Fokas method

The Fokas method uses a global relation for a linear boundary value problem between transformed boundary traces, complex spectral symmetries, and contour integration to eliminate unknown traces or construct a numerical boundary scheme. The distinction between the domains of analyticity of the transforms and the decay sectors of spectral exponentials controls valid contour deformations.

For $q_t+q_x-q_{xxx}=0$ on $x>0$, its local relation has flux $(1+k^2)q-ik{q}_x-q_{xx}$. With $Q_0(k)={\int }_0^{\infty }{e}^{-ikx}q(x,0)dx$ and $G_j(k,T)={\int }_0^T{e}^{w(k)s}{\partial }_x^{j}q(0,s)ds$, its global relation is $e^{wT}\hat{q}=Q_0+(1+k^2)G_0-ik{G}_1-G_2$. In $D_{+}=\{\mathrm{Im}k>0,\mathrm{Re}w(k)<0\}$, choose the other cubic root $\nu (k)$ in the lower half-plane. Then the known spectral density is $-Q_0(\nu )+(k^2-{\nu }^2)G_0-i(k-\nu )G_1$. The remaining final-state transform has zero integral around $\partial D_{+}$ by the Cauchy integral theorem and exponential decay. The full boundary includes the real axis and the hyperbola $\mathrm{Im}k=\sqrt{1+3(\mathrm{Re}k{)}^2}$, oriented with the domain on the left.

For a holomorphic function on $x>0$, $0<y<l$, define boundary spectra by counterclockwise traversal: ${\rho }_L={\int }_{il}^0{e}^{-ikw}f(w)dw$, ${\rho }_B={\int }_0^{\infty }{e}^{-ikx}f(x)dx$, and ${\rho }_T=-e^{kl}{\int }_0^{\infty }{e}^{-ikx}f(x+il)dx$. The displayed reconstruction follows from the Cauchy integral formula and the ray identity ${\int }_0^{e^{i\theta }\infty }{e}^{ik(z-w)}dk=i/(z-w)$ whenever that exponential decays. The three rays are selected by the inward normals of the three sides.

// Target: analysis.bigb

For $\lambda >0$, put $a=k-\lambda /k$, $b=k+\lambda /k$. The spectral integral uses the positive real and positive imaginary rays, both oriented outward. On the first ray its density is ${\rho }_x=i{\int }_0^{\infty }{e}^{-iax}[bq(x,0)-q_y(x,0)]dx$; on the second it is ${\rho }_y={\int }_0^{\infty }{e}^{by}[aq(0,y)-i{q}_x(0,y)]dy$. The global relation ${\rho }_x+{\rho }_y=0$ holds in the third spectral quadrant. Primitives from the corner and from each infinite boundary end give an additive Riemann-Hilbert problem. Their limits are $-q$ at infinity and $q$ at zero, which explains the factor $1/(4\pi i)$.

The reflected global relation expresses the unknown value trace in terms of the prescribed derivative trace. Multiplying the half-line drift null contour identity by $(ik+\alpha )/(ik)$ removes the reflected time-dependent transform. Its pole at $k=0$ lies below the contour and is not crossed. The formula concerns the derivative $q_x(0,t)$, rather than the outward derivative $-q_x(0,t)$.

Let $I_C[F]=(2\pi {)}^{-1}{\int }_C{e}^{ikx-(k^2-i\alpha k)t}F(k)dk$. The drift-diffusion half-line global relation and the half-line drift null contour identity give the formula in the secondary title. Its real-space kernel is the difference of two drifted Gaussian heat kernels, and its boundary-forcing kernel is $x(2\sqrt{\pi }{t}^{3/2}{)}^{-1}\exp [-(x+\alpha t{)}^2/(4t)]$. This kernel is an approximate identity in time as $x\downarrow 0$, which verifies the Dirichlet boundary condition.

A spectral-parameter-dependent divergence identity equivalent to a partial differential equation. Integrating it over a space-time region gives a global relation between transformed initial and boundary data. For the half-line free Schrödinger equation, take $P=e^{-ikx+i{k}^{2}t}q$ and $Q=e^{-ikx+i{k}^{2}t}(-kq+i{q}_x)$.

To eliminate an unknown boundary transform by a dispersion relation symmetry, evaluate the global relation at $\nu (k)$ only where the spatial transform remains analytic. Substitute the resulting trace identity into the contour representation. An unwanted transform vanishes only if its transformed argument and exponential admit a valid contour deformation and decay estimate. For constant drift, $\nu (k)=i\alpha -k$ maps the upper domain $\mathrm{Re}(k^2-i\alpha k)<0$ into the lower transform half-plane.

For the backward-sign Airy half-line global relation, suppose exactly one trace of derivative order $j\in \{0,1,2\}$ is unknown. Its spectral multiplier is a constant times $k^r$, $r=2-j$. Use the rotated null contours for half-range Fourier inversion with the displayed coefficients. The unknown contribution becomes an integral over the boundary of the middle upper sector $\pi /3<\arg k<2\pi /3$. There $\mathrm{Im}{k}^3<0$, sois bounded, while $e^{ikx}$ decays exponentially. The Cauchy integral theorem therefore removes that entire unknown contribution. The resulting solution depends on the initial transform and the two prescribed boundary transforms only.

// Target: analysis.bigb

For the linear dispersive Stokes equation, $\omega (\nu )=\omega (k)$ factors as $(\nu -k)({\nu }^2+k\nu +k^2-1)=0$. In $D_{+}=\{\mathrm{Im}k>0,\mathrm{Re}\omega (k)<0\}$ the two displayed other roots lie in the lower half-plane, so both can be used in the global relation. To check the root locations, at $k=ib$ they have imaginary part $-b/2$. A root could reach the real axis only where $\mathrm{Re}\omega (k)=0$, so continuity preserves their signs throughout this connected domain. The branch points $\pm 2/\sqrt{3}$ lie outside its closure. The weights $A_1=(k-{\nu }_2)/({\nu }_1-{\nu }_2)$ and $A_2=({\nu }_1-k)/({\nu }_1-{\nu }_2)$ obey $A_1+A_2=1$ and ${\nu }_j^{\prime }=-A_j$.

If $P(k)$ is analytic in the upper half-plane with suitable growth, paired integrals along the positive real and imaginary rays cancel when their exponential is $e^{ik(x+iy)}$ with $x,y>0$. For $0<y<\ell$, the analogous pair along the imaginary and negative real rays cancels with $e^{ik(x+i(y-\ell ))}$. These two applications of the Cauchy integral theorem eliminate reflected unknown transforms from a strip global relation; an individual ray integral need not vanish.