# Endless Transformation in finding Particular Solution of Riccati Equation

I have a question regarding "Endless Transformation. I'm actually working on Riccati Equation and I found a Post here on MO (Looking for the solution of first order non-linear differential equation ($y ′+y^{2}=f(x)$) without knowing a particular solution.), wherein the equation was solved using Endless transformation. I'm really curious about that method but I can't find any references wherein Endless Transform was elaborated.

Is "Endless Transform" really an existing Method? I can't search it on Google. Does it have another name? Who established the Method?

I really need your help. Thanks!

• My guess is that "endless" is an idiosyncratic translation. (One possibility is 'unendlich' = infinite was translated as 'endless'.) Looking at what was written, and allowing for the fact this is not an area of research for me, it looks like the OP means to find a fixed point $f$ of a differential or integral operator by means of a convergent infinite sequence in a function space, or something like that. Googling for "endless" something or other is almost certainly hopeless. Why do you really need to know? – Todd Trimble Jul 5 '16 at 14:10
• Specifically, as @ToddTrimble mentions, the idea of an "endless integral transformation" seems to be just the classical proof of Picard–Lindelöf. I don't know anything about the convergence of the analogous "endless derivative transformation", but I suspect that there are problems (since differentiating makes things worse in general). – LSpice Jul 5 '16 at 19:18
• can someone help me apply it on this questn y'=2x(1-y^2) , y(0)=0 – user98309 Sep 11 '16 at 10:24
• @stephen Apply $y=1/2x(u+1/2x)$ on your equation and you will get $u'+u^2=4x^2+3/4x^2$ and so you got $f(x)=4x^2+3/4x^2$. Then follow the method in answer to find u now. – Mathlover Sep 17 '16 at 10:17

It is too long for a comment , Thus I wrote an answer

I am the person who wrote the transform that you mentioned in your question. I did not see anywhere the term of endless transform, I just noticed if we apply the variable transform $$y=\frac{f(x)}{\frac{f'(x)}{2f(x)}+y_1}$$ on Riccati Differential Equation $y'+y^{2}=f(x)$,
We get same type Riccati Differantial Equation after the transform. $$y_1 '+y_1^{2}=f_1(x)=f(x)+(\frac{-f'(x)}{2f(x)})^{2}+(\frac{-f'(x)}{2f(x)})'$$

Then, if we apply the transform $$y_1=\frac{f_1(x)}{\frac{f_1'(x)}{2f_1(x)}+y_2}$$ on $y_1 '+y_1^{2}=f_1(x)$ , we get the same type equation again $$y_2 '+y_2^{2}=f_2(x)=f_1(x)+(\frac{-f_1'(x)}{2f_1(x)})^{2}+(\frac{-f_1'(x)}{2f_1(x)})'$$ It never ends , we always get the same type equation ($y'+y^{2}=f(x)$) , Thus I wrote it as endless transform.

Finally, the particular solution can be written in infinite terms as

$$y_p(x)=\frac{f(x)}{\frac{f'(x)}{2f(x)}+\frac{f_1(x)}{\frac{f_1'(x)}{2f_1(x)}+\frac{f_2(x)}{\frac{f_2'(x)}{2f_2(x)}+.....}} }$$

Where $$f_{n+1}(x)=f_n(x)+(\frac{-f_n'(x)}{2f_n(x)})^{2}+(\frac{-f_n'(x)}{2f_n(x)})'$$ and $$f_0(x)=f(x)$$

I do not know the name of this transform in literature, So I named it as endless transform. Maybe someone else can help to find what the name of the transform is in literature.

I hope it helps to enlighten you

• @J.M. You are welcome I am happy if I could help you..I have been working on the Riccati differential equation for many years.It is very important equation because the change of variable $y=\psi'/\psi$ gives as Schrodinger equation $$\psi''=f(x)\psi,$$ with general potential $f(x)$. It is real chestnut to find the exact solution without particular solution. I am very interested in what your research is about. Please update me about your work and progress details. I would like to be happy to hear about it. Best Regards – Mathlover Jul 13 '16 at 11:23