How to solve the functional equation $f(s,t) - sf(t,1) - tf(s,1) + tsf(1,1) = s(1-t)$ I am interested in solving in the real function $f:[0,1]\times [0,1] \rightarrow \mathbb R$ the equation
$$f(s,t) - sf(t,1) - tf(s,1) + tsf(1,1) = s(1-t)$$
I have no clue how to attack such a problem. Does this kind of functional equation have a name? What are some techniques to solve such a problem? Can we know if a solution is unique?
I am not interested in a solution, but in how to find a solution of the equation (to be honest, I guessed a solution, but it is luck, not mathematics. Besides, I do not know if it is unique).
Optimally, I would like to study the class of equations
$$f(s,t) - sf(t,1) - tf(s,1) + tsf(1,1) = g_1(s) g_2(t)$$
where $g_1$ and $g_2$ are given real continuous functions such that $g_1(0)=0, g_2(1)=0$. I feel that for any solution $f$, it must have separate variables, i.e, $f(s,t)=f_1(s)f_2(t)$? Is it true?
 A: Observe that $f$ is defined on a square, and the relation:
$$f(s,t) = sf(t,1) + tf(s,1)-stf(1,1) + s(1-t)$$
implies that the behaviour of $f$ is completely determined by how its defined at a subset of the boundary of that square, that is, at points $(x,1)$. There is no constraint as to what does there, so there are as many solutions as functions on the interval $g: [0,1]\rightarrow \mathbb{R}$.
For each one of those functions, you obtain a solution of your equation, with the boundary condition $f(x,1) = g(x)$, and its expression is
$$f(s,t) = sg(t) + tg(s) - stg(1) + s(1-t)$$.
For example, taking $g(x) = x^2$ gives the solution:
$$f(s,t) = st^2 + ts^2 + s(1-t) = st^2 + s^2t -st +s = s(t^2 + st -t+1) = $$
Which has not separated variables.
Everything I said applies to the general case.
A: Assuming $f(x,y)$ is polynomial and proposing
$$
f(x,y) = a x + b y + c x y
$$
after substitution we have
$$
(a-1-b)s(t-1)=0
$$
and thus
$$
f(x,y) = a x+(a-1)y+c x y
$$
is a solution.
The functional equation also admits separable solutions as
$$
\cases{
f_1(x) = \frac{(c_1c_2+1)x-1}{c_1}\\
f_2(y) = c_2 y\\
f(x,y) = f_1(x)f_2(y)
}
$$
