0
$\begingroup$

The equation \begin{equation} (r^2)R''+2rR'+[(kr)^{2}-l(l+1)]R=0, \end{equation} where $R=R(r)$, $R'=\frac{\mathrm{d}R}{\mathrm{d}r}$, etc. and $k, l$ are constants, is the Spherical Bessel equation of order $l$. I added the correction of the factor r^2 which I had forgot to include on the second derivative as someone commented.

Its solutions are not too hard to find, but how would you evaluate the inner product of one of its solutions with itself? If you had a partial differential equation in spherical coordinates to solve, say Schrödinger equation, and the initial conditions were of the polar and azimuthal angle, then to get the coefficients of the series solution, you would have to take an inner product with $R,$ and thus it would be necessary to know $\langle R|R \rangle.$

I ran into some functions like "The Sine Integral" $Si(x)$ etc. while looking for an answer so I am not sure there really is an answer to this question of a general and well-defined nature that could be used to compute answers for specific problems.

$\endgroup$
  • $\begingroup$ So, what you are after is the squared norm (integral over all space of squared absolute value)? Of one of the types of spherical Bessel function? $\endgroup$ – Whit3rd Aug 31 '16 at 0:57
1
$\begingroup$

You can related the spherical Bessel functions to the ordinary, cylindrical, ones using an identity:$$ j_\nu(x) = \sqrt{\frac{\pi}{2r}} J_{\nu+1/2} (x).$$

Then orthogonality follows from equation 53 of the MathWorld article on the Bessel functions. The DLMF also has a number of integral identities.

$\endgroup$
  • $\begingroup$ Ah! So when you seek the norm of j_v(r) you can just do the same thing as with Bessel functions, and get around the 1/sqrt(r) factor by just taking the weight function to be r^2 instead of r. I feel silly for not seeing it before, thank you. $\endgroup$ – Ben Ionescu Sep 1 '16 at 17:57
1
$\begingroup$

Correction: Your 2nd derivative term is missing a factor of $r^2$.

The general framework for understanding the orthogonality and normalization relationships you are interested in is Sturm-Liouville theory. It deals with a general class of differential eigenvalue eqs. of the form $$ \frac{1}{w(x)}\left(-\frac{d}{dx}\left[p(x)\frac{df}{dx}\right] + q(x) f(x) \right) = \lambda f(x) $$ with $p(x), w(x) > 0$. If $p$, $dp/dx$, $w$, $q$ are continuous on the interval of interest $[a, b]$, and the boundary conditions are of the form $\alpha\; df/dx + \beta f = 0$, the operator on the lhs is self-adjoint and the solutions $f_\lambda$ are its eigenfunctions. Then the $f_\lambda$ are orthogonal in the sense of the weight $w(x)$: $$ \int_a^b{dx\; w(x)\; f_\lambda(x)\; f_{\lambda'}(x)} \sim \delta_{\lambda\lambda'} $$

In the case of the spherical Bessel functions there are two ways to look at their eq. as an eigenvalue eq.: in terms of $k$ or in terms of $l(l+1)$.

If the chosen eigenvalue parameter is $k$, the weight is $w(r )= r^2$ and the corresponding orthogonality relation reads $$ \int_0^\infty{dr\; r^2 \; j_l(k_1r)j_l(k_2r)} \sim \delta(k_1 - k_2) $$ This is also known as the "closure relation" for spherical Bessel functions, and is the result on which the previous answer zeroed in.

If however the eigenvalue is chosen as $l(l+1)$, then the weight is trivially $w(r ) = 1$ and the corresponding orthogonality relation is $$ \int_0^\infty{dr \; j_{l_1}(kr)j_{l_2}(kr)} \sim \delta_{l_1, l_2} $$ This is simply known as the "orthogonality relation" of the spherical Bessel functions. Compare for instance with eqs.(4.8-4.9) in this paper on The Plane Wave Expansion, Infinite Integrals And Identities Involving Spherical Bessel Functions, where it is derived in a different way.

Left as exercise: verify the weights in the two cases above.

$\endgroup$
0
$\begingroup$

There's several issues at play here, and some of them can get rather subtle.

First of all, to start with clear air, the definite and indefinite integrals of special functions are typically handled using pretty different tools. The sine integral function $\mathrm{Si}(x)$ is an indefinite integral, and the norm squared $⟨R|R⟩$ of a radial solution is a definite integral. As such, they mostly belong to rather different domains.

In general, the indefinite integral of a special function is just another special function, but there is very little more that you can say about it. The statement "$f$ is a special function" is a bit tricky to pin down, but it usually implies that $f$ is some specific function, usually not elementary, which is useful in some sense, and which can be calculated, among other methods,

  • as a series representation,
  • as an integral representation,
  • as some specific solution of a differential equation,
  • via some recursion relation,

and so on. Very often, if you know any of these representations for $f$, you can transform it into an analogous representation for its indefinite integral.

This sort of ties in with one of the things you said about the Bessel equation - that "its solutions are not too hard to find". While this is true, by "finding" those solutions you probably mean finding their Taylor series representation, via the Frobenius method. This works, but it is the clunkiest, clumsiest way to define the Bessel function.

More importantly, the Frobenius method is quite apt at giving the impression that the Bessel function $J_0$ "is" its Taylor series, and that's not a great way to approach special functions in general. The Bessel function is just the mapping $z\mapsto J_0(z)$, and there is a host of different representations for its value, all of which are equivalent. As an example, I personally think of $J_0$ most often as "being" its integral representation: $$ J_0(z) = \frac{1}{2\pi} \int_0^{2\pi} e^{i z\cos(\theta)}\mathrm d\theta. $$ Then, if you insist, I can build its Taylor series (easy), or show that it satisfies the Bessel differential equation (easy), or build recursion relations between it and other $J_\nu$s. Moreover, I can show it's an entire function of $z$, estimate its growth at imaginary arguments, and provide good asymptotic estimates both at large and small $z$.

Similarly, the Taylor series is only used for numerical computations a fraction of the time. The DLMF has good Computation sections in its various chapters; have a gander at the Bessel function one to see what's used in practice.

All of this is to say, really, that special functions are relatively complex objects, and that there isn't generally a silver bullet for dealing with them. With questions about integrals, for example, you're mostly left with doing each computation individually and hoping for a relatively simple result.

For indefinite integrals, for example, for broad classes of special functions and their combinations you can express the integrand as some form of generalized hypergeometric function, or a Meijer $G$ function, and those generally integrate within their class. The challenge is then expressing the resulting monster in a form that's useful. For definite integrals, you mostly have to sit down and work out which representation is more likely to yield results, but even then you're never guaranteed a specific, elementary answer.

The specific case of $⟨R|R⟩$, is slightly tricky because you haven't specified completely whether they're the exact same solution of different solutions of the same equation. If they are different solutions, i.e. integrals of the form $$ \int_0^\infty J_l(x)J_{l'}(x) x\mathrm dx, $$ then the integral is zero by orthogonality, and that is most easily shown using the Sturm-Liouville properties of the differential equation they satisfy.

If it is the same solution, on the other hand, the solution is easy: the integral diverges, $$ \int_0^\infty J_l(x)^2 x\mathrm dx = \infty, $$ because Bessel waves (like plane waves) have infinite energy. Even though they decay at large argument, they do so too slowly for the integral to converge. (This is, of course, mostly by construction: they're built to have roughly the same energy in any disk slice between radii $r$ and $r+\Delta r$, regardless of $r$, so they need to have an infinite norm.)

I can clarify things further but that's as far as I can go on the nebulous current form of the question; if you ask something more specific then I'm happy to expand on that.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.