Away from $x=1/2$, the <outer solution> is obtained by setting $\varepsilon=0$:
$$
y_0(x)=\left|x-\frac12\right|.
$$
It already satisfies both endpoint conditions, but its derivative jumps at $x=1/2$. Introduce the <inner variable>
$$
\xi=\frac{x-\frac12}{\varepsilon}
$$
and write $y=\varepsilon Y(\xi)$. The inner equation is
$$
Y''-Y=-|\xi|.
$$
Matching to the outer cusp requires $Y\sim|\xi|$ as $|\xi|\to\infty$. The even solution is
$$
Y(\xi)=|\xi|+e^{-|\xi|}.
$$
Subtracting the common part $\varepsilon|\xi|$ gives the <composite asymptotic expansion>
$$
\boxed{
y(x;\varepsilon)\sim
\left|x-\frac12\right|
+\varepsilon
\exp\left(-\frac{|x-\frac12|}{\varepsilon}\right)
}.
$$
Its endpoint errors are exponentially small.

At $x=1/2$, the first and second derivatives from the two sides agree, but the third derivatives have opposite signs. The composite expansion is therefore $C^2$ but not $C^3$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 336 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-336/2/a/solution