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Suppose $f:\mathbb{R}\to\mathbb{R}$ is a continuous differentiable function such that $f(r)=r,$ for some $r.$ Then how to show that

If $f'(r) < 1,$ then the problem $$x'=f(x/t)$$ has no other solution tangent at zero to $\phi(t)=rt, t>0$.

Tangent here means

$$\lim_{t\to 0^{+}}\frac{\psi(t)-\phi(t)}{t}=0$$

I could only prove that $\psi(0^+)=0,$ and $\psi'(0^+)=r.$ The problem was to use the fact that $f'(r) < 1.$

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    Is $x'$ supposed to mean $x'(t)$?2012-03-22
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    The wording is rather strange; $x$ should be proven to be tangent to $\phi$ at $0$?2012-03-22
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    Yeah, @PeterT.off!2012-03-22
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    @math What is $ψ$?2012-03-22
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    If $f'(r)<1$ then $f$ is greater than $r$ in $r-\epsilon$ and smaller than $r$ in $r+\epsilon$2012-03-22
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    Well my approach was supposing that there was a solution $\psi$ tangent to $\psi$ then I would infer that this solution was equal to $\phi$.2012-03-22
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    I added a long-ish title, but more informative I guess.2012-03-22

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