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This question is a sequel to this previous question. As before, some background information is needed first as follows from my textbook:

The standard form of Bessel's differential equation is $$x^2y^{\prime\prime}+xy^{\prime} + (x^2 - p^2)y=0\tag{1}$$ where $(1)$ has a first solution given by $$\fbox{$J_p(x)=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+p)}\left(\frac{x}{2}\right)^{2n+p}$}\tag{2}$$ and a second solution given by $$\fbox{$J_{-p}(x)=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1-p)}\left(\frac{x}{2}\right)^{2n-p}$}\tag{3}$$ where $J_p(x)$ is called the Bessel function of the first kind of order $p$.

Although $J_{−p}(x)$ is a satisfactory second solution when $p$ is not an integer, it is customary to use a linear combination of $J_p(x)$ and $J_{−p}(x)$ as the second solution. Any combination of $J_p(x)$ and $J_{−p}(x)$ is a satisfactory second solution of Bessel’s equation. The combination which is used is called the Neumann (or the Weber) function and is denoted by $N_p(x)$ where $$N_p(x)=\frac{\cos(\pi p)J_p(x)-J_{-p}(x)}{\sin(\pi p)}\tag{4}$$

Full details on the derivation of $(2)$ as a solution to $(1)$ can be found here in my previous question.

Many differential equations occur in practice that are not of the standard form $(1)$ but whose solutions can be written in terms of Bessel functions. It can be shown that the differential equation: $$\fbox{$y^{\prime\prime}+\left(\frac{1-2a}{x}\right)y^{\prime}+\left[\left(bcx^{c-1}\right)^2+\frac{a^2-p^2c^2}{x^2}\right]y=0$}\tag{5}$$ has the solution $$\fbox{$y=x^aZ_p\left(bx^c\right)$}\tag{6}$$ where $Z_p$ stands for $J_p$ or $N_p$ or any linear combination of them, and $a,b,c,p$ are constants.

To see how to use this, let us “solve” the differential equation: $$y^{\prime\prime}+9xy=0\tag{7}$$ If $(7)$ is of the type $(5)$, then we must have $$1-2a=0$$ $$2(c-1)=1$$ $$(bc)^2=9$$ $$a^2-p^2c^2=0$$ from these $4$ equations we find $$a=\dfrac12$$ $$c=\dfrac32$$ $$b=2$$ $$p=\dfrac{a}{c}=\dfrac13$$

Then the solution of $(7)$ is $$y=x^{1/2}Z_{1/3}\left(2x^{3/2}\right)\tag{8}$$ This means that the general solution of $(7)$ is $$y=x^{1/2}\left[AJ_{1/3}\left(2x^{3/2}\right)+BN_{1/3}\left(2x^{3/2}\right)\right]\tag{9}$$ where $A$ and $B$ are arbitrary constants.


My goal is to show that $(9)$ is a solution to $(7)$.

I will be writing out almost all intermediate steps; so if an error exists it will be easier to locate.


Starting from $(9)$ and substituting $(2)$, $(3)$ and $(4)$: \begin{align}y &=x^{1/2}\left[AJ_{1/3}\left(2x^{3/2}\right)+BN_{1/3}\left(2x^{3/2}\right)\right]\\&=x^{1/2}\left[AJ_{1/3}\left(2x^{3/2}\right)+\frac{\frac12 B J_{1/3}-B J_{-1/3}\left(2x^{3/2}\right)}{\left(\sqrt{3}/2\right)}\right]\\&= x^{1/2}\left[A J_{1/3}\left(2x^{3/2}\right)+\frac{ B J_{1/3}-2B J_{-1/3}\left(2x^{3/2}\right)}{\sqrt{3}}\right]\\&=x^{1/2}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}\left(\frac{2x^{3/2}}{2}\right)^{2n+1/3}\quad-\frac{2Bx^{1/2}}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac23)}\left(\frac{2x^{3/2}}{2}\right)^{2n-1/3}\\&\fbox{$y=x^{1/2}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n+1/2}-\frac{2Bx^{1/2}}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac23)}x^{3n-1/2}$}\end{align}

Now taking the derivative of $y$ with respect to $x$ by use of the product rule: $$\fbox{$y^{\prime}=\frac12x^{-1/2}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n+1/2}\quad+x^{1/2}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n\left(3n+\frac12\right)x^{3n-1/2}}{\Gamma(n+1)\Gamma(n+\frac43)}\,\,\,-\,\frac{Bx^{-1/2}}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac23)}x^{3n-1/2}-\frac{2Bx^{1/2}}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n\left(3n-\frac12\right)x^{3n-3/2}}{\Gamma(n+1)\Gamma(n+\frac23)}$}$$

Now taking the derivative of $y^{\prime}$ with respect to $x$ by use of the product rule again:

\begin{align}y^{\prime\prime}=-\frac14x^{-3/2}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n+1/2}\quad+\frac12x^{-1/2}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n\left(3n+\frac12\right)x^{3n-1/2}}{\Gamma(n+1)\Gamma(n+\frac43)}\quad+\frac12x^{-1/2}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n\left(3n+\frac12\right)x^{3n-1/2}}{\Gamma(n+1)\Gamma(n+\frac43)}\quad+x^{1/2}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n\left(3n+\frac12\right)\left(3n-\frac12\right)x^{3n-3/2}}{\Gamma(n+1)\Gamma(n+\frac43)}\quad+\frac{Bx^{-3/2}}{2\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac23)}x^{3n-1/2}-\,\,\,\frac{Bx^{-1/2}}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n\left(3n-\frac12\right)x^{3n-3/2}}{\Gamma(n+1)\Gamma(n+\frac23)}-\frac{Bx^{-1/2}}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n\left(3n-\frac12\right)x^{3n-3/2}}{\Gamma(n+1)\Gamma(n+\frac23)}-\frac{2Bx^{1/2}}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n\left(3n-\frac12\right)\left(3n-\frac32\right)x^{3n-5/2}}{\Gamma(n+1)\Gamma(n+\frac23)}\end{align}


Substitution of $y^{\prime\prime}$ and $y$ into $$y^{\prime\prime}+9xy=0\tag{7}$$ gives

\begin{align}-\frac14x^{-3/2}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n+1/2}\quad+\frac12x^{-1/2}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n\left(3n+\frac12\right)x^{3n-1/2}}{\Gamma(n+1)\Gamma(n+\frac43)}\quad+\frac12x^{-1/2}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n\left(3n+\frac12\right)x^{3n-1/2}}{\Gamma(n+1)\Gamma(n+\frac43)}\quad+x^{1/2}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n\left(3n+\frac12\right)\left(3n-\frac12\right)x^{3n-3/2}}{\Gamma(n+1)\Gamma(n+\frac43)}\quad+\frac{Bx^{-3/2}}{2\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac23)}x^{3n-1/2}-\,\,\,\frac{Bx^{-1/2}}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n\left(3n-\frac12\right)x^{3n-3/2}}{\Gamma(n+1)\Gamma(n+\frac23)}-\frac{Bx^{-1/2}}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n\left(3n-\frac12\right)x^{3n-3/2}}{\Gamma(n+1)\Gamma(n+\frac23)}-\frac{2Bx^{1/2}}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n\left(3n-\frac12\right)\left(3n-\frac32\right)x^{3n-5/2}}{\Gamma(n+1)\Gamma(n+\frac23)}+9x^{3/2}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n+1/2}-\frac{18Bx^{3/2}}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac23)}x^{3n-1/2}\end{align}

Simplifying a little gives

\begin{align}-\frac14x^{-1}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n}+\frac12x^{-1}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n\left(3n+\frac12\right)x^{3n}}{\Gamma(n+1)\Gamma(n+\frac43)}\quad+\frac12x^{-1}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n\left(3n+\frac12\right)x^{3n}}{\Gamma(n+1)\Gamma(n+\frac43)}\quad+x^{-1}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n\left(3n+\frac12\right)\left(3n-\frac12\right)x^{3n}}{\Gamma(n+1)\Gamma(n+\frac43)}+\frac{Bx^{-2}}{2\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac23)}x^{3n}-\frac{Bx^{-2}}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n\left(3n-\frac12\right)x^{3n}}{\Gamma(n+1)\Gamma(n+\frac23)}-\frac{Bx^{-2}}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n\left(3n-\frac12\right)x^{3n}}{\Gamma(n+1)\Gamma(n+\frac23)}\quad-\frac{2Bx^{-2}}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n\left(3n-\frac12\right)\left(3n-\frac32\right)x^{3n}}{\Gamma(n+1)\Gamma(n+\frac23)}+9x^{2}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n}-\frac{18Bx}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac23)}x^{3n}\end{align}

Expanding out some of the summations gives

\begin{align}-A\frac14x^{-1}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n}-\frac{Bx^{-1}}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n}\quad+x^{-1}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n\left(3n+\frac12\right)x^{3n}}{\Gamma(n+1)\Gamma(n+\frac43)}\quad+x^{-1}\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n\left(9n^2-\frac14\right)x^{3n}}{\Gamma(n+1)\Gamma(n+\frac43)}+\frac{Bx^{-2}}{2\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac23)}x^{3n}-2\sqrt{3}Bx^{-2}\sum_{n=0}^\infty\frac{n(-1)^nx^{3n}}{\Gamma(n+1)\Gamma(n+\frac23)}\quad+\frac{Bx^{-2}}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^nx^{3n}}{\Gamma(n+1)\Gamma(n+\frac23)}\quad-2\sqrt{3}Bx^{-2}\sum_{n=0}^\infty\frac{(-1)^n\left(3n^2-2n+\frac14\right)x^{3n}}{\Gamma(n+1)\Gamma(n+\frac23)}\quad+9Ax^{2}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n}\quad+\frac{B}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n}\quad-\frac{18Bx}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac23)}x^{3n}\end{align}

Does anyone have any ideas on how I can show this is equal to zero?

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  • $\begingroup$ Impressive! :-) (+1) $\endgroup$ – Markus Scheuer Jun 13 '16 at 8:25
  • $\begingroup$ @Markus Thanks! It's quite a messy problem here and I still have to show that entire lot is equal to zero. $\endgroup$ – BLAZE Jun 13 '16 at 12:23
  • $\begingroup$ @Markus Now I'm stuck on another question :-( Would you mind taking a look at this Airy functions question? Many thanks! $\endgroup$ – BLAZE Jun 14 '16 at 3:06
  • $\begingroup$ I will take a look at it after work. In the meantime I was busy with another nice question :-) $\endgroup$ – Markus Scheuer Jun 14 '16 at 6:51
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    $\begingroup$ @Markus You know it's funny I was looking at that very same question myself yesterday and tried to answer it. Congrats on getting that bounty (another one to your collection!) (+1) :-) $\endgroup$ – BLAZE Jun 14 '16 at 12:04
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Here are some aspects to master such expressions:

Let's denote the last expression with $G(x)$.

\begin{align*} G(x)&=-A\frac14\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n-1} -\frac{B}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n-1}\\ &\quad+\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n\left(3n+\frac12\right)x^{3n-1}}{\Gamma(n+1)\Gamma(n+\frac43)} +\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n\left(9n^2-\frac14\right)x^{3n-1}}{\Gamma(n+1)\Gamma(n+\frac43)}\\ &\quad+\frac{B}{2\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac23)}x^{3n-2}\ -2\sqrt{3}B\sum_{n=0}^\infty\frac{n(-1)^nx^{3n-2}}{\Gamma(n+1)\Gamma(n+\frac23)}\\ &\quad+\frac{B}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^nx^{3n-2}}{\Gamma(n+1)\Gamma(n+\frac23)} -2\sqrt{3}B\sum_{n=0}^\infty\frac{(-1)^n\left(3n^2-2n+\frac14\right)x^{3n-2}}{\Gamma(n+1)\Gamma(n+\frac23)}\\ &\quad+9A\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n+2} +\frac{B}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n}\\ &\quad-\frac{18B}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac23)}x^{3n+1}\tag{1} \end{align*}

We see $G(x)$ has a representation as sum of eleven series. The powers of the series of the right hand side have the following general term from first to last:

\begin{align*} &3n-1, 3n-1, 3n-1, 3n-1, \\ &3n-2, 3n-2, 3n-2, 3n-2,\tag{2} \\ &3n+2, 3n, 3n+1 \end{align*}

This suggests a split of the investigation of $G(x)$ in powers of $x^n$ according to $n$ modulo $3$.

We have to show that $$G(x)=\sum_{n=-2}^\infty g_n x^n\equiv 0$$ This means that each coefficient $g_n=0$ for $n\geq -2$. The series starts from $n=-2$, since the term with the smallest power can be found in the fifth to eight series, each starting with $x^{-2}$.

It's convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of a series. We can write this way e.g. \begin{align*} [x^{k}]G(x)=[x^{k}]\sum_{n=-2}^\infty g_n x^n=g_{k} \end{align*}

We see in (2) we have to consider $3n-1$ together with $3n+2$ when analysing the coefficients of $G(x)$ which are congruent $2$ modulo $3$ and we have to consider $3n-2$ together with $3n+1$ when analysing the coefficients of $G(x)$ which are congruent $1$ modulo $3$.

Attention: A series with general power $3n$ occurs only once. So, these coefficients will not vanish indicating some calculation error. In fact it seems a factor $9x^2$ is missing there.

Let's check e.g. the coefficients with general power $3k-2$, $k\geq 0$.

\begin{align*} [x^{3k-2}]G(x)&=\frac{B}{2\sqrt{3}}[x^{3k-2}]\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac23)}x^{3n-2}\\ &\quad-2\sqrt{3}B[x^{3k-2}]\sum_{n=0}^\infty\frac{n(-1)^nx^{3n-2}}{\Gamma(n+1)\Gamma(n+\frac23)}\\ &\quad+\frac{B}{\sqrt{3}}[x^{3k-2}]\sum_{n=0}^\infty\frac{(-1)^nx^{3n-2}}{\Gamma(n+1)\Gamma(n+\frac23)}\\ &\quad-2\sqrt{3}B[x^{3k-2}]\sum_{n=0}^\infty\frac{(-1)^n\left(3n^2-2n+\frac14\right)x^{3n-2}}{\Gamma(n+1)\Gamma(n+\frac23)}\\ &\quad-\frac{18B}{\sqrt{3}}[x^{3k-2}]\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac23)}x^{3n+1}\tag{3}\\ &=\frac{B}{2\sqrt{3}}\frac{(-1)^k}{\Gamma(k+1)\Gamma(k+\frac23)}\ -2\sqrt{3}B\frac{k(-1)^k}{\Gamma(k+1)\Gamma(k+\frac23)}\\ &\quad+\frac{B}{\sqrt{3}}\frac{(-1)^k}{\Gamma(k+1)\Gamma(k+\frac23)} -2\sqrt{3}B\frac{(-1)^k\left(3k^2-2k+\frac14\right)}{\Gamma(k+1)\Gamma(k+\frac23)}\\ &\quad-\frac{18B}{\sqrt{3}}\frac{(-1)^{k-1}}{\Gamma(k)\Gamma(k-\frac13)}\tag{4}\\ &=\frac{(-1)^kB}{\Gamma(k+1)\Gamma(k+\frac23)} \left(\frac{1}{2\sqrt{3}}-2\sqrt{3}k+\frac{1}{\sqrt{3}}\right.\\ &\qquad\qquad\left.-2\sqrt{3}\left(3k^2-2k+\frac14\right) +\frac{18}{\sqrt{3}}k\left(k+\frac{2}{3}\right)\right)\\ &=\frac{(-1)^kB}{\Gamma(k+1)\Gamma(k+\frac23)} \left(-\frac{5}{\sqrt{3}}k^2+\frac{44}{3\sqrt{3}}k\right)\tag{5} \end{align*}

Comment:

  • In (3) we consider only the series which contribute to coefficients with powers congruent $1$ modulo $3$.

  • In (4) we select the coefficients with power $3k-2$. This implies that in the last series we have to put $n=k-1$ in order to get the coefficient of $x^{3k-2}$.

  • In (5) we simplify and collect corresponding terms.

Warning: Since the simplification does not reduce to zero there is another calculation error somewhere (either in my answer or in OPs calculation).

But this is not that important since I only want to demonstrate a convenient approach.


Please note, that you can considerably simplify the calculations by avoiding the product rule. Just write $y$ in the form \begin{align*} y=\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n+1}-\frac{2B}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac23)}x^{3n}\tag{6} \end{align*}

Differentiation of (6) results in two summands instead of four. A following differentiation of $y^\prime$ by avoiding the product rule again reduces the number of summands considerably.

The above approach could be applied afterwards if it seems to be convenient.

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