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Use a time Laplace transform and a spatial Green function. The first-order transport term must be retained: the Airy resolvent kernel for ${\partial }_x^3+p$ alone would solve a different equation. The governing linear partial differential equation is the linear dispersive Stokes equation, distinct from the viscous-fluid Stokes equation.

For the time interval of interest, extend the given boundary signal by zero after $T$ and putThe possible discontinuity at $T$ does not affect a solution at $t<T$. Equivalently, use any sufficiently regular continuation beyond $T$: the resulting causal Green function produces the same earlier solution. We seek the usual solution with spatial decay and a time transform, rather than an unrestricted spatially growing solution on the half-line.

Writing $Q(x,p)$ for the time Laplace transform, the Laplace transform of a derivative givesThe corresponding characteristic roots are the three roots of a polynomialFor $\mathrm{Re}p>0$, label the unique root of negative real part by $r_{-}(p)$, and the two roots of positive real part by $r_1(p),r_2(p)$. This spatial root splitting for the dispersive Stokes resolvent is easily checked: if $r=ik$, then $p=-i(k-k^3)$ is imaginary, so no root crosses the imaginary axis as $p$ varies in the right half-plane. For real $p>0$ the cubic is strictly increasing on the real line and has one negative real root, while its conjugate pair has positive real parts. The root count remains the same throughout that connected half-plane. A repeated root would satisfy $3r^2+1=0$ and again put $p$ on the imaginary axis. All roots are therefore simple there, and $r_{-}(p)$ is a nonzero holomorphic function.

Set $d_j=3r_j^2+1$, including $j=-$. Define the Stokes resolvent kernel with advectionThe sign on the negative half-line is essential. To verify the Green function, use the partial fraction decompositionIts expansion at infinity yields the reciprocal-polynomial root identitiesConsequently $R_p$ and $R_p^{\prime }$ are continuous at zero, and $R_p^{\prime \prime }$ has jump one. Each piece solves the homogeneous ordinary differential equation, so the jump conditions for a third-order Green function imply, in the sense of distributional derivatives,Both pieces decay in their respective spatial directions.

Now defineIt satisfies the transformed inhomogeneous equation. The difference $Q-V$ solves the homogeneous linear ordinary differential equation; spatial decay leaves only $e^{r_{-}x}$. Imposing the Neumann boundary condition fixes its coefficient, giving the half-line Stokes solution with a boundary derivativeNo division by a vanishing root is involved, since $r_{-}=0$ would require $p=0$. Moreover the remaining boundary trace is explicitly known from $q_0$:Thus introduce the completely explicit initial-data kernelFor any upward Bromwich contour $\mathrm{Re}p=\sigma >0$ to the right of any transform singularities, the requested integral representation isThe contour integral is understood in the standard Bromwich inversion formula sense, with truncation limits if it is not absolutely convergent. Every term is expressed in the supplied data and algebraic characteristic roots; there are no unknown Dirichlet or second-derivative boundary traces.

The transformed equation verifies the linear partial differential equation and the initial term, while taking the spatial derivative gives $Q_x(0,p)=G(p)$. Bromwich inversion formula therefore supplies the prescribed derivative for $0<t<T$ and the initial value for $x>0$. For zero initial and boundary data, the transformed solution is a decaying homogeneous mode with zero boundary derivative, hence vanishes; this proves uniqueness in the stated transform/decay class. A change of the boundary continuation after $T$ has transform $e^{-pT}$ times another causal transform and contributes only after $T$, by the Laplace transform time-shift rule. The printed dot on $q_0$ is interpreted as its derivative in its sole variable $x$, so the stated corner compatibility is $q_0^{\prime }(0)=g_1(0)$. Additional differentiability at the corner requires the corresponding higher-order compatibility, but is not needed for the representation at positive $x,t$.

Interpret the malformed printed domain inequality as the first quadrant $x>0,y>0$, as indicated by the two axis boundary conditions. There is also a genuine missing hypothesis: decay of the boundary data does not require the harmonic function itself to be bounded. For zero data, both $q=0$ and $q=xy$ satisfy the Laplace equation and both edge values, but the Wirtinger derivative of the latter isThus the data do not determine an arbitrary solution's derivative without a growth restriction. We derive the integral representation for the bounded solution, which is also the standard decaying solution when the continuous edge data decay. Any other admissible unbounded solution can differ by a zero-boundary harmonic function.

The conformal map $w=z^2$ sends the quadrant bijectively to the complex upper half-plane: the positive real axis maps to $s>0$, and the positive imaginary axis maps to $s<0$. Define its real-axis boundary function byThe compatibility at the corner makes $f$ continuous there. The Dirichlet Poisson integral in a quadrant follows from the upper-half-plane Poisson kernel:It is bounded by the supremum of the bounded edge data, and the kernel's approximate identity property supplies both Dirichlet boundary conditions. More explicitly, for $z^2=a+ic$, where $a=x^2-y^2$ and $c=2xy>0$, this integral isIt displays directly which part of the real-line boundary corresponds to each original axis.

To obtain its derivative, first take real data and use the Schwarz integral formula. A holomorphic function whose real part is the Poisson solution in the $w$-plane isThe regularizing subtraction is independent of $w$ and affects only the imaginary normalization; it permits bounded data without an artificial integrability assumption. Then $q=\mathrm{Re}F(z^2)$. By the chain rule and the Wirtinger derivative,The factor is $z$, not $2z$: taking the real part contributes the factor $1/2$ in the complex derivative. Differentiation under the integral sign is justified uniformly on compact interior subsets because the denominator stays away from the real boundary and decays quadratically at infinity. Splitting the real integral at zero and substituting $s=r^2$ or $s=-r^2$ gives the holomorphic derivative of a quadrant Dirichlet solution:This converges for bounded data, and in particular for the stated sufficiently decaying data. No interior pole is encountered since $\mathrm{Im}(z^2)=2xy>0$. For complex-valued data the same formula holds by applying the real-data construction separately to their real and imaginary parts and using linearity.

An equivalent form, often convenient when the data's derivatives decay, comes from integration by parts. Sincethe two endpoint terms are $-g_2(0)/z^2$ and $+g_1(0)/z^2$. They cancel by the specified compatibility, leavingThis is the harmonic quadrant solution from tangential boundary derivatives applied to the actual derivatives $g_1^{\prime },g_2^{\prime }$; the supplied $g_1,g_2$ in this question are boundary values, not derivatives. Omitting those primes would solve a different boundary problem.

For completeness, bounded Dirichlet uniqueness in a quadrant selects the solution just constructed. The difference of two bounded solutions is zero on both axes. Odd reflection across each axis gives a bounded harmonic extension to the punctured plane; the removable singularity for a bounded harmonic function fills the corner. The harmonic Liouville theorem makes the extension constant, and oddness makes it zero. Decaying data give a decaying Poisson extension: split the real-axis integral into a fixed compact interval and a tail with arbitrarily small data; the compact contribution tends to zero as $|z^2|\to \infty$, while the tail is bounded by that small supremum.

As literally printed without boundedness or a growth condition, the most one can assert isThe zero-boundary harmonic growth in a quadrant includes $v=cxy$ and more generally $v=\mathrm{Im}(z^{2m})$. This is an actual nonuniqueness of the printed problem, not a defect in the integral for the normalized bounded solution.

At the absolute wavenumber, the group velocity $d\omega /dk$ is zero. The corresponding accessible saddle therefore governs the disturbance seen at a fixed position. Its exponential growth rate isSaddle accessibility can be checked directly here. With $D=1+i{c}_d$, whose real part is positive, Fourier inversion gives the Green function of the linear complex Ginzburg-Landau equation,At fixed $x$ its exponential rate is $\mathrm{Im}{\omega }_0$, while along the packet centre $x=Ut$ it is $\mu$. For real wavenumbers of Fourier modes, the temporal growth rate is $\mathrm{Im}\omega (k)=\mu -k^2$, so the flow is temporally unstable exactly when $\mu >0$.

Consequently the classifications for a localized disturbance in this laboratory frame areFor convective hydrodynamic instability, a travelling wave packet amplifies but the response at each fixed position decays. For absolute hydrodynamic instability, that fixed-position response amplifies. The equality is the marginal absolute threshold: its exponential rate is zero and the impulse response has a $t^{-1/2}$ prefactor. For $\mu <0$ there is no temporal growth; $\mu =0$ is temporally marginal. When $U=0$ the convective window is empty. These conclusions use the infinite-line impulse problem; a general dispersion relation requires its own spatial-branch or saddle selection.

The cubic amplitude saturation equation $R^{\prime }=R(Q^2-R^2)$ and the Picard-Lindelof theorem imply that positive solutions cannot cross the equilibrium $Q$. If $0<R_0<Q$, then $R$ stays below $Q$ and increases. If $R_0>Q$, then it stays above $Q$ and decreases. The explicit cubic amplitude saturation solution tends to $Q$ in both cases, soThe same relation makes the distance a decreasing monotone function in any spatial norm for which the plane wave has finite norm, for example on a periodic domain. At large time, $R-Q\sim -\frac{Q}{2}(Q^2/R_0^2-1)e^{-2Q^{2}t}$.

The printed monotonicity wording must be interpreted as monotone amplitude relaxation, or monotone decay of the distance between the waves. Complex numbers have no ordinary monotone ordering, and the real part at a fixed point can oscillate. For instance, $k=U=1$, $\mu =2$ and $R_0=1/2$ give $\mathrm{Re}{\psi }_R(0,t)=R(t)\cos t$, which is not monotone despite the monotone amplitude. This result concerns the common-phase single-mode family, not stability against arbitrary spatial perturbations.

Use $G=c=1$ and the physical real part of the printed plane wave. Its null wavevector obeys $k^{\alpha }{k}_{\alpha }=0$, so each component satisfies $\square h_{\alpha \beta }=0$. Direct differentiation of the defined stress tensor givesThe trace is proportional to $k^{\nu }{k}_{\nu }$ for a single plane wave and hence vanishes. ThereforeThis is local conservation of energy and momentum conservation; spatial integration gives conserved total charges when the boundary flux vanishes. For a finite-amplitude complex phasor, one takes the real field before quadratic products and may time-average: $⟨{\tau }_{\mu \nu }⟩=k_{\mu }{k}_{\nu }{H}^{\alpha \beta }{\overline{H}}_{\alpha \beta }/(64\pi )$ in transverse-traceless gauge. Its energy density is nonnegative and its flux travels in the positive $z$ direction.

Assume incompressible inviscid layers, gravity $-g{\mathbf{e}}_y$, no surface tension, and irrotational perturbations of the uniform base flows. The full bulk equations are $\nabla \cdot {\mathbf{u}}_j=0$ and ${\partial }_t{\mathbf{u}}_j+{\mathbf{u}}_j\cdot \nabla {\mathbf{u}}_j=-{\rho }_j^{-1}\nabla p_j-g{\mathbf{e}}_y$. At the rigid walls $u_{1y}(h)=u_{2y}(-h)=0$. At the material interface $y=\eta (x,t)$ each layer obeys $u_{jy}={\eta }_t+u_{jx}{\eta }_x$, and $p_1=p_2$. There is no tangential-velocity matching condition across this vortex sheet.

Write ${\mathbf{u}}_j=(U_j,0)+\nabla {\varphi }_j$, with $U_1=U$, $U_2=-U$. Then ${\nabla }^2{\varphi }_j=0$, and the wall conditions are ${\partial }_y{\varphi }_1(h)={\partial }_y{\varphi }_2(-h)=0$. Linearizing the interface conditions at zero givesThe second follows from linear Bernoulli pressure $p_j^{\prime }=-{\rho }_j({\partial }_t+U_j{\partial }_x){\varphi }_j$ and the hydrostatic pressure shift $-{\rho }_{j}g\eta$.

For $k>0$, use amplitudes ${\varphi }_1=B_1\cosh k(y-h)e^{ikx+\sigma t}$ and ${\varphi }_2=B_2\cosh k(y+h)e^{ikx+\sigma t}$. Kinematics gives $B_1=-(\sigma +ikU)A/[k\sinh kh]$ and $B_2=(\sigma -ikU)A/[k\sinh kh]$. The dynamic condition becomesCompleting the square yieldsA positive real part occurs precisely whenThe function $\tanh (kh)/k$ decreases strictly from $h$ at $k=0$ to zero at infinity. Indeed its derivative has numerator $kh{\mathrm{sech}}^2(kh)-\tanh (kh)<0$, since $\sinh (2kh)>2kh$. Therefore the minimum speed producing instability at every finite nonzero wavenumber isAt equality every $k>0$ still satisfies the strict growth inequality, since $\tanh (kh)<kh$; the limiting infinite-wavelength mode is marginal. A strict uniform long-wave margin requires $U>U_{\mathrm{crit}}$.

The Mazur theorem states that the weak closure and norm closure of a convex set $C$ in a normed vector space coincide:Norm closure is contained in weak closure because the weak topology is coarser. Conversely, if $x\notin {\overline{C}}^{\Vert \cdot \Vert }$, the Hahn-Banach separation theorem provides a continuous real linear functional strictly separating $x$ from that closed convex set. In a complex space this is the real part of a continuous complex linear functional. A weak neighborhood of $x$ then misses $C$, so $x\notin {\overline{C}}^w$. This proves the equality. It also gives the usual Mazur lemma: if $x_n\rightharpoonup x$, then $x$ lies in the norm closure of the convex hull of each tail, so one can choose tail convex combinations $y_n$ with $\Vert y_n-x\Vert <1/n$.

Now let $K$ be a weakly compact set in a normed space $X$. Each $\phi \in X^{\ast }$ is bounded on $K$ because it is weakly continuous. The family $\{Jx:x\in K\}$ in $\mathcal{B}(X^{\ast },\mathbb{F})$, where $J$ is the canonical embedding into the bidual, is therefore pointwise bounded. The space $X^{\ast }$ is Banach even if $X$ is not. Apply the Uniform boundedness principle and use $\Vert Jx\Vert =\Vert x\Vert$ to obtainThis proves that a weakly compact set is norm bounded without assuming completeness of the original space.

For the real-valued dual and integral formulas that follow, take $X$ to be real, as in the PDF. In a complex space the norming formula uses real parts, and the integral identities use complex-valued functionals instead.

If the separable Banach space $X$ is nonzero, choose a norm-dense sequence $(u_n)$ in its unit sphere. By the Hahn-Banach theorem, choose ${\phi }_n\in X^{\ast }$ with $\Vert {\phi }_n\Vert =1$ and ${\phi }_n(u_n)=1$. For any unit vector $u$, arbitrarily close $u_n$ satisfyScaling gives the countable norming family identityFor $X=\{0\}$ use the constant sequence of zero functionals. If $f:\Omega \to X$ is norm-Borel measurable, every ${\phi }_n\circ f$ is measurable, so its countable supremum $\Vert f(\cdot )\Vert$ is measurable. Equivalently, this also follows directly from continuity of the norm.

For any $\phi \in X^{\ast }$, continuity makes $\phi \circ f$ measurable, andThus the assumed integrability of the norm implies scalar integrability, anddefines a bounded linear functional on $X^{\ast }$. Use the granted weak-star continuity of $T_f$. By the continuous dual of a weak-star topology, $T_f$ is evaluation at a vector of $X$. Indeed, continuity gives finitely many $x_1,\dots ,x_m$ and $\epsilon >0$ such that $|T_f(\phi )|<1$ whenever $|\phi (x_j)|<\epsilon$ for all $j$. Scaling shows that $T_f$ vanishes on the common kernel of these evaluations. It therefore factors through their finite-dimensional coordinate map, so $T_f(\phi )={\sum }_j{a}_j\phi (x_j)=\phi ({\sum }_j{a}_j{x}_j)$. The Hahn-Banach theorem makes this representing vector unique. HenceIn this separable setting the vector is the Bochner integral.

Return to a weakly compact set $K\subseteq X$ and its inclusion $f:K\to X$. For each fixed $a\in X$, the identity $\Vert x-a\Vert ={\sup }_n{\phi }_n(x-a)$ makes $x\mapsto \Vert x-a\Vert$ weakly Borel measurable. Norm balls are consequently weakly Borel measurable. Separability gives a countable base of such balls, so every norm-open set is weakly Borel measurable. This proves measurability of $f$, and establishes the equality of the weak and norm Borel sigma-algebras in a separable Banach space.

Put $M={\sup }_{x\in K}\Vert x\Vert <\infty$. For every finite signed Borel measure $\mu$ on $K$,For positive measures this is the integral in the question. For signed measures, the correct integrability condition uses the variation measure; define the integral by taking the difference of the positive and negative integrals. The printed $\Omega$ in this clause should be $K$, the domain of the inclusion.

The Riesz-Markov-Kakutani representation theorem now defines the bounded linear mapFor each $\phi \in X^{\ast }$ the restriction $\phi {|}_K$ is in $C(K)$, andThe right side is weak-star continuous in $\mu$. The defining property of the weak topology therefore proves that $T$ is weak-star-to-weak continuous, for arbitrary nets. For a Dirac measure, $\boxed{T({\delta }_x)=x}$.

If $K\ne \varnothing$, let $\mathcal{P}(K)$ be its regular probability measures. This is a weak-star closed subset of $B_{C(K{)}^{\ast }}$: its conditions are $\mu (1)=1$ and $\mu (g)\ge 0$ for every nonnegative $g\in C(K)$. It is compact by Banach-Alaoglu theorem. Thus $T(\mathcal{P}(K))$ is weakly compact and convex, and contains $K$ because it contains all $T({\delta }_x)$. It is weakly closed, hence norm closed, so it contains $C={\overline{\mathrm{conv}K}}^{\Vert \cdot \Vert }$. By Mazur theorem, $C$ is weakly closed. Therefore it is a closed subset of the weakly compact set $T(\mathcal{P}(K))$, provingThe empty case is immediate. In fact $T(\mathcal{P}(K))=C$: a barycenter of a measure on a Banach space outside $C$ would be strictly separated by a functional $\phi$, contradicting $\phi (T(\mu ))={\int }_K\phi d\mu \le {\sup }_K\phi$.

The Cauchy integral formula says that if $f$ is holomorphic on a neighbourhood of the closed disc $\overline{D(z_0;R)}$, then for $z\in D(z_0;R)$,In particular, at the centre,for every $0<r<R$.

If $f(z_0)=0$, the assumed bound gives $|f(z)|\le 0$, so $f=0$. Otherwise define $g=f/f(z_0)$. Then $g(z_0)=1$ and $|g|\le 1$. Applying the centre formula and taking real parts givesThe continuous nonnegative function $1-\mathrm{Re}g$ has integral zero, so it vanishes around the circle. Equality $\mathrm{Re}g=1$ together with $|g|\le 1$ forces $g=1$. This holds for every $r<R$, hence $f(z)=f(z_0)$ throughout the disc. This is the equality case underlying the maximum modulus principle.

The Wirtinger derivative $u_z=(u_x-i{u}_y)/2$ is holomorphic, since ${\partial }_{\bar{z}}{u}_z=\Delta u/4=0$ by the Laplace equation. The conformal map $w=z^2$ takes the quadrant to the complex upper half-plane. Integrating the prescribed tangential boundary derivatives, choose a common corner value $c$ and define the real boundary functionThe decaying data are bounded under the usual regularity at the corner, so $f(s)=O(1+\sqrt{|s|})$. Thus its Poisson integral converges. A holomorphic function whose real part is that Poisson integral is supplied by the regularized Schwarz integral formula:For $u_{\ast }(z)=\mathrm{Re}W(z^2)$, the chain rule gives $(u_{\ast }{)}_z=z{W}^{\prime }(z^2)$. Apply integration by parts to obtainSubstituting $s=r^2$ and $s=-r^2$ gives the requested particular integral representation:Both integrals converge absolutely for an interior point of the quadrant. The Sokhotski–Plemelj formula gives $\mathrm{Re}(u_{\ast }{)}_z(x)=g_2(x)/2$ on the horizontal edge and $\mathrm{Im}(u_{\ast }{)}_z(iy)=-g_1(y)/2$ on the vertical edge, verifying the prescribed derivatives.

The printed conditions alone do not determine $u_z$ uniquely. The general answer adds a holomorphic function $H$ satisfyingEquivalently, write $H(z)=ih(z^2)/z$, where $h$ is holomorphic in the complex upper half-plane with real boundary values on the nonzero real axis. An antiderivative of $2H$ has a harmonic real part with zero tangential boundary derivatives. For example, $u=xy$ contributes $H=-iz/2$ without changing either datum. If the two edge constants differ by $c_2-c_1$, the angular harmonic function contributes $i(c_2-c_1)/(\pi z)$. The boxed expression selects the Poisson integral representative with equal edge constants; regularity and suitable growth conditions can be used to select that representative.

Apply the time Laplace transform to the Airy equation. Extend $g$ past $T$ in any suitable way and writeFor example, setting $g=0$ after $T$ makes $G(p)={\int }_0^T{e}^{-pt}g(t)dt$; the final solution for $t<T$ is independent of this extension. The Laplace transform of a derivative givesTake the principal cube root $\lambda =p^{1/3}$ for $\mathrm{Re}p>0$. The three characteristic roots of the spatial ordinary differential equation areSince $|\arg \lambda |<\pi /6$, only $-\lambda$ has negative real part. Thus spatial decay leaves just one homogeneous exponential, which the single prescribed Neumann boundary condition determines.

The Airy resolvent kernel on the whole real line isIt decays at both ends, is continuous together with its first derivative, and satisfies $R_p^{\prime \prime }(0+)-R_p^{\prime \prime }(0-)=1$. Consequently $({\partial }_s^3+p)R_p={\delta }_0$ as a distribution. A particular solution is the Green-function representationAdding the decaying homogeneous mode to impose the boundary derivative givesEvery term is known. If ${\tilde{u}}_0(r)={\int }_0^{\infty }{e}^{-ry}{u}_0(y)dy$ denotes the spatial Laplace transform, thenThe Bromwich inversion formula now gives the required integral representation:Here $\gamma >0$ is to the right of any singularities required by the growth of the data. The usual decay or growth hypotheses are understood for this Laplace transform construction; when absolute inversion is unavailable, the vertical integral is interpreted as the limit of truncated Bromwich contours. The transformed ordinary differential equation verifies the partial differential equation and initial condition, while differentiating at $x=0$ gives exactly $G(p)$ and hence $u_x(0,t)=g(t)$. The derivative compatibility $u_0^{\prime }(0)=g(0)$ makes the two data agree at the corner.

Boundedness of $(z_n)$ makes both real sequences $(x_n)$ and $(y_n)$ bounded. The Bolzano-Weierstrass theorem first gives a subsequence $(z_{n_j})$ for which $(x_{n_j})$ converges. Applying it again to $(y_{n_j})$ gives a subsubsequence for which the imaginary parts also converge. Since convergence of complex numbers is equivalent to convergence of their real parts and imaginary parts, this subsubsequence converges in $\mathbb{C}$.

The linearized equations are homogeneous and have constant coefficients in $x$ and $t$. Consequently, spatial Fourier modes evolve independently, and the superposition principle reconstructs arbitrary sufficiently regular initial disturbances from them. It is therefore enough to seek the normal modesHere the real $k$ is the wavenumber, $\sigma$ is the possibly complex temporal exponent, and taking the real part produces a physical displacement.

Suppose a $C^1$ classical solution existed and set $w=u+iv$. The system is preciselywhich are the Cauchy-Riemann equations for $w$ as a function of the complex number $z=x+it$. Hence $w$ is holomorphic near $z=0$. Every holomorphic function is real analytic, so its restrictionto the real axis is real analytic near zero. Its real part and imaginary part show that both $f$ and $g$ must be real analytic, contradicting the hypothesis. Therefore no such $C^1$ solution exists.

The principal cube root is $z^{1/3}=\exp (\mathrm{Log}z/3)$ using the principal complex logarithm. In the right half-plane, its complex argument lies between $-\pi /6$ and $\pi /6$. Consequently the roots of $r^3=-z$ consist of one root with negative real part, $-z^{1/3}$, and two with positive real part, $z^{1/3}{e}^{\pm i\pi /3}$.