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Operational Solutions to Differential EquationsQ: If is the electrostatic potential on the axis of a cylindrically symmetric system, then the potential at the point , where is the perpendicular distance from the axis, is given by the following (see, e.g., [6])
How can I implement the operator ? A: The definition of is given at functions.wolfram.com/03.01.02.0001.01.
Then, formally, one has
For an arbitrary potential , the operator formalism can be used to obtain the Taylor series expansion of about using NestList. For example, here are the first four terms.
For an axially symmetric potential, the Laplacian in cylindrical coordinates reads,
and the potential satisfies Laplace's equation, . Verifying that the operator expansion produces a formal powerseries solution is immediate.
Now for a concrete example. For a disk of radius , with uniform surface charge density , oriented with its normal vector along the axis, here is the potential at .
Using the operator formalism one obtains
There is another approach to this problem: In the case of azimuthal symmetry, the general solution to Laplace's equation is (the multipole expansion),
where and are the radial and polar spherical coordinates, respectively. Here is the truncated solution.
Now, if the potential is known on the axis, that is , then one can use equation (1) to determine and by series expansion of and termbyterm comparison. For , here is the series expansion of the axial potential.
Now, equate this to and solve.
Hence we obtain the (truncated series expansion of the) potential of the disk off the axis in spherical coordinates.
To compare this solution to that obtained earlier, we expand into a series in , valid for . See functions.wolfram.com/01.01.06.0002.01 and functions.wolfram.com/01.01.06.0003.01.
Check that we have the correct expansion for .
Hence the potential of the disk off the axis is given by
Changing coordinates from polar to cylindrical coordinates, and , we verify that the two expansions are consistent.
Elmar Zeitler (zeitler@fhiberlin.mpg.de) submitted another example of an operator expansion. Using the integral definition (functions.wolfram.com/03.01.07.0005.01),
and the identity (functions.wolfram.com/01.07.16.0096.01),
then the change of variables yields
Now, is a polynomial in and, since , we see that
Implementation of this operator expansion is direct.


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