Dtyjlui

I am interested in the following model: $$y'=frac{y^alpha}{e^{y}}$$ with $alpha>1, y(0)=1$ (this is a growth model of knowledge with self-amplification and exponentially increasing difficulty a la Rescher's law). Numerically it is easy to see that it follows a logistic-like curve of initial accelerating growth that levels off.

My question is whether $y$ properly asymptotes?

The RHS is always positive and nonzero, but declines strongly with increasing $y$. Is the positivity enough to prove lack of asymptoticity?

Trying to solve the equation gives me the expression$$t+C = int_1^y e^u u^{-alpha} du$$ where symbolic integrators happily tell me the RHS is $(-1)^alpha [Gamma(1-alpha, -y) - Gamma(1-alpha,-1)]$ (with the incomplete gamma function) or $E_alpha(-1)-y^{1-alpha}E_alpha(-y)$ (with the exponential integral $E_n(x)=int_1^infty e^{-xu}u^{-alpha}du$). Both answers are confusing since the exponential integral looks divergent for negative $x$ and at least Matlab refuses to calculate the incomplete gamma function for doubly negative arguments. Numerically the integral is of course well behaved.

At the same time, squinting hard, if $t+C approx Gamma(y)$ then we have an answer since we know $y$ grows as the inverse of the factorial of time, i.e. absurdly slowly yet without bound.

I have a feeling that there is either a simple way of proving lack of asymptote, or that one can juggle the special functions into something more illuminating, that I am simply not seeing.

We have $y(t)>1$ for all $t>0$. Then $y'ge e^{-y}$. Multiplying by $e^y$ we get $(ey)'ge1$, and integrating $y(t)gelog(e+t)$. You can get better growth estimates now from $y'gelog(e+t)^alpha e^{-y}$.

With the condition $y(0)=1$ or equivalently $t(1)=0$ :
$$t(y) = int_1^y e^u u^{-alpha} du tag 1$$
Of course a closed form for $t(y)$ is :
$$t(y)=(-1)^alphaGamma(1-alpha,-y)-(-1)^alphaGamma(1-alpha,-1)$$
For $alpha>1$ $Gamma(1-alpha,-y)$ is complex, but with the complex coefficient $(-1)^alpha$ one obtain real $t(y)$ on a convenient branch. This seems complicated, but there is no need to use the Incomplete Gamma function to compute and draw $t(y)$ and $y(t)$. Numerical calculus of the integral is much simpler.

To draw $y(t)$, along numerical integration after each increment of $y$ one know the computed value of $t$. Then simply plot $(t,y)$.

ASYMPTOTIC STUDY :

We could use more or less complicated set formulas issued from the knowledge of the properties of the Incomplete Gamma function. More directly by successive partial integration of $(1)$ :
$$t(y)sim frac{e^y}{y^alpha}left(1+frac{alpha}{y}+frac{alpha(alpha+1)}{y^2}t(y)+...+frac{alpha(alpha+1)(alpha+2)…(alpha+n-1)}{y^n} right) qquad ygg (alpha+n)$$

Due to $e^y$, smaller terms related to the low limit of the integral are not included.

The effect of the number or terms is shown on the next figure, for example in the case $alpha=2$ .

Of course, all those asymptotic approximates are false for small values of $t$ .

Inverting the asymptotic formulas for $y(t)$ instead of $t(y)$ is arduous.

Even with the leading term only $t(y)sim frac{e^y}{y^alpha}$ the inverse function is a special function :
$$y(t)sim -alpha: W_{-1}left(-frac{1}{alpha:t^{1/alpha}} right)$$
This involves the Lambert W function. The function $W(X)$ is multi valuated for $X<0$. In the present case the branch $W_{-1}(X)$ where $W_{-1}(Xto 0^-)to -infty$ , so that $yto+infty$ .

The asymptotic formula $y(t)sim -alpha: W_{-1}left(-frac{1}{alpha:t^{1/alpha}} right)$ is accurate only for very large values of $t$ , as shown on the next figure :

IN ADDITION :

After the Julián Aguirre's comment, consider his proposed equation :
$$y'geq left(log (e+t)right)^alpha$$
which leads to :
$$y(t)geq lnleft(e+int_0^t left(ln(e+tau) right)^a dtauright)$$
Again a closed form for the integral requires the Incomplete Gamma function. Without using it and more simply the numerical integration allows to draw the lower limit for the asymptotic curve (red curve added on the last figure).