D.2 Some Green’s func­tions

D.2.1 The Pois­son equa­tion

The so-called Pois­son equa­tion is

$$- \nabla^2 u = f \qquad \nabla \equiv {\hat\imath}\frac{\p... ...ac{\partial}{\partial y} + {\hat k}\frac{\partial}{\partial z}$$

Here $f$ is sup­posed to be a given func­tion and $u$ an un­known func­tion that is to be found.

The so­lu­tion $u$ to the Pois­son equa­tion in in­fi­nite space may be found in terms of its so-called Green’s func­tion $G({\skew0\vec r})$. In par­tic­u­lar:

$$u({\skew0\vec r}) = \int_{{\rm all\ }{\underline{\skew0\vec... ...ac{1}{4\pi\vert{\skew0\vec r}-{\underline{\skew0\vec r}}\vert}$$

Loosely speak­ing, the above in­te­gral so­lu­tion chops func­tion $f$ up into spikes $f({\underline{\skew0\vec r}}){\rm d}^3{\underline{\skew0\vec r}}$. A spike at a po­si­tion ${\underline{\skew0\vec r}}$ then makes a con­tri­bu­tion $G({\skew0\vec r}-{\underline{\skew0\vec r}})f({\underline{\skew0\vec r}}){\rm d}^3{\underline{\skew0\vec r}}$ to $u$ at ${\skew0\vec r}$. In­te­gra­tion over all such spikes gives the com­plete $u$.

Note that of­ten, the Pois­son equa­tion is writ­ten with­out a mi­nus sign. Then there will be a mi­nus sign in $G$.

The ob­jec­tive is now to de­rive the above Green’s func­tion. To do so, first an in­tu­itive de­riva­tion will be given and then a more rig­or­ous one. (See also chap­ter 13.3.4 for a more phys­i­cal de­riva­tion in terms of elec­tro­sta­t­ics.)

The in­tu­itive de­riva­tion de­fines $G({\skew0\vec r})$ as the so­lu­tion due to a unit spike, i.e. a delta func­tion, lo­cated at the ori­gin. That means that $G$ $\vphantom0\raisebox{1.5pt}{$=$}$ $G({\skew0\vec r})$ is the so­lu­tion to

$$\nabla^2 G = \delta^3 \qquad \mbox{with}\quad G({\skew0\vec r})\to0 \quad\mbox{when}\quad {\skew0\vec r}\to\infty$$

Here $\delta^3$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\delta^3({\skew0\vec r})$ is the three-di­men­sion­al delta func­tion, de­fined as an in­fi­nite spike at the ori­gin that in­te­grates to 1.

By it­self the above de­f­i­n­i­tion is of course mean­ing­less: in­fin­ity is not a valid num­ber. To give it mean­ing, it is nec­es­sary to de­fine an ap­prox­i­mate delta func­tion, one that is merely a large spike rather than an in­fi­nite one. This ap­prox­i­mate delta func­tion $\delta^3_\varepsilon$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\delta^3_\varepsilon({\skew0\vec r})$ must still in­te­grate to 1 and will be re­quired to be zero be­yond some small dis­tance $\varepsilon$ from the ori­gin:

$$\int\delta^3_\varepsilon({\skew0\vec r}){\,\rm d}^3{\skew0\... ...\vec r}\vert \mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon$$

In the above in­te­gral the re­gion of in­te­gra­tion should at least in­clude the small re­gion of ra­dius $\varepsilon$ around the ori­gin. The ap­prox­i­mate delta func­tion will fur­ther be as­sumed to be non­neg­a­tive. It must have large val­ues in the small vicin­ity around the ori­gin where it is nonzero; oth­er­wise the in­te­gral over the small vicin­ity would be small in­stead of 1. But the key is that the val­ues are not in­fi­nite, just large. So nor­mal math­e­mat­ics can be used.

The cor­re­spond­ing ap­prox­i­mate Green’s func­tion $G_\varepsilon$ $\vphantom0\raisebox{1.5pt}{$=$}$ $G_\varepsilon({\skew0\vec r})$ of the Pois­son equa­tion sat­is­fies

$$-\nabla^2 G_\varepsilon = \delta^3_\varepsilon \qquad \mbo... ..._\varepsilon\to0 \quad\mbox{when}\quad {\skew0\vec r}\to\infty$$

In the limit $\varepsilon\to0$, $\delta^3_\varepsilon({\skew0\vec r})$ be­comes the Dirac delta func­tion $\delta^3({\skew0\vec r})$ and $G_\varepsilon({\skew0\vec r})$ be­comes the ex­act Green’s func­tion $G({\skew0\vec r})$.

To find the ap­prox­i­mate Green’s func­tion, it will be as­sumed that $\delta^3_\varepsilon({\skew0\vec r})$ only de­pends on the dis­tance $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert{\skew0\vec r}\vert$ from the ori­gin. In other words, it is as­sumed to be spher­i­cally sym­met­ric. Then so is $G_\varepsilon$. (Note that this as­sump­tion is not strictly nec­es­sary. That can be seen from the gen­eral so­lu­tion for the Pois­son equa­tion given ear­lier. But it should at least be as­sumed that $\delta^3_\varepsilon({\skew0\vec r})$ is non­neg­a­tive. If it could have ar­bi­trar­ily large neg­a­tive val­ues, then $G_\varepsilon$ could be any­thing.)

Now in­te­grate both sides of the Pois­son equa­tion over a sphere of a cho­sen ra­dius $r$:

$$- \int_{\vert{\underline{\skew0\vec r}}\vert\,\mathrel{\rai... ...,r} \delta^3_\varepsilon {\,\rm d}^3{\underline{\skew0\vec r}}$$

As noted, the delta func­tion in­te­grates to 1 as long as the vicin­ity of the ori­gin is in­cluded. That means that the right hand side is 1 as long as $r$ $\raisebox{-.5pt}{$\geqslant$}$ $\varepsilon$. This will now be as­sumed. The left hand side can be writ­ten out. That gives

$$- \int_{\vert{\underline{\skew0\vec r}}\vert\,\mathrel{\rai... ...x{if}\quad r \mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon$$

Ac­cord­ing to the [di­ver­gence] [Gauss] [Os­tro­grad­sky] the­o­rem, the left hand side can be writ­ten as a sur­face in­te­gral to give

$$- \int_{\vert{\underline{\skew0\vec r}}\vert=r} {\vec n}\cd... ...x{if}\quad r \mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon$$

Here $S$ stands for the sur­face of the sphere of ra­dius $r$. The to­tal sur­face is $4{\pi}r^2$. Also ${\vec n}$ is the unit vec­tor or­thog­o­nal to the sur­face, in the out­ward di­rec­tion. That is the ra­dial di­rec­tion. The to­tal dif­fer­en­tial of cal­cu­lus then im­plies that ${\vec n}\cdot\nabla{G}_\varepsilon$ is the ra­dial de­riv­a­tive $\partial{G}_\varepsilon$$\raisebox{.5pt}{$/$}$​$\partial{r}$. So,

$$- \frac{\partial G_\varepsilon}{\partial r} 4\pi r^2 = 1 \... ...x{if}\quad r \mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon$$

Be­cause $G_\varepsilon$ is re­quired to van­ish at large dis­tances, this in­te­grates to

$$G_\varepsilon = \frac{1}{4\pi r} \quad\mbox{if}\quad r \mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon$$

The ex­act Green’s func­tion $G$ has $\varepsilon$ equal to zero, so

$$G = \frac{1}{4\pi r} \quad\mbox{if}\quad r \ne 0$$

Fi­nally the rig­or­ous de­riva­tion with­out us­ing poorly de­fined things like delta func­tions. In the sup­posed gen­eral so­lu­tion of the Pois­son equa­tion given ear­lier, change in­te­gra­tion vari­able to $\vec\rho$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline{\skew0\vec r}}-{\skew0\vec r}$

$$u({\skew0\vec r}) = \int_{{\rm all\ }{\skew0\vec r}} G({\sk... ...vec\rho \qquad G(\vec\rho) = \frac{1}{4\pi\vert\vec\rho\vert}$$

It is to be shown that the func­tion $u$ de­fined this way sat­is­fies the Pois­son equa­tion $\nabla^2u({\skew0\vec r})$ $\vphantom0\raisebox{1.5pt}{$=$}$ $f({\skew0\vec r})$. To do so, ap­ply $\nabla$ twice:

$$\nabla^2 u({\skew0\vec r}) = \int_{{\rm all\ }\vec\rho} G(... ...) \nabla_\rho^2 f({\skew0\vec r}+\vec\rho) {\,\rm d}^3\vec\rho$$

Here $\nabla_\rho$ means dif­fer­en­ti­a­tion with re­spect to the com­po­nents of $\vec\rho$ in­stead of the com­po­nents of ${\skew0\vec r}$. Be­cause $f$ de­pends only on ${\skew0\vec r}+\vec\rho$, you get the same an­swer whichever way you dif­fer­en­ti­ate.

It will be as­sumed that the func­tion $f$ is well be­haved, at least con­tin­u­ous, and be­comes zero rea­son­ably quickly at in­fin­ity. In that case, you can get a valid ap­prox­i­ma­tion to the in­te­gral above if you ex­clude very small and very large val­ues of ${\skew0\vec r}$:

$$\nabla^2 u({\skew0\vec r}) \approx \int_{\varepsilon<\vert\... ...) \nabla_\rho^2 f({\skew0\vec r}+\vec\rho) {\,\rm d}^3\vec\rho$$

In par­tic­u­lar, this ap­prox­i­ma­tion be­comes ex­act in the lim­its where the con­stants $\varepsilon\to0$ and $R\to\infty$. The in­te­gral can now be rewrit­ten as

$$\nabla^2 u({\skew0\vec r}) \approx \int_{\varepsilon<\vert\... ...ec r}+\vec\rho) \nabla_\rho^2 G(\vec\rho) {\,\rm d}^3\vec\rho$$

as can be ver­i­fied by ex­plic­itly dif­fer­en­ti­at­ing out the three terms of the in­te­grand. Next note that the third term is zero, be­cause as seen above $G$ sat­is­fies the ho­mo­ge­neous Pois­son equa­tion away from the ori­gin. And the other two terms can be writ­ten out us­ing the [di­ver­gence] [Gauss] [Os­tro­grad­sky] the­o­rem much like be­fore. This pro­duces in­te­grals over both the bound­ing sphere of ra­dius $R$, as well as over the bound­ing sphere of ra­dius $\varepsilon$. But the in­te­grals over the sphere of ra­dius $R$ will be van­ish­ingly small if $f$ be­comes zero suf­fi­ciently quickly at in­fin­ity. Sim­i­larly, the in­te­gral of the first term over the small sphere is van­ish­ingly small, be­cause $G$ is 1$\raisebox{.5pt}{$/$}$​$4\pi\varepsilon$ on the small sphere but the sur­face of the small sphere is $4\pi\varepsilon^2$. How­ever, in the sec­ond term, the de­riv­a­tive of $G$ in the neg­a­tive ra­dial di­rec­tion is 1/$4\pi\varepsilon^2$, which mul­ti­plies to 1 against the sur­face of the small sphere. There­fore the sec­ond term pro­duces the av­er­age of $f({\skew0\vec r}+\vec\rho)$ over the small sphere, and that be­comes $f({\skew0\vec r})$ in the limit $\vert\vec\rho\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\varepsilon\to0$. So the Pois­son equa­tion ap­plies.

D.2.2 The screened Pois­son equa­tion

The so-called screened Pois­son equa­tion is

$$- \nabla^2 u + c^2 u = f \qquad \nabla \equiv {\hat\imath}... ...ac{\partial}{\partial y} + {\hat k}\frac{\partial}{\partial z}$$

Here $f$ is sup­posed to be a given func­tion and $u$ an un­known func­tion that is to be found. Fur­ther $c$ is a given con­stant. If $c$ is zero, this is the Pois­son equa­tion. How­ever, nonzero $c$ cor­re­sponds to the in­ho­mo­ge­neous steady Klein-Gor­don equa­tion for a par­ti­cle with nonzero mass.

The analy­sis of the screened Pois­son equa­tion is al­most the same as for the Pois­son equa­tion given in the pre­vi­ous sub­sec­tion. There­fore only the dif­fer­ences will be noted here. The ap­prox­i­mate Green’s func­tion must sat­isfy, away from the ori­gin,

$$- \nabla^2 G_\varepsilon + c^2 G_\varepsilon = 0 \quad\mbox{if}\quad r \mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon$$

The so­lu­tion to this that van­ishes at in­fin­ity is of the form

$$G_\varepsilon = C \frac{e^{-cr}}{r} \quad\mbox{if}\quad r \mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon$$

where $C$ is some con­stant. To check this, plug it in, us­ing the ex­pres­sion (N.5) for $\nabla^2$ in spher­i­cal co­or­di­nates. To iden­tify the con­stant $C$, in­te­grate the full equa­tion

$$- \nabla^2 G_\varepsilon + c^2 G_\varepsilon = \delta^3_\varepsilon$$

over a sphere of ra­dius $\varepsilon$ around the ori­gin and ap­ply the di­ver­gence the­o­rem as in the pre­vi­ous sub­sec­tion. Tak­ing the limit $\varepsilon\to0$ then gives $C$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1/$4\pi$, which gives the ex­act Green’s func­tion as

$$G({\skew0\vec r}) = \frac{e^{-cr}}{4\pi r} \quad\mbox{if}\quad r = \vert{\skew0\vec r}\vert \ne 0$$

The rig­or­ous de­riva­tion is the same as be­fore save for an ad­di­tional $c^2Gf$ term in the in­te­grand, which drops out against the $-f\nabla_\rho^2G$ one.