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Take real $\kappa$ and use the L2 norm on the spatial interval. Existence, uniqueness and continuous dependence are the three requirements of Hadamard well-posedness. An energy method supplies the decisive estimate. For a smooth solution with homogeneous Dirichlet boundary conditions, integration by parts givesThe drift contributes only a boundary term, which vanishes. The Poincare inequality $\Vert u_x{\Vert }_2^2\ge {\pi }^2\Vert u{\Vert }_2^2$ further givesApply the same argument to the difference of two solutions to obtain uniqueness and continuous dependence on the initial data.

For existence, use the Dirichlet gauge transform for constant drift: $w=e^{\kappa x/2}u$ satisfies $w_t=w_{xx}-{\kappa }^{2}w/4$ with zero boundary values. Expanding $w_0$ in its Fourier sine series givesFor $u_0\in L^2(0,1)$ the series defines a solution continuous in $L^2$ down to $t=0$ and smooth for positive time; multiplication by the fixed bounded exponentials preserves this interpretation. Its energy estimate follows by approximation with smooth initial data. For a classical solution at the initial corners, require the usual smoothness and boundary compatibility instead. The problem is well posed in $L^2$, with a contraction estimate independent of the initial data.

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.

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 $w_t=w_{xx}-a^{2}w$. 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 areThe method of images gives an alternative, often better at short times: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 isThis 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 $-{\partial }_x^2+q^2$ isIt 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 solutionThe Bromwich inversion formula therefore yieldsThe 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.

After the Dirichlet gauge transform for constant drift, put $q=\sqrt{p+a^2}$. The Dirichlet Green function of $p-{\partial }_x^2+a^2$ isThe original unweighted resolvent kernel is $e^{-ax}{R}_q(x,y)e^{ay}$. Its poles are $p=-a^2-(n\pi /L{)}^2$. The kernel is even in $q$, so the apparent square-root branch point is removable.