Consider the following system:
$$
\begin{cases}
x\_1 + 3 x\_3 = 4a, \\
f(x\_1) + 3 f(x\_3) = 8 f(a), \\
f'(x\_1) = 3 f'(x\_3).
\end{cases}
$$


I want to find all functions (or at least learn some properties that hold for all of them) $f : [0,1] \to [0,1]$ that are continuous, differentiable on $[0,1]$, monotonically decreasing on $[0,1]$ and the aforementioned system has a solution for every $a \in (0,1)$. In other words, if $a$ is fixed, there should exist $x\_1, x\_3 \in [0,1]$ that satisfy the system.


Or, a bit re-phrased: find $f$ such that for all $a \in (0,1)$, the aforementioned system has at least one pair $(x\_1, x\_3) \in [0,1]^2$ satisfying the system. And the general question is of course to find all such $f$.


UPD: Apparently, the solution is trivial only. I created another question that is hopefully more of interest: [Solve differential system with a parameter](https://mathoverflow.net/q/387315)



Solve differential system of equations
Assume such an $f$ exists. For each $a\in(0,1)$ fix some solution $x\_1(a),x\_3(a)$ of the system. The first equality can be restated as
$$ \frac{1}{4}x\_1(a) + \frac{3}{4}x\_3(a) = a.$$
In other words, $a$ is a convex combination of $x\_1(a),x\_3(a)$. Consider a sequence $a\_k \searrow 0$. As $x\_1(a\_k),x\_3(a\_k) \geq 0$ it follows that $x\_1(a\_k),x\_3(a\_k) \to 0$. Taking the limit $k\to \infty$ in the second equation, we obtain from the continuity of $f$ that
$$ 4 f(0) = 8 f(0).$$
Hence $f(0) = 0$. As $f$ is monotonically decreasing and takes nonnegative values, it follows that $f$ is constant.


