A simple quadratic flow model leads to the following apparently simple equation

$$y(t)=e^{-\frac{t}{\tau y(t)}}$$

where the flow, $y$ is a function of time, $t$ and $\tau $ is a constant.

But is there a closed form solution for $y(t)$ just in terms of $t$ and $\tau$ ?

Or can this only be solved by numerical means ?

  • 1
    $\begingroup$ It's not a differential equation. I would try to solve $$x=e^{a\over x}$$ for $x$ in terms of $a$. In the solution substitute $-{t\over\tau}$ for $a$ and that's the function $y(t)$ you're after. However, I don't see how to tackle this equation apart from numerically. Equivalent equation: $$x\ln x=a$$ $\endgroup$ – Heimdall May 20 '16 at 0:49

There is a "closed-form" expression in terms of the Lambert W Function.

We begin with the equation

$$y(t)=e^{-t/\tau y(t)} \tag 1$$

Let $z=-t/\tau$ and let $W=\frac{z}{y(t)}$. Then, upon rearranging $(1)$ we find that

$$z=We^W \tag 2$$

Noting that $(2)$ defines the Lambert W, we have immediately that

$$\bbox[5px,border:2px solid #C0A000]{y(t)-\frac{-t/\tau}{W\left(-t/\tau\right)}}$$

Special attention is required to determine the domain for which $y(t)$ is real and single valued.

  • $\begingroup$ Impressive solution, thank you -very much. And I get it, but trying to wrap my head around what it might mean physically in my application, and how I'm going to work with it. For my practical purpose $y(t)$ is a real valued function, $\tau$ is a real number, and $t$ is time $>0$, so then the values of $W$ must be real, right? $\endgroup$ – docscience May 20 '16 at 14:58
  • $\begingroup$ Yes, $W$ must be real $\endgroup$ – Mark Viola May 20 '16 at 15:51
  • $\begingroup$ You're welcome. My pleasure. -Mark $\endgroup$ – Mark Viola May 20 '16 at 15:58

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