= Solution
{c}
With the negative coefficient of $y''$, the reduced equation is unchanged, but the fast mode now decays away from the right endpoint. Apply the left <boundary condition> to the <outer solution>, giving
$$
\boxed{y_{\mathrm{out},0}(x)=e^{-x^2/2}}.
$$
The <boundary layer> still has width $O(\epsilon)$. Put $X=(1-x)/\epsilon$ and $Y(X)=y(1-\epsilon X)$. The rescaled equation is
$$
-Y_{XX}-Y_X+\epsilon(1-\epsilon X)Y=0.
$$
Its leading <inner expansion> has $Y_0=A+Be^{-X}$. The <asymptotic matching condition> is $Y_0(\infty)=e^{-1/2}$, while $Y_0(0)=0$. Thus
$$
\boxed{Y_0(X)=e^{-1/2}(1-e^{-X})}.
$$
Subtracting the overlap $e^{-1/2}$ in the <additive composite expansion> gives
$$
\boxed{y_{\mathrm{comp},0}(x)=e^{-x^2/2}-e^{-1/2}e^{-(1-x)/\epsilon}}.
$$
Its graph starts exponentially close to one, follows the gently decreasing <outer solution>, and drops from near $e^{-1/2}$ to zero in a right <boundary layer> of width $O(\epsilon)$. The right endpoint value is exactly zero; the left endpoint error is $-e^{-1/2-1/\epsilon}$. Its right endpoint slope is $-e^{-1/2}(1+1/\epsilon)$, showing the rapid final descent.
\Image[/past-exam-of-the-differential-equations-2-course-of-the-university-of-oxford/2023-boundary-layers.png]
{title=Leading composite profiles for the two signs of the highest derivative}
{description=Original sketches from the derived <additive composite expansions>, for $\epsilon=0.04$ and $0.08$. Left: the positive sign gives a left <boundary layer> and zero <outer solution>. Right: the negative sign gives a right <boundary layer> and the <outer solution> $e^{-x^2/2}$. Dashed curves show the corresponding <outer solutions>.}
Back to article page
