The solution and its first <derivative> are <continuous> at $x=1$: a jump would create a <Dirac delta distribution> or its derivative, absent from the forcing. Put $r=x-1$, $A=a+1$ and $B=b$. The <outer expansions>, fixed by the respective endpoint <boundary conditions>, are
$$
y_L=A+r+\varepsilon\bigl[r+1+A\log|r|\bigr]+\cdots\quad(r<0),
\qquad y_R=B+\varepsilon B\log r+\cdots\quad(r>0).
$$
Balancing diffusion and advection near $r=0$ gives the <interior layer at a simple zero of advection> with <inner variable> $z=r/\varepsilon$. The leading <inner expansion> solves $Y_0''+2zY_0'=0$, so matching to $A$ and $B$ gives
$$
Y_0=c+d\operatorname{erf}z,\qquad c=\frac{A+B}{2},\qquad d=\frac{B-A}{2}.
$$
For the next term define the <Gaussian drift primitives>
$$
E_0(z)=\int_0^z e^{-t^2}\int_0^t e^{u^2}\,du\,dt,
\qquad E_1(z)=\int_0^z e^{-t^2}\int_0^t e^{u^2}\operatorname{erf}u\,du\,dt.
$$
Writing $H$ for the <Heaviside step function>, let $F(z)=zH(-z)+(\sqrt\pi/4)\operatorname{erf}|z|$. Its value and first derivative match at zero and $(D^2+2zD)F=2zH(-z)$. Thus
$$
Y_1=F+2cE_0+2dE_1+\alpha+\beta\operatorname{erf}z,
$$
where the logarithmic overlaps fix
$$
\alpha=\frac12-\frac{\sqrt\pi}{4}+c\log\varepsilon-2cC_1,
\qquad\beta=-\frac12+d\log\varepsilon-2dC_2.
$$
Here $C_1,C_2$ are the constants in the supplied large-positive-$z$ limits of $E_0,E_1$. The terms $\varepsilon\log\varepsilon$ are essential <switchback terms>; discarding them would not give accuracy through $O(\varepsilon)$.

Subtracting the common overlap from the <inner expansion> and <outer expansions> gives the <additive composite expansion>
$$
\boxed{y_{\rm comp}(x)=Y_0(z)+\varepsilon Y_1(z)+\frac{\varepsilon r}{2}(1-\operatorname{erf}z),\qquad z=\frac{x-1}{\varepsilon}.}
$$
The standard overlap subtraction leaves $\varepsilon rH(-r)$. Replacing $H(-r)$ by $(1-\operatorname{erf}z)/2$ changes the value only by $O(\varepsilon^2)$ uniformly and gives a composite with continuous first derivative. The last term restores the endpoint values through the retained order. When $b=a+1$, $d=0$ and the order-one error-function jump disappears. An $O(\varepsilon)$ interior adjustment remains to accommodate the leading outer derivative mismatch, together with logarithmic matching when $A\ne0$.

Reversing the diffusion sign changes the leading inner equation to $-Y_0''+2zY_0'=0$. Its nonconstant solution grows like the integral of $e^{z^2}$, so bounded matching forces the same leading value $C$ on both sides. The bulk is $C+r$ on the left and $C$ on the right; both endpoint conditions are instead supplied by decaying <endpoint layers for reversed diffusion>, of width $\varepsilon^2$. A weaker width-$\varepsilon$ interior adjustment matches derivatives and selects $C$. Indeed, its first-order equation has a non-growing solution only if
$$
\int_{-\infty}^{\infty}e^{-z^2}\bigl(2C+2zH(-z)\bigr)\,dz=0,
\qquad C=\frac1{2\sqrt\pi}.
$$
The leading structure is consequently
$$
y\sim C+rH(-r)+(a-C+1)e^{-2x/\varepsilon^2}+(b-C)e^{-2(2-x)/\varepsilon^2},
$$
with the smaller interior correction understood. Unlike the positive-diffusion case, the endpoint values are not transported into two distinct order-one inner limits.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 336 / 2 / i / Solution

For $-\varepsilon^2y''+2(x-x_*)y'+\cdots=f$ on an interval straddling $x_*$, the bounded leading interior solution cannot connect unequal constants. A common bulk value is instead selected by an inner <solvability condition>, and width-$\varepsilon^2$ endpoint layers enforce the two boundary values. At endpoints $x=0,2$ with $x_*=1$, their leading decaying factors are $e^{-2x/\varepsilon^2}$ and $e^{-2(2-x)/\varepsilon^2}$. The interior layer still corrects derivative mismatches at smaller amplitude.

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 Incoming links: Endpoint layers for reversed diffusion