# $\varphi (t)$ is a maximal solution of $x'=f(x) \implies \varphi (t+c)$ is a maximal solution

Consider a function $f:]a,b[\to\mathbb R$ satisfying the Lipschitz condition.

1. Let $\varphi : ]\alpha , \beta[ \to ]a,b[$ be a maximal solution, $c\in\mathbb R$. Show the function $$\varphi ^ c: ]\alpha -c,\beta -c[ \to ]a,b[$$ $$\varphi ^c (t):= \varphi (t+c)$$ is also a maximal solution.
2. Let $\varphi,\psi$ be maximal solutions of $x'=f(x)$ with a common image. Prove one is a translation of the other.

For 1. I firstly proved "$\varphi$ is a solution $\implies$ $\varphi ^c$ is a solution". To prove it's also a maximal solution I tried the contrapositive:

Suppose $\varphi ^ c$ isn't a maximal solution. Then there exists an extension $\varphi _I^c$ of $\varphi ^c$ defined on an interval $I$ where $]\alpha - c,\beta -c[\subset I$. Now I need to build a function $\varphi_I$ defined on the interval $I+c$ such that $\varphi_I (t)$=$\varphi (t)$ for all $t\in ]\alpha,\beta[$ to conclude but it's not clear to me how should I do this.

For the second question, I don't understand the intuition behind it. If both solutions have a common image then there exists $t,s$ such that $\varphi(t)=\psi (s)$. Rewriting we have $\varphi(t)=\psi (t+c)$ where $c = s-t$. I don't know where to go from here. Any help is appreciated.

For 1. I think it is enough to observe that, if $\varphi\colon I \to \mathbb{R}$ is any solution of $x'=f(x)$, then $\varphi^c\colon I^c := I-c \to\mathbb{R}$ is also a solution.
For 2., let $x_0 := \varphi(t_0) = \psi(s_0)$ (in your notation). Since $f$ is Lipschitz continuous, $\varphi$ is the unique maximal solution of the Cauchy problem $x'=f(x)$, $x(t_0) = x_0$. On the other hand, also $\psi^c$, for $c:= t_0 - s_0$, is a solution of the same Cauchy problem, hence $\psi^c = \varphi$.
• I did prove if $\varphi$ is a solution then $\varphi ^c$ is also a solution. From this can I conclude it's also true for maximal solutions? – AndreGSalazar May 16 '17 at 9:13
• I'd say yes: if $\varphi$ is a maximal solution, then $\psi := \varphi^c$ is a solution that can't be prolonged, for otherwise $\psi^{-c}$ would be a proper extension of $\varphi$. – Rigel May 16 '17 at 10:12