Solution of Inhomogeneous Wave Equation

Equation (14), as well as the three Cartesian components of Equation (15), are inhomogeneous three-dimensional wave equations of the general form

$$\left(\frac{1}{c^{\,2}}\frac{\partial^{\,2}}{\partial t^{\,2}}-\nabla^{\,2}\right)u = v,$$ (30)

where $u({\bf r}, t)$ is an unknown potential, and $v({\bf r}, t)$ a known source function. Let us investigate whether it is possible to find a unique solution of this type of equation.

Let us assume that the source function $v({\bf r}, t)$ can be expressed as a Fourier integral,

$$v({\bf r}, t) = \int_{-\infty}^{\infty} v_\omega({\bf r})\,{\rm e}^{-{\rm i}\, \omega\, t}\,d\omega.$$ (31)

The inverse transform is

$$v_\omega({\bf r}) = \frac{1}{2\pi} \int_{-\infty}^{\infty} v({\bf r}, t) \,{\rm e}^{+{\rm i}\, \omega \,t}\,d t.$$ (32)

Similarly, we can write the general potential $u({\bf r}, t)$ as a Fourier integral,

$$u({\bf r}, t) = \int_{-\infty}^{\infty} u_\omega({\bf r})\,{\rm e}^{-{\rm i}\, \omega \,t}\,d\omega,$$ (33)

with the corresponding inverse

$$u_\omega({\bf r}) = \frac{1}{2\pi} \int_{-\infty}^{\infty} u({\bf r}, t) \,{\rm e}^{+{\rm i}\, \omega \,t}\,d t.$$ (34)

Fourier transformation of Equation (30) yields

$$(\nabla^{\,2} + k^{\,2})\,u_\omega = - v_\omega,$$ (35)

where $k=\omega/c$ .

Equation (35), which reduces to Poisson's equation (see Section 2.3),

$$\nabla^{\,2}\,u_\omega = -v_\omega,$$ (36)

in the limit $k \rightarrow 0$ , is known as Helmholtz's equation. Because Helmholtz's equation is linear, it is appropriate to attempt a Green's function method of solution. Let us try to find a Green's function, $G_\omega({\bf r}, {\bf r}')$ , such that

$$(\nabla^{\,2} + k^{\,2})\,G_\omega ({\bf r}, {\bf r}')= -\delta({\bf r} - {\bf r}').$$ (37)

The general solution to Equation (35) is then [cf., Equation (142)]

$$u_\omega({\bf r}) = \int v_\omega({\bf r}')\,G_\omega ({\bf r}, {\bf r}')\,dV'.$$ (38)

Let us adopt the spatial boundary condition $G_\omega({\bf r}, {\bf r}')\rightarrow 0$ as $\vert{\bf r} -{\bf r}'\vert\rightarrow \infty$ , so as to ensure that the potential goes to zero a long way from the source. Because Equation (37) is spherically symmetric about the point ${\bf r}'$ , it is plausible that the Green's function itself is spherically symmetric: that is, $G_\omega({\bf r}-{\bf r}')=G_\omega(\vert{\bf r}-{\bf r}'\vert)$ . In this case, Equation (37) reduces to

$$\frac{1}{R}\frac{d^{\,2}(R\, G_\omega)}{dR^{\,2}} + k^{\,2} \,G_\omega = -\delta({\bf R}),$$ (39)

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

$$G_\omega (R) = \frac{A \,{\rm e}^{+{\rm i}\, k\,R} + B\, {\rm e}^{-{\rm i}\, k\,R}}{4\pi\, R}.$$ (40)

However, we know that Helmholtz's equation tends towards Poisson's equation in the limit $k \rightarrow 0$ . It stands to reason that the Green's function for Helmholtz's equation much tend toward that for Poisson's equation in the same limit. Now, the Green's function for Poisson's equation, (36), satisfies

$$\nabla^{\,2} G({\bf r}, {\bf r}') = -\delta({\bf r}-{\bf r}'),$$ (41)

as well as the usual constraint that $G({\bf r}, {\bf r}')\rightarrow 0$ as $\vert{\bf r} -{\bf r}'\vert\rightarrow \infty$ . It follows from Equation (25) that

$$G({\bf r}, {\bf r}')=G(\vert{\bf r}-{\bf r}'\vert) = \frac{1}{4\pi\,\vert{\bf r}-{\bf r'}\vert}= \frac{1}{4\pi\, R}.$$ (42)

Thus, the condition that $G_\omega(R)\rightarrow G(R)$ as $k \rightarrow 0$ implies that $A+B=1$ .

Reconstructing $u({\bf r}, t)$ from Equations (33), (38), and (40), we obtain

$$u({\bf r}, t) = \frac{1}{4\pi} \int\! \int\frac{ v_\omega({\bf r'... ...mega\,(t-R/c)}+ B \,{\rm e}^{-{\rm i}\, \omega\,(t+R/c)} \right]\,d\omega\,dV'.$$ (43)

It follows from Equation (31) that

$$u({\bf r}, t) = \frac{A}{4\pi} \int \frac{v({\bf r}', t- R/c)}{R}\, dV' + \frac{B}{4\pi} \int \frac{v({\bf r}', t+ R/c)}{R} \,dV'.$$ (44)

Now, the real-space Green's function for the inhomogeneous three-dimensional wave equation, (30), satisfies

$$\left(\frac{1}{c^{\,2}}\frac{\partial^{\,2}}{\partial t^{\,2}}-\n... ...right) G({\bf r}, {\bf r'}; t, t') = \delta({\bf r} - {\bf r}')\, \delta(t-t').$$ (45)

Hence, the most general solution of Equation (30) takes the form

$$u({\bf r}, t) =\int\!\int v({\bf r'}, t')\, G({\bf r}, {\bf r'}; t, t') \,dV' dt'.$$ (46)

Comparing Equations (44) and (46), we obtain

$$G({\bf r}, {\bf r}'; t, t') = A\,G^{(+)} ({\bf r}, {\bf r}'; t, t') + B\, G^{(-)} ({\bf r}, {\bf r}'; t, t'),$$ (47)

where

$$G^{(\pm)} ({\bf r}, {\bf r}'; t, t') = \frac{\delta(t'-[t\mp \vert{\bf r}- {\bf r'}\vert/c])}{4\pi \,\vert{\bf r}-{\bf r}'\vert},$$ (48)

and $A+B=1$ .

The real-space Green's function specifies the response of the system to a point source located at position ${\bf r}'$ that appears momentarily at time $t'$ . According to the retarded Green's function, $G^{(+)}$ , this response consists of a spherical wave, centered on the point ${\bf r}'$ , that propagates forward in time. In order for the wave to reach position ${\bf r}$ at time $t$ , it must have been emitted from the source at ${\bf r'}$ at the retarded time $t_r = t - \vert{\bf r}-{\bf r'}\vert/c$ . According to the advanced Green's function, $G^{(-)}$ , the response consists of a spherical wave, centered on the point ${\bf r}'$ , that 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 that is consistent with our experience is

$$G ({\bf r}, {\bf r}'; t, t')= G^{(+)} ({\bf r}, {\bf r}'; t, t') ... ...ta(t'-[t - \vert{\bf r}- {\bf r'}\vert/c])}{4\pi \,\vert{\bf r}-{\bf r}'\vert}.$$ (49)

Incidentally, we are able to find solutions of the inhomogeneous wave equation, (30), that 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, (30), that satisfies sensible boundary conditions at infinity, and is consistent with causality, is

$$u({\bf r}, t) = \int \frac{ v({\bf r}', t - \vert{\bf r} -{\bf r}'\vert/c)} {4\pi\,\vert{\bf r} - {\bf r}'\vert}\,dV'.$$ (50)

This expression is sometimes written

$$u({\bf r}, t) = \int \frac{ [v({\bf r}')]} {4\pi\,\vert{\bf r} - {\bf r}'\vert}\,dV',$$ (51)

where the rectangular bracket symbol $[\,]$ denotes that the terms inside the bracket are to be evaluated at the retarded time $t-\vert{\bf r} -{\bf r}'\vert/c$ . Note, in particular, from Equation (50), that if there is no source [i.e., if $v({\bf r}, t) =0$ ] then there is no field [i.e., $u({\bf r}, t)=0$ ]. But, is expression (50) really the only solution of Equation (30) that satisfies sensible boundary conditions at infinity? In other words, is this solution really unique? Unfortunately, there is a weak link in our derivation--between Equations (38) and (39)--where we assumed, without proof, that the Green's function for Helmholtz's equation, subject to the boundary condition $G_\omega({\bf r}, {\bf r}')\rightarrow 0$ as $\vert{\bf r} -{\bf r}'\vert\rightarrow \infty$ , is spherically symmetric. Let us try to fix this problem.

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

$$G_\omega = \frac{{\rm e}^{+{\rm i}\,k\,R}}{4\pi \,R}$$ (52)

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

$$G_\omega = \frac{{\rm e}^{-{\rm i}\,k\,R}}{4\pi \,R}$$ (53)

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

$$\lim_{R\rightarrow\infty} R \left(\frac{\partial G_\omega}{\partial R} -{\rm i}\,k\, G_\omega \right) = 0.$$ (54)

This is called the Sommerfeld radiation condition, and basically ensures that infinity is an absorber of radiation, but not a source. But, does this boundary condition uniquely select the spherically symmetric Green's function (52) as the solution of

$$(\nabla^{\,2} + k^{\,2})\, G_\omega(R, \theta, \varphi) = - \delta({\bf R}) ?$$ (55)

Here, $(R, \theta, \varphi)$ are spherical polar coordinates. If it does then we can be sure that Equation (50) represents the unique solution of the inhomogeneous wave equation, (30), that is consistent with causality.

Let us suppose that there are two different solutions of Equation (55), both of which satisfy the boundary condition (54), and revert to the unique (see Section 2.3) Green's function for Poisson's equation, (42), 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,

$$(\nabla^{\,2} + k^{\,2}) \,w =0,$$ (56)

throughout $V$ . According to the generalized (to deal with complex potentials) Green's theorem (see Section 2.9),

$$\int_V (w\, \nabla^{\,2} w^\ast- w^\ast\, \nabla^{\,2} w)\,dV = \... ...partial w^\ast}{\partial n} -w^\ast\, \frac{\partial w}{\partial n}\right)\,dS,$$ (57)

where $\partial/\partial n$ denotes a derivative normal to the surface in question. It is clear from Equation (56) that the volume integral is zero. It is also clear that the first surface integral is zero, because both $u_1$ and $u_2$ must revert to the Green's function for Poisson's equation in the limit $R\rightarrow 0$ . Thus,

$$\int_{\Sigma_\infty} \left( w\,\frac{\partial w^\ast}{\partial n} -w^\ast \,\frac{\partial w}{\partial n}\right)\,dS =0.$$ (58)

Equation (56) can be written

$$\frac{\partial^{\,2}(R\,w)}{\partial R^{\,2}} + \frac{D\,(R\,w)}{R^{\,2}} + k^{\,2}\,R\,w=0,$$ (59)

where $D$ is the spherical harmonic operator

$$D= \frac{1}{\sin\theta} \frac{\partial}{\partial\theta}\left( \si... ...ght) +\frac{1}{\sin^{\,2}\theta}\,\frac{\partial^{\,2}}{\partial\varphi^{\,2}}.$$ (60)

The most general solution to Equation (59) takes the form

$$w(R,\theta,\varphi) =\sum_{l,m=0,\infty} \left[ C_{l,m}\, h_l^{(1)}(k\,R) + D_{l,m}\, h_l^{(2)}(k\,R) \right] Y_{l,m}(\theta,\varphi).$$ (61)

Here, the $C_{l,m}$ and $D_{l,m}$ are arbitrary coefficients, the $Y_{l,m}$ are spherical harmonics (see Section 3.3), and

$$h_l^{(1,2)}(\rho) = \sqrt{\frac{\pi}{2\,\rho}}\,H_{l+1/2}^{1,2} (\rho),$$ (62)

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

$$H_n^1(\rho) = \sqrt{\frac{2}{\pi \,\rho}}\, {\rm e}^{+{\rm i}\,[\rho-(n+1/2)\,\pi/2]} \sum_{m=0, 1, 2, \cdots}\frac{(n,m)}{(-2\,{\rm i}\,\rho)^{\,m}},$$ (63)

$$H_n^{\,2}(\rho) = \sqrt{\frac{2}{\pi\, \rho}}\, {\rm e}^{-{\rm i}\,[\rho-(n+1/2)\,\pi/2]} \sum_{m=0, 1, 2, \cdots}\frac{(n,m)}{(+2\,{\rm i}\,\rho)^{\,m}},$$ (64)

where

$$(n,m) = \frac{(4\,n^{\,2}-1)\,(4\,n^{\,2}-9)\cdots (4\,n^{\,2}-\{2\,m -1\}^2)}{2^{\,2\,m}\, m!}$$ (65)

and $(n,0) =1$ . Note that the summations in Equations (63) and (64) terminate after $n+1/2$ terms.

The large-$R$ behavior of the $h_l^{(2)}(k\,R)$ functions is clearly inconsistent with the Sommerfeld radiation condition, (54). It follows that all of the $D_{l,m}$ in Equations (61) are zero. The most general solution can now be expressed in the form

$$w(R,\theta,\varphi) = \frac{{\rm e}^{+{\rm i}\,k\,R}}{R}\sum_{n=0,\infty} \frac{f_n(\theta,\varphi)}{R^{\,n}},$$ (66)

where the $f_n(\theta,\varphi)$ are various weighted sums of the spherical harmonics. Substitution of this solution into the differential equation (59) yields

$${\rm e}^{+{\rm i}\,k\,R} \sum_{n=0,\infty} \left[-\frac{2\,{\rm i... ...n}{R^{\,n+1}} +\frac{n\,(n+1)}{R^{\,n+2}} + \frac{D}{R^{\,n+2}}\right] f_n = 0.$$ (67)

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

$${\rm e}^{+{\rm i}\,k\,R} \sum_{n=0\,\infty} \frac{-2\,{\rm i}\,k\, (n+1) \,f_{n+1} + [n\,(n+1)+D]\,f_n }{R^{\,n+2}} =0,$$ (68)

which yields the recursion relation

$$2\,{\rm i} \,k\,(n+1) f_{n+1} = [n\,(n+1)+D]\,f_n.$$ (69)

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

Let us now consider the surface integral (58). Because we are interested in the limit $R\rightarrow\infty$ , we can replace $w$ by the first term of its expansion in (66), so that

$$\int_{\Sigma_\infty} \left(w\,\frac{\partial w^\ast}{\partial n} ... ... n}\right) \,dS = -2\,{\rm i} \,k \int \vert f_0\vert^{\,2}\,d{\mit\Omega} = 0,$$ (70)

where $d{\mit\Omega}$ is an element 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 Equation (55) that is consistent with the Sommerfeld radiation condition, and this is given by Equation (52). We can now be sure that Equation (50) is the unique solution of Equation (30), subject to the boundary condition (54). This boundary condition ensures that infinity is an absorber of electromagnetic radiation, but not an emitter, which seems entirely reasonable.