Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 76 / 3

Assume $ϵ\downarrow 0$ and fixed $A>0$. Away from the degenerate endpoint $x=0$, the reduced equation is $A{y}_0^{\prime }+y_0=0$. The positive drift makes the rapidly varying homogeneous mode decay as one moves away from the left boundary, so the outer solution is selected by the right boundary condition:It does not satisfy the left value: its limit there is $L=e^{1/A}$.

If the layer width is $\delta$, diffusion is $O(ϵ/{\delta }^2)$, while $Ax{y}^{\prime }$ is $O(A)$ because the drift vanishes linearly. Their dominant balance gives the square-root boundary layer at a vanishing driftWriting $y=Y(\xi )$ gives $Y^{\prime \prime }+\xi Y^{\prime }+[\sqrt{ϵ}/A^{3/2}]\xi Y=0$. The first nonzero inner expansion satisfies $Y_0^{\prime \prime }+\xi Y_0^{\prime }=0$, hence $Y_0^{\prime }=B{e}^{-{\xi }^2/2}$. Imposing $Y_0(0)=1$ and matching $Y_0(\infty )=L$ givesThe overlap region is $\sqrt{ϵ/A}\ll x\ll 1$ for fixed $A$. The additive composite expansioncombines the two leading approximations and satisfies the left value exactly and the right value up to an exponentially small term. Only the first nonzero terms were requested; higher matching terms need not be added.

For fixed $A<0$, the decaying fast mode is at the other endpoint. The outer solution is now fixed by $y(0)=1$, giving $e^{-x/A}$. At $x=1$ the drift does not vanish, so the right layer has width $O(ϵ/|A|)$, with leading transient proportional to $e^{-|A|(1-x)/ϵ}$. In particular its leading inner profile is $e^{-1/A}+(1-e^{-1/A})e^{-\eta }$, where $\eta =|A|(1-x)/ϵ$. The vanishing drift at the left endpoint does not require an order-one mismatch layer for these data.

For $A=0$, the reduced equation is $x{y}_0=0$, which cannot supply the nonzero endpoint values. The equation becomes an Airy ordinary differential equation: $x={ϵ}^{1/3}\xi$ gives $Y^{\prime \prime }+\xi Y=0$. The exact solutions are combinations of $\mathrm{Ai}(-x/{ϵ}^{1/3})$ and $\mathrm{Bi}(-x/{ϵ}^{1/3})$. For $x>0$ they oscillate, with local wavelength of order $\sqrt{ϵ/x}$, rather than forming a decaying exponential layer matched to a smooth nonzero outer profile. The $O({ϵ}^{1/3})$ region resolves the turning endpoint, but oscillations extend throughout the interval. Boundary-value resonances can occur at isolated parameter values. Thus setting $A=0$ is not a continuous substitution into either fixed-sign layer approximation.

Interpret the Fourier transform as a distributional Fourier transform. Exponentially regularize the Heaviside step function:As $a\downarrow 0$, the first term tends to $\pi \delta (k)$ in the tempered distribution sense, since its integral is $\pi$ and it concentrates at zero. The second tends to $-i\mathrm{PV}(1/k)$. ThereforeThe principal-value convention is essential: the printed $1/(ik)$ is not an ordinary integrable function at zero. This normalization also satisfies $ik\hat{H}=1$, the Fourier transform of a derivative version of $H^{\prime }=\delta$.

The real-space expression $1/x$ means the principal-value reciprocal distribution. Apply even exponential damping. Oddness givesDifferentiate the integral with respect to $k$; its derivative is ${\int }_0^{\infty }{e}^{-ax}\cos (kx)dx=a/(a^2+k^2)$. Its value at $k=0$ is zero, so the integral equals $\arctan (k/a)$. Taking $a\downarrow 0$ givesThe limit is a tempered distribution limit: the regularized transforms are uniformly bounded and converge away from zero, allowing dominated convergence against a Schwartz function. This Fourier transform of a principal-value reciprocal argument also establishes the needed Dirichlet integral ${\int }_0^{\infty }\sin u/udu=\pi /2$ in its Abel interpretation.

The locally integrable logarithm has distributional derivative $(\log |x|{)}^{\prime }=\mathrm{PV}(1/x)$. By the Fourier transform of a derivative and the previous result,For $k\ne 0$, this gives $-\pi /|k|$. At zero the expression must be extended as a Hadamard finite-part integral, not treated as a locally integrable function. For example, define the finite-part inverse-absolute-value distribution byThen $kT=\mathrm{sgn}k$. Different reference cutoffs change $T$ only by a multiple of the Dirac delta distribution. The general solution of the transform equation is consequentlyIndeed multiplication by $k$ annihilates only a delta term among distributions supported at zero; delta derivatives would generate additional nonzero terms. This supplies the precise interpretation of the paper's formula, and the convention-dependent constant $C$ need not be found.

For a locally integrable periodic function, its Fourier series coefficients are $c_m={\int }_0^{1}f(x)e^{-2\pi imx}dx$. Window the function to one period by $w(x)=H(x)-H(x-1)$. With the stated Fourier transform convention,There is no extra factor of $2\pi$: the Fourier coefficient already integrates over a period of length one.

For a completely arbitrary periodic distribution, multiplication by a discontinuous window need not be defined, for example at a derivative of a delta supported on a period boundary. The sharp-window assertion therefore presupposes a windowing convention or enough regularity at the endpoints. The smooth-window Fourier coefficients of a periodic distribution use a smooth compactly supported partition-of-unity window $\chi$ with ${\sum }_{j\in \mathbb{Z}}\chi (x+j)=1$; then $c_m=\widehat{\chi f}(2\pi m)$ in the distribution sense. The logarithmic example below is locally integrable, so the printed sharp window is legitimate and no ambiguity affects it.

For $f(x)=\log |x-1/2|$, put $u=x-1/2$. Symmetry gives, for integer $m\ne 0$,An integration by parts has no endpoint term: $u\log u\to 0$ at zero and $\sin (\pi m)=0$ at the upper endpoint. It follows thatwhere the sine integral is $\mathrm{Si}(z)={\int }_0^z\sin t/tdt$. Its limiting value $\pi /2$ follows from the Abel-regularized calculation in part (ii); integration by parts of its tail gives $\mathrm{Si}(\pi m)=\pi /2-(-1{)}^m/(\pi m)+O(m^{-3})$ for positive integer $m$. Hence the Fourier coefficients of a periodic logarithmic singularity haveThe first term is the requested leading behaviour. Its alternating sign records the singularity at half a period, and its $m^{-1}$ decay records the logarithmic cusp. The $m^{-2}$ correction comes from the derivative mismatch across the periodically joined endpoints. Also $c_0=-1-\log 2$ and $c_{-m}=c_m$, consistent with the real symmetric profile.