# On reducing complex ODE's to Bessel's form using Kummer's series

I am trying to reduce the following ODE to Bessel's ODE form and solve it: $$x^{2}y''(x)+x(4x^{3}-3)y'(x)+(4x^{8}-5x^{2}+3)y(x)=0\tag{1} \, .$$

I tried to solve it via the standard method, i.e., by comparing it with a generalised ODE form and finding the solution from then on. The general form (as given in Mary L. Boas- Mathematical Methods in Physical Sciences) is:

$$y''(x)+\frac{1-2a}{x}y'(x)+\left((bcx^{c-1})^{2}+\frac{a^{2}-p^{2}c^{2}}{x^{2}}\right)y(x)=0\tag{2} \, ,$$ and the solution:$$y(x)=x^{a}Z_{p}(bx^{c})\tag{3} \, .$$

But I am unable to get the answer via this method. The solution which is as follows: $$y(x)=x^{2}e^{-\frac{x^{4}}{2}}[AI_{1}(\sqrt{5}x)+BK_{1}(\sqrt{5}x)]\tag{4}$$ Is obtained using comparison with another standard form which is given as follows:

$$x^{2}y''(x)+x(a+2bx^{p})y'(x)+[c+dx^{2q}+b(a+p-1)x^{p}+b^{2}x^{2p})y(x)=0\tag{5} \, ,$$ and the solution as: $$y(x)=x^{\alpha}e^{-\beta x^{p}}[AJ_{\nu}(\lambda x^{q})+BY_{\nu}(\lambda x^{q})]\tag{6} \, .$$

Where: $\alpha=\frac{1-a}{2}$, $\beta=\frac{b}{p}$, $\lambda=\frac{\sqrt{d}}{q}$, $\nu=\frac{\sqrt{(1-a)^{2}-4c)}}{2q}$

If I divide through the ode by $x^{2}$, I would get the Fuchasian form: $$y''(x)+f(x)y'(x)+g(x)y(x)=0$$

The terms of $xf(x)$ and $x^{2}g(x)$ are expandable in convergent power series $\sum_{n=0}^{\infty}a_{n}x^{n}$, hence there exists a nonessential singularity at the origin. But I am unable to solve via the Frobenius method.

Hence, my question- How is the generalised form of equation $(5)$ arrived at and why can't I use $(2)$ instead? Rather than bringing this ODE to a non-standard form as given in equation $(5)$, is there a way to derive the equation itself (and deduce the general solution)? Any help is appreciated.

Edit: I found the following form in a book:

$$x^{2}y''(x)+x(a+2bx^{p})y'(x)+[c+dx^{2q}+fx^{q}+b(a+p-1)x^{p}+b^{2}x^{2p})y(x)=0\tag{7} \, .$$

The only difference between the above and equation $(6)$ is the extra term:$fx^{q}$ Now if I substitute $y=we^{-\frac{bx^{p}}{p}}$ in equation $(7)$, it simplifies to the following linear equation:

$$x^{2}w''(x)+axw'(x)+(dx^{2q}+fx^{q}+c)w(x)=0\tag{8}\,.$$

Now using the transformation $z=x^{q}$, and $y=wz^{k}$, where $k$ is the root of the following quadratic equation: $q^{2}k^{2}+q(a-1)k+c=0$; leads to a further simplified and linear form:

$$q^{2}zy''(z)+[qbz+2kq^{2}+q(q-1+a)]y'(z)+(dz+kqb+f)y(z)=0\tag{9}\,.$$ This equation has the solution: $y(x)=e^{kx}w(z)$, where $w(z)$ is the solution to the hypergeometric equation as given below

Now, let a function $\Omega(b,a;x)$ be an arbitrary solution to the degenerate hypergeometric equation:

$$xy''(x)+(a-x)y'(x)-by(x)=0\tag{10}\,.$$

And $Z_{\nu}(x)$ be an arbitrary solution of the Bessel equation. Now in equation $(10)$, if $b\neq0,-1,-2,-3,...$, the solution is given by the Kummer's series as: $$\Phi(b,a;x)=1+\sum_{k=1}^{\infty}\frac{(b)_{k}x^{k}}{(a)_{k}k!}$$ Where:$(b)_{k}=b(b+1)...(b+k-1)$ When $a$ is not an integer, the solution can be written as: $$y=C_{1}\Phi(b,a;x)+C_{2}x^{1-a}\Phi(b-a+1,2-a;x)$$ Make the following replacements: $b=2n$ and $a=n$ Now the series becomes:

$$\Phi(n,2n;x)=\Gamma\left(n+\frac{1}{2}\right)e^{\frac{x}{2}}\left(\frac{x}{4}\right)^{(-n+\frac{1}{2})}I_{n-\frac{1}{2}}(\frac{x}{2})$$ And $$\Phi(-n,-2n;x)=\frac{1}{\sqrt{\pi}}e^{\frac{x}{2}}\left(x\right)^{(-n+\frac{1}{2})}K_{n-\frac{1}{2}}(x)$$

Substituting the above in the solution of equation $(9)$, the general solution becomes: $$y=e^{x(k+\frac{1}{2})}\left[C_{1}\Gamma\left(n+\frac{1}{2}\right)\left(\frac{x}{4}\right)^{(-n+\frac{1}{2})}I_{n+\frac{1}{2}}(x)+C_{2}\frac{1}{\sqrt{\pi}}\left(x\right)^{(-n+\frac{1}{2})}K_{n+\frac{1}{2}}(x)\right]$$ Which simplifies to:

$$y(x)=\left(x\right)^{(-n+\frac{1}{2})}e^{x(k+\frac{1}{2})}\left[C_{1}\Gamma\left(n+\frac{1}{2}\right)\left(\frac{1}{4}\right)^{(-n+\frac{1}{2})}I_{n+\frac{1}{2}}(x)+C_{2}\frac{1}{\sqrt{\pi}}K_{n+\frac{1}{2}}(x)\right]$$

Which is the final solution. I tried to do the same for equation $(6)$, but did not get the solution. Any help is appreciated.

• I don't understand how you expect that (1) is in the form of (2). If you divide both sides by $x^2$ (to get $y''(x)$), you'll end up with a $4xy'$ term, which is not in the form of (2). – Alex R. Apr 25 '17 at 17:41
• @AlexR. that is my question, this ode does not comply with the standard form, I could divide with the factor that you have mentioned and go on to simplify by completing the squares, but I would end up with 'x' terms in my constants a, b,c and p. Hence I ask if there is a way to bring the ode to form 2 ( any trick) and how does one derive the generalised form as given in 5 – Spoilt Milk Apr 25 '17 at 17:46
• The coefficients of $y'(x),y(x)$ in your ODE are linear combinations of $1,x^3$ and $1,x^2,x^8$ respectively. By contrast, the coefficients of $y'(x)$ and $y(x)$ in (7) are linear combinations of $1,x^p$ and $1,x^p,x^q,x^{2p},x^{2q}$ respectively. But there's no combination of $p,q$ that will make these two match, so your ODE is not of the form (7) and any conclusions based on that are dubious; in particular, one can't conclude that (4) solves (1). – Semiclassical Apr 28 '17 at 12:59
• @Semiclassical But that's the answer given in the book. My prof. too says that is the right answer. – Spoilt Milk Apr 28 '17 at 13:34

Since for the step that pointly eliminate the $y$ term of coefficient $x^8$ respect to the $y''$ term of coefficient $x^2$ ,

Let $y=e^{mx^4}u$ ,

Then $y'=e^{mx^4}u'+4mx^3e^{mx^4}u$

$y''=e^{mx^4}u''+4mx^3e^{mx^4}u'+4mx^3e^{mx^4}u'+(16m^2x^6+12mx^2)e^{mx^4}u=e^{mx^4}u''+8mx^3e^{mx^4}u'+(16m^2x^6+12mx^2)e^{mx^4}u$

$\therefore x^2(e^{mx^4}u''+8mx^3e^{mx^4}u'+(16m^2x^6+12mx^2)e^{mx^4}u)+x(4x^3-3)(e^{mx^4}u'+4mx^3e^{mx^4}u)+(4x^8-5x^2+3)e^{mx^4}u=0$

$x^2(u''+8mx^3u'+(16m^2x^6+12mx^2)u)+x(4x^3-3)(u'+4mx^3u)+(4x^8-5x^2+3)u=0$

$x^2u''+8mx^5u'+(16m^2x^8+12mx^4)u+x(4x^3-3)u'+(16mx^7-12mx^4)u+(4x^8-5x^2+3)u=0$

$x^2u''+x(8mx^4+4x^3-3)u'+((16m^2+4)x^8+16mx^7-5x^2+3)u=0$

Choose $m=\pm\dfrac{i}{2}$ , the ODE becomes $x^2u''+x(\pm4ix^4+4x^3-3)u'+(\pm8ix^7-5x^2+3)u=0$

The more tedious $u$ term of coefficient $x^7$ also come out.

Also, checking in WolframAlpha, it does not take any comments about its general solution, does it reasonable for just the simple linking with Bessel equation?

I admit that if the ODE is $x^2y''(x)+x(4x^4-3)y'(x)+(4x^8-5x^2+3)y(x)=0$ instead the situation will becomes much more simpler.

Since for the step that pointly eliminate the $y$ term of coefficient $x^8$ respect to the $y''$ term of coefficient $x^2$ ,

Let $y=e^{mx^4}u$ ,

Then $y'=e^{mx^4}u'+4mx^3e^{mx^4}u$

$y''=e^{mx^4}u''+4mx^3e^{mx^4}u'+4mx^3e^{mx^4}u'+(16m^2x^6+12mx^2)e^{mx^4}u=e^{mx^4}u''+8mx^3e^{mx^4}u'+(16m^2x^6+12mx^2)e^{mx^4}u$

$\therefore x^2(e^{mx^4}u''+8mx^3e^{mx^4}u'+(16m^2x^6+12mx^2)e^{mx^4}u)+x(4x^4-3)(e^{mx^4}u'+4mx^3e^{mx^4}u)+(4x^8-5x^2+3)e^{mx^4}u=0$

$x^2(u''+8mx^3u'+(16m^2x^6+12mx^2)u)+x(4x^4-3)(u'+4mx^3u)+(4x^8-5x^2+3)u=0$

$x^2u''+8mx^5u'+(16m^2x^8+12mx^4)u+x(4x^4-3)u'+(16mx^8-12mx^4)u+(4x^8-5x^2+3)u=0$

$x^2u''+x((8m+4)x^4-3)u'+((16m(m+1)+4)x^8-5x^2+3)u=0$

Choose $m=-\dfrac{1}{2}$ , the ODE becomes $x^2u''-3xu'-(5x^2-3)u=0$

Which is very lucky that the $y'$ term of coefficient $x^5$ can also eliminate.

Now for the step that adjust the $u'$ term of coefficient $x$ ,

Let $u=x^nv$ ,

Then $u'=x^nv'+nx^{n-1}v$

$u''=x^nv''+nx^{n-1}v'+nx^{n-1}v'+n(n-1)x^{n-2}v=x^nv''+2nx^{n-1}v'+n(n-1)x^{n-2}v$

$\therefore x^2(x^nv''+2nx^{n-1}v'+n(n-1)x^{n-2}v)-3x(x^nv'+nx^{n-1}v)-(5x^2-3)x^nv=0$

$x^{n+2}v''+2nx^{n+1}v'+n(n-1)x^nv-3x^{n+1}v'-3nx^nv-(5x^2-3)x^nv=0$

$x^{n+2}v''+(2n-3)x^{n+1}v'-(5x^2-n(n-4)-3)x^nv=0$

$x^2v''+(2n-3)xv'-(5x^2-n(n-4)-3)v=0$

Choose $n=2$ , the ODE becomes $x^2v''+xv'-(5x^2+1)v=0$

Which relates to Bessel equation.

• Thanks for answering. The substitution works for a standard Bessel type ode. Rather than bringing this ode to non-standard type as given in equation (5), is there a way to derive the equation itself (and deduce the general solution)? – Spoilt Milk Apr 26 '17 at 5:53
• Mathematica also fails to find a closed-form solution of the given ODE. So it seems dubious that there is a nice solution. – Semiclassical Apr 27 '17 at 6:11
• @Semiclassical Please see edit – Spoilt Milk Apr 28 '17 at 9:36
• @doraemonpaul Thank you for that fantastic answer! Do you have any suggestions for ODE and PDE handbook that encompasses all transformations and solutions? Apart from Polyanin. – Spoilt Milk Apr 29 '17 at 5:38

Working in Mathematica, I find that (3) is the general solution to the differential equation $$x^{2}y''(x)+x(4x^{4}-3)y'(x)+(4x^{8}-5x^{2}+3)y(x)=0.$$ This almost entirely matches (1), but differs in that the linear term contains $x^4$ rather than $x^3$. So you should check the source of the problem to see if (1) was transcribed improperly: If it was, then that fixes things; if the book indeed had $x^3$, though, then it's a typo and the prof should be informed.

To determine the ODE which the general solution satisfies, I defined

y1[x_] := x^2 Exp[-x^4/2] BesselI[1, x Sqrt[5]]

as one of the two linearly-independent solutions to the ODE. Assuming that the relevant desired ODE is of the form $$x^2y''(x)+x A(x) y'(x)+B(x)y(x)=0$$ for some appropriate $A,B$, I plugged the above solution into the left-hand side and simplify:

x^2 y1''[x] + x A[x] y1'[x] + B[x] y1[x] // FullSimplify

==> E^(-(x^4/2)) x^2 (Sqrt[5]x (3 - 4 x^4 + A[x]) BesselI[0,Sqrt[5] x]
+ (5 x^2 - 10 x^4 + 4 x^8 + A[x] - 2 x^4 A[x] + B[x]) BesselI[1, Sqrt[5] x])


From this we can read off that both terms will vanish identically when \begin{align} A(x)&=4x^4-3\\\\ B(x)&=(2 x^4 - 1) A(x) + 10 x^4 - 4 x^8 - 5 x^2\\&=4x^8 - 5 x^2 + 3 \end{align} which is the result cited above. This can be further checked by asking Mathematica to solve the ODE directly:

 DSolve[x^2 y''[x]+x (4x^4-3)y'[x]+(3-5 x^2+4 x^8) y[x]==0,y[x],x]

==> {{y[x]->E^(-(x^4/2)) x^2 (-I BesselI[1,Sqrt[5] x] C[1]
+BesselY[1,-I Sqrt[5] x] C[2])}}
`

Trading Bessel functions $J,Y$ for modified Bessel functions $I,K$ then gives the indicated general solution up to integration constants.

• could you please edit your answer by including the mathematical code you have used to obtain the solution. I have tried with Wolfram Mathematica, but got a super crazy answer. Btw did you obtain the solution expressed in terms of Bessel functions or was it expressed in terms of a sum? – Spoilt Milk Apr 28 '17 at 15:25
• Sure, I can do that in a few hours. But in the meantime, I'll point out that WolframAlpha is capable of solving the ODE as I've stated it (link). – Semiclassical Apr 28 '17 at 16:12
• Thanks a lot for that! One last thing though-equation $(3)$ actually does not satisfy the general solution as there is an exponential term missing. I got the generalized equation $(7)$ from Andrei Polyanin (Exact Solutions to ODE's) [page 252, equation 147-leads to equation 146-leads to equation 108- leads to equation 70 for solution]. So, although I do know I have got the right answer and the question is right (just checked, no typos), finally I did not succeed at obtaining equation $(6)$ and it's solution via substitution and simplifications. Any closing remarks on this note? – Spoilt Milk Apr 28 '17 at 18:22
• How does the Wolfram Mathematica's result super crazy for that for $x^2y''+x(4x^3-3)y'+(4x^8-5x^2+3)y=0$ ? Note that for e.g. $x^2y''+x(4x^4+n)y'+(4x^8-5x^2+3)y=0$ is also still have luck to solve, provide that we should consider more types of transformations. – doraemonpaul Apr 30 '17 at 22:37