Classroom Tips and Techniques:
Eigenvalue Problems for ODEs - Part 2
Robert J. Lopez
Emeritus Professor of Mathematics and Maple Fellow
Maplesoft
Initializations
Introduction
In Part 1 of this series of articles on solving eigenvalue problems for ODEs, we discussed equations for which the general solution readily yielded eigenvalues and eigenfunctions without the need for detailed knowledge of any of the special functions of applied mathematics. In Part 2 of this series, we concentrate on eigenvalue problems for Bessel's equation whose solution requires some knowledge of Bessel functions of the first and second kinds.
Since Bessel's equation readily arises when separating variables in Laplace's equation in cylindrical coordinates, we have allowed our discusion to be colored by references to calculations arising from boundary value problems posed in a cylinder.
Steady-State Temperatures in a Cylinder
At steady state, the temperature in a cylinder satisfies Laplace's equation and some conditions on the boundary of the cylinder. For example, if the cylinder is described in cylindrical coordinates by we could impose the condition or on the top surface, or on the curved wall of the cylinder, and or on the bottom surface. (Subscripts denote partial derivatives.)
If the temperature on the curved wall of the cylinder is fixed at , we say that a homogeneous Dirichlet condition has been imposed. Alternatively, if the curved wall is insulated so the heat flux across this surface is zero, we say that a homogeneous Dirichlet condition has been imposed. The flux across the curved wall is the normal derivative given by evaluated at .
If the prescribed temperature on the top surface is a function of alone, the temperature in the cylinder will be axially symmetric so that Alternatively, if this prescribed temperature is then the temperature in the cylinder will be axially asymmetric so that
Axial Symmetry
Separation of Variables
Steady-state temperatures in a cylinder obey Laplace's equation. Under the assumption of axial symmetry, Laplace's equation in cylindrical coordinates becomes
Assuming the separated solution
leads to
and
as the variable-separated equation. Introducing the Bernoulli separation constant , we have the ordinary differential equation
for the radial component . Manipulating this to the form
and then
we obtain the solution as
The "self-adjoint" form of the equation for the radial component is
The radical in the solution suggests the substitution
in which case the solution is given by
The solution is a linear combination of two Bessel functions, one of the "first kind" and one of the "second."
The radial equation can be transformed to Bessel's equation by the change of variables
Factoring leads to Bessel's equation of order zero, that is, to
The solution of this equation is
so that Figure 1 contains a graph of (in black) and (in red).
Figure 1 In black, , the Bessel function of the first kind, and in red, the Bessel function of the second kind.
Consistent with the graph in Figure 1, we have
so that the -axis in a vertical asymptote for , and is unbounded in any interval that includes .
The function is well approximated by
as we confirm in Figure 2, where is graphed in black, and is in red.
Figure 2 In black, and in red,
Ignoring Maple's automatically simplifications of to and to , we see from
that is asymptotic to .
Homogeneous Dirichlet Condition
The solution of the singular Sturm-Liouville eigenvalue problem
bounded on
consists of the eigenvalues (the zeros of ) and the eigenfunctions
The numbers are represented symbolically by
and in floating-point form by
That Maple recognizes the eigenvalues as exact is shown by
If, for example , the eigenvalues would be
Figure 3 shows the first three eigenfunctions for the case
Figure 3 For the first three eigenfunctions in black, red, and green, respectively
The eigenfunctions are orthogonal with respect to the weight function , as can be seen from the integral
which vanishes if and are distinct eigenvalues for the interval Making use of Maple's built-in representation of the zeros of we also have
Unfortunately, Maple does not yet detect that this result is not valid if in which case the integral evaluates to
Since this result simplifies to Thus, under suitable conditions, a function can be represented by the Fourier-Bessel series where
and the eigenvalue is the th zero of .
Homogeneous Neumann Condition
where we have used the relationship Note carefully that but for positive integers .
Figure 4 shows the first three eigenfunctions for the case
Figure 4 For the first three eigenfunctions in black, red, and green, respectively
which vanishes if and are distinct eigenvalues for the interval Of course, Maple has used for
Making use of Maple's built-in representation of the zeros of we also have
As seen earlier, Maple does not yet detect that this result is not valid if in which case the integral evaluates to
Axial Asymmetry
If the temperature prescribed on top of the cylinder is given by then obeys Laplace's equation in the form
The separation assumption
Moving the term in to the right-hand side leaves us with the equation
Introducing the Bernoulli separation constant then gives
The first of these equations is of immediate interest. We first multiply through by so that the term containing contains only the independent variable This gives the equation
Next, we would like to move the term to the left, and the term containing to the right. The first transformation is given by
and the second, by
Introducing a new separation constant we have the two ordinary differential equations
The second equation can be put into the form
and has general solution
Continuity of implies the periodic boundary conditions
Imposing these conditions leads to
from which we see that Notice that for , the eigenspace has dimension 1 and basis , but for the eigenspaces have dimension 2 and bases .
The equation for the radial component now takes the form
After moving all terms to the left, we obtain the self-adjoint form as
The solution of this equation is given by
suggesting first, the substitution so that the equation for the radial component becomes
Since the solution of this form of the equation will be
we make the change of variables so that This change is implement in Maple via
Under the assumption that this equation simplifies to
a Bessel equation of order , and has general solution
Continuity on implies that where is constant. The homogeneous Dirichlet condition implies so that must be a zero of If we denote the th zero of by then represents a doubly-indexed set of eigenvalues for which the corresponding eigenfunctions are
To establish orthogonality of these eigenfunctions, we will have to evaluate the integral
where the superscript has been dropped from the distinct eigenvalues and For this integral Maple gives
which clearly vanishes by the definition of the eigenvalues and
The function will be given by a Fourier-Bessel series in which a double sum is taken over both and . The coefficients in this series require us to evaluate the integral
which we enter into Maple as
We call this the "Norm Relation" since it is related to the -norm of the eigenfunction.
Now it is indeed unfortunate that Maple incorrectly evaluates this integral to
an error that has been corrected for the next release of the product. The correct value of this integral is
as can be found, for example, in the text Boundary Value Problems, Ladis D. Kovach, Addison-Wesley Publishing Company, 1984. Consequently, we have
=
There is one additional subtlety to confront. Since the first eigenvalue in the sequence is not zero, so corresponds to the indexing available via Maple's BesselJZeros command. Indeed, we see that
is the first zero for . However, for , so the "first zero" is , but this cannot be an eigenvalue because by definition, eigenvalues are numbers for which there are nontrivial solutions. Consequently, for the homogeneous Dirichlet condition, each sequence starts with for any value of . This indexing is consistent with Maple's BesselJZeros command where "1" in the second argument always gives the first positive zero.
The homogeneous Neumann condition implies so that must be a zero of If we denote the th zero of by then represents a doubly-indexed set of eigenvalues for which the corresponding eigenfunctions are
To establish orthogonality of these eigenfunctions, we will have to show
vanishes. To this end, we seek to eliminate and introduce . A generalization of the identity is
Solving this for leads to
Replacing in this equation with its value from the previous equation gives
Finally, substituting for gives
from which it is evident that the functions are orthogonal, even when are zeros of
Since but then is an eigenvalue and the indexing starts at The corresponding eigenfunction is 1.
Maple computes from the formula
Consequently, a division-by-zero error occurs in, for example
However, an examination of the series expansions for shows that this error is spurious. Indeed, we have
Each of these functions can be readily evaluated at . Surprisingly, for but . Therefore, is not even a candidate for being an eigenvalue when , and for greater values of , although is a zero, so that would not be an eigenvalue. Consequently, for the homogeneous Neumann condition, zero is an eigenvalue only for .
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