I am working on a PDE problem. The goal is to connect the higher order energies of the Gross-Pitaevskii equation to those of the Korteweg-de-Vries equation. As these higher order energies are recursively defined polynomials, combinatorics come into play. I have determined a key combinatorial problem, a certain cancellation of recursively defined polynomials, and I know all the right coefficients to cancel them out in the way I want, having checked up to n = 11. I just can't to prove it. If one of you guys finds a solution, perhaps we can collaborate on a paper.
I will now present the problem and some of my approaches. I use $\land$ to denote the minimum and $\lor$ the maximum. I write $n \text{ mod } 2$ to mean $1$ if $n$ is odd, and $0$ otherwise.
The Problem.
We first define some coefficients:
\begin{align*} A(n, j) &= C_{n-j} \sqrt{2}^{1+j-n} (-1)^{\lfloor \frac{n+1}{2} \rfloor + \lfloor \frac{j}{2} \rfloor} , \end{align*} \begin{align*} P(l, n) &= (-1)^{\lfloor \frac{l}{2} \rfloor + \lfloor \frac{n}{2} \rfloor} (2 \sqrt{2})^{l - n} \begin{pmatrix} - \frac12 + \lfloor \frac{l}{2} \rfloor - n \\ l - n \end{pmatrix} , \end{align*} \begin{align*} Q(l, n) &= (-1)^{\lfloor \frac{l-1}{2} \rfloor + \lfloor \frac{n+1}{2} \rfloor} (2 \sqrt{2})^{l - n} \begin{pmatrix} - \frac12 - \lfloor \frac{n}{2} \rfloor \\ l - n \end{pmatrix} . \end{align*}
Here \begin{equation*} C_n = \frac{(2n)!}{n!(n+1)!} \end{equation*} are the Catalan numbers.
Now consider the polynomial ring $\mathcal{R} = \mathbb{C}[G_0^0, G_1^0, G_1^2]$. We define polynomials $G_n^k \in \mathcal{R}$ for $n \geq 0$ and $k \in [0, 2n]$. If $k$ is odd, then we also set $G_n^k = 0$. For even $k \geq 4$ and $n \geq 1$, we set \begin{align*} G_{n+1}^k &= \sum_{j=0}^n \sum_{m=0 \lor (k+2j-2-2n)}^{(2j) \land (k-2)} (n \text{ mod } 2 - (-1)^j) G_j^m G_{n-j}^{k-2-m} + \sum_{j=0}^n A(n, j) G_j^k \\ \end{align*}
The hypothesis is now any of the three following statements, which are more or less equivalent: Let $l \geq 0$ and $k \in [0, 2 l]$ be an even number.
- Define \begin{align*} E_l^k = \sum_{n = \frac{k}{2}}^l P(l, n) \, G_n^k. \end{align*} Then \begin{equation*} l \geq k+2 \Longrightarrow E_l^k = 0 \end{equation*}
- We may define the same $E_l^k$ by \begin{align*} E_l^k &= G_l^k + \sum_{n = \frac{k}{2}}^{(l - 1) \land (k + 1)} Q(l, n) \, E_n^k. \end{align*} Then \begin{equation*} l \geq k+2 \Longrightarrow E_l^k = 0 \end{equation*}
- Combining 1. and 2. leads to the following claim: If we fix a $k$, then the recursion in $n$ is solved by this exact formula, which depends on the first $k+1$ terms: \begin{equation*} l \geq k + 2 \Longrightarrow G_l^k = - \sum_{j = \frac{k}{2}}^{k + 1} G_j^k \sum_{n = j}^{k + 1} Q(l, n) P(n, j) \,. \end{equation*} Note also that \begin{equation*} \sum_{n = j}^l Q(l, n) P(n, j) = \delta_{l, n} \end{equation*} This means that the formula also holds trivially for $l < k + 2$, it is "consistent".
My Approaches.
The first thing I tried was to show 3. by induction. I used the definition of the $G_n^k$, applied formula 3. to the terms that appeared and tried to regroup everything to obtain formula 3. for my current term. I could not make this work, as there are many terms and they can't obviously be regrouped into what you want.
Then I tried generating functions and actually had some success.
Generating Functions.
Define \begin{align*} U_n^k = (-1)^n G_{2n}^{2k} \qquad \qquad k \in [0, 2n] \qquad \qquad U_0^0 \text{ given} \end{align*} \begin{align*} V_n^k = (-1)^n G_{2n+1}^{2k} \qquad \qquad k \in [0, 2n+1] \qquad \qquad V_0^0, V_0^1 \text{ given} \end{align*}
Then this solves
\begin{align*} U_{n+1}^k &= - 2 \sum_{j=0}^n \sum_{m=0 \lor (k-1-2n+2j)}^{(2j+1) \land (k-1)} V_j^m U_{n-j}^{k-1-m} + \sum_{j=0}^n C_{2n-2j} \sqrt{2}^{1+2j-2n} V_j^k + \sum_{j=0}^n C_{1+2n-2j} \sqrt{2}^{2j-2n} U_j^k \end{align*} \begin{align*} V_{n+1}^k &= - \sum_{j=0}^{n+1} \sum_{m=0 \lor (k-3-2n+2j)}^{(2j) \land (k-1)} U_j^m U_{n-j+1}^{k-1-m} - \sum_{j=0}^n \sum_{m=0 \lor (k-2-2n+2j)}^{(2j+1) \land (k-1)} V_j^m V_{n-j}^{k-1-m} + \sum_{j=0}^{n+1} C_{2n+2-2j} \sqrt{2}^{2j-2n-1} U_j^k + \sum_{j=0}^n C_{2n+1-2j} \sqrt{2}^{2j-2n} V_j^k \end{align*}
We want to rewrite this in terms of the generating functions \begin{align*} u(y, z) = \sum_{n, k} U_n^k y^n z^k \qquad \qquad v(y, z) = \sum_{n, k} V_n^k y^n z^k . \end{align*}
I haven't checked this computation many times yet, but I obtained:
\begin{align*} u(y, z) = u(y, z) \Big(2 - \sqrt{1 + 2 \sqrt{2} \sqrt{y}} - \sqrt{1 - 2 \sqrt{2} \sqrt{y}}\Big) + v(y, z) \sqrt{y} \Big(\sqrt{1 + 2 \sqrt{2} \sqrt{y}} - \sqrt{1 - 2 \sqrt{2} \sqrt{y}}\Big) - 2 u(y, z) v(y, z) y z \end{align*} \begin{align*} v(y, z) = v(y, z) \Big(2 - \sqrt{1 + 2 \sqrt{2} \sqrt{y}} - \sqrt{1 - 2 \sqrt{2} \sqrt{y}}\Big) + u(y, z) \frac{1}{\sqrt{y}} \Big(\sqrt{1 + 2 \sqrt{2} \sqrt{y}} - \sqrt{1 - 2 \sqrt{2} \sqrt{y}}\Big) - u(y, z)^2 z - v(y, z)^2 y z \end{align*} Unfortunately, I don't see how to rewrite one of the hypotheses 1., 2., or 3. in terms of these generating functions.
I would be very happy if anyone can tell me how difficult one should expect this problem to be, or maybe even solve it.
