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A homogeneous mode can be order one inside an interval and exponentially small at an endpoint. Adjusting an endpoint value beyond all algebraic orders can then change the interior solution by order one. In the nested layers at an essential singular endpoint example, the mode $\exp (-x/ϵ-2{ϵ}^2/x)$ has this behavior: prescribed exact endpoint values give a unique solution for each fixed positive $ϵ$, but leading asymptotic endpoint data do not select a unique bounded family.

Multiply the equation by $x^2$. Its left side is an exact derivative, yielding the first integralFor a solution continuous at the singular endpoint, the limiting value is $C=-2{ϵ}^3\gamma$. This can also be established from the exact integrating-factor formula below, without assuming uniform bounds on derivatives near zero.

Away from zero, the outer expansion has leading term $y\sim C/x^2$. The endpoint value at one requires $C\sim {ϵ}^3$ if an exponentially amplified homogeneous term is to be avoided. It follows already thatTo exhibit the matching and explain the required boundedness, first set $x=ϵX$, $y=ϵY(X)$ for a representative without an order-one homogeneous contribution. The leading inner expansion obeysIntegration giveswhere $\mathrm{Ei}$ is the exponential integral. For $X\to \infty$, $Y\sim X^{-2}$, matching the outer term ${ϵ}^3/x^2$. For $X\to 0$, $Y\sim -1/X$, so the intermediate solution behaves as $-{ϵ}^2/x$ and reaches order one on a second scale $x={ϵ}^2\xi$.

On this second scale, the first integral becomes, at leading order, ${\xi }^2{Y}_0^{\prime }-2Y_0=1$. Equivalently its differentiated equation is $Y_0^{\prime \prime }+(2/\xi -2/{\xi }^2)Y_0^{\prime }=0$, as in the hint. The matching solution that tends to zero as $\xi \to \infty$ isIt has $Y_0\sim -1/\xi$ at large $\xi$, matching the intermediate region, and $Y_0(0)=-1/2$. Thus the nested layers at an essential singular endpoint connect an order-${ϵ}^3$ outer solution to an order-one endpoint value without an unbounded overshoot.

The issue of uniqueness is clarified by exact integration. PutFor every fixed $ϵ>0$ and every prescribed exact value of $\gamma$, the solution satisfying the right endpoint isAs $x\to 0$, the integral is asymptotic to $\mu (x)/(2{ϵ}^2)$, proving $y(0)=\gamma$. Since $h(1)>0$, the two exact endpoint values fix the two constants: for a prescribed exact $\gamma$, this finite-$ϵ$ boundary-value problem is unique.

However, specifying only the leading value $\gamma \sim -1/2$ and uniform order-one size does not select a unique asymptotic family. To see this constructively, choose $x_m=\sqrt{2}{ϵ}^{3/2}$ and an arbitrary bounded real constant $A$. WriteThe right endpoint fixesEndpoint Laplace's method gives $I=ϵ\mu (1)[1+2ϵ+O({ϵ}^2)]$. Therefore every bounded choice of $A$ haswith changes in $A$ affecting $\gamma$ only by $O({ϵ}^{-3}{e}^{-1/ϵ})$. These solutions are uniformly order one: near zero their particular part has the finite limit $\gamma$, near one it follows the small outer solution, and the freely chosen value at $x_m$ is $Ah(x_m)=O(1)$. The inner-layer homogeneous contribution is also bounded. Indeed $h$ vanishes at zero, is order one near $x_m$, and is exponentially small at one.

Consequently the order-one selected value is $-1/2$, but algebraic asymptotic endpoint data leave an order-one interior freedom. This is beyond-all-orders boundary-value sensitivity. If “constant” were interpreted as fixing $\gamma =-1/2$ exactly and independently of $ϵ$, the omitted algebraic corrections would excite an exponentially large interior homogeneous term; a genuinely uniformly bounded exact family requires the higher-order tuning of $\gamma$. Constancy in $x$ does not preclude that parameter dependence.