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For a Laplace integral with local phase $\varphi (t;\delta )=\delta t+b{t}^3/3+\cdots$, $b>0$, an interior minimum approaches the endpoint as $\delta \uparrow 0$. The distinguished limit is $t=O({\nu }^{-1/3})$, $\delta =O({\nu }^{-2/3})$. With $s=(\nu b{)}^{1/3}t$, its leading local integral isThe cubic Laplace transition integral joins the ordinary endpoint estimate for positive $\delta$, the cubic endpoint estimate at zero, and the interior Laplace's method estimate for negative $\delta$. Separate fixed-parameter formulas fail to be uniform when the minimum is within its own width of the endpoint.

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.

A matched asymptotic expansion constructs approximations in regions with different distinguished limits and fixes their free constants by requiring agreement in an overlap region. The inner expansion resolves rapid variation; the outer expansion describes the surrounding region. Matching compares their common limiting behavior, rather than imposing distant boundary conditions directly on a local approximation.^[ref]

For the singular perturbation problem $ϵy^{\prime \prime }+y^{\prime }=0$, $y(0)=0$, $y(1)=1$, the leading outer solution is $y_0=1$. The inner variable $X=x/ϵ$ gives $Y_0^{\prime \prime }+Y_0^{\prime }=0$ and $Y_0(0)=0$, hence $Y_0=A(1-e^{-X})$. In the overlap region $ϵ\ll x\ll 1$, the inner variable tends to infinity while the outer coordinate tends to zero. The asymptotic matching condition isso $A=1$. One concrete path through the overlap region is $x=\sqrt{ϵ}$: then $x\to 0$ and $X={ϵ}^{-1/2}\to \infty$. Matching selects the free constant that the left boundary condition alone could not determine.

Why choose the right boundary condition for the outer solution? If instead one selects $y_0=0$ from the left, a proposed right boundary layer with $Z=(1-x)/ϵ$ satisfies $U_0^{\prime \prime }-U_0^{\prime }=0$. Its general solution is $A+B{e}^Z$. Matching to zero as $Z\to \infty$ forces $A=B=0$, contradicting $U_0(0)=1$. The direction in which the exponential function decays determines which endpoint can carry the boundary layer; endpoint selection is part of the calculation.

Putting $\epsilon =0$ gives the local regular perturbation result $y_0=(x^2+1)/2$. It cannot describe the whole half-line: its derivative grows like $x$, making $\epsilon y^{\prime }$ order one at $x=O({\epsilon }^{-1})$.

The useful exact change of variables isDirect substitution turns the equation into two autonomous differential equations,The initial values here are $z(0)=w(0)=\epsilon$. Near the origin, linearizing this system gives eigenvalues $1$ and $-1-\epsilon$. Equivalently, neglecting only the derivative-square nonlinearity in the original equation gives the two modes $x^2$ and $x^{-\epsilon }$. Their matching coefficients areThus the growing derivative has $w\sim Cx$. Keeping these unexpanded linear-mode coefficients is convenient for matching; it does not assert that the linearized solution is the exact nonlinear one.

At the distinguished limit $X=Cx=O(1)$, set $R=\log X$. To leading order the derivative equation becomes $w_R=w(1-w)$, the logistic differential equation. Matching $w\sim e^R$ as $R\to -\infty$ selectsThen $z_R+z=w$, and the growing-mode matching $z\sim X/2$ givesThis proves the logistic crossover in a weakly nonlinear Euler equation. In particular the nonlinear-region leading solution is $y\sim H(Cx)/(\epsilon C)$, or ${\epsilon }^{-2}H(\epsilon x)$ with the constants taken at their lowest order.

A composite asymptotic expansion retaining the small-distance stable mode and satisfying the initial value exactly isSince $H(X)=X^2/2+O(X^3)$, this matches $A{x}^2+B{x}^{-\epsilon }$ when $Cx\ll 1$ and hence $(x^2+1)/2$ at fixed $x$ as $\epsilon \to 0$. Its derivative at one differs from the prescribed value only by $O(\epsilon )$, within leading-order accuracy. For $Cx\gg 1$ it gives $y\sim x/\epsilon$. Thus the composite covers the initial region, the nonlinear transition and the arbitrarily large-$x$ regime; uniform leading accuracy is understood relative to the growing solution, not as an absolute error independent of unbounded $x$.

There is no later hidden growth regime. The exact positive system keeps $z,w$ in $[0,1+\epsilon {]}^2$, because its vector field points inward at every boundary. Its positive equilibrium point is $z=w=1+\epsilon$, with Jacobian matrix having trace $-2-3\epsilon$ and determinant $1+\epsilon$. The origin has no attracting direction inside the positive quadrant, while the divergence $-\epsilon -2w<0$ excludes cycles by the Bendixson-Dulac theorem. The Poincaré-Bendixson theorem therefore gives convergence to the positive stable equilibrium. At fixed small positive $\epsilon$, the exact far-field slope is $y/x\to (1+\epsilon )/\epsilon$, whose difference from $1/\epsilon$ is only relative $O(\epsilon )$, consistent with the global leading composite.

Let $d=U^{\prime }(0)>0$, $h=U^{\prime \prime \prime }(0)$, $p={\varphi }_0(0)\ne 0$, and $q={\varphi }_0^{\prime }(0)$. At a regular neutral Rayleigh equation for inviscid shear flow mode, multiply the equation by $U-c_0$ and evaluate it at the critical layer in a shear flow. The term multiplying ${\varphi }_0^{\prime \prime }-k_0^2{\varphi }_0$ vanishes, leaving $U^{\prime \prime }(0)p=0$. Therefore $U^{\prime \prime }(0)=0$. If this condition failed, the coefficient $U^{\prime \prime }/(U-c_0)$ would have a simple pole and a nonzero $p$ would force a logarithmically singular slope instead of a regular eigenfunction.

There is a sign error in the printed first-order equation. Direct expansion of the original Rayleigh equation for inviscid shear flow givesThe coefficient on the left must have a minus sign. Indeed, $1/(U-c)=1/(U-c_0)+\epsilon c_1/(U-c_0{)}^2+\cdots$, and moving both first-order parameter corrections to the right gives exactly this equation. The plus sign printed there is incompatible with the original equation and its neutral eigenfunction. The later matching and integral formulas follow from the corrected equation. The PDF also has $y^{3}h/6$ in the Taylor expansion of $U$, and $U^{\prime \prime \prime }(0)$ in the requested logarithmic coefficient; these are mistranscribed in the converted TeX.

Near zero, $U-c_0=dy+h{y}^3/6+\cdots$ and $U^{\prime \prime }=hy+O(y^2)$. Thus $U^{\prime \prime }/(U-c_0)\to h/d$, whereas $U^{\prime \prime }/(U-c_0{)}^2=h/(d^{2}y)+O(1)$. The singular part of the first-order equation isIntegrating twice on either side givesThe constant $a$ is common because the inner solution is continuous. The coefficient necessarily also contains $p$: rescaling the neutral eigenfunction rescales its correction. One may choose the normalization $p=1$ if a formula only in terms of the flow derivatives and $c_1$ is wanted.

The outer expansion of $1/(U-c)$ fails when $|dy|$ is comparable with $\epsilon |c_1|$. Therefore the distinguished limit isIn this inner variable, the original equation becomesAt orders one and $\epsilon$, its right side is zero. Matching the outer expansion therefore selects ${\Phi }_0=p$ and ${\Phi }_1=q\eta +a$. At order ${\epsilon }^2\log \epsilon$, ${\theta }_2^{\prime \prime }=0$. The outer term $\epsilon \beta y\log |y|$ becomes ${\epsilon }^2\beta \eta (\log \epsilon +\log |\eta |)$, so the required switchback term is ${\theta }_2=\beta \eta$. At order ${\epsilon }^2$,Integrating, or using the supplied hint, givesThe constants $B,C$ are homogeneous matching freedoms; they do not affect the slope jump. The quadratic term agrees with ${\varphi }_0^{\prime \prime }(0)=p(k_0^2+h/d)$, another check on the sign of $L_0$.

For $\mathrm{Im}{c}_1>0$, $d\eta -c_1$ runs below the real axis. Choose the continuous branch of the complex logarithm there. At positive infinity its argument tends to zero; at negative infinity it tends to $-\pi$. Thus the linear parts of the two large-$|\eta |$ expansions of ${\Phi }_2$ areThese expressions omit the already matched common quadratic term and terms only logarithmic or constant in $\eta$. Matching the linear coefficients to $b_{\pm }$ proves the neutral-mode critical-layer matching relationThe sign comes from passing below the pole; taking $\mathrm{Im}{c}_1<0$ would reverse it.

Finally use the Wronskian $W={\varphi }_1{\varphi }_0^{\prime }-{\varphi }_0{\varphi }_1^{\prime }$. Since $L_0{\varphi }_0=0$ and $L_0{\varphi }_1=H{\varphi }_0$, subtraction gives $W^{\prime }=-H{\varphi }_0^2$. The wall boundary conditions give $W(a)=W(b)=0$, soThis is the requested excised-interval identity. As $\delta \downarrow 0$, the equal logarithmic slope terms cancel andThe remaining real singular integral is interpreted as a Cauchy principal value, consistent with excision on both sides of the critical layer in a shear flow. DefineThen the neutral Rayleigh-mode dispersion correction iswhere the last form assumes a nonzero denominator. For real $k_1$, $\mathrm{Im}{c}_1=2k_0{k}_{1}I\pi M/(J^2+{\pi }^2{M}^2)$, so the assumed growing branch must have the corresponding sign. If $h=0$, the displayed logarithmic coefficient and jump vanish; further degeneracy can require higher-order analysis. No nonzero growth rate follows from that degenerate first-order calculation alone.

For fixed $0\le \alpha <1$, the phase derivative is $-1+\alpha \cos x\le -(1-\alpha )<0$ on the integration interval. Its maximum is at $x=0$, with phase zero and derivative $-(1-\alpha )$. The Laplace endpoint estimate, or the substitution $x=t/[\lambda (1-\alpha )]$, givesThis fixed-parameter formula is not uniform as $\alpha \uparrow 1$; its diverging coefficient signals the distinguished limit derived in the parent solution.

One convenient version of Watson's lemma is the following. Suppose $G(t)$ is integrable away from zero and has, as $t\downarrow 0$, an asymptotic expansion $G(t)\sim {\sum }_{j\ge 0}{c}_j{t}^{\beta +j-1}$, where $\beta >0$. On a finite interval, or on an infinite interval with suitable exponential growth control,To prove a finite truncation, split the integral at a fixed small $\delta >0$. The part away from zero is exponentially small. On $(0,\delta )$ subtract the first $N$ local terms and bound the remainder by $C{t}^{\beta +N-1}$; its integral is $O({\lambda }^{-\beta -N})$. Extending each polynomial integral to infinity changes it only exponentially, and the substitution $u=\lambda t$ produces the Gamma integral. A little-o local remainder gives the corresponding little-o asymptotic remainder. This is a proof by localization and remainder bounds, rather than merely formal integration of a series.

For the monotone-phase integral, assume the derivatives needed below exist and are bounded near $a$, and the rest of the integral is controlled. Put $t=\varphi (a)-\varphi (x)$. Since ${\varphi }^{\prime }(a)<0$, this is a regular endpoint coordinate andDifferentiating with $dx/dt=-1/{\varphi }^{\prime }$ gives $G(0)=-g(a)/{\varphi }^{\prime }(a)$ and $G^{\prime }(0)=g^{\prime }(a)/{\varphi }^{\prime }(a{)}^2-g(a){\varphi }^{\prime \prime }(a)/{\varphi }^{\prime }(a{)}^3$. Watson's lemma therefore gives the two-term monotone-endpoint Laplace expansionAn $O(e^{\lambda \varphi (a)}{\lambda }^{-3})$ remainder follows, for example, from a sufficiently smooth transformed amplitude with bounded second derivative near zero. The displayed second coefficient can vanish for special amplitudes without changing the expansion.

The unheaded transition request concerns a cubic endpoint-to-saddle transition. Near $x=0$ its exponent has the expansionAt the cubic endpoint $x$ has size ${\lambda }^{-1/3}$. Requiring the linear term to contribute on the same scale gives $1-\alpha =O({\lambda }^{-2/3})$. Thus the distinguished limit and its uniform leading integral areFor fixed $\nu$, the substitution $x={\lambda }^{-1/3}t$ makes the omitted exponent terms $O({\lambda }^{-2/3})$ on bounded $t$ intervals. Cubic decay controls the remaining tail, while the original interval away from zero is exponentially negligible. The estimate is uniform for $\nu$ in compact sets. In the printed range $\alpha \le 1$, $\nu \ge 0$; the same transition integral also makes sense for negative fixed $\nu$.

At $\nu =0$, the Gamma integral gives $h(0)=6^{1/3}\Gamma (1/3)/3$, recovering the cubic endpoint regime. For $\nu \to +\infty$, scale $t=u/\nu$ and expand $e^{-u^3/(6{\nu }^3)}$ to obtain $h(\nu )={\nu }^{-1}-{\nu }^{-4}+O({\nu }^{-7})$. Its leading term gives ${\lambda }^{-1/3}/\nu =1/[\lambda (1-\alpha )]$, recovering the ordinary endpoint regime in their overlap. Both matching limits are therefore checked explicitly.

Use the characteristic polynomial in the convention $p(\lambda )=\det (\lambda I-A)$:For fixed $\alpha >0$ with $\alpha \ne 1$, all three limiting eigenvalues are simple. Expanding this polynomial at each root, or applying first-order perturbation of a simple eigenvalue, givesThese are fixed-$\alpha$ asymptotic expansions, not uniform through $\alpha =1$. At $\alpha =1/3$ the displayed correction to ${\lambda }_{\alpha }$ vanishes because $\lambda =1/3$ is an exact root for every $ϵ$; there is no second nonzero term for that branch.

The two roots near $1$ interact when the correction $ϵ/(\alpha -1)$ is comparable to their unperturbed gap $\alpha -1$. Thus the distinguished limit is $\alpha =1+d\sqrt{ϵ}$, with bounded real $d$. Set $h=\sqrt{ϵ}$ and $\lambda =1+hL+h^{2}N+\cdots$. The polynomial becomesAt orders $h^2$ and $h^3$,With $D=\sqrt{d^2+4}$, the two interacting eigenvalues areThe $h^2$ correction to the pair is included to resolve the next perturbative term as well as the leading splitting. In particular the previously excluded fixed case $\alpha =1$ is $d=0$:The half-integer powers reflect square-root splitting of a defective double eigenvalue: the limiting matrix has a nontrivial Jordan block at $1$.

For matching, write $\Delta =\alpha -1$ and take the overlap $\sqrt{ϵ}\ll |\Delta |\ll 1$. For $d\to +\infty$, $L_{+}=d+d^{-1}+O(d^{-3})$, $N_{+}=1+O(d^{-2})$, while $L_{-}=-d^{-1}+O(d^{-3})$, $N_{-}=O(d^{-2})$. Consequently the distinguished branches reduce toThe regular coefficient satisfies $(3\alpha -1)/[(\alpha +1)(\alpha -1)]=1/\Delta +2/(2+\Delta )=1/\Delta +1+O(\Delta )$, proving agreement. For $d\to -\infty$ the labels $+$ and $-$ exchange roles and the same matching holds. The isolated negative branch also matches because $-2/(\alpha +1)=-1+O(\Delta )$. As a spectral check, positive diagonal symmetrization of a matrix transforms this tridiagonal matrix to a real symmetric matrix with off-diagonal entries $\sqrt{2ϵ}$ and $\sqrt{ϵ}$; its eigenvalues are real for positive $ϵ$.

For fixed $\gamma >1$, the linear term $1-\gamma +\gamma x$ vanishes at $x_{\ast }=1-1/\gamma$. Put $L_{\ast }=\log (1+1/\gamma )>0$ and resolve its neighborhood using $x=x_{\ast }+(\epsilon L_{\ast }/\gamma )s$. The integral is dominated by the resulting narrow peak:The local Taylor expansion of the logarithm changes it by a vanishing relative amount on this width, while both endpoints are many peak widths away for fixed $\gamma >1$.

There is no additional breakdown when $\gamma \gg 1$: the peak approaches $x=1$, but its distance from that endpoint is $d=1/\gamma$, whereas its width is $w=\epsilon L_{\ast }/\gamma \sim \epsilon /{\gamma }^2$. Thus $w/d=\epsilon L_{\ast }\sim \epsilon /\gamma \ll 1$. The relative logarithm variation over a width is $O(\epsilon /\gamma )$, also improving as $\gamma$ grows. For $\gamma \ge 1+c$ with fixed $c>0$ these separation conditions hold uniformly, including arbitrarily large $\gamma$. In particularThe only endpoint transition requiring a new distinguished limit is the earlier one near $\gamma =1$, where the peak approaches $x=0$. Moving toward an endpoint is not by itself a reason for nonuniformity; the peak width must be compared with its endpoint distance.

The three distinguished limits are the outer expansion at $x=O(1)$, an intermediate asymptotic region at $x=O(ϵ)$, and an inner expansion at $x=O({ϵ}^2)$. The middle scale resolves the coefficient change from $x$ to $x+ϵ$; the narrower scale resolves the unmet left boundary condition. The original equation is understood on $x>0$ with a continuous extension to the boundary. As explained below, imposing a classical second derivative at $x=0$ would be too strong for this degenerate ordinary differential equation.

In the outer expansion, set $y=y_0+ϵy_1+\cdots$. The leading equation is $x(x+1)y_0^{\prime }-x{y}_0=0$. Enforcing the right boundary condition gives $y_0=x+1$. Because $y_0^{\prime \prime }=0$, the next equation is particularly simple:An integrating factor, or the substitution $y_1=(x+1)u$, gives $u^{\prime }=-1/[x(x+1)]$. ThusThe error statement is for $x$ bounded away from zero. The left limiting form is $1+x-ϵ\log (2x)$ through the terms needed for matching. A logarithm in the correction makes a pure integer-power inner ansatz inadequate.

For the intermediate asymptotic region, put $x=ϵs$ and $y=V(s)$. The transformed equation isThe leading $V_0$ is a constant fixed to $1$ by matching. There is no order-$\sqrt{ϵ}$ forcing, so that homogeneous constant is zero after matching. At order $ϵ$, $(s+1)V_1^{\prime }=s$, giving $V_1=s-\log (1+s)+C(ϵ)$. Matching for $1\ll s\ll 1/ϵ$ with the outer expansion fixes $C=-\log (2ϵ)$. HenceThe switchback term $-ϵ\log ϵ$ is larger than an ordinary $O(ϵ)$ constant correction and must be retained. The highest derivative term first changes this middle solution at order ${ϵ}^{3/2}$, so it does not alter the displayed result.

To locate the inner expansion, compare the derivative terms for $x\ll ϵ$. Their ratio at a layer width $\delta$ is $ϵ/\sqrt{\delta }$, so the square-root-degenerate endpoint layer has $\delta ={ϵ}^2$. Put $x={ϵ}^{2}X$ and $y=U(X)$. After dividing by the common leading factor, the equation through order $ϵ$ isBoth orders therefore have the same homogeneous operator $\sqrt{X}{d}^2/d{X}^2+d/dX$. Its derivative mode is proportional to $e^{-2\sqrt{X}}$. Using $U(0)=0$ and the supplied elementary integral, defineThe intermediate asymptotic region gives the overlap value $1-ϵ\log (2ϵ)$ as $s\to 0$. ThereforeIt satisfies the left boundary condition exactly to the retained orders, and its large-$X$ limit matches the middle result in $1\ll X\ll 1/ϵ$. The outer and middle limits match with $ϵ\ll x\ll 1$. These three formulae include all terms through $O(ϵ)$, including the logarithmic switchback; the omitted terms in the middle and inner regions need not be $O({ϵ}^2)$.

For a check on that last point, continue the middle equation by one order. Its ${ϵ}^{3/2}$ term obeysThe integration constant is selected by the outer limit. Its value at $s=0$ is $\pi /2$, so the inner matching amplitude has a next correction $\pi {ϵ}^{3/2}/2$. This independently explains why an $o(ϵ)$ remainder is appropriate.

If a single additive composite expansion is useful, combine the three formulae and subtract their two common overlaps. Writing $A=1-ϵ\log (2ϵ)$ and interpreting $x\log x=0$ at zero givesIt is zero at $x=0$, gives $2+O({ϵ}^2)$ at $x=1$, and reproduces all three retained expansions. It does not assert an $O({ϵ}^2)$ error uniformly through the nested regions.

Finally, near zero $H(X)=2X-\frac{8}{3}{X}^{3/2}+O(X^2)$. Thus the solution has a finite nonzero first derivative but generally a second derivative diverging like $x^{-1/2}$. This is compatible with the equation for $x>0$: the vanishing coefficient of $y^{\prime \prime }$ multiplies that singular derivative to give a finite limit. If one instead required $y\in C^2[0,1]$ and the differential equation pointwise at $0$, the left condition would force $y^{\prime }(0)=0$. The normalized equation's other coefficients have only integrable $x^{-1/2}$ singularities, so uniqueness for the initial value problem with $y(0)=y^{\prime }(0)=0$ would give the zero solution, contradicting the right condition. Explicitly, write the normalized equation as a first-order system for $(y,y^{\prime })$ with an integrable coefficient matrix; its integral equation and the Gronwall inequality force a solution with zero initial vector to vanish. The continuous-endpoint interpretation is therefore essential.

The ordinary saddle-point approximation with a nearby pole loses uniformity when the pole lies within the width of the saddle's Gaussian function profile. The distinguished limit isUse the exact quadratic phase coordinate $s=2i\sqrt{k}(u-1)$ and defineOn the descending contour, $s$ runs along the real axis from $-\infty$ to $+\infty$. The transformed differential has the useful partial fraction decompositionThe first term contains the nearby pole; the second is regular at the saddle and contributes $O(e^{-k}{k}^{-1/2})$. Approaching from below the descending contour means $\mathrm{Im}{s}_d<0$. The residue theorem then gives the Gaussian saddle-pole transitionThe exponent is $-s^2/4$, as in the original PDF; the converted TeX's $-s^2/2$ is a transcription error. The exact identity $k\varphi (z_0)=-k-s_d^2/4$ recovers the pole exponential in the requested expression.

The Gaussian pole integral expresses the bracket as $i\pi w(s_d/2)$, where $w$ is the Faddeeva function. Thus a convenient uniform leading answer isThe error estimate applies for bounded scaled pole position on the indicated side. In the overlap $1\ll |s_d|\ll \sqrt{k}$ below the real axis, the integral contributes $-2\sqrt{\pi }/s_d$ to leading order while the residue remains $2\pi i{e}^{-s_d^2/4}$. Since $s_d\sim \sqrt{k}(z_0-i)$, this reproduces the near-saddle limit of the separate saddle and residue terms. Boundary values on the descending contour are limits of this combined expression; the isolated real-axis pole integral itself needs a Cauchy principal value prescription.

Three distinguished limits are needed.

In the origin layer, put $x=\epsilon X$ and $y(x;\epsilon )=Y(X)$. Since $\mathrm{sech}(\epsilon X)=1+O({\epsilon }^2)$, the leading equation isIt has the first integralSolving this first-order linear differential equation givesRegularity at $X=0$ requires $D=-C$, and $Y(0)=1$ then fixes $C=1$. Thus Region I, $x=O(\epsilon )$, hasIts matching limit is $Y\sim X^{-1}$ as $X\to \infty$.

In Region II, $x=O(1)$ with $x\gg \epsilon$, setting $\epsilon =0$ reduces the second-order equation toHence $y_0=C{e}^{-x}/x$. Matching $C/x$ with the inner limit $\epsilon /x$ fixes $C=\epsilon$, so

The correction generated by $\epsilon x^{3}y$ changes the local decay rate when $x=O({\epsilon }^{-1/2})$. For Region III setThe terms of order ${\epsilon }^{-1/2}$ give the eikonal equationand the next balance gives $Z{G}^{\prime }+G=0$. Matching to Region II selectsThereforeThis solution is already beyond all algebraic orders in $\epsilon$ in the distant region. A leading multiplicative composite expansion that contains all three balances and satisfies the boundary value exactly iswith its value at $x=0$ understood by continuity.

Finally, the coefficient of $y^{\prime \prime }$ vanishes at the origin. Taking the regular limit of the original equation there givesso the prescribed value also fixes $y^{\prime }(0)=-1/(2\epsilon )$. Equivalently, the second inner solution behaves as $1/X$ and is excluded by regularity. This regular singular point is why one boundary condition determines the unique regular solution.

The model matrix $B=\left(\begin{array}{cc}0& 1\\ ϵ& d\end{array}\right)$ has a defective double eigenvalue when $d=ϵ=0$. Its characteristic polynomial is ${\lambda }^2-d\lambda -ϵ$, giving the displayed roots. For fixed nonzero $d$ the small root is $-ϵ/d+O({ϵ}^2)$ and the other is $d+ϵ/d+O({ϵ}^2)$. These expansions cease to be ordered at $d=O(\sqrt{ϵ})$, the distinguished limit in which both roots split on a square-root scale. This is the generic leading behavior when a perturbation closes a two-dimensional Jordan block through a nonzero lower-left entry; it need not occur for a perturbation that leaves that entry zero.