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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 $x'(t)$? – Pedro Tamaroff Mar 22 '12 at 0:45
The wording is rather strange; $x$ should be proven to be tangent to $\phi$ at $0$? – Pedro Tamaroff Mar 22 '12 at 0:47
Yeah,! – checkmath Mar 22 '12 at 0:47
@math What is $ψ$? – Pedro Tamaroff Mar 22 '12 at 0:48
If $f'(r)<1$ then $f$ is greater than $r$ in $r-\epsilon$ and smaller than $r$ in $r+\epsilon$ – Pedro Tamaroff Mar 22 '12 at 0:50

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) $x(t)<rt$ 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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