FOURIER METHODS AND DISTRIBUTIONS MATP16 PARTIAL ... · FOURIER METHODS AND DISTRIBUTIONS MATP16 PARTIAL DIFFERENTIAL EQUATIONS SPRING 2016 ... and recall some facts about Fourier

FOURIER METHODS AND DISTRIBUTIONSMATP16 PARTIAL DIFFERENTIAL EQUATIONS

SPRING 2016

Erik Wahlén

Copyright c© 2016 Erik Wahlén

Preface

The following notes are intended to serve as a complement to Evans’ book [3]. In Chapter1 we discuss separation of variables and recall some facts about Fourier series. In Chapter 2we discuss the action of the Fourier transform on the Schwartz space S (Rn), a topic which isnot discussed by Evans. One of the main advantages of using the Schwartz space comparedto Lp spaces is that we don’t have to worry about justifying different operations. Schwartzfunctions are infinitely differentiable and decay rapidly, as do their Fourier transforms. Thismeans that we can differentiate under the integral sign as many times as we like. In Chapter3 we then go on to discuss the notion of distributions. Distributions are a generalization offunctions with the advantage that they are always infinitely differentiable. They are rigorouslydefined as continuous linear functionals on a space of test functions. While the space C∞

c (Rn)of smooth, compactly supported functions is the most commonly used space of test functions,it turns out that S (Rn) is more convenient if one wants to define the Fourier transform ofa distribution. We also discuss Sobolev spaces of negative index. Finally, in Chapter 4 wediscuss applications to PDE. In particular, we discuss how to interpret fundamental solutionsin the context of distribution theory. At the end of the notes is a short collection of exercises.I have tried to be as consistent with the notation in Evans book as possible, but in some placesthe notation differs. In particular, the Fourier variable will be denoted ξ instead of y.

3

Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1 Separation of variables and Fourier series 71.1 Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 Applications to PDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2 The Fourier transform 172.1 The Schwartz space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 The Fourier transform on S . . . . . . . . . . . . . . . . . . . . . . . . . . 192.3 * Additional notes on the Fourier transform on Lp . . . . . . . . . . . . . . . 21

3 Distribution theory 233.1 Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.2 Operations on distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.3 Distributional solutions of PDE . . . . . . . . . . . . . . . . . . . . . . . . . 273.4 Tempered distributions and the Fourier transform . . . . . . . . . . . . . . . 283.5 * Approximation with smooth functions . . . . . . . . . . . . . . . . . . . . 323.6 * Sobolev spaces with negative index . . . . . . . . . . . . . . . . . . . . . . 33

4 Linear PDE with constant coefficients 354.1 Classification of PDE and the symbol of a partial differential operator . . . . 354.2 Fundamental solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5 Exercises 45

5

6

Chapter 1

Separation of variables and Fourier series

1.1 Fourier seriesWe recall here some basic facts about Fourier series. For the proofs we refer to any of the text-books Dym and McKean [2], Folland [4], Stein and Shakarchi [10], Vretblad [12] or Zygmund[13], although the results are not always stated in precisely the same form as here.

In this section we will consider functions u defined on R which are periodic with period2π , that is u(x+2π) = u(x) for all x∈R. We can also think of such functions as defined just onthe interval [−π,π] and then extended periodically. Therefore we will e.g. speak of functionsu∈ L1(−π,π) which are 2π-periodic. Note that such a function will belong to L1

loc(R), but notL1(R) unless it is identically zero (or rather, zero almost everywhere). A third way of thinkingof a 2π-periodic function is as a function defined on the unit circle. The argument x in u(x) isthen simply the angle.

The main goal of this section is to discuss under what conditions a 2π-periodic functioncan be expanded in a Fourier series

u(x) =∞

∑k=−∞

ukeikx. (1.1)

The conditions depend on which meaning we give to the equality sign. The Fourier coefficientsuk are defined by the formula

uk =1

2π

∫π

−π

u(x)e−ikx dx. (1.2)

One can show that if the identity (1.1) is to hold with any reasonable definition of convergence,then uk must be given by (1.2). To see this, begin by equipping the space L2(−π,π) with theinner product

(u,v) =1

2π

∫π

−π

u(x)v(x)dx.

Recall that L2(−π,π) is a Hilbert space with this inner product. Using this inner product, (1.2)can be rewritten as uk = (u,eikx). Straightforward calculations prove the following result.

7

8 Separation of variables and Fourier series

Lemma 1.1. The functions eikx, k ∈ Z, form an orthonormal set in L2(−π,π), that is,

(ei jx,eikx) =

{1, j = k,0, j 6= k.

Assume now that

limN→∞

N

∑k=−N

ckeikx = u

in L2(−π,π) for some sequence {ck}∞k=−∞

. Then

c j = (N

∑k=−N

ckeikx,ei jx)→ (u,e j)

as N→ ∞, so that c j = (u,ei jx) = u j.

In fact, we can say much more.

Theorem 1.2. {eikx}∞k=−∞

is an orthonormal basis for L2(−π,π). That is, in addition to itbeing an orthonormal set, the identity (1.1) holds with convergence in L2(−π,π) for everyu ∈ L2(−π,π). Moreover, Parseval’s formula

12π

∫π

−π

u(x)v(x)dx =∞

∑k=−∞

ukvk

holds for each u,v ∈ L2(−π,π).

This result is sufficient for many purposes. However, one sometimes wants to know if(1.1) holds pointwise or even uniformly. The following results can then be used.

Theorem 1.3. Assume that u∈ L1(−π,π) is 2π-periodic. Assume also that the limits from theleft u(a−) and right u(a+) of u exist at x = a and that the right and left derivatives, u′(a+) =limh→0+(u(a+h)−u(a+))/h and u′(a−) = limh→0−(u(a+h)−u(a−))/h exist. Then

limN→∞

N

∑k=−N

ukeika =u(a+)+u(a−)

2.

In particular, the (symmetric) Fourier series converges to u(a) at x = a if u is continuousat x = a and has derivatives from the right and left there.

Recall that a function u is called locally absolutely continuous if it can be written in theform

u(x) = c+∫ x

0v(y)dy,

for some function v ∈ L1loc(R). Such a function is automatically continuous on R and has

derivative v for almost every x. In particular, any continuously differentiable function is locallyabsolutely continuous. So is any piecewise continuously differentiable function.

Separation of variables and Fourier series 9

Theorem 1.4. Assume that u is 2π-periodic and locally absolutely continuous with derivativein L2(−π,π). Then the Fourier series ∑

∞k=−∞

ukeikx converges to u uniformly on R. Moreover,the Fourier coefficients satisfy ∑k∈Z |k|2|uk|2 < ∞.

In fact, the class of functions appearing in this theorem can be identified with

H1per(R) = {u ∈ H1

loc(R) : u(x+2π) = u(x)},

where the Sobolev space H1(U) is defined in Evans [3, Chapter 5] and ‘loc’ means that u ∈H1(U) for any open, bounded subset of R. For the interested reader, we reveal that H1(U)consists of functions u ∈ L2(U) whose distributional (or weak) derivative lies in L2(U) (seeChapter 3 below). Note that we cannot use H1(−π,π) instead of H1

per(R) here since theperiodic extension of a function in H1(−π,π) need not be locally absolutely continuous onR (it will be absolutely continuous on each interval (−(2m− 1)π,(2m+ 1)π), m ∈ Z). Thenotations H1(R/(2πZ)) and H1(T) are sometimes used instead of H1

per(R), when one thinksof the functions as being defined on the unit circle (which can be seen as a one-dimensionaltorus).

We also record some useful properties of Fourier series. Here we use the notation F (u)k :=uk.

Theorem 1.5. Assume that u,v ∈ L1(−π,π) are 2π-periodic. Then for every j,k ∈ Z anda ∈ R, we have that

(1) F (ei jxu(x))k = uk− j,

(2) F (u)k = u−k,

(3) F (u(−x))k = u−k,

(4) F (u(x+a))k = eikauk,

(5) F (u′)k = ikuk, if u is locally absolutely continuous,

(6) F (u∗ v)k = ukvk, where (u∗ v)(x) = 12π

∫π

−πu(x− y)v(y)dy.

In many applications, one encounters other types of Fourier series. First of all, note thatwe can use Euler’s identities to express eikx as cos(kx)+ i sin(kx) and obtain the expansion

u(x) =a0

2+

∞

∑k=1

(ak cos(kx)+bk sin(kx))

withak =

1π

∫π

−π

u(x)cos(kx)dx

andbk =

1π

∫π

−π

u(x)sin(kx)dx.


This has the advantage that ak and bk are real if u is real.We can also obtain pure sine and cosine series on the interval (0,π) by using even and odd

extensions. If u is originally defined on (0,π) and we extend it to an odd function on (−π,π)by setting u(x) = −u(−x) for x ∈ (−π,0), and then extend it periodically to R, then in theabove formulas we see that ak = 0 for all k. Hence, if e.g. u ∈ L2(0,π), we obtain that

u(x) =∞

∑k=1

bk sin(kx) (1.3)

with convergence in L2(0,π), where we can now compute bk using the formula

bk =2π

∫π

0u(x)sin(kx)dx

since the extended function is odd. Similarly, by doing an even extension, we obtain that

u(x) =a0

2+

∞

∑k=1

ak cos(kx)

in L2(0,π), where

ak =2π

∫π

0u(x)cos(kx)dx.

Whether or not these series converge pointwise/uniformly of course depends on whetherthe so extended functions satisfy the conditions for pointwise/uniform convergence. E.g. ifu ∈ C1[0,π] and u(0) = u(π) = 0, then the series (1.3) converges uniformly on [0,π] (anduniformly to the extended function on R).

We also remark that the choice of the intervals (−π,π) and (0,π) was just for convenience.One obtains analogous results for the interval (−`,`), ` > 0, by using the functions eikxπ/` andfor the interval (0, `) by using the functions cos(kxπ/`), sin(kxπ/`). The precise formulaswill of course depend on `.

Finally, one can consider Fourier series of functions of several variables, which are pe-riodic in each variable. We can think of these as being defined on the n-dimensional torusTn which can be identified with (R/(2πZ))n = Rn/(2πZ)n. It’s not difficult to show that thefunctions eik·x = eik1xeik2x · · ·eiknx, with k ∈ Zn, form an orthonormal basis for L2(Tn). Hence,the L2 theory is precisely the same as in one dimension. On the other hand, the conditions forpointwise and uniform convergence become more and more restrictive with increasing n.

1.2 Applications to PDEIn this section we discuss how we can use separation of variables together with the resultsin the previous section to solve linear PDE with constant coefficients (the method also worksoccasionally in other situations — see Section 4.1 in Evans [3]).


Consider a vibrating string described by the wave equation

utt(x, t) = c2uxx (1.4)

(u describes the transverse motion). We assume that the string has length ` in its equilibriumsituation and that it is then parametrized by arc length. Without loss of generality, we canassume that u : [0, `]→R. We also assume that the string is held fixed at the endpoints so thatu satisfies the boundary conditions

u(0, t) = u(`, t) = 0, (1.5)

for t > 0. The initial conditions are {u(x,0) = g(x),ut(x,0) = h(x),

(1.6)

Separation of variables works as follows. We first make an Ansatz for a solution of theequation and the boundary conditions. We find an infinite family of solutions using thisAnsatz. We then try to find a solution of the whole initial/boundary-value problem by forminga linear combination of the previous solutions. The Ansatz that we make is u(x, t) = ϕ(x)ψ(t),that is, u is a product of a function which only depends on x and another function which onlydepends on t (we have ‘separated the variables’). Substituting this into the wave equation, wefind that

ϕ(x)ψ ′′(t) = c2ϕ′′(x)ψ(t).

For x and t such that ϕ(x)ψ(t) 6= 0 we obtain that

ϕ ′′(x)ϕ(x)

= c−2 ψ ′′(t)ψ(t)

.

Since the left hand side is independent of t and the right hand side is independent of x, thewhole expression must in fact be independent of both x and t. In other words, it equals someconstant −λ . Thus,

ψ′′(t) =−λc2

ψ(t)

while {−ϕ

′′(x) = λϕ(x), 0≤ x≤ `,

ϕ(0) = ϕ(`) = 0,(1.7)

where we have also used the boundary conditions on u. Problem (1.7) has a non-trivial solutionif and only if λ = (kπ/`)2, k = 1,2,3, . . ., in which case

ϕk(x) = Bk sin(

kπx`

).

(check this!). Moreover, we find that

ψk(t) = ak cos(

kπct`

)+bk sin

(kπct`

).


Hence, we have obtained the solutions

u(x, t) =(

ak cos(

kπct`

)+bk sin

(kπct`

))sin(

kπx`

), k = 1,2,3, . . .

through this Ansatz (note that Bk can be absorbed into ak and ck). Note that this has the formof a standing wave which is time-periodic with frequency kc/(2`). Although we assumed thatϕ(x)ψ(t) 6= 0 in the above derivation, one can verify directly that the formula gives a solutionof the wave equation satisfying the boundary conditions.

In order to satisfy the initial conditions (1.6), we make an Ansatz in the form of a linearcombination of standing waves:

u(x, t) =∞

∑k=1

(ak cos

(kπct`

)+bk sin

(kπct`

))sin(

kπx`

). (1.8)

We then obtain that

g(x) = u(x,0) =∞

∑k=1

ak sin(

kπx`

).

If we assume that series for u can be differentiated termwise, we obtain that

h(x) = ut(x,0) =∞

∑k=1

kπcbk

`sin(

kπx`

).

Thus, by expanding g and h in sine series, we can determine the constants ak and bk.

If g and h are sufficiently regular, one can differentiate the series termwise, in whichcase u is a solution of (1.4)–(1.6). Assume for example that g is C3 and h is C2 and thatg(0) = g′′(0) = h(0) = h′′(0) = g(`) = g′′(`) = h(`) = h′′(`) = 0. Then one finds by extendingg first to an odd function on (−`,`) and then to a periodic function on R, that the sine seriesfor g converges uniformly and that one can differentiate it termwise twice and still get uni-form convergence. Similarly, the series for h converges uniformly and can be differentiatedtermwise once. In fact, by integrating by parts in the formulas for ak and bk, one obtains that

∞

∑k=1

k6(|ak|2 + |bk|2)< ∞

and the tail of the sine series for g′′(x) can then be estimated by

∞

∑k=N+1

(kπ

`

)2

|ak| ≤(

π

`

)2(

∞

∑k=N+1

k6|ak|2)1/2(

∞

∑k=N+1

k−2

)1/2

→ 0

as N→∞ (the other series can be estimated similarly). The series for u(x, t) can be estimated ina similar way to show that u is C2 and solves (1.4)–(1.6). It follows from Section 2.4 in Evans[3] that the series solution (1.8) is the only solution of the problem. Note that we actuallyneeded a bit more regularity than was used to derive d’Alembert’s formula (cf. Theorem 1 inEvans [3, Section 2.4]).

We also consider an explicit example, which doesn’t satisfy the above regularity condi-tions.


Example 1.6. A string of length π (in equilibrium) is fixed at the ends. We wish to determinethe position of the string u(x, t) if ut(x,0)≡ 0 and u(x,0) = g(x), where

g(x) =

{x, 0≤ x≤ π/2,π− x, π/2 < x≤ π.

x

y

π

In order to do this, we expand g in a sine series,

g(x) =∞

∑k=1

ak sin(kx),

where

ak =2π

∫π

0u(x)sin(kx)dx =

{0, k even,4(−1) j

πk2 , k = 2 j+1, odd.

Since h(x) = 0 we obtain that

u(x, t) =∞

∑k=1

ak cos(kct)sin(kx) =4π

∞

∑j=0

(−1) j

(2 j+1)2 cos((2 j+1)ct)sin((2 j+1)x). (1.9)

One can show that this defines a distributional solution of the wave equation (see Exercise 8).

We also consider the heat equation equation with Dirichlet boundary conditions (corre-sponding to fixed temperature) in an interval of length π

ut = uxx,

u(0, t) = u(π, t) = 0,u(x,0) = g(x).

(1.10)

Making the Ansatz u(x, t) = ϕ(x)ψ(t) we see that u solves the heat equation and satisfies theboundary conditions precisely if

ψ′(t) =−λψ(t),


and {−ϕ

′′(x) = λϕ(x), 0≤ x≤ π,

ϕ(0) = ϕ(π) = 0,(1.11)

for some λ . The only non-trivial solutions are obtained for λ = k2, k = 1,2, . . .. These are

u(x, t) = bke−k2t sin(kx).

We make an Ansatz for the solution of (1.13) in the form

u(x, t) =∞

∑k=1

bke−k2t sin(kx), (1.12)

where

g(x) =∞

∑k=1

bk sin(kx),

so that bk are the coefficients in the sine series for g. If e.g. g ∈ C1[0,π] with g(0) = g(π),the expansion of g converges uniformly. One can then check that u ∈C∞(R× (0,∞))∩C(R×[0,∞)) (note that ∑

∞k=1 |k|2|bk|2 < ∞, so that in particular bk = o(k) as k→ ∞). Moreover, the

solution is unique by Evans [3, Section 2.3, Theorem 5]. Note finally that

|u(x, t)| ≤∞

∑k=1|bk|e−k2t ≤ e−t

∞

∑k=1|bk| ≤ e−t

(∞

∑k=1

k2|bk|2)1/2(

∞

∑k=1

1k2

)1/2

so that the solution converges to zero exponentially fast as t → ∞. This makes sense in viewof that that the endpoints are kept fixed at zero temperature.

Example 1.7. We solve the problem

ut = uxx, 0 < x < π, t > 0,u(0, t) = u(π, t) = 0, t > 0,u(x,0) = sinxcos(4x), 0 < x < π.

The solution has the form

u(x, t) =∞

∑k=1

bke−k2t sin(kx),

where bk are the coefficients in the sine series for g(x) = sinxcos(4x). We find that

sinxcos(4x) =12(sin(x+4x)+ sin(x−4x)) =

12(sin(5x)− sin(3x)).

Hence, b3 =−1/2, b5 = 1/2 and bk = 0 for all other k. We obtain that

u(x, t) =−12

e−9t sin(3x)+12

e−25t sin(5x).


We also present a different way of viewing the separation of variables method. To be con-crete, we consider the heat equation on the interval (0,π) with Neumann boundary conditions

ut = uxx, 0≤ x≤ π,

ux(0, t) = ux(π, t) = 0,u(x,0) = g(x).

(1.13)

Physically we can e.g. think of this as modeling diffusion of a substrate in thin pipe which isclosed at the ends. Instead of using separation of variables, we use directly that a sufficientlynice solution can be expanded in a cosine series for each t, that is

u(x, t) =a0(t)

2+

∞

∑k=1

ak(t)cos(kx).

Plugging this into the equation and differentiating termwise, we formally find that

a′k(t) =−k2ak(t)

for each k. Alternatively, one differentiate under the integral sign in the formula

ak(t) =2π

∫π

0u(x, t)cos(kx)dx,

use the heat equation and integrate by parts to derive the same equation (do this!). In eithercase, we find that

ak(t) = ak(0)e−k2t ,

so that

u(x, t) =a0(0)

2+

∞

∑k=1

ak(0)e−k2t cos(kx),

where ak(0) are the cosine coefficients of g. Arguing as in the Dirichlet case, we obtain asmooth solution on R× (0,∞) which is continuous on R× [0,∞) if g ∈ C1[0,π]. Note thatwe don’t have to require that g satisfies the boundary conditions! These will automatically besatisfied by u for t > 0 anyway (however, if we want the solution to be C1 up to t = 0, we doneed to impose the boundary conditions). Uniqueness is discussed in Exercise 4. Note that

limt→∞

u(x, t) =a0(0)

2=

1π

∫π

0g(x)dx.

in this case.

Finally, we remark that the above method can also be used in higher dimensions. E.g. ifwe wish to solve the heat equation in a bounded domain U ⊂Rn with Dirichlet conditions, weobtain by separation of variables the two problems

ψ′(t) =−λψ(t),


and {−∆ϕ = λϕ in U,

ϕ = 0 on ∂U.(1.14)

One can show that there is an increasing sequence of eigenvalues {λk}∞k=0, repeated according

to multiplicity, for which the latter problem has a nontrivial solution, and associated eigen-functions {ϕk}∞

k=0. Using these instead of the sine and cosine functions, we obtain (at leastformally) the solution

u(x, t) =∞

∑k=0

ake−λktϕk(x),

where the ak are determined by expanding the initial datum g using the eigenfunctions. SeeSections 4.1.1 and 6.5 in Evans [3] for more information. In the case of periodic boundaryconditions, one can use the multivariable Fourier series mentioned above.

Chapter 2

The Fourier transform

2.1 The Schwartz spaceIn Evans [3, Section 4.3.1] the Fourier transform is defined and studied on L1(Rn) and L2(Rn).Here we will instead study it on a class of very nice functions. This means that we don’t have toworry much about technical details when discussing derivatives. All functions in this chapterare assumed to be complex-valued.

Definition 2.1 (Schwartz space). The Schwartz space1 S (Rn) is the vector space of rapidlydecreasing smooth functions,

S (Rn) :={

f ∈C∞(Rn) : supx∈Rn|xαDβ

x f (x)|< ∞ for all α,β ∈ Nn0

}.

S (Rn) contains the class of C∞ functions with compact support, C∞c (Rn). The main reason

for using S instead of C∞c is that, as we will see, the former class is invariant under the Fourier

transform. This is not the case for C∞c .

The topology on S (Rn) is defined by the family of semi-norms

‖ f‖α,β := supx∈Rn|xαDβ

x f (x)|, α,β ∈ Nn0.

In other words, we say thatϕk→ ϕ

in S (Rn) iflimk→∞‖ϕk−ϕ‖α,β = 0

for all α,β ∈ Nn0. One can in fact show that

ρ( f ,g) := ∑α,β

12|α|+|β |

‖ f −g‖α,β

1+‖ f −g‖α,β,

1Named after the French mathematician Laurent Schwartz (1915–2002).

17

18 The Fourier transform

defines a metric on S (Rn). The topology defined by this metric is the same as the one definedby the semi-norms and the metric space (S (Rn),ρ) is complete. S (Rn) is a so-called Frechétspace, but we will not use this in any significant way. Notice, though, that the Schwartz spaceis not a Banach space, i.e. it cannot be equipped with a norm making it complete (in particular,f 7→ ρ( f ,0) does not define a norm).

Proposition 2.2.

(1) For all N ≥ 0 and α ∈Nn0 there exists a C > 0 such that |Dα

x f (x)| ≤C(1+ |x|)−N , x∈Rn.

(2) S (Rn) is closed under multiplication by polynomials.

(3) S (Rn) is closed under differentiation.

(4) S (Rn) is an algebra: f g ∈S (Rn) if f ,g ∈S (Rn).

(5) C∞c (Rn) is a dense subspace of S (Rn).

(6) S (Rn) is a dense subspace of Lp(Rn) for 1≤ p < ∞.

Proof. The first property is basically a restatement of the definition of S (Rn). Properties (2)and (3) follow from the definition of the Schwartz space. Property (4) is a consequence ofLeibniz’ formula for derivatives of products.

It is clear that C∞c (Rn) ⊂ S (Rn). To show that it is dense, let f ∈ S (Rn) and define

fk(x) = ϕ(x/k) f (x), where ϕ ∈C∞c (Rn) with ϕ(x) = 1 for |x| ≤ 1 (see Appendix C.5 in Evans

[3]). Thenfk− f = (ϕ(x/k)−1) f (x)

with support in Rn \B0(0,k). From property (1) it follows that | f (x)| ≤CN(1+ |x|)−N for anyN ≥ 0. Therefore

‖ fk− f‖L∞(Rn) ≤CN(1+‖ϕ‖L∞(Rn))(1+ k)−N → 0

as k→∞. Since N is arbitrary, the same is true if we multiply fk− f by a polynomial. Since thepartial derivatives of ϕ(x/k)−1 all have support in Rn \B0(0,k) and are bounded uniformlyin k, the convergence remains true after differentiation. This proves (5).

By estimating| f (x)| ≤C(1+ |x|)−n−1

and using the fact that∫Rn(1+ |x|)−p(n+1) dx =C

∫∞

0

rn−1

1+ rp(n+1)dr < ∞,

it follows that f ∈ Lp(Rn) for 1 ≤ p < ∞ if f ∈S (Rn). Since C∞c (Rn) is dense in Lp(Rn),

property (6) now follows from property (5) (we only use the part that C∞c (Rn)⊂S (Rn)).

Example 2.3. The Gaussian function e−|x|2

is an example of a function in S (Rn) which isnot in C∞

c (Rn).

The Fourier transform 19

2.2 The Fourier transform on S

Since S (Rn)⊂ L1(Rn), the Fourier transform f , of a function f ∈S (Rn) is defined just likein Evans [3, Section 4.3.1],

f (ξ ) :=1

(2π)n/2

∫Rn

f (x)e−ix·ξ dx (2.1)

while the inverse Fourier transform is defined by

f (x) := f (−x) =1

(2π)n/2

∫Rn

f (ξ )eix·ξ dξ . (2.2)

We also use the notation f = F ( f ) and f = F−1( f ).

Proposition 2.4. For f ∈S (Rn), we have that f ∈C∞(Rn) and

f (x+ y) F7→ eiy·ξ f (ξ ),

eiy·x f (x) F7→ f (ξ − y),

f (Ax) F7→ |detA|−1 f (A−Tξ )

f (λx) F7→ |λ |−n f (λ−1ξ ),

Dαx f (x) F7→ (iξ )α f (ξ ),

xα f (x) F7→ (i∂ξ )α f (ξ ),

where y ∈ Rn, λ ∈ R\{0} and A is an invertible real n×n matrix.

Proof. The first four properties follow by changing variables in the integral in (2.1). The fifthproperty follows by partial integration and the sixth by differentiation under the integral sign.This also proves that f ∈C∞(Rn).

Example 2.5. Let us compute the Fourier transform of the Gaussian function e−|x|2/2, x ∈

Rn (see also Evans [3, Section 4.3, Example 1]). We begin by considering the case n = 1.The function u(x) = e−x2/2 satisfies the differential equation u′(x) = −xu(x). Taking Fouriertransforms we therefore find that iξ u(ξ ) =−iu′(ξ ), or equivalently u′(ξ ) =−ξ u(ξ ). But thismeans that

u(ξ ) =Ce−ξ 22 ,

whereC = u(0) =

1√2π

∫R

e−x22 dx = 1.

Hence,

F (e−x22 )(ξ ) = e−

ξ 22 ,


that is, u is invariant under the Fourier transform!This can be generalized to Rn, since

e−|x|2/2 =

n

∏j=1

e−x2j/2, x = (x1, . . . ,xn) ∈ Rn.

From this it follows that

F (e−|x|2/2)(ξ ) =

n

∏j=1

F (e−x2j/2)(ξ j) =

n

∏j=1

e−ξ 2j /2 = e−|ξ |

2/2.

Note in particular that the Fourier transform of a Gaussian function is in S (Rn). This isno coincidence.

Theorem 2.6. For any f ∈S (Rn), F ( f ) ∈S (Rn), and F is a continuous transformationof S (Rn) to itself.

Proof. We estimate the S (Rn) semi-norms of f as follows:

‖ f‖α,β = ‖ξ αDβ

ξf‖L∞(Rn)

= ‖F (Dαx (x

β f ))‖L∞(Rn)

≤ (2π)−n/2‖Dαx (x

β f )‖L1(Rn)

= (2π)−n/2‖(1+ |x|)−(n+1)(1+ |x|)n+1Dαx (x

β f )‖L1(Rn)

≤C‖(1+ |x|)n+1Dαx (x

β f )‖L∞(Rn).

The last expression can be estimated by a finite number of semi-norms of f using Leibniz’formula. It follows that F maps S (Rn) to itself continuously.

Theorem 2.7 (Fourier’s inversion formula). The Fourier transform is an isomorphism onS (Rn) with inverse equal to F−1. That is,

F−1F = FF−1 = Id on S (Rn).

Proof. We have already shown that F maps S (Rn) to itself. The identity F−1F followsfrom Evans [3, Section 4.3, Theorem 2 (iv)]. We must also show that FF−1 = Id. Definethe reflection operator R : S (Rn)→S (Rn) by R( f )(x) = f (−x). R commutes with F andF−1 (Proposition 2.4), is its own inverse and satisfies F−1 = RF . It follows that

FF−1 = FRF = F−1F = Id .

Recall that the convolution of two functions f and g is defined by

( f ∗g)(x) :=∫Rn

f (x− y)g(y)dy, (2.3)

whenever this makes sense. Since S (Rn)⊂ L1(Rn)∩L2(Rn), we find that f ∗g = (2π)n/2 f gby Evans [3, Section 4.3, Theorem 2 (iii)]. Since S (Rn) is an algebra and F−1 preservesS (Rn), the following result is now obvious.

Corollary 2.8. Let f ,g ∈S (Rn). Then f ∗g ∈S (Rn).

The Fourier transform 21

2.3 * Additional notes on the Fourier transform on Lp

Evans discusses the Fourier transform on L1 and L2, but we mention two additional resultshere.

Theorem 2.9 (Riemann-Lebesgue lemma). Let f ∈ L1(Rn). Then, f ∈ L∞(Rn)∩C(Rn),

‖ f‖L∞(Rn) ≤ (2π)−n/2‖ f‖L1(Rn), (2.4)

and limx→∞ f (x) = 0.

Proof. The estimate (2.4) follows directly from the definition. To prove that f (x)→ 0 asx→ ∞, we take ε > 0 and g ∈ S (Rn) with ‖ f − g‖L∞(Rn) ≤ (2π)−n/2‖ f − g‖L1(Rn) < ε/2.Choosing R large enough, we find that |g(x)|< ε/2 for |x| ≥ R. Hence,

| f (x)| ≤ ‖ f − g‖L∞(Rn)+ |g(x)|< ε

for |x| ≥ R.

Since the Fourier transform is defined for L1(Rn) and L2(Rn), one can in fact extend itto Lp(Rn), 1 ≤ p ≤ 2, by a method called interpolation. The following result is a directconsequence of the Riesz-Thorin interpolation theorem (see e.g. Folland [5, Theorem 6.27]).

Theorem 2.10 (Hausdorff-Young inequality). Let p∈ [1,2] and q∈ [2,∞] with 1p +

1q = 1. The

Fourier transform extends uniquely to a bounded linear operator F : Lp(Rn)→ Lq(Rn), with

‖F ( f )‖Lq(Rn) ≤ (2π)d2

(1− 2

p

)‖ f‖Lp(Rn), f ∈ Lp(Rn).

The extension to Lp(Rn) for p > 2 is a completely different matter. We will return to thisin Chapter 3.


Chapter 3

Distribution theory

The notion of weak solutions is an indispensable tool in the theory of partial differential equa-tions. The key is to multiply the equation by a ‘test function’ ϕ ∈ C∞

c (Rn) and integrate byparts to put all the derivatives on ϕ . The name test function refers to the fact that the equationis ‘tested’ against ϕ . This gives a suitable definition of weak solutions in which the solutionu doesn’t need to be differentiable. If u is a sufficiently regular weak solution, one shows thatit is a classical solution, meaning that the PDE holds pointwise. What is involved here is theconcept of ‘weak’ or ‘distributional’ derivatives. The weak derivative of a non-differentiablefunction is in general not a function, but a distribution. The purpose of this chapter is partlyto clarify what this means. Distributions or ‘generalized functions’ have a long prehistory.The theory was made rigorous in the middle of the 20th century, in particular by LaurentSchwartz. Distributions are central to the modern theory of linear partial differential equa-tions. This chapter is only a brief introduction; for more details we refer to Friedlander [6],Duistermaat and Kolk [1], Hörmander [7], Rauch [9] and Strichartz [11].

3.1 DistributionsAny locally integrable function u∈ L1

loc(Rn) gives rise to a linear functional1 ù : C∞c (Rn)→C

defined by ù(ϕ) = 〈u,ϕ〉, where

〈u,ϕ〉 :=∫Rn

u(x)ϕ(x)dx. (3.1)

Note that this differs from the inner product in L2(Rn) in that there is no complex conjugateon ϕ and that we don’t require u,ϕ ∈ L2(Rn). It is clear at least if u is continuous that ùdetermines u uniquely, since if ù(ϕ) = 0 for all ϕ ∈ C∞

c (Rn) then u ≡ 0. Below we shallprove that this is also the case if u ∈ L1

loc(Rn).

In keeping with the notation of Laurent Schwartz, we denote the space of test functionsC∞

c (Rn) by D(Rn).

1A linear functional is just a linear map with values in C.

23

24 Distribution theory

Definition 3.1. A distribution ` is a linear functional on D(Rn) which is continuous in thefollowing sense. If {ϕk} ⊂D(Rn) has the following properties:

(1) there exists a compact set K ⊂ Rn such that suppϕk ⊆ K for all k, and

(2) there exists a function ϕ ∈D(Rn) such that for all α ∈ Nn0, Dαϕk→ Dαϕ uniformly,

then `(ϕk)→ `(ϕ) as k→ ∞. The space of distributions is denoted D ′(Rn).

Remark 3.2. D ′(Rn) is simply the space of continuous linear functionals on D(Rn) whenD(Rn) is given the topology defined by (1) and (2) above. When (1) and (2) hold, we say thatϕk→ ϕ in D(Rn). One can define D(U) and D ′(U) if U is an open subset of Rn in a similarfashion and much of the theory below also holds in this setting.

A function f ∈ L1loc(Rn) gives rise to a distribution through the formula (3.1). The fact that

this expression defines a distribution follows from the estimate

|〈u,ϕ〉| ≤(∫

K|u(x)|dx

)‖ϕ‖L∞(Rn),

if ϕ ∈D(Rn) with suppϕ ⊆K. Let us prove that (3.1) determines the distribution ù uniquely.By linearity it suffices to show that that u = 0 (as an element of L1

loc(Rn)) if ù = 0.

Theorem 3.3. Let u ∈ L1loc(Rn). If 〈u,ϕ〉= 0 for all ϕ ∈D(Rn), then u(x) = 0 a.e.

Proof. Let η be the standard mollifier as in Evans [3, Appendix C.5]. Then ηε ∗ u→ u inL1

loc(Rn) by Evans [3, Appendix C, Theorem 7]. On the other hand, since ηε(x−·) ∈D(Rn),it follows that ηε ∗u =

∫Rn u(y)ηε(x− y)dy = 0 for all ε . Hence, u = 0 in L1

loc(Rn).

The above theorem implies that we can identify u with the linear functional ù. We will dothis without further mention in what follows. We shall also use the notation 〈u,ϕ〉 for u(ϕ)even when u is just an element of D ′(Rn). Let us give an example of a distribution which doesnot correspond to a locally integrable function.

Example 3.4. The Dirac2 distribution δa at a ∈ Rn is defined by

〈δa,ϕ〉 := ϕ(a).

It is clear that 〈δa,ϕk〉 = ϕk(a)→ ϕ(a) = 〈δa,ϕ〉 if ϕk → ϕ in D(Rn), so that δa really is adistribution. It is also clear that δa doesn’t correspond to a locally integrable function, since〈δa,ϕ〉 = 0 if ϕ(a) = 0. If 〈δa,ϕ〉 = 〈u,ϕ〉 for some u ∈ L1

loc(Rn), this would imply thatu(x) = 0 a.e. and hence that 〈δa,ϕ〉 = 〈u,ϕ〉 = 0 for all ϕ ∈ D(Rn). The reader who isfamiliar with measure theory will notice that we can think of δa as a point measure in a (orrather as integration with respect to this measure).

We also need to know what is meant by convergence of a sequence of distributions.

2After the physicist Paul Dirac (1902–1984).

Distribution theory 25

Definition 3.5. We say that uk→ u in D ′(Rn) if 〈uk,ϕ〉 → 〈u,ϕ〉 for all ϕ ∈D(Rn)

Example 3.6. Consider the fundamental solution of the heat equation

Φ(x, t) =

1(2πt)n/2 e−

|x|24t , t > 0,

0, t < 0.

We have that

〈Φ(·, t),ϕ〉=∫Rn

Φ(x, t)ϕ(x)dx =∫Rn

Φ(0− x, t)ϕ(x)dx→ ϕ(0)

as t→ 0+ for any ϕ ∈D(Rn) by Evans [3, Section 2.3, Theorem 1 (iii)]. Hence, Φ(·, t)→ δ0 inD ′(Rn) as t→ 0+. The front cover shows Φ(x, t) at three different points in time. Intuitively,all the mass concentrates at the origin as t → 0+. This is one way of thinking of the Diracdistribution.

More generally, one can show that Kε → δ0 in D ′(Rn) if K ∈ L1(R) with∫Rn K(x)dx = 1

and Kε(x) := ε−nK(ε−1x), ε > 0, by a similar argument as for Φ (ε corresponds to√

t above).

3.2 Operations on distributionsWe now discuss how to extend various operations on functions to distributions. To explain themethod, we consider translations in x. Given a function ϕ ∈ D(Rn) and a vector h ∈ Rn, wedefine (τhϕ)(x) = ϕ(x−h). If ϕ,ψ ∈D(Rn) we have that

〈τhϕ,ψ〉=∫Rn

ϕ(x−h)ψ(x)dy =∫Rn

ϕ(y)ψ(y+h)dx = 〈ϕ,τ−hψ〉.

It therefore makes sense to define

〈τhu,ϕ〉 := 〈u,τ−hϕ〉 (3.2)

if u is just a distribution. Note that τ−hϕk → τ−hϕ if ϕk → ϕ in D(Rn), so that (3.2) indeeddefines a distribution τhu. It is clear that the new definition agrees with the old one if u ∈D(Rn) in the sense that the distribution τhu defined by (3.2) can be represented by the functionu(·−h).

Example 3.7. We have τhδ0 = δh, since

〈τhδ0,ϕ〉= 〈δ0,τ−hϕ〉= ϕ(0+h) = ϕ(h).

The above extension of the translation operator is an example of a general principle.

Proposition 3.8. Suppose that L is a linear operator on D(Rn) which is continuous in thesense that ϕk → ϕ in D(Rn) implies L(ϕk)→ L(ϕ). Suppose, moreover, that there exists acontinuous linear operator LT on D(Rn), with the property that 〈L(ϕ),ψ〉 = 〈ϕ,LT (ψ)〉 forall ϕ,ψ ∈D(Rn). Then L extends to a continuous linear operator on D ′(Rn) given by

〈L(u),ϕ〉= 〈u,LT (ϕ)〉 for all u ∈D ′(Rn) and ϕ ∈D(Rn). (3.3)


Proof. The continuity of LT shows that L(u) defined in (3.3) is a distribution. It is also clearfrom the definition of the transpose that (3.3) defines an extension of the original operator L.Finally, if uk→ u in D ′(Rn), then

〈L(uk),ϕ〉= 〈uk,LT (ϕ)〉 → 〈u,LT (ϕ)〉= 〈L(u),ϕ〉,

so the extension is continuous on D ′(Rn).

We call the operator LT the transpose of L. Note that this might differ from the L2 adjointsince there is no complex conjugate on ψ in 〈ϕ,ψ〉. We now give a list of common operatorsand their transposes. Here (σλ ϕ)(x) = ϕ(λx) is dilation by λ ∈ R \ {0} and (Mψϕ)(x) =ψ(x)ϕ(x) multiplication by ψ ∈C∞(Rn).

L LT

τh τ−hσλ |λ |−dσ1/λ

Mψ Mψ

Dα (−1)|α|Dα

The reader is encouraged to verify these formulas. As a corollary to Proposition 3.8 we there-fore have the following result.

Proposition 3.9. Each of the operators τh, σλ , Mψ and Dα on D(Rn) have extensions tocontinuous maps on D ′(Rn). The extensions are given by 〈Lu,ϕ〉= 〈u,LT ϕ〉 where L and LT

are as in the above table.

Note the restriction to multiplication with smooth functions. In general we can’t define theproduct of a distribution and a non-smooth function ψ since then ϕψ 6∈D(Rn). In particular,one can in general not define the product of two distributions.

Example 3.10. The Heaviside3 function

H(x) =

{1, x > 0,0, x≤ 0

is locally integrable and hence H ∈D ′(R). The distributional derivative of H is calculated asfollows:

〈H ′,ϕ〉=−〈H,ϕ ′〉=−∫

∞

0ϕ′(x)dx = ϕ(0) = 〈δ0,ϕ〉.

It follows that H ′ = δ0 ∈ D ′(R). Note that the pointwise derivative of H is zero everywhereexcept at 0, where it is undefined. We can continue the process and calculate higher derivativesof H, e.g.

〈H ′′,ϕ〉= 〈δ ′0,ϕ〉=−〈δ0,ϕ′〉=−ϕ

′(0).

3After the physicist Oliver Heaviside (1850–1925).


3.3 Distributional solutions of PDEWe can now define what we mean by a distributional solution of a PDE4. We restrict ourselvesto linear equations with constant coefficients. Suppose that the equation takes the form

P(D)u = f , (3.4)

whereP(D) = ∑

|α|≤maαDα

and f ∈D ′(Rn) is given. Then u∈D ′(Rn) is a distributional solution if (3.4) holds as an iden-tity in D ′(Rn). We also say that u solves the equation in the sense of distributions. Concretely,this means that

〈u,PT (D)ϕ〉= 〈 f ,ϕ〉

for every ϕ ∈D ′(Rn), where

PT (D) = P(−D) = ∑|α|≤m

aα(−1)|α|Dα

is the transpose of P(D). Note that if f is actually given by a continuous function and u by aCm function, then the distributional derivatives of u of order less than or equal to m are equalto the classical derivatives of u and the equation is satisfied in the usual sense.

Example 3.11. The Heaviside function H is a distributional solution of the equation

u′ = δ0

in D ′(R). One can show that the general solution is u = H(x)+C, where C is an arbitraryconstant.

Example 3.12. Let f ∈ L1loc(R) and set u(x, t) = f (x−t). Then u∈ L1

loc(R2) is a distributionalsolution of the one-dimensional wave equation. To see this, note that

〈∂ 2t u−∂

2x u,ϕ〉= 〈u,∂ 2

t ϕ−∂2x ϕ〉

=∫∫

R2f (x− t)(∂ 2

t ϕ(x, t)−∂2x ϕ(x, t))dxdt

for ϕ ∈D(R2). Making the change of variables u = x− t, v = x+ t in the last integral, we findthat f (x− t) = f (u), while ϕ(x, t) = ϕ(u+v

2 , v−u2 ) = ψ(u,v), with ψ ∈D(R2). Moreover,

∂2t ϕ = ∂

2u ψ +∂

2v ψ−2∂u∂vψ

and∂

2x ϕ = ∂

2u ψ +∂

2v ψ +2∂u∂vψ

4Distributional solutions are sometimes called weak solutions, but we will reserve that terminology for solu-tions which are actually given by functions.


so that∂

2t ϕ−∂

2x ϕ =−4∂u∂vψ.

Hence,

〈∂ 2t u−∂

2x u,ϕ〉=−2

∫∫R2

f (u)∂v∂uψ(u,v)dvdu

=−2∫R

f (u)(∫

R∂v∂uψ(u,v)dv

)︸︷︷︸

=0

du

= 0.

3.4 Tempered distributions and the Fourier transformOne operation which can not be defined on D ′(Rn) is the Fourier transform. As the followinglemma shows, F is its own transpose.

Lemma 3.13. For ϕ,ψ ∈S (Rn), we have that

〈F (ϕ),ψ〉= 〈ϕ,F (ψ)〉.

Proof. By Fubini’s theorem we have that

〈F (ϕ),ψ〉= (2π)−n/2∫Rn

(∫Rn

ϕ(x)e−ix·ξ dx)

ψ(ξ )dξ

= (2π)−n/2∫Rn

ϕ(x)(∫

Rnψ(ξ )e−ix·ξ dξ

)dx

= 〈ϕ,F (ψ)〉.

The problem with defining 〈F (u),ϕ〉= 〈u,F (ϕ)〉 is that F doesn’t map D(Rn) to itself(compact support is lost). The key is to instead use S (Rn) as the space of test functions.

Definition 3.14. A continuous linear functional on S (Rn) is called a tempered distribution.The space of tempered distributions is denoted S ′(Rn).

Convergence in S ′(Rn) is defined analogously to convergence in D ′(Rn).

Definition 3.15. We say that uk→ u in S ′(Rn) if uk(ϕ)→ u(ϕ) for all ϕ ∈S (Rn)

Continuity of an element u ∈S ′(Rn) means that

〈u,ϕk〉 → 〈u,ϕ〉

if ϕk→ ϕ in S (Rn). If u ∈S ′(Rn), then 〈u,ϕ〉 is well-defined if ϕ ∈ D(Rn). Moreover, ifϕk→ ϕ ∈ D(Rn), then clearly ϕk→ ϕ in the topology of S (Rn). Consequently, the restric-tion of a tempered distribution to D(Rn) is an element of D ′(Rn). Moreover, u is uniquely


determined by its restriction to D(Rn), since D(Rn) is a dense subspace of S (Rn) (Proposi-tion 2.2 (5)). It therefore makes sense to write S ′(Rn) ⊂ D ′(Rn). The inclusion is strict, inthe sense that there exist elements u ∈D ′(Rn) which cannot be extended to continuous linearfunctionals on S (Rn).

Example 3.16.

(1) The Dirac distribution δa belongs to S ′(Rn).

(2) Any locally integrable function u with (1+ | · |2)−Nu ∈ L1(Rn) for some N belongs toS ′(Rn) since∫

Rn|u(x)ϕ(x)|dx≤

∫Rn(1+ |x|2)−N |u(x)|(1+ |x|2)N |ϕ(x)|dx

≤ ‖(1+ | · |2)−Nu‖L1(Rn)‖(1+ | · |2)Nϕ‖L∞(Rn).

(3) In particular, any u ∈ Lp(Rn), p ∈ [1,∞], belongs to S ′(Rn) since (1 + | · |2)−Nu ∈L1(Rn) if N is sufficiently large by Hölder’s inequality.

(4) u(x) = e|x| is in D ′(Rn) but not in S ′(Rn). Strictly speaking this means that the func-tional ϕ 7→ 〈u,ϕ〉, ϕ ∈D(Rn), can not be extended to an element of S ′(Rn). This cane.g. be seen by approximating a function ϕ ∈S (Rn) satisfying ϕ(x) = e−|x|/2 for large|x| with a sequence in D(Rn) in the topology of S (Rn).

The proof of the following result is exactly the same as the proof of Proposition 3.8.

Proposition 3.17. The result in Proposition 3.8 remains true if D(Rn) is replaced by S (Rn)and D ′(Rn) by S ′(Rn).

As a consequence, Proposition 3.9 holds also for S ′(Rn) if we in the definition of themultiplication operator assume that ψ ∈S (Rn), so that ψϕ ∈S (Rn) if ϕ ∈S (Rn). Moregenerally, one can assume that ψ and its derivatives grow at most polynomially.

Proposition 3.18. Let ψ ∈C∞(Rn) and assume that ψ and its derivative grow at most poly-nomially. Each of the operators τh, σλ , Mψ and Dα on S (Rn) extend to continuous operatorson S ′(Rn).

Alternatively, one can view these operators as restrictions of the corresponding operators onD ′(Rn) to S ′(Rn).

The main purpose of introducing tempered distributions is to extend the Fourier trans-form. The fact that F : S (Rn)→S (Rn) is continuous shows that following definition makessense.


Definition 3.19. The Fourier transform on S ′(Rn) is the continuous operator F : S ′(Rn)→S ′(Rn) defined by

〈F (u),ϕ〉 := 〈u,F (ϕ)〉,

for u ∈ S ′(Rn) and ϕ ∈ S (Rn). The inverse Fourier transform is the continuous operatorF−1 : S ′(Rn)→S ′(Rn) defined by

〈F−1(u),ϕ〉 := 〈u,F−1(ϕ)〉.

It follows easily from the definition of F−1 on S (Rn) that F−1 = RF = FR. Asusual, we write u for F (u) also if u ∈S ′(Rn). The following properties are analogous to theones in Proposition 2.4.

Proposition 3.20. For u ∈S ′(Rn), we have that

τ−yu F7→ eiy·ξ u,

eiy·xu F7→ τyu,

σλ u F7→ |λ |−dσ1/λ u,

Dαx u F7→ (iξ )α u,

xα u F7→ (i∂ξ )α u,

where y ∈ Rn and λ ∈ R\{0}.

Proof. The identities follow by combining the definitions of these operators in terms of theirtransposes with the corresponding properties for Schwartz class functions. We illustrate themethod by proving the first identity and leave the others to the reader. Considering ϕ as afunction of ξ we have that

〈F (τ−yu),ϕ〉= 〈τ−yu, ϕ〉= 〈u,τyϕ〉= 〈u,F (eiy·ξϕ)〉= 〈u,eiy·ξ

ϕ〉= 〈eiy·ξ u,ϕ〉.

Theorem 3.21. F : S ′(Rn)→S ′(Rn) is an isomorphism with inverse F−1.

Proof. We simply note that

〈F−1F (u),ϕ〉= 〈F (u),F−1(ϕ)〉= 〈u,FF−1(ϕ)〉= 〈u,ϕ〉.

Similarly, 〈FF−1(u),ϕ〉= 〈u,F−1F (ϕ)〉= 〈u,ϕ〉.

Example 3.22. We have 〈δ0,ϕ〉 = 〈δ0, ϕ〉 = ϕ(0) = (2π)−n/2 ∫Rn ϕ(x)dx, so δ0 = (2π)−n/2.

By Fourier’s inversion formula it follows that F (1)=RF−1(1)= (2π)n/2R(δ0)= (2π)n/2δ0.


In order to extend convolutions to S ′(Rn) we first prove that ϕ ∗ is a continuous linearoperator on S (Rn) for a fixed ϕ ∈S (Rn).

Lemma 3.23. Assume that ϕ ∈S (Rn) and that ψk→ ψ in S (Rn). Then ϕ ∗ψk→ ϕ ∗ψ inS (Rn).

Proof. From the convolution theorem it follows that F (ϕ ∗ψk) = (2π)n/2ϕψk. The resultfollows since Mϕ , F and F−1 are continuous operators on S (Rn).

The transpose of ϕ ∗ is computed by changing the order of integration in 〈ϕ ∗u,ψ〉:∫Rn

(∫Rn

ϕ(x− y)u(y)dy)

ψ(x)dx =∫Rn

u(y)(∫

Rnϕ(x− y)ψ(x)dx

)dy

if u ∈S (Rn), so that (ϕ ∗)T = R(ϕ)∗, where as usual R(ϕ)(x) = ϕ(−x).

Proposition 3.24. For ϕ ∈ S (Rn), the operator ϕ ∗ extends to a continuous operator onS ′(Rn) given by 〈ϕ ∗u,ψ〉 := 〈u,R(ϕ)∗ψ〉. The extension satisfies

F (ϕ ∗u) = (2π)n/2ϕ u,

Dα(ϕ ∗u) = (Dαϕ)∗u = ϕ ∗ (Dαu).

Proof. The existence of the extension is a consequence of Proposition 3.17 and Lemma 3.23.The identity F (ϕ ∗u) = (2π)n/2ϕ u follows from the calculation

〈F (ϕ ∗u),ψ〉= 〈ϕ ∗u, ψ〉= 〈u,R(ϕ)∗ ψ〉= 〈u,FF−1(R(ϕ)∗ ψ)〉= 〈u,F−1(R(ϕ)∗ ψ)〉= 〈u,(2π)n/2

ϕψ〉= 〈(2π)n/2

ϕ u,ψ〉,

in which the convolution theorem for S (Rn) and the relation F−1 = FR = RF have beenused. The second identity follows from similar manipulations and its proof is left to thereader.

Since convolution is commutative on S (Rn) we shall also write u ∗ϕ instead of ϕ ∗ u.Keeping u fixed and varying ϕ we obtain a continuous linear operator from S (Rn) to S ′(Rn).Note that u∗ v is in general not well-defined if u,v ∈S ′(Rn).

Example 3.25. We have 〈δ0 ∗ϕ,ψ〉= 〈δ0,R(ϕ)∗ψ〉=∫Rn ϕ(x)ψ(x)dx = 〈ϕ,ψ〉. Thus, δ0 ∗

is the identity on S (Rn). Similarly, (Dαδ0)∗= Dα on S (Rn).

Example 3.26. If Kε is as in Example 3.6, we find that Kε ∗ϕ → δ0 ∗ϕ = ϕ in S ′(Rn).


3.5 * Approximation with smooth functionsThere is an important question that we have avoided. In Propositions 3.9 and 3.17 we didn’tsay anything about the uniqueness of the extensions. The extensions are in fact unique, butto prove this one has to show that D(Rn) is dense in D ′(Rn) in the first case and that S (Rn)is dense S ′(Rn) in the second case. This is done by the usual procedure of mollification andmultiplication with smooth ‘cut-off’ functions, but there are some technical details. We willpresent the proofs for the interested reader. The section may be skipped on a first reading sincethe results are not used elsewhere in the notes.

We begin by noting that the definition 〈u∗ϕ,ψ〉 := 〈u,R(ϕ)∗ψ〉 also makes sense whenu ∈ D ′(Rn) and ϕ,ψ ∈ D(Rn), since then R(ϕ) ∗ψ ∈ D(Rn). There is however anothernatural way of defining convolutions. We show that these two definitions agree.

Lemma 3.27. For u ∈D ′(Rn) and ϕ ∈D(Rn) we have that u∗ϕ ∈C∞(Rn) with

(u∗ϕ)(x) = 〈u,τxR(ϕ)〉= 〈u,ϕ(x−·)〉.

Proof. Since x 7→ ϕ(x−·) is continuous from Rn to D(Rn) it follows that x 7→ 〈u,ϕ(x−·)〉 ∈C(Rn). Moreover, limh→0 h−1(ϕ(x+he j−·)−ϕ(x−·)) = ∂ jϕ(x−·) in D(Rn) for each x andj, and x 7→ ∂ jϕ(x−·) is continuous from Rn to D(Rn). From this it follows that x 7→ 〈u,ϕ(x−·)〉 ∈C1(Rn). Proceeding by induction, one shows that x 7→ 〈u,ϕ(x−·)〉 ∈C∞(Rn). The resulttherefore follows if we prove the equality (u∗ϕ)(x) = 〈u,ϕ(x−·)〉. Let v(x) = 〈u,ϕ(x−·)〉.Using Riemann sums we find that

〈v,ψ〉=∫Rn

v(x)ψ(x)dx

= limk→∞

∑α∈Zn

ψ(α/k)〈u,ϕ(α/k−·)〉k−n

=

⟨u, lim

k→∞∑

α∈Znψ(α/k)ϕ(α/k−·)k−n

⟩= 〈u,R(ϕ)∗ψ〉

for ψ ∈D(Rn), where we have used the fact that ∑α∈Zd ψ(α/n)ϕ(α/k−·)k−n→R(ϕ)∗ψ inD(Rn) (left as an exercise for the reader). The expression on the last line is simply 〈u∗ϕ,ψ〉,so this proves the equality u∗ϕ = v.

Theorem 3.28. D(Rn) is dense in S ′(Rn) and D ′(Rn).

Proof. We first approximate u ∈ D ′(Rn) with the sequence ψ(·/k)u in D ′(Rn), where ψ ∈D(Rn) with ψ(x)= 1 for |x| ≤ 1. It is clear that ψ(·/k)ϕ→ϕ in D(Rn) if ϕ ∈D(Rn) (they areequal for large k) and hence ψ(·/k)u→ u in D ′(Rn). Similarly, if u ∈S ′(Rn), the sequenceconverges to u in S ′(Rn). We can therefore replace u by ψu, for some function ψ ∈D(Rn), inwhat follows. We define uk = (ψu)∗η1/k, where η is the standard mollifier. By the previouslemma it follows that uk ∈C∞(Rn) as an element of D ′(Rn). Moreover, uk has compact supportsince uk(x) = 〈ψu,η1/k(x−·)〉= 〈u,ψη1/k(x−·)〉= 0 if dist(x,suppψ)≥ 1/k. Finally,

〈uk,ϕ〉= 〈ψu,η1/k ∗ϕ〉= 〈ψu, ψ(η1/k ∗ϕ)〉,


where ψ ∈ D(Rn) with ψ = 1 on suppψ . We have that Dα(η1/k ∗ϕ) = η1/k ∗Dαϕ → Dαϕ

uniformly. Hence ψ(η1/k ∗ϕ)→ ψϕ in D(Rn) and uk→ ψu in D ′(Rn). To prove the sameconvergence in the topology of S ′(Rn), we first note that 〈(ψu)∗η1/k,ϕ〉 is given by 〈w,ϕ〉,where w(x) = 〈u,η1/k(x−·)〉 is in D(Rn), whenever ϕ ∈ D(Rn). However, since D(Rn) isdense in S (Rn) and w has compact support, the same is true by an approximation argumentwhenever ϕ ∈S (Rn). We still have that Dα(η1/k ∗ϕ) = η1/k ∗Dαϕ → Dαϕ uniformly andhence ψ(η1/k ∗ϕ)→ ψϕ in S (Rn). It follows that uk→ ψu in S ′(Rn).

Corollary 3.29. The extensions in Propositions 3.9 and 3.17 are unique.

3.6 * Sobolev spaces with negative indexEvans [3, Section 5.8, Theorem 8] characterizes the Sobolev space Hk, k ∈N0, as the space ofu ∈ L2(Rn) such that

(1+ |ξ |k)u ∈ L2(Rn)

and shows that the norm ‖u‖Hk(Rn) is equivalent to the norm

‖(1+ |ξ |k)u‖L2(Rn).

He then defines the fractional Sobolev space Hs(Rn), s ≥ 0, as the space of u ∈ L2(Rn) suchthat

‖(1+ |ξ |s)u‖L2(Rn) < ∞.

We remark that the function (1+ |ξ |s) above is often replaced by (1+ |ξ |2)s/2, which is smoothat the origin. This makes no real difference for s > 0 since the corresponding norms areequivalent. The latter turn out to be more convenient when working with Sobolev spaces withnegative index, though. We define these using tempered distributions

Definition 3.30. Let s ∈ R with s < 0. Hs(Rn) is the space of tempered distributions u suchthat u ∈ L1

loc(Rn) with

‖u‖Hs(Rn) :=(∫

Rn(1+ |ξ |2)s|u(ξ )|2 dξ

)1/2

< ∞. (3.5)

Note that S (Rn)⊂Ht(Rn)⊂Hs(Rn)⊂S ′(Rn) when s < t, each inclusion being contin-uous. From now on, we let ‖u‖Hs(Rn) = ‖(1+ |ξ |2)s/2u‖L2(Rn) for all real s. We already knowthat the spaces Hk, k ∈ N0 are complete, and therefore Hilbert spaces with an appropriatelydefined inner product. The same is true for Hs.

Theorem 3.31. Hs(Rn) is a Hilbert space with inner product

(u,v)Hs(Rn) :=∫Rn(1+ |ξ |2)su(ξ )v(ξ )dξ .


Proof. It is clear that (·, ·)Hs(Rn) defines an inner product on Hs(Rn) with (u,u)Hs(Rn) =

‖u‖2Hs(Rn), so we only need to prove the completeness. This follows since the map T : u 7→

(1+ |ξ |2)s/2u is an isometric isomorphism from Hs(Rn) to L2(Rn).

Note that the elements of Hs(Rn) need not be functions when s < 0.

Example 3.32. The distribution δ0 ∈ S ′(Rn) satisfies δ0 = (2π)−n/2. Using polar coordi-nates, we find that ∫

Rn(1+ |ξ |2)s dξ < ∞

if and only if s <−n/2. Hence δ0 ∈ Hs(Rn) if and only if s <−n/2.

Evans also defines H−1(U) in a different way in Section 5.9.1, namely as the dual spaceof H1

0 (U). We should check that these definitions are equivalent. Note first that when U =Rn,H1

0 (U) coincides with H1(U). Thus, we should check that H−1(Rn) is the dual space ofH1(Rn). It follows from the Riesz representation theorem that for any s ∈ R, the dual ofHs(Rn) can be identified with itself if an element v ∈ Hs(Rn) is identified with the boundedlinear functional

Hs(Rn) 3 u 7→ (u,v)Hs(Rn) =∫Rn(1+ |ξ |2)su(ξ )v(ξ )dξ . (3.6)

The dual can however also be identified with H−s(Rn) using an extension of the usual pairingbetween S ′(Rn) and S (Rn). Rewriting

〈w,ϕ〉=∫Rn

w(ξ )ϕ(−ξ )dξ =∫Rn(1+ |ξ |2)−

s2 w(ξ )(1+ |ξ |2)

s2 ϕ(−ξ )dξ ,

we see that 〈·, ·〉 extends to a bilinear map H−s(Rn)×Hs(Rn)→ C. It is also clear that thefunctional defined by v ∈ Hs(Rn) according to (3.6) is identical to the functional u 7→ 〈w,u〉where w :=Λ2sC v, in which C v= v for v∈S (Rn) and is extended by duality to S ′(Rn), andΛ2s := F−1M(1+|ξ |2)sF : Hs(Rn)→ H−s(Rn). Whenever one identifies the dual with eitherHs(Rn) or H−s(Rn) one has to keep in mind how the identification is done.

Chapter 4

Linear PDE with constant coefficients

In this chapter we see how distribution theory can be used to study linear PDE with con-stant coefficients. In particular we will see how fundamental solutions can be understoodnaturally in terms of distributions. We will also introduce various classes of PDE which gen-eralize the equations we have studied so far in the course. The definitive source on linear PDEwith constant coefficients treated using distribution theory is Hörmander [8], but the booksby Friedlander [6], Duistermaat and Kolk [1], Rauch [9] and Strichartz [11] are maybe moreaccessible.

4.1 Classification of PDE and the symbol of a partial differ-ential operator

Previously in the course, we have encountered three different equations with very differentbehavior: Laplace’s equation, the heat equation and the wave equation. These are examplesof three different types of equations called elliptic, parabolic and hyperbolic. A general linear,homogeneous, second order PDE with real coefficients can be written as

n

∑i, j=1

ai j∂ 2u

∂xi∂x j+

n

∑i=1

ai∂u∂xi

+a0u = 0. (4.1)

The coefficients ai j, ai are real numbers. We assume that not all ai j = 0. Without loss ofgenerality we can assume that ai j = a ji. Consider a linear change of variables y = Bx, whereB is an n×n matrix. That is,

yk = ∑l

bklxl.

Using the chain rule, we obtain that

∂u∂xi

= ∑k

∂u∂yk

∂yk

∂xi= ∑

kbki

∂u∂yk

and∂ 2u

∂xi∂x j=

(∑k

bki∂

∂yk

)(∑

lbl j

∂

∂yl

)u = ∑

k,lbkibl j

∂ 2u∂yk∂yl

.

35

36 Linear PDE with constant coefficients

The second order part of the PDE is therefore converted as follows:

n

∑i, j=1

ai j∂ 2u

∂xi∂x j= ∑

k,l

(∑i, j

bkiai jbl j

)∂ 2u

∂yk∂yl.

We associate to the second order part the quadratic form

q(ξ ) = ∑i, j

ai jξiξ j = ξT Aξ ,

where ξ ∈ Rn is thought of as a column vector and ξ T is the transpose of ξ . Let q(η) be theform corresponding to the transformed second order part. Then

q(η) = ηT BABT

η .

Note that this is just the change of variables formula associated with the transformation ξ =BT η . Recall that the number of positive (m+) and negative (m−) eigenvalues of the symmetricmatrix associated with a quadratic form is invariant under linear transformations and that wecan find a transformation B so that

q(η) = q(BTη) =

n

∑j=1

σ jη2j ,

in which σ j ∈ {−1,0,1} with m+ positive and m− negative squares. We now make a classi-fication by considering the quadratic form q (or q). The names correspond to the character ofthe quadratic surface q(η) = c.

Definition 4.1.

• We say that the PDE (4.1) is elliptic if the quadratic form q is positive or negativedefinite, that is, if all the eigenvalues are positive or if all are negative. In other wordsthis means that after linear change of variables, we can transform the PDE to the Laplaceequation plus terms of lower order.

• If one eigenvalue is positive and the others negative (or vice versa) we say that theequation is hyperbolic. This means that we can transform the equation to the waveequation plus lower order terms.

• Finally, if one eigenvalue is zero and the others have the same sign, we say that theequation is parabolic. This is a degenerate case in which the lower order terms areimportant.

Roughly speaking, we expect elliptic equations to behave like Laplace’s equation, hy-perbolic equations to behave like the wave equation and parabolic equations like the heatequation.

A large portion of the history of PDE has been devoted to finding suitable generalizationsof these definitions for operators of arbitrary order and for systems. However, note that even

Linear PDE with constant coefficients 37

the above classification for second order equations with real coefficients is not exhaustive.Recall that a general linear partial differential operator of order m with constant coefficientscan be written

P(D) = ∑|α|≤m

aαDα ,

where aα ∈C. There is no need to assume that the coefficients are real, but note that a complexequation always can be written as a system of two real equations. We associate with such anoperator the complex polynomial

p(ξ ) = ∑|α|≤m

aα(iξ )α

called the symbol of the operator1. If u ∈S (Rn) or u ∈S ′(Rn) it follows from Propositions2.4 and 3.20 that

F (P(D)u)(ξ ) = p(ξ )u(ξ ).

Thus, by using the Fourier transform, the study of a linear partial differential operator withconstant coefficients is replaced by the study of a complex polynomial in n variables. We alsodefine the principal symbol pm(ξ ) by

pm(ξ ) = ∑|α|=m

aα(iξ )α ,

and call Pm(D) = ∑|α|=m aαDα the principal part of the operator.

Example 4.2. The Laplace operator ∆ has the form

P(D) = ∆ = ∂2x1+ · · ·+∂

2xn

Its symbol is thusp(ξ ) =−|ξ |2,

and p2(ξ ) = p(ξ ). The minus sign in the formula explains why one usually looks at −∆

instead of ∆.

Example 4.3. The heat operator ∂t−∆ has the symbol

p(ξ ) = iξn+1 + |ξ ′|2,

where ξ = (ξ ′,ξn+1) ∈ Rn+1. Its principal symbol is p2(ξ ) = |ξ ′|2.

Example 4.4. The wave operator �= ∂ 2t −∆ has the symbol

p(ξ ,τ) =−ξ2n+1 + |ξ ′|2

and this is also its principal symbol.1In many books the notation Dx j is used for ∂x j/i, while D = (Dx1 , . . . ,Dxn). This makes the correspondence

between the operator and its symbol somewhat easier: P(D) has the symbol P(ξ ).


Example 4.5. The bi-Laplace operator (or bi-Laplacian) ∆2 = ∑ni, j=1 ∂ 2

xi∂ 2

x jhas the sybmol

p(ξ ) = |ξ |4

and p4(ξ ) = p(ξ ).

Of the four different types of equations mentioned above, the easiest to generalize is theelliptic kind.

Definition 4.6 (Ellipticity). We say that the operator P(D) is elliptic if pm(ξ ) 6= 0 for ξ ∈Rn \{0}.

Note that this agrees with the definition for second order equations. In addition to the Laplaceoperator, we see that also the bi-Laplace operator is elliptic. The idea behind the notion ofellipticity is essentially that one can solve the equation P(D)u = f by dividing with p(ξ ) onthe Fourier side. This idea doesn’t quite work since p(ξ ) might vanish for small |ξ |. However,ellipticity implies that |p(ξ )| ≥ c|ξ |m for large |ξ | and this implies e.g. that u is smooth if f issmooth, since the regularity of u has to do with the decay of u at infinity. We show a regularityassumption under a hypothesis which is a bit stronger than ellipticity to give the idea.

Theorem 4.7. Assume that P is elliptic and p(ξ ) 6= 0 for all ξ ∈ Rn. If f ∈ S (Rn) andu ∈S ′(Rn) solves P(D)u = f in the sense of distributions, then u ∈S (Rn).

Proof. Since pm(ξ ) 6= 0 when ξ 6= 0 it follows that c := inf|ξ |=1 |pm(ξ )|> 0. Hence,

|pm(ξ )|= |pm(|ξ ||ξ |−1ξ )|= |ξ |m|pm(|ξ |−1

ξ )| ≥ c|ξ |m

for all ξ . On the other hand, p(ξ )− pm(ξ ) only contains terms of degree m−1 or less. Hence,there is a constant b > 0 such that |p(ξ )− pm(ξ )| ≤ b|ξ |m−1 for all ξ . But this means that

|p(ξ )| ≥ c|ξ |m−b|ξ |m−1 ≥ c2|ξ |m

if |ξ | ≥ 2bc . On the compact set B(0, 2b

c ), |p(ξ )| is bounded from below by a positive constant.Hence, there exists a constant C > 0 such that

|p(ξ )| ≥C(|ξ |m +1)

for all ξ ∈ Rn. Using this, one finds from the quotient rule that 1/p(ξ ) is smooth withDα

(1

p(ξ )

)≤Cα(1+ |ξ |)−m−|α| for any α ∈ Nn

0Taking the Fourier transform of the equation, we have that

p(ξ )u = f .

and henceu =

1p(ξ )

f ∈S (Rn)

by Leibniz rule.


The result is still true if we just assume ellipticity and replace the conclusion by u∈C∞(Rn)(Laplace’s equation has infinitely many smooth polynomial solutions, which are clearly notSchwartz functions). Note that this is not at all true for the wave equation, which in one spacedimension has the distributional solution f (x− t) where f is an arbitrary locally integrablefunction (see Example 3.12)

One can also generalize the notion of hyperbolic equations. This definition is intimatelyconnected with the initial-value problem. The equation

pm(ξ ) = 0

appearing above is called the characteristic equation. A smooth (n−1)-dimensional surfaceΣ in Rn is called non-characteristic if at each point of the surface, pm(ν) 6= 0, where ν isa normal vector. Suppose that we are given m smooth functions u0, . . . ,um−1 defined on thesurface and try to solve the equation P(D)u = 0 with the initial conditions

u|Σ = u0,

∂νu|Σ = u1,

...

(∂ν)m−1u|Σ = um−1.

Assume for simplicity that Σ = {x ∈ Rn : xn = 0} and that ν = (0, · · · ,0,1). From the initialconditions, we know all the derivatives ∂ k

xnu, k = 0, . . . ,m− 1 at Σ, but also all mixed partial

derivatives Dαu with αn < m, since these can be computed from the partial derivatives of thefunctions u0, . . . ,um−1. The condition pm(ν) 6= 0 now means precisely that the coefficientin front of the partial derivative ∂ m

xnu in the equation is non-zero. This means that we can

also compute ∂ mxn

u|Σ by using the partial differential equation. This procedure can now beiterated in order to compute all the partial derivatives of u at Σ. Thus, the initial data andthe equation determine u and its partial derivatives completely at the initial surface if thesurface is non-characteristic. Heuristically, this tells us that we should be able to solve theinitial-value problem. Under the additional assumptions that the surface and the data are real-analytic, one can indeed show that this is the case by using a power series approach. This leadsto the famous Cauchy-Kovalevskaya theorem, which is discussed in Evans [3, Section 4.6].The assumption of real-analyticity is often too strong for the theorem to be of any practicalrelevance. Moreover, the solution may not depend continuously on the data. However, if weadd some additional assumptions, we get a well-posed initial-value problem.

Definition 4.8 (Hyperbolicity). A partial differential operator P(D) of order m is said to behyperbolic in the direction ν ∈ Rn if its principal part satisfies

pm(ν) 6= 0,

and there exists τ0 ∈ R such that

p(ξ + τν) 6= 0, ξ ∈ Rn, Imτ ≤ τ0.


One can show that the initial-value problem P(D)u = 0 with data on a non-characteristicsurface is well-posed in a natural sense if and only if P(D) is hyperbolic in the normal direc-tion2. The definition of hyperbolicity is unfortunately unstable with respect to perturbationsof the lower order terms. This also turns out to be a problem when considering equations withvariable coefficients. There is a stronger notion which doesn’t share these problems.

Definition 4.9 (Strict hyperbolicity). Let ν ∈ Rn be a unit vector. P(D) is said to be strictlyhyperbolic in the direction ν if pm(ν) 6= 0 and for every ξ ∈ Rn not parallel with ν , theequation

pm(ξ + τν) = 0

only has simple real roots τ .

One can show that strict hyperbolicity implies hyperbolicity.

Example 4.10. Consider the wave operator P(D) = ∂ 2t −∆ in Rn+1 with principal symbol

p2(ξ ) =−ξ 2n+1 + |ξ ′|2. Let ν = (0, . . . ,0,1). Then p2(ν) =−1. Moreover,

p2(ξ + τν) =−(ξn+1 + τ)2 + |ξ ′|2,

sop2(ξ + τν) = 0

if and only ifτ =−ξn+1±|ξ ′|.

Hence the roots are real and they are simple as long as ξ and ν are not parallel. It follows thatthe wave operator is strictly hyperbolic (and hence hyperbolic) in the (positive) t-direction.Since p2(ξ ) = p(ξ ) in this case, one can also use Definition 4.8 directly to show that theoperator is hyperbolic.

Finally, we remark that there also characteristic initial-value problems which are well-posed, such as for the heat equation or the Schrödinger equation. The main difference betweenthese and the hyperbolic initial-value problem is that the hyperbolic initial-value problem iswell-posed without assuming any growth conditions on the initial data, whereas this is nec-essary in the characteristic case. An example of this which we have already encountered isTychonoff’s counterexample for the heat equation. This example can be generalized to manyother equations (including the Schrödinger equation). The reason for this difference is thathyperbolic equations have finite propagation speed, meaning that the solution at a specificpoint in space-time is determined by the initial data only in a bounded set. This property isnot shared by well-posed, characteristic initial-value problems. We will return to this issue inthe next section.

2It actually doesn’t matter which normal vector one considers since one can show that if P(D) is hyperbolicin the direction ν , then it is also hyperbolic in the direction −ν .


4.2 Fundamental solutions

The concept of fundamental solutions has been mentioned several times throughout the courseand it has already been mentioned in Evans that this is related to the Dirac distribution. Wenow give the formal definition.

Definition 4.11. A distribution Φ ∈ D ′(Rn) is called a fundamental solution of a partial dif-ferential operator P(D) if P(D)Φ = δ0.

Theorem 4.12. Let Φ be a fundamental solution of the partial differential operator P(D) andlet f ∈D(Rn). Then Φ∗ f is a solution of the equation P(D)u = f .

Proof. Note that Φ∗ f ∈C∞(Rn) by Lemma 3.27. Moreover, the identity Dα(u∗ϕ) = (Dαu)∗ϕ from Proposition 3.24 continues to hold when u ∈D ′(Rn) and ϕ ∈D(Rn). It follows that

P(D)(Φ∗ f ) = (P(D)Φ)∗ f = δ0 ∗ f = f

(cf. Example 3.25).

Example 4.13. Let us check that the fundamental solution

Φ(x) =

{− 1

2πlog |x|, n = 2,1

n(n−2)α(n)1|x|n−2 , n≥ 3

for the Laplace equation from Section 2.2 in Evans [3] actually is a fundamental solutionaccording to the above definition. For ϕ ∈D(Rn) we have that

〈−∆Φ,ϕ〉= 〈Φ,−∆ϕ〉=−∫Rn

Φ(x)∆ϕ(x)dx =−∫Rn

Φ(y)∆ϕ(0− y)dy

where we have made the change of variables y =−x. Investigating the proof of Theorem 1 inSection 2.2 of Evans [3], we see that the last integral equals ϕ(0). It follows that −∆Φ = δ0in the sense of distributions, so that Φ is a fundamental solution for −∆.

Note that fundamental solutions are in general not unique since we can always add a solu-tion of the homogeneous equation P(D)u = 0 in order to obtain a new fundamental solution.For the Laplace operator, we could e.g. simply add a constant to Φ.

We have also discussed fundamental solutions for initial-value problems, such as Φ(x, t)for the heat equation. As pointed out by Evans, Φ formally solves the problem

∂tΦ(x, t)−∆Φ(x, t) = 0, t > 0,Φ(x,0) = δ0(x).


In fact, it turns out that these two types of fundamental solutions are linked and that Φ is alsoa fundamental solution in the sense that ∂tΦ−∆Φ = δ0(x, t) in Rn+1. To see this, note that

〈(∂t−∆)Φ,ϕ〉= 〈Φ,(−∂t−∆)ϕ〉=∫

∞

0

∫Rn

Φ(x, t)(−∂t−∆)ϕ(x, t)dxdt

=∫

ε

0

∫Rn

Φ(x, t)(−∂t−∆)ϕ(x, t)dxdt +∫

∞

ε

∫Rn

Φ(x, t)(−∂t−∆)ϕ(x, t)dxdt

= O(ε)+∫Rn

Φ(x,ε)ϕ(x,ε)dx+∫

∞

ε

∫Rn

(∂t−∆)Φ(x, t)︸︷︷︸=0

ϕ(x, t)dxdt

= O(ε)+∫Rn

Φ(x,ε)ϕ(x,0)dx

since supx∈R |ϕ(x,ε)−ϕ(x,0)| → 0 as ε → 0+ and∫Rn Φ(x,ε)dx = 1. Letting ε → 0+, we

see that〈(∂t−∆)Φ,ϕ〉= ϕ(0,0).

More generally, consider the initial-value problem

∂tu+ P(Dx)u = 0, t > 0,u = g, t = 0,

for some partial differential operator P(Dx). Suppose that u : Rn× [0,∞)→ R is a smoothsolution and extend it to a function on Rn+1 by setting

v(x, t) =

{u(x, t), t ≥ 0,0, t < 0.

Then

〈(∂t + P(Dx))v,ϕ〉=∫

∞

0

∫Rn

u(x, t)(−∂t + P(−Dx))ϕ(x, t)dxdt

=∫Rn

u(x,0)ϕ(x,0)dxdt +∫

∞

0

∫Rn

(∂t + P(Dx))u(x, t)︸︷︷︸=0

ϕ(x, t)dxdt

= 〈g(x)δ0(t),ϕ〉.

Hence,(∂t + P(Dx))v = g(x)δ0(t)

in the sense of distributions. Thus, the initial-value problem can be transformed into a nonho-mogeneous equation in Rn+1. If we formally take g(x) = δ0(x) above, we find that

(∂t + P(Dx))v = δ0(x)δ0(t) = δ0(x, t),

showing that we will get a fundamental solution for the operator ∂t + P(Dx) by solving theinitial-value problem with δ0(x) as initial data. These formal considerations can be maderigorous for certain types of operators (this is linked to the discussion below about evolution


operators). Note that the expressions g(x)δ0(t) and δ0(x)δ0(t) above are actually not well-defined. They should be replaced by the tensor products g(x)⊗δ0(t) and δ0(x)⊗δ0(t), whichare well-defined distributions (see Exercise 12).

From the above discussion it follows that we already know several fundamental solutions.In fact, any linear partial differential operator with constant coefficients (except the zero oper-ator) has a fundamental solution.

Theorem 4.14 (Malgrange-Ehrenpreis). Every non-zero linear partial differential operatorP(D) with constant coefficients has a fundamental solution Φ ∈D ′(Rn).

For (different) proofs of this remarkable theorem we refer to [1, 6] and [8]. One can in factshow that there always is a fundamental solution Φ ∈S ′(Rn).

As we saw above, fundamental solutions supported in a half-space are very useful whenconsidering initial-value problems. An operator which such a fundamental solution is usuallycalled an evolution operator with respect to the half-space3. One can show that if P(D) ishyperbolic with respect to a vector ν then P(D) is an evolution operator with respect to thehalf-space Hν = {x ∈ Rn : x · ν ≥ 0}. As remarked previously, there are many importantexamples of initial-value problems in which the operator P(D) is not hyperbolic. Fortunately,it turns out that hyperbolicity is not necessary for P(D) to be an evolution operator. Anothersufficient condition is the following.

Definition 4.15 (Petrowsky’s condition). We say that P(D) satisfies Petrowsky’s conditionwith respect to ν if the coefficient of the highest power of τ in p(ξ + τν) is independent of ξ

and there exists τ0 ∈ R such that p(ξ + τν) 6= 0 if Imτ ≤ τ0.

Theorem 4.16. If P(D) satisfies Petrowsky’s condition with respect to ν , then it is an evolutionoperator with respect to the half-space Hν .

One can also show that this condition is close to being necessary in a certain specific sense(see [8, Section 12.8]).

Example 4.17. Consider the heat equation as an equation in Rn+1, for which

p(ξ ) = iξn+1 + |ξ ′|2

Let ν = (0,0, . . . ,0,1) ∈ Rn+1 and H = {(x, t) ∈ Rn+1 : (x, t) ·ν ≥ 0}= Rn× [0,∞). We findthat

p(ξ + τν) = i(ξn+1 + τ)+ |ξ ′|2 = iτ + iξn+1 + |ξ ′|2

which clearly show that the coefficient of the highest (and only) power of τ is independent ofξ . Moreover p(ξ + τν) = 0 if and only if

τ =−ξn+1 + i|ξ ′|2.

3For a distribution u ∈ D ′(Rn), we say that suppu ⊆ Hν if 〈u,ϕ〉 = 0 for any ϕ ∈ D(Rn) with suppϕ ⊂Rn \Hν .


We therefore see that p(ξ + τν) 6= 0 if Imτ ≤ τ0 for any τ0 < 0. Hence, the heat operatoris an evolution operator with respect to Hν . If we change ν to (0, . . . ,0,−1), the roots of theequation p(ξ + τν) = 0 are given by τ = ξn+1− i|ξ ′|2, whence the heat equation is not anevolution operator with respect to the half-space {xn+1 ≤ 0}.

Example 4.18. Let P(D) be hyperbolic with respect to ν . Then p(ξ + τν) = τm pm(ν) +O(τm−1), with pm(ν) 6= 0, showing that the first part of Petrowsky’s condition is satisfied. Thesecond part of Petrowsky’s condition is just the second part of the definition of hyperbolicity.Hence, a hyperbolic operator satisfies Petrowsky’s condition and is therefore an evolutionoperator with respect to Hν .

There is however a very important difference between these two cases, which we havealluded to earlier. One can show that a hyperbolic operator has a fundamental solution whosesupport is not only contained in the half-space Hν , but in a closed subcone of Hν which onlyintersects ∂Hν in the origin. Using this, one can show that the equation has finite speed ofpropagation. In the characteristic case, there is no such fundamental solution.

Chapter 5

Exercises

1. Use separation of variables to solve Laplace’s equation in the unit square [0,1]× [0,1] withthe following boundary conditions u(0,y) = u(1,y) = uy(x,0) = 0, u(x,1) = x2− x.

2. Solve Laplace’s equation ∆u = 0 in the unit ball (or disc) in R2 with Dirichlet boundarycondition u= g on ∂B(0,1) by expressing the equation in polar coordinates (r,θ) and usingseparation of variables. You may assume that g = g(θ) is as smooth as you like. Show thatyou recover Poisson’s formula for the ball (see Evans [3, Section 2.2, Theorem 15]).

3. Use separation of variables to solve the initial-value problem for Schrödinger’s equationwith periodic boundary conditions:

iut +uxx = 0 in R× (0,∞),

u(x+2π, t) = u(x, t) in R× [0,∞),

u = g on R×{t = 0}.

Here u and g are complex-valued and g is assumed to be 2π-periodic and smooth. Showthat the L2 norm

‖u(·, t)‖L2(−π,π)

of the solution is independent of time.

Remark: As stated here, the periodic ‘boundary conditions’ are actually no boundary con-ditions, but just saying that we solve the equation on T = R/(2πZ). One could replacethese boundary conditions by the conditions u(−π, t) = u(π, t) and ux(−π, t) = ux(π, t),solve the equation on (−π,π)× (0,∞) and prove that the solution can be extended period-ically instead.

4. Assume that u ∈C21([0,π]× (0,T ])∩C([0,π]× [0,T ]) with ut = uxx, ux(0, t) = ux(π, t) = 0

for all t ∈ (0,T ] and u(x,0) = 0 for all x ∈ [0,π]. Show that u(x, t)≡ 0.

Hint: There are several ways of doing this. You can either use an energy method or amaximum principle argument. For the latter, it might be useful to extend u to a solution onR× (0,T ] which is periodic in x.

45

46 Exercises

5. Prove thatF (e−|x|) =

1√2π

21+ξ 2

in R. Use this to calculate∫R

dx1+x2 and

∫R

dx(1+x2)2 .

6. Use Fourier transformation in x to solve the problem

uxx +uyy = 0

in the upper half-plane with boundary condition u(x,0) = g(x). Relate this to Poisson’sformula for the upper half-plane.

Hint: Use the previous exercise.

7. One says that an initial-value problem is well posed if for every initial datum g in someclass X , there exists a unique solution u in some other class Y and, moreover, the mapS : X → Y , S : g 7→ u, is continuous. In order to make this precise, one has to equip X andY with some topologies. We will discuss this in some specific cases later in the course.

Show that for the heat equation on Rn× [0,T ], T < ∞, one obtains for every g ∈S (Rn)a unique smooth solution u ∈ C∞(Rn× [0,T ])∩L∞(Rn× [0,T ]) satisfying u(x,0) = g(x)(you may use results from Evans [3, Section 2.3]). Show, moreover, that u(·, t) ∈S (Rn)for every t ≥ 0 and that the map g 7→ u is continuous if we use the topology of S (Rn) ong and the semi-norms sup0≤t≤T ‖u(·, t)‖α,β , α,β ∈ Nn

0, on u.

Remark: The number T is introduced for convenience and can be replaced by ∞, but youdon’t need to show this.

8. Show that (1.9) defines a distributional solution of the wave equation.

9. Show that a linear functional u on S (Rd) belongs to S ′(Rd) if and only if there exists anatural number n and a constant C such that

|〈u,ϕ〉| ≤C‖ϕ‖n

for all ϕ ∈S (Rd), where ‖ϕ‖n = ∑|α|+|β |≤n ‖ϕ‖α,β (‖ ·‖α,β denote the usual semi-normson S (Rd)).

10. Consider the equation −u′′+u = δ0, u ∈D ′(R).

a. Find a solution up ∈S ′(R) using the Fourier transform.

b. Show that u = up +Aex +Be−x ∈D ′(R) is also a solution for any A,B ∈ C. For whichA and B is u ∈S ′(R)?

11. Show that

Φ(x, t) =

{12 , t ≥ |x|,0 t < |x|

is a fundamental solution for the wave operator ∂ 2t −∂ 2

x on R2.

Exercises 47

12. Given two test functions ϕ1 ∈ D(Rn1) and ϕ2 ∈ D(Rn2), their tensor product ϕ1⊗ϕ2 ∈D(Rn1+n2) is defined by (ϕ1⊗ϕ2)(x,y) = ϕ1(x)ϕ2(y), x ∈ Rn1 , y ∈ Rn2 . Show that giventwo distributions u1 ∈D ′(Rn1) and u2 ∈D ′(Rn2), one can define a tensor product u1⊗u2 ∈D ′(Rn1+n2) either by

(u1⊗u2)(ψ) = 〈u1,〈u2,ψ(x,y)〉〉

or(u1⊗u2)(ψ) = 〈u2,〈u1,ψ(x,y)〉〉,

for ψ ∈ D(Rn1+n2), where the notation means that we keep x fixed in 〈u2,ψ(x,y)〉 and letu2 operate in the y variable only, and keep y fixed in 〈u1,ψ(x,y)〉 and let u1 operate in thex variable. This means that you should show that x 7→ 〈u2,ψ(x,y)〉 ∈ D(Rn1) and that themap ψ 7→ 〈u1,〈u2,ψ(x,y)〉〉 is continuous (a similar result holds for the other definition ofu1⊗u2 by symmetry). Show that with either definition,

(u1⊗u2)(ϕ1⊗ϕ2) = u1(ϕ1)u2(ϕ2).

for ϕ1 ∈ D(Rn1) and ϕ2 ∈ D(Rn2). One show there is a unique distribution with thisproperty and use this to show that the two definitions of u1⊗u2 agree, but you don’t needto do this.

13. Show that any second order linear partial differential equation with constant coefficients ishyperbolic according to Definition 4.1 if and only if it is strictly hyperbolic in some direc-tion according to Definition 4.9. Find the characteristic and strictly hyperbolic directions.

48 Exercises

Bibliography

[1] J. J. DUISTERMAAT AND J. A. C. KOLK, Distributions, Cornerstones, BirkhäuserBoston Inc., Boston, MA, 2010. Theory and applications, Translated from the Dutchby J. P. van Braam Houckgeest.

[2] H. DYM AND H. P. MCKEAN, Fourier series and integrals, Academic Press, New York-London, 1972. Probability and Mathematical Statistics, No. 14.

[3] L. C. EVANS, Partial differential equations, vol. 19 of Graduate Studies in Mathematics,American Mathematical Society, Providence, RI, second ed., 2010.

[4] G. B. FOLLAND, Fourier analysis and its applications, The Wadsworth & Brooks/ColeMathematics Series, Wadsworth & Brooks/Cole Advanced Books & Software, PacificGrove, CA, 1992.

[5] G. B. FOLLAND, Real analysis, Pure and Applied Mathematics (New York), John Wiley& Sons Inc., New York, second ed., 1999. Modern techniques and their applications, AWiley-Interscience Publication.

[6] F. G. FRIEDLANDER, Introduction to the theory of distributions, Cambridge UniversityPress, Cambridge, second ed., 1998. With additional material by M. Joshi.

[7] L. HÖRMANDER, The analysis of linear partial differential operators. I, Classics inMathematics, Springer-Verlag, Berlin, 2003. Distribution theory and Fourier analysis,Reprint of the second (1990) edition.

[8] , The analysis of linear partial differential operators. II, Classics in Mathematics,Springer-Verlag, Berlin, 2005. Differential operators with constant coefficients, Reprintof the 1983 original.

[9] J. RAUCH, Partial differential equations, vol. 128 of Graduate Texts in Mathematics,Springer-Verlag, New York, 1991.

[10] E. M. STEIN AND R. SHAKARCHI, Fourier analysis, vol. 1 of Princeton Lectures inAnalysis, Princeton University Press, Princeton, NJ, 2003. An introduction.

[11] R. S. STRICHARTZ, A guide to distribution theory and Fourier transforms, River Edge,NJ: World Scientific., 2003.

49

50 BIBLIOGRAPHY

[12] A. VRETBLAD, Fourier analysis and its applications, vol. 223 of Graduate Texts inMathematics, Springer-Verlag, New York, 2003.

[13] A. ZYGMUND, Trigonometric series. Vol. I, II, Cambridge Mathematical Library, Cam-bridge University Press, Cambridge, third ed., 2002. With a foreword by Robert A.Fefferman.

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