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For sufficiently regular integrable functions on a half-line, Fourier sine transform inversion recovers the odd extension at continuity points. A zero endpoint value is required when demanding continuity at the endpoint.

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.

The unknown Neumann boundary condition can be removed by the Fokas method. Evaluate the global relation for the half-line free Schrodinger equation at $-k$:Consequently $k{G}_0-i{G}_1=2k{G}_0-{\hat{q}}_0(-k)+e^{i{k}^{2}t}\hat{q}(-k,t)$. In the representation from part (a), the last term has integral ${\int }_{\partial D_{+}}{e}^{ikx}\hat{q}(-k,t)dk=0$, by the Cauchy integral theorem and Jordan lemma. It is analytic in the upper half-plane, and the exponential decays on the closing first-quadrant arc. Hence an expression involving only the given data isOne may replace $t$ by $T$ in the upper limit defining $G_0$: the contribution of boundary times $s>t$ closes to zero in $D_{+}$, since $e^{-i{k}^2(t-s)}$ then decays there. Using $t$ makes causality transparent.

For an explicit proof of uniform convergence, it is useful to apply boundary lifting before inversion. Set $h(x)=e^{-x}$, $c=g_0(0)$, $w_0(x)=q_0(x)-ch(x)$, andThe compatibility condition gives $w_0(0)=0$. The Fourier sine transform of $w=q-g_{0}h$ satisfies $S_t+i{k}^{2}S=Hf$, with initial value $S_0$. The integrating factor therefore gives the equivalent representationThis last integral is absolutely and uniformly convergent for $x\ge 0$, $0\le t\le T$ under concrete sufficient hypotheses $w_0,w_0^{\prime },w_0^{\prime \prime }\in L^1$, decay of the boundary terms, and $g_0\in C^2[0,T]$. Indeed, two integrations by parts give $|S_0(k)|\le \Vert w_0^{\prime \prime }{\Vert }_1/k^2$ for $k\ge 1$. Another integration by parts, this time in $s$, givesThus the initial term is uniformly $O(k^{-2})$ and the forcing term uniformly $O(k^{-3})$. For $0<k<1$, use $|S_0(k)|\le \Vert w_0{\Vert }_1$ and the bounded time integral. An integrable majorant proves the claimed uniform convergence and permits evaluation at both boundaries.

At $t=0$, Fourier sine inversion gives $q(x,0)=c{e}^{-x}+w_0(x)=q_0(x)$. At $x=0$, the integral vanishes, giving $q(0,t)=g_0(t)$, including the compatible corner. To verify the equation, note that $h^{\prime \prime }=h$ and the transformed equation impliesAdding the lifted part gives $i{q}_t+q_{xx}=0$. Under the stated smoothness, this holds classically in the interior; differentiated spectral integrals can first be Gaussian-regularized, or read in the sine-transform $L^2$ sense and then identified with the smooth solution. Uniform convergence of $q$ itself does not require claiming uniform convergence of every differentiated integral at the corner.

For the free Schrodinger equation with compatible smooth initial and boundary data, subtract $g(t)e^{-x}$ before taking a Fourier sine transform. The lifted initial datum vanishes at the endpoint, making its sine transform $O(k^{-2})$ if its second derivative is integrable. The transformed forcing is $k(ig-g^{\prime })/(1+k^2)$; time integration followed by integration by parts makes its solution contribution uniformly $O(k^{-3})$. An integrable spectral majorant proves uniform convergence even at the compatible initial-boundary corner.