Let $\dot{x} = f(t,x)$, such that $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ is continuous. Prove that if for every solution to the equation, $x(t)$ and for every $c \in \mathbb{R}$, $x(t + c)$ is also a solution, then $f$ is autonomous.
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Let's assume that the existence and uniqueness theorem holds throughout the domain $\Omega:={\rm dom}(f)\subset{\mathbb R}^2$. Consider two points $(t_0,x_0)$, $(t_0+c,x_0)\in\Omega$. There is a unique solution $\phi:\ t\mapsto\phi(t)$ of the initial value problem $$\dot x(t)=f\bigl(t, x(t)\bigr)\ ,\quad x(t_0)=x_0\ ,$$ valid in some $t$-interval $I$ with midpoint $t_0$. By assumption the function $$\psi:\quad t\mapsto \psi(t):=\phi(t-c)\ ,$$ defined in an interval $I'$ with midpoint $t_0+c$, is also a solution of the differential equation; furthermore $\psi(t_0+c)=\phi(t_0)=x_0$. It follows that $\psi$ is the solution of the initial value problem $$\dot x(t)=f\bigl(t, x(t)\bigr)\ ,\quad x(t_0+c)=x_0\ .$$ Therefore we have $$f(t_0+c,x_0)=f\bigl(t_0+c,\psi(t_0+c)\bigr)=\dot\psi(t_0+c)=\dot\phi(t_0)=f(t_0,x_0)\ .$$ This shows that $f$ is constant on horizontal lines in $\Omega$. |
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Hint: What differential equation does $x_c(t) = x(t+c)$ satisfy? |
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