4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – Alex R. Apr 25 '17 at 17:41
  • $\begingroup$ @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 $\endgroup$ – Spoilt Milk Apr 25 '17 at 17:46
  • 1
    $\begingroup$ 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). $\endgroup$ – Semiclassical Apr 28 '17 at 12:59
  • $\begingroup$ @Semiclassical But that's the answer given in the book. My prof. too says that is the right answer. $\endgroup$ – Spoilt Milk Apr 28 '17 at 13:34
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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)? $\endgroup$ – Spoilt Milk Apr 26 '17 at 5:53
  • $\begingroup$ Mathematica also fails to find a closed-form solution of the given ODE. So it seems dubious that there is a nice solution. $\endgroup$ – Semiclassical Apr 27 '17 at 6:11
  • $\begingroup$ @Semiclassical Please see edit $\endgroup$ – Spoilt Milk Apr 28 '17 at 9:36
  • $\begingroup$ @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. $\endgroup$ – Spoilt Milk Apr 29 '17 at 5:38
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Spoilt Milk Apr 28 '17 at 15:25
  • 1
    $\begingroup$ 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). $\endgroup$ – Semiclassical Apr 28 '17 at 16:12
  • $\begingroup$ 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? $\endgroup$ – Spoilt Milk Apr 28 '17 at 18:22
  • $\begingroup$ 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. $\endgroup$ – doraemonpaul Apr 30 '17 at 22:37

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.