TITLE: Double Sequences solving method
QUESTION [11 upvotes]: I am searching some information on what I think is called a real double sequence. What I call double sequence is a function from $\mathbb{N} \times \mathbb{N} \to \mathbb{R}$. By opposition to this, I call simple sequence (just a regular sequence), a function $\mathbb{N}  \to \mathbb{R}$.
Recursion relation
I learnt that for a simple sequence, there are methods for solving recursion definitions of a sequence. For example, the simple function $u(n)$ defined as $\forall n\in\mathbb{N}, u(n+1)= u(n)+ u(n-1)$, we solve the characteristic polynomial associated with this recursion relation, and we find that the space of solution $S$ is the span of the Fibonacci numbers $F_n$ and Lucas numbers $L_n$ : $S\in Span(F_n,L_n)=Vect(F_n,L_n)$.
My problem
However, I have some interests in double sequences $u(i,j)$ defined by a recursion relation so that $\forall (i,j)\in \mathbb{N}^2,u(i,j) = f_1(i,j) \cdot u(i,j-1) + f_2(i,j) \cdot u(i-1,j-1)$, with the first term $u(0,0)$ being equal to a real constant, let's say $u(0,0) = \alpha$; and that $\forall (i,j)\in \mathbb{Z}^2, (i<0 \vee j<0),u(i,j) = 0$ so the terms with negative index in the recursion relation cancels out nicely.
A change of variable?
Someone suggested me to perform a change of variable so that we get a double sequence $v(i,j)$ with the form $\forall (i,j)\in \mathbb{N}^2,v(i,j) = g_1(i,j) \cdot v(i,j-1) + g_2(i,j) \cdot v(i-1,j)$. So, it makes sense for me because it seems so much easier to compute.
My questions
My questions are:

Is there a general method for solving
$ \forall (i,j)\in \mathbb{N}^2.$
$$ v(i,j) = g_1(i,j) \cdot v(i,j-1) + g_2(i,j) \cdot v(i-1,j) ?$$
Is there a method if $g_1$ and $g_2$ are polynomials of second order of $i$ and $j$ : $g_1(i,j) = a + bi + ci^2 + dj + dj^2 + eij + fi^2j + gi^2j^2 + \dots $ ?
Is there a matrix representation of this system that could be used to compute this?

Edit due to Yuval Filmus' comment
Edit: The comment turned out to be a little bit helpful with the user named Yuval Filmus telling me about generating functions and linking this book. Basically the idea of this clue is to replace the study of $u(i,j)$ by the study of $\sum \sum u(i,j) x^i y^j$; then, you replace the terms with partial derivatives as described on the book, and you turn your recursion problem in a PDE problem. However, I the problem is that the functions $g_1$ and $g_2$ are polynomials of order 2, and I don't know how to make it appear. So I would like to know if someone could explicitly turn my problem in a PDE problem then solve it? Then, once the function $\sum \sum u(i,j) x^i y^j$ has been found, you just do the do product with the family of functions $x^i$ and $y^j$ to get $u(i,j)$.
Side note: I didn't know about this method to solve recursion problem, but I knew the other way around: replace a PDE with $\sum \sum u(i,j) x^i y^j$ then identify the recursion relation, then solve it (assuming it was an easy one); then try to recognize it with the well-known Taylor expansions.

REPLY [2 votes]: The key word in this question is solve. What does it mean to "solve" a recursion
relation? Looking at a few simple examples, it is clear that this is like asking to find
integrals of general functions "in closed form". The answer is that you can not do this
except for some special cases.
Considering just a simple one dimensional recursion relation
$\ u(n+1) = f(u(n)),\ $ with initial value $\ u(0)=z,\ $already you can see that although you can compute
any value $\ u(n)\ $ by starting from $\ u(0),\ $ in general, there is no better way to find
the value. Thus, for your first question about general methods for solving recursion
relations even in one dimension or higher, the answer is no. For your second question,
even restricting the functions $\ g_1()\ $ and $\ g_2()\ $ to be polynomials at most of
second degree in $\ i,j\ $ the answer is still no.
For your "double sequences" a few preliminary remarks. Consider the simplest example
of such sequences which is the binomial coefficients, also known as Pascal's triangle. In one
version they are defined by $\ u(n,k) := (n+k)!/(n!k!)\ $ and are the entries of a symmetric infinite square matrix
with recursion relation $\ u(n,k) = u(n-1,k) + u(n,k-1).\ $ In a second version they are defined by
$\ u(n,k) := n!/(k!(n-k)!)\ $ and are the entries of a symmetric infinite triangular array with
recursion relation $\ u(n,k) = u(n-1,k) + u(n-1,k-1).\ $ In either version, the indexing could either
begin with $\ n=k=0\ $ or with $\ n=k=1.\ $ These are merely matters of personal choice and convenience.
Let us try solving a few simple examples of the recursion
$$ v(i,j) = g_1(i,j) \cdot v(i,j-1) + g_2(i,j) \cdot v(i-1,j) $$
with initial value $\ v(1,1) = 1.\ $
First, let $\ g_1(i,j) = 1, \ g_2(i,j) = i.\ $
The array of numbers is OEIS sequence A132159. Explicitly, the solution is $$ v(i,j) = i\ (i+j-2)!/(j-1)!. $$
Using the information in the OEIS entry, the exponential generating function (egf) is given by
$$ \frac{\exp(y z)}{(1 - xz)^2} = \sum_{n=0}^\infty \frac{z^n}{n!}\: \Big(\sum_{i+j=n+2} v(i,j)\ x^i y^j\Big). $$
Next, let $\ g_1(i,j) = j, \ g_2(i,j) = i.\ $
The array of numbers is OEIS sequence A091441. Explicitly, the solution is $$ v(i,j) = i\ j\ (i+j-2)!. $$
Using the information in the OEIS entry, the egf is given by
$$ \frac1{((1 - xz)(1-yz))^2} = \sum_{n=0}^\infty \frac{z^n}{n!}\: \Big(\sum_{i+j=n+2} v(i,j)\ x^i y^j\Big). $$
Next, let $\ g_1(i,j) = i, \ g_2(i,j) = 1.\ $
The array of numbers is OEIS sequence A106800. Explicitly, the solution is $$ v(i,j) = S2(i+j-1,i) $$ where $S2()$ is the Stirling numbers of the second kind.
Using the information in the OEIS entry, the egf is given by
$$ \exp((\exp(y\ z)-1)\ x/y)-1 = \sum_{n=1}^\infty \frac{z^n}{n!}\: \Big(\sum_{i+j=n} v(i,j)\ x^i y^j\Big). $$
Next, let $\ g_1(i,j) = i, \ g_2(i,j) = j.\ $
The array of numbers is OEIS sequence A008292. These are the Eulerian numbers.
Using the information in the OEIS entry, the egf is given by
$$ \frac{x-y}{1-y/x\ \exp(xz-yz)}-x = \sum_{n=1}^\infty \frac{z^n}{n!}\: \Big(\sum_{i+j=n+1} v(i,j)\ x^i y^j\Big). $$
There are several other simple examples of this kind that I have worked on so far, but my usual method is to compute the first few numbers and look them up in the OEIS and use the information there to find an explicit expression for the numbers and/or an exponential generating function for them. This only works in certain special cases, of course.
For example, let $\ g_1(i,j) = 1, \ g_2(i,j) = i^2.\ $ The array of numbers is currently not in the OEIS, but despite this, the explicit expression is
 $\ v(i,j) = i\ i!\ (i+j-2)!/(j-1)!\ $ but so far I have not found an expression for the egf.
My experience strongly suggests that quadratic functions $\ g_1(),\ g_2()\ $ are much harder to solve recursions for. I don't know of any example of PDE that are associated with these quadratic functions, and even then, the PDE may not be any easier to solve for.