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Rescale variables so that the terms responsible for the leading behaviour have comparable size. The resulting distinguished limit exposes a governing equation that may differ from the naive zero-parameter equation. Examples are dominant balance for algebraic roots and the derivative balance in a square-root boundary layer at a vanishing drift.

An algebraic root can depend on a parameter through fractional powers even when the polynomial coefficients have ordinary power series. For example, the roots of $t^3={ϵ}^2$ are on the scale ${ϵ}^{2/3}$. Dominant balance for algebraic roots identifies the exponent, and local expansion of the scaled root supplies successive coefficients.

Use dominant balance for algebraic roots to separate the three scales. On writing $x=ϵX$, the equation becomes $X^3-X^2+2ϵX+2{ϵ}^3=0$. Its nonzero simple limiting root is $X=1$, so one root has $x_1\sim ϵ$.

For the two remaining roots put $x={ϵ}^{2}Y$. Division by ${ϵ}^5$ gives $ϵY^3-Y^2+2Y+2ϵ=0$. The nonzero limiting root is $Y=2$, giving $x_2\sim 2{ϵ}^2$. Finally $x={ϵ}^{3}Z$ gives, after division by ${ϵ}^6$,Its simple limiting root is $Z=-1$. These three distinct balances account for all roots of the cubic, andFor the root of smallest magnitude, set $Z=-1+bϵ+O({ϵ}^2)$. The order-$ϵ$ coefficient is $-1+2b$, so $b=1/2$. HenceEach rescaled root is simple at $ϵ=0$, justifying the regular perturbation of a simple algebraic root rather than treating the original triple root as an ordinary regular perturbation.

The dominant balance for algebraic roots separates one small root of a polynomial from three large ones. For the small root of a polynomial, write $z=\epsilon +a{\epsilon }^3+O({\epsilon }^5)$ in $z=\epsilon (1+z^2{)}^2$. Comparing coefficients gives $a=2$, soThis is a real analytic function of real $\epsilon$ near zero, since the derivative with respect to $z$ of $z-\epsilon (1+z^2{)}^2$ is one at $(0,0)$.

For the large roots of a polynomial, use the real cube root $s=\mathrm{sgn}(\epsilon )|\epsilon {|}^{1/3}$ and set $z=w/s$. The transformed equation isIts three nonzero leading roots of a polynomial are $w=\omega$, with ${\omega }^3=1$. Writing $w=\omega +a{s}^2+O(s^4)$ gives $a=4a+2{\omega }^2$, hence $a=-2{\omega }^2/3$. ThereforeThe choice $\omega =1$ is the second real root of a polynomial; the other two form a complex conjugate pair for either sign of $\epsilon$. The nonzero transformed roots of a polynomial are simple, so their Taylor series in $s^2$ exist locally and exhaust the three large roots of a polynomial. Together with the small one these give all four roots of a polynomial for sufficiently small nonzero $\epsilon$. At $\epsilon =0$ itself the rational equation has only the finite root zero; the other three escape to infinity. Multiplying by $(1+z^2{)}^2$ has not introduced roots at $z=\pm i$, since those points do not solve the multiplied equation.

A regular perturbation admits an asymptotic expansion about the limiting problem on the relevant scale without losing its solutions or conditions. A singular perturbation changes the limiting structure, so some solutions require a different scale: a differential equation may lose an order, or a polynomial may lose its degree.

The two finite roots approach $s=\pm \sqrt{2}$. Set $x=s+ϵa+O({ϵ}^2)$. Substitution into the polynomial gives $s^2-2=0$ at leading order and $s^3+2sa=0$ at the next order. Since $s^2=2$, $a=-1$ for either sign. Each branch is a regular perturbation of a simple algebraic root, withThe limiting quadratic has only two roots, whereas the original equation has three. The missing root is a divergent algebraic root in a singular perturbation. The dominant balance for algebraic roots sets $x={ϵ}^{-1}X$, giving $X^3+X^2=2{ϵ}^2$. Its nonzero leading root is $X=-1$. Put $X=-1+a{ϵ}^2+O({ϵ}^4)$; the left side is $a{ϵ}^2+O({ϵ}^4)$, so $a=2$. Hence the second nonzero term occurs at order $ϵ$, not order one:For small nonzero real $ϵ$ these three branches are real and distinct; the divergent branch changes sign with $ϵ$. The two bounded branches are regular even though the full root problem is singular.

The small root and the root near $-2$ are regular perturbation roots: they stay finite as the parameter vanishes. Substitution of power series givesFor the third root, dominant balance for algebraic roots requires $\lambda =\mu /\epsilon$. The largest terms give ${\mu }^2(\mu +1)=0$, and the missing branch is $\mu =-1$. The sum of the three roots then yieldsThus the leading roots are $-\epsilon /2$, $-2$, and $-1/\epsilon$; only the last is a divergent perturbation root. The first regular branch happens to have a zero limit, so retaining its first nonzero term is essential on the long spatial scale.

First take $\epsilon \to 0^{+}$, the decaying half-line regime. If the three exact roots are $r_j$, the exact initial value problem solution isThis follows either by solving the three initial-value equations or from the Laplace transform $\hat{y}(s)=\epsilon /(\epsilon s^3+s^2+2s+\epsilon )$. The coefficient formula has sums $\sum A_j=\sum r_j{A}_j=0$ and $\sum r_j^2{A}_j=1$.

Retain the three leading roots but compute their coefficients without expanding their denominators. Set $a=\epsilon /2$, $b=2$, $d=1/\epsilon$. A convenient composite asymptotic expansion isIt satisfies all three initial conditions exactly and retains the fast transient, the ordinary decay, and the slow decay. Its first two coefficients are $\epsilon /2+O({\epsilon }^2)$ and $-\epsilon /2+O({\epsilon }^2)$; the fast coefficient is ${\epsilon }^2+O({\epsilon }^3)$. Consequently the simpler bulk expression is $\frac{\epsilon }{2}(e^{-\epsilon x/2}-e^{-2x})$, but that expression alone does not reproduce the initial derivative layer.

The absolute error estimate isTo see its uniformity, the slow-root error is $O({\epsilon }^2)$, while its magnitude is $O(\epsilon )$ and its coefficient is $O(\epsilon )$. The uniform error bound for nearby decaying exponentials therefore gives an $O({\epsilon }^2)$ contribution even for $x=O({\epsilon }^{-1})$. The middle-root error is $O(\epsilon )$ with coefficient $O(\epsilon )$, again giving $O({\epsilon }^2)$. The fast-root error is $O(1)$ with magnitude $O({\epsilon }^{-1})$ and coefficient $O({\epsilon }^2)$, giving only $O({\epsilon }^3)$. Coefficient errors are $O({\epsilon }^2)$ or smaller. This estimate concerns $y$ itself, rather than asserting the same uniform order for all derivatives or a relative error at its zero.

For the sketch, on $x=\epsilon X$ one obtainsThus $y\sim x^2/2$ at the origin and rises smoothly after the fast layer. For $x=O(1)$ it rises towards a plateau of height approximately $\epsilon /2$, then decays on the much longer scale $x=O({\epsilon }^{-1})$. The bulk maximum lies near $x=\log (4/\epsilon )/(2-\epsilon /2)$ and has height asymptotic to $\epsilon /2$.

The sign of the parameter matters on an infinite interval. If $\epsilon \to 0^{-}$ is allowed, the singular mode grows rather than decays. For each fixed $x>0$, its dominant contribution is $y(x)\sim {\epsilon }^2\exp [(-1/\epsilon +2)x]$. The extra $2$ in that exponent is needed for relative leading accuracy at fixed $x$. The positive-parameter uniform absolute bound and decaying sketch do not extend to that regime.

For $ϵ>0$, dominant balance for algebraic roots gives two roots on the scale $\sqrt{ϵ}$ and a third on the scale ${ϵ}^{2/3}$. On the first scale put $t=\sqrt{ϵ}s$. The equation becomes $s^5-s^3+\sqrt{ϵ}=0$. The derivative of $s^5-s^3$ at either $s=1$ or $s=-1$ is two, so the first correction is $-\sqrt{ϵ}/2$. On the second scale put $t={ϵ}^{2/3}v$; then $1-v^3+{ϵ}^{1/3}{v}^5=0$. Writing $v=1+a{ϵ}^{1/3}+\cdots$ gives $a=1/3$. Thus all three real-root expansions for positive $ϵ$ areThese are fractional-power root expansions. To check that none is missing, the derivative of the polynomial is $t^2(5t^2-3ϵ)$. Its nonzero critical points are $\pm \sqrt{3ϵ/5}$; for sufficiently small positive $ϵ$ the polynomial is positive at the negative one and negative at the positive one. Its three monotone ranges therefore contain exactly these three real roots. The remaining two small roots correspond to the nonreal cube roots of unity in the second balance.

The printed limit does not specify the sign of $ϵ$. If a two-sided real limit is intended, also put $ϵ=-\eta$ with $\eta >0$. The polynomial $t^5+\eta t^3-{\eta }^3$ is strictly increasing and has just one real root. The same dominant balance for algebraic roots gives the negative-parameter branch