There are many particular solutions.

- Consider

where is a constant, so that

This corresponds to an isolated*line charge*on the -axis in electrostatics and an isolated*line current*on the -axis in MHD. For a fluid this gives a*line source*() and a*line sink*() - The above solution may be modified to place the line source at
a position
by using

and . - A uniform field or flow, in the -direction at large
distances, is

and

- A
*line dipole*in either electrostatics or MHD of moment or a*line doublet*in fluids has a potential

and

These are just particular solutions of a more general class given by separable solutions to (5.17).

__Example 5. .3__More general solutions can be obtained by using separation of variables.
Hence, we set

Substituting into (5.17) we obtain

Rearranging to into functions of and alone gives

The separation constant is and it must be the square of an integer. Solving for we obtain,

and so must be an integer to ensure that . The radial variations are given by the solution to

with solutions

The case with gives (5.18). Therefore, we have four types of potential solutions (also called harmonic functions)

and the general solution is a linear combination of all types. Note the following points.

- For , and . Hence,

- From above the solution gives the following possible
symmetric,
solutions for ,

- The solutions involving and are
linearly independent for different values of ,

__Example 5. .4__Find the incompressible, irrotational flow of an ideal fluid or
plasma, uniform at infinity, past a cylinder of radius .

The situation is shown in Figure 5.2.

The boundary conditions areand on the surface of the cylinder there can be no radial flow

To proceed, we note that

Thus,

The potential must satisfy

and the most general solution to (5.24), that is even in , is

The condition as shows that all coefficients of the positive powers of must be zero, except for . Hence,

The coefficients are determined by the condition on the surface of the cylinder. Here .

On we must have

Using (5.23) each term is orthogonal to each other and so the coefficient of each term in must be zero. This implies that

From this we find that the solution is

Another way of thinking of this solution is to say that the solution giving a uniform flow at large distances is

To this we must add on a term that is going to make on and this must have the same form in . Hence the extra term to add on is of the form

Then we can choose so that we do indeed have on .

In this last example we can calculate the pressure from Bernoulli's
equation (2.20). Hence, in the absence of gravity,

On the surface of the cylinder and

Therefore,

Thus, the pressure on the surface of the cylinder varies as we move round the cylinder.

__Example 5. .5__Consider a cylinder rotating with a constant angular speed, .
In this case the potential must generate a velocity component have
on . The potential is

Hence,