product of these solutions and the general solution to Laplace's equation is given by a linear combination of these solutions: where the are constants with respect to the cylindrical coordinates and the limits of the summation and integration are determined by the boundary conditions of the problem. Note that the integral may be replaced by a sum for appropriate boundary conditions. The orthogonality of the is often very useful when finding a solution to a particular problem. The and functions are essentially Fourier or Laplace expansions, and form a set of orthogonal functions. When is simply , the orthogonality of , along with the orthogonality relationships of and allow the constants to be determined. If is the sequence of the positive zeros of then: In solving problems, the space may be divided into any number of pieces, as long as the values of the potential and its derivative match across a boundary which contains no sources. Example: Point source inside a conducting cylindrical tube As an example, consider the problem of determining the potential of a unit source located at inside a conducting cylindrical tube (e.g. an empty tin can) which is bounded above and below by the planes and and on the sides by the cylinder . (In MKS units, we will assume ). Since the potential is bounded by the planes on the z axis, the Z(k,z) function can be taken to be periodic. Since the potential must be zero at the origin, we take the function to be the ordinary Bessel function , and it must be chosen so that one of its zeroes lands on the bounding cylinder. For the measurement point below the source point on the z axis, the potential will be: where is the r-th zero of and, from the orthogonality relationships for each of the functions: Above the source point: It is clear that when or , the above function is zero. It can also be easily shown that the two functions match in value and in the value of their first derivatives at . Point source inside cylinder Removing the plane ends (i.e. taking the limit as L approaches infinity) gives the field of the point source inside a conducting cylinder: Point source in open space As the radius of the cylinder (a) approaches infinity, the sum over the zeroes of becomes an integral, and we have the field of a