Repeated rescaling of differential equations

Question

This is a follow-up question to Gauge transformation of differential equations. . As we know by repeatedly transforming the variables of a linear ODE we can generate a whole sequence of new ODE's whose solutions are related to those of the original one.

Now what we have done in here is the following. We started from the Bessel ODE and then changed the abscissa $x\rightarrow {\bf f}(x)$ and after that we reduced the ODE to its normal form (meaning such form where the coefficient at the first derivative is zero) by changing the ordinate appropriately $y \rightarrow {\bf m} \cdot y$. Having done all this we realized that the resulting ODE is rather simple so we repeated the whole process by using a different scaling function $f_1(x)$ and then again reducing the resulting ODE to its normal form by $y \rightarrow m_1 \cdot y$. This leads to the following results.

Let $A$,$B$,$C$,$D$ and $m$ be parameters.

Now define:

\begin{eqnarray} {\bf f}(x) &:=& \frac{A x+B}{C x+D}\\ f_1(x)&:=& x^m \end{eqnarray}

Then the ODE: \begin{eqnarray} \!\!\!\!\!\!\!\!\!\!\!\!y^{''}(x) + \left( m^2(B C-A D)^2\frac{ x^{2 m-2} \left(\left(1-4 a^2\right) \left(C x^m+D\right)^2+4 \left(A x^m+B\right)^2\right)}{4 \left(A x^m+B\right)^2 \left(C x^m+D\right)^4}+\frac{1-m^2}{4 x^2}\right) y(x)=0 \end{eqnarray} is solved by \begin{eqnarray} y(x) = \sqrt{\frac{{\bf f}(f_1(x))}{{\bf f}^{'}(f_1(x))\cdot f_1^{'}(x)}}\cdot \left( C_1 J_a({\bf f}(f_1(x))) + C_2 Y_a({\bf f}(f_1(x))) \right) \end{eqnarray}

Likewise let us define: \begin{eqnarray} {\bf f}(x) &:=& x^m \\ f_1(x)&:=& \frac{A x+B}{C x+D} \end{eqnarray}

Then the ODE: \begin{eqnarray} \!\!\!\!\!\!\!\!\!\!\!\!y^{''}(x) + \left( (B C-A D)^2\cdot \frac{ \left(\left(1-4 a^2 m^2\right) (C x+D)^{2 m}+4 m^2 (A x+B)^{2 m}\right)}{4 (A x+B)^2 (C x+D)^{2 m+2}}\right) y(x)=0 \end{eqnarray} is also solved by \begin{eqnarray} y(x) = \sqrt{\frac{{\bf f}(f_1(x))}{{\bf f}^{'}(f_1(x))\cdot f_1^{'}(x)}}\cdot \left( C_1 J_a({\bf f}(f_1(x))) + C_2 Y_a({\bf f}(f_1(x))) \right) \end{eqnarray}

Here $J_a$ and $Y_a$ are the Bessel functions of the first and the second kind respectively.

In[171]:= Clear[f]; Clear[f1]; A =.; B =.; CC =.; DD =.; m =.; x =.;
f[x_] = (A x + B)/(CC x + DD);
f1[x_] = x^m;
FullSimplify[(D[#, {x, 
       2}] + ((1 - m^2) /(
        4  (x^2) ) + (B CC - A DD)^2 m^2 ( 
         x^(2 m - 
           2) ((1 - 4 a^2) (DD + CC x^m)^2 + 4 (B + A x^m)^2))/(
         4 (B + A x^m)^2 (DD + CC x^m)^4)) #) & /@ {Sqrt[f[f1[x]]]/(
    Sqrt[Derivative[1][f1][x]] Sqrt[
     f'[f1[x]]]) (BesselJ[a, f[f1[x]]] C[1] + 
      BesselY[a, f[f1[x]]] C[2])}]
Clear[f]; Clear[f1]; A =.; B =.; CC =.; DD =.; m =.; x =.;
f[x_] = x^m;
f1[x_] = (A x + B)/(CC x + DD);
FullSimplify[(D[#, {x, 
        2}] + (B CC - A DD)^2  ((1 - 4 m^2 a^2) (DD + CC x)^(2 m) + 
         4 m^2 (B + A x)^(2 m))/(
       4 (B + A x)^2 (DD + CC x)^(2 + 2 m)) #) & /@ {Sqrt[f[f1[x]]]/(
     Sqrt[Derivative[1][f1][x]] Sqrt[
      f'[f1[x]]]) (BesselJ[a, f[f1[x]]] C[1] + 
       BesselY[a, f[f1[x]]] C[2])}] // PowerExpand

Out[174]= {0}

Out[178]= {0}

My question is now are there other examples where a repeated application of variable transformations leads to rather simple looking ODEs? In other words can I always choose the scaling functions $f(x)$ in a clever way so that the resulting ODE is pretty neat rather than some unwieldy expression?

Answer 1

In fact you need to find some "representative special cases".

For example $y''(x)+\left((BC-AD)^2\dfrac{\left(1-4a^2m^2\right)(Cx+D)^{2m}+4m^2(Ax+B)^{2m}}{4(Ax+B)^2(Cx+D)^{2m+2}}\right)y(x)=0$ ,

The "representative special cases" appeared for example when $m=-\dfrac{1}{2}$ , $m=\dfrac{1}{2}$ , $m=1$ , etc.

i.e. at least for example this types of ODEs

$y''(x)+\left((BC-AD)^2\dfrac{\left(1-a^2\right)(Ax+B)+Cx+D}{4(Ax+B)^3(Cx+D)^2}\right)y(x)=0$

$y''(x)+\left((BC-AD)^2\dfrac{\left(1-a^2\right)(Cx+D)+Ax+B}{4(Ax+B)^2(Cx+D)^3}\right)y(x)=0$

$y''(x)+\left((BC-AD)^2\dfrac{\left(1-4a^2\right)(Cx+D)^2+4(Ax+B)^2}{4(Ax+B)^2(Cx+D)^4}\right)y(x)=0$

you find that they are relativity easy to solve.

Some of them are belongs to the ODE of the form $p(x)y''+q(x)y=0$ for odd order polynomials $p(x)$ and polynomials $q(x)$ at least give some interchange points.

Some "representative special cases" for example mentioned in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=268 are quite interesting:

$x^2(ax^n+b)^2y''(x)+(n+1)x(a^2x^{2n}-b^2)y'(x)+cy(x)=0$ , it suggest the change of abscissa $t=\dfrac{1}{nb}\ln\dfrac{ax^n}{ax^n+b}$ .

$2(ax^m+bx^n+c)y''(x)+(amx^{m-1}+bnx^{n-1})y'(x)+dy(x)=0$ , it suggest the change of abscissa $t=\int\dfrac{dx}{\sqrt{ax^m+bx^n+c}}$ .

And you are recommended to find more examples e.g. $y''-\dfrac{(1+x^2)+ax}{b(1+x^2)^3}y=0$ , $a,b\neq0$ in https://math.stackexchange.com/questions/2744228 to practice yourself.