 Solution

The partial differential equation. Each term in $K_{\alpha }$ is a translated heat kernel and satisfies $({\partial }_t-{\partial }_x^2-\alpha {\partial }_x)K_{\alpha }=0$ for $t>0$. The same holds for $P_{\alpha }$, either by direct differentiation or because it is $-(2{\partial }_x+\alpha )H(x+\alpha t,t)$. For $x>0$, $P_{\alpha }(x,t)$ and its derivatives vanish faster than any power as $t\downarrow 0$. Therefore differentiating the boundary convolution creates no extra upper-endpoint term. Gaussian domination justifies differentiating both integrals on compact subsets of $x,t>0$, proving the advection-diffusion equation.

The initial condition. For fixed $x>0$, the first Gaussian in $K_{\alpha }$ is an approximate identity centered at $y=x+\alpha t\to x$. Its integral tends to $u_0(x)$. The reflected Gaussian is exponentially small as $t\downarrow 0$, because its center lies outside the half-line. The boundary convolution also tends to zero for $x>0$. Thus $u(x,t)\to u_0(x)$.

The Dirichlet boundary condition. The identityshows that $K_{\alpha }(0,y,t)=0$, so the initial-data contribution vanishes at zero. The boundary contribution must be evaluated as a limit, not by substituting $x=0$ inside its singular integral. The positive kernel satisfiesfor every $ϵ>0$. Hence it is a one-sided approximate identity at zero time, and continuity of $g_0$ givesThe mass formula follows from the Gaussian Laplace integral, or from the decaying solution of the corresponding constant-coefficient ordinary differential equation. Compatibility $u_0(0)=g_0(0)$ gives the continuous corner value. These arguments also verify the equivalent contour solution in (iii). In the decaying energy class the solution is unique: the difference $v$ of two solutions has zero data and $\frac{d}{dt}{\int }_0^{\infty }{v}^{2}dx=-2{\int }_0^{\infty }{v}_x^{2}dx$ for real solutions, with the analogous modulus identity for complex solutions.

The unheaded sine transform question. The direct classical Fourier sine transform does not close on $u$. Ifthen integration by parts givesThe drift introduces an unknown cosine transform; the usual scalar sine-transform solution of the heat equation is therefore unavailable directly.

A Dirichlet gauge transform for constant drift does provide a qualified alternative. Set $v=e^{\alpha x/2+{\alpha }^{2}t/4}u$. Then $v_t=v_{xx}$, with $v_0=e^{\alpha x/2}{u}_0$ and $v(0,t)=e^{{\alpha }^{2}t/4}{g}_0(t)$. If these weighted data have the decay needed for an ordinary Fourier sine transform, it solves the transformed problem and produces exactly the kernel above. Mere decay of $u_0$ does not ensure this weighted integrability. Thus not directly by the classical sine transform of $u$; yes after a gauge transformation when the required weighted-transform hypotheses hold, or after a justified cutoff/limit argument.