We know that solutions exist for equations of the following variety: $$ye^y=x \iff y=W(x)$$ Where W is the Lambert W function. We can augment the problem slightly, and ask if there exist solutions for equations of the following form: $y^2e^y=x$ Taking the square root and dividing by 2 on both sides allows us to obtain a form in which we can use Lambert W once more.

However, do there exist any solutions for equations of the following form: $$(y^2+\epsilon)e^y=x$$ Where $\epsilon$ is some constant.

Scott and Man have authored a paper on equations of the form $$(y-a)(y-b)+e^y=0$$ and so this may be of some help.

Is anyone aware of a paper that shows solutions to this problem?

  • $\begingroup$ Check this. $\endgroup$ – Mhenni Benghorbal Aug 27 '13 at 22:18
  • $\begingroup$ @MhenniBenghorbal That link is to a linear expression equal to an exponential. This question is different, involving setting a quadratic times an exponential (in $y$) to $x$ and looking to find $y$ as a function of $x$. Seems it would be more difficult... $\endgroup$ – coffeemath Aug 27 '13 at 23:25
  • $\begingroup$ As far as I can tell the equation you wrote is the definition of the W function. It is the inverse of the function f(w) = we^w. Like many inverse functions it is not single branched, but can be on a properly restricted domain. So on that domain it does indeed have solutions. How to find them is a different question. $\endgroup$ – Betty Mock Aug 28 '13 at 4:28

Consider the function $f(y) = (y^2 + c)e^y$. Then $f$ is continuous and has limiting behaviour $\lim_{y \to -\infty} f(y) = 0$ and $\lim_{y \to \infty} f(y) = \infty$. By the intermediate value theorem, there is a solution to $f(y) = x$ for any $x \in (0, \infty)$.

This misses a little bit of information, as if $c < 0$ then there are solutions for some $x < 0$. The same reasoning will work for us. In particular, when $c < 0$ then $f$ takes its minimum at $y_m = -1 + \sqrt{1 - c}$.

We can see this by taking the derivative and setting equal to $0$. The solutions to $f'(y) = 0$ are $y = -1 \pm \sqrt{1 - c}$. The left point $y_M = -1 - \sqrt{1 - c}$ is a maximum. The right point $y_m = -1 + \sqrt{1 - c}$ is a minimum. There are no other extrema. For ease, denote the local minimum $f(y_m) = m$ and the local maximum $f(y_M) = M$. With this, we can classify all solutions.

When $c < 0$, there are solutions to $f(y) = x$ for all $x \geq m$.

  1. There is exactly one solution to $f(y) = m$.
  2. For $x \in (m, 0]$ there are exactly two solutions to $f(y) = x$. Further, one solution is in $(y_M, y_m)$ and the other is in $(y_m, \infty)$.
  3. For $x \in (0, M)$ there are exactly three solutions to $f(y) = x$. One solution is in $(-\infty, y_M)$, another in $(y_M, y_m)$, and the third in $(y_m, \infty)$.
  4. There are exactly two solutions to $f(y) = M$. One is $y_M$, and the other is in $(y_m, \infty)$.
  5. There is exactly one solution to $f(y) = x$ for all $x > M$.

One can state analogous analysis for $c \geq 0$.

This has all been for the existence of solutions. Numerically, it is relatively easy to find solutions, especially if we restrict the domain in $y$ to $[y_m, \infty)$. In this region, $f(y)$ is strictly increasing and one can approximate $f(y)$ by $y^2 e^y$, use Lambert's $W$ function, and proceed by bisection to compute solutions.


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