# How can I solve this system of functional equations?

I would like to use Mathematica to find maps a, b, c, d, e, f such that

eq1 = a[t] a[s] + b[t] d[s] == a[t + s]
eq2 = a[t] b[s] + b[t] e[s] == b[t + s]
eq3 = d[t] a[s] + e[t] d[s] == d[t + s]
eq6 = d[t] b[s] + e[t] e[s] == e[t + s]
eq4 = a[t] c[s] + b[t] f[s] + c[t] == c[t + s]
eq5 = d[t] c[s] + e[t] f[s] + f[t] == f[t + s]


Can anybody can help me with this?

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These are more naturally expressed as matrix equations $\mathbb{A}(s)\mathbb{A}(t)=\mathbb{A}(s+t)$ and $\mathbb{A}(t)x(s)=x(s+t)$ for a $2$ by $2$ matrix $\mathbb{A}$ and $2$ by $1$ vector $x$. It is immediate that $x(t)=\mathbb{A}(t)x(0)$ and almost as clear (assuming the map $t\to\mathbb{A}(t)$ is differentiable) that $\mathbb{A}(t)=\exp(a(t))$ for some $2$ by $2$ matrix $a$. Having observed that, how much help do you really need from Mathematica? –  whuber Apr 30 '13 at 14:47
Maybe replace every f_[t_] with exponent f Exp[t] then first four are just algebraic, then substitute its solution to two last equations and solve them? –  swish Apr 30 '13 at 15:00
@swish That isn't likely to work, because the matrix of exponentials of coefficients is not the same as the exponential of a matrix. As a simple example, $\exp{\left(t \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)\right)}$ is the rotation matrix $\left( \begin{array}{cc} \cos (t) & -\sin (t) \\ \sin (t) & \cos (t) \end{array} \right)$. (I meant to write "$\exp(a t)$" instead of "$\exp(a(t))$" in my previous comment.) –  whuber Apr 30 '13 at 16:14