On a differential inequality

Question

Let $A>0$. If $f$ satisfies the differential inequality $f^{\prime\prime}(t)+f(t)≥A$, and $y$ is the solution to the ODE $y^{\prime\prime}(t)+y(t)=A$ with $y(0)=f(0)$ and $y^{\prime}(0)=f^{\prime}(0)$, then \begin{equation}\label{GP} \begin{cases} f(t)\leq y(t) \quad \text{for all $t<0$}\\ f(t)\geq y(t) \quad \text{for all $t>0$}. \end{cases} \end{equation}

Question: How can I prove this result?

Answer 1

I don't think the result is true...

The general solution for the function $y(t)$ is $$y(t)=C_1\sin(t) + C_2 \cos(t) +A$$

Let $$g(t) \equiv f(t)-y(t)$$ Then we have $$ g''(t)+g(t) \ge 0 $$ with initial conditions $$ g(0)=g'(0)=0 $$

if $f(t) $ was a function satisfying $$ f''(t)+f(t)=A+B $$ for some $B\ge 0$

Then its general solution would be $$f(t)=C_3\sin(t) + C_4 \cos(t) +A+B$$

So that $$g(t) = (C_3-C_1)\sin(t) + (C_4 -C_2)\cos(t) +B $$

$$ g'(0)=0 \implies C_3-C_1=0 $$

$$ g(0)=0 \implies C_4-C_2=B $$

So

$$ g(t)=B \Big( 1+cos(t) \Big)$$

Notice that $g(t)\ge 0$ so that $$ f(t) \ge y(t) $$

for all $t$.

Answer 2

Starting from $g''+g\ge 0$, which had been shown by @WW1 and @amsmath.

The roots of characteristic equation of corresponding equation are $\lambda_\pm \pm i$, thus the inequality can be rearranged by \begin{equation} g''-\lambda_- g' - \lambda_+ (g'-\lambda_- g)\ge 0 \end{equation} Then by defining $G=g'-\lambda_- g$, we obtain \begin{equation} G'-\lambda_+ G\ge 0\to \frac{d}{d x}(e^{-x\lambda_+}G)\ge0 \end{equation} which gives \begin{equation} G\ge G(0)e^{x\lambda_+},\qquad G(0)=g'(0)-\lambda_- g(0) \end{equation} namely, $G(0)$ depends on initial conditions. Now we shall solve \begin{equation} g'-\lambda_- g \ge G(0)e^{x\lambda_+}\to \frac{d}{d x}(e^{-x\lambda_-}g)\ge G(0)e^{x(\lambda_+-\lambda_-)} \end{equation} which gives \begin{equation} g\ge \frac{G(0)}{\lambda_+-\lambda_-} (e^{x\lambda_+}-e^{x \lambda_-}) =G(0) \sin(x). \end{equation} Therefore we arrive at \begin{equation} f\ge G(0) \sin(x)+y,\qquad y=A +[f(0) -A]\cos(x)+f'(0)\sin(x) \end{equation} This should be the final answer.