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

Away from $x=1/2$, the outer solution is obtained by setting $\epsilon =0$:It already satisfies both endpoint conditions, but its derivative jumps at $x=1/2$. Introduce the inner variableand write $y=\epsilon Y(\xi )$. The inner equation isMatching to the outer cusp requires $Y\sim |\xi |$ as $|\xi |\to \infty$. The even solution isSubtracting the common part $\epsilon |\xi |$ gives the composite asymptotic expansionIts 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 $C^3$.

For $\sigma =-1$, the reduced first-order outer equation isThe condition at $x=0$ selectsIndeed $z_0\to 1$ as $x\to 0^{+}$, while $z_0\to 0$ as $x\to 1^{-}$. The boundary condition at $x=1$ must therefore be supplied by a boundary layer.

SetSince $x{\log }^{2}x=O({\epsilon }^2{X}^2)$ in this layer, the convection term is lower order. The leading inner equation and matching conditions areThusThe leading uniformly valid expression isNear $x=0$ diffusion regularizes the divergent derivative of the outer approximation on the thinner scale $x|\log x|=O(\epsilon )$, but the leading value there remains the already matched constant $1$.

For $\sigma =1$, the formal outer family isEvery nonzero member diverges as $x\to 1^{-}$, so bounded matching selects the outer solution $z_0=0$. Boundary layers are now required at both endpoints.

The $x=1$ layer still has width $O(\epsilon )$ and leading profileNear $x=0$, diffusion and the logarithmically vanishing convection balance on the thinner scaleAt leading order the reaction term is smaller there. Using the supplied first integral with $\mu ={\epsilon }^2$, defineThe left layer that equals $1$ at the endpoint and matches zero isConsequently a leading composite description isIts sketch has value $1$ at each endpoint, drops sharply to an almost-zero outer plateau just to the right of $x=0$, and rises through an $O(\epsilon )$ layer just before $x=1$.