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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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    I added a long-ish title, but more informative I guess.2012-03-22

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Peter Tamaroff gave a very good hint in comments. Here is what comes out of it.

Suppose that $x$ is a solution tangent to $rt$ and not equal to it. Since solution curves do not cross, either (i) $x(t)>rt$ for all $t>0$, or (ii) $x(t) for all $t>0$. I will consider (i), the other case being similar.

By assumption, $x/t\to r$ as $t\searrow 0$. From $f(x/t)=r+f'(r)(x/t-r)+o(x/t-r)$ and $f'(r)<1$ we obtain $t(x/t)'=f(x/t)-x/t = (f'(r)-1)(x/t-r)+o(x/t-r)$ which is negative for small $t$. This means that $x/t$ increases as $t\searrow 0$, contradicting $x/t\to r$.

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    I did not quite understand your solution. Do you suppose x a rt solution? This diagonal arrow means what?2017-04-12