Solution to $f(x) = e^{axf(x)}$ ,where $ax$ is a constant

Question

I am trying to make an analytical expression of exponential taper.

$f(x) = e^{axf(x)}$

$a$ = constant

Even though $x$ is a variable of $f(x)$, can I deal with $x$ as a constant to solve the Lambert W function? then, set $ax=k$ and $f(x) = -\frac{1}{k}W(-k) $ ?

How can $f(x)$ be defined using lambert W function?

Thank you in advance,

Answer 1

You are almost there.

Consider $$y=e^{a x y} \implies y=-\frac{W(-a x)}{a x}=f(x)$$

Answer 2

Part $-1$.

Unfortunately, you can't always solve this equation for the variable $x$ using the Lambert -W function.

You can consider the following counterexample.

Let, $a=3, f(x)=\sin x$, then you have

$$ e^{3x\sin x}=\sin x$$

This is an example of a Transcendental equation. It is impossible to solve the roots of this equation in closed form using the Lambert $W$ function. Roots can only be expressed as approximate numbers.

---

Part $-2$.

Based on what you said, we can form the following one variable equation. $(f(x)\mapsto x, ax \mapsto k)$

We have,

$$\begin{align}&e^{k x}=x\\ &ke^{kx}=kx\\ &k=kx e^{-kx}\\ &-k=-kxe^{-kx}\\ &W(-k)=W\left(-kxe^{-kx}\right)\\ &W(-k)=-kx\\ &\color{red}{\boxed{\color{black}{x=-\frac {W(-k)}{k}}}}\end{align}$$

Did this answer help you?