Green's Functions

Chapter 12

Green’s Functions

12.1 One-dimensional Helmholtz Equation

Suppose we have a string driven by an external force, periodic with frequency ω. The diﬀerential equation (here f is some prescribed function)

∂2 1 ∂2 U(x, t)= f(x)cos ωt (12.1) ∂x2 − c2 ∂t2 represents the oscillatory motion of the string, with amplitude U, which is tied down at both ends (here l is the length of the string):

U(0,t)= U(l,t)=0. (12.2)

We seek a solution of the form (thus we are ignoring transients)

U(x, t)= u(x)cos ωt, (12.3) so u(x) satisﬁes d2 + k2 u(x)= f(x), k = ω/c. (12.4) dx2 The solution to this inhomogeneous Helmholtz equation is expressed in terms of ′ the Green’s function Gk(x, x ) as

l ′ ′ ′ u(x)= dx Gk(x, x )f(x ), (12.5) Z0 where the Green’s function satisﬁes the diﬀerential equation

d2 + k2 G (x, x′)= δ(x x′). (12.6) dx2 k − 125 Version of November 23, 2010 126 Version of November 23, 2010 CHAPTER 12. GREEN’S FUNCTIONS

As we saw in the previous chapter, the Green’s function can be written down in terms of the eigenfunctions of d2/dx2, with the speciﬁed boundary conditions, d2 λ u (x)=0, (12.7a) dx2 − n n un(0) = un(l)=0. (12.7b) The normalized solutions to these equations are

2 nπx nπ 2 u (x)= sin , λ = , n =1, 2,.... (12.8) n l l n − l r The factor 2/l is a normalization factor. From the general theorem about eigenfunctions of a Hermitian operator given in Sec. 11.5, we have p 2 l nπx mπx dx sin sin = δ . (12.9) l l l nm Z0 Thus the Green’s function for this problem is given by the eigenfunction expansion ∞ 2 nπx nπx′ ′ l sin l sin l Gk(x, x )= . (12.10) 2 nπ 2 n=1 k X − l But this form is not usually very convenient for calculation . Therefore we solve the diﬀerential equation (12.6) directly. When x = x′ the inhomogeneous term is zero. Since 6

′ ′ Gk(0, x )= Gk(l, x )=0, (12.11) we must have

′ ′ ′ x < x : Gk(x, x )= a(x ) sin kx, (12.12a) x > x′ : G (x, x′)= b(x′) sin k(x l). (12.12b) k − We determine the unknown functions a and b by noting that the derivative of G must have a discontinuity at x = x′, which follows from the diﬀerential equation (12.6). Integrating that equation just over that discontinuity we ﬁnd

′ x +ǫ 2 d 2 ′ dx 2 + k Gk(x, x )=1, (12.13) ′ dx Zx −ǫ or x=x′+ǫ d ′ Gk(x, x ) =1, (12.14) dx ′ x=x −ǫ because 2ǫG (x′, x′) 0 as ǫ 0. Although d G (x, x′) is discontinuous at k dx k x = x′, G(x, x′) is continuous→ there:→

′ x +ǫ d G(x′ + ǫ, x′) G(x′ ǫ, x′) = dx G(x, x′) − − ′ dx Zx −ǫ 12.1. ONE-DIMENSIONAL HELMHOLTZ EQUATION127 Version of November 23, 2010

′ ′ x d x +ǫ d = dx G(x, x′) + dx G(x, x′) ′ dx ′ dx Zx −ǫ Zx d d = ǫ G(x, x′) + G(x, x′) , (12.15) dx ′ dx ′ x=x −ξ x=x +ξ¯

where by the mean value theorem, 0 < ξ ǫ, 0 < ξ¯ ǫ. Therefore ≤ ≤ x=x′+ǫ G(x, x′) = (ǫ) 0 as ǫ 0. (12.16) ′ O → → x=x −ǫ

′ Now using the continuity of G and the discontinuity of G , we ﬁnd two equations for the coeﬃcient functions a and b:

a(x′) sin kx′ = b(x′) sin k(x′ l), (12.17a) − a(x′)k cos kx′ +1 = b(x′)k cos k(x′ l). (12.17b) − It is easy to solve for a and b. The determinant of the coeﬃcient matrix is sin kx′ sin k(x′ l) D = − − = k sin kl, (12.18) k cos kx′ k cos k(x′ l) − − −

′ independent of x . Then the solutions are 1 0 sin k(x′ l) sin k(x′ l) a(x′) = − − = − , (12.19a) D 1 k cos k(x′ l) k sin kl − − − 1 sin kx′ 0 sin kx′ b(x′) = = . (12.19b) D k cos kx′ 1 k sin kl −

Thus we ﬁnd a closed form for the Green’s function in the two regions:

sin k(x′ l) sin kx x < x′ : G (x, x′)= − , (12.20a) k k sin kl sin kx′ sin k(x l) x > x′ : G (x, x′)= − , (12.20b) k k sin kl or compactly, 1 G (x, x′)= sin kx sin k(x l), (12.21) k k sin kl < > − where we have introduced the notation

′ x< is the lesser of x, x , ′ x> is the greater of x, x . (12.22)

′ ′ Note that Gk(x, x )= Gk(x , x) as is demanded on general grounds, as a conse- quence of the reciprocity relation (11.110). ′ Let us analyze the analytic structure of Gk(x, x ) as a function of k. We see that simple poles occur where

kl = nπ, n = 1, 2,.... (12.23) ± ± 128 Version of November 23, 2010 CHAPTER 12. GREEN’S FUNCTIONS

There is no pole at k = 0. For k near nπ/l, we have

sin kl = sin nπ + (kl nπ)cos nπ + . . . = (kl nπ)( 1)n. (12.24) − − − If we simply sum over all the poles of Gk, we obtain

∞ nπx< nπ ′ n sin l sin l (x> l) Gk(x, x ) = ( 1) nπ − n=−∞ − l (kl nπ) Xn6=0 − ∞ nπx nπx′ sin l sin l = nπ n=−∞ nπ k l Xn6=0 − ∞ nπx nπx′ 1 1 1 = sin sin l l nπ k nπ − k + nπ n=1 − l l X∞ 2 nπx nπx′ 1 = sin sin . (12.25) l l l 2 nπ 2 n=1 k X − l This is in fact equal to Gk, as seen in the eigenfunction expansion (12.10), because the diﬀerence is an entire function vanishing at inﬁnity, which must be zero by Liouville’s theorem, see Sec. 6.5.

12.2 Types of Boundary Conditions

Three types of second-order, homogeneous differential equations are commonly encountered in physics (the dimensionality of space is not important): 1 ∂2 Hyperbolic: 2 u(r,t)=0, (12.26a) ∇ − c2 ∂t2 Elliptic: 2 + k2 u(r)=0, (12.26b) ∇ 1 ∂ Parabolic: 2 T (r,t)=0. (12.26c) ∇ − κ ∂t The first of these equations is the wave equation, the second is the Helmholtz equation, which includes Laplace’s equation as a special case (k = 0), and the third is the diffusion equation. The types of boundary conditions, specified on which kind of boundaries, necessary to uniquely specify a solution to these equations are given in Table 12.1. Here by Cauchy boundary conditions we means that both the function u and its normal derivative ∂u/∂n is specified on the boundary. Here ∂u = ˆn ∇u, (12.27) ∂n · where ˆn is a(n outwardly directed) normal vector to the surface. As we have seen previously, Dirichlet boundary conditions refer to specifying the function u on the surface, Neumann boundary conditions refer to specifying the normal derivative ∂u/∂n on the surface, and mixed boundary conditions refer to 12.3. EXPRESSION OF FIELD IN TERMS OF GREEN’S FUNCTION129 Version of November 23, 2010

Type of Equation Type of Boundary Condition Type of Boundary Hyperbolic Cauchy Open Elliptic Dirichlet, Neumann, or mixed Closed Parabolic Dirichlet, Neumann, or mixed Open

Table 12.1: Boundary conditions required for the three types of second-order differential equations. The boundary conditions referred to in the first and third cases are actually initial conditions. specifying a linear combination, αu + β∂u/∂n, on the surface. If the specified boundary values are zero, we say that the boundary conditions are homogeneous; otherwise, they are inhomogeneous.

Example.

To determine the vibrations of a string, described by

∂2 1 ∂2 u =0, (12.28) ∂x2 − c2 ∂t2 we must specify ∂u u(x, 0), (x, 0) (12.29) ∂t at some initial time (t = 0). The line t = 0 is an open surface in the (ct,x) plane.

12.3 Expression of Field in Terms of Green’s Function

Typically, one determines the eigenfunctions of a diﬀerential operator subject to homogeneous boundary conditions. That means that the Green’s functions obey the same conditions. See Sec. 11.8. But suppose we seek a solution of

(L λ)ψ = S (12.30) − subject to inhomogeneous boundary conditions. It cannot then be true that

ψ(r)= (dr′) G(r, r′)S(r′). (12.31) ZV To see how to deal with this situation, let us consider the example of the three-dimensional Helmholtz equation,

( 2 + k2)ψ(r)= S(r). (12.32) ∇ 130 Version of November 23, 2010 CHAPTER 12. GREEN’S FUNCTIONS

We seek the solution ψ(r) subject to arbitrary inhomogeneous Dirichlet, Neu- mann, or mixed boundary conditions on a surface Σ enclosing the volume V of interest. The Green’s function G for this problem satisﬁes

( 2 + k2)G(r, r′)= δ(r r′), (12.33) ∇ − subject to homogeneous boundary conditions of the same type as ψ satisﬁes. Now multiply Eq. (12.32) by G, Eq. (12.33) by ψ, subtract, and integrate over the appropriate variables:

(dr′) G(r, r′)( ′2 + k2)ψ(r′) ψ(r′)( ′2 + k2)G(r, r′) ∇ − ∇ ZV = (dr′) [G(r, r′)S(r′) ψ(r′)δ(r r′)] . (12.34) − − ZV Here we have interchanged r and r′ in Eqs. (12.32) and (12.33), and have used the reciprocity relation, G(r, r′)= G(r′, r). (12.35) (We have assumed that the eigenfunctions and hence the Green’s function are real.) Now we use Green’s theorem to establish

dσ G(r, r′)∇′ψ(r′) ψ(r′)∇′G(r, r′) − · − ZΣ ψ(r), r V, + (dr′) G(r, r′)S(r′)= (12.36) 0, r ∈ V, ZV 6∈ where in the surface integral dσ is the outwardly directed surface element, and r′ lies on the surface Σ. This generalizes the simple relation given in Eq. (12.31). How do we use this result? We always suppose G satisfies homogeneous boundary conditions on Σ. If ψ satisfies the same conditions, then for r V Eq. (12.31) holds. But suppose ψ satisfies inhomogeneous Dirichlet boundary∈ conditions on Σ, r′ r′ ψ( ) r′∈Σ = ψ0( ), (12.37) a specified function on the surface. Then we impose homogeneous Dirichlet conditions on G, r r′ G( , ) r′∈Σ =0. (12.38) Then the first surface term in Eq. (12.36) is zero, but the second contributes.

For example if S(r) = 0 inside V , we have for r V ∈ ψ(r)= dσ [∇′G(r, r′)]ψ (r′), (12.39) · 0 ZΣ which express ψ in terms of its boundary values. If ψ satisﬁes inhomogeneous Neumann conditions on Σ, ∂ψ r′ r′ ′ ( ) = N( ), (12.40) ∂n ′ r ∈Σ

12.4. HELMHOLTZ EQUATION INSIDE A SPHERE131 Version of November 23, 2010 a speciﬁed function, then we use the Green’s function which respects homogeneous Neumann conditions,

∂ r r′ ′ G( , ) =0, (12.41) ∂n ′ r ∈Σ

so again if S = 0 inside V , we have within V

ψ(r)= dσ G(r, r′)N(r′). (12.42) − ZΣ Finally, if ψ satisﬁes inhomogeneous mixed boundary conditions,

∂ r′ r′ r′ r′ ′ ψ( )+ α( )ψ( ) = F ( ), (12.43) ∂n ′ r ∈Σ

then when G satisﬁes homogeneous boundary conditions of the same type ∂ r′ r r′ ′ + α( ) G( , ) =0, (12.44) ∂n ′ r ∈Σ we have for r V ∈ ψ(r)= (dr′) G(r, r′)S(r′) dσ G(r, r′)F (r′). (12.45) − ZV Zσ 12.4 Helmholtz Equation Inside a Sphere

Here we wish to ﬁnd the Green’s function for Helmholtz’s equation, which sat- isﬁes ( 2 + k2)G (r, r′)= δ(r r′), (12.46) ∇ k − in the interior of a spherical region of radius a, with homogeneous Dirichlet boundary conditions on the surface,

′ Gk(r, r ) =0. (12.47) |r|=a

We will use two methods.

12.4.1 Eigenfunction Method We know that the eigenfunctions of the Laplacian are

m jl(kr)Yl (θ, φ), (12.48) in spherical polar coordinates, r, θ, φ; that is,

( 2 + k2)j (kr)Y m(θ, φ)=0. (12.49) ∇ l l 132 Version of November 23, 2010 CHAPTER 12. GREEN’S FUNCTIONS

Here jl is the spherical Bessel function, π j (x)= J (x), (12.50) l 2x l+1/2 r m and the Yl are the spherical harmonics,

2l +1 (l m)! 1/2 Y m(θ, φ)= − P m(cos θ)eimφ, (12.51) l 4π (l + m)! l m where Pl is the associated Legrendre function. Here l is a nonnegative integer, and m is an integer in the range l m l. For example, the ﬁrst few spherical Bessel functions (which are− simpler≤ ≤ than the cylinder functions, the Bessel functions of integer order) are sin x j (x) = , (12.52a) 0 x sin x cos x j (x) = , (12.52b) 1 x2 − x 3 1 3 j (x) = sin x cos x, (12.52c) 2 x3 − x − x2 and in general 1 d l sin x j (x)= xl . (12.53) l −x dx x The associated Legrendre function is given by

d l+m (cos2 θ 1)l P m(cos θ) = ( 1)m sinm θ − . (12.54) l − d cos θ 2l l! For example, the ﬁrst few spherical harmonics are 1 Y 0 = , (12.55a) 0 √4π

1 3 iφ Y1 = sin θe , (12.55b) −r8π 0 3 Y1 = cos θ, (12.55c) r4π 1 3 −iφ Y1 = sin θe , (12.55d) r8π 2 15 2 2iφ Y2 = sin θe , (12.55e) r32π 1 15 iφ Y2 = cos θ sin θe , (12.55f) −r8π 0 5 2 Y2 = (3 cos θ 1), (12.55g) r16π − 12.4. HELMHOLTZ EQUATION INSIDE A SPHERE133 Version of November 23, 2010

−1 15 −iφ Y2 = cos θ sin θe , (12.55h) r8π −2 15 2 −2iφ Y2 = sin θe . (12.55i) r32π

The eigenfunctions must vanish ar r = a, so if βln is the nth zero of jl,

jl(βln)=0, n =1, 2, 3,..., (12.56) the desired eigenfunctions are r ψ (r, θ, φ)= A j β Y m(θ, φ), (12.57) nlm nl l ln a l and the eigenvalues are

β 2 λ = k2 = ln . (12.58) ln − ln − a

The normalization constant Anl is determined by the requirement that

r2 dr dΩ ψ (r, θ, φ) 2 =1, (12.59) | nlm | Z where dΩ = sin θ dθ dφ is the element of solid angle. Since the spherical harmonics are normalized so that [Ω = (θ, φ) represents a point on the unit sphere]

m′∗ m dΩ Yl′ (Ω)Yl (Ω) = δll′ δmm′ , (12.60) Z the normalization constant is determined by the requirement

a 2 2 2 r Anl r dr jl βln =1. (12.61) | | 0 a Z h i Now a 1 r2 dr j (β r/a)j (β r/a)= δ a3j2 (β ), (12.62) l ln l lm nm 2 l+1 ln Z0 which for n = m follows from the orthogonality property (11.68). So 6 2 1 Anl = 3 , (12.63) | | ra jl+1(βln) and the Green’s function has the eigenfunction expansion

2 1 Y m(Ω)Y m∗(Ω′)j (β r/a)j (β r′/a) r r′ l l l ln l ln Gk( , )= 3 2 2 2 , (12.64) a jl+1(βln) k (βln/a) nlmX − where Ω = (θ, φ), Ω′ = (θ′, φ′). 134 Version of November 23, 2010 CHAPTER 12. GREEN’S FUNCTIONS

This result can be simpliﬁed by carrying out the sum on m, using the addition theorem for spherical harmonics,

l 4π Y m∗(Ω′)Y m(Ω) = P (cos γ), (12.65) 2l +1 l l l mX=−l 0 where Pl(cos γ)= Pl (cos γ) is Legendre’s polynomial, and γ is the angle between the directions represented by Ω and Ω′, or

cos γ = cos θ cos θ′ + sin θ sin θ′ cos(φ φ′). (12.66) − Then we obtain 2 2l +1 1 j (β r/a)j (β r′/a) r r′ l ln l ln Gk( , )= 3 Pl(cos γ) 2 2 2 . (12.67) a 4π jl+1(βln) k (βln/a) Xnl − This leads us to the second method.

12.4.2 Discontinuity (Direct) Method Let us adopt the angular dependence found above:

∞ 2l +1 G (r, r′)= P (cos γ)g (r, r′), (12.68) k 4π l l Xl=0 m where we will call gl the reduced Green’s function. Because Yl is an eigenfunction of the angular part of the Laplacian operator,

l(l + 1) 2Y m(Ω) = Y m(Ω), (12.69) ∇ l − r2 l and the delta function can be written as 1 δ(r r′)= δ(r r′)δ(Ω Ω′), (12.70) − rr′ − − we see that, because of the orthonormality of the spherical harmonics, Eq. (12.60), the Green’s function equation (12.46) corresponds to the following equation satisﬁed by the reduced Green’s function, the inhomogeneous “spherical Bessel equation,”

d2 2 d l(l + 1) 1 + + k2 g (r, r′)= δ(r r′). (12.71) dr2 r dr − r2 l rr′ − We solve this equation directly. For (0 < r′

0 r < r′ : g (r, r′)= a(r′)j (kr), (12.72a) ≤ l l r′ < r a : g (r, r′)= b(r′)j (kr)+ c(r′)n (kr). (12.72b) ≤ l l l 12.4. HELMHOLTZ EQUATION INSIDE A SPHERE135 Version of November 23, 2010

Only jl appears in the ﬁrst form because the solution must be ﬁnite at r = 0, and the second solution to the spherical Bessel equation,

π n (x)= N (x), (12.73) l 2x l+1/2 r where Nν is the Neumann function, is singular at x = 0. For example, cos x n (x)= , (12.74) 0 − x and in general 1 d l cos x n (x)= xl . (12.75) l − −x dx x To determine the functions a, b, and c, we proceed as follows. The boundary ′ condition at r = a, gl(a, r ) = 0, implies

′ ′ 0= b(r )jl(ka)+ c(r )nl(ka), (12.76) or ′ b(r ) nl(ka) ′ = . (12.77) c(r ) − jl(ka) Thus we can write in the outer region,

a r > r′ : g (r, r′)= A(r′)[j (kr)n (ka) n (kr)j (ka)]. (12.78) ≥ l l l − l l ′ The next condition we impose is that of the continuity of gl at r = r :

a(r′)j (kr′)= A(r′)[j (kr′)n (ka) n (kr′)j (ka)]. (12.79) l l l − l l ′ On the other hand, the derivative of gl is discontinuous at r = r , as we may see by integrating Eq. (12.71) over a tiny interval around r = r′:

r=r′+ǫ d ′ 1 gl(r, r ) = ′2 , (12.80) dr ′ r r=r −ǫ

which implies 1 ka(r′)j′(kr′) kA(r′)[j′(kr′)n (ka) n′(kr′)j (ka)] = . (12.81) l − l l − l l −r′2 ′ ′ ′ Now multiply Eq. (12.79) by kjl(kr ), and Eq. (12.81) by jl(kr ), and subtract: j (kr′) l = kA(r′)j (ka)[j (kr′)n′(kr′) n (kr′)j′(kr′)]. (12.82) r′2 − l l l − l l

Now jl, nl are the independent solutions of the spherical Bessel equation 1 d d l(l + 1) r2 + k2 u =0, (12.83) r2 dr dr − r2 136 Version of November 23, 2010 CHAPTER 12. GREEN’S FUNCTIONS the Wronskian of which,

∆(r) j (kr)n′ (kr) n (kr)j′(kr) (12.84) ≡ l l − l l has the form const. ∆(r)= , (12.85) r2 as we saw in Problem 4 of Assignment 8. We can determine the constant by considering the asymptotic forms of jl, nl, sin(kr lπ/2) j (kr) − , kr 1, (12.86a) l ∼ kr ≫ cos(kr lπ/2) n (kr) − , kr 1, (12.86b) l ∼ − kr ≫ which imply 1 ∆(r) = [sin2(kr lπ/2)+ cos2(kr lπ/2)] k2r2 − − 1 = . (12.87) (kr)2

Thus since the right-hand side of Eq. (12.82) is proportional to the Wronskian, we ﬁnd the function A: j (kr′) A(r′)= k l , (12.88) − jl(ka) and then from Eq. (12.79) we ﬁnd the function a:

′ k ′ ′ a(r )= [jl(kr )nl(ka) nl(kr )jl(ka)]. (12.89) −jl(ka) − Hence the Green’s function is explicitly

′ ′ jl(kr) ′ ′ r < r : gl(r, r )= k [jl(kr )nl(ka) nl(kr )jl(ka)], (12.90a) − jl(ka) − ′ ′ ′ jl(kr ) r > r : gl(r, r )= k [jl(kr)nl(ka) nl(kr)jl(ka)], (12.90b) − jl(ka) − or n (ka) n (kr ) g (r, r′)= kj (kr )j (kr ) l l > , (12.91) l − l < l > j (ka) − j (kr ) l l > ′ where r< is the lesser of r, r , and r> is the greater. From this closed form we may extract the eigenvalues and eigenfunctions of the spherical Bessel diﬀerential operator appearing in Eq. (12.83). The poles of gl occur where jl(ka) has zeroes, all of which are real, at ka = βln, the nth zero of jl, or β 2 k2 = ln . (12.92) a 12.5. HELMHOLTZ EQUATION IN UNBOUNDED SPACE137 Version of November 23, 2010

In the neighborhood of this zero,

j (ka) = (ka β )j′(β ). (12.93) l − ln l ln But at the zero the Wronskian is

1 ′ 2 = nl(βln)jl(βln). (12.94) (βln) −

Now from the recursion relation λ J ′ (z)= J (z) J (z), (12.95) λ z λ − λ+1 we see that the derivative of the spherical Bessel function (12.50) satisﬁes, at the zero, j′(β )= j (β ). (12.96) l ln − l+1 ln Thus the residue of the pole of gl at k = βln/a is

1 jl(βlnr/a) 2 2 . (12.97) a βln jl+1(βln)

Now jl is an even or odd function of z depending on whether n is even or odd. So if β is a zero of j , so is β , and hence if we add the contributions of ln l − ln these two poles, we get the corresponding contribution to gl:

1 j (β r/a)j (β r′/a) 1 1 g (r, r′) l ln l ln . (12.98) l ∼ a2β [j (β )]2 k β /a − k + β /a ln l+1 ln − ln ln Summing up the contribution of all such pairs of poles, we obtain

2 ∞ j (β r/a)j (β r′/a) 1 g (r, r′)= l ln l ln , (12.99) l a3 [j (β )]2 k2 (β /a)2 n=1 l+1 ln ln X − which is the eigenfunction expansion displayed in Eq. (12.67).

12.5 Helmholtz Equation in Unbounded Space

Again we are solving the equation

( 2 + k2)G (r, r′)= δ(r r′), (12.100) ∇ k − but now in unbounded space. The solution to this equation is an outgoing spherical wave: ′ 1 eik|r−r | G (r, r′)= G (r r′)= . (12.101) k k − −4π r r′ | − | 138 Version of November 23, 2010 CHAPTER 12. GREEN’S FUNCTIONS

This may be directly veriﬁed. Consider a small sphere S, of radius ǫ, centered on r′: 1 eikρ (dr)( 2 + k2)G (r r′) (dρ) 2 ∇ k − ≈ ∇ρ −4π ρ ZS ZS d 1 eikρ = dΩ ρ2 dρ −4π ρ Z ρ=ǫ

1, (12.102) → ′ 2 as ǫ 0. Evidently, for r = r , Gk satisﬁes the Helmholtz equation, ( + 2 → 6 ∇ k )Gk = 0. Alternatively, we may construct Gk from the eigenfunction expansion (11.109), ψ∗ (r′)ψ (r) G (r r′)= n n (12.103) k − λ λ n n X − where λ = k2, λ = k′2, where the eigenfunctions are solutions of − n − 2 ′2 ( + k )ψk′ (r)=0, (12.104) ∇ that is, they are plane waves,

1 ik′·r ψk′ (r)= e , (12.105) (2π)3/2 Here the (2π)−3/2 factor is for normalization:

′ ∗ ′ ′ (dk ) ψk′ (r) ψk′ (r ) = δ(r r ), (12.106a) − Z ∗ ′ (dr) ψk′ (r) ψk(r) = δ(k k ), (12.106b) − Z where we have noted that the spectrum of eigenvalues is continuous,

(dk). (12.107) → n X Z Thus the eigenfunction expansion for the Green’s function has the form

′ ′ ′ (dk′) e−ik ·r eik ·r G (r r′)= . (12.108) k − (2π)3 k2 k′2 Z − Let us evaluate this integral in spherical coordinates, where we write (dk′)= k′2 dk′ dφ′ dµ′, µ′ = cos θ′, (12.109) where we have chosen the z axis to lie along the direction of r r′. The integration over the angles is easy: −

′ ′ ′ 1 ∞ 2π 1 eik |r−r |µ G (r r′) = dk′ k′2 dφ′ dµ′ k − (2π)3 k2 k′2 Z0 Z0 Z−1 − ∞ ′ ′2 1 1 dk k 1 ik′ρ −ik′ρ = 2 2 ′2 ′ e e , (12.110) (2π) 2 −∞ k k ik ρ − Z − 12.5. HELMHOLTZ EQUATION IN UNBOUNDED SPACE139 Version of November 23, 2010

k - § ¤ - ¦ ¥ ••k −

Figure 12.1: Contour in the k′ plane used to evaluate the integral (12.110). The integral is closed in the upper (lower) half-plane if the exponent is positive (negative). The poles in the integrand are avoided by passing above the one on the left, and below the one on the right. deﬁning ρ = r r′ , where we have replaced ∞ by 1 ∞ because the inte- | − | 0 2 −∞ grand is even in k′2. We evaluate this integral by contour methods. Because R R now k can coincide with an eigenvalue k′, we must choose the contour appropri- ately to deﬁne the Green’s function. Suppose we choose the contour as shown in Fig. 12.1, passing below the pole at k and above the pole at k. We close the contour in the upper half plane for the eikρ and in the lower half− plane for the e−ikρ term. Then by Jordan’s lemma, we immediately evaluate the integral:

1 1 2πi keikρ 2πi keikρ G (r r′) = + k − (2π)2 2 − 2k iρ 2k iρ − 1 eikρ = , (12.111) −4π ρ which coincides with Eq. (12.101). If a different contour defining the integral had been chosen, we would have obtained a different Green’s function, not one corresponding to outgoing spherical waves. Boundary conditions uniquely determine the contour. Note that ′ ′ Gk(r, r )= Gk(r , r), (12.112) even though Gk is complex. The self-adjointness property (11.110) implied by the eigenfunction expansion is only formal, and is spoiled by the contour choice. 140 Version of November 23, 2010 CHAPTER 12. GREEN’S FUNCTIONS

12.6 Green’s Function for the Scalar Wave Equa- tion

The inhomogeneous scalar wave equation, 1 ∂2 2 ψ(r,t)= ρ(r,t), (12.113) ∇ − c2 ∂t2 requires boundary and initial conditions. The boundary conditions may be Dirichlet, Neumann, or mixed. The initial conditions are Cauchy (see Sec. 12.2). ∂ Thus, we might specify at an initial time t = t0 both ψ(r,t0) and ∂t ψ(r,t0) at every point r in the region being considered. The corresponding Green’s function G(r,t; r′,t′) satisfies 1 ∂2 2 G(r,t; r′,t′)= δ(r r′)δ(t t′). (12.114) ∇ − c2 ∂t2 − − It must satisfy the homogeneous form of the boundary conditions satisfied by ψ. Thus, if ψ has a specified value everywhere on the bounding surface, the corresponding Green’s function must vanish on the surface. In classical physics it is customary to adopt as initial conditions G(r,t; r′,t′) = 0 if tt′,t′′: −∞

T dt (dr) G(r,t; r′,t′) 2G(r, t; r′′, t′′) ∇ − − Z−∞ ZV G(r, t; r′′, t′′) 2G(r,t; r′,t′) − − − ∇ 1 ∂2 G(r,t; r′,t′) G(r, t; r′′, t′′) − c2 ∂t2 − − 1 ∂2 + G(r, t; r′′, t′′) G(r,t; r′,t′) − − c2 ∂t2 = G(r′, t′; r′′, t′′)+ G(r′′,t′′; r′,t′). (12.117) − − − 12.6. GREEN’S FUNCTION FOR THE SCALAR WAVE EQUATION141 Version of November 23, 2010

Now use Green’s theorem, together with the corresponding identity,

∂ ∂ ∂ ∂2 ∂2 A B B A = A B B A, (12.118) ∂t ∂t − ∂t ∂t2 − ∂t2 to conclude that

G(r′′,t′′; r′,t′) G(r′, t′; r′′, t′′) T − − − = dt dσ G(r,t; r′,t′)∇G(r, t; r′′, t′′) · − − Z−∞ ZΣ G(r, t; r′′, t′′)∇G(r,t; r′,t′) − − − 1 ∂ (dr) G(r,t; r′,t′) G(r, t; r′′, t′′) − c2 ∂t − − ZV ∂ t=T G(r, t; r′′, t′′) G(r,t; r′,t′) . (12.119) − − − ∂t t=−∞

The surface integral vanishes, since both Green’s function s satisfy the same homogeneous boundary conditions on Σ. (The boundary conditions are time independent.) The second integral is also zero because from Eq. (12.115)

G(r, ; r′,t′) −∞ =0, (12.120a) ∂G (r, ; r′,t′) ∂t −∞ since

1 ∂2 ′2ψ(r′,t) ψ(r′,t′) = ρ(r′,t′), (12.122a) ∇ − c2 ∂t′2 1 ∂2 ′2G(r,t; r′,t′) G(r,t; r′,t′) = δ(r r′)δ(t t′). (12.122b) ∇ − c2 ∂t′2 − − Note that the diﬀerentiations on G are with respect to the second set of argu- ments (this equation follows from the reciprocity relation). Again multiply the ﬁrst equation by G(r,t; r′,t′), the second by ψ(r′,t′), subtract, integrate over the volume, and over t′ from t

t+ ψ(r,t) = dt′ (dr′) G(r,t; r′,t′)ρ(r′,t′) Zt0 ZV t+ dt′ dσ G(r,t; r′,t′)∇′ψ(r′,t′) ψ(r′,t′)∇′G(r,t; r′,t′) − · − Zt0 IΣ 1 ∂ ∂ (dr′) G(r,t; r′,t ) ψ(r′,t ) ψ(r′,t ) G(r,t; r′,t ) . − c2 0 ∂t 0 − 0 ∂t 0 ZV 0 0 (12.124)

This is our result. The interpretation is as follows: 1. The first integral represents the effect of the sources ρ distributed through- out the volume V . 2. The second integral represents the boundary conditions. If, for example, ψ satisfies inhomogeneous Neumann boundary conditions on Σ,

ˆn ∇ψ = f(r′) (12.125) · Σ

is speciﬁed, then we use homogeneous Neumann boundary conditions for G, ˆn ∇G(r,t; r′,t′) =0. (12.126) · Σ

Then the second integral reads

t+ dt′ dσ G(r,t; r′,t)∇′ψ(r′,t′). (12.127) − · Zt0 IΣ That is, ˆn ∇′ψ(r′,t′) represents a surface source distribution. Other types of boundary− · conditions are as discussed earlier. 3. The third integral represents the eﬀect of the initial conditions, where

′ ∂ ′ ψ(r ,t0), ψ(r ,t0) (12.128) ∂t0 12.7. WAVE EQUATION IN UNBOUNDED SPACE143 Version of November 23, 2010

are speciﬁed. They correspond to impulsive sources at t = t0: 1 ∂ ρ (r′,t′)= ψ(r′,t )δ(t′ t )+ ψ(r′,t )δ′(t′ t ) . (12.129) init −c2 ∂t 0 − 0 0 − 0 0 We verify this statement by integrating by parts, and letting the lower limit of the t′ integration be t ǫ. 0 − 12.7 Wave Equation in Unbounded Space

We now wish to solve Eq. (12.114)

1 ∂2 2 G(r,t; r′,t′)= δ(r r′)δ(t t′), (12.130) ∇ − c2 ∂t2 − − in unbounded space by noting that then G is a function of R = r r′ and T = t t′ only, − − G(r,t; r′,t′)= G(r r′,t t′)= G(R,T ). (12.131) − − Then we can introduce a Fourier transform in space and time,

k R g(k,ω)= (dR) dT ei · e−iωT G(R,T ). (12.132) Z The Fourier transform of the Green’s function equation is (we have set c = 1 temporarily for convenience),

( k2 + ω2)g(k,ω)=1, (12.133) − where we write k2 = k k, which has the immediate solution · 1 g(k,ω)= . (12.134) ω2 k2 − Thus the Green’s function has the formal representation

(dk) dω k R 1 G(R,T )= e−i · eiωT . (12.135) (2π)3 2π ω2 k2 Z − The ω integral here is not well deﬁned until we impose the boundary condition (12.115) G(R,T ) = 0 if T < 0. (12.136) This will be true if the poles are located above the real axis, as shown in Fig. 12.2. Here the contour is closed in the upper half plane if T > 0, and in the lower half plane if T < 0. In both cases, by Jordan’s lemma, the inﬁnite semicircle gives no contribution. We have

ikT −ikT ∞ dω 1 i e e ,T> 0, eiωT = 2k − 2k (12.137) −∞ 2π (ω k)(ω + k) 0 T < 0. Z − ( 144 Version of November 23, 2010 CHAPTER 12. GREEN’S FUNCTIONS

k + iǫ k + iǫ − -••-

Figure 12.2: Contour in the ω plane used to evaluate the integral (12.135).

Thus, if T > 0,

1 ∞ 1 i G(R,T ) = k2 dk 2π dµ e−ikRµ eikT e−ikT (2π)3 2k − Z0 Z−1 i ∞ k dk = eikR e−ikR eikT e−ikT (2π)2 2ikR − − Z0 1 1 1 ∞ = dk eik(R+T ) + e−ik(R+T ) eik(R−T ) eik(T −R) (2π)2 2R 2 − − Z−∞ 1 1 = [δ(R + T ) δ(R T )] . (12.138) 2π 2R − − But R and T are both positive, so R + T can never vanish. Thus we are left with 1 1 G(R,T )= δ(R T ), (12.139) −4π R − or restoring c,

1 1 r r′ G(r r′,t t′)= δ | − | (t t′) . (12.140) − − −4π r r′ c − − | − | The eﬀect at the observation point r at time t is due to the action at the source point r′ at time r r′ t′ = t | − |. (12.141) − c Physically, this means that the “signal” propagates with speed c. 12.7. WAVE EQUATION IN UNBOUNDED SPACE145 Version of November 23, 2010

Let us make this more concrete by considering a simple example, a point d “charge” moving with velocity v(t)= dt r(t), ρ(r,t)= qδ(r r(t)). (12.142) − There are no effects from the infinite surface, nor from the infinite past, so we have from Eq. (12.124)

t+ ψ(r,t) = dt′ (dr′) G(r r′,t t′)ρ(r′,t′) −∞ V − − Z Z+ q t 1 r r(t′) = dt′ δ | − | (t t′) . (12.143) −4π r r(t′) c − − Z−∞ | − | If we let R(t′)= r r(t′), the distance from the source to the observation point −′ at time t′ = t R(t ) , we write this as − c + q t 1 R(t′) ψ(r,t)= dt′ δ (t t′) . (12.144) −4π R(t′) c − − Z−∞ Let τ = R(t′)/c + t′, where τ = t determines the “retarded time” t′ so

1 dR dτ = dt′ 1+ , (12.145) c dt′ where dR 1 d R v = R R = · , (12.146) dt′ 2R dt′ · − R that is R v dτ = dt′ 1 · . (12.147) − Rc Thus the ﬁeld is evaluated as

q t+R(t)/c dτ 1 ψ(r,t) = δ(τ t) −4π R(τ) 1 R·v (τ) − Z−∞ − Rc q 1 = . (12.148) −4π R(t) R(t) v(t)/c − ·