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For $u_t+u_{xxx}=0$ on $x>0$, define $V(x,p)={\int }_0^{\infty }{R}_p(x-y)u_0(y)dy$ and let $G$ be the time Laplace transform of $u_x(0,t)$. The transformed solution with spatial decay isOnly one characteristic root has negative real part, so one scalar boundary derivative fixes the remaining homogeneous mode. The Bromwich inversion formula supplies an integral representation involving only the initial and boundary data. For general linear differential equations on a half-line, the number and form of the necessary boundary data depend on the decaying spatial roots; Fokas and Wang study the corresponding boundary maps for linear dispersive equations.

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.