I'm studying for an admission test for the PhD in Mathematical Analysis. I'm stuck to solve this exercise:
Consider the ODE $x'(t)+a(t)x(t)=b(t)$ where $a(t)$ and $b(t)$ are continuous functions from $\mathbb{R}$ to itself. Prove that this Equation cannot have 3 periodic non-constant solutions with pairwise incommensurable periods.
I recall that if $k\in\mathbb{R}$ and $q\in\mathbb{R}-\{0\}$, then they are called incommensurable when $\frac{k}{q}\notin\mathbb{Q}$.
I solved the first part of the exercise that asked to prove that if a continuous function on $\mathbb{R}$ is periodic and has two periods that are incommensurable then it's a constant function. But I can't see how the first part of the exercise can help with the second part. I also tried to do some computations with the solutions (subtracting and adding them etc.) and to write the general integral of the ODE and see if it could help. In both cases all I did seemed useless.
Let $x_1,x_2,x_3:\mathbb{R} \to\mathbb{R}$ be non-constant solutions of $x'+a(t)x=b(t)$ with pairwise incommensurable periods $p_1,p_2,p_3>0$. Set $y_1:=x_1-x_2$ and $y_2:=x_2-x_3$. Note that $y_1$ cannot be the zero function since otherwise $x_1=x_2$ has period $p_1$ and $p_2$ and would be constant. The same way $y_2 \not=0$. Now, $y_1$ and $y_2$ are solutions of the homogeneous equation $x'+a(t)x=0$, hence linear dependent: $\alpha_1 y_1 + \alpha_2y_2= 0$ for some $\alpha_1,\alpha_2 \in \mathbb{R}$ with with $\alpha_1\not=0\not=\alpha_2$. So $$ 0=\alpha_1(x_1(t)-x_2(t)) + \alpha_2(x_2(t)-x_3(t)) = \alpha_1x_1(t) + (\alpha_2-\alpha_1)x_2(t)-\alpha_2 x_3(t) \quad (t \in \mathbb{R}). $$ Thus $$ 0= \alpha_1x_1(t) + (\alpha_2-\alpha_1)x_2(t+p_1)-\alpha_2 x_3(t+p_1) \quad (t \in \mathbb{R}), $$ which yields $$ (\alpha_2-\alpha_1)x_2(t+p_1)-\alpha_2 x_3(t+p_1)= -\alpha_1x_1(t)=(\alpha_2-\alpha_1)x_2(t)-\alpha_2 x_3(t) \quad (t \in \mathbb{R}), $$ that is $$ \frac{\alpha_2-\alpha_1}{\alpha_2}(x_2(t+p_1)-x_2(t))= x_3(t+p_1)-x_3(t)\quad (t \in \mathbb{R}). $$ This shows that and $t \mapsto x_3(t+p_1)-x_3(t)$ is $p_2$-periodic. Since it is also $p_3$-periodic it is constant. So, there is some constant $c$ such that $$ x_3(t+np_1+mp_3)-x_3(t)=nc \quad (t \in \mathbb{R}, n,m \in \mathbb{Z}). $$ Since $x_3$ is a continuous periodic function it is uniformly continuous. Let $\varepsilon > 0$ and choose $\delta >0$ such that $|x_3(t)-x_3(s)| < \varepsilon$ $(|t-s| < \delta)$. Since $p_1$ and $p_3$ are incommensurable we find $n,m \in \mathbb{Z}$, $n \not=0$ such that $|np_1+mp_3| < \delta$. Then $$ |nc| =|x_3(t+np_1+mp_3)-x_3(t)| < \varepsilon \quad (t \in \mathbb{R}), $$ hence $|c|< \varepsilon/|n| \le \varepsilon$. Now $\varepsilon \to 0$ shows $c=0$. Summing up, $x_3$ is $p_1$ and $p_3$ periodic, hence constant. A contradiction.