# Controlling subsolutions of a second order linear ODE

Let $f:[0,\infty) \to \mathbb{R}$ obey the differential inequality $$f'' - 2\alpha f' + 2\alpha f \leq 0$$ where $0 < \alpha < 2$ is some constant. If $f(0) = 0$ and $f'(0) = 1$, can I say that $f(x) < e^x - 1$ for some $x$?

Note that the solution to the corresponding differential equation oscillates since the characteristic equation has complex roots (call this solution $g$). Thus we can certainly say $g(x) < e^x -1$ for infinitely many $x$. My first thought was to try to control $f$ by $g$ a la Gronwall's inequality. However, I was recently shown that the analogue to Gronwall for degree two differential equations doesn't hold.

Any ideas would be welcome. Also, any good references for differential inequalities that might help me solve this problem are equally welcome.

• Courtesy of google: see Theorem 2 of google.com/… Commented Jan 11, 2018 at 14:39
• This theorem might be useful, but it doesn't address the question. It says that the solution $g$ dominates the sub-solution $f$ on neighborhood of 0. However, the solution $g$ lies about $e^x - 1$ for awhile before it dips below, so it's not enough to control $f$ by $g$ on a small neighborhood of 0. In fact, the length of the interval on which $g$ lies above $e^x - 1$ seems to go to infinity as $\alpha$ goes to 2, so we would need to control the length of the interval on which $f$ is dominated by $g$.
– H_R
Commented Jan 11, 2018 at 17:06

Let us actually show more than requested, namely, that $f(x_n)\le0$ for some sequence $(x_n)$ converging to $\infty$ and all natural $n$. Moreover, the initial conditions, $f(0) = 0$ and $f'(0) = 1$, will not be used or needed.

Indeed, suppose that, to the contrary, the statement in the first sentence of this answer is false. Then $$\text{there is some real x_*>0 such that f(x)>0 for all x>x_*.} \tag{1}$$ Let $a:=\alpha$. Let $h:=f'' - 2a f' + 2a f$, so that $h\le0$. Let $$r(x):=f(x)/e^{ax}. \tag{2}$$ Then the equation $h=f'' - 2a f' + 2a f$ can be rewritten as \begin{equation*} r''+b^2r=g,\quad\text{where}\ g(x):=h(x)/e^{ax}\le0 \end{equation*} and $b:=\sqrt{(2-a)a}>0.$ In view of (1) and (2), $r>0$ on $(x_*,\infty)$. So, $r''=-b^2r+g<0$, so that $r$ is a strictly positive concave function on $(x_*,\infty)$. It follows that $r'\ge0$ and hence $r$ is nondecreasing on $(x_*,\infty)$; indeed, if $r'(x_{**})<0$ for some real $x_{**}>x_*$, then the concavity of $r$ implies $r(x)\le r(x_{**})+r'(x_{**})(x-x_{**})\to-\infty$ as $x\to\infty$, which contradicts the positivity of $r$. So, $r(x)\uparrow R$ and $r'(x)\downarrow k$ as $x\to\infty$, for some $R\in(0,\infty]$ and $k\in[0,\infty)$.

Take now any $\rho\in(0,R)$ and let \begin{equation*} d:=r-s,\quad\text{where }s(x):=\rho\sin bx. \end{equation*} Then $d>0$ on $[x_{***},\infty)$ for some real $x_{***}>0$. Also, $d''+b^2d=g$, so that $d''=-b^2d+g<0$ and hence $d$ is a positive concave function on $[x_{***},\infty)$, so that $d'(x)\downarrow \ell$ as $x\to\infty$, for some $\ell\in[0,\infty)$. Therefore, $b\rho\cos bx=s'(x)=r'(x)-d'(x)\to k-\ell$ as $x\to\infty$, which is absurd. Thus, (1) is false, and hence the statement in the first sentence of this answer is true.

• It looks like I have been able to improve the previous partial affirmative answer to a complete one now Commented Jan 11, 2018 at 20:06
• Why does $r(x) \to R$ as $x \to \infty$?
– H_R
Commented Jan 12, 2018 at 1:32
• @H_R : Because $r$ is positive and concave and hence nondecreasing (make a picture). I have also added a detail on this. Commented Jan 12, 2018 at 4:39
• Oh I'm sorry, I didn't see that $R$ could be infinite.
– H_R
Commented Jan 12, 2018 at 5:36
• I have updated the answer to show more clearly that actually more is proved than requested. Commented Jan 12, 2018 at 14:51