Finding all solutions to the functional equation $-\frac{1}{2}f(t,x) = f(2t, -x)$ I'm trying to find the class of differential equations $\frac{\mathrm dx}{\mathrm dt} = f(t,x)$ with $f$ continuous that are invariant under the change of variables $s=2t$ and $y=-x$.
I have come up with this functional equation:
$$-\frac{1}{2}f(t,x) = f(2t,-x) \text.$$
Letting $f(t,x) = \frac{q(x)}{p(t)}$ with $p$ and $q$ multiplicative and $p(t) \ne 0$ for all $t$, we get a family of solutions; because
$$
f(2t, -x) = \frac{q(-x)}{p(2t)} = \frac{-q(x)}{2p(t)} = -\frac{1}{2} f(t,x) \text.
$$
$f(t,x) = \log_a\left(k^{\frac{q(x)}{p(t)}}\right)$ with $k \in \mathbb{R}^+$ gives another type of solutions; because
$$
f(2t, -x) = \log_a\left(k^{\frac{-q(x)}{2p(t)}}\right) = -\frac{1}{2} \log_a\left(k^{\frac{q(x)}{p(t)}}\right)= -\frac{1}{2} f(t,x) \text.
$$
However, are they all the solutions? Is there a standard method to solve this?
 A: The only continuous solution to the functional equation
$$ - \frac 1 2 f ( t , x ) = f ( 2 t , - x ) \tag 0 \label 0 $$
is the constant zero function (which obviously is a solution). To see that, substitute $ \frac t 2 $ for $ t $ and $ - x $ for $ x $ in \eqref{0} to get
$$ f \left( \frac t 2 , - x \right) = - 2 f ( t , x ) \text . \tag 1 \label 1 $$
Again, substitute $ \frac t 2 $ for $ t $ and $ - x $ for $ x $ in \eqref{1} and use \eqref{1} itself to see that
$$ f \left( \frac t 4 , x \right) = - 2 f \left( \frac t 2 , - x \right) = 4 f ( t , x ) \text . \tag 2 \label 2 $$
Now, using \eqref{2} and mathematical induction, you can prove
$$ f \left( \frac t { 4 ^ n } , x \right) = 4 ^ n f ( t , x ) \tag 3 \label 3 $$
for any nonnegative integer $ n $. The left-hand side of \eqref{3} has the limit $ f ( 0 , x ) $ as $ n $ tends to infinity, by continuity of $ f $ at $ ( 0 , x ) $. But if $ f ( t , x ) \ne 0 $, then the right-hand side of \eqref{3} becomes unbounded as $ n $ tends to infinity, and cannot have a finite limit. Therefore, we must have $ f ( t , x ) = 0 $. As this could be argued for arbitrary $ t $ and $ x $, the value of $ f $ must be constantly zero.
