The one-dimensional SUPG parameter is positive and bounded

Put $q(s)=s\cosh s-\sinh s$. Then $q(0)=0$ and $q^{\mathrm{\prime }}(s)=s\sinh s>0$ for $s>0$, so $q(s)>0$ and $m(s)=q(s)\mathrm{/}(s\sinh s)>0$.

The identity $\coth s-1=2\mathrm{/}(e^{2s}-1)$ and $e^{2s}-1={\int }_0^{2s}{e}^{t}dt>2s$ give $m(s)<1$.

For $0<s\le 1$, $q(s)={\int }_0^{s}t\sinh tdt\ge s^3\mathrm{/}3$. The function $\sinh s\mathrm{/}s$ is increasing because its derivative is $q(s)\mathrm{/}{s}^2>0$, hence $\sinh s\le s\sinh 1$ on this range. Therefore $m(s)\ge s\mathrm{/}(3\sinh 1)$. For every $s>0$, monotonicity of $\sinh$ also gives $q(s)\le \sinh (s){\int }_0^{s}tdt=s^2\sinh (s)\mathrm{/}2$, whence $m(s)\le s\mathrm{/}2$.

Since $\sinh s>s$ for $s>0$, $m^{\mathrm{\prime }}(s)=s^{-2}-(\sinh s{)}^{-2}>0$. Thus for $s\ge 1$, $m(s)\ge m(1)=\coth 1-1$. Together with step 2.2 this proves the two-sided bound. Substituting $s=\mathrm{P}\mathrm{e}$ into [F2] gives the stated lower and first upper bounds on $\tau$, using $h(2\mathrm{\mid }b\mathrm{\mid }{)}^{-1}\min \{\mathrm{P}\mathrm{e},1\}=\frac{1}{2}\min \{h^2\mathrm{/}(2ϵ),h\mathrm{/}\mathrm{\mid }b\mathrm{\mid }\}$. The last upper bound follows from $m(\mathrm{P}\mathrm{e})\le \mathrm{P}\mathrm{e}\mathrm{/}2$.