In order for the question that I have to make any sense I must first include some background information as given in 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.

Finally,$\color{#180}{\text{ my goal is to show that }}$${(6)}$ $\color{#180}{\text{is a solution to }}$$(5)$.

However to gain some insight, I must first be able to show that $(8)$ or $(9)$ is a solution to $(7)$.

So my attempt goes as follows:

So I need to compute $y^{\prime\prime}$ or at least to begin with, $y^{\prime}$; It is at this point where I am immediately stuck as I do not understand how to differentiate $$y=x^{1/2}Z_{1/3}\left(2x^{3/2}\right)\tag{8}$$ as I'm confused as to how to take the derivative of the $Z_{1/3}\left(2x^{3/2}\right)$ factor.

Could someone please provide some hints or advice on how I would go about carrying out this differentiation?

  • 2
    $\begingroup$ It seems like you have to differentiate the power series term by term. $\endgroup$ – Chee Han May 29 '16 at 8:17
  • $\begingroup$ @CheeHan Well I've just done exactly what you suggested here. Any ideas on how I should proceed? Thanks. $\endgroup$ – BLAZE Jun 14 '16 at 3:01

[2016-06-07] Note: General solution added to provide a comparison with the example part.

Here we show that according to OPs example \begin{align*} y(x)=x^{\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) \end{align*} is a solution of the differential equation \begin{align*} y^{\prime\prime}+9xy=0\tag{1} \end{align*}

In the following we use the prime notation $f^\prime$ to denote the derivative of $f$. We will often use the product rule for derivatives \begin{align*} (f\cdot g)^\prime=f^\prime \cdot g+f\cdot g^\prime\qquad\qquad (f(x)g(x))^\prime=f^\prime(x) g(x)+f(x)g^\prime(x) \end{align*} and the chain rule \begin{align*} (f\circ g)^\prime=(f^\prime \circ g)\cdot g^\prime\qquad\qquad\quad (f(g(x)))^\prime=f^\prime(g(x))g^\prime(x) \end{align*}

We start with \begin{align*} y(x)&=x^{\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}})\\ \end{align*} and obtain the first derivative \begin{align*} y^\prime(x)&=\left(x^{\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}})\right)^\prime\\ &=\left(x^{\frac{1}{2}}\right)^\prime Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +x^\frac{1}{2}\left(Z_{\frac{1}{3}}(2x^{\frac{3}{2}})\right)^\prime\tag{2}\\ &=\frac{1}{2}x^{-\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +x^{\frac{1}{2}}{Z^\prime_{\frac{1}{3}}}(2x^{\frac{3}{2}})\cdot\left(2x^{\frac{3}{2}}\right)^\prime\tag{3}\\ &=\frac{1}{2}x^{-\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +x^{\frac{1}{2}}Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}})\cdot3x^{\frac{1}{2}}\\ &=\frac{1}{2}x^{-\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +3xZ^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}})\tag{4}\\ \end{align*}


  • In (2) we apply the product rule

  • In (3) we apply the chain rule

The next step is getting the second derivative. We obtain from (4) \begin{align*} y^{\prime\prime}(x)&=\left(\frac{1}{2}x^{-\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}})\right)^\prime +\left(3xZ^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}})\right)^\prime\\ &=\left(\frac{1}{2}x^{-\frac{1}{2}}\right)^\prime Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\frac{1}{2}x^{-\frac{1}{2}}\left(Z_{\frac{1}{3}}(2x^{\frac{3}{2}})\right)^\prime\\ &\qquad+\left(3x\right)^\prime Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}}) +3x \left(Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}})\right)^\prime\tag{5}\\ &=-\frac{1}{4}x^{-\frac{3}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\frac{1}{2}x^{-\frac{1}{2}}Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}})\cdot\left(2x^{\frac{3}{2}}\right)^\prime\\ &\qquad+3 Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}}) +3xZ^{\prime\prime}_{\frac{1}{3}}(2x^{\frac{3}{2}})\cdot\left(2x^{\frac{3}{2}}\right)^\prime\tag{6}\\ &=-\frac{1}{4}x^{-\frac{3}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\frac{1}{2}x^{-\frac{1}{2}}Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}})\cdot 3x^{\frac{1}{2}}\\ &\qquad+3 Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}}) +3xZ^{\prime\prime}_{\frac{1}{3}}(2x^{\frac{3}{2}})\cdot 3x^{\frac{1}{2}}\\ &=-\frac{1}{4}x^{-\frac{3}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\frac{9}{2}Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}}) +9x^{\frac{3}{2}}Z^{\prime\prime}_{\frac{1}{3}}(2x^{\frac{3}{2}})\tag{7}\\ \end{align*}


  • In (5) we apply the product rule for both terms

  • In (6) we apply the chain rule for both terms

So, let's put $y(x)=x^{\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}})$ and the result (7) of $y^{\prime\prime}(x)$ into the differential equation (1). We obtain \begin{align*} y^{\prime\prime}&+9xy\\ &=-\frac{1}{4}x^{-\frac{3}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\frac{9}{2}Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}}) +9x^{\frac{3}{2}}Z^{\prime\prime}_{\frac{1}{3}}(2x^{\frac{3}{2}}) +9x\left(x^{\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}})\right)\\ &=9x^{\frac{3}{2}}Z^{\prime\prime}_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\frac{9}{2}Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\left(9x^{\frac{3}{2}}-\frac{1}{4}x^{-\frac{3}{2}}\right)Z_{\frac{1}{3}}(2x^{\frac{3}{2}})\tag{8} \end{align*}

On the other hand we know that $Z_p(x)$ is a solution of \begin{align*} x^2y^{\prime\prime}(x)+xy^{\prime}(x)+(x^2-p^2)y(x)=0 \end{align*} This means $Z_{\frac{1}{3}}(x)$ fulfils \begin{align*} x^2Z_{\frac{1}{3}}^{\prime\prime}(x)+xZ_{\frac{1}{3}}^{\prime}(x)+\left(x^2-\frac{1}{9}\right)Z_{\frac{1}{3}}(x)=0\tag{9} \end{align*}

Substituting \begin{align*} x\rightarrow 2x^{\frac{3}{2}} \end{align*} in (9) results in \begin{align*} \left(2x^{\frac{3}{2}}\right)^2Z_{\frac{1}{3}}^{\prime\prime}(2x^{\frac{3}{2}})+2x^{\frac{3}{2}}Z_{\frac{1}{3}}^{\prime}(2x^{\frac{3}{2}})+\left((2x^{\frac{3}{2}})^2-\frac{1}{9}\right)Z_{\frac{1}{3}}(2x^{\frac{3}{2}})=0\\ 4x^3Z_{\frac{1}{3}}^{\prime\prime}(2x^{\frac{3}{2}})+2x^{\frac{3}{2}}Z_{\frac{1}{3}}^{\prime}(2x^{\frac{3}{2}})+\left(4x^3-\frac{1}{9}\right)Z_{\frac{1}{3}}(2x^{\frac{3}{2}})=0\\ \end{align*} and multiplying this equation with $\frac{9}{4}x^{-\frac{3}{2}}$ finally gives

\begin{align*}9x^{\frac{3}{2}}Z^{\prime\prime}_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\frac{9}{2}Z^\prime_{\frac{1}{3}}(2x^{\frac{3}{2}}) +\left(9x^{\frac{3}{2}}-\frac{1}{4}x^{-\frac{3}{2}}\right)Z_{\frac{1}{3}}(2x^{\frac{3}{2}})=0 \end{align*}

We see the left-hand side is equal to (8) and we conclude that $x^{\frac{1}{2}}Z_{\frac{1}{3}}\left(2x^{\frac{3}{2}}\right)$ is a solution of \begin{align*} y^{\prime\prime}+9xy=0 \end{align*}

Note: In the same way we can show that \begin{align*} y(x)=x^aZ_p(bx^c) \end{align*} is a solution of \begin{align*} y^{\prime\prime}+\left(\frac{1-2a}{x}\right)y^\prime+\left[(bcx^{c-1})^2+\frac{a^2-p^2c^2}{x^2}\right]y=0 \end{align*}

Note: Here as small supplement to the already given nice answers by @JJacquelin and @Bacon the general derivation just to provide a comparison with the example part of my answer.

We show that
\begin{align*} y(x)=x^{a}Z_{p}(bx^{c}) \end{align*} is a solution of the differential equation \begin{align*} y^{\prime\prime}+\left(\frac{1-2a}{x}\right)y^\prime+\left[(bcx^{c-1})^2+\frac{a^2-p^2c^2}{x^2}\right]y=0\tag{10} \end{align*}

We start with calculating the first derivative \begin{align*} y^\prime(x)&=\left(x^{a}Z_{p}(bx^{c})\right)^\prime\\ &=\left(x^{a}\right)^\prime Z_{p}(bx^{c}) +x^a\left(Z_{p}(bx^{c})\right)^\prime\tag{11}\\ &=ax^{a-1}Z_{p}(bx^{c}) +x^{a}Z^\prime_p(bx^{c})\cdot\left(bx^{c}\right)^\prime\tag{12}\\ &=ax^{a-1}Z_{p}(bx^{c}) +x^{a}Z^\prime_p(bx^{c})\cdot bcx^{c-1}\\ &=ax^{a-1}Z_{p}(bx^{c}) +bcx^{a+c-1}Z^\prime_p(bx^{c})\tag{13}\\ \end{align*}


  • In (11) we apply the product rule

  • In (12) we apply the chain rule

The next step is getting the second derivative. We obtain from (13) \begin{align*} y^{\prime\prime}(x)&=\left(ax^{a-1}Z_{p}(bx^{c})\right)^\prime +\left(bcx^{a+c-1}Z^\prime_{p}(bx^{c})\right)^\prime\\ &=(ax^{a-1})^\prime Z_{p}(bx^{c})+ax^{a-1}\left(Z_{p}(bx^{c})\right)^\prime\\ &\qquad+(bcx^{a+c-1})^\prime Z_{p}^\prime(bx^{c})+bcx^{a+c-1}\left(Z_{p}^\prime(bx^{c})\right)^\prime\tag{14}\\ &=a(a-1)x^{a-2} Z_{p}(bx^{c})+ax^{a-1}Z_{p}^\prime(bx^{c})\cdot\left(bx^{c}\right)^\prime\\ &\qquad+bc(a+c-1)x^{a+c-2} Z_{p}^\prime(bx^{c}) +bcx^{a+c-1}Z_{p}^{\prime\prime}(bx^{c})\cdot\left(bx^{c}\right)^\prime\tag{15}\\ &=a(a-1)x^{a-2} Z_{p}(bx^{c})+ax^{a-1}Z_{p}^\prime(bx^{c})\cdot bcx^{c-1}\\ &\qquad+bc(a+c-1)x^{a+c-2} Z_{p}^\prime(bx^{c}) +bcx^{a+c-1}Z_{p}^{\prime\prime}(bx^{c})\cdot bcx^{c-1}\\ &=a(a-1)x^{a-2} Z_{p}(bx^{c})+(2a+c-1)bcx^{a+c-{2}}Z_{p}^\prime(bx^{c})\\ &\qquad+b^2c^2x^{a+2c-2}Z_{p}^{\prime\prime}(bx^{c})\tag{16}\\ \end{align*}


  • In (14) we apply the product rule for both terms

  • In (15) we apply the chain rule for both terms

So, let's put $y(x)=x^{a}Z_{p}(bx^{c})$ and (13), the result of $y^\prime(x)$ and (16), the result of $y^{\prime\prime}(x)$ into the differential equation (10). We obtain \begin{align*} y^{\prime\prime}(x)&+\left(\frac{1-2a}{x}\right)y^\prime(x)+\left[(bcx^{c-1})^2+\frac{a^2-p^2c^2}{x^2}\right]y(x)\\ &=a(a-1)x^{a-2} Z_{p}(bx^{c})+(2a+c-1)bcx^{a+c-2}Z_{p}^\prime(bx^{c})\\ &\qquad+b^2c^2x^{a+2c-2}Z_{p}^{\prime\prime}(bx^{c})\\ &\qquad+\left(\frac{1-2a}{x}\right)\left(ax^{a-1}Z_{p}(bx^{c}) +bcx^{a+c-1}Z^\prime_p(bx^{c})\right)\\ &\qquad+\left[(bcx^{c-1})^2+\frac{a^2-p^2c^2}{x^2}\right]x^{a}Z_{p}(bx^{c})\\ &=c^2x^{a-2}\left(b^2x^{2c}Z_{p}^{\prime\prime}(bx^{c})+bx^cZ_{p}^{\prime}(bx^{c}) +(b^2x^{2c}-p^2)Z_{p}(bx^{c})\right)\tag{17} \end{align*}

On the other hand we know that $Z_p(x)$ is a solution of \begin{align*} x^2y^{\prime\prime}(x)+xy^{\prime}(x)+(x^2-p^2)y(x)=0 \end{align*} This means $Z_{p}(x)$ fulfils \begin{align*} x^2Z_{p}^{\prime\prime}(x)+xZ_{p}^{\prime}(x)+\left(x^2-p^2\right)Z_{p}(x)=0\tag{18} \end{align*}

Substituting \begin{align*} x\rightarrow bx^{c} \end{align*} in (18) results in \begin{align*} (bx^{c})^2Z_{p}^{\prime\prime}(bx^{c})+bx^{c}Z_{p}^{\prime}(bx^{c})+\left((bx^{c})^2-p^2\right)Z_{p}(bx^{c})=0\\ b^2x^{2c}Z_{p}^{\prime\prime}(bx^{c})+bx^{c}Z_{p}^{\prime}(bx^{c})+\left(b^2x^{2c}-p^2\right)Z_{p}(bx^{c})=0 \end{align*} and multiplying this equation with $c^2x^{a-2}$ finally gives

\begin{align*}c^2x^{a-2}\left(b^2x^{2c}Z_{p}^{\prime\prime}(bx^{c})+bx^{c}Z_{p}^{\prime}(bx^{c})+\left(b^2x^{2c}-p^2\right)Z_{p}(bx^{c})\right)=0 \end{align*}

We see the left-hand side is equal to (17) and we conclude that $x^{a}Z_{p}\left(bx^{c}\right)$ is a solution of \begin{align*} x^2y^{\prime\prime}(x)+xy^{\prime}(x)+(x^2-p^2)y(x)=0 \end{align*}

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  • $\begingroup$ @BLAZE: By the way, it was a good approach to start with the specific, somewhat simpler example, which already contains all the essentials we need to attack the general problem. This strategy is often convenient. $\endgroup$ – Markus Scheuer Jun 8 '16 at 5:11
  • $\begingroup$ I have now read all of your answer, and understand everything. This has lead me to question something else: In your proof you use the notation $\left(Z_{p}(bx^{c})\right)^\prime$; My question is why is $\left(Z_{p}(bx^{c})\right)^\prime\ne Z_{p}^{\prime}(bx^{c})$? For example, if I had $f(x)=3x+1$ and $g(x)=2x$ then $f(g(x))'=\left(3(2x)+1\right)'=\left(6x+1\right)^\prime=6$ On the other hand $f^\prime(g(x))=f^\prime(2x)=f^\prime(6x+1)=6$ as before. I am unable to find a counterexample. I know this is very basic. But I cannot seem to understand why $(Z_p(bx^{c}))' \ne Z_{p}'(bx^{c})$? $\endgroup$ – BLAZE Jun 8 '16 at 6:04
  • $\begingroup$ @BLAZE: Many thanks for your very nice comment. Practicing Tai Chi prevents me becoming bigheaded when considering your overwhelming compliments. :-) Many thanks for granting the bounty and accepting my answer. New challenging questions are welcome! But only, if you can't solve it by your own, of couse! :-) $\endgroup$ – Markus Scheuer Jun 10 '16 at 7:12
  • 1
    $\begingroup$ @BLAZE: Joriki has already provided a very nice answer. Two small aspects: Since one of the solutions is $r^{-l-1}$ which has a negative power, the pure power series approach can't be successful. The power Ansatz mentioned by Joriki was my first idea and seems to be more natural: When reading the solutions and since according to the problem text we are free to use any power series method, a simple one is to consider $r^\alpha$ with unknown $\alpha$ and check for possible values of $\alpha$. Regards, $\endgroup$ – Markus Scheuer Aug 9 '16 at 18:55

This answer is too long for a comment and refers to BLAZE's comment above as to how to compute $$\frac{d}{dx}\frac{dY}{dX}$$

Setting the Scene $y(x)=x^{\alpha}Y(X),\quad X=\beta x^{\gamma}$ Now it is obvious that \begin{align} \frac{dy}{dx} &= \alpha x^{\alpha - 1}Y+x^{\alpha}\frac{dY}{dx} \\ &= \alpha x^{\alpha-1}Y+x^{\alpha}\frac{dY}{dX}\frac{dX}{dx} \end{align} We could expand as JJacquelin does but let's leave the calculation here. Now computing the second derivative \begin{align} \frac{d^{2}y}{dx^{2}} &= \frac{d}{dx}\left(\alpha x^{\alpha-1}Y+x^{\alpha}\frac{dY}{dX}\frac{dX}{dx} \right) \\ &= \alpha (\alpha - 1)x^{\alpha-1}Y+\alpha x^{\alpha-1}\frac{dY}{dx}+\alpha x^{\alpha-1}\frac{dY}{dX}\frac{dX}{dx} + x^{\alpha}\frac{d}{dx}\left(\frac{dY}{dX}\frac{dX}{dx}\right) \\ &= \alpha (\alpha - 1)x^{\alpha-1}Y+\alpha x^{\alpha-1}\frac{dY}{dX}\frac{dX}{dx}+\alpha x^{\alpha-1}\frac{dY}{dX}\frac{dX}{dx}+x^{\alpha}\frac{d^{2}Y}{dX^{2}}\left(\frac{dX}{dx}\right)^{2}+x^{\alpha}\frac{dY}{dX}\frac{d^{2}X}{dx^{2}}\\ &=\alpha (\alpha - 1)x^{\alpha-1}Y+2\alpha x^{\alpha-1}\frac{dY}{dX}\frac{dX}{dx}+x^{\alpha}\frac{d^{2}Y}{dX^{2}}\left(\frac{dX}{dx}\right)^{2}+x^{\alpha}\frac{dY}{dX}\frac{d^{2}X}{dx^{2}} \end{align} I hope this clears things up a little for BLAZE.

Apologies to the moderators with regards to this not being a full answer to the question, but I thought the level of computation was too long for a comment.

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  • $\begingroup$ Thanks, that was incredibly helpful; I now understand that the product rule along with the chain rule was needed to compute $\dfrac{d}{dx}\dfrac{dY}{dX}$. I appreciate your time (+1). $\endgroup$ – BLAZE Jun 2 '16 at 12:21
  • $\begingroup$ @BLAZE No problem at all, glad to be of assistance! $\endgroup$ – vectorbundle Jun 2 '16 at 13:23
  • $\begingroup$ @Bacon : Thanks for your contribution. This avoid me to add more intermediate steps in my main answer. $\endgroup$ – JJacquelin Jun 3 '16 at 8:15
  • $\begingroup$ @JJacquelin You're very welcome my friend! I love this type of calculation and DEs involving Bessel functions are a personal joy! $\endgroup$ – vectorbundle Jun 3 '16 at 8:22
  • $\begingroup$ @Bacon : I am always pleased to meet people liking special functions. They are not many todays. The role played in mathematics by the Bessel functions and more generally by special functions is especially of interest for me. Some years ago I wrote a paper for general public on this subject : fr.scribd.com/doc/14623310/… $\endgroup$ – JJacquelin Jun 3 '16 at 8:43

$$\fbox{$y^{\prime\prime}+\left(\frac{1-2a}{x}\right)y^{\prime}+\left[\left(bcx^{c-1}\right)^2+\frac{a^2-p^2c^2}{x^2}\right]y=0$}\tag{5}$$ Let $y(x)=x^\alpha Y(X)\quad$ with $X=\beta x^\gamma$ $\qquad \alpha, \beta , \gamma$ are constants.

$\frac{dX}{dx}=\gamma\beta x^{\gamma-1} \quad;\quad \frac{d^2 X}{dx^2}=\gamma(\gamma-1)\beta x^{\gamma-2}$

$\frac{dy}{dx}=\alpha x^{\alpha-1}Y + x^\alpha \frac{dY}{dX}\frac{dX}{dx} = \alpha x^{\alpha-1}Y + \gamma\beta x^{\alpha +\gamma-1}\frac{dY}{dX}$

$\frac{d^2y}{dx^2}=\alpha(\alpha-1) x^{\alpha-2}Y + 2\alpha x^{\alpha-1}\frac{dY}{dX}\frac{dX}{dx}+x^\alpha \frac{d^2 Y}{dX^2}(\frac{dX}{dx})^2 +x^\alpha \frac{dY}{dX}\frac{d^2 X}{dx^2}$

$\frac{d^2y}{dx^2}=\alpha(\alpha-1) x^{\alpha-2}Y + (2\alpha+\gamma-1) \gamma\beta x^{\alpha+\gamma-2}\frac{dY}{dX} + \gamma^2\beta^2 x^{\alpha+2\gamma-2} \frac{d^2 Y}{dX^2} $

Putting them into (5) :

$$y^{\prime\prime}+\left(\frac{1-2a}{x}\right)y^{\prime}+\left[\left(bcx^{c-1}\right)^2+\frac{a^2-p^2c^2}{x^2}\right]y = 0 =\\ \alpha(\alpha-1) x^{\alpha-2}Y + (2\alpha+\gamma-1) \gamma\beta x^{\alpha+\gamma-2}\frac{dY}{dX} + \gamma^2\beta^2 x^{\alpha+2\gamma-2} \frac{d^2 Y}{dX^2} +(1-2a)\left(\alpha x^{\alpha-2}Y + \gamma\beta x^{\alpha+\gamma-2}\frac{dY}{dX}\right) +\left[\left(bcx^{c-1}\right)^2+\frac{a^2-p^2c^2}{x^2}\right]x^\alpha Y$$

With $X=\beta x^\gamma$ and after simplification : $$ X^2\frac{d^2 Y}{dX^2} + \frac{(2\alpha+\gamma-2a)}{ \gamma} X\frac{dY}{dX} +\left[\frac{ b^2c^2 }{\gamma^2\beta^{2c/\gamma}}X^{2c/\gamma} + \frac{a^2-p^2c^2+\alpha(\alpha-2a)}{\gamma^2} \right]Y =0$$

In order to obtain the standard form of Bessel equation : $X^2Z_p''+XZ_p'+(X^2-p^2)Z_p=0$ , the relationships are : $$\begin{cases} \frac{(2\alpha+\gamma-2a)}{ \gamma}=1 \\ 2c/\gamma=2 \\ \frac{ b^2c^2 }{\gamma^2\beta^{2c/\gamma}}=1\\ \frac{a^2-p^2c^2+\alpha(\alpha-2a)}{\gamma^2}=-p^2 \\ Y(X)=Z_p(X) \end{cases} \quad\to\quad \begin{cases} \alpha=a \\ \beta=b \\ \gamma=c \end{cases} $$ Thus $y(x)=x^a Y(X)=x^a Z_p(X)\quad$ with $X=b x^c$

Hense, the basic solutions of eq.(5) are $y(x)=x^a Z_p(b x^c)$

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  • $\begingroup$ Thanks for your answer sorry for late response. I have read your answer carefully and assuming that $\dfrac{dy}{dx}$ is equal to yellow highlighted expression I find that $\dfrac{d^2y}{dx^2}=\alpha(\alpha-1) x^{\alpha-2}Y + \alpha\beta\gamma x^{\alpha+\gamma-2}\dfrac{dY}{dX}+\beta\gamma x^{\alpha + \gamma - 2} \dfrac{dY}{dX} +\beta\gamma x^{\alpha+\gamma -1} \dfrac{d}{dx}\dfrac{dY}{dX}$. Where I do not understand how to compute $\dfrac{d}{dx}\dfrac{dY}{dX}$. Please let me know if you agree with my expression for $\dfrac{d^2y}{dx^2}$ and how to compute $\dfrac{d}{dx}\dfrac{dY}{dX}$? $\endgroup$ – BLAZE Jun 2 '16 at 8:04
  • $\begingroup$ @BLAZE : I corrected the typing mistake of sign. As you can see, there was no mistake latter which shows that the typing mistake was only in this equation, not in the following ones. Of course, I could add some intermediate steps in the calculus of the second derivative. But Bacon did it meanwhile. This avoid me to spend time. I thank him for that. $\endgroup$ – JJacquelin Jun 3 '16 at 8:10

I will here show how one could go about to derive the solution of your ODE without knowing a priori what the answer should be.

Your ODE $(5)$ can after multiplication by $x^2$ be written


This looks very close to the Bessel ODE so we will try to see if we can bring it closer to that form. The term with $x^c$ on the right-hand side is troublesome so lets try to simplify this bringing it closer to the $x^2$ form we have in the Bessel ODE. This motivates a change of variables $x\to x^c$. Another reason for why this is a natural thing to try is the fact that the derivatives are all on the (logarithmic) form $x^ny^{(n)}(x)$. This means that a change of variables $x\to x^n$ will not change the basic structure of the derivative-terms in the ODE - just the coefficient in front of them.

We therefore take $z = x^c$ and compute

$$x\frac{d}{dx} = x\frac{dz}{dx}\frac{d}{dz} = cz\frac{d}{dz}$$

$$x^2\frac{d^2}{dx^2} = c^2 z^2\frac{d^2}{dz^2} + c(c-1) z \frac{d}{dz}$$

to get that the ODE transforms into

$$z^2 y''(z) + z\left(1-\frac{2a}{c}\right)y'(z) + [(bz)^2 + \frac{a^2}{c^2} - p^2]y(z) = 0$$

where a prime is now a differential with respect to $z$. The $z^2$ term is now close to the desired form and we just need a simple scaling of the variables $w = bz$ to get it right

$$w^2 y''(w) + w\left(1-\frac{2a}{c}\right)y'(w) + [w^2 + \frac{a^2}{c^2} - p^2]y(w) = 0$$

Now this is starting to look more and more like the Bessel ODE, however we still have the troublesome $-\frac{2a}{c}$ term. There is no simple change of variables that can solve this, but a transformation $y(w) = w^n g(w)$ could potentially work as this has the effect of modifying the relative coefficents of the $w^nf^{(n)}(w)$ terms. Another way that could have lead us to try this is the fact that if we did not have the $w^2y(w)$ term in the ODE then $y(w) = w^{\frac{a}{c} \pm p}$ would be a solution to the ODE so it would be natural to try to factor out the power-law growth by trying an ansatz on the form $y(w) = w^n g(w)$.

Substituting $y(w) = w^n g(w)$ into the ODE above leads to

$$w^2g''(w) + \left(1-\frac{2a}{c} + 2n\right)wg'(x) + \left[w^2 + \frac{a^2}{c^2} -2n\frac{a}{c} + n^2 - p^2\right]g(w) = 0$$

We see that the choice $n = \frac{a}{b}$ does the job and brings the term $wg'(w)$ to the desired form (relative to $w^2g''(w)$) and as a bonus the last terms also happens to simplify for this choice of $n$ leaving us with $$w^2g''(w) + wg'(w) + \left[w^2 - p^2\right]g(w) = 0$$

which is exactly the ODE for the Bessel-functions. Backtracking our steps we get the solution in terms of the original variables

$$f(x) = w^nZ_p(w) = (bz)^nZ_p(bz) = b^{\frac{a}{c}} x^{\frac{a}{c}\cdot c} Z_p(bx^c) \propto x^a Z_p(bx^c)$$

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  • $\begingroup$ Thank you for your answer it's helpful to see a method without actually knowing the solution first hand, regards $\endgroup$ – BLAZE Jun 9 '16 at 22:28

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