91Ó°ÊÓ

Equation 11.65is given by:

$G=\frac{e^{ikr}}{4\mathrm{π}r}$ …….. (1)

We need to check that satisfy equation which is given by:

$\left({∇}^2+k^2\right)G\left(r\right)={\mathrm{δ}}^3\left(r\right)$ ……… (2)

We have:

$∇G=-\frac{1}{4\mathrm{π}}\left(\frac{1}{r}∇e^{ikr}+e^{ikr}∇\frac{1}{r}\right)$

So, the first term of equation (2) is:

role="math" localid="1658314246252" ${∇}^{2}G=∇.\left(∇G\right).......\left(3\right)$

role="math" localid="1658314207228" ${∇}^{2}G=-\frac{1}{4\mathrm{π}}\left[2\left(∇\frac{1}{r}\right).\left(∇e^{ikr}\right)+\frac{1}{r}{∇}^2\left(e^{ikr}\right)+e^{ikr}{∇}^2\left(\frac{1}{r}\right)\right]......\left(4\right)$

But,

role="math" localid="1658314351283" $$\begin{gathered} ∇\frac{1}{r}=-\frac{1}{r^2}\hat{r}∇\left(e^{ikr}\right) \\ ∇\frac{1}{r}=ik{e}^{ikr}\hat{r} \end{gathered}$$

And,

$$\begin{gathered} {∇}^2{e}^{ikr}=ik∇.\left(e^{ikr}\hat{r}\right) \\ =ik\frac{1}{r^2}\frac{d}{dr}\left(r^2{e}^{ikr}\right) \\ =\frac{ik}{r^2}\left(2r{e}^{ikr}+ik{r}^2{e}^{ikr}\right) \\ =ik{e}^{ikr}\left(\frac{2}{r}+ik\right) \end{gathered}$$

Plug all these values into equation (3), so we get (note that ${∇}^2\left(1/r\right)=-4\mathrm{π}{\mathrm{δ}}^3\left(r\right)$):

$$\begin{gathered} {∇}^{2}G=-\frac{1}{4\mathrm{π}}\left[2\left(-\frac{1}{r^2}\hat{r}\right).\left(ik{e}^{ikr}\hat{r}\right)+\left(\frac{2}{r}+ik\right)-4\mathrm{π}{e}^{ikr}{\mathrm{δ}}^3\left(r\right)\right] \\ ={\mathrm{δ}}^3\left(r\right)-\frac{1}{4\mathrm{π}}{e}^{ikr}\left[-\frac{2ik}{r^2}+\frac{2ik}{r^2}-\frac{k^2}{r}\right] \\ ={\mathrm{δ}}^3\left(r\right)+k^2\frac{e^{ikr}}{4\mathrm{π}r} \\ ={\mathrm{δ}}^3\left(r\right)-k^{2}G \end{gathered}$$

Note that we used$e^{ikr}{\mathrm{δ}}^3\left(r\right)={\mathrm{δ}}^3\left(r\right)$in the second line. Thus:

$\left({∇}^2+k^2\right)G\left(r\right)={\mathrm{δ}}^3\left(r\right)$.