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Now $h_1=g_1$ and $h_2=g_2$. Define their Fourier sine transformsFirst make the elimination from the global relations explicit. Write $H_j(p)={\int }_0^{\infty }{e}^{-ips}{h}_j(s)ds$ and $N_j(p)={\int }_0^{\infty }{e}^{-ips}{n}_j(s)ds$. For $p>0$, set $\kappa =(p+B)/2$. Apply the relation of part (a) at the two negative real spectral points $k=-\kappa$ and $k=-\lambda /\kappa$. They have the same $b=-B$ and opposite $a=-p,p$. Subtracting cancels the left derivative transform, and givesLikewise use $k=-i\kappa ,-i\lambda /\kappa$, which have the same $a=-iB$ and opposite $b=-ip,ip$. Their difference givesThus both unknown normal traces have sine transforms determined by the given data. The reciprocal symmetries of the dispersion curve, rather than an arbitrary assumption about an unknown trace, perform the elimination.

Here is a direct way to finish that elimination in the integral representation of part (b). Its full-plane kernel has Fourier representationand solves $(-\Delta +4\lambda )\Gamma =\delta$. Form the four-image kernelFor $x,y>0$, the three reflected source points are outside the quadrant. Their boundary integrals against the solution vanish by Green second identity; spectrally these are superpositions of the same reflected global relations used above. Thus replacing $\Gamma$ by $G_Q$ in part (b) does not change $q$. But $G_Q$ vanishes on either boundary, so all terms multiplying unknown normal traces disappear. The remaining inward kernel derivatives reconstruct the value:Differentiate the displayed spectral kernel and combine positive and negative $p$. The bottom derivative is $(2/\pi ){\int }_0^{\infty }\sin (px)\sin (p\xi )e^{-B(p)y}dp$, and the left derivative is the corresponding expression with $x,y$ interchanged. This derives the quadrant modified Helmholtz Dirichlet Poisson formula, entirely in known data:It is an integral on the same complex dispersion curve. To display the requested complex $k$-plane contours explicitly, let $a=k-\lambda /k$ and $b=k+\lambda /k$ as before. On the real ray above $\sqrt{\lambda }$, $p=a$, $B=b$ and $dp=bdk/k$. On the imaginary ray above $i\sqrt{\lambda }$, $p=-ib$, $B=-ia$ and $dp=-iadk/k$. ConsequentlyBoth contours are oriented outward, and the sine-transform arguments on them are positive real numbers. No unknown boundary transform remains.

Every integrand solves $q_{xx}+q_{yy}-4\lambda q=0$, because $B^2-p^2=4\lambda$. As $y\downarrow 0$ with $x>0$, the first term tends to $g_2(x)$ by Fourier sine inversion and the second tends to zero by exponential convergence; the left trace follows symmetrically. Although the two separate contributions may have different corner limits, their sum is continuous at the corner when $g_1(0)=g_2(0)$. For example, the local constant-data corner contributions are the complementary angular harmonic weights, whose sum is one; the mass term is lower order under corner rescaling. Equivalently, subtract a smooth local lifting with the common corner value and use the zero-trace quadrant Green function to obtain a continuous remainder. Decay and uniqueness follow from the positive mass and the weak maximum principle, or from the positive energy identity for a difference with zero boundary data.

Take the negative-exponential Half-range Fourier transform and setThe advection-diffusion equation has the local relation with flux $X=-q_x-(ik+\alpha )q$. Integration over the half-line gives the half-line drift global relationFor the upper decay domainwrite $k=u+iv$. It is the region above $v=(\alpha +\sqrt{{\alpha }^2+4u^2})/2$, in particular above $v=\alpha$. Its boundary is oriented from upper-left infinity through $i\alpha$ to upper-right infinity, with the domain on the left. Fourier inversion and deformation of the boundary terms giveUse the dispersion symmetry elimination of a boundary trace $\nu (k)=i\alpha -k$. It preserves $\omega$ and maps $D_{+}$ into the lower half-plane. Because $i\nu +\alpha =-ik$, the global relation at $\nu$ impliesThe last term contributes an holomorphic upper-domain integral multiplied by $e^{ikx}$, so it vanishes upon closing the contour for $x>0$. The solution representation with only known data is consequentlyFor the boundary value $\cos t$, direct integration givesThis quotient is an entire function of $\omega$ when its apparent singularities at $\omega =\pm i$ are filled by their limits. Those points are not genuine forcing poles of the finite-time transform.

To obtain an exponentially decaying spectral integrand, choose $0<\theta <\pi /4$ and the exponential-decay contour for half-line drift diffusionDeform $\partial D_{+}$ to $\Gamma$. No pole is crossed: the initial-data denominator in that boundary integral has its only pole at $k=i(\alpha -a)$, which lies below $i\alpha$ since $a>0$. The apparent singularities of the forcing transform are removable. In the swept outer sectors, $e^{-\omega t}{G}_0={\int }_0^t{e}^{-\omega (t-s)}\cos sds$ has the appropriate bounded growth for the deformation. The contour formula becomesEvaluate the second quotient as a removable expression at $\omega =\pm i$, rather than splitting it into artificial residue terms. On either ray,Since $\cos 2\theta >0$, the transient factors have Gaussian decay as $r\to \infty$. The nontransient forcing term is $O(r^{-1})e^{-rx\sin \theta }$ and thus also decays exponentially for every $x>0$. The real-axis integral has the Gaussian factor $e^{-k^{2}t}$. These estimates provide the requested improvement over integration along $\mathrm{Re}\omega =0$.

The first panel shows the cubic domain from Question 1. The second uses $\alpha =1$ and $\theta =\pi /8$; its marked initial-data pole corresponds to $a=0.6$ and lies below both boundary contours.

An independent real-kernel form also makes the initial and boundary traces explicit. Apply the Dirichlet gauge transform for constant drift, $u=e^{\alpha x/2}q$, and then multiply by $e^{{\alpha }^{2}t/4}$ to obtain the heat equation. The odd-reflected heat kernel for initial data and its Dirichlet boundary-forcing heat-kernel formula yieldThe first kernel tends to the initial delta kernel on $x>0$ as $t\downarrow 0$ and vanishes at $x=0$. For the second kernel, the substitution $v=x/(2\sqrt{\tau })$ gives the unit-mass approximate boundary delta as $x\downarrow 0$, so its trace is $\cos t$; it vanishes initially for fixed $x>0$. Differentiating the kernels for $x,t>0$ verifies the advection-diffusion equation. The integrals converge for every positive $a,\alpha$, even when the exponentially weighted initial datum grows, because the spatial Gaussian controls that growth. Thus this gives the same decaying solution without introducing an additional restriction on $a$.

For a general decaying initial datum, an unweighted Fourier sine transform does not directly diagonalize the drift operator. DefineTwo integrations by parts for $q_{xx}$ and one for $q_x$ giveThe cosine transform remains unknown, so this is not a closed sine-transform evolution. In contrast, the weighted sine transform for half-line drift diffusion of $u=e^{\alpha x/2}q$ is closed:When the weighted initial transform exists, Fourier sine inversion gives $q(x,t)=e^{-\alpha x/2}(2/\pi ){\int }_0^{\infty }\sin (kx)\tilde{S}(k,t)dk$. This method requires suitable decay of $e^{\alpha x/2}{q}_0$, not merely decay of $q_0$. For the exponential datum it is an ordinary integrable sine transform only when $a>\alpha /2$. At $a\le \alpha /2$ one would need a generalized or regularized transform instead; the contour and real-kernel formulas above already work without that restriction. Therefore the ordinary sine transform of $q$ is not a closed solution method; a weighted sine transform works under the stated extra weighted-decay condition.

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.