Subsections

### 2.3 The Poisson or Laplace equation in a finite region

#### 2.3.1 Overview

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.

#### 2.3.2 Intro to the solution procedure

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.

#### 2.3.3 Derivation of the integral solution

(book, example 8.2)

The desired integral solution for the finite-region Poisson solution is a generalization of the infinite domain solution

where is the infinite domain Green’s function derived in the previous section. In the two-dimensional case discussed here:

Since , the expression above represents an important relationship between the infinite-domain solution and the actual, finite-domain, solution :

where

is the Laplacian with respect to and . George Green discovered that the integral in the right hand side could be simplified into surface integrals using the divergence theorem, and that doing so directly relates to .

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:

Move a out in front in the second integral to create another divergence integral, adding another correction term:

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:

 (2.9)

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

That is just the average of on the small circle, and it becomes at the point , (used here as origin,) in the limit that the radius of the small circle becomes zero. So, since this integral simplifies to , all the other integrals in equation (2.9) merely describe the difference between the true solution and the infinite-domain solution .

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

and that becomes zero in the limit .

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

In three or more dimensions, the constant is zero. The two integrals over the big circle become, noting that ,

which becomes in the limit .

Collecting the results together, the solution for the temperature at any point is:

 (2.10)

where is the infinite domain Green’s function, , with the distance between the point of integration and the point at which the temperature is desired. The first integral is therefor the infinite domain solution , which has the right values for the added heat , but does not satisfy the correct boundary condition on

2.3.3 Review Questions
1

Perform the equivalent analysis in the three dimensional case.

#### 2.3.4 Boundary integral (panel) methods

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

of these sources are not known, since even if is given on the boundary, is not. So the strength of each source is an unknown, and an equally large number of equations is needed. These equations can be found from requiring that at that many points, the error in the boundary condition as computed from (2.10) is zero. Put all these equations on a computer and solve. And with the source strength now known, can then be evaluated at any arbitrary point.

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:

 (2.11)

This shows that a dipole corresponds to an infinitely large source of heat and an infinitely large sink of heat infinitely close together.

#### 2.3.5 Poisson’s integral formulae

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:

where is a given function, physically the temperature on the boundary in heat conduction problems, and is the radius of the ball.

In two dimensions, using polar coordinates, the solution is

 (2.12)

and in three dimensions, using spherical coordinates, the solution is
 (2.13)

These results are known as “Poisson’s integral formula” in two, respectively three dimensions.

#### 2.3.6 Derivation

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,

It is a simple matter of rescaling to get back to the formulae for a ball of arbitrary 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

 (2.14)

Here is a constant still to be selected. Note that if then : these rules turn solutions inside the ball into solutions outside the ball. The transformation is called an inversion with respect to the surface of the unit ball.

The first thing to show is that satisfies the Laplace equation. The integral solution (2.10) does not apply otherwise. The Laplacian,

must be zero.

To show that this is so, first differentiate (2.14) once, using the chain rule to convert the derivatives of to derivatives:

Differentiate this once more to get the second derivative. Note that you now have to use the product rule of differentiation to differentiate the factors. And you need again the chain rule for differentiating the first factor. You get

Also,

If you plug these derivatives into the Laplacian given above, you get

Since , you recognize the Laplacian of inside the square brackets. That is zero because satisfies the Laplace equation. Then you see that so does .

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,

Note that in the final term, the first independent variable in has been renamed simply . It does not make a difference what you call the independent variable of a function; we just used a bar on it when we were treating at one location to define at another location. The bar was merely to keep the two locations apart.

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

 (2.15)

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

for the point at which the temperature is desired and the point of integration respectively. Then the “surface” element in the integral over the circle perimeter is , and on the circle.

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

 (2.16)

whose derivative with respect to equals

or getting rid of the ugly dot product term using the expression (2.16) for ,

So you can write using the chain rule that

Plug in the expression for the two-dimensional Green’s function, and note that on the circle to get:

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

The mean value theorem, proved in {D.2}, says that equals the average of on the circle.

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.

2.3.6 Review Questions
1

Find a suitable solution outside the sphere in three dimensions. Show that it satisfies the Laplace equation.

2

Derive the Poisson integral formula in three dimensions as given in the previous subsection.

#### 2.3.7 The integral formula for the Neumann problem

The Neumann problem in two dimensions is:

This corresponds physically to a problem where the heat flux instead of the temperature is described on the boundary. The solution is
 (2.17)

Note that there is only a proper solution for if

If you put in an invalid , you will get a , but it will not have heat flux through the boundary. In particular, putting in a nonzero constant for will produce and no heat flux. It can be seen from the above expression that the undetermined constant is the temperature at the center of the circle.

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:

and use of the expression (2.16) for gives the stated result.

#### 2.3.8 Smoothness of the solution

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)

If you take derivatives of with respect to the components of , you will be differentiating inside the integral. And has infinitely many finite derivatives away from the singular point , in other words, away from the boundary.

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.)