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The negative drift moves the outflow boundary layer to $x=1$. The first-order outer equation is $-(1+x)y_0^{\prime }+y_0=0$. It must satisfy the left boundary condition, giving $y_0=1+x$. This tends to two at the right endpoint.

Use $X=(1-x)/ϵ$. The leading inner equation is $Y_0^{\prime \prime }+2Y_0^{\prime }=0$, and impose $Y_0(0)=1$ and $Y_0\to 2$ on matching. Thus $Y_0=2-e^{-2X}$, and a leading uniform asymptotic approximation isFor higher accuracy, expand the drift as $2-ϵX$ in the inner equation, solve the successive inhomogeneous equations, and match to outer corrections satisfying zero left endpoint data. There is no need for a left endpoint layer. Choosing the growing exponential at the wrong endpoint would prevent matching and is the reason the layer location matters.

Away from the thin layer, set $ϵ=0$. The outer equation isand the boundary condition selected by this first-order problem is the right-end condition. Thuswhich satisfies $y_{\mathrm{out}}(1)=1$ but has $y_{\mathrm{out}}(0)=e^{3/8}$.

The missing left condition is supplied by an outflow boundary layer. Put $X=x/ϵ$ and $y=Y(X)$. After multiplication by $ϵ$, the equation isAt leading order,The conditions $Y_0(0)=1$ and $Y_0\to y_{\mathrm{out}}(0)=e^{3/8}$ as $X\to \infty$ give

The common overlap is $e^{3/8}$. The additive composite expansion is thereforeIt satisfies both endpoint values up to exponentially small or higher-order errors and is uniformly valid to $O(1)$ on $0\le x\le 1$.

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.