Appel W. Mathematics for Physics and Physicists

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Convolution algebras 237

The convolution equation is then solved by putting

X = A

∗−1

∗ B,

and this solution is unique.

To summariz e, we can state the following theorem:

THEOREM 8.30 Let A be a convolution algebra. The convolution equation

A ∗ X = B

admits a solution in A for arbitrary B ∈ A if and only if A has a convolution

inverse A

∗−1

in this algebra. I n this case, the solution is unique and is given by

X = A

∗−1

∗ B.

Remark 8.31 It is i mport ant to notice that we have always spoken of the inverse of a distribu-

tion in a given algebra A . In fact, a distribution A may very well have a certain convolution

inverse in one algebra and another inverse in a different algebra. Take, for instance, the case of

a harmonic oscillator characterized by a convolution equation of the type

(δ

′′

+ ω

δ) ∗ X (t) = B(t),

where B(t) is a distribution characterizing the input signal. The reader is invited to check that

(δ

′′

+ ω

δ) ∗

H (t) sinωt = δ.

Since the distribution

H (t) sin ωt is an element of D

′

, it follows that, for any B ∈ D

′

, that

is, for any input which started at a ﬁnite instant t ∈ R and was zero previously, the equation

has a solution in D

′

which is unique and is gi ven by

X (t) = B(t) ∗

H (t) sin ωt causal solution.

But it is also possible to check that

(δ

′′

+ ω

δ) ∗



−

H (−t) sinωt



= δ,

and therefore that −

H (−t) sinωt is a convolution inverse of (δ

′′

+ ω

δ) in D

′

−

. If th e

input signal B(t) belongs to D

′

−

, it is therefore necessary, to get the solution, to carry out the

convolution of B by −

H (−t) sinωt:

X (t) = B(t) ∗

−1

H (−t) sinωt anti-causal solution.

We also deduce that this convolution operator does not have an inverse in E

′

One of the great computational problems that can confront a physicist

is to ﬁnd the convolution inverse of an operator (its Green function). For

instance, the Green function of the Laplace operator in R

is given by the

Coulomb potential 1/r, as we have seen by performing t he relevant convo-

lution product. But how, given an operator, can we ﬁnd it s inverse? If the

238 Distributions II

convolution operator is a differential operator (more precisely, a differential

operator applied to δ), there exist tools which are particularly ad apted to com-

puting its inverse, namely the Fourier and Laplace transforms (see Chapters 10

and 12). Inﬁnitely many technical details for t he inversion of the laplacian

(with reﬁnements th ereof) are given in the reference [24], which is somewhat

hard to read, however.

8.4

Solving a differential equation

with initial conditions

Most students in physics remember very well that the evolution of a system

is given by a differential equation, but forget that the real problem is to solve

this equation for given initial conditions. It is true th at, in the linear case,

these initial conditions are often used to pa rameterize families of solutions;

however, this is not a general rule.

The Cauchy problem (an equation or s ystem of differential equations,

with initial conditions) is often very difﬁcult to solve, as every one of you

knows. Sometimes, it is very hard to s how that a solution exists (without even

speaking of computing it!), or to show that it is unique.

In the cases below, we have autonomous linear differential equations with

constant coefﬁcients. The theoretical issues are therefore much simpler, and we

are just presenting a convenient computational method . Some cases w hich are

more complicated (and more useful in physics) will be presented on page 348.

8.4.a First order equations

Consider a differential equation of the type

˙u + αu = 0, (e)

where we only wish to know u(t) for positive values of t, and u : R

→ R is

a function satisfying t he initial condition

u(0) = u

∈ R.

It is possible to use the tools of the theory of distributions to solve this

equation with it s initial conditions.

The most spectacular counterexamples are given by chaotic systems.

For instance, the “old” problem of an electric charge in an electromagnetic ﬁeld is only

partly solved [11]. The difﬁculty of this system is that it is described by highly nonlinear

equations whe re the charge i nﬂuences the ﬁelds, which in turn inﬂuence the charge. The

reader interested in these problems of mathematical physics can read [84 ].

Solving a differential equation with initial conditions 239

We then slightly change our point of view, by looking no longer for a

solution in the sense of functions to this dif ferential equation, but for a

solution in the sense of d istributions, of the type U (t) = H (t) u(t). (This

solution will then be zero for negative time, which is of no importance since

we are interested in positive time only.)

If u is a solution of (e), which equation will the distribution U s atisfy?

Since

U = ˙u + u

δ, we see that U satisﬁes the equation

U + αU = u

δ,

which may be put in the form of a convolution

[δ

′

+ αδ] ∗ U = u

δ. (E)

There only remains to ﬁnd the unique solution to this equation (in the sense

of distributions) which lies in the convolution algebra D

′

of distributions

with support bounded on the left.

THEOREM 8.32 The convolution inverse of [δ

′

+ αδ] in D

′

G(t) = H (t) e

−αt

(The inverse in D

′

−

is t 7→ −H (−t) e

−αt

Proof. Indeed, we have G

′

(t) = −α e

−αt

+δ(t), w h ich shows that [δ

′

+αδ]∗G = δ.

Taking the convolution of (E) with G(t) = H (t) e

−αt

, we ﬁnd that

U = G ∗(u

δ), or U = u

H (t) e

−αt

It may seem that we used something like a hammer to crush a very small

ﬂy. Quite. But this systematic procedure becomes more interesting when

considering a differential equation of higher order, or more complicated initial

conditions, for instance on surfaces (see Section 15.4, page 422).

8.4.b The case of the harmonic oscillator

Consider now the case of a harmonic oscillator, controlled by the equation

+ ω

u = 0, (e

′

)

where we seek a function u(t) satisfying t his differential equation for t ¾ 0,

with given initia l conditions u(0) and u

′

(0). For this purpose, we will look

for a distribution U (t), of the type U (t) = H (t) u(t), which satisﬁes the

differential equation in the sens e of functions. W hat differential equation in

the sense of distributions will it satisfy? This will be, taking the discontinuities

into account,



′′

+ ω



∗ U = u(0) δ

′

+ u

′

(0) δ. (E

′

)

240 Distr i butions II

THEOREM 8.33 (Harmonic oscillator) The convolution inverse of [δ

′′

+ ω

δ]

in D

′

(t) =

H (t) sinωt.

The convolution inverse of [δ

′′

+ ω

δ] in D

′

−

(see Remark 8.31 on page 237) is

−

(t) =

−1

H (−t) sinωt.

Remark 8.34 There exist methods to rediscover this result, the proof of which is immediate:

such convolution inverses, or Green functions, of course do not come out of thin air. These

methods are explained in Section 15.2 on page 409.

Then we ﬁnd directly, by ta king the convolution of equation (E

′

) with G

that

U (t) =

H (t) sinωt ∗



u(0) δ

′

(t) + u

′

(0) δ(t)



= H (t)



u(0) cos ωt +

′

(0)

sin ωt



To summariz e, the same problem can be seen from two different angles:

1. the ﬁrst, classical, shows a differential equation with given initial con-

ditions at the point 0;

2. the second shows a differential equations valid in the sense of distribu-

tions, without conditions at 0, for w hich a solution is sought in a g iven

convolution algebra (for inst ance, that of distributions with support

bounded on the left).

The right-hand s ide of (E

′

) shows what distribution is necessary to move

the system at rest (for t < 0) “instantaneously” to the initial state



u(0), u

′

(0)



8.4.c Other equations of physical origin

We will see in the following chapters, devoted to the Fourier and Laplace

transforms, how to compute the convolution inverses (also called the Green

functions) of a differential operator.

Here we merely state a n interesting result:

THEOREM 8.35 (Heat equation) The convolution inverse of the heat operator



∂

∂ t

−κ

∂

∂ x



in D

′

, called the heat kernel, is

G(x, t) =

H (t)

κπc t

−c x

/4πt

Exercises 241

that is,



cδ

′

(t) −κδ

′′

(x)



∗G(x, t) = δ

(2)

(x, t),

∂G

∂ t

−κ

∂

∂ x

= δ(x) δ(t).

Proof. This result is proved ab initio in Section 15.4, page 42 2 .

EXERCISES

Examples of distributions

 Exercise 8.4 (Cauchy principal value) Show that the map deﬁned on D by

: ϕ ∈ D 7−→ lim

ǫ→0

|x|>ǫ

ϕ(x)

which deﬁnes the Cauchy principal value is the derivative, in the sense of distributions, of

the regular distribution associated to the locally integrable fu nct ion x 7→ log |x|.

 Exercise 8.5 (Beyond principal va lue) Show that the map

ϕ 7−→ lim

ǫ→0

Z

∞

ϕ(x)

dx −

ϕ(0)

+ ϕ

′

(0) log ǫ



deﬁnes a distribution, that is, that it is linear and continuous.

 Exercise 8.6 (Finite part of 1/ x

) We denote by fp(1/x

) the opposite of the distribution

derivative of the distribution pv(1/x) (“fp” for ﬁnite part):

def

= −





′

Show th at



, ϕ

= −

+∞

−∞

log |x|ϕ

′′

(x) dx = lim

ǫ→0

|x|>ǫ

ϕ(x) −ϕ(0)

dx,

and that, if 1 denotes as usual the constant function equal to 1, or the associated distribution,

we have

·fp

= 1 .

Generalize these results to deﬁne distributions related to the (not locally integrable) functions

x 7→ 1/x

for any n ∈ N.

 Exercise 8.7 Show that we have, for all a ∈ R,

δ(ax) =

|a|

δ(x),

whereas, if a > 0,

H (ax) = H (x).

242 Distributions II

 Exercise 8.8 Prove the relation : xδ

′

= −δ. Compute similarly xδ

′′

. Generalize by com-

puting f (x) δ

′

, where f is a (locally integrable) function of C

∞

class. Is it possible to relax

those assumptions?

Composition of a δ distribution and a function

 Exercise 8.9 (Special relativity) We consider the setting of special relativity, with the mo-

mentum vector p = (E, p) = (p

, p), wit h a Minkowski metri c of signature (+, −, −, −) and

the convention c = 1. Thus, we have p

= E

− p

Recall that for a classical particle, we have p

= m

, where m is the mass o f the particle. In

ﬁeld theory, one is often led to compute integrals of the type

δ(p

−m

) H (p

) f (p) d

What is the interpretation of the functions δ and H ? These integrals have the nice property

of being invariant under Lorentz transformations. Show that, if we put

E( p)

def

+ m

then we have

δ(p

−m

) H (p

) d

p =

2E( p)

which is the usual way of writing that for any sufﬁciently smooth function (indicate precisely

the necessary regularity) we have

δ(p

−m

) H (p

) f (p) d

p =



E( p), p



2E( p)

The important point to remember is the following:

THEOREM 8.36 The differential element d

p/2E, whe re E

= p

+ m

, is Lorentz invariant.

 Exercise 8.10 Let S denote the sphere centered at 0 with radius R. Can we write δ

the form δ(r −R), where r =

+ y

+ z

? Can we write it δ(r

− R

Convergence in DD

 Exercise 8.11 Let n ∈ N

∗

and α ∈ R. We consider the function

: R \πZ −→ R,

x 7−→

sin

πnx/2

sin

πx/2

Show that f

can be extended by continuit y to R. Compute

−1

(t) dt and show that the

result is independent of n.

Show that the sequence ( f

)

n∈N

tends to X in the sense of distributions.

 Exercise 8.12 Show that



x 7→







′

−−→ δ(x) [a → 0

Exercises 243

 Exercise 8.13 Let α be a real number. Show that, in the sense of distributions, we have

n∈Z

sin α |n| =

sin α

1 −cos α

Differentiation in higher dimensions

 Exercise 8.14 We co nsider a two-dimensional space-time setting. D eﬁne

E (x, t) =

1 if c t > |x|,

0 if c t < |x|,

or, in short-hand notation, E (x, t) = H (c t − x). Compute E in the sense of distributions,

where the d’Alembertian operator  is deﬁned by



def

∂

∂ t

−

∂

∂ x

 Exercise 8.15 I n the plane R

, we deﬁne th e function E(x, t) = H (x) H (t), and its asso-

ciated regular distribution, also denoted E. Compute, in the sense of distributions,

∂

∂ x ∂ t

and △E.

 Exercise 8.16 (The wages of fear) We model a truck and its suspension by a spring with

strength k carrying a mass m at one of its extremities. The wheels at the bottom part of the

spring are forced to move on an irregular road, the height of which with respect to a ﬁxed

origin is described by a function of the distance E(x). The h orizontal speed of the truck is

assumed to be constant equal to v, so that at any time t, the wheels are at height E(vt). Denote

by R(t) the height of the truck, compared to its height at rest (it is the response of the system).

i) Show that



+ ω



R(t) = ω

E(vt), where ω

ii) In order to generalize the problem, we assume that E is a distribution. What is then

the equation linking the response R(t) of the truck to a given road proﬁle E(x) ∈ D

′

where D

′

is th e space of distributions with support bounded on the left? Comment

on the choice of D

′

iii) Compute the response of the truck to the proﬁle of a road wit h the following irregu-

larities:

• a very narrow bump E(x) = δ(x);

• a ﬂat road E(x) = ℓ H (x);

• a wavy road E(x) = ℓ H (x) sin(x/λ).

Compute the amplitude of the movements of the truck. How does it depend on the

speed?

iv) In the movie The Wages of Fear by Henri-Georges Clouzot (1953, original title Le salaire

de la peur), a truck full of nitroglycerin must cross a very rough segment of road. One

of the characters, who knows the place, explains that the truck can only cross safely at

very low speed, or on the contrary at high speed, but not at moderate speed. Explain.

244 Distributions II

PROBLEM

 Problem 3 (Kirchhoff integrals) The wave equation in three dimensions, for a scalar quan-

tity ψ, is



△−

∂

∂ t



ψ( r, t) = 0,

where we denote r

def

= (x, y, z). Monochromatic waves with pulsation ω are given by

ψ( r, t) = ϕ( r) e

−iωt

1. Show that the complex amplitude ϕ th en satisﬁes t h e equation called the Helmoltz

equation:



△+ k



ϕ( r ) = 0, with k

def

Let S be an arbitrary closed surface. We are looking for a distribution Φ( r ) such that

Φ( r) =

ϕ( r ) inside S ,

0 outside S ,

and which is a solution in the se nse of distributions of the Helmoltz equation.

2. Show that differentiating in the sense of distribution leads to



△+ k



δ( r )

∗Φ( r ) = ϕ

( r) +



dϕ



( r ),

where n is the interior normal to S (the choi ces of orientation followed he re are those

in the book Principles of Optics by Born and Wolf [14]).

3. Show that a Green function of the operator [△ + k

], that is, its convolution inverse,

is given by

G( r) = −

4π

ikr

, where r

def

= kr k.

4. Deduce that

Φ( r) =

4π



ikq

−

ikq



dϕ



x with q = kr − xk.

5. Interpretation: the complex amplitude inside a close d surface S , with given limit con-

ditions, i s the same as that produced by a certain surface density of ﬁctive sources

located on S . This result is the basis of what is called the method of images in electro-

magnetism and it is, in the situation considered here, the founding principle of t h e

scalar theory of diffraction in optics.

A more complete version of Kirchhoff’s Theorem, for an arbitrary dependence on time,

is presented in the book of Born and Wolf [14, p. 420].

Possibly very difﬁcult to compute explicitly! Note that this computation requires knowing

both the amplitude ϕ and its normal derivative with respect to the surface. In some cases only

we may hope to exploit this exact formula, where symmetries yield additional information.

Solutions of exercises 245

SOLUTIONS

 Solution of exercise 7.2 on page 198

〈δ

′

(ax),ϕ(x)〉 =

|a|

′

(x), ϕ



E

= −

a |a|

′

(0)

a |a|

〈δ

′

, ϕ〉.

 Solution of exercise 7.4 on page 211. Put h(x) =

f (t) g(x −t) dt and make the substitu-

tion y = x − t, with jacobian |dy/dt| = 1, which yields

h(x) =

f (x − y) g( y ) dy =

g( y) f (x − y) dy = [g ∗ f ](x).

 Solution of exercise 7.5 on page 211. A simple but boring computation shows that

−|ax|

∗e

−|b x|

− b

[a e

−b|x|

− b e

−a|x|

 Solution of exercise 8 .1 on page 229. Notice that sin x = Im(e

i x

) and that z 7→ e

i z

holomorphic on C and decays in the upper half-plane (at least in th e sense of Jordan’s second

lemma). Using then the Kramers-Kronig formula from Theorem 8.2, we obtain

∞

sin x

dx =

+∞

−∞

sin x

lim

ǫ→0

−ǫ

−∞

+∞

sin x

dx (continuous function)

+∞

−∞

Ime

i x

dx =

Re(e

) =

 Solution of exercise 8 .4. Let ϕ ∈ D. We then have



log |x|



′

, ϕ

= −〈log |x|,ϕ

′

(x)〉 = −

log |x|ϕ

′

(x) dx

= lim

ǫ→0+

−ǫ

−∞

+∞

+ǫ



−log |x|ϕ(x)



dx,

since x 7→ log |x| is locally integrable (the integral is well deﬁned around 0). Integrating each

of the two integrals by parts, we get



log |x|



′

, ϕ

= lim

ǫ→0

log(ǫ) ϕ(ǫ) −log(ǫ) ϕ(−ǫ) +

|x|>ǫ

ϕ(x)

= pv

+∞

−∞

ϕ(x)

dx,

which shows that



log |x|



′

= pv

 Solution of exercise 8 .5. First, we show that the stated limit does exist. Let ϕ ∈ D. Inte-

grating by parts twice, we get

+∞

ϕ(x)

dx = −

+∞

′′

(x) log x dx −ϕ

′

(ǫ) log ǫ +

ϕ(ǫ)

246 Distributions II

and therefore

+∞

ϕ(x)

dx −

ϕ(0)

+ ϕ

′

(0) log ǫ =

−

+∞

′′

(x) log x dx +



′

(0) −ϕ

′

(ǫ)



log ǫ +

ϕ(ǫ) −ϕ(0)

The second term tends to 0, and the third to ϕ

′

(0) as [ǫ → 0

]. The ﬁrst term is integrable

when ǫ = 0. Hence the expression given has a limit, which we denote by 〈T , ϕ〉.

The linearity is obvious, and hence we have to prove the continuity, and it is sufﬁcient to

show the continuity at 0 (the zero function of D).

Let (ϕ

)

n∈N

be a sequence of funct ions in D, co nverging to 0 in D. All the functions ϕ

have support included in a ﬁxed interval [−A, A], where we can assume for convenience and

without loss of generality that A > e. Then

〈T , ϕ

〉 = −

′′

(x) log x dx + ϕ

′

(0),

and hence



〈T , ϕ

〉



¶ kϕ

′′

∞

(A log A − A) + kϕ

′

∞

Choose ǫ > 0. There exi sts an integer N such that for n ¾ N we have kϕ

′′

∞

¶ ǫ/(A log A −

A) and kϕ

′

∞

¶ ǫ, so that |〈T , ϕ

〉| ¶ 2ǫ for any n ¾ N . This proves that 〈T ,ϕ

〉 tends to 0

as [n → ∞].

Since we have proved the continuity of T , this li near functional is indeed a distribution.

 Solution of exercise 8 .6. To generalize, deﬁne:

(−1)

n−1

(n −1)!

log |x| =

(−1)

n−1

(n −1)!

n−1

 Solution of exercise 8 .7. Let a 6= 0. Then for any ϕ ∈ D we have



δ(ax), ϕ



|a|



δ(x), ϕ(x/a)



|a|

ϕ(0) =

|a|

〈δ, ϕ〉.

For any a > 0, the functions x 7→ H (x) and x 7→ H (ax) are equal. Hence the associated

regular distributions are also equal. If a proof by substitution really is required, simply compute,

for any test function ϕ



H (ax), ϕ



|a|



H , ϕ(x/a)



|a|

+∞

ϕ(x/a) dx =

+∞

ϕ(x) dx = 〈H , ϕ〉.

For a < 0, the reader can check that H (ax) = 1 − H (x).

 Solution of exercise 8 .8. For any test function ϕ ∈ D we have

〈xδ

′

, ϕ〉 = 〈δ

′

, xϕ〉 = −[xϕ]

′

(0) = −ϕ(0) −0 ·ϕ

′

(0) = −〈δ, ϕ〉.

Similarly, by letting it act on any test function ϕ, one shows that xδ

′′

= −2δ

′

. Note that this

relation can also be recovered directly by differentiating the relation xδ

′

= −δ, which yields

′

+ xδ

′′

= −δ

′

and thus xδ

′′

= −2δ

′

If f is a locally integrable functio n of C

∞

class, we can deﬁne the product f δ

′

. By the

same method as before, we ﬁnd that

f (x) δ

′

(x) = f (0 ) δ

′

− f

′

(0) δ.

The assumptions can be weakened by asking only that f be continuous and differentiable

at 0. Without a more complete theory of distributions, the relation above can be taken as a

deﬁnition for f (x) δ

′

(x) in this case.