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Find the minimum value of $a$ if there's a differentiable function in $\mathbb{R}$ for which :

$$e^{f'(x)}= a {\frac{|(f(x))|}{|(1+f(x)^2)|}}$$ for every $x$

pretty much stuck. I think the minimum value should be $1$ but not sure.

Mhenni Benghorbal
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Plom
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  • "I think the minimum value should be 1 but not sure"... Why do you think so? – Did Jun 17 '13 at 06:44
  • If a<1, then f'(x)<0 for every x. If u differentiate again u will get e^f'(x)f''(x)= af'(x)(1-f^2(x)/(1+f^2(x))^2 . Now if f(x)<1 for some values then its descending and its derivative is also descending which means f(x)=0 for some x, but f(x)=/=0 . I was trying to find a value of f(x) which is <1 so that this works somehow.. – Plom Jun 17 '13 at 06:50
  • Plom: I see. Thanks. – Did Jun 17 '13 at 07:16

2 Answers2

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Taking logarithms, $f'(x) = \log(a) + \log |f(x)| - \log (1 + f(x)^2)$. This is an autonomous differential equation. WLOG we can look at $f > 0$. An equilibrium solution would be a positive constant $c$ such that $\dfrac{ac}{1+c^2} = 1$. This has real solutions iff $a \ge \ldots$. On the other hand, if $a < \ldots$, consider what the phase-plane looks like.

Robert Israel
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  • How are these considerations showing whether $f$ is defined for every $x$ or not? – Did Jun 17 '13 at 06:49
  • @Did: I think that was the point. For $a<?$, $f$ crosses zero in finite time. – copper.hat Jun 17 '13 at 06:55
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    cant somebody explain to me whats going on? ._. – Plom Jun 17 '13 at 06:59
  • @copper.hat Sure, the fact that $f$ crosses zero in finite time is what should be shown. Following the lead in this answer, I arrive at $f'(x)\lt0$ if $a\lt2$ (this is all the phase-plane can give). And? This does not imply that $f$ meets zero in finite time. – Did Jun 17 '13 at 07:14
  • Sorry; I wasn't thinking of the phase portrait. Since $\frac{|y|}{1+y^2} \le \frac{1}{2}$ for all $y$, we have $f'(x) \le \ln a + \ln \frac{1}{2} = \ln \frac{a}{2}$. Hence we hit zero in finite time (if $a<2$). – copper.hat Jun 17 '13 at 07:18
  • @Plom: Your equation is equivalent to a differential equation. If $a \ge 2$ there is a nice constant solution. If $a < 2$ there is no solution (that is, once which remains positive). – copper.hat Jun 17 '13 at 07:21
  • @copper.hat This is a different argument (not visible in the phase-plane), which indeed implies the result (and is the basis of achille's answer). – Did Jun 17 '13 at 07:24
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Let $f(x)$ be any solution of the DE for any $a$ over all $\mathbb{R}$:

$$e^{f'(x)} = a \left|\frac{f(x)}{1+f(x)^2}\right|\tag{*}$$

Notice the L.H.S $> 0$ for any choice of $f(\cdot)$ and $x$. This means we must have $a > 0$ and $f(x) \ne 0$ for any $x$ and hence $f(\cdot)$ is having the same sign over all $\mathbb{R}$. Let us consider the case $f(0) > 0$ first. We have:

$$\begin{align} &1 + f'(x) \le e^{f'(x)} = a \frac{f(x)}{1+f(x)^2} = \frac{a}{f(x) + f(x)^{-1}} \le \frac{a}{2}\\ \implies & f'(x) \le \frac{a-2}{2}, \text{ for all } x \in \mathbb{R}\\ \implies & f(x) \le f(0) + \frac{a-2}{2} x, \text{ for all } x \in [0,\infty) \end{align}$$

If $a < 2$, this leads to the contradiction that $f(x)$ changes sign before $2 f(0)/(2 - a)$.

When $f(0) < 0$ and $a < 2$, apply the same argument above to $-f(-x)$, we see $f(x)$ changes sign somewhere between $0$ and $2 f(0)/(2-a)$ again. From these we can conclude if the DE $(*)$ has a solution, we must have $a \ge 2$.

If $a$ is indeed $\ge 2$, it is trivial to see $(*)$ has a constant solution. This implies the desired minimum value of $a$ is $2$.

achille hui
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  • i dont get why it changes sign b4 2f(0)/(2-a) :S – Plom Jun 17 '13 at 07:26
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    $$f(\frac{2f(0)}{2-a} + \epsilon) < f(0) + \frac{a-2}{2}\frac{2f(0)}{2-a} = 0.$$ The implies $f(x)$ change sign before or at $2f(0)/(2-a)$. The inequality is actually strict unless $f'(x) = 0$ and $f(x) = 1$. So we know $f(x)$ actually change sign before $2f(0)/(2-a)$. – achille hui Jun 17 '13 at 07:33
  • aaaaa, thats the value for which RHS becomes 0, okay, i got it, thanks a bunch – Plom Jun 17 '13 at 07:35