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Suppose that a function $f:[0,a]\to\Bbb{R}$ is given. We are interested to find its Bessel-Expansion and study the convergence of this expansion specially at the end points $r=0$ and $r=a$.

Consider the following singular Sturm-Liouville systems for zero and first order Bessel functions

\begin{align} A.\,&\frac{d}{dr}\left[r\frac{dR}{dr}\right]+\lambda r R=0 & B.\,&\frac{d}{dr}\left[r\frac{dR}{dr}\right]+\left[\lambda r + \frac{1}{r}\right]R=0\\ &R(0)\lt\infty & &R(0)\lt\infty\\ &\frac{dR}{dr}(a)=0 & &R(a)=0 \tag{1} \end{align}

I know that the eigen-functions of systems $A$ and $B$ can form a basis for all $C^1[0,a]$ functions. Then we can construct the Bessel expansions corresponding to systems $A$ and $B$ as

\begin{align} S(r):=\sum_{i=1}^{\infty}C_iR(\lambda_i,r) \tag{2} \end{align}

I know that $S$ converges to $\frac{f(r^+)+f(r^-)}{2}$ at the points in the interval $(0,a)$. Consequently, $S$ will converge to $f$ at the points of coninuity of $f$ inside the aforementioned interval but I don't know what happens at the end points. Here is my question

  1. What are the necessary, sufficient, necessary and sufficient conditions that $S$ converges to $f$ at $r=0$ and $r=a$?

  2. Suppose we answered question $1$. Can this be generalized to any Sturm-Liouville expansions like Fourier, Chebyshev, Hermit, etc?

This animation shows the convergence of the eigen-function expansion of system $A$ to the modified Bessel function of order zero $f(r)=I_0(r)$. In this case convergence at the end points is achieved although $f(r)$ does not satisfy the second BC mentioned in Sturm-Liouville system A.

enter image description here

and this one shows the convergence of the eigen-function expansion of system $B$ to $f(r)=I_0(r)$. It seems that the series does not converge to the function at the end points in this case and $f(r)$ is not satistying the second BC of Sturm-Liouville system B.

enter image description here

Here is the mathematica code for making the animations.

ClearAll["Global`*"]

f[r_] := BesselI[0, r]

a = 1;

Subscript[N, max] = 40;

A[n_] = Simplify[\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)]\(r*f[r]*
    BesselJ[0, \[Alpha][n]*r] \[DifferentialD]r\)\)/\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)]\(r*
\*SuperscriptBox[\(BesselJ[
      0, \[Alpha][n]*r]\), \(2\)] \[DifferentialD]r\)\), 
  Assumptions -> {BesselJ[1, \[Alpha][n]*a] == 0}]

(2 BesselI[1, 1])/(BesselJ[0, \[Alpha][n]] (1 + \[Alpha][n]^2))

Eig = Table[{i, N[BesselJZero[1, i]/a]}, {i, 1, Subscript[N, max]}];

Eig = Prepend[Eig, {0, 0}];

(\[Alpha][#] = #2) & @@@ Eig;

BesselSeries[r_, n_] := \!\(
\*SubsuperscriptBox[\(\[Sum]\), \(i = 0\), \(n\)]\(A[i]*
   BesselJ[0, \[Alpha][i]*r]\)\)

ConvAnim = 
  Table[Plot[Evaluate[{f[r], BesselSeries[r, n]}], {r, 0, a}, 
    PlotRange -> {0.9*f[0], 1.1*f[a]}, ImageSize -> Large, 
    AspectRatio -> Automatic], {n, 1, Subscript[N, max]}];

(*Export["Convergence.gif",ConvAnim,"DisplayDurations"\[Rule]{0.25}]*)
  • 1
    On the inside of the interval, you won't get convergence at every point of continuity; that is too much to expect even for the usual Fourier series. If the function is of bounded variation on a closed interval $[a,b] \subset (0,1)$ or if there is some smoothness condition on $[a,b]$, then you'll get pointwise convergence in $(a,b)$, but continuity is not enough for the classical case. Convergence at $r=0$ is problematic because $J_0(0)=1$, which is what makes this an interesting question. :) – 2017-01-14
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    @TrialAndError: Thanks for the attention. In fact, I said that $f$ is piece-wise continuous after $(1)$. I saw in my book that being piece-wise continuous is enough for point wise convergence inside the interval! Is this false? :) – 2017-01-14
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    Continuous is not enough for convergence of the classical Fourier series. If it is piecewise smooth, then you get convergence. So I doubt that continuity is enough for the Fourier-Bessel series either. I know that the Fourier-Bessel series on $[a,b] \subset (0,\infty)$ converges at a point in (a,b)$ iff the trigonometric series with the same endpoint conditions converges; this is proved with asymptotics for Bessel functions. – 2017-01-14
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    @TrialAndError: I got your point. Will check that. :) Thanks again. Please let me know if you got anything about this question. :) – 2017-01-14
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    @TrialAndError: Can you give me any hint for this question? No answers after two days! :) – 2017-01-17
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    Unfortunately, I don't know of a good answer at the singular endpoint, which is where you want to know about convergence. Even for $J_0$, I don't know how to answer your question. Your pictures are better than anything I've got. – 2017-01-17
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    As I mentioned, the left endpoint at $r=0$ is problematic for the $J_0$ case because all of the functions in the expansion are $1$ at $0$. So you really need to look at the normalization constants and consider what happens when you expand an $L^2_{r}[0,1]$ function. I don't know the answer, but I suspect that divergence is more common at $r=0$ for the Fourier-Bessel series than it is for other points. – 2017-01-17
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    @TrialAndError: I found in a book that a sufficient condition is that the function $f$ should satisfy the BCs in the Sturm-Liouville system. According to the animation for system $A$, it seems that it shouldn't be a necessary condition! – 2017-01-17
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    The singular condition at an endpoint is a Wronskian condition with respect to an eigenfunction. That will certainly improve convergence because the convergence is strongly affected by the largest power of the operator in which the function is found in the domain. I would not except such conditions to be necessary, any more than they are for the ordinary Fourier series. – 2017-01-17

0 Answers 0