Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2009 / A / 2

Set $x=\delta X$ and write $y(x)=Y(X)$. The equation becomesA dominant balance must compare all three coefficients, rather than just the two derivative terms.

For $-1<\alpha <1$, balancing diffusion and drift gives $\delta ={ϵ}^{1/(\alpha +1)}$. Their common coefficient tends to infinity, so the reaction term is smaller and the leading inner equation isThis derivative is integrable at infinity, allowing an ordinary localized endpoint adjustment. Thus the thickness is ${ϵ}^{1/(\alpha +1)}$ in this range.

For $\alpha =1$, $\delta =\sqrt{ϵ}$ makes all three terms comparable. Keep all of them:One decaying solution is $e^{-X^2/2}$, and reduction of order gives the other. This is a square-root-thickness layer with all terms retained.

For $\alpha >1$, the first variation scale is $\delta =\sqrt{ϵ}$. Keep diffusion and reaction, and drop drift at this scale:This leading motion is oscillatory, not a monotone decaying boundary-layer profile. To describe localization one must also resolve the slower damping envelope. Removing the first derivative in the original equation bygivesNear the endpoint the wavelength is $O(\sqrt{ϵ})$, while the envelope changes on $x=O({ϵ}^{1/(\alpha +1)})$. Hence an oscillatory endpoint region is possible, but it has distinct wavelength and envelope scales; calling the diffusion-drift balance alone its leading equation would incorrectly discard the reaction term. This generalizes the oscillatory endpoint with a quadratically vanishing drift.

For $\alpha \le -1$, the diffusion-drift balance would require ${\delta }^{\alpha +1}=ϵ$, which gives no shrinking positive power scale. At every shrinking scale the singular drift dominates diffusion, so no regular finite-amplitude diffusion boundary layer of the preceding kind occurs. The coefficient itself is singular at zero; singular endpoint solutions are a different issue from a bounded boundary-layer correction.

Put $p=y^{\prime }$. The dynamical system isThe nullclines are $p=0$ and $p=e^y$. There are no equilibria, since they never intersect. To the right in the $(y,p)$ plane means $p>0$, and to the left means $p<0$. The arrows point upwards above $p=e^y$ and downwards below it.

Near the repelling critical manifold $p=e^y$, an orbit can move on the slow $x=O(1)$ scale, with $y^{\prime }\simeq e^y$. More precisely the slow trajectory has $p=e^y+ϵe^{2y}+\cdots$. Away from this narrow region, $p$ changes rapidly on $x=O(ϵ)$; an $O(1)$ change of $y$ in a fast layer has $p=O({ϵ}^{-1})$.

For example, take a fixed $A<-\log 2$. A trajectory starting at $y=A$ can shadow the slow curve, then depart upwards so that $y$ rapidly increases to zero at the right endpoint. This gives a terminal outflow boundary layer at $x=1$, of width $O(ϵ)$. For $-\log 2<A<0$, the outer trajectory reaches a positive value near one, and the terminal adjustment is downwards: the orbit turns below the nullcline, crosses $p=0$, and returns to $y=0$. The boundary layer is still at $x=1$. These directions follow from the phase portrait and agree with the matched solution below.

In the outer expansion, omit $ϵy^{\prime \prime }$ to get $y_0^{\prime }=e^{y_0}$. Integrating and imposing the left boundary condition givesThis is regular throughout $[0,1]$. Its right-end value is $B=-\log (C-1)$, generally different from zero.

At $x=1$ set $X=(x-1)/ϵ$. The equation is $Y^{\prime \prime }-Y^{\prime }+ϵe^Y=0$, so the leading inner expansion is $Y=C_0+C_1{e}^X$. Matching as $X\to -\infty$ gives $C_0=B$, while $Y(0)=0$ gives $C_1=-B$. ThusThe common part is $B$. Subtract it once when forming the additive composite expansion:There is one generic boundary layer, at the right endpoint, of width $ϵ$. A left-end exponential would grow into the interval and cannot supply a decaying matching correction. At $A=-\log 2$ the displayed layer amplitude vanishes; the leading outer solution already meets the right condition, though a smaller correction can still be needed at higher order. The composite meets $y(1)=0$ exactly and $y(0)=A$ up to an exponentially small term.