The last equation suggests that E should be homogeneous of degree 2 - n. Recallthat u [member of] S' is homogeneous of degree σ if <u, fλ>= λσ, f [member of] S, where fλ(x) =λ-n f(x/λ), λ > 0 is the λ L1-normalized dilate of f. This reasoning turns out tobe correct for n ≥ 3, but not for n = 2 since then 2 - n = 0 and constant functionsare the ones on R+ which are homogeneous of degree 0, and then cannot give usour fundamental solution of the Laplacian on R2. But since ln r also solves (1.1.6)when n = 2 and r2-n is a solution of the equation for n ≥ 3, perhaps

f(r) = anr2-n, n ≥ 3, and f(r) = an ln r, n = 2,

will work for the appropriate constants an.

Specifically, we claim that we can choose the an so that if

E(x) = anr2-n, n ≥ 3, and E(x) = an ln |x|, n = 2, (1.1.7)

then for g [member of] S we have

g(0) = <E, Δg>, g [member of] S, (1.1.8)

since g(0) = <δ0, g> = <ΔE, g> = <ΔE, Δg>, for all g [member of] Sif and only if δ0 = ΔE.

To verify (1.1.8), we shall need to use the divergence theorem. Note that ν =-x|x| is the outward unit normal for a point x on the complement of the sphereof radius ε > 0 centered at the origin. Thus, for n ≥ 3, if dσ denotes the induced

Lebesgue measure on this sphere,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

The first term in the right vanishes since, as noted above, Δ|x|-n+2 = 0, whenx ≠ 0. For a given ε > 0, the second term is bounded by

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

and so its limit is 0. The last term is

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

which, as ε -> 0+, tends to

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Therefore, if An = |Sn-1| = ∫|x|=1dσ denotes the area of the unit sphere in Rn, wehave the desired identity (1.1.8) for n ≥ 3 if

an = -1/(n - 2)An, n ≥ 3. (1.1.9)

We leave it as an exercise for the reader that we also have (1.1.8) for n = 2 if weset

a2 = 1/2π. (1.1.10)

The minus signs in (1.1.9) are due to the fact that the Laplacian has a negativesymbol, -|[xi]|2.

Let us record what we have done in the following.

Theorem 1.1.1If n ≥ 3 and E(x) = an|x|2-n or n = 2 with E(x) = an ln |x|,with an de_ned by (1.1.9){(1.1.10), then E is a fundamental solution of Δ, i.e.,ΔE = δ0.

The fundamental solution for Δ, (1.1.7), that we have just computed is notunique since others are given by E(x) + h(x) if h(x) is a harmonic function, i.e.,Δh(x) = 0 for all x [member of] Rn. E, though, for n ≥ 3 is the unique fundamental solutionvanishing at in_nity.

Remark 1.1.2 To help motivate computations that we shall carry out in the nextchapter and throughout the text, let us see how the fundamental solution E for Δin Theorem 1.1.1 is pulled back via a linear bijection T : Rn -> Rn. If y = Tx then

Tt [partial derivative]/[partial derivative]y = [partial derivative]/[partial derivative]x

and so

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (1.1.11)

Then Δyu(y) = F(y) is equivalent to Δgu(Tx) = F(Tx). In other words, thepullback of Δy via T is Δg. If n ≥ 3,since an|y|2-n = an|x|2-ng if

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

we conclude that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

if

|g| = det (gjk).

Note that, by a theorem of Jacobi (Sylvester's law of inertia), every symmetricpositive definite matrix gjk can be written in the above form, gjk = LLt, where L isreal and invertible. Consequently, if we take T = L-1 and TtT = (gjk) = (gjk)-1,then the above argument shows that if n ≥ 3, then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

is a fundamental solution for [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

This decomposition gjk = LLt is not unique, however, it is if we require thatL is a lower triangular matrix with strictly positive diagonal entries (the Choleskydecomposition from linear algebra). We shall make this assumption in all suchfactorizations to follow. Note that the matrix L in this decomposition is, roughlyspeaking, the matrix analog of taking the square root of a positive number.

Let us conclude this section by reviewing one more equation involving the Laplacian.

As we pointed out before, we can use E to solve Laplace's equation Δu = F. Letus briey study one more important equation involving Δ, the Dirichlet problemfor R1+n = [0, ∞) × Rn, since it will also have some relevance in our calculation offundamental solutions for the d'Alembertian, [??] = [partial derivative]2t - Δ.

If (y, x) [member of] [0, ∞) × Rn, the Dirichlet problem for this upper half-space is to showthat for a given f [member of] S(Rn) on the boundary we can find a function u(y, x) that isharmonic in R1+n+ with boundary values f, i.e., a solution of

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (1.1.13)

By using the fact that in polar coordinates

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

where r = |x| and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]is the induced Laplacian on Sn-1 (see § 3.4), one easilychecks that if

P(y, x) = y/(y2 + |x|2)(n+1)/2,

then

([partial derivative]2/[partial derivative]y2 + Δ) P(y, x) = 0,(y, x) [member of] (0, ∞) × Rn.

Therefore, if bn is a constant and f [member of] S,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1.1.14)

is harmonic in (0, ∞) × Rn, i.e., it satisfies the first part of (1.1.13). Since

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

we also have that as y -> 0+

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

provided that the constant bn is chosen so that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

in other words

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (1.1.15)

with

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (1.1.16)

being Euler's gamma function. Thus we have shown that the Poisson integral off, (1.1.14), with bn given by (1.1.15) solves the Dirichlet problem for the upperhalf-space, (1.1.13). Note also that the constant bn in (1.1.15) is also given by theformula

bn = 2/An+1, (1.1.17)

where, as before, Ad = 2πd/2/Γ(d/2) denotes the areaof the unit sphere, Sd-1, in Rd.

1.2 FUNDAMENTAL SOLUTIONS OF THE D'ALEMBERTIAN

In this section we shall compute fundamental solutions for the d'Alembertian inR1+n. Thus we seek distributions E for which we have

[??]E = δ0,0(t, x), [??] = [partial derivative]2t- Δ, (t, x) [member of] R1+n. (1.2.1)

Here δ0,0 is the Dirac delta distribution centered at the origin in space-time.

Recall that the Lorentz transformations are the linear maps from R1+n to itselfpreserving the Lorentz form

Q(t, x) = t2 - |x|2, (1.2.2)

which is the natural quadratic form associated with [??}. Since both [??] and δ0,0 areinvariant under these transformations, we expect E to also enjoy this property.Thus, we expect it to be of the form E = f(t2 - |x|2) where f is some distribution.If we plug this into our equation (1.2.1) and we use the polar coordinates formulafor Δ, we can see that if E were of this form then we would have to have that

f"(ρ) + n + 1/2ρ f'(ρ) = 0, whenρ = t2 - |x|2 ≠ 0. (1.2.3)

From this we expect f to be homogeneous of degree (1 - n)=2. If

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (1.2.4)

denotes the Heaviside step function on R, then we can write down solutions of(1.2.3) with this homogeneity. Specifically, when n ≤ 3 the equation has the solution

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

Thus, we expect that for appropriate constants cn the following are fundamentalsolutions for [??] in spatial dimensions n = 1, 2, 3

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (1.2.5)

We added the factor H(t) since, as we shall see below when we solve the Cauchyproblem for [??], it is natural to want our fundamental solution to be supported inR1+n+ = [0, ∞) × Rn. When n = 3 our guessinvolves δ(t2 - |x|2), which is the Leraymeasure in R1+3 associated with the function t2 - |x|2 (see Theorem A.4.1 in theappendix).

Let us show that our guess is correct when n = 2, since this will serve as amodel for arguments to follow. Thus, we wish to see that c2 can be chosen so thatwhenever F [member of] S(R1+2) we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (1.2.6)

To simplify the integration by parts arguments, we note that we can regularize ourguess by extending it into the complex plane and taking limits. Specifically, insteadof truncating the distribution about its singularity as we did for the fundamentalsolution of the Laplacian, we shall use the fact that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

due to the fact that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is real when |x|2 >t2, and ofpositive imaginary part if (t, x) is fixed with |x|2< t2 and ε > 0 small. Therefore,(1.2.6) is equivalent to showing that c2 can be chosen so that we always have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (1.2.7)

In addition to motivating what is to follow, the advantage of (1.2.7) over (1.2.6) isthat the integration by parts arguments that we require are easy for the latter. Ifwe use the polar coordinates formula for the Laplacian it is not di_cult to checkthat

([partial derivative]2t - Δ)|x|2- (t + iε)2))-1/2 = 0, ε > 0, (t, x) [member of] R1+2,

and of course we also have that for ε > 0, Im (|x|2 - (t+iε)2)-1/2 = 0 when t = 0.Therefore, if we integrate by parts, we find that when ε > 0 we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]