How was this $y\left(y’’+\frac{1-p}vy’\right)-(y+1-p)y’^2=0$ power series recurrence derived?

Question

In György Steinbrecher’s and William Shaw’s Quantile mechanics $(47)$ to $(51)$, it can be found that:

$$\begin{aligned}y\left(y’’+\frac{1-p}vy’\right)-(y+1-p)y’^2=0,y(0)=0,y (0)=1\iff (1-p)y’^2=\frac{1-p}vyy’+yy’’-yy’^2\\\implies y(v)=\sum_{n=1}^\infty a_n v^n,n(n+p)a_n=\sum_{k=1}^n\sum_{j=1}^{n-k+1}a_ka_ja_{n-k-j+2}j(n-k-j+2)-\Delta(n)\sum_{k=2}^n a_ka_{n-k+2}k(k-p+(p-1)(n-k+2)),a_1=1,\Delta(n)=\begin{cases}0,&n<2\\1,&n\ge2\end{cases}\end{aligned}$$

Appendix A $(93)$ to $(97)$ gives steps for finding the non linear recurrence. First, one substitutes in the power series:

$$\begin{aligned}y=\sum_{n=1}^\infty a_n x^n:(1-p)\sum_{n=1}^\infty a_n n v^{n-1}\sum_{k=1}^\infty a_k k v^{k-1}=\frac{1-p}v\sum_{s=1}^\infty a_sv^s\sum_{q=1}^\infty a_q q v^{q-1}+\sum_{s=1}^\infty a_s v^s\sum_{q=1}^\infty a_qq(q-1)v^{q-2}-\sum_{s=1}^\infty a_sv^s\sum_{q=1}^\infty a_q q v^{q-1}\sum_{r=1}^\infty a_rrv^{r-1}\\\implies \sum_{n=1}^\infty\sum_{k=1}^\infty (1-p)a_na_knkv^{n+k-2}=\sum_{s=1}^\infty\sum_{q=1}^\infty a_sa_qq(q-p)v^{s+q-2}-\sum_{s=1}^\infty\sum_{q=1}^\infty\sum_{r=1}^\infty a_sa_qa_rqrv^{s+q+r-2}\end{aligned}$$

and equates $v$’s powers using Kronecker delta.

$$\mathop\implies^?\sum_{k=1}^\infty (1-p)a_na_knk=\sum_{s=1}^\infty\sum_{q=1}^\infty a_sa_qq(q-p)\delta_{n+k-2,s+q-2}-\sum_{s=1}^\infty\sum_{q=1}^\infty\sum_{r=1}^\infty a_sa_qa_rqr\delta_{n+k-2,s+q+r-2} $$

However, it is not clear how to do this, so we stop. The $(1-p)y’^2$, a product of $2$ functions, makes one equate a double sum’s terms. The final result will have $a_n$ in terms of a single and a double sum implying that both $\sum_\limits{n=1}^\infty\sum\limits_{k=1}^\infty$ disappear. Maybe a modified differential equation was used.

How could the blockquoted recurrence relation have been derived?

Answer 1

Suppose given a power series

$$ y(v) := \sum_{n=1}^\infty a_n v^n. \tag1 $$

Expand in power series

$$ t := s - u = \sum_{n=1}^\infty b_n v^n,\;\; s: =-(1-p)y’^2+\frac{1-p}vyy’+yy’’,\;\;u: = yy’^2. \tag2 $$

This expansion leads to

$$ c_n := \sum_{k=2}^{n} c_{n,k},\;\; c_{n,k} := a_k a_{n-k+2}\,k(k-p+(p-1)(n-k+2)),\\ e_n := c_{n,1} + c_{n,n+1},\;\; d_n := \sum_{k=1}^n\sum_{j=1}^{n-k+1} a_ka_ja_{n-k-j+2}\,j(n-k-j+2). \tag3 $$

For example, $\,vy' = \sum_{n=0}^\infty na_nv^n\,$ and thus

$$ v^2u = v^2yy'y' = \sum_{k=1}^\infty \sum_{j=1}^\infty\sum_{m=1}^\infty a_k\,ja_j\,ma_m v^{k+j+m} = \sum_{n=1}^\infty d_{n+2}v^{n+2} \tag4 $$

where $\,n+2 = k+j+m\,$ or $\,m = n-k-j+2.\,$ Divide both sides of equation $(4)$ by $\,v^2\,$ to get the triple sum in equation $(3)$.

Then

$$ b_n = e_n + c_n - d_n,\quad e_n = n(p+n)a_1a_{n+1}. \tag5 $$

Note that equation $(3)$ implies

$$ s = \sum_{n=1}^\infty (e_n + c_n) v^n = \sum_{n=1}^\infty \left( \sum_{k=1}^{n+1} c_{n,k}\right) v^n. \tag6 $$

Note that $\,c_n-d_n\,$ is a polynomial that only involves $a_1,\dots,a_n.$ Thus, given the values of $\,b_1,b_2,\dots\,$ and $\,a_1\,$ then the rest of the values of $\,a_n\,$ can be computed.