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Let $f: (a,b) \times \mathbb{R} \rightarrow \mathbb{R}$ be of class $C^1$ in $D:=(a,b) \times \mathbb{R}$ and satisfies condition $| f(t,x)| \leq A+B|x| \textrm{ for } (t,x) \in D,$ where $A,B$ are fixed real constants and let $t_0 \in (a,b)$.

How to prove using the fixed point method that for arbitrary $x_0\in \mathbb{R}$ there exist exactly one solution $x: (a,b)\rightarrow \mathbb{R}$ of differential equation $\frac{dx}{dt}=f(t,x) $ with condition $x(t_0)=x_0$ ?

Thanks.

Added.

Maybe it would be. Let $X=\{x:(a,b) \rightarrow \mathbb{R}: \sup_{t\in (a,b)} e^{-B\gamma|t-t_0|} |x(t)| <\infty, x(t_0)=x_0 \}$, $d(x,y)=\sup_{t\in (a,b)} e^{-B\gamma|t-t_0|} |x(t)-y(t)|$ for $x,y \in X$, where $\gamma$ is a suitable positive constant. Then $(X,d)$ is a complete metric space and $Tx(t):=x_0+\int_{t_0}^t f(s,x(s))ds$, for $x \in X$ and $t\in (a,b)$, maps X into itself (because $|f(s,x(s))|\leq A+Be^{B\gamma|t-t_0|}\cdot sup_{t\in (a,b)} |x(s)|e^{-B\gamma|t-t_0|} |x(t)|$ and $| \int_{t_0}^t e^{B \gamma |s-t_0|} ds| \leq \frac{1}{B \gamma} e^{B\gamma|t-t_0|}$). However I don't know is it $T$ a contraction with some $\gamma>0$ and whether or not each solution of the differential equation belongs to $X$.

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    @Richard: if $x_0\neq 0$, $X$ is not a vector space, and $f$ may not be Lipschitz continuous, for example with $f(t,x)=\sin (x^2)$.2012-01-06

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By the Picard-Lindelof theorem there is a unique local solution, and it is global if you can show that the solution does not blow up. The latter is easy to show by using the given bound on f.

Namely, we have $ \frac12\frac{dx^2}{dt}=xf(x,t)\leq |x|(A+B|x|)\leq A+(A+B)x^2, $ giving an exponential bound on $x^2$.

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    Very thanks for help.2012-01-11