I am trying to find the solutions of the following functional equation:
$-f(a+b, c, d) e^{a \cdot b}+f(a, b+c, d) e^{b \cdot c}-f(a, b, c+d) e^{c \cdot d}+f(a+d, b, c) e^{-a \cdot d}=0$
where $f(a, b, c)$ is a function of three vectors and is assumed to be Taylor expandable in terms of various scalar products of its arguments a, b, c (the output of $f$ is a scalar). In the equation, a, b, c, d are any vectors in $\mathbb{R}^{2}$.
In addition, we define the scalar product as
$a \cdot b=a_{1} b_{2}-a_{2} b_{1}$
this means that $a \cdot b=-b \cdot a$, and $a \cdot a=0$ for all a, b in $\mathbb{R}^{2}$.
My ideas so far:
- I managed to find some semi-trivial solutions which I am not interested in. Namely,
$f(a, b, c)=-g(a+b, c) e^{a \cdot b}+g(a, b+c) e^{b \cdot c}-g(a+c, b) e^{-a \cdot c}$
solves the above equation for any Taylor expandable function $g(a, b)$, but I am looking for other solutions.
- I noticed that if I change the sign of $d$ in the last term, i.e., $-f(a+b, c, d) e^{a \cdot b}+f(a, b+c, d) e^{b \cdot c}-f(a, b, c+d) e^{c \cdot d}+f(a-d, b, c) e^{a \cdot d}=0$
then a simple solution is $f(a, b, c)=\exp [a \cdot b+a \cdot c+b \cdot c]$ which I am not sure if it is useful.
- In two-dimension any three vectors are linearly dependent, in particular, we have the following identity for any a, b, c, d:
$(a \cdot b)(c \cdot d)+(b \cdot c)(a \cdot d)-(a \cdot c)(b \cdot d) \equiv 0$