply the above results for sequences.In particular we can differentiate a series of distributions term-by-term, i.e.,*Ã ∞X

k=1

Tk

!0; Á

+=

∞Xk=1

hT 0k; Ái (1.69)

Using a Riemann approach to integration as a limit of a summation, we may use linearity to writedistributional equations such as

f(t) =

Z ∞

−∞f(¿)±(t¡ ¿) d¿ (1.70)


where f is now regarded as a locally integrable function. What this means is that for all Á 2 D,¿Z ∞

−∞f(¿)±(t¡ ¿) d¿; Á(t)

À= hf(t); Á(t)i (1.71)

Notice the difference between this distributional equality and the previous pointwise relationship(1.47).

We may regard (1.70) as expressing f(t) as a linear combination of impulses. It may be helpful tothink of the analogy of writing a vector as a linear combination of basis vectors, e.g.,

v =X

viei

The delta functions are the basis vectors, indexed by their location ¿ . The sum is over the index ofthese basis vectors ¿ which correspond to i. The numbers f(¿) d¿ are analogous to the coefficientsof the expansion vi, and the resulting function f is analogous to the vector v.

1.5 Differential equations involving distributions

We wish to consider differential equations in which the solutions and driving terms are general-ized functions. We shall mainly be concerned with linear differential equations with homogeneousboundary and/or initial conditions.

As a speciÞc example, consider the problem of a string of length L under tension T constrained sothat its transverse displacement u(x) is zero at the endpoints. i.e., u(0) = u(l) = 0. A distributionof transverse �pressure� p(x) (actually a force per unit length) is applied to the string and we wishto determine the resulting displacement. The differential equation is

Lu = ¡Td2u

dx2= p(x); u(0) = u(l) = 0 (1.72)

where L denotes the differential operator

L = ¡Td2

dx2(1.73)

For each distribution of pressure p(x) there is a resulting displacement u(x). We can treat theequation like a physical system with p(x) as an input function and u(x) as the output function.We see that the system is linear, i.e., if u1(x) is the displacement for pressure p1(x) and u2(x) isthe displacement for pressure p2(x), the displacement if we apply the pressure c1p1(x) + c2p2(x)is c1u1(x) + c2u2(x). This follows from the linearity of L and the homogeneity of the boundaryconditions. Linearity allows us to apply the following important idea:

We know that for any locally integrable function p(x), we can write p(x) as a linear combinationof impulses i.e.,

p(x) =

Z ∞

−∞p(»)±(x¡ ») d» (1.74)

So if we know how the physical system responds to the input ±(x¡ »), we can use this to work outhow it will respond to p(x) which is simply a linear combination of these delta function inputs.

Let us suppose that we can Þnd the solution h(xj») to the problemLh(xj») = ±(x¡ »); h(0j») = h(lj») = 0 (1.75)


where the input to the system is a delta function. Then by linearity the output for input p(x) mustbe given by

u(x) =

Z ∞

−∞p(»)h(xj») d» (1.76)

which is simply (1.74) with the delta function replaced by the response that it produces when it is�processed� by the system.

The function h(xj») is called the impulse response of the system. It is also known as theGreen�sfunction of the boundary value problem.

Example: Calculating the Green�s function for

¡Td2h(xj»)

dx2= ±(x¡ »); h(0j») = h(lj») = 0 (1.77)

This corresponds to the application of a concentrated force at x = ». In this case h(xj») can befound by direct integration (Þgure 1.7). We see that

d2h

dx2= ¡ 1

T±(x¡ ») (1.78)

dh

dx= c1 ¡ 1

Tu(x¡ ») (1.79)

h(xj») =½

c2 + c1x for x < »c2 + c1x¡ 1

T (x¡ ») for x > »(1.80)

The quantities c1 and c2 are constants of integration. For all values of c1 and c2; the function h (xj»)satisÞes the differential equation, but the boundary conditions are only satisÞed for a special choiceof the constants. Applying the boundary conditions, we Þnd that c2 = 0 and c1 = (l ¡ »)=(lT ).Hence

h(xj») =

8><>:x (l ¡ »)

lTfor x < »

» (l ¡ x)

lTfor x > »

(1.81)

Notice that

1. In regions x < » and x > », the Green�s function satisÞes the homogeneous equation

d2h

dx2= 0 (1.82)

2. At x = », h(xj») is continuous and @h=@x has a jump discontinuity to give the delta functionin the second derivative

dh

dx(»+j»)¡ dh

dx(»−j») = ¡ 1

T(1.83)

3. The solution of the boundary value problem for an arbitrary p(x) is given by

u(x) =

Z l

0p(»)h(xj») d» (1.84)

=

Z x

0p(»)

» (l ¡ x)

lTd» +

Z l

xp(»)

x (l ¡ »)

lTd» (1.85)

Exercise: Check that this is a solution of the boundary value problem. You may Þnd the followingresult (which you should also prove if you have not seen it before) useful:

d

dx

Z b(x)

a(x)f(t; x) dt =

Z b(x)

a(x)

@f

@x(t; x) dt¡ f(a(x); x) a0(x) + f(b(x); x) b0(x) (1.86)


Figure 1.7 Calculation of Green�s function by direct integration. Values of constants are found bysatisfying the boundary conditions.

1.5.1 Green�s function for regular n�th order linear differential equations with homo-geneous initial/boundary conditions

The above example may be generalized to the following. Consider the linear n�th order differentialequation on the interval I

Lg = an(x)dng

dxn+ an−1(x)

dn−1gdxn−1

+ :::+ a1(x)dg

dx+ a0(x)g = f(x) (1.87)

together with n homogeneous boundary or initial conditions (which constrain u and/or its deriv-atives at the endpoints of I). We assume that an(x) 6= 0 for all x 2 I so that the equation isregular.

The Green�s function h(xj») is the solution of

Lh(xj») = ±(x¡ ») (1.88)

where

1. h(xj») satisÞes the homogeneous equation (with zero right-hand side) for x < » and for x > »,

2. h(xj») satisÞes all the boundary or initial conditions for each »,

3. The functions h, @h=@x,..., @n−2h=@xn−2 are continuous throughout I and in particular atx = ».

4. @n−1h=@xn−1 has a jump discontinuity at x = » and

dn−1hdxn−1

(»+j»)¡ dn−1hdxn−1

(»−j») = 1

an(»)(1.89)


The solution to the original problem can be found for any f(x) and is

g(x) =

ZIf(»)h(xj») d» (1.90)

Example: Calculate the Green�s function for the boundary value problem

d2u

dx2+ k2u = f(x); u(0) = u(l) = 0 (1.91)

and use the Green�s function to Þnd the solution for f(x) = x.

The Green�s function h (xj») is a solution of

d2h(xj»)dx2

+ k2h(xj») = ±(x¡ »); h(0j») = h(lj») = 0 (1.92)

Where the �spike� of the delta function is located at x = »: In order to Þnd h; we consider the tworegions x < » and x > » separately. In each of the two regions, h is a solution of the homogeneousdifferential equation, but with different constants. Hence

h (xj») =½

A cos kx+B sin kx for x < »C cos kx+D sin kx for x > »

(1.93)

We have to choose the constants A; B, C and D (which all depend on ») to satisfy the boundaryconditions and the �joining conditions� at x = »: The boundary condition at zero tells us that

A = 0; (1.94)

since x = 0 < »; and we have to use the upper expression for h: The boundary condition at x = l > »imposes constraints on the lower expression for h: In particular

C cos kl +D sin kl = 0: (1.95)

The joining conditions tell us Þrstly that h (xj») is continuous at x = » since h is differentiablethere. Approaching x = » from below and above and equating the results shows us that

A cos k» +B sin k» = C cos k» +D sin k»: (1.96)

The Þrst derivative @h=@x is however discontinuous at x = » since the second derivative has a deltafunction at »: The size of the jump discontinuity is equal to the area under the delta function, sothat @h (»+j») =@x¡ @h (»−j») =@x = 1: In terms of the expressions above

(¡kC sin k» + kD cos k»)¡ (¡kA sin k» + kB cos k») = 1 (1.97)

Solving equations (1.94) through (1.97) simultaneously yields

A = 0; B = ¡sin k (l ¡ »)

k sin kl; C = ¡sin k»

kand D =

sin k» cos kl

k sin kl(1.98)

Substituting back into the expression for h (xj») and simplifying, we Þnd

h (xj») =

8><>:¡sin k (l ¡ ») sin kx

k sin klfor x < »

¡sin k» sin k (l ¡ x)

k sin klfor x > »

: (1.99)


0 0.2 0.4 0.6 0.8 1−0.05

0

0.05

0.1

0.15

x

h(x|

0.6)

Figure 1.8 Green�s function h (xj») for » = 0:6; k = 7:5 and l = 1: Notice that the function iscontinuous at x = »; but that the derivative has a jump of unit size there.

In this case, we note that the Green�s function is symmetric, i.e., that h (xj») = h (»jx) : This is thecase if the boundary value problem is self-adjoint. Symmetric Green�s functions can be writtenmore compactly in terms of the variables x< = min (x; ») and x> = max (x; ») : In the example,

h (xj») = ¡ sin kx< sin k (l ¡ x>)

k sin kl(1.100)

This Green�s function is plotted in Figure 1.8.

We now wish to solve the boundary value problem Lu = f with f (x) = x: By linearity, we havethat

u (x) =

Z l

0h (xj») f (») d» =

Z l

0»h (xj») d» (1.101)

Since h (xj») has different functional forms depending on whether x < » or x > »; we split the rangeof integration into two. Notice carefully that the integration variable is »; so that the interval needsto be split into [0; x] and [x; l] : Thus,

u (x) =

Z x

0»h (xj») d» +

Z l

x»h (xj») d»: (1.102)

Consider the Þrst integral. In the interval » 2 [0; x] ; we have that » < x and so we must usethe lower of the two expressions for h (xj»). For the second integral, in the interval » 2 [x; l] ; wehave that » > x and so we must use the upper of the two expressions of h (xj») : Substituting theappropriate formulae gives

u (x) = ¡Z x

0»sin k» sin k (l ¡ x)

k sin kld» ¡

Z l

x»sin k (l ¡ ») sin kx

k sin kld»

= ¡sin k (l ¡ x)

k sin kl

Z x

0» sin k» d» ¡ sin kx

k sin kl

Z l

x» sin k (l ¡ ») d»

=x sin kl ¡ l sin kx

k2 sin kl:

It is easy to check that this satisÞes the boundary conditions u (0) = u (l) = 0 and u00 (x)+k2u (x) =x; as required.


Exercise: Calculate the Green�s function for the following problems

1.d2u

dx2+ k2u = f(x); u(0) = u0(l) = 0

and use the Green�s function to Þnd the solution for f(x) = 1.

2.

RCdvCdt

+ vC = vI ; vC(¡1) = 0

This is the impulse response of a RC quasi-integrator circuit. Calculate the response ifvI(t) = cos 2¼ºt.

3.

LCd2vCdt2

+RCdvCdt

+ vC = vI ; vC(¡1) = v0C(¡1) = 0

This is the impulse response of a series LCR circuit when the output is taken across thecapacitor. Calculate the response when the circuit is excited with a unit step input voltage.

References:

Distribution theory is covered in many texts with classiÞcation number 515.782. Two readableaccounts are

Theory of Distributions, A non-technical introduction by J.I. Richards and H.K. Youn, (CambridgeUniversity Press, 1990).

Distribution Theory and Transform Analysis by A.H. Zemanian, (McGraw-Hill, 1965).

A useful book for mathematical background is

Mathematical Analysis by T.M. Apostol, (Addison-Wesley, 1974).