# An existence of global solution of differential equation of first order

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.

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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On first glance this looks like a candidate for the Banach fixed-point theorem. – Alex Becker Jan 6 '12 at 14:51
I agree. I suppose that it could be done by taking operator $Tx(t)=x_0+\int_{x_0}^t f(s,x(s))ds$ for $x\in X$ and $t\in(a,b)$, where $X$ is the space continuous functions such that $sup_{t\in (a,b)} |x(t)| e^{-L|t-t_0|}< \infty$ and $x(t_0)=x_0$ with Bielecki's norm $\|x\|=sup_{t\in (a,b)} |x(t)| e^{-L|t-t_0|}$, where $L$ is Lipschitz constant for $f$. – Richard Jan 6 '12 at 15:24
@Richard: if $x_0\neq 0$, $X$ is not a vector space, and $f$ may not be Lipschitz continuous, for example with $f(t,x)=\sin (x^2)$. – Davide Giraudo Jan 6 '12 at 20:17

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 $x^2$.
Thanks. But could you write more detaily why a local solution can be extended to global one on $(a,b)$? – Richard Jan 7 '12 at 11:46