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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 $x(t_0)=x_0$ ?



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
up vote 6 down vote accepted

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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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
@Richard: I updated the answer. – timur Jan 10 '12 at 0:15
Very thanks for help. – Richard Jan 11 '12 at 19:50

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