(Book, section 8.2)
This section works out the Green’s function idea for the Poisson
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 for the above problem is by definition a solution when function is a delta function. A delta function is an infinitely narrow spike that integrates to one. We will write for the point at which the temperature is desired. Further is the position of the delta function.
In two dimensions and
. Also, the
two-dimensional delta function can be written in terms of
one-dimensional ones as
In any number of dimensions, the Green’s function solution of the
Poisson equation with a delta function as right hand side will be
In terms of the Green’s function, the solution to the Poisson
equation with an arbitrary right hand side can be written as
It will be found that the Green’s function for the two-dimensional
infinite-domain Poisson problem is:
Similarly, you will find in the homework that the Green’s function for
the three-dimensional infinite-domain Poisson problem is:
Therefore the two-dimensional temperature distribution
corresponding to a general distribution of added heat
Given an added-heat distribution , you can find the temperature distribution 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.
To verify the two-dimensional Green’s function given in the previous section, the solution to the Poisson equation must be found in which is a delta function spike at some point .
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 . (And so we are, really.) We then mentally cut up this function into spikes. That idea is sketched in two-dimensions in figure fig:2dspike.
The problem for such a narrow, but finite spike can be solved with some physical intuition. The solution will again be called . The total solution will then be the sum of the solutions for all the spikes. Of course, each solution will not just depend on the position at which you want to know the temperature. It will also depends on where the spike is, as indicated by its center point .
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 . The dimensions of the little rectangle to which the heat is added by the considered spike will be indicated by . The amount of heat added is , since variations in over the small rectangle can be ignored. Since the Green’s function is the solution for unit added heat flux, the final Green’s function will be obtained by dividing the solution by the amount of heat . (Formally speaking, you would then still need to take the limit , but that becomes trivial under the approximations to be made.)
The mathematical problem being solved is:
For convenience, for now use a polar coordinate system centered around the point of heat addition, as indicated in figure 2.7. Further, since the rectangle is assumed to be very small, almost a single point, you can reasonably assume that the temperature distribution depends only on the distance from the point where the heat is added, not on .
Under those assumptions, it is easiest to simply integrate the
mathematical problem above over the inside of a circle of radius :
In this two-dimensional problem the “surface” is the
perimeter of the circle. And is the radial polar
, which makes the total derivative
equal to the derivative with respect
. So you have:
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.
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 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
The first thing to check is that you do at least get some function by doing the integration. That is not automatic, since the integrand is infinite when . And integration over an infinite region is not proper either. What you must do is exclude the inside of a very small circle around and the outside of a very large circle from the integration. Then you define to be the limit of the integral when the radius of the small circle becomes zero, and the radius of the the large circle becomes infinite. (Assuming that those limits exist.) See figure 2.8.
A local polar coordinate system will be used centered at
the point at which the temperature is desired. Then is
the distance that the heat addition point is away from
the point at which the temperature is desired. Note that the
variables in the integration are and ; is just a
fixed point in this entire story. The two-dimensional Green’s
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
, is large, since it was assumed that function vanishes
sufficiently quickly at large distances. The effect of the excluded
area inside the small circle is can be estimated as
But does it satisfy the Poisson equation ? Now what you cannot do here is simply differentiate the Green’s function (the logarithm) within the integral
Instead you need to go back to the basic definition of the partial
derivatives. As an example, take
The integral for
Plug this, and the expression for itself, into the limit
above to get:
The net result is that
Now you need to show that the integral in the right hand side is equal
to to finish the proof that
. To shorten the
notations, the Green’s function will again be written as , and
the question is whether
If you add a second term,
Green’s second identity now says that the expression that should equal
is the “surface” integral
So finally, you conclude that is indeed .