= Solution
\b[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 $t\downarrow0$. 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>.
\b[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 $t\downarrow0$, because its center lies outside the half-line. The boundary <convolution> also tends to zero for $x>0$. Thus $u(x,t)\to u_0(x)$.
\b[The <Dirichlet boundary condition>.] The identity
$$
H(-y+\alpha t,t)=e^{\alpha y}H(y+\alpha t,t)
$$
shows 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 satisfies
$$
\int_0^\infty P_\alpha(x,s)\,ds=e^{-\alpha x}\longrightarrow1,
\qquad
\int_\epsilon^\infty P_\alpha(x,s)\,ds\longrightarrow0
\quad(x\downarrow0)
$$
for every $\epsilon>0$. Hence it is a one-sided <approximate identity> at zero time, and continuity of $g_0$ gives
$$
\boxed{\lim_{x\downarrow0}u(x,t)=g_0(t),\qquad 0<t<T.}
$$
The 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 $\frac d{dt}\int_0^\infty v^2dx=-2\int_0^\infty v_x^2dx$ for real solutions, with the analogous modulus identity for complex solutions.
\b[The unheaded <sine transform> question.] The direct classical <Fourier sine transform> does not close on $u$. If
$$
S(k,t)=\int_0^\infty\sin(kx)u(x,t)\,dx,\qquad
C(k,t)=\int_0^\infty\cos(kx)u(x,t)\,dx,
$$
then <integration by parts> gives
$$
S_t=-k^2S+kg_0-\alpha kC.
$$
The 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 $v=e^{\alpha x/2+\alpha^2t/4}u$. Then $v_t=v_{xx}$, with $v_0=e^{\alpha x/2}u_0$ and $v(0,t)=e^{\alpha^2t/4}g_0(t)$. 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 \b[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.
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