11.1 One-dimensional Helmholtz Equation

Transcription

1 Chapter Green s Functions. One-dimensiona Hemhotz Equation Suppose we have a string driven by an externa force, periodic with frequency ω. The differentia equation here f is some prescribed function) 2 x 2 2 ) c 2 t 2 Ux, t) = fx)cosωt.) represents the osciatory motion of the string, with ampitude U, which is tied down at both ends here is the ength of the string): U0, t) = U, t) = 0..2) We seek a soution of the form thus we are ignoring transients) Ux, t) = ux)cos ωt,.3) so ux) satisfies ) d 2 dx 2 + k2 ux) = fx), k = ω/c..4) The soution to this inhomogeneous Hemhotz equation is expressed in terms of the Green s function G k x, x ) as ux) = 0 dx G k x, x )fx ),.5) where the Green s function satisfies the differentia equation ) d 2 dx 2 + k2 G k x, x ) = δx x )..6) 23 Version of December 3, 20

2 24 Version of December 3, 20 CHAPTER. GREEN S FUNCTIONS As we saw in the previous chapter, the Green s function can be written down in terms of the eigenfunctions of d 2 /dx 2, with the specified boundary conditions, ) d 2 dx 2 λ n u n x) = 0,.7a) u n 0) = u n ) = 0..7b) The normaized soutions to these equations are 2 nπx nπ ) 2 u n x) = sin, λ n =, n =, 2,.....8) The factor 2/ is a normaization factor. From the genera theorem about eigenfunctions of a Hermitian operator given in Sec. 0.5, we have 2 0 dxsin nπx sin mπx = δ nm..9) Thus the Green s function for this probem is given by the eigenfunction expansion 2 nπx G k x, x sin sin nπx ) = n= k 2 ) nπ 2..0) But this form is not usuay very convenient for cacuation. Therefore we sove the differentia equation.6) directy. When x x the inhomogeneous term is zero. Since we must have G k 0, x ) = G k, x ) = 0,.) x < x : G k x, x ) = ax )sin kx,.2a) x > x : G k x, x ) = bx )sin kx )..2b) We determine the unknown functions a and b by noting that the derivative of G must have a discontinuity at x = x, which foows from the differentia equation.6). Integrating that equation just over that discontinuity we find or x +ǫ x ǫ ) d 2 dx dx 2 + k2 G k x, x ) =,.3) d dx G kx, x ) x=x +ǫ x=x ǫ =,.4) because 2ǫG k x, x ) 0 as ǫ 0. Athough d dx G kx, x ) is discontinuous at x = x, Gx, x ) is continuous there: Gx + ǫ, x ) Gx ǫ, x ) = x +ǫ x ǫ dx d dx Gx, x )

3 .. ONE-DIMENSIONAL HELMHOLTZ EQUATION25 Version of December 3, 20 x x +ǫ = dx d x ǫ dx Gx, x ) + dx d x dx Gx, x ) [ d = ǫ dx Gx, x ) + d ] dx Gx, x ),.5) x=x ξ x=x + ξ where by the mean vaue theorem, 0 < ξ ǫ, 0 < ξ ǫ. Therefore x=x +ǫ Gx, x ) = Oǫ) 0 as ǫ 0..6) 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: ax )sinkx = bx )sin kx ), ax )k coskx + = bx )k coskx )..7a).7b) It is easy to sove for a and b. The determinant of the coefficient matrix is D = sinkx sinkx ) k coskx k coskx ) = k sin k,.8) independent of x. Then the soutions are ax ) = D 0 sinkx ) k coskx ) = sin kx ),.9a) k sin k bx ) = D sinkx 0 sin kx k coskx = k sin k..9b) Thus we find a cosed form for the Green s function in the two regions: x < x : G k x, x ) = sin kx )sinkx,.20a) k sink x > x : G k x, x ) = sin kx sin kx ),.20b) k sink or compacty, G k x, x ) = k sin k sin kx < sin kx > ),.2) where we have introduced the notation x < is the esser of x, x, x > is the greater of x, x..22) Note that G k x, x ) = G k x, x) as is demanded on genera grounds, as a consequence of the reciprocity reation 0.0). Let us anayze the anaytic structure of G k x, x ) as a function of k. We see that simpe poes occur where k = nπ, n = ±, ±2, )

4 26 Version of December 3, 20 CHAPTER. GREEN S FUNCTIONS There is no poe at k = 0. For k near nπ/, we have sin k = sin nπ + k nπ)cos nπ +... = k nπ) ) n..24) If we simpy sum over a the poes of G k, we obtain G k x, x ) = = = = n= n 0 n= n 0 n= n= ) sin nπx sin nπx nπx< n sin nπ sin nπx nπ k nπ ) sin nπx 2 nπx sin sin nπx sin nπ x > ) k nπ) [ nπ k nπ k 2 nπ ] k + nπ ) 2..25) This is in fact equa to G k, as seen in the eigenfunction expansion.0), because the difference is an entire function vanishing at infinity, which must be zero by Liouvie s theorem, see Sec Types of Boundary Conditions Three types of second-order, homogeneous differentia equations are commony encountered in physics the dimensionaity of space is not important): Hyperboic: 2 c 2 ) 2 t 2 ur, t) = 0,.26a) Eiptic: 2 + k 2) ur) = 0,.26b) Paraboic: 2 ) Tr, t) = 0..26c) κ t The first of these equations is the wave equation, the second is the Hemhotz equation, which incudes Lapace s equation as a specia case k = 0), and the third is the diffusion equation. The types of boundary conditions, specified on which kind of boundaries, necessary to uniquey specify a soution to these equations are given in Tabe.. Here by Cauchy boundary conditions we means that both the function u and its norma derivative u/ n is specified on the boundary. Here u = ˆn u,.27) n where ˆn is an outwardy directed) norma vector to the surface. As we have seen previousy, Dirichet boundary conditions refer to specifying the function u on the surface, Neumann boundary conditions refer to specifying the norma derivative u/ n on the surface, and mixed boundary conditions refer to

5 .3. EXPRESSION OF FIELD IN TERMS OF GREEN S FUNCTION27 Version of December 3, 20 Type of Equation Type of Boundary Condition Type of Boundary Hyperboic Cauchy Open Eiptic Dirichet, Neumann, or mixed Cosed Paraboic Dirichet, Neumann, or mixed Open Tabe.: Boundary conditions required for the three types of second-order differentia equations. The boundary conditions referred to in the first and third cases are actuay initia conditions. specifying a inear combination, αu + β u/ n, on the surface. If the specified boundary vaues are zero, we say that the boundary conditions are homogeneous; otherwise, they are inhomogeneous. Exampe. To determine the vibrations of a string, described by 2 x 2 2 ) c 2 t 2 u = 0,.28) we must specify ux, 0), u x, 0).29) t at some initia time t = 0). The ine t = 0 is an open surface in the ct, x) pane..3 Expression of Fied in Terms of Green s Function Typicay, one determines the eigenfunctions of a differentia operator subject to homogeneous boundary conditions. That means that the Green s functions obey the same conditions. See Sec But suppose we seek a soution of L λ)ψ = S.30) subject to inhomogeneous boundary conditions. It cannot then be true that ψr) = dr )Gr,r )Sr )..3) V To see how to dea with this situation, et us consider the exampe of the three-dimensiona Hemhotz equation, 2 + k 2 )ψr) = Sr)..32)

6 28 Version of December 3, 20 CHAPTER. GREEN S FUNCTIONS We seek the soution ψr) subject to arbitrary inhomogeneous Dirichet, Neumann, or mixed boundary conditions on a surface Σ encosing the voume V of interest. The Green s function G for this probem satisfies 2 + k 2 )Gr,r ) = δr r ),.33) subject to homogeneous boundary conditions of the same type as ψ satisfies. Now mutipy Eq..32) by G, Eq..33) by ψ, subtract, and integrate over the appropriate variabes: dr ) [ Gr,r ) 2 + k 2 )ψr ) ψr ) 2 + k 2 )Gr,r ) ] V = dr )[Gr,r )Sr ) ψr )δr r )]..34) V Here we have interchanged r and r in Eqs..32) and.33), and have used the reciprocity reation, Gr,r ) = Gr,r)..35) We have assumed that the eigenfunctions and hence the Green s function are rea.) Now we use Green s theorem to estabish dσ [Gr,r ) ψr ) ψr ) Gr,r ) ] Σ { + dr )Gr,r )Sr ψr), r V, ) =.36) 0, r V, V where in the surface integra dσ is the outwardy directed surface eement, and r ies on the surface Σ. This generaizes the simpe reation given in Eq..3). How do we use this resut? We aways suppose G satisfies homogeneous boundary conditions on Σ. If ψ satisfies the same conditions, then for r V Eq..3) hods. But suppose ψ satisfies inhomogeneous Dirichet boundary conditions on Σ, ψr ) r Σ = ψ 0r ),.37) a specified function on the surface. Then we impose homogeneous Dirichet conditions on G, Gr,r ) r = 0..38) Σ Then the first surface term in Eq..36) is zero, but the second contributes. For exampe if Sr) = 0 inside V, we have for r V ψr) = dσ [ Gr,r )]ψ 0 r ),.39) Σ which express ψ in terms of its boundary vaues. If ψ satisfies inhomogeneous Neumann conditions on Σ, ψ n r ) = Nr ),.40) r Σ

7 .4. HELMHOLTZ EQUATION INSIDE A SPHERE29 Version of December 3, 20 a specified function, then we use the Green s function which respects homogeneous Neumann conditions, n Gr,r ) = 0,.4) r Σ so again if S = 0 inside V, we have within V ψr) = dσ Gr,r )Nr )..42) Σ Finay, if ψ satisfies inhomogeneous mixed boundary conditions, [ r n ψr ) + αr )ψr )] = Fr ),.43) Σ then when G satisfies homogeneous boundary conditions of the same type [ ] n + αr ) Gr,r ) = 0,.44) r Σ we have for r V ψr) = V dr )Gr,r )Sr ) dσ Gr,r )Fr )..45) σ.4 Hemhotz Equation Inside a Sphere Here we wish to find the Green s function for Hemhotz s equation, which satisfies 2 + k 2 )G k r,r ) = δr r ),.46) in the interior of a spherica region of radius a, with homogeneous Dirichet boundary conditions on the surface, G k r,r ) = 0..47) r =a We wi use two methods..4. Eigenfunction Method We know that the eigenfunctions of the Lapacian are in spherica poar coordinates, r, θ, φ; that is, j kr)y m θ, φ),.48) 2 + k 2 )j kr)y m θ, φ) = 0..49)

8 30 Version of December 3, 20 CHAPTER. GREEN S FUNCTIONS Here j is the spherica Besse function, π j x) = 2x J +/2x),.50) and the Y m are the spherica harmonics, Y m θ, φ) = [ 2 + 4π ] /2 m)! P m cos θ)e imφ,.5) + m)! where P m is the associated Legrendre function. Here is a nonnegative integer, and m is an integer in the range m. For exampe, the first few spherica Besse functions which are simper than the cyinder functions, the Besse functions of integer order) are and in genera j 0 x) = sin x x, j x) = sin x x 2 cosx x, 3 j 2 x) = x 3 ) sinx 3 x x 2 cosx,.52a).52b).52c) j x) = x ) d sinx x dx x..53) The associated Legrendre function is given by ) +m d P m cos θ) = ) m sin m cos 2 θ ) θ d cosθ 2..54)! For exampe, the first few spherica harmonics are Y0 0 =, 4π 3 Y = 8π sin θ eiφ, Y 0 = 3 4π cosθ, 3 Y = 8π sin θ e iφ, 5 Y2 2 = 32π sin2 θ e 2iφ, 5 Y2 = 8π cosθ sin θ eiφ, Y 0 2 = 5 6π 3 cos2 θ ),.55a).55b).55c).55d).55e).55f).55g)

9 .4. HELMHOLTZ EQUATION INSIDE A SPHERE3 Version of December 3, 20 Y 2 = Y 2 2 = 5 8π cosθ sin θ e iφ, 5 32π sin2 θ e 2iφ. The eigenfunctions must vanish ar r = a, so if β n is the nth zero of j,.55h).55i) j β n ) = 0, n =, 2, 3,...,.56) the desired eigenfunctions are r ) ψ nm r, θ, φ) = A n j β n Y m θ, φ),.57) a and the eigenvaues are ) 2 λ n = kn 2 = βn..58) a The normaization constant A n is determined by the requirement that r 2 dr dω ψ nm r, θ, φ) 2 =,.59) where dω = sin θ dθ dφ is the eement of soid ange. Since the spherica harmonics are normaized so that [Ω = θ, φ) represents a point on the unit sphere] dω Y m Ω)Y m Ω) = δ δ mm,.60) the normaization constant is determined by the requirement Now a 0 A n 2 a 0 r 2 r )] 2 dr [j β n =..6) a r 2 dr j β n r/a)j β m r/a) = δ nm 2 a3 j 2 +β n ),.62) which for n m foows from the orthogonaity property 0.68). So 2 A n = a 3 j + β n ),.63) and the Green s function has the eigenfunction expansion G k r,r ) = nm 2 a 3 j 2 + β n) where Ω = θ, φ), Ω = θ, φ ). Y m Ω)Y m Ω )j β n r/a)j β n r /a) k 2 β n /a) 2,.64)

10 32 Version of December 3, 20 CHAPTER. GREEN S FUNCTIONS This resut can be simpified by carrying out the sum on m, using the addition theorem for spherica harmonics, 4π 2 + m= Y m Ω )Y m Ω) = P cos γ),.65) where P cosγ) = P 0 cos γ) is Legendre s poynomia, and γ is the ange between the directions represented by Ω and Ω, or Then we obtain cosγ = cosθ cosθ + sin θ sin θ cosφ φ )..66) G k r,r ) = a 3 4π P cosγ) j+ 2 β n) This eads us to the second method. n.4.2 Discontinuity Direct) Method Let us adopt the anguar dependence found above: G k r,r ) = =0 j β n r/a)j β n r /a) k 2 β n /a) 2..67) 2 + 4π P cos γ)g r, r ),.68) where we wi ca g the reduced Green s function. Because Y m is an eigenfunction of the anguar part of the Lapacian operator, and the deta function can be written as 2 Y m + ) Ω) = r 2 Y m Ω),.69) δr r ) = rr δr r )δω Ω ),.70) we see that, because of the orthonormaity of the spherica harmonics, Eq..60), the Green s function equation.46) corresponds to the foowing equation satisfied by the reduced Green s function, the inhomogeneous spherica Besse equation, [ d 2 dr r ] d + ) dr r 2 + k 2 g r, r ) = rr δr r )..7) We sove this equation directy. For 0 < r < a) 0 r < r : g r, r ) = ar )j kr),.72a) r < r a : g r, r ) = br )j kr) + cr )n kr)..72b)

11 .4. HELMHOLTZ EQUATION INSIDE A SPHERE33 Version of December 3, 20 Ony j appears in the first form because the soution must be finite at r = 0, and the second soution to the spherica Besse equation, π n x) = 2x N +/2x),.73) where N ν is the Neumann function, is singuar at x = 0. For exampe, and in genera n 0 x) = cosx x,.74) n x) = x ) d cosx x dx x..75) To determine the functions a, b, and c, we proceed as foows. The boundary condition at r = a, g a, r ) = 0, impies or Thus we can write in the outer region, 0 = br )j ka) + cr )n ka),.76) br ) cr ) = n ka) j ka)..77) a r > r : g r, r ) = Ar )[j kr)n ka) n kr)j ka)]..78) The next condition we impose is that of the continuity of g at r = r : ar )j kr ) = Ar )[j kr )n ka) n kr )j ka)]..79) On the other hand, the derivative of g is discontinuous at r = r, as we may see by integrating Eq..7) over a tiny interva around r = r : which impies d dr g r, r ) r=r +ǫ r=r ǫ =,.80) r 2 kar )j kr ) kar )[j kr )n ka) n kr )j ka)] =..8) r 2 Now mutipy Eq..79) by kj kr ), and Eq..8) by j kr ), and subtract: j kr ) r 2 = kar )j ka)[j kr )n kr ) n kr )j kr )]..82) Now j, n are the independent soutions of the spherica Besse equation [ d r 2 r 2 d ) ] + ) dr dr r 2 + k 2 u = 0,.83)

12 34 Version of December 3, 20 CHAPTER. GREEN S FUNCTIONS the Wronskian of which, r) j kr)n kr) n kr)j kr).84) has the form r) = const. r 2,.85) as we saw in Probem 4 of Assignment 8. We can determine the constant by considering the asymptotic forms of j, n, which impy sinkr π/2) j kr), kr kr,.86a) coskr π/2) n kr), kr kr,.86b) r) = k 2 r 2 [sin2 kr π/2) + cos 2 kr π/2)] = kr) 2..87) Thus since the right-hand side of Eq..82) is proportiona to the Wronskian, we find the function A: Ar ) = k j kr ) j ka),.88) and then from Eq..79) we find the function a: ar ) = k j ka) [j kr )n ka) n kr )j ka)]..89) Hence the Green s function is expicity or r < r : g r, r ) = k j kr) j ka) [j kr )n ka) n kr )j ka)],.90a) r > r : g r, r ) = k j kr ) j ka) [j kr)n ka) n kr)j ka)],.90b) [ g r, r n ka) ) = kj kr < )j kr > ) j ka) n ] kr > ),.9) j kr > ) where r < is the esser of r, r, and r > is the greater. From this cosed form we may extract the eigenvaues and eigenfunctions of the spherica Besse differentia operator appearing in Eq..83). The poes of g occur where j ka) has zeroes, a of which are rea, at ka = β n, the nth zero of j, or ) 2 k 2 βn =..92) a

13 .5. HELMHOLTZ EQUATION IN UNBOUNDED SPACE35 Version of December 3, 20 In the neighborhood of this zero, But at the zero the Wronskian is Now from the recursion reation j ka) = ka β n )j β n)..93) β n ) 2 = n β n )j β n)..94) J λz) = λ z J λz) J λ+ z),.95) we see that the derivative of the spherica Besse function.50) satisfies, at the zero, j β n) = j + β n )..96) Thus the residue of the poe of g at k = β n /a is j β n r < /a)j β n r > /a) a 2 β n j+ 2 β..97) n) Now j is an even or odd function of z depending on whether n is even or odd. So if β n is a zero of j, so is β n, and hence if we add the contributions of these two poes, we get the corresponding contribution to g : g r, r ) j β n r/a)j β n r /a) a 2 β n [j + β n )] 2 k β n /a k + β n /a Summing up the contribution of a such pairs of poes, we obtain g r, r ) = 2 a 3 n= )..98) j β n r/a)j β n r /a) [j + β n )] 2 k 2 β n /a) 2,.99) which is the eigenfunction expansion dispayed in Eq..67)..5 Hemhotz Equation in Unbounded Space Again we are soving the equation 2 + k 2 )G k r,r ) = δr r ),.00) but now in unbounded space. The soution to this equation is an outgoing spherica wave: G k r,r ) = G k r r ) = e ik r r 4π r r..0)

14 36 Version of December 3, 20 CHAPTER. GREEN S FUNCTIONS This may be directy verified. Consider a sma sphere S, of radius ǫ, centered on r : dr) 2 + k 2 )G k r r ) dρ) 2 ρ e ikρ ) S S 4π ρ = dω ρ 2 d e ikρ ) dρ 4π ρ ρ=ǫ,.02) as ǫ 0. Evidenty, for r r, G k satisfies the Hemhotz equation, 2 + k 2 )G k = 0. Aternativey, we may construct G k from the eigenfunction expansion 0.09), G k r r ) = n ψ n r )ψ n r) λ n λ where λ = k 2, λ n = k 2, where the eigenfunctions are soutions of that is, they are pane waves,.03) 2 + k 2 )ψ k r) = 0,.04) ψ k r) = Here the 2π) 3/2 factor is for normaization: dk )ψ k r) ψ k r ) = δr r ), dr)ψ k r) ψ k r) = δk k ), r eik,.05) 2π) 3/2.06a).06b) where we have noted that the spectrum of eigenvaues is continuous, dk)..07) n Thus the eigenfunction expansion for the Green s function has the form dk G k r r ) e ik r e ik r ) = 2π) 3 k 2 k 2..08) Let us evauate this integra in spherica coordinates, where we write dk ) = k 2 dk dφ dµ, µ = cosθ,.09) where we have chosen the z axis to ie aong the direction of r r. The integration over the anges is easy: G k r r ) = 2π) 3 = 2π) π dk k 2 dφ 0 dk k 2 k 2 k 2 ik ρ dµ eik r r µ k 2 k 2 e ik ρ e ik ρ ),.0)

15 .5. HELMHOLTZ EQUATION IN UNBOUNDED SPACE37 Version of December 3, 20 k k Figure.: Contour in the k pane used to evauate the integra.0). The integra is cosed in the upper ower) haf-pane if the exponent is positive negative). The poes in the integrand are avoided by passing above the one on the eft, and beow the one on the right. defining ρ = r r, where we have repaced by 0 2 because the integrand is even in k 2. We evauate this integra by contour methods. Because now k can coincide with an eigenvaue k, we must choose the contour appropriatey to define the Green s function. Suppose we choose the contour as shown in Fig.., passing beow the poe at k and above the poe at k. We cose the contour in the upper haf pane for the e ikρ and in the ower haf pane for the e ikρ term. Then by Jordan s emma, we immediatey evauate the integra: G k r r ) = [ 2π) 2 2πi 2 2k k e ikρ iρ + 2πi 2k k e ikρ ] = e ikρ 4π ρ,.) which coincides with Eq..0). If a different contour defining the integra had been chosen, we woud have obtained a different Green s function, not one corresponding to outgoing spherica waves. Boundary conditions uniquey determine the contour. Note that G k r,r ) = G k r,r),.2) even though G k is compex. The sef-adjointness property 0.0) impied by the eigenfunction expansion is ony forma, and is spoied by the contour choice. iρ

16 38 Version of December 3, 20 CHAPTER. GREEN S FUNCTIONS.6 Green s Function for the Scaar Wave Equation The inhomogeneous scaar wave equation, 2 c 2 ) 2 t 2 ψr, t) = ρr, t),.3) requires boundary and initia conditions. The boundary conditions may be Dirichet, Neumann, or mixed. The initia conditions are Cauchy see Sec..2). Thus, we might specify at an initia time t = t 0 both ψr, t 0 ) and t ψr, t 0) at every point r in the region being considered. The corresponding Green s function Gr, t;r, t ) satisfies 2 c 2 ) 2 t 2 Gr, t;r, t ) = δr r )δt t )..4) It must satisfy the homogeneous form of the boundary conditions satisfied by ψ. Thus, if ψ has a specified vaue everywhere on the bounding surface, the corresponding Green s function must vanish on the surface. In cassica physics it is customary to adopt as initia conditions Gr, t;r, t ) G t r, t;r, t ) } = 0 if t < t..5) These then define the so-caed retarded Green s functions. They ensure that an effect occurs after its cause. In fact, however, this time asymmetry of the Green s function, which is not present in the wave equation, is not necessary; and in fact it is impossibe to maintain in reativistic quantum mechanics.) With such a Green s function, what takes the pace of the sef-adjointness property given in Sec. 0.8? Since the second time derivative is invariant under t t, we have in addition to the inhomogeneous equation.4) 2 c 2 ) 2 t 2 Gr, t;r, t ) = δr r )δt t )..6) Mutipy Eq..6) by Gr, t;r, t), Eq..4) by Gr, t;r, t ), subtract, and integrate over the voume being considered, and over t from to T, where T > t, t : T { dt dr) Gr, t;r, t ) 2 Gr, t;r, t ) V Gr, t;r, t ) 2 Gr, t;r, t ) Gr, t;r, t ) 2 c 2 t 2 Gr, t;r, t ) + Gr, t;r, t ) 2 } c 2 t 2 Gr, t;r, t ) = Gr, t ;r, t ) + Gr, t ;r, t )..7)

17 .6. GREEN S FUNCTION FOR THE SCALAR WAVE EQUATION39 Version of December 3, 20 Now use Green s theorem, together with the corresponding identity, to concude that A t t B B t ) A = A 2 t 2 B B 2 A,.8) t2 = Gr, t ;r, t ) Gr, t ;r, t ) { dt dσ Gr, t;r, t ) Gr, t;r, t ) Σ } Gr, t;r, t ) Gr, t;r, t ) dr) {Gr, V c 2 t;r, t ) t Gr, t;r, t ) Gr, t;r, t ) } t=t t Gr, t;r, t )..9) t= T The surface integra vanishes, since both Green s functions satisfy the same homogeneous boundary conditions on Σ. The boundary conditions are time independent.) The second integra is aso zero because from Eq..5) since < t, and } Gr, ;r, t ) G t r, ;r, t = 0,.20a) ) } Gr, T;r, t ) G t r, T;r, t = 0,.20b) ) since T < t. Thus the reciprocity reation here is Gr, t;r, t ) = Gr, t ;r, t).2) How do we express a soution to the wave equation.3) in terms of the Green s function? The procedure is the same as that given earier. The fied, and the Green s function, satisfy 2 ψr, t) c Gr, t;r, t ) c 2 2 t 2 ψr, t ) = ρr, t ),.22a) t 2 Gr, t;r, t ) = δr r )δt t )..22b) Note that the differentiations on G are with respect to the second set of arguments this equation foows from the reciprocity reation). Again mutipy the first equation by Gr, t;r, t ), the second by ψr, t ), subtract, integrate over the voume, and over t from t 0 < t to t +, where t + means t + ǫ, ǫ 0 through

18 40 Version of December 3, 20 CHAPTER. GREEN S FUNCTIONS positive vaues. Then for r V, t + dt t 0 V { dr ) Gr, t;r, t ) 2 ψr, t ) ψr, t ) 2 Gr, t;r, t ) c 2 [ Gr, t;r, t ) 2 t + = ψr, t) + dt t 0 ] } t 2 ψr, t ) ψr, t ) 2 t 2 Gr, t;r, t ) dr )Gr, t;r, t )ρr, t )..23) V Now we again use Green s theorem and the identity.8) to concude ψr, t) = t + dt dr )Gr, t;r, t )ρr, t ) t 0 V t + dt dσ [Gr, t;r, t ) ψr, t ) ψr, t ) Gr, t;r, t ) ] t 0 Σ [ c 2 dr ) Gr, t;r, t 0 ) ψr, t 0 ) ψr, t 0 ) ] Gr, t;r, t 0 ). t 0 t 0 V This is our resut. The interpretation is as foows:.24). The first integra represents the effect of the sources ρ distributed throughout the voume V. 2. The second integra represents the boundary conditions. If, for exampe, ψ satisfies inhomogeneous Neumann boundary conditions on Σ, ˆn ψ = fr ).25) Σ is specified, then we use homogeneous Neumann boundary conditions for G, ˆn Gr, t;r, t ) = 0..26) Σ Then the second integra reads t + dt dσ Gr, t;r, t) ψr, t )..27) t 0 Σ That is, ˆn ψr, t ) represents a surface source distribution. Other types of boundary conditions are as discussed earier. 3. The third integra represents the effect of the initia conditions, where ψr, t 0 ), t 0 ψr, t 0 ).28)

19 .7. WAVE EQUATION IN UNBOUNDED SPACE4 Version of December 3, 20 are specified. They correspond to impusive sources at t = t 0 : ρ init r, t ) = [ ] c 2 ψr, t 0 )δt t 0 ) + ψr, t 0 )δ t t 0 )..29) t 0 We verify this statement by integrating by parts, and etting the ower imit of the t integration be t 0 ǫ..7 Wave Equation in Unbounded Space We now wish to sove Eq..4) 2 c 2 ) 2 t 2 Gr, t;r, t ) = δr r )δt t ),.30) in unbounded space by noting that then G is a function of R = r r and T = t t ony, Gr, t;r, t ) = Gr r, t t ) = GR, T)..3) Then we can introduce a Fourier transform in space and time, gk, ω) = dr)dte ik R e iωt GR, T)..32) The Fourier transform of the Green s function equation is we have set c = temporariy for convenience), k 2 + ω 2 )gk, ω) =,.33) where we write k 2 = k k, which has the immediate soution gk, ω) = ω 2 k 2..34) Thus the Green s function has the forma representation dk) dω GR, T) = e ik R 2π) 3 e iωt 2π ω 2 k 2..35) The ω integra here is not we defined unti we impose the boundary condition.5) GR, T) = 0 if T < 0..36) This wi be true if the poes are ocated above the rea axis, as shown in Fig..2. Here the contour is cosed in the upper haf pane if T > 0, and in the ower haf pane if T < 0. In both cases, by Jordan s emma, the infinite semicirce gives no contribution. We have dω 2π eiωt ω k)ω + k) = { e i ikt 2k e ikt 2k ), T > 0, 0 T < 0..37)

20 42 Version of December 3, 20 CHAPTER. GREEN S FUNCTIONS k + iǫ k + iǫ Figure.2: Contour in the ω pane used to evauate the integra.35). Thus, if T > 0, GR, T) = 2π) 3 = i 2π) 2 0 = 2π) 2 k 2 dk 2π dµ e ikrµ i 2k k dk 0 2ikR 2R 2 e ikt e ikt) e ikr e ikr) e ikt e ikt) dk e ikr+t) + e ikr+t) e ikr T) ikt e R)) = [δr + T) δr T)]..38) 2π 2R But R and T are both positive, so R + T can never vanish. Thus we are eft with GR, T) = δr T),.39) 4π R or restoring c, Gr r, t t ) = r r ) 4π r r δ t t )..40) c The effect at the observation point r at time t is due to the action at the source point r at time t = t r r..4) c Physicay, this means that the signa propagates with speed c.

21 .7. WAVE EQUATION IN UNBOUNDED SPACE43 Version of December 3, 20 Let us make this more concrete by considering a simpe exampe, a point charge moving with veocity vt) = d dt rt), ρr, t) = qδr rt))..42) There are no effects from the infinite surface, nor from the infinite past, so we have from Eq..24) ψr, t) = t + dt V t + = q 4π dr )Gr r, t t )ρr, t ) dt r rt ) δ r rt ) c ) t t )..43) If we et Rt ) = r rt ), the distance from the source to the observation point at time t = t Rt ) c, we write this as ψr, t) = q t + dt Rt ) 4π Rt ) δ ) t t )..44) c Let τ = Rt )/c + t, where τ = t determines the retarded time t so dτ = dt + ) dr c dt,.45) where that is Thus the fied is evauated as ψr, t) = q 4π dr dt = 2R = q 4π d v dt R R = R R,.46) dτ = dt R v )..47) Rc t+rt)/c dτ Rτ) R v Rc ) τ) δτ t) Rt) Rt) vt)/c..48)