POISSON’S EQUATION

8. POISSON’S EQUATION

We are going to derive Poisson’s equation (1.2) for a simple problem whose answer we know in advance. Using our known solution, we shall be able to see a method of solving more difficult problems.

Recall from Chapter 6, Section 8, that the gravitational field is conservative, that is, curl F = 0, and there is a potential function V such that F = −∇V . If we consider the gravitational field at a point P due to a point mass m a distance r away, we have

where u is a unit vector along r toward P . It is straightforward to show that div F = 0 and V satisfies Laplace’s equation (Problem 1), that is,

Now suppose there are many masses m i at distances r i from P . The total potential at P is the sum of the potentials due to the individual m i , that is,

and the total gravitational field at P is the vector sum of the fields F i , that is,

F=−

∇ V i = −∇V.

Note that we are taking it for granted that none of the masses m i are at P , that is, that no r i is zero. Since

∇· F i

= −∇ 2 V i = 0,

Section 8 Poisson’s Equation 653

we have also

Instead of a number of masses m i , we can consider a continuous distribution of mass inside a volume τ (Figure 8.1). Let ρ be the mass density of the distribution; then the mass in an element dτ is ρ dτ . The gravita- tional potential at P due to this mass ρdτ is −(Gρ/r) dτ. Then the total gravitational po- tential at P due to the whole mass distribution is the triple integral over the volume τ :

As before, the contribution to V at P due to each bit of mass satisfies Laplace’s equation and therefore V satisfies Laplace’s equation. Also the total field F at P is the vector sum of the fields due to the elements of mass, and as before we have

Again note that we are implicitly assuming that none of the mass distribution the region τ .

Now let us investigate what happens if P is a point of τ . Can we find V from (8.3) and does V satisfy Laplace’s equation? Let S be a small sphere of radius a about P ; imagine all the mass removed from inside S (Figure 8.2). Then our previous dis- cussion holds at points inside S since these points are not in the mass distri- bution. If F ′ and V ′ are the new field and potential (with the matter inside

S removed), then ∇ · F ′ = −∇ 2 V ′ =0

at points inside S. Now restore the mass inside S; let F and V represent the field and potential due to the whole distri- bution and let F S and V S represent the field and potential due to just the mass inside S. Then F = F ′ +F S and at

Figure 8.2 points inside S

∇· F=∇·F ′ +∇·F S =∇·F S

since ∇ · F ′ = 0 inside S. By the divergence theorem (see Figure 8.2 and Chapter 6, Section 10)

volume of S

surface of S

If we let the radius a of S tend to zero, the density ρ of matter inside S tends to its value at P ; thus for small a, S contains a total mass M approximately equal to

654 Partial Differential Equations Chapter 13

3 πa ρ, where ρ is evaluated at P . The gravitational field at the surface of S due to this mass is of magnitude