Prove $-(\frac{kv+g}{k^2})e^{-kt}-\frac{g}{k}t+(\frac{kv+g}{k^2})0$

Question

How do you prove the inequality $-(\frac{kv+g}{k^2})e^{-kt}-\frac{g}{k}t+(\frac{kv+g}{k^2})<vt-\frac{1}{2}gt^2$ for $k,v,g,t>0$ and $-(\frac{kv+g}{k^2})e^{-kt}-\frac{g}{k}t+(\frac{kv+g}{k^2})>0$ ? An elementary proof is preferred.

This left side of the inequality represents the height of a particle travelling under air resistance. The right side of the inequality represents the height of a particle identical in all ways, except it does not have air resistance.$k$ is a parameter determining the force of air resistance. $v$ is the initial velocity of the particle. $g$ is the gravitational acceleration of the particle. $t$ represents the time that the particles have travelled.

Thanks in advance.

Answer 1

Rearranging the inequality yields

$$\left(\frac{kv+g}{k^2}\right) \left(1 - e^{-kt}\right) < \left(\frac{kv+g}{k}\right)t - \frac{1}{2}gt^2.\tag{1}$$

So it's clear that it can only hold for small enough $t$, since the right hand side becomes negative for large $t$.

For small positive $t$, rewriting both sides as the integral of their derivatives shows the inequality follows from

$$\frac{kv+g}{k}e^{-kt} < \frac{kv+g}{k} - gt$$

or

$$gt < \frac{kv+g}{k}(1 - e^{-kt}).$$

Now, $1 - e^{-x} > x - \frac{x^2}{2}$ for $x > 0$, so

$$\frac{kv+g}{k}(1-e^{-kt}) > \frac{kv+g}{k}\left(kt - \frac{k^2t^2}{2}\right) = gt + kvt -\frac{(kv+g)k}{2}t^2$$

and $$kv - \frac{(kv+g)k}{2}t > 0$$ yields the inequality for

$$t < \frac{2kv}{(kv+g)k}.$$

The shape of the functions in $(1)$ makes it clear that there is exactly one $t_1 > 0$ at which equality holds, so $(1)$ holds for $0 < t < t_1$. Finding the (approximate) value of $t_1$, if desired, would probably best be done by a root-finding algorithm like Newton's method.

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In response to the edit that the inequality should hold for

$$\varphi(t) := \frac{kv+g}{k^2}\left(1 - e^{-kt}\right) - \frac{g}{k}t > 0,$$

we see that the positive zero of $\psi(t) := vt - \frac12 gt^2$ is $t_0 = \frac{2v}{g}$. A computation (see below) shows that $\varphi(t_0) < 0$, so the positive zero $t_1$ of $\varphi$ is smaller than $t_0$, whence $\psi(t_1) > 0$.

Since $\psi - \varphi$ has only one positive zero⁽¹⁾, it follows that $\varphi(t) < \psi(t)$ for $0 < t < t_1$.

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⁽¹⁾ $(\psi - \varphi)''(t) = (kv+g)e^{-kt} - g$. In particular, $(\psi-\varphi)''(0) = kv > 0$, and $(\psi-\varphi)''$ is strictly decreasing with asymptote $-g$.

So $\psi - \varphi$ is strictly convex and strictly increasing for $t < t_2 := \frac1k \log \frac{kv+g}{g}$, and strictly concave for $t > t_2$ (still strictly increasing for a while). A strictly concave function that is positive at $x$ can have at most one zero larger than $x$.

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Verification of $\varphi(t_0) < 0$:

$$\begin{align} \varphi(t_0) &< 0\\ \iff \frac{kv+g}{k^2}\left(1 - e^{-2kv/g}\right) &< \frac{2v}{k}\\ \iff 1 - e^{-2kv/g} &< \frac{2kv}{kv+g}\\ \iff 1-\frac{2kv}{kv+g} &< e^{-2kv/g}\\ \iff \frac{g-kv}{g+kv} &< e^{-2kv/g}. \end{align}$$

Now, if $kv \geqslant g$, that is obviously satisfied. Otherwise, let $x := \frac{kv}{g}$. We then need to see that

$$\frac{1-x}{1+x} < e^{-2x}.$$

This holds for $0 < x < 1$, since

$$\begin{align} \frac{d}{ds}\left((1+s)e^{-s} - (1-s)e^s\right) =s(e^s - e^{-s}) &> 0\\ \Rightarrow \int_0^x s(e^s-e^{-s})\,ds &> 0\\ \Rightarrow \left[(1+s)e^{-s} - (1-s)e^s\right]_0^x = (1+x)e^{-x}-(1-x)e^x &> 0\\ \iff (1+x)e^{-x} &> (1-x)e^x\\ \iff e^{-2x} &> \frac{1-x}{1+x}. \end{align}$$