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There are two positive roots for $u>1$: the derivative of $x{e}^{1/x}$ is $e^{1/x}(1-1/x)$, so the function decreases to its minimum $e$ at $x=1$ and then increases. Negative $x$ cannot solve the equation. For the larger root, expansion of the exponential function in $1/x$ givesso series reversion yieldsFor the smaller root, set $y=1/x$ and take a logarithm. Then $y-\log y=u$, so successive substitution gives $y=u+\log u+O(\log u/u)$ andThese are the first two nonzero terms of the two asymptotic expansions. Equivalently, the roots are $-1/W_0(-e^{-u})$ and $-1/W_{-1}(-e^{-u})$, from the real branches of Lambert W.

Let $\tau$ be the time at which a right-running characteristic curve leaves the piston. The boundary velocity and the characteristic it emits areWiththese relations becomeBecause $d\varphi /d\theta =1-ϵ\cos \theta >0$ and $\varphi (\theta +2\pi )=\varphi (\theta )+2\pi$, $U$ is single-valued and $2\pi$-periodic behind the wavefront. Ahead of the first characteristic, where $\varphi <0$, the fluid remains undisturbed and $U=0$. The implicit relation is Kepler equation; unless $ϵ=0$, its inverse is not a pure sinusoid.

Series reversion givesExpanding the cosine then givesand hence

On $-e^{-1}<z<0$, the Lambert W function has two real branches: $-1<W_0(z)<0$ and $W_{-1}(z)<-1$. The derivative of $w{e}^w$ is $e^w(1+w)$, so this function decreases on $(-\infty ,-1)$ and increases on $(-1,\infty )$, proving the two-branch structure. As $u\to \infty$, series reversion givesConsequently the two positive roots of $x{e}^{1/x}=e^u$ are $-1/W_0(-e^{-u})$ and $-1/W_{-1}(-e^{-u})$.