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A polynomial whose degree decreases when its parameter vanishes may have roots escaping to infinity. Such a singular perturbation root is found by balancing powers after rescaling $\lambda ={\epsilon }^{-p}\mu$. If the root enters an exponential, even a lower-order additive root correction can matter on a fixed interval when its leading exponential grows. This phenomenon is different from a repeated root of a fixed polynomial.

Define $E_0(z)={\int }_0^z{e}^{-t^2}{\int }_0^t{e}^{u^2}dudt$ and $E_1(z)={\int }_0^z{e}^{-t^2}{\int }_0^t{e}^{u^2}\mathrm{erf}ududt$. They satisfy $(D^2+2zD)E_0=1$ and $(D^2+2zD)E_1=\mathrm{erf}z$. The first is an even function, the second an odd function, and their real tails are respectively $\frac{1}{2}\log |z|+C_1+o(1)$ and $\mathrm{sgn}z[\frac{1}{2}\log |z|+C_2+o(1)]$. Their derivatives have inverse-linear tails, explaining the logarithmic matching constants in a singular perturbation.

Here the singular perturbation has an exact conservative structure:The integrating factor for this first-order linear differential equation givesAt $x=1$ and $x=-1$ the integral terms have opposite signs while the exponential factors agree. Equal boundary data therefore force $C=0$, and normalization gives the unique exact solutionThis exposes the Gaussian interior layer of a conservative drift equation. Its central scale is $x=O(\sqrt{ϵ})$, where $x=\sqrt{ϵ}\xi$ gives $Y^{\prime \prime }+\xi Y^{\prime }+Y=0$. The central Gaussian function profile is $e^{-{\xi }^2/2}$ with amplitude $e^{1/(2ϵ)}$.

Away from the center, $|x|\gg \sqrt{ϵ}$, use exponential WKB approximation branches rather than an assumed bounded algebraic outer solution. The reduced ordinary differential equation $x{y}_0^{\prime }+y_0=0$ would give $C/x$; matching both equal positive endpoint values with that algebraic branch is impossible. The exact solution selects its zero coefficient and the exponentially large branch with a Gaussian function profile instead. Near either endpoint, a distance $O(ϵ)$ changes $y$ by order one relative to its endpoint value: with $r=(1-|x|)/ϵ$, $y=e^{r-ϵr^2/2}$. These endpoint scales describe normalization of the same global exponential solution.

Thus a central $O(\sqrt{ϵ})$ turning region, exponential outer regions on both sides, and $O(ϵ)$ endpoint normalization scales give the complete regional description. The peak is exponentially large, so no uniformly bounded regular outer expansion should be presumed. The exact expression already provides a uniform solution without further matching calculations.

Use enzyme conservation to writeSince $\tau =k_1{e}_{0}t$, $u=s/s_0$, $v_i=c_i/e_0$, and $ϵ=e_0/s_0$, substitution into the mass-action equations givesandwhereThe factors of $ϵ$ multiplying $d{v}_i/d\tau$ identify the complex concentrations as fast variables in a singular perturbation.

The singular perturbation equation $ϵ\sqrt{x}{u}^{\prime \prime }+u^{\prime }=0$ balances its derivatives on the width $x={ϵ}^{2}X$. The inner equation is $\sqrt{X}{U}^{\prime \prime }+U^{\prime }=0$, so $U^{\prime }=C{e}^{-2\sqrt{X}}$. Its solution with $U(0)=0$ and $U(\infty )=1$ is the displayed $H$, since ${\int }_0^{\infty }{e}^{-2\sqrt{q}}dq=1/2$. Near zero $H=2X-8X^{3/2}/3+O(X^2)$, so the first derivative is finite while the second is singular. Treat the differential equation on the open interval and impose the endpoint value by continuous extension.