= Solution
Put $a=\alpha/2$ and apply the <Dirichlet gauge transform for constant drift> $w(x,t)=e^{ax}u(x,t)$. Taking <derivatives> directly removes the <advection> term and gives the damped <heat equation> $w_t=w_{xx}-a^2w$. Thus $w_0(y)=e^{ay}u_0(y)$, while the two <Dirichlet boundary conditions> become $w(0,t)=g_0(t)$ and $w(L,t)=e^{aL}h(t)$.
The <Dirichlet heat kernel on an interval> and its damped version are
$$
H_L(x,y,\tau)=\frac2L\sum_{n=1}^{\infty}e^{-q_n^2\tau}\sin(q_nx)\sin(q_ny),\qquad q_n=\frac{n\pi}{L},\qquad K_a=e^{-a^2\tau}H_L.
$$
The <method of images> gives an alternative, often better at short times:
$$
H_L(x,y,\tau)=\sum_{j\in\mathbb Z}\left[\frac{e^{-(x-y+2jL)^2/(4\tau)}}{\sqrt{4\pi\tau}}-\frac{e^{-(x+y+2jL)^2/(4\tau)}}{\sqrt{4\pi\tau}}\right].
$$
To determine the boundary signs, multiply the equation for $w$ by the backward <heat kernel> $K_a(x,y,t-s)$ and use <integration by parts> in $y$. Since the kernel vanishes at $y=0,L$, the surviving boundary expression is $K_{a,y}(x,0,t-s)g_0(s)-K_{a,y}(x,L,t-s)e^{aL}h(s)$. Consequently an <integral representation> containing only the given data is
$$
\boxed{u(x,t)=e^{-ax}\left\{\int_0^L K_a(x,y,t)e^{ay}u_0(y)\,dy+\int_0^t\left[K_{a,y}(x,0,t-s)g_0(s)-K_{a,y}(x,L,t-s)e^{aL}h(s)\right]ds\right\}.}
$$
This is a <Dirichlet boundary-forcing heat-kernel formula>. For $0<x<L$ the short-time Gaussian decay makes the forcing integral well defined. The endpoint values are interior limits of the complete formula: evaluating a termwise sine series at an endpoint before taking the time integral loses the nonzero boundary values. At the initial corners a continuous classical solution requires $u_0(0)=g_0(0)$ and $u_0(L)=h(0)$; otherwise the same formula describes the solution away from those corners.
Here is a second <integral representation>, useful when the <integral transforms> are explicit. Extend $g_0,h$ by zero after $T$; the values of the extension cannot affect a solution at $t<T$. Write their <Laplace transforms> as $G(p),H(p)$ and set $q=\sqrt{p+a^2}$ using the <principal square root>. The transformed <Dirichlet Green function> for $-\partial_x^2+q^2$ is
$$
R_q(x,y)=\frac{\sinh(q\min(x,y))\sinh(q[L-\max(x,y)])}{q\sinh(qL)}.
$$
It vanishes at both endpoints and its first <derivative> in $x$ jumps by $-1$, which verifies the sign of its unit source. The <resolvent kernel for Dirichlet advection-diffusion on an interval> gives the transformed solution
$$
W(x,p)=\int_0^L R_q(x,y)e^{ay}u_0(y)\,dy+\frac{\sinh(q[L-x])}{\sinh(qL)}G(p)+e^{aL}\frac{\sinh(qx)}{\sinh(qL)}H(p).
$$
The <Bromwich inversion formula> therefore yields
$$
\boxed{u(x,t)=\frac{e^{-ax}}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{pt}W(x,p)\,dp,\qquad\gamma>0.}
$$
The square-root notation creates no genuine branch singularity here: all three kernels are even in $q$. Their only spatial <resolvent operator> poles are $p=-a^2-q_n^2$. This transformed formula and the causal <heat kernel> formula represent the same solution in the usual smooth-data class.
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