 Solution

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.