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.

Use the Wirtinger derivatives $q_z=(q_x-i{q}_y)/2$ and $q_{\bar{z}}=(q_x+i{q}_y)/2$. The Laplace equation gives $q_{z\bar{z}}=0$, so writewhere $\Phi$ is holomorphic in the first quadrant and $\Psi$ is holomorphic in the fourth quadrant. They are independent functions: the data may be complex, so a complex-conjugation relation between them cannot be assumed.

Let $ϵ=(-1{)}^n$, where ${\beta }_1+{\beta }_2=n\pi /2$. The oblique derivative boundary conditions becomeThe hinted complementary traces would giveHere $u_2$ is a function of $x>0$ on the horizontal side; the printed $y$ range in that hint is a variable typo. We shall eliminate the same unknown traces by reflection of the holomorphic derivatives.

The polygonal ray formula can first be viewed as the Cauchy integral formula. For an oriented edge point $\zeta$, choose its spectral ray so that $e^{ik(z-\zeta )}$ decays for interior $z$. ThenInterchanging the edge and ray integrals in the supplied formula gives $\Phi (z)=(2\pi i{)}^{-1}{\int }_{\partial \Omega }\Phi (\zeta )/(\zeta -z)d\zeta$, with counterclockwise edge orientation. Thus a quadrant limit of that formula can be evaluated by the corresponding holomorphic boundary jumps, provided the additional arcs at infinity and at the corner give no contribution.

Define an holomorphic function separately in the four quadrants byThe angle condition gives $e^{2i({\beta }_1+{\beta }_2)}=e^{-2i({\beta }_1+{\beta }_2)}=ϵ$. It makes the reflected definitions consistent, with parity $H(-z)=-ϵH(z)$. On each axis oriented outward from zero, let $J=H_{\mathrm{left}}-H_{\mathrm{right}}$. The two known boundary equations give all four jumps:For example the vertical equation gives $\Phi (iy)-e^{-2i{\beta }_1}\Psi (-iy)=-i{e}^{-i{\beta }_1}{g}_1(y)$; reversing the jump order on the upward axis accounts for its displayed sign. The horizontal equation similarly supplies the positive real-axis jump, and reflection supplies the other two.

Apply the Cauchy integral formula in the four quadrants and combine their common edges. The edge contributions are exactly the known jumps of this scalar Riemann-Hilbert problem. If the corner and large-arc integrals vanish, the representation in the first quadrant isThis is the rational-angle oblique derivative problem on a quadrant representation, with all unknown traces removed. The Cauchy integrals have precisely the displayed jumps by the Sokhotski–Plemelj formula. Their reflection parity then reconstructs $\Psi$ from the fourth-quadrant value of $H$, so the two original boundary equations hold. Integrating $\Phi (z)$ and $\Psi (\bar{z})$ gives a harmonic function $q$ up to an additive constant.

To put the answer back on the spectral rays of the question, defineUse $1/(s-z)=i{\int }_0^{\infty }{e}^{ik(z-s)}dk$, $1/(s+z)=-i{\int }_0^{\infty }{e}^{ik(z+s)}dk$, and the same identities with $s$ replaced by $is$ and the ray $0\to i\infty$. The spectral representation isAll transforms used on these rays are ordinary boundary values of convergent half-line transforms. For $z=x+iy$ with $x,y>0$, the real-ray integrand decays through $e^{-ky}$ and the imaginary-ray integrand through $e^{-\mathrm{Im}kx}$.

There is an admissibility qualification in passing from a bounded polygon to this unbounded quadrant. Sufficient conditions are $H(z)\to 0$ at infinity uniformly away from the axes, with the corresponding large-arc integral tending to zero, and $|z|H(z)\to 0$ uniformly on small corner arcs. The latter allows logarithmic gradient singularities but excludes a pole with nonzero corner contribution. Smooth decaying data alone, as printed, do not impose these conditions on $q$.

For a concrete homogeneous corner ambiguity in a quadrant Laplace problem, with ${\beta }_1={\beta }_2=0$ and zero data, $q=xy$ satisfies both prescribed tangential derivative conditions but has $q_z=-iz/2$, while the displayed particular representation is zero. With ${\beta }_1={\beta }_2=\pi /2$, the zero-data solution $q=\log |z|$ has $q_z=1/(2z)$: it decays at infinity but has an excluded corner pole. Thus no unique representation of every unrestricted solution follows from the printed data alone. Without the arc conditions, one must add a homogeneous holomorphic term $P(z)$ to the displayed $q_z$, with reflected parity $P(-z)=-ϵP(z)$; permitted corner singularities are included if the admissible class allows them. With the stated decay and corner conditions that term vanishes, giving the boxed formulas. This identifies the missing solution-class conditions without assuming real-valued data.

Put $x=R^2>0$. Substituting the plane wave into the quintic real Ginzburg-Landau equation givesso the two nonzero branches areThe upper branch has $x_{+}\ge \alpha /2$ and exists for $Q^2\le \mu +{\alpha }^2/4$. The lower branch is positive when $\mu <Q^2\le \mu +{\alpha }^2/4$; at $Q^2=\mu$ it reaches the zero solution. The two branches meet at the fold $x=\alpha /2$, $\mu =Q^2-{\alpha }^2/4$. Thus for a given $\mu$, nonzero waves exist only if $\mu \ge -{\alpha }^2/4$, withThe zero solution exists for every $\mu$ independently of $Q$, because its spatial phase has no meaning.

For stability, write $A=e^{iQX}(R+u+iv)$ with real small $u,v$. Linearizing and using the steady-wave condition givesDefine $\Lambda =2R^2(2R^2-\alpha )$. A sideband of wavenumber $p$ has matrixand exact growth ratesOn the lower branch $\Lambda <0$, the uniform amplitude perturbation already grows at rate $-\Lambda >0$, so that branch is unstable. On the upper branch $\Lambda >0$, the fast amplitude mode decays and the slow phase mode hasThe coefficient of phase diffusion gives the Eckhaus boundary for a subcritical quintic amplitude equation:for robust stability to long wavelengths. At equality, ${\sigma }_{+}=-p^4/\Lambda +O(p^6)$; the leading diffusion coefficient is marginal, while nonzero sufficiently long-wave linear perturbations still decay at the next spatial order. Indeed when $4Q^2\le \Lambda$, the matrix trace is negative and its determinant is $p^2(p^2+\Lambda -4Q^2)\ge 0$, proving linear non-growth at every $p$, not just asymptotically small $p$. The neutral $p=0$ phase mode reflects the constant phase symmetry.

At the fold, $\Lambda =0$. If $Q\ne 0$, ${\sigma }_{+}=-p^2+2|Qp|$ is positive for small $p$, so the fold wave is unstable. At $Q=0$, the fold is a marginal linear point, but uniform amplitude perturbations have the one-sided nonlinear instability of a saddle-node; it is not a robust stable state. For the trivial state the linear growth rate is $\mu -p^2$, so it is stable for $\mu <0$ and only linearly marginal at $\mu =0$.

The requested regions can be expressed without eliminating $R$. In the $(\mu ,R)$ plane, nonzero waves exist wherever $\mu \ge R^4-\alpha R^2$. The robust stable part isThe left boundary is the zero-detuning wave $Q=0$; the right boundary is the marginal Eckhaus curve. Both meet at $(\mu ,R)=(-{\alpha }^2/4,\sqrt{\alpha /2})$. At fixed $\mu >-{\alpha }^2/4$, an equivalent description iswith the lower endpoint added if only marginal linear stability is required.

In the $(\mu ,Q)$ plane, eliminate $R$ from the marginal condition. It gives $R^2=[\alpha +\sqrt{{\alpha }^2+16Q^2}]/4$, henceThe equality curve is the marginal boundary; the existence boundary is $\mu =Q^2-{\alpha }^2/4$. They touch at $Q=0,\mu =-{\alpha }^2/4$ but separate for nonzero detuning, leaving an unstable wave band. Equivalently, for fixed $\mu >-{\alpha }^2/4$ the robust stable interval isThe stable amplitude is the upper root throughout this interval. The sketch uses scaled coordinates $\mu /{\alpha }^2,R/\sqrt{\alpha },Q/\alpha$, so it applies to every fixed positive $\alpha$. Green denotes stable nonzero waves; the marginal curves form its boundaries, while the surrounding gray region contains existing but unstable nonzero waves.

The six critical Fourier modes are at $\pm {\mathbf{k}}_1,\pm {\mathbf{k}}_2,\pm {\mathbf{k}}_3$, with ${\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3=0$. Reality of the pattern pairs each positive-wavevector coefficient with the conjugate coefficient at the negative wavevector. The leading critical part of the extended centre manifold parametrization is consequentlyThe displayed six-mode expression is its leading-order projection; nonlinear harmonics on the full centre manifold need not vanish identically.

A translation $\mathbf{d}$ acts by $A_i\mapsto e^{i{\mathbf{k}}_i\cdot \mathbf{d}}{A}_i$. Each monomial in the equation for $A_i$ must therefore carry wavevector ${\mathbf{k}}_i$. Since $-{\mathbf{k}}_{i+1}-{\mathbf{k}}_{i+2}={\mathbf{k}}_i$, the allowed quadratic term is $A_{i+1}^{\ast }{A}_{i+2}^{\ast }$. At cubic order the allowed terms are $A_i|A_i{|}^2,A_i|A_{i+1}{|}^2,A_i|A_{i+2}{|}^2$. Rotation through $60$ degrees acts, with one choice of orientation, asIts square cyclically permutes the amplitudes, equating the coefficients across the three equations. Its cube conjugates every amplitude; equivariance under this half-turn makes all coefficients real. It does not interchange the two cross-coupling coefficients. A reflection would do that and force $b=c$, but that symmetry is expressly absent. Thus the chiral hexagonal amplitude equations areat the retained order, with cyclic indices and real coefficients. The missing opening modulus bar in the last cubic term of the printed formula must be supplied; phase equivariance requires $|A_{i+2}{|}^2$.

Write $A_i=r_i{e}^{i{\varphi }_i}$, with all three amplitudes nonzero, and let $\Phi ={\varphi }_1+{\varphi }_2+{\varphi }_3$. Separating real and imaginary parts givesHence the phase locking of a resonant hexagonal triad obeysFor $\alpha >0$, $\Phi =0$ is phase stable and $\Phi =\pi$ is phase unstable. Two independent translation coordinates remove two phases; since the wavevectors sum to zero, the remaining phase is exactly $\Phi$. Thus a stable fully active phase-locked pattern can be represented by three positive real amplitudes after choosing the spatial origin. The statement is modulo translations, not a restriction on every representative. Boundary states must also be retained: the zero solution is stable for $\mu <0$, and rolls have two zero amplitudes. These are the necessary qualifications to a literal claim that all stable solutions are strictly positive.

In the real nonnegative subspace, a roll isor a cyclic permutation. For hexagons let $s=a+b+c$, which is positive under the assumptions. The equal-amplitude branches solve $\mu +\alpha r-s{r}^2=0$, givingwith only positive roots retained. The upper branch exists for $\mu \ge -{\alpha }^2/(4s)$; the lower branch is positive for $-{\alpha }^2/(4s)\le \mu <0$. They meet at a fold.

For a purported rectangle $(p,p,q)$ with $p,q>0$, subtract its first two equilibrium equations after dividing each by its own amplitude. The quadratic terms agree, and the difference is $(c-b)(p^2-q^2)=0$. Since $(b-a)(c-a)<0$ implies $b\ne c$, it forces $p=q$. A state with exactly two nonzero amplitudes is also impossible because the quadratic term drives the zero third amplitude. Thus there are no nontrivial rectangles.

In fact no other unequal positive equilibria are possible under the straddling cross-couplings. Here is the uniqueness of positive hexagon equilibria with straddling cross-couplings argument. Put $x_i=r_i^2$, $p=r_1{r}_2{r}_3>0$. The equilibrium equations implyConsider $b>a>c$, write $B=b-a>0$, $C=c-a<0$, and relabel cyclically so $x_1$ is maximal. If $x_2\ge x_3$, subtracting equations one and two givesThe left side is positive unless all three are equal, while the right side is nonpositive. If $x_2\le x_3$, subtract equations two and three instead:Now the left side is negative unless all are equal and the right side is nonnegative. Both cases force equality. For $c>a>b$, reverse the cyclic orientation and exchange $b,c$. Thus the only nonzero equilibria in this subspace are the stated rolls and hexagons.

At a roll the active radial eigenvalue is $-2\mu$. The two inactive amplitudes have real-part linearizationIts determinant is ${\mu }^2(a-c)(a-b)/a^2-{\alpha }^2\mu /a<0$. It therefore has a positive eigenvalue: the rolls are unstable for every $\mu >0$, which is stronger than the large-$\mu$ conclusion. In the complex system there is also a neutral translation phase, and the imaginary transverse block has the same determinant.

At a positive hexagon the real-amplitude Jacobian is circulant, with diagonal $d=-2a{r}^2-\alpha r$ and off-diagonals $p=\alpha r-2b{r}^2$, $q=\alpha r-2c{r}^2$. Its uniform eigenvector $(1,1,1)$ has eigenvalueThe other two eigenvectors lie in the zero-sum subspace; using the two nontrivial cube roots of unity givesFor sufficiently large $\mu$, $r\sim \sqrt{\mu /s}$, and the positive term $(b+c-2a)r^2$ dominates the quadratic-coupling term $2\alpha r$. Thus the hexagons are unstable at large $\mu$, while the uniform radial direction on the upper branch is stable. The two translation phases are neutral and the total phase has eigenvalue $-3\alpha r$, so the amplitude instability is genuine.

More precisely, the upper branch is amplitude stable between its fold and the chiral hexagon Hopf thresholdAt this threshold the conjugate pair crosses with nonzero frequency because $b\ne c$, permitting a Hopf bifurcation. The linear calculation locates the crossing; the nonlinear normal form decides the stability and direction of its periodic branch.

The dynamics are bounded in the retained model. For $S=r_1^2+r_2^2+r_3^2$ in the positive phase-locked subspace,Since $r_1{r}_2{r}_3\le (S/3{)}^{3/2}$, the negative quadratic-in-$S$ term dominates at large amplitude. For large positive $\mu$, the origin, rolls and hexagons are all unstable, so generic bounded dynamics cannot end at an asymptotically stable steady pattern. The chiral difference $b-c$ favors successive changes in dominant roll orientation, in the spirit of the Küppers–Lortz instability. Oscillatory competition or periodic switching is the natural behavior near the Hopf transition; linear stability alone does not establish a unique global limit cycle. With $\alpha =0$, coordinate planes are invariant and heteroclinic roll switching is possible. For $\alpha >0$ a missing third mode is regenerated when two others are present, so that exact boundary-plane heteroclinic picture is broken. Finally, sufficiently large $\mu$ is a conclusion of the cubic truncation; far from onset, omitted higher-order terms can change the physical pattern dynamics.