There is a problem in Quantum Mechanics where this differential equation arises:
$( l (l + 1) + b x^2 \left(\frac{x}{-1+x}\right)^{7/4} - 2 b x^2 ) y - 2 x y' + x (-1 + x) y'' = 0$
with the border conditions:
$\lim_{x\rightarrow 1}y[x]=0$
and
$\lim_{x\rightarrow \infty}y[x]=0$
and we can set $y[x]=0$ for all $x<1$ (that means, for $x<1$ a solution other than the trivial solution $y[x]=0$ makes no physical sense)
The term $l(l+1)$ corresponds to an angular momentum with $l=\dots -3,-2,-1,0,1,2,3\dots$
And $b$ is an energy related Eigenvalue, perhaps depending on $l$ and additional integer numbers (quantum numbers).
This differential equation has an irregular singularity at $x=1.$ That means, there is no standard method, for example like the Frobenius method, to solve this equation, because the solution will provide terms, which can not be developed in a Taylor series at x=1. A term with this property would be for example $e^{\frac{-1}{x-1}}$.
The problem of this differential equation for me is, that I could not find any transformation to a known and solved differential equation. The numerical solution shows a shape of the function as expected. The ground state will provide a function starting from $y[1]=0$ (boundary condition) rising to a maximum and will finally fall down to $0$ (limit to Infinity). The first excited state will show a similar behaviour at x=1 and infinity, but the function will cross the x-axis. The second excited state will cross the x-axis two times and so on.
This differential equation describes matter in a new kind of understanding. I am convinced, that this problem will be solved by intuition. Mathematica and Wolfram Alpha failed. Solving this differential equation means, to find the right trick. For this I give it a try in this forum. Perhaps there is someone who is a lucky person and is able "to see" the right way.
I did not see any option to upload a file. I prepared a "Notebook" file to be used in Mathematica containing some improvements to this problem. How do I provide a file to the community?
Regards
Robert
I would like to provide some improvements:
The problem of the differential equation obviously is the term
$\left(\frac{x}{x-1}\right)^{7/4}$
I made some investigations and I could find, that if I put the following term
$\text{a1} \sqrt[4]{\frac{x}{x-1}}+\text{a2} \sqrt{\frac{x}{x-1}}+\text{a3} \left(\frac{x}{x-1}\right)^{3/4}$
into the differential equation, I will get
$\frac{\sqrt[4]{\frac{x}{x-1}} \left(16 \text{a2} b x^5-\text{a1} (x-1) \left(32 b x^4-32 b x^3+16 \left(l^2+l-1\right) x-16 l (l+1) x^2+3\right)\right)}{16 (x-1)^2 x}+\frac{\text{a1} b x^4}{(x-1)^2}+\frac{\sqrt{\frac{x}{x-1}} \left(4 \text{a3} b x^5-\text{a2} (x-1) \left(8 b x^4-8 b x^3+4 \left(l^2+l-2\right) x-4 l (l+1) x^2+1\right)\right)}{4 (x-1)^2 x}+\frac{\text{a3} \left(\frac{x}{x-1}\right)^{3/4} \left(-32 b x^4+32 b x^3-16 \left(l^2+l-3\right) x+16 l (l+1) x^2-3\right)}{16 (x-1) x}=0$
and we can see the repeating pattern of the terms with the rational exponents
But there is still the irregular singularity at x=1. The following function shows a behaviour, that the function will be zero at x=1:
$\frac{e^{-\frac{1}{x-1}}}{x-1}$
plot of the above mentioned function
With the combination of the exponential function term with the sum of terms with rational exponents, it is possible to make the Ansatz
$e^{\frac{\text{c1}}{x-1}} \left(\text{a1} \sqrt[4]{\frac{x}{x-1}}+\text{a2} \sqrt{\frac{x}{x-1}}+\text{a3} \left(\frac{x}{x-1}\right)^{3/4}\right)$
The result does not show the exponential function, because it was canceled
$\frac{\sqrt[4]{\frac{x}{x-1}} \left(\text{a1} \left(16 x^3 \left(2 b+4 \text{c1}-3 l^2-3 l+1\right)+16 x^4 \left(-6 b+l^2+l\right)-32 b x^6+96 b x^5+x^2 \left(16 \text{c1}^2-56 \text{c1}+48 l^2+48 l-35\right)-2 x \left(4 \text{c1}+8 l^2+8 l-11\right)-3\right)+16 \text{a2} b (x-1) x^5\right)}{16 (x-1)^3 x}+\frac{\text{a1} b x^4}{(x-1)^2}+\frac{\sqrt{\frac{x}{x-1}} \left(\text{a2} \left(4 x^3 \left(2 b+4 \text{c1}-3 l^2-3 l+2\right)+4 x^4 \left(-6 b+l^2+l\right)-8 b x^6+24 b x^5+x^2 \left(4 \text{c1}^2-12 \text{c1}+12 l^2+12 l-17\right)-2 x \left(2 \text{c1}+2 l^2+2 l-5\right)-1\right)+4 \text{a3} b (x-1) x^5\right)}{4 (x-1)^3 x}+\frac{\text{a3} \left(\frac{x}{x-1}\right)^{3/4} \left(16 x^3 \left(2 b+4 \text{c1}-3 \left(l^2+l-1\right)\right)+16 x^4 \left(-6 b+l^2+l\right)-32 b x^6+96 b x^5+x^2 \left(16 \text{c1}^2-40 \text{c1}+48 l^2+48 l-99\right)-2 x \left(12 \text{c1}+8 l^2+8 l-27\right)-3\right)}{16 (x-1)^3 x}=0$
The result shows, that the whole equation is collected in the above mentioned terms with rational exponents. To solve the differential equation, the Ansatz should contain additional arguments to make the coefficients of the terms with rational exponents zero. For this it is possible to extend the Ansatz as follows
$e^{\frac{\text{c1}}{x-1}} \left(\sqrt[4]{\frac{x}{x-1}} \text{a1}(x)+\sqrt{\frac{x}{x-1}} \text{a2}(x)+\left(\frac{x}{x-1}\right)^{3/4} \text{a3}(x)+\frac{x \text{a4}(x)}{x-1}\right)$
The solution of this Ansatz provides a set of 4 coupled ordinary differential equations of second order. It is possible to write it down as a matrix equation:
R a'' + P a' + Q == 0
with a as a vector
$\left( \begin{array}{c} \text{a1}(x) \\ \text{a2}(x) \\ \text{a3}(x) \\ \text{a4}(x) \\ \end{array} \right)$
Matrix R
$\left( \begin{array}{cccc} (x-1) x & 0 & 0 & 0 \\ 0 & (x-1) x & 0 & 0 \\ 0 & 0 & (x-1) x & 0 \\ 0 & 0 & 0 & x^2 \\ \end{array} \right)$
Matrix P
$\left( \begin{array}{cccc} -\frac{4 x^2+4 \text{c1} x-3 x-1}{2 (x-1)} & 0 & 0 & 0 \\ 0 & -\frac{2 x^2+2 \text{c1} x-x-1}{x-1} & 0 & 0 \\ 0 & 0 & -\frac{4 x^2+4 \text{c1} x-x-3}{2 (x-1)} & 0 \\ 0 & 0 & 0 & -\frac{2 x \left(x^2+\text{c1} x-1\right)}{(x-1)^2} \\ \end{array} \right)$
Matrix Q
$\left( \begin{array}{cccc} -\frac{32 b x^6-96 b x^5-16 l^2 x^4+96 b x^4-16 l x^4+48 l^2 x^3-32 b x^3-64 \text{c1} x^3+48 l x^3-16 x^3-16 \text{c1}^2 x^2-48 l^2 x^2+56 \text{c1} x^2-48 l x^2+35 x^2+16 l^2 x+8 \text{c1} x+16 l x-22 x+3}{16 (x-1)^3 x} & \frac{b x^4}{(x-1)^2} & 0 & 0 \\ 0 & -\frac{8 b x^6-24 b x^5-4 l^2 x^4+24 b x^4-4 l x^4+12 l^2 x^3-8 b x^3-16 \text{c1} x^3+12 l x^3-8 x^3-4 \text{c1}^2 x^2-12 l^2 x^2+12 \text{c1} x^2-12 l x^2+17 x^2+4 l^2 x+4 \text{c1} x+4 l x-10 x+1}{4 (x-1)^3 x} & \frac{b x^4}{(x-1)^2} & 0 \\ 0 & 0 & -\frac{32 b x^6-96 b x^5-16 l^2 x^4+96 b x^4-16 l x^4+48 l^2 x^3-32 b x^3-64 \text{c1} x^3+48 l x^3-48 x^3-16 \text{c1}^2 x^2-48 l^2 x^2+40 \text{c1} x^2-48 l x^2+99 x^2+16 l^2 x+24 \text{c1} x+16 l x-54 x+3}{16 (x-1)^3 x} & \frac{b x^4}{(x-1)^2} \\ \frac{b x^4}{(x-1)^2} & 0 & 0 & -\frac{x \left(2 b x^5-6 b x^4-l^2 x^3+6 b x^3-l x^3+3 l^2 x^2-2 b x^2-4 \text{c1} x^2+3 l x^2-4 x^2-\text{c1}^2 x-3 l^2 x+2 \text{c1} x-3 l x+8 x+l^2+2 \text{c1}+l-4\right)}{(x-1)^4} \\ \end{array} \right)$
To remove all denominators from the matrix equation, we multiply with x(-1+x)^4
Then all components of matrix Q are polynomes of grade 7. Matrix Q has a digonal and a hyper diagonal.
All components of the diagonal matrix P are polynomes of grade 6
All components of the diagonal matrix R are polynomes of grade 7
If the a-vector consists of polynomes of finite grade, the Q matrix shows a term with x^7, which can not be removed by terms from P a' or R a''. For this, the x^7 term has to be removed by the substraction of the highest terms of the a-vector, which is possible, because the Q matrix has a diagonal and a hyper diagonal. But this will only work, if all components of the a-vector will have the same grade.
For the a-vector we can make the following Ansatz:
$\left\{\sum _{i=1}^n d_{1i} x^{i+k-1},\sum _{i=1}^n d_{2i} x^{i+k-1},\sum _{i=1}^n d_{3i} x^{i+k-1},\sum _{i=1}^n d_{4i} x^{i+k-1}\right\}$
For n=1 we get only the trivial solution (all $d_{i1}=0$). For n=2,3,4 also b=0 and c1=0 and for n>4 no solution.
That seems to mean, that I need a different Ansatz for the a-vector
$\left( \begin{array}{c} \text{a1}(x) \\ \text{a2}(x) \\ \text{a3}(x) \\ \text{a4}(x) \\ \end{array} \right)$