Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 86 / 1

Extend $q$ by zero to $x<0$. Its ordinary Fourier transform is precisely the half-line transform $\hat{q}$, so Fourier inversion gives the real-axis term for every interior point $x>0$. At $x=0$ the zero extension can have a jump; the interior inversion is the assertion needed here.

Write $k=u+iv$. The contour is the upper branch of a hyperbola,with $u$ increasing from $-\infty$ to $\infty$. In the region above it, $\nu (k)=i\alpha -k$ has nonpositive imaginary part. Therefore $\hat{q}(\nu (k))$ is analytic there: its defining integral has factor $e^{iux-(v-\alpha )x}$ and converges for a sufficiently decaying $q$. Smoothness and integration by parts also give $\hat{q}(\nu )=O(|\nu {|}^{-1})$ away from bounded spectral sets.

For an interior spatial point $x>0$, the factor $e^{ikx}$ decays exponentially as $\mathrm{Im}k\to +\infty$. Truncate the hyperbolic contour and close it above, using the Cauchy integral theorem; the closing segments vanish by this decay and the transform estimate. No poles occur in the enclosed region. Consequently the half-line drift null contour identity isAdding any constant multiple of zero to Fourier inversion proves the proposed inversion, for every $c$. Oscillatory real-axis inversion may be understood by a Gaussian cutoff followed by its usual limit. The same contour argument allows a spectral multiplier analytic above $L$ with polynomial growth, because the exponential still controls the closing paths. We will need this slight extension for derivative boundary data.

Set $\omega (k)=k^2-i\alpha k$ and $\nu (k)=i\alpha -k$, so $\omega (\nu )=\omega (k)$. For boundary traces $h_j(s)={\partial }_x^{j}q(0,s)$, defineThe Half-range Fourier transform of the equation is obtained by two integrations by parts:Solving this scalar equation gives the drift-diffusion half-line global relationIn particular the reflected relation is valid on $L$, since $\mathrm{Im}\nu (k)\le 0$ there.

In the region between the real axis and $L$, $\mathrm{Re}\omega \ge 0$. Thus a boundary contribution $e^{-\omega t}{G}_j={\int }_0^t{e}^{-\omega (t-s)}{h}_j(s)ds$ admits an upward contour deformation from the real axis to $L$. Large-contour estimates follow from the spatial exponential and the time integral; at the upper time endpoint one can integrate in time once or use a small contour displacement. The initial-data transform is retained on the real axis. Hence, if $I_C[F]=(2\pi {)}^{-1}{\int }_C{e}^{ikx-\omega (k)t}F(k)dk$, ordinary inversion first givesFor the Dirichlet boundary condition, take $h_0=g_0$. Subtract the null integral of the reflected time-dependent transform, using the inversion from part (a) with $c=-1$. The unknown derivative transform cancels because $i\nu +\alpha =-ik$. This derives the Dirichlet spectral formula for half-line constant drift:Only initial and prescribed boundary data occur in this expression.

For the Neumann boundary condition, take $h_1=g_1$. The reflected global relation reads $e^{\omega t}\hat{q}(\nu ,t)=Q(\nu )-G_1+ik{G}_0$. Solve for $G_0$ and substitute it in the preliminary representation. The leftover term is a null integral with multiplier $(ik+\alpha )/(ik)$, analytic above $L$; its pole at zero is below the contour. Thus the Neumann spectral formula for half-line constant drift isThere is no residue from the origin, since this elimination never moves the contour across it. The given derivative is the inward-coordinate derivative $q_x$, not the outward derivative $-q_x$. The initial compatibility for this case is $q_0^{\prime }(0)=g_1(0)$.

The Dirichlet spectral formula for half-line constant drift can be checked both spectrally and through its heat kernel. For $x>0,t>0$, each factor $e^{ikx-\omega t}$ solves the differential equation, since $-\omega =(ik{)}^2+\alpha ik$. Differentiating the upper limit in $G_0$ introduces onlyThis vanishes by closing upward for $x>0$, or by recognizing the polynomial Fourier multiplier as a distribution supported at $x=0$. Thus differentiation under regularized contours, followed by the smooth interior limit, proves the PDE. At $t=0$ the boundary transform vanishes and the reflected initial-data integral is zero by part (a), leaving Fourier inversion of $q_0$.

For a direct verification of the boundary limit, evaluate the contour integrals. LetThe real-axis Gaussian Fourier transform gives this kernel. In the reflected initial term, $e^{-i\nu (k)\xi }=e^{ik\xi +\alpha \xi }$, so a valid contour deformation gives $e^{\alpha \xi }K(x+\xi ,t)$. The boundary integral gives $-(2{\partial }_x+\alpha )K(x,t)=xK(x,t)/t$. Consequently the same solution iswhereInterchange of the space or time integrals with the spectral integral can first be performed with cutoffs; the displayed Gaussian kernels justify their removal for the stated smooth decaying data. The possible exponential $e^{\alpha \xi }$ is compensated by Gaussian decay, so exponential decay of $q_0$ is not an extra assumption.

At $x=0$, $K(-\xi ,t)=e^{\alpha \xi }K(\xi ,t)$, so the initial-data contribution has zero boundary trace. The kernel $P$ is a time approximate identity as $x\downarrow 0$. Indeed it is the standard positive heat boundary kernel multiplied by $e^{-\alpha x/2-{\alpha }^{2}u/4}$. The standard kernel has total mass one on $u>0$, as the substitution $r=x/(2\sqrt{u})$ shows; its mass on $u\ge \delta$ is $O(x/\sqrt{\delta })$. The additional factor tends to one uniformly on sufficiently short time intervals as $x\downarrow 0$, so ${\int }_0^{t}P(x,u)du\to 1$ for fixed $t>0$. Continuity of $g_0$ and the vanishing tail then giveFor fixed $x>0$, the first kernel tends to a point mass at $\xi =x$, the reflected kernel is exponentially small as $t\downarrow 0$, and the boundary-forcing integral is also exponentially small. This independently proves $q(x,t)\to q_0(x)$. The compatibility $q_0(0)=g_0(0)$ supplies continuity at the initial corner.