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Consider $f\in C^2$ so that $$f''(x)+f(x)=-x\,g(x)f'(x), \ \forall x\ge0 $$ where $g(x)\ge 0$. Then ($\forall x\ge 0$)

  • (A) $f(x)^2+f'(x)^2$ is non-increasing,
  • (B) $f(x)^2<3f(0)^2+(2f'(0))^2$,
  • (C) $|f(x)|\le\alpha$, where $\alpha$ is a fixed real constant.
  • (D) $\lim_{x\to\infty} f(x)\sin\left(\dfrac{1}{x}\right)$ exists.

original task description

Answer Given: (A),(B),(C),(D)

What I have tried : I differentiated option (A) and then with the help of the given equation in the question I was able to show that it is less than zero . Hence I was able to infer that option (A) is correct .

Now coming to option (B). I assumed the function in option (A) to be $h(x)$.

$$ h(x) = f(x)^2 + f'(x)^2 $$

Since $h'(x)<0$ ,therefore $h(x)$ is a decreasing function

hence for all $x\ge 0$

$h(x)< h(0)$

which implies $(f(x))^2 + (f'(x))^2<(f(0))^2 + (f'(0))^2$

From the above expression we can get the second option .

Is my method correct ?. If not , please show the right method .

I am unable to get the the 3rd and 4th option .

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  • $\begingroup$ Non-increasing doesn't mean decreasing, so you should actually have $h'(x) \le 0$. This should still work for your solution in (B) $\endgroup$ – Dylan Jun 26 '18 at 14:03
  • $\begingroup$ (B) is not true for the zero solution $f\equiv0$. It may be a nitpick, but it is not excluded in the task. --- If $g$ falls fast enough to zero, perhaps with $g(x)=\exp(-x^2)$, is it possible that the amplitude does not fall to zero for $x\to\infty$? $\endgroup$ – Lutz Lehmann Jun 27 '18 at 11:40
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$h(x) = f(x)^2 + f'(x)^2$ is non-increasing (but not necessarily decreasing) on $[0, \infty) $ because $$ h'(x) = 2f(x) f'(x) + 2f'(x) f''(x) = -2 x g(x)f'(x)^2 \le 0 \, , $$ so (A) is true. Then $$ f(x)^2 \le h(x) \le h(0) = f(0)^2 + f'(0)^2 \le 3f(0)^2+(2f'(0))^2 \, , $$ which is (B) with $\le$ instead of $<$. As already stated in the comments, the example $f(x) = 0$ shows that equality can hold in (B).

(C) and (D) are true because $$ |f(x)| \le \sqrt{h(0)} =: \alpha $$ and $$ \left|\, f(x) \sin \left(\frac 1x\right) \right| \le \frac{\alpha}{x} \to 0 \text{ for } x \to \infty \, . $$

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  • $\begingroup$ This maybe a bit crass but I still want to ask . It is clear that f(x)^2 <=h(0) from the expression. How did you get the modulus ? . Thank you for the answering the question . $\endgroup$ – Alphanerd Jun 28 '18 at 5:02
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    $\begingroup$ @Alphanerd: $f(x)$ is a real number, therefore $|f(x)|^2 = f(x)^2 \le h(0)^2$. $\endgroup$ – Martin R Jun 28 '18 at 5:06
  • $\begingroup$ (Of course it should be $|f(x)|^2 = f(x)^2 \le h(0)$.) $\endgroup$ – Martin R Jun 28 '18 at 6:46

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