Is there an upper bound on the Lambert function $$W(-\frac{k}{e})$$ for $0 < k < 1$? Or should the condition be $0 < k \leq 1$?
I know that $W(-\frac{1}{e}) = -1$; would it be okay to say $W(-\frac{k}{e}) \leq -1$?
I just recently learned about Lambert functions and am having a hard time understanding all these. This came from an approximation of an expression given by:
$$\ln \phi_\infty = \mu + \lambda \phi_\infty – \mu e^{-\lambda \phi_\infty},$$
which by letting $z=\lambda \phi_\infty$, yields
$$z = -\frac{1}{1+\mu}W(-\lambda(1+\mu))$$
which only exists if
$$\lambda (1+\mu) \leq \frac{1}{e}.$$
If we let $\lambda = ke^{-1}\frac{1}{1+\mu}$ for $0 < k \leq 1$, we can have
$$ z = -\frac{1}{1+\mu}W\bigg(-\frac{k}{e}\bigg). $$
What I need is to represent this last equation in terms of $\phi_\infty$ alone (and these parameters $\lambda, \mu$) so I was wondering if I could get an upper bound for the Lambert function and do just that.
Thank you so much for your help!
Best Answer
$W$ without a subscript usually means $W_0$.
Let $f(k)=W\left(-\frac{k}{e}\right)$ for $0<k<1$.
See Wolfram Alpha for the graph of $f$, extended by the points $(0,0)$ and $(1,-1)$.
$$\lim_{x\to 0}f(x)=0,\ \ \lim_{x\to 1}f(x)=-1$$
See at Wikipedia: Upper and lower bounds, Bounded function and Bounded set.
$f$ is a bounded function: bounded above by $0$ and bounded below by $-1$.
$$-1<f(k)<0$$