# Maximal Solutions

Let $I\subseteq \boldsymbol{R}$ be a non-degenerated interval and $f:I\to \boldsymbol{R}$ a continuous function and $\forall x\in I$, $f(x)\neq 0$. Given an $a\in I$ let

\begin{align} F_a:&I\to\boldsymbol{R} \\ x&\mapsto \int_{a}^{x}\frac{ds}{f(s)}\end{align}.

If $\varphi_a$ is the inverse of $F_a$, how to prove that, for a $c\in\boldsymbol{R}$, $\varphi_{a,c}(t) = \varphi_a(t-c)$ is a maximal solution of the differential equation

$$\frac{dx}{dt} = f\big(x(t)\big)?$$

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In the last equality is $f(x)$ supposed to be $f(x(t))$? – Git Gud Sep 4 '13 at 15:01
@GitGud: Yes, is a autonomous equation. I will edit it. – user34870 Sep 4 '13 at 15:19
With 'for a $c\in \Bbb R$', do you mean 'for some $c\in \Bbb R$'? Just wanna make sure that $c$ is existentially quantified rather than universally. – Git Gud Sep 4 '13 at 17:05
@GitGud, In my attempt, I choose $c\in\boldsymbol{R}$ such that $J_{a,c} = (J_a + c)\cap J_a\neq\emptyset$ and, thus, define $\varphi_{a,c}:J_{a,c}\to I$. Here, $J_a$ is the domain of $\varphi_a$ – user34870 Sep 4 '13 at 17:13

First: using the derivative of the inverse function and the fundamental theorem of calculus, you get:
$$\frac{\partial \varphi_a}{\partial t}(t)=\frac{1}{\frac{\partial F_a}{ \partial x} (\varphi_a(t))}=f(\varphi_a(t))$$ Therefore $\varphi_a(t)$ is a solution of the differential equation.

I believe you did not ask for first part of the problem, but just in case.

Second: We can suppose without loss of generality that $f>0$ and say that $I=(A,B)$. Note that

$$Dom(\varphi_a)=Im(F_a)=\left(-\int_A^a\frac{1}{f}\,,\,\int_a^B \frac{1}{f}\right),$$ and then $$Dom (\,\varphi_{a,c}\,)=\left(c-\int_A^a\frac{1}{f}\,,\,c+\int_a^B \frac{1}{f}\right)$$

Now consider the initial condition, say $x(t_0)=\alpha \in I$. Then we need that $t_0$ belong to the domain of $\varphi_{a,c}$. For any $a$ we always can choose $c$ such that $t_0 \in Dom (\varphi_{a,c})$:

$$\varphi_{a,c}(t_0)=\varphi_a(t_0-c)=\alpha \;\;\mbox{ iff }\;\; t_0-F_{a}(\alpha)=c$$ Finally, $$Dom (\,\varphi_{a,c}\,)=\left(t_0-\int_A^\alpha\frac{1}{f}\,,\,t_0+\int_\alpha^B \frac{1}{f}\right)$$

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