I'm interested in solving the differential equation

$$ a\frac{d y}{dx} = -\frac{1}{y(x)} e^{-\frac{1}{y(x)}} $$ where $a>0$.

$Mathematica$ can solve this but it gives the answer in terms of InverseFunctions of exponential integrals. I'm only interested in the asymptotic behavior of $y(x)$ as $x \to \infty$ but I'm not sure how to proceed with this sort of a problem.

Edit: The suggestion was to transform to $z(x) = \frac{1}{y(x)}$. In terms of this new variable, the problem becomes $$ a \frac{dz}{dx} = z(x)^3 e^{-z(x)} $$ As far as initial conditions are concerned, let's say $z(0) = 1$. Even in these new variables, I don't really know how to proceed.

  • 1
    $\begingroup$ 1) Do you have an initial condition in mind? Without that there's not a single solution but a family of them. 2) The most obvious thing to do with an ODE like this is to consider $z(x)=1/y(x)$ as the dependent variable instead. That gives $-\frac{a}{z^2}\frac{dz}{dx}=-ze^{-z}$ which is separable. $\endgroup$ Sep 28, 2016 at 0:06
  • 1
    $\begingroup$ At first glance, $\lim_{x\to\infty}y$ is finite, for if $y(x)$ was very large, the derivative would be near $0$. So I'm thinking the asymptotic behavior is $$y(x)\sim c$$ $\endgroup$ Sep 28, 2016 at 0:16
  • $\begingroup$ As Semiclassical suggests, you can just use separation of variables on the updated equation. $a\frac{dz}{dx}=z^3e^{-z}$, so $dx = az^{-3}e^zdz$; now integrate. You'll get an implicit equation for $z$ (and thus $y$), but it's hard to do much better than that. $\endgroup$ Sep 28, 2016 at 0:25
  • $\begingroup$ I understand that part but it doesn't really give any further insights into the asymptotic behavior. It seems to me that the behavior is finite as @SimpleArt suggested, but I still don't see a formal way of showing that. $\endgroup$
    – Aegon
    Sep 28, 2016 at 0:30
  • $\begingroup$ @Aegon I'll check my answer tomorrow because it's too late over here. Wait... $\endgroup$ Sep 28, 2016 at 4:29

2 Answers 2


Motivated by Felix Marin's response yesterday, I was able to figure this out for a more general class of problems. I am interested in solving $$ a \frac{dy}{dt} = - y(t)^\alpha e^{-1/y(t)} $$ for $\alpha \in \mathbb{Z}$, $a>0$, and in the limit $t\to \infty$. Let us also consider the initial condition $y(0) = y_0$. The solution proceeds as follows $$ a \frac{d y}{d t} = - y^\alpha e^{-1/y} \implies \int_{y(0)}^{y(t)} dz\,\frac{e^{1/z}}{z^\alpha} = -\frac{t}{a} $$ Change variables, $z = -\frac{1}{x}$ $$ (-1)^\alpha \int_{-\frac{1}{y_0}}^{-\frac{1}{y(t)}} dx\, x^{\alpha - 2} e^{-x} = - \frac{t}{a} \implies \int_{-\frac{1}{y_0}}^{\infty} dx\, x^{\alpha - 2} e^{-x} - \int_{-\frac{1}{y(t)}}^{\infty} dx\, x^{\alpha - 2} e^{-x} = (-1)^{\alpha + 1}\frac{t}{a} \implies \Gamma\left(\alpha - 1,-\frac{1}{y_0}\right) - \Gamma\left(\alpha - 1,-\frac{1}{y(t)}\right) = (-1)^{\alpha + 1}\frac{t}{a} $$

Now, we are interested in the behavior of $y(t)$ as $t\to \infty$. We note that $\forall n \in \mathbb{Z}$, $$ \Gamma(n,z) \sim \frac{e^{-z}}{z^{1-n}} \quad z \to -\infty $$

Since we want the LHS to blow up as well, we can then see that the appropriate limit is $y(t) \to 0^+$ so that $$ \Gamma\left(\alpha - 1, - \frac{-1}{y(t)} \right) \sim (-1)^{\alpha - 2} y(t)^{\alpha - 2} e^{1/y(t)} $$ So, asymptotically, $$ y(t)^{\alpha - 2} e^{1/y(t)} \sim \frac{t}{a} $$

We can invert this to find

$$ y(t) \sim \frac{-1}{W\left(- \frac{\left(\frac{t}{a} \right)^{\frac{1}{2 - \alpha}}}{\alpha - 2}\right)} $$

where $W(z)$ is the Lambert-W function. We can push this even further by considering that

$$ W(x) \sim \log(x) - \log(\log(x)) \quad x\to \infty $$

So, $$ y(t) \sim \frac{1}{\log(\frac{t}{a}) + (\alpha - 2)\log(\log(\frac{t}{a}))} $$

  • $\begingroup$ very well done (+1) $\endgroup$
    – tired
    Sep 29, 2016 at 9:20

Here's a possible start, where I wander around for a while and don't come to any satisfactory conclusion.

From $a\frac{d y}{dx} = -\frac{1}{y(x)} e^{-\frac{1}{y(x)}} $ we get $yy' =-e^{-1/y} $ or $-yy' =e^{-1/y} $.

Let $z = -1/y$, or $y = -1/z$ so $y' = z'/z^2$ and $yy' =(-1/z)(z'/z^2) =-z'/z^3 $.

This becomes $z'/z^3 =e^z $, so $z' = z^3 e^z $.


$\begin{array}\\ z'' &=3z^2z'e^z + z^3z'e^z\\ &=z'z^2e^z(3+z)\\ &=z'z^2(z'z^{-3})(3+z)\\ &=z'^2z^{-1}(3+z)\\ \text{or}\\ zz'' &=z'^2(3+z)\\ \end{array} $

At this point, we could try a power series for $z$ and get a recurrence for the coefficients. I'm tired, so I'll try something else.

Let $z = x^aw$ where $w(0) \ne 0$. $z' = x^aw'+ax^{a-1}w$ and

$\begin{array}\\ z'' &=x^aw''+ax^{a-1}w'+a((a-1)x^{a-2}w+x^{a-1}w')\\ &=x^{a-2}(x^2w''+axw'+a((a-1)w+xw')\\ &=x^{a-2}(x^2w''+2axw'+a(a-1)w)\\ \end{array} $

From $zz'' =z'^2(3+z) $ we get $(x^aw)x^{a-2}(x^2w''+2axw'+a(a-1)w) =(x^aw'+ax^{a-1}w)^2(3+x^aw) $ or $x^{2a-2}w(x^2w''+2axw'+a(a-1)w) =x^{2a-2}(xw'+aw)^2(3+x^aw) $ or $w(x^2w''+2axw'+a(a-1)w) =(xw'+aw)^2(3+x^aw) $.

I'm not sure what to do from here, and everything looks messy, so I'll stop.


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