2.2 The Poisson equation in infinite space

(Book, section 8.2)

2.2.1 Overview

This section works out the Green’s function idea for the Poisson equation

$$\nabla^2 u = f$$

in infinite space. In two dimensions, this can be understood to be heat conduction in an infinite plate, with $u$ the temperature and $-f$ the heat added to the plate per unit surface area. In three dimensions, the heat could be generated internally, due to chemical or nuclear reactions, say, or be due to absorbed radiation passing through the body. Or $f$ can be understood to be charge density in electro-statics, with $u$ being the potential. In the fluid dynamics of viscous unidirectional flows, $f$ would be the pressure gradient. In various numerical schemes for viscous incompressible flows, $u$ could be pressure and $f$ velocity terms, or in two dimensions $u$ could be streamfunction and $f$ vorticity.

In this section, the Green’s function in infinite two-dimensional space will be derived. For discussion purposes, the problem will be assumed to be heat conduction in a plate, but the mathematical solution does not depend on what the physical meaning is. In the homework you will derive the Green’s function for the Poisson equation in infinite three-dimensional space; the analysis is similar but the result will be quite different.

First of all, a Green’s function $G$ for the above problem is by definition a solution when function $f$ is a delta function. A delta function is an infinitely narrow spike that integrates to one. We will write $\vec{x}$ for the point at which the temperature is desired. Further $\vec\xi$ is the position of the delta function.

In two dimensions $\vec{x}=(x,y)$ and $\vec\xi=(\xi,\eta)$. Also, the two-dimensional delta function can be written in terms of one-dimensional ones as

$$\delta^2(\vec x-\vec\xi) = \delta(x-\xi)\delta(y-\eta)$$

In any number of dimensions, the Green’s function solution of the Poisson equation with a delta function as right hand side will be written as

$$G(\vec x;\vec\xi)$$

(Since it is a constant coefficient problem, the Green’s function will simplify to $G(\vec{x}-\vec\xi)$, and even further to $G(\vert\vec{x}-\vec\xi\vert)$ because the problem is rotationally symmetric.)

In terms of the Green’s function, the solution $u$ to the Poisson equation with an arbitrary right hand side $f$ can be written as

$$u(\vec x) = \int_{{\rm all }\xi} G(\vec x;\vec\xi) f(\vec\xi){ \rm d}V_{\vec\xi}$$ (2.3)

Here ${ \rm d}{V}_{\vec\xi}$ stands for a “volume” integral over all components of vector $\vec\xi$. In two dimensions that becomes

$$u(x,y) = \int_{-\infty}^\infty \int_{-\infty}^\infty G(x,y;\xi,\eta) f(\xi,\eta){ \rm d}\xi{\rm d}\eta$$ (2.4)

The “volume” is here really an area.

It will be found that the Green’s function for the two-dimensional infinite-domain Poisson problem is:

$$\fbox{$\displaystyle G(\vec x;\vec\xi) = \frac{1}{2\pi}\... ...ec x-\vec\xi\right\vert = \sqrt{(x-\xi)^2+(y-\eta)^2} $} %$$ (2.5)

The physical meaning of $d$ is the distance between the point $\vec\xi=(\xi,\eta)$ where the heat is added and the point $\vec{x}=(x,y)$ at which the temperature is desired.

Similarly, you will find in the homework that the Green’s function for the three-dimensional infinite-domain Poisson problem is:

$$\fbox{$\displaystyle G(\vec x;\vec\xi) = \frac{-1}{4\pi ... ...right\vert = \sqrt{(x-\xi)^2+(y-\eta)^2+(z-\theta)^2} $} %$$ (2.6)

Therefore the two-dimensional temperature distribution $u(x,y)$ corresponding to a general distribution of added heat $f(\xi,\eta)$ is:

$$\fbox{$\displaystyle u(x,y) = \int\!\!\int \frac{f(\x... ...\ln\sqrt{(x-\xi)^2+(y-\eta)^2}{ \rm d}\xi{\rm d}\eta $} %$$ (2.7)

while the three-dimensional temperature distribution is

$$\fbox{$\displaystyle u(x,y,z) = \int\!\!\int\!\!\int ... ...^2+(z-\theta)^2}}{ \rm d}\xi{\rm d}\eta{\rm d}\theta $} %$$ (2.8)

Given an added-heat distribution $f$, you can find the temperature distribution $u$ by doing the corresponding integral. Especially if you only want the temperature at a few points, this can be quite effective. (If you want the temperature at essentially all points, a “multigrid” numerical method that directly solves the partial differential equation is far more efficient. However, there are so-called “fast-summation” methods, like the one by Van Dommelen and Ründensteiner, that can do the integrals very fast too, especially if the region of heat addition is limited.)

The physical meaning of the Green‘s function varies with setting. In heat transfer, it is the solution for a point heat source, in electrostatics a point charge, in gravitation a point mass, in potential flows a point source of fluid, in two-dimensional vortex flows a point vortex, etcetera.

2.2.2 Loose derivation

To verify the two-dimensional Green’s function $G$ given in the previous section, the solution to the Poisson equation must be found in which $f$ is a delta function spike at some point $\vec\xi$.

However, dealing with infinite functions like delta functions is a very abstract and fishy problem. Therefore an approach like in the first section will be used. It will assumed that we are really trying to solve the Poisson equation for an arbitrary function $f$. (And so we are, really.) We then mentally cut up this function $f$ into spikes. That idea is sketched in two-dimensions in figure fig:2dspike.

Figure 2.6: One of the spikes out of which an arbitrary two-dimensional function $f$ consists is shown in outline.

The problem for such a narrow, but finite spike can be solved with some physical intuition. The solution will again be called $\Delta{u}$. The total solution will then be the sum of the solutions $\Delta{u}$ for all the spikes. Of course, each solution will not just depend on the position $(x,y)$ at which you want to know the temperature. It will also depends on where the spike is, as indicated by its center point $(\xi,\eta)$.

Figure 2.7: Sketch of the problem to be solved: heat is added only to the small dark rectangle around a point $(\xi,\eta)$.

Figure 2.7 shows a two-dimensional top view equivalent to figure 2.6. In other words, it shows just the plate, not the function $f$. The dimensions of the little rectangle to which the heat is added by the considered spike will be indicated by $\Delta\xi\Delta\eta$. The amount of heat added is $f(\xi,\eta)\Delta\xi\Delta\eta$, since variations in $f$ over the small rectangle can be ignored. Since the Green’s function $G$ is the solution for unit added heat flux, the final Green’s function will be obtained by dividing the solution $\Delta{u}$ by the amount of heat $f(\xi,\eta)\Delta\xi\Delta\eta$. (Formally speaking, you would then still need to take the limit $\Delta\xi,\Delta\eta\to 0,0$, but that becomes trivial under the approximations to be made.)

The mathematical problem being solved is:

$$\nabla^2 \Delta u = \left\{ \begin{array}{l} f(\xi,... ...ta)$} \\ 0 \mbox{ everywhere else} \end{array} \right.$$

For convenience, for now use a polar coordinate system $r,\vartheta$ centered around the point $(\xi,\eta)$ of heat addition, as indicated in figure 2.7. Further, since the rectangle $\Delta\xi\Delta\eta$ is assumed to be very small, almost a single point, you can reasonably assume that the temperature distribution $\Delta u$ depends only on the distance $r$ from the point where the heat is added, not on $\vartheta$.

Under those assumptions, it is easiest to simply integrate the mathematical problem above over the inside of a circle of radius $d$:

$$\int_{r=0}^d\int_{\vartheta=0}^{2\pi} \nabla^2 \Delta u r{ \rm d}r{\rm d}\vartheta = f(\xi,\eta)\Delta\xi\Delta\eta$$

But $\nabla^2{\Delta}u={\rm div}\left({\rm grad}\;{\Delta}u\right)$, so you can apply the divergence theorem here:

$$\int_S \vec n \cdot \nabla \Delta u { \rm d}S = f(\xi,\eta)\Delta\xi\Delta\eta.$$

In this two-dimensional problem the “surface” $S$ is the perimeter $2\pi d$ of the circle. And $\vec n$ is the radial polar unit vector ${\hat\imath}_r$, which makes the total derivative $\vec{n}\cdot\nabla{{\Delta}u}$ equal to the derivative with respect to radius, ${\rm d}{\Delta}u/{\rm d}d$. So you have:

$$\frac{{\rm d}\Delta u}{{\rm d}d} 2\pi d = f(\xi,\eta)\Delta\xi\Delta\eta$$

Take the $2\pi d$ to the other side and integrate to get ${\Delta}u$:

$$\Delta u = \frac{f(\xi,\eta)\Delta\xi\Delta\eta}{2\pi}\ln d$$

The Green’s function is the solution for unit heat added:

$$G = \frac{1}{2\pi}\ln d$$

This derivation used a polar coordinate system centered around the heat addition point $(\xi,\eta)$. To get the expression for whatever the origin of the coordinate system is, substitute

$$d = \vert\vec x - \vec\xi\vert = \sqrt{(x-\xi)^2+(y-\eta)^2}.$$

That gives the Green’s function as stated in the overview.

2.2.2 Review Questions

1

Do an analysis similar to either this subsection, or the next one, to derive the Green’s function of the Poisson equation in three dimensional infinite space.

Solution pninfl-a

2.2.3 Rigorous derivation

For students who do not like the above derivation with infinitesimal regions, and the assumption that their temperature distribution only depends on distance, here is a mathematically solid derivation.

It will be assumed that a suitable heat distribution $f(x,y)$ is given with at least continuous low-order derivatives. Also that it disappears sufficiently quickly at large distances that you do not have to worry about that region. Then it is to be shown that if you do the Green’s function integration

$$u(x,y) = \int\!\!\int G(x,y;\xi,\eta) f(\xi,\eta){ \rm d}... ...\eta \qquad G = \frac{1}{2\pi}\ln\vert\vec x - \vec\xi\vert$$

you get a function $u(x,y)$ satisfying the Poisson equation:

$$\nabla^2 u(x,y) = f(x,y)$$

Figure 2.8: Region of integration for the Green’s function integral. Excluded regions are left blank. The point $(x,y)$ at which the temperature $u$ is to be found is in the center of the excluded small circle.

The first thing to check is that you do at least get some function $u(x,y)$ by doing the integration. That is not automatic, since the integrand is infinite when $(\xi,\eta)=(x,y)$. And integration over an infinite region is not proper either. What you must do is exclude the inside of a very small circle around $(x,y)$ and the outside of a very large circle from the integration. Then you define $u(x,y)$ to be the limit of the integral when the radius of the small circle $\varepsilon$ becomes zero, and the radius $R$ of the the large circle becomes infinite. (Assuming that those limits exist.) See figure 2.8.

A local polar coordinate system $\rho,\varphi$ will be used centered at the point $(x,y)$ at which the temperature is desired. Then $\rho$ is the distance that the heat addition point $(\xi,\eta)$ is away from the point $(x,y)$ at which the temperature is desired. Note that the variables in the integration are $\xi$ and $\eta$; $(x,y)$ is just a fixed point in this entire story. The two-dimensional Green’s function is

$$G = \frac{1}{2\pi}\ln(\rho)$$

in these terms.

The effect of the excluded area outside the large circle may be taken to be vanishingly small if the radius of the large circle, call it $R$, is large, since it was assumed that function $f$ vanishes sufficiently quickly at large distances. The effect of the excluded area inside the small circle is can be estimated as

$$\left\vert\int\!\!\int \frac{1}{2\pi}\ln(\rho) f(\xi,\eta)... ...}^\varepsilon \frac{1}{2\pi}\ln(\rho)\; 2\pi\rho{ \rm d}\rho$$

where $\vert f_{\rm max}\vert$ is the maximum value of $f$ within the circle and $\varepsilon$ is the radius of the small circle. Now $\vert f_{\rm {max}}\vert$ is finite since $f$ is continuous, and the integral becomes vanishingly small when $\epsilon\to0$. So the effect of the small circle too can be ignored if it is small enough. In particular, the limit when the radius $\epsilon$ of the small circle tends to zero and the radius of the large circle $R$ tends to infinity exists. So the temperature $u(x,y)$ exists.

But does it satisfy the Poisson equation $\nabla^2u=f$? Now what you cannot do here is simply differentiate the Green’s function (the logarithm) within the integral

$$\begin{eqnarray*} u(x,y) & = &\int\!\!\int G(x,y;\xi,\eta) f(\xi,\eta) { \rm ... ...ta)}{2\pi}\ln\sqrt{(x-\xi)^2+(y-\eta)^2}{ \rm d}\xi{\rm d}\eta \end{eqnarray*}$$

a couple of times with respect to $x$ and $y$. If you differentiate the logarithm, it becomes a more singular function, and you get into trouble.

Instead you need to go back to the basic definition of the partial derivatives. As an example, take

$$\frac{\partial u}{\partial x} \equiv \lim_{\Delta x \to 0} \frac{u(x + \Delta x,y) - u(x,y)}{\Delta x}.$$

The integral for $u(x+\Delta x,y)$;

$$u(x+\Delta x,y) = \int\!\!\int \frac{f(\xi,\eta)}{2\pi} ... ...qrt{(x+\Delta x-\xi)^2+(y-\eta)^2} { \rm d}\xi{\rm d}\eta,$$

can be manipulated by defining $\bar\xi=\xi-\Delta x$ to become

$$u(x+\Delta x,y) = \int\!\!\int \frac{f(\bar\xi+\Delta x,... ...sqrt{(x-\bar\xi)^2+(y-\eta)^2} { \rm d}\bar\xi{\rm d}\eta,$$

You can now again drop the bar on $\bar\xi$ since it is just an integration variable whose name makes no difference.

Plug this, and the expression for $u(x,y,z)$ itself, into the limit above to get:

$$\frac{\partial u}{\partial x} = \lim_{\Delta x \to 0} ... ...} \ln\sqrt{(x-\xi)^2+(y-\eta)^2} { \rm d}\xi{\rm d}\eta,$$

In the limit $\Delta x\to 0$, the term in the numerator becomes the partial derivative $\partial f/\partial\xi$! So $x$-derivatives transform to $\xi$-derivatives on $f$ inside the integral. And the equivalent thing happens to $y$-derivatives. Since the derivatives of $f$ are assumed to be continuous just like $f$, there are no problems with these integrals.

The net result is that

$$\nabla^2 u = \int\!\!\int \frac{\nabla^2_{\vec\xi} f}... ...} \ln\sqrt{(x-\xi)^2+(y-\eta)^2} { \rm d}\xi{\rm d}\eta,$$

using the notations

$$\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\pa... ...tial^2}{\partial\xi^2} + \frac{\partial^2}{\partial\eta^2}.$$

Now you need to show that the integral in the right hand side is equal to $f(x,y)$ to finish the proof that $\nabla^2 u = f$. To shorten the notations, the Green’s function will again be written as $G$, and the question is whether

$$\int\!\!\int G \nabla^2_{\vec\xi} f { \rm d}\xi{\rm d}\eta \mbox{ equals $f$?}$$

To show this, exclude again the inside of a small circle of radius $\varepsilon$ and the outside of a large circle of radius $R$ around $(x,y)$ from the integration, as in figure 2.8. In that case, you do not have to worry about infinite quantities, and you can again take the limit $\varepsilon\to0$ and $R\to\infty$ later to get the final result.

If you add a second term,

$$\int\!\!\int G \nabla^2_{\vec\xi} f { \rm d}\xi{\rm d}\et... ...- \int\!\!\int f \nabla^2_{\vec\xi} G { \rm d}\xi{\rm d}\eta$$

you can use Green’s second integral identity from section 1.2 in the book. The added term is zero, since $\nabla^2_{\vec\xi}G=0$ away from the singular point $\vec x=\vec\xi$. (Remember, $G$ is the solution of the Poisson problem where the forcing is a delta function, zero everywhere except at the singular point. So $\nabla^2G=0$ away from the point, and since the components of $\vec\xi$ appear in $G$ in exactly the same way as those of $\vec x$, if $\nabla^2G=0$, then so is $\nabla^2_{\vec\xi}G$.)

Green’s second identity now says that the expression that should equal $f$ is the “surface” integral

$$\int G \frac{\partial f}{\partial n_{\vec\xi}} { \rm d}S_... ... \frac{\partial G}{\partial n_{\vec\xi}} { \rm d}S_{\vec\xi}$$

where $\partial/\partial n_{\vec\xi}=\vec n\cdot\nabla_{\vec\xi}$ is the derivative normal to the surface with respect to the components of $\vec\xi$. The “surface” $S_{\vec\xi}$ is in this two-dimensional case the perimeter of the region of integration. Part of it is the large circle $\rho=R$, but you can ignore that since it is assumed that $f$ vanishes sufficiently rapidly at large distances. The other part is the perimeter of the little circle $\rho=\varepsilon$ that was excluded from the integration. The integral over this little circle cannot be ignored. On it, the normal derivative $\partial/\partial n_{\vec\xi}$ is minus the radial derivative $\partial/\partial\rho$, (minus since the normal vector must stick out of the region of integration, which here is into the excluded little circle.) So you have on the little circle

$$G = \frac{1}{2\pi}\ln(\rho) = \frac{1}{2\pi}\ln(\varepsilo... ...partial G}{\partial n_{\vec\xi}} = -\frac{1}{2\pi\varepsilon}$$

The perimeter $S_{\vec\xi}$ of the little circle is $2\pi\varepsilon$. So you can estimate the two integrals above as

$$- \varepsilon \ln(\varepsilon) \oint_{\rho=\varepsilon} ... ...t_{\rho=\varepsilon} f \frac{{\rm d}S_{\vec\xi}}{S_{\vec\xi}}$$

The first term is certainly no larger than $\varepsilon\ln(\varepsilon)$ times the maximum value of the gradient of $f$ on the spherical surface and disappears when you take the limit $\varepsilon\to0$. The second term however is the average value of $f$ on the perimeter, and that becomes simply $f$ when the circle becomes so small that the variations in $f$ can be ignored.

So finally, you conclude that $\nabla^2 u$ is indeed $f$.