Okay, I've figured out the solution. All we need is a second order difference equation.
\[
\varphi_3(y,s) = \left(1-\rho_1\right) C(y-2,s)+ \left(\rho_1 + \rho_2\right)\varphi_3(y-1,s)-\rho_1\rho_2 \varphi_3(y-2)\\
\]
Where:
\[
\varphi_3(y-1,s) = \varphi_2\\
\]
And \(\varphi_1\) equals a different equation hidden in the above equation. The values \(\rho_1,\rho_2\) are differentials of the form:
\[
\begin{align}
\rho_1 = \frac{\partial}{\partial \varphi_1} \varphi_3\\
\rho_2 = \frac{\partial}{\partial \varphi_2} \varphi_3\\
\end{align}
\]
Where we're looking at \(\varphi_3\) as a tangent plane.
And again,
\[
C(y,s) = \varphi_3 \Big{|}_{\varphi_1',\varphi_2'}
\]
This turns the really hard problem, into an infinite composition problem. I'm going to refrain from posting for a while, until I have a well working theory. And at least a confirmatory program. So, give me a week or two. I see the math, but there's a lot of work to solve this.
\[
\varphi_3(y,s) = \left(1-\rho_1\right) C(y-2,s)+ \left(\rho_1 + \rho_2\right)\varphi_3(y-1,s)-\rho_1\rho_2 \varphi_3(y-2)\\
\]
Where:
\[
\varphi_3(y-1,s) = \varphi_2\\
\]
And \(\varphi_1\) equals a different equation hidden in the above equation. The values \(\rho_1,\rho_2\) are differentials of the form:
\[
\begin{align}
\rho_1 = \frac{\partial}{\partial \varphi_1} \varphi_3\\
\rho_2 = \frac{\partial}{\partial \varphi_2} \varphi_3\\
\end{align}
\]
Where we're looking at \(\varphi_3\) as a tangent plane.
And again,
\[
C(y,s) = \varphi_3 \Big{|}_{\varphi_1',\varphi_2'}
\]
This turns the really hard problem, into an infinite composition problem. I'm going to refrain from posting for a while, until I have a well working theory. And at least a confirmatory program. So, give me a week or two. I see the math, but there's a lot of work to solve this.