This section will derive the solution of the Poisson equation in a finite region as sketched in figure 2.9. The region will be denoted as , and its boundary by . It will again be assumed that the region is two-dimensional, leaving the three-dimensional case to the homework. As shown in figure 2.9, inside the region the Poisson equation applies. In case , this becomes the Laplace equation. On the boundary there is some boundary condition that will for now be left arbitrary.
The big idea is to relate the finite domain solution to the infinite domain solution derived earlier.
For the given heat addition , you can still do the integral over the domain to get the infinite domain solution. That solution will now be called , as indicated in figure 2.10.
The infinite domain solution satisfies the Poisson equation, but it does not satisfy the boundary conditions. It will turn out that some surface integrals must be added to it to get the boundary conditions right.
The precise form of these integrals can vary. The different versions all give the same, correct, solution inside the domain . However, they give different answers for the continuation of this solution to outside the domain . So a meaningful discussion of the various possibilities requires consideration of the solution outside the domain. Even though the solution outside domain is not actually a part of the problem.
The solution outside will be indicated as , so the picture becomes as shown in figure 2.10. Since you surely do not want to just make up an arbitrary function outside , it will be assumed that outside. So satisfies the homogeneous Poisson equation, the Laplace equation. And so does the infinite space solution outside , for that matter. Only integrating over the domain is the same as setting to zero outside the domain.
(book, example 8.2)
The desired integral solution for the finite-region Poisson
solution is a generalization of the infinite domain solution
Since , the expression above represents an important
relationship between the infinite-domain solution and
the actual, finite-domain, solution :
Some caution is needed, however. The Green’s function is infinite when , and integrals of infinite functions are not proper. And neither are integrals over infinite regions. You must exclude a very small circle around the point at which is desired from the integration, and also the outside of a very large circle, as indicated in figure 2.11. The correct value for can then be obtained as the limit in which the radius of the small circle becomes zero and the radius of the large circle becomes infinite. Also, you must think of the integral as really consisting of two separate integrations; one over the dark grey region and one over the light grey exterior of . The reason is that and its derivatives are not normally continuous on the surface , and the divergence theorem can only be used for fairly smooth functions.
To simplify the remaining discussion, the origin of the coordinate
system will be shifted towards the point at which
is desired. The integration coordinate can
then be described by polar coordinates and centered
around this point. That simplifies the expression to be evaluated to
To get a divergence integral, move a
out in front,
adding a correction term to make up for the error in doing so:
This final correction term, however, is zero. To see why, remember that the Green’s function is the temperature distribution due to a spike of heat at point . So everywhere except at the singular point . And since appears exactly the same way in the Green’s function as , then so is zero.
The remaining two terms become “surface” (actually,
contour in 2D,) integrals using the divergence theorem. In
particular:
To verify this expression, note that the “surfaces” include the small and big circles, and that counts as both part of the “surface” of the dark grey region in figure 2.11 as well as part of the “surface” of the light grey region. The normal vector on was taken to stick out of region , which accounts for the additional minus sign in the corresponding terms. Also is according to the total differential of calculus the derivative in the direction normal to the surface. On the big circle, that is the same as , and on the small circle it is since there the outward normal points towards the origin.
The second integral over the small circle is particularly interesting:
since
, its derivative is , which
is the inverse of the “surface” (perimeter) of the
circle. So you get
The first integral over the small circle in (2.9) is
vanishingly small and can be ignored. To see why, note that it is no
larger than the maximum value of the gradient of on the small
circle times
There is little that can be done about the integrals over
“surface” . However, the integrals over
the big circle in (2.9) still must be evaluated. To do
so, you must know something about the behavior of the solution
for large values of . In general, it is
described by
Collecting the results together, the solution for the temperature
at any point is:
Perform the equivalent analysis in the three dimensional case.
The previous subsection derived the solution to the Poisson equation in a finite domain. It was given by equation (2.10). This subsection will examine how this solution may be evaluated.
Except for , all other quantities in the right hand side of equation (2.10) are evaluated at the point of integration . For example, stands for the normal derivative evaluated at the boundary point of integration. That means that if you merely know and the normal derivative on the boundary, you can find in the interior by taking to be zero and doing the integrals above. Unfortunately, a priori at most only one of (Dirichlet boundary condition) or (Neumann boundary condition) will be known on the boundary.
Various solutions for this problem are possible. A panel method might decide to compute the particular solution where is not zero, but has the same values as on the boundary. The big advantage is then that the second integral in (2.10) drops out, leaving only the last integral as a problem.
A simple panel method will now discretize the boundary in a large
number of densely spaced points, and then put a Green’s function
at each point. Since each Green’s function corresponds to the
addition of a spike of heat at that point, this is called a surface
“source” distribution. The problem remains that the
strengths
Alternatively, a panel method might decide to compute the solution for
the case that and have the same normal derivatives
on the boundary. That kills off the source integral, leaving the
second integral in (2.10). The quantity
in this integral is called a
“dipole.” The reason for that name can be understood by
writing the definition of the derivative:
The previous subsection showed that the Poisson equation can be solved by using suitable source and/or dipole distributions on the boundary of the domain. However, the strengths of these distributions are not usually known, since they involve both and its normal derivative on the boundary, and there is only one boundary condition. And if an exterior solution is chosen to eliminate one of them, that has the effect of introducing the unknown values of or its derivative into the problem. So at least one distribution strength must be found using brute numerical force. Or by brute analytical force, maybe, if the domain is simple.
There is an exception, however, and it occurs for the Dirichlet problem inside a ball (a circle in two dimensions, a sphere in three-dimensions, etcetera.) In that case, suitable distribution strengths can be found by simple means.
The following discussion will restrict itself to the Laplace equation,
since the Poisson equation can always be turned into the Laplace
equation by subtracting the unbounded space solution .
This only produces an unimportant change in the inhomogeneous term of
the boundary condition. The problem to be solved is then:
In two dimensions, using polar coordinates, the solution is
This subsection will derive the two-dimensional formula above, leaving
the three-dimensional one for the homework. For simplicity, from now
on it will be assumed that the ball (i.e. circle in two-dimensions)
has unit radius,
The integral formula can be derived by a clever selection for the
solution outside the circle in the integral solution
(2.10). In particular, the trick is to take
The first thing to show is that satisfies the Laplace
equation. The integral solution (2.10) does not apply
otherwise. The Laplacian,
To show that this is so, first differentiate (2.14) once,
using the chain rule to convert the derivatives of to
derivatives:
Now the idea is to try to choose the constant so that the integral
solution (2.10) only involves the given values of on
the boundary. In particular, the normal derivative of
must be eliminated. Now for a spherical boundary, the
normal derivative is the radial derivative. And on the surface of the
ball, . So on the boundary, using the expressions above,
For
to vanish on the surface of
the sphere. according to the above equations you need to take .
In that case,
on the sphere equals , and is
the given function on the surface of the sphere. So the integral
solution (2.10) becomes
The above solution is completely in terms of the given function . So the Dirichlet problem has been solved.
But of course you want to clean it up. You would like the solution of
a problem in a circle to be in terms of polar coordinates. So set
Also, the derivative
normal to the
circle is simply
. G is a function of the
distance
between the points and
; in particular in two dimensions. You can
write
Plug that into the integral expression (2.15) for
, taking from (2.16) with
equal to
, to get
The two final terms are just constants, and they cancel each other.
The reason is that
Also, to allow for the case that the radial coordinate is not normalized with the circle radius , you want to replace in the above result with . That produces the Poisson integral as stated in the previous subsection.
Find a suitable solution outside the sphere in three dimensions. Show that it satisfies the Laplace equation.
Derive the Poisson integral formula in three dimensions as given in the previous subsection.
The Neumann problem in two dimensions is:
The derivation of the formula above is similar to the one in the
previous subsection. You will be disappointed to learn that you must
miss doing it in the homework. The same story does not work in three
dimensions since you cannot get rid of the unknown surface values of
in both the source and dipole distributions. In two dimensions,
however, if you take
, the
dipole strength is zero and only the source integral remains:
One important qualitative conclusion that can be drawn from the various results of the previous subsections is that the solution of a Laplace equation problem is infinitely smooth in the interior of the region in which it applies.
For example, consider the derived expression for if the exterior
solution is zero:
(2.18) |
So, if and are merely integrable on the boundary, which still allows them to be quite singular, the solution at every point in the interior will have infinitely many continuous derivatives.
It is somewhat different for the Poisson equation, since if the forcing has a singularity at some point, then so will the solution . But still the solution for will be less singular than is. For example, in two-dimensions a delta function in , whose square is not integrable, produces a logarithmic Green‘s function, for which every power is integrable over the singular point. In general, it can be seen from Fourier solution of the Poisson problem that will in general have two more square integrable derivatives than . (Assuming that lack of decay of at large distances is not a factor or subtracted out first.)