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Let $f$ be :

$f: [t_0,t_1] \times [x_0 - b, x_0+ b] \longrightarrow \mathbb{R}$

continous and such that $(f(t,x_2) - f(t,x_1))(x_2-x_1) \leq 0$ $\forall t\in [t_0,t_1] $ and $\forall x_1,x_2 \in [x_0-b, x_0+b]$. Prove that the Cauchy problem:

$x' = f(t,x)$

$x(t_0) = x_0$

has unique solution.

My attemp: I know about a theorem that allows to prove that the Cauchy problem has unique solution if $f$ is Lipschitz but I don't think i can show that with the hipothesis I have.

Any hints?

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Existence of solution follows from the continuity of $f$. Suppose that $x_1(t)$ and $x_2(t)$ are solutions and let $h(t)=(x_2(t)-x_1(t))^2$. Then $$ h'(t)=2(x_2(t)-x_1(t))(x_2'(t)-x_1'(t))=2(x_2(t)-x_1(t))\bigl(f(t,x_2(t))-f(t,x_1(t))\bigr)\le0. $$ Then $h$ is decreasing, non-negative and $h(t_0)=0$. The only possibility is that $h(t)=0$ for $t\in[t_0,t_1]$.

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  • $\begingroup$ I need the prove not only that a solution exist, but that it is unique $\endgroup$ – José D. Jun 4 '13 at 16:26
  • $\begingroup$ I have read it again and now it makes sense. Thanks. $\endgroup$ – José D. Jun 4 '13 at 16:49

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