Recently in my problem solving class we had the question: If $f: [0,\infty) \to \Bbb R$ is a twice differentiable function such that $f''(x) + e^x f(x) = 0$, then prove that $f$ is bounded.

The solution I found was to create a function $g(x)=f(x)^2 + e^{-x}f'(x)^2$. Looking at $g'$, we have $$g'=2ff' -e^{-x}f'^2 + 2e^{-x}f'f''$$ $$g'=-e^{-x}f'^2+2ff'+2e^{-x}f'f''$$ $$g' = -e^{-x}f'^2 +2e^{-x}f' \left(e^xf + f'' \right) = -e^{-x}f'^2 \leq 0$$ Thus we have that $g$ is a decreasing function. This means that for $x>0$, we have $$g(x) \leq g(0)$$ Since $g(0)=R$ for some constant $R$, we have $$R \geq g(x)=f(x)^2+e^{-x}f'(x)^2 > f(x)^2$$ thus $f^2$ is bounded, so $f$ itself must be bounded.

It is not so hard to show that this works if we replace $e^x$ with any differentiable function $k(x)$ that is positive and increasing on $[0,\infty)$. Thus we have the more general result that if $f: [0,\infty) \to \Bbb R$ is a twice differentiable function such that $f''(x) + k(x) f(x) = 0$ on $[a,\infty)$ for some $a \geq 0$ and $k(x)$ is positive, increasing, and differentiable on $[a,\infty)$, then $f$ is bounded.

However, my teacher also gave us a similar problem: if $f: [0,\infty) \to \Bbb R$ is a twice differentiable function such that $f''(x) + \frac{1}{x} f(x) = 0$ on $[a,\infty)$ for some $a > 0$, then $f$ is bounded above.

Unfortunately for us, our above result does not apply, since $\frac{1}{x}$ is decreasing, rather than increasing on $[a,\infty)$. In fact, defining $g(x)$ analogously as $g(x)=f(x)^2+xf'(x)^2$ gives us that $g' \geq 0$, so the technique fails. All similarly defined functions in terms of $f(x)$ and $f'(x)$ that I've tried have also failed to tell me anything about the boundedness of $f$. However, since there is a degree of intuition required for competition type problems like this, it's certainly very very likely that I missed one that does work.

I will mention that defining $h(x) = \frac{1}{x}f(x)^2 + f'(x)^2$ does tell us that $f'$ must be bounded, but I couldn't get anything about the boundedness of $f$ out of this.

I've been a member of stackexchange for a couple of months now and I really wanted my first posted question to be something really interesting and I believe that I found one! Since the problem involving $e^x$ generalizes very nicely to all increasing, positive, differentiable functions, I imagine that a solution of the problem involving $\frac{1}{x}$ would also generalize to all decreasing, positive, differentiable functions, so a proof of this could possibly provide a very interesting result concerning boundedness of functions that satisfy relatively simple differential equations. I've hit a bit of an impasse in solving this problem and I would like to know if anyone has any suggestions on how to prove this.


It's difficult to prove because $f$ can be unbounded from above. One of the solutions of the equation $f''(x) + \frac{1}{x} f(x) = 0\ $ is $f(x)=\sqrt{x} J_1\left(2 \sqrt{x}\right)$, where $J_1$ is the Bessel function of the first kind. From the asymptotic $$ J_\alpha(x)\approx \sqrt{\frac{2}{\pi x}} \cos \left( x-\frac{\alpha\pi}{2} - \frac{\pi}{4} \right),\quad x\to+\infty, $$ it follows that $$ f(x) \approx x^{1/4}\sqrt{\frac{1}{\pi}}\cos \left( 2\sqrt{x}-\frac{3\pi}{4} \right). $$

  • $\begingroup$ That certainly explains why proving it was so hard! I've asked my teacher and we've both examined the problem in a bit more detail and eventually determined that the site he got the problem from put the $frac{1}{x}$ in front of the wrong term. Thanks for the help Andrew! $\endgroup$ – Michael Banaszek Oct 23 '11 at 3:56

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