Some special cases of the Lambert W function

Suppose \(f(x)\) is given by the functional relation \begin{equation} e^{-a(x) f(x) + b(x)} = c(x) f(x) + d(x) \end{equation} then \begin{equation} f(x) = \frac{1}{a(x)}W\left( \frac{a(x)}{c(x)} e^{b(x)+\frac{a(x)d(x)}{c(x)} } \right) - \frac{d(x)}{c(x)} \end{equation} where \(W\) is the Lambert W Function. You can click through to Wikipedia for a proof. Here I merely list some special cases which every good citizen should instantly recognize (and it can be mildly tedious to reproduce them).

Recovering the definition of \(W(x)\)

Just to check we are on the right planet set \(a=1\), \(c=1\), \(b=\ln(x)\) and \(d=0\) to get \begin{equation} x e^{-f(x)} = f(x) \end{equation} or equivalently \( f(x)e^{f(x)} = x\) which is the definition of the Lambert W function. Sure enough those substitutions recover $$ f(x) = W(x) $$ as expected. A minor modification uses \(a(x)=1\), \(c(x)=1\), \(b(x)=0\) but leaves \(d(x)\) general.

The solution to \( e^{-f(x)} - f(x) = d(x) \)

is \begin{equation} f(x) = W \left( e^{d(x)} \right) \end{equation} Only slightly more generally, set \(a(x)=a\), \(b=\ln(g(x)), c=1, d=1\) for those cases where \(g(x)\) is on the wrong side, as it were.

The solution to \( e^{-af(x)} = f(x) g(x) \)

or equivalently \( f(x) e^{af(x)} g(x) = 1 \) is \begin{equation} f(x) = \frac{1}{a} W\left( a g(x) \right) \end{equation} In particular,

The solution to \( x^k e^{-af(x)} f(x) = 1 \)

which seems to crop up a fair bit for your author is \begin{equation} f(x) = \frac{1}{a} W\left( a x^{-k} \right) \end{equation}

Similarly setting \(a=-1\) ...

The solution to \( x e^{f(x)} f(x) = 1 \)

(i.e. where \(x\) is on the wrong side but we otherwise have the Lambert W definition) must be \begin{equation} f(x) = - W\left( -\frac{1}{x} \right) \end{equation} We might also take \(b = \ln(g(x))\) and thus

The solution to \( g(x) e^{-af(x)} = f(x) \)

or equivalently \( f(x)e^{af(x)} = g(x) \) is \begin{equation} f(x) = \frac{1}{a} W\left( a g(x) \right) \end{equation} which reduces to the Lambert W function for \(a=1\), as we expect. It is also pretty obvious from first principles, because if we multiply both sides by \(a\) we have \begin{equation} (af) e^{(af)} = ag \end{equation} and thus \(af = W(ag)\). Next suppose we want a power of \(f\) to appear. Let \(b=\beta(x)/k\), \(c = \gamma^{1/k}\), \(a = k/\alpha\) and \(d=0\). And then raise both sides to the power \(k\). It follows that...

The solution to \( e^{\alpha f(x) + \beta(x)} = \gamma f(x)^k \)

is \begin{equation} f(x) = \frac{\alpha}{k}W\left( \frac{k e^{\frac{1}{k}\beta(x)} }{\alpha \gamma^{1/k}} \right) \end{equation} and if, in particular, \(\beta(x) = -\ln(g(x))\) then

The solution to \( e^{\alpha f(x)} = g(x) \gamma f(x)^k \)

is \begin{equation} f(x) = \frac{\alpha}{k}W\left( \frac{k }{\alpha g(x)^{1/k} \gamma^{1/k}} \right) \end{equation} and if we take \(c = \frac{1}{\gamma}\) and \(g(x)=x\) and \(\alpha = \frac{s}{2}\) and \(k=2\) then p>

The solution to \( e^{-\frac{s}{2}f(x)} x f(x)^2 = c \)

is \begin{equation} f(x) = \frac{s}{2k}W\left( \frac{2k}{s} \sqrt{ \frac{c}{x}} \right) \end{equation}