Solution of the inhomogeneous wave equation

Equations (2.93) all have the general form

$$\Box \psi({\bfm r}, t) = g({\bfm r}, t).$$ (139)

Can we find a unique solution to the above equation? Let us assume that the source function $g({\bfm r}, t)$ can be expressed as a Fourier integral

$$g({\bfm r}, t) = \int_{-\infty}^{\infty} g_\omega({\bfm r})\,e^{-{\rm i}\, \omega t}\,d\omega.$$ (140)

The inverse transform is

$$g_\omega({\bfm r}) = \frac{1}{2\pi} \int_{-\infty}^{\infty} g({\bfm r}, t) \,e^{\,{\rm i}\, \omega t}\,d t.$$ (141)

Similarly, we may write the general potential $\psi({\bfm r}, t)$ as a Fourier integral

$$\psi({\bfm r}, t) = \int_{-\infty}^{\infty} \psi_\omega({\bfm r})\,e^{-{\rm i}\, \omega t}\,d\omega,$$ (142)

with the corresponding inverse

$$\psi_\omega({\bfm r}) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \psi({\bfm r}, t) \,e^{\,{\rm i}\, \omega t}\,d t.$$ (143)

Fourier transformation of Eq. (2.103) yields

$$(\nabla^2 + k^2)\psi_\omega = - g_\omega,$$ (144)

where $k=\omega/c$.

The above equation, which reduces to Poisson's equation in the limit $k \rightarrow 0$, and is called Helmholtz's equation, is linear, so we may attempt a Green's function method of solution. Let us try to find a function $G_\omega({\bfm r}, {\bfm r}')$ such that

$$(\nabla^2 + k^2)G_\omega ({\bfm r}, {\bfm r}')= -\delta({\bfm r} - {\bfm r}').$$ (145)

The general solution is then

$$\psi_\omega({\bfm r}) = \int g_\omega({\bfm r}')\,G_\omega ({\bfm r}, {\bfm r}')\,dV'.$$ (146)

The ``sensible'' spatial boundary conditions which we impose are that $G_\omega({\bfm r}, {\bfm r}')\rightarrow 0$ as $\vert{\bfm r} -{\bfm r}'\vert\rightarrow \infty$. In other words, the field goes to zero a long way from the source. Since the system we are solving is spherically symmetric about the point ${\bfm r}'$ it is plausible that the Green's function itself is spherically symmetric. It follows that

$$\frac{1}{R}\frac{d^2(R\, G_\omega)}{dR^2} + k^2 G_\omega = -\delta({\bfm R}),$$ (147)

where ${\bfm R} = {\bfm r - r}'$ and $R=\vert{\bfm R}\vert$. The most general solution to the above equation in the region $R>0$ is¹

$$G_\omega (R) = \frac{A \,e^{\,{\rm i}\, kR} + B\, e^{-{\rm i}\, kR}}{4\pi\, R}.$$ (148)

We know that in the limit $k \rightarrow 0$ the Green's function for Helmholtz's equation must tend towards that for Poisson's equation, which is

$$G_\omega(R) = \frac{1}{4\pi\, R}.$$ (149)

This is only the case if $A+B=1$.

Reconstructing $\psi({\bfm r}, t)$ from Eqs. (2.106), (2.110), and (2.112), we obtain

$$\psi({\bfm r}, t) = \frac{1}{4\pi} \int \int\frac{ g_\omega(... ...R/c)}+ B \,e^{-{\rm i}\, \omega(t+R/c)} \right]\,d\omega\,dV'.$$ (150)

It follows from Eq. (2.104) that

$$\psi({\bfm r}, t) = \frac{A}{4\pi} \int \frac{g({\bfm r}', t... ...V' + \frac{B}{4\pi} \int \frac{g({\bfm r}', t+ R/c)}{R} \,dV'.$$ (151)

Now, the real space Green's function for the inhomogeneous wave equation (2.103) satisfies

$$\Box G({\bfm r}, {\bfm r'}; t, t') = \delta({\bfm r} - {\bfm r}')\, \delta(t-t').$$ (152)

Hence, the most general solution of this equation takes the form

$$\psi({\bfm r}, t) =\int\int g({\bfm r'}, t')\, G({\bfm r}, {\bfm r'}; t, t') \,dV' dt'.$$ (153)

Comparing Eqs. (2.115) and (2.117) we obtain

$$G({\bfm r}, {\bfm r}'; t, t') = A\,G^{(+)} ({\bfm r}, {\bfm r}'; t, t') + B\, G^{(-)} ({\bfm r}, {\bfm r}'; t, t'),$$ (154)

where

$$G^{(\pm)} ({\bfm r}, {\bfm r}'; t, t') = \frac{\delta(t'-[t\... ...r}- {\bfm r'}\vert/c])}{4\pi \,\vert{\bfm r}-{\bfm r}'\vert},$$ (155)

and $A+B=1$.

The real space Green's function specifies the response of the system to a point source at position ${\bf r}'$ which appears momentarily at time $t'$. According to the retarded Green's function $G^{(+)}$ the response consists of a spherical wave, centred on ${\bfm r}'$, which propagates forward in time. In order for the wave to reach position ${\bfm r}$ at time $t$ it must have been emitted from the source at ${\bfm r'}$ at the retarded time $t_r = t - \vert{\bfm r}-{\bfm r'}\vert/c$. According to the advanced Green's function $G^{(-)}$ the response consists of a spherical wave, centred on ${\bfm r}'$, which propagates backward in time. Clearly, the advanced potential is not consistent with our ideas about causality, which demand that an effect can never precede its cause in time. Thus, the Green's function which is consistent with our experience is

$$G ({\bfm r}, {\bfm r}'; t, t')= G^{(+)} ({\bfm r}, {\bfm r}'... ...r}- {\bfm r'}\vert/c])}{4\pi \,\vert{\bfm r}-{\bfm r}'\vert}.$$ (156)

We are able to find solutions of the inhomogeneous wave equation (2.103) which propagate backward in time because this equation is time symmetric (i.e., it is invariant under the transformation $t\rightarrow -t$).

In conclusion, the most general solution of the inhomogeneous wave equation (2.103) which satisfies sensible boundary conditions at infinity and is consistent with causality is

$$\psi({\bfm r}, t) = \int \frac{ g({\bfm r}', t - \vert{\bfm ... ...\bfm r}'\vert/c)} {4\pi\,\vert{\bfm r} - {\bfm r}'\vert}\,dV'.$$ (157)

This expression is sometimes written

$$\psi({\bfm r}, t) = \int \frac{ [g({\bfm r}')]} {4\pi\,\vert{\bfm r} - {\bfm r}'\vert}\,dV',$$ (158)

where the rectangular bracket symbol $[\,]$ denotes that the terms inside the bracket are to be evaluated at the retarded time $t-\vert{\bfm r} -{\bfm r}'\vert/c$. Note, in particular, from Eq. (2.122) that if there is no source (i.e., $g({\bfm r}, t) =0$) then there is no field (i.e., $\psi({\bfm r}, t)=0$). But, is the above solution really unique? Unfortunately, there is a weak link in our derivation, between Eqs. (2.110) and (2.111), where we assume that the Green's function for the Helmholtz equation subject to the boundary condition $G_\omega({\bfm r}, {\bfm r}')\rightarrow 0$ as $\vert{\bfm r} -{\bfm r}'\vert\rightarrow \infty$ is spherically symmetric. Let us try to fix this problem.

With the benefit of hindsight, we can see that the Green's function

$$G_\omega(R) = \frac{e^{{\rm i}\,kR}}{4\pi \,R}$$ (159)

corresponds to the retarded solution in real space and is, therefore, the correct physical Green's function. The Green's function

$$G_\omega(R) = \frac{e^{-{\rm i}\,kR}}{4\pi \,R}$$ (160)

corresponds to the advanced solution in real space and must, therefore, be rejected. We can select the retarded Green's function by imposing the following boundary condition at infinity

$$\lim_{R\rightarrow\infty} R\!\left(\frac{\partial G}{\partial R} -{\rm i}\,k G \right) = 0.$$ (161)

This is called the Sommerfeld radiation condition; it basically ensures that sources radiate waves instead of absorbing them. But, does this boundary condition uniquely select the spherically symmetric Green's function (2.123) as the solution of

$$(\nabla^2 + k^2) G_\omega(R, \theta, \varphi) = - \delta({\bfm R}) ?$$ (162)

Here, $(R, \theta, \varphi)$ are spherical polar coordinates. If it does then we can be sure that Eq. (2.122) represents the unique solution of the wave equation (2.103) which is consistent with causality.

Let us suppose that there are two solutions of Eq. (2.126) which satisfy the boundary condition (2.125) and revert to the unique Green's function for Poisson's equation (2.113) in the limit $R\rightarrow 0$. Let us call these solutions $u_1$ and $u_2$, and let us form the difference $w=u_1-u_2$. Consider a surface $\Sigma_0$ which is a sphere of arbitrarily small radius centred on the origin. Consider a second surface $\Sigma_\infty$ which is a sphere of arbitrarily large radius centred on the origin. Let $V$ denote the volume enclosed by these surfaces. The difference function $w$ satisfies the homogeneous Helmholtz equation,

$$(\nabla^2 + k^2) w =0,$$ (163)

throughout $V$. According to Green's theorem

$$\int_V (w \nabla^2 w^\ast- w^\ast \nabla^2 w)\,dV = \left( \... ...{\partial n} -w^\ast \frac{\partial w}{\partial n}\right)\,dS,$$ (164)

where $\partial/\partial n$ denotes a derivative normal to the surface in question. It is clear from Eq. (2.127) that the volume integral is zero. It is also clear that the first surface integral is zero, since both $u_1$ and $u_2$ must revert to the Green's function for Poisson's equation in the limit $R\rightarrow 0$. Thus,

$$\int_{\Sigma_\infty} \left( w\frac{\partial w^\ast}{\partial n} -w^\ast \frac{\partial w}{\partial n}\right)\,dS =0.$$ (165)

Equation (2.127) can be written

$$\frac{\partial^2(R\,w)}{\partial R^2} + \frac{D(R\,w)}{R^2} + k^2\,Rw=0,$$ (166)

where $D$ is the spherical harmonic operator

$$D= \frac{1}{\sin\theta} \frac{\partial}{\partial\theta}\!\le... ...) +\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}.$$ (167)

The most general solution of Eq. (2.130) takes the form (see Section 7)

$$w(R,\theta,\varphi) =\sum_{l,m=0}^\infty \left[ C_{lm}\, h_l... ...kR) + D_{lm}\, h_l^{(2)}(kR)\, \right] Y_{lm}(\theta,\varphi).$$ (168)

Here, the $C_{lm}$ and $D_{lm}$ are arbitrary coefficients, the $Y_{lm}$ are spherical harmonics,² and

$$h_l^{(1,2)}(\rho) = \sqrt{\frac{\pi}{2\rho}}\,H_{l+1/2}^{1,2} (\rho),$$ (169)

where $H_n^{1,2}$ are Hankel functions of the first and second kind.³ It can be demonstrated that⁴

$$H_n^1(\rho) = \sqrt{\frac{2}{\pi \rho}}\, e^{\,{\rm i}\,(\rho-(n+1/2)\pi/2)} \sum_{m=0, 1, 2, \cdots}\frac{(n,m)}{(-2{\rm i}\,\rho)^m},$$ (170)

$$H_n^2(\rho) = \sqrt{\frac{2}{\pi \rho}}\, e^{-{\rm i}\,(\rho-(n+1/2)\pi/2)} \sum_{m=0, 1, 2, \cdots}\frac{(n,m)}{(+2{\rm i}\,\rho)^m},$$ (171)

where

$$(n,m) = \frac{(4n^2-1)(4n^2-9)\cdots (4n^2-\{2m -1\}^2)}{2^{2m}\, m!}$$ (172)

and $(n,0) =1$. Note that the summations in Eqs. (2.314) terminate after $n+1/2$ terms.

The large $R$ behaviour of the $h_l^{(2)}$ is clearly inconsistent with the Sommerfeld radiation condition (2.125). It follows that all of the $D_{lm}$ in Eq. (2.132) are zero. The most general solution can now be expressed in the form

$$w(R,\theta,\varphi) = \frac{e^{\,{\rm i}\,kR}}{R}\sum_{n=0}^{\infty} \frac{f_n(\theta,\varphi)}{R^n},$$ (173)

where the $f_n(\theta,\varphi)$ are various weighted sums of the spherical harmonics. Substitution of this solution into the differential equation (2.130) yields

$$e^{\,{\rm i}\,kR} \sum_{n=0}^{\infty} \left(-\frac{2{\rm i}\... ...}} +\frac{n(n+1)}{R^{n+2}} + \frac{D}{R^{n+2}}\right) f_n = 0.$$ (174)

Replacing the index of summation $n$ in the first term of the parentheses by $n+1$ we obtain

$$e^{\,{\rm i}\,kR} \sum_{n=0}^{\infty} \frac{-2{\rm i}k\, (n+1) f_{n+1} + [n(n+1)+D]f_n }{R^{n+2}} =0,$$ (175)

which gives us the recursion relation

$$2{\rm i} \,k(n+1) f_{n+1} = [n(n+1)+D]f_n.$$ (176)

It follows that if $f_0=0$ then all of the $f_n$ are equal to zero.

Let us now consider the surface integral (2.129). Since we are interested in the limit $R\rightarrow\infty$ we can replace $w$ by the first term of its expansion in (2.136), so

$$\int_{\Sigma_\infty} \left(w\frac{\partial w^\ast}{\partial ... ...ight) \,dS = -2{\rm i} \,k \int \vert f_0\vert^2\,d\Omega = 0,$$ (177)

where $d\Omega$ is a unit of solid angle. It is clear that $f_0=0$. This implies that $f_1=f_2=\cdots = 0$ and, hence, that $w=0$. Thus, there is only one solution of Eq. (2.126) which is consistent with the Sommerfeld radiation condition, and this is given by Eq. (2.123). We can now be sure that Eq. (2.122) is a unique solution of Eq. (2.103) subject to the boundary condition (2.125). This boundary condition basically says that infinity is an absorber of radiation but not an emitter, which seems entirely reasonable.