Evaluating decay rate with trigonometric explanation According to the equation 4, 
$$\phi(0,t)= \frac{A_0}{(1+\frac{2t^2}{R^4})^{3/4}}\cos \left(\sqrt2 t+ \frac{3}{2}\tan^{-1}\left[\frac{\sqrt2 t}{R^2}\right]\right)\tag{1}$$
what conditions makes, $$\cos \left(\sqrt2 t+ \frac{3}{2}\tan^{-1}\left[\frac{\sqrt2 t}{R^2}\right]\right)=1$$
so the equation (1) will be 
$$\phi(0,t)= \frac{A_0}{(1+\frac{2t^2}{R^4})^{3/4}}$$
The author used the article reference to establish the equation 
$$\frac{1}{2} \Gamma_{lin}= \frac{1}{\tau_{linear}} \approx \frac{1.196}{\omega_{mass}} \approx \frac{.846}{R^2}$$
but I didn't get any argument there, can you explain this a bit please.
 A: You don't need to actually find such $t$ to make the inside equal to points where $cos(\theta)=1.$ You only need that as $t\to \infty$ there will be a sequence of $t$ values for which $\theta =2n\pi$. From your expression,
$$\theta=\sqrt2 t+ \frac{3}{2}\tan^{-1}\left[\frac{\sqrt2 t}{R^2}\right],$$
and the arctan part added is bounded, while $\sqrt2t\to \infty$ This means $\theta \to \infty$ as $t \to \infty$, and $\theta$ is a continuous function of $t$, which forces $\theta$ to take on infinitely many values of the form $\theta =2n\pi$, so that your cosine term will be $1$ for all those values. It will similarly be $-1$ infinitely often, making the whole $\cos \theta$ oscillate in $[-1,1]$ (hitting both extremes infinitely often as $t \to \infty$) So the damping factor in front is definitely the best for the absolute value of your function $\phi(0,t).$
ADDED: The OP requests in comments more explanation about $\cos \theta=1$ happening infinitely often as $t \to \infty$. For each $t$ there is a value of $\theta(t)$ given by the above formula (which was the inside of the function cosine in the OP). This function $\theta(t)$ is a continuous function of $t$, and the above argument shows it goes to $\infty$ as $t \to \infty.$ Now imagine the number line with a "tick mark" at every point where the cosine is $1$ (these are the points $2\pi n$ since cosine is periodic and is $1$ only at $0$ in the interval $[0,2\pi)$. Now $\theta(t)$ starts out on this line at some point $\theta(0)$, and as $t$ increases, $\theta(t)$ may move forward or backward at any one time (because of the arctangent in the formula), but eventually $\theta(t)$ moves all the way from $\theta(0)$ to $+\infty$. So in the process, $\theta(t)$ will cross through all the tick marks to the right of $\theta(0)$, so that at each such $t$ value where $\theta(t)$ crosses a tick mark we will have $\cos(\theta(t))=1.$
A: Well, if you read carefully, the authors clearly state that

$$
\phi(0,t) = \frac{A_0}{\left(1 + \frac{2 t^2}{R^4}\right)^{3/4}}\cos\left(\sqrt{2}t + \frac{3}{2} \tan^{-1} \left[\frac{\sqrt{2}t}{R^2}\right]\right), \tag{4}
$$
  whos envelope of oscillation is given by $\phi(0,t) = A_0/(1 + 2t^2/R^4)^{3/4}$, which reaches $1/e$ of its initial value in a time given by $\mathcal{T}_{linear} \simeq .836\omega_{mass} R^2$. This yields the linear decay width
  $$
\frac{1}{2}\Gamma_{lin} = \frac{1}{\mathcal{T}_{linear}} \simeq \frac{1.196}{\omega_{mass}R^2} \simeq \frac{.846}{R^2}. \tag{5}
$$

The key here is the statement "which reaches $1/e$ of its initial value at time given by..."
The time where the oscillon reaches a fraction $1/e$ of its initial value is
$$
\frac{A_0}{e} = \frac{A_0}{\left(1 + \frac{2 t^2}{R^4}\right)^{3/4}}
$$
Solving for $t$,
$$
t = \left(\frac{e^{4/3} - 1}{2}\right)^{1/2} R^2.
$$
Now, using that $\omega_{mass} = \sqrt{2}$, we have
$$
\mathcal{T}_{linear} = \frac{1}{\sqrt{2}} \left(\frac{e^{4/3} - 1}{2}\right)^{1/2} \omega_{mass} R^2 \simeq 0.836 \omega_{mass}R^2,
$$
and eq. (5) follows.
EDIT
Lets take a look of the graph of $\phi(0,t)$:
$\hskip1in$!
It's clear that the oscillation decays, with a decay rate of $A_0/(1 + 2t^2/R^2)^{3/4}$. If we plot this decay rate,
$\hskip1in$!
The dashed curves are the envelopes of the field $\phi(0,t)$. What the authors are definining as $\mathcal{T}_{linear}$, is the point in time where the envelope has reduced it's amplitude by $1/e$. In this case, $2/e \simeq 0.736$:
$\hskip1in$
