Global relation for a linear boundary value problem ($\mathcal{G}(\lambda )=0$)

A global relation is an identity connecting transforms or integrals of the boundary traces of a solution. For a formally self-adjoint operator $L=\Delta -k^2$, Green second identity givesfor every homogeneous adjoint solution $Lv=0$. Exponential adjoint families turn this into a spectral identity between the Dirichlet boundary condition and unknown normal derivative.

Let $F(k)={\int }_0^{\infty }{e}^{-ikx}{q}_0(x)dx$ and $G_0(k,t)={\int }_0^t{e}^{w(k)s}{g}_0(s)ds$. Using the roots of cubic dispersion symmetry with negative transport, put $\mathcal{T}F=[(k-{\nu }_2)F({\nu }_1)+({\nu }_1-k)F({\nu }_2)]/({\nu }_1-{\nu }_2)$. The Fokas method formula is an initial real-line integral minus an integral over $\partial D_{+}$ of $e^{ikx-wt}[\mathcal{T}F+(3k^2+1)G_0]$, both divided by $2\pi$. The boundary is oriented with $D_{+}$ on the left. Its unknown final-state transform has zero integral by analyticity and upper-half-plane decay.

For the Airy equation $q_t+q_{xxx}=0$ on $0<x<L$, define $F_j={\int }_0^t{e}^{-i{k}^{3}s}{\partial }_x^{j}q(0,s)ds$ and $G_j$ analogously at $L$. Set $\mathcal{F}=k^2{F}_0-ik{F}_1-F_2$ and $\mathcal{G}=k^2{G}_0-ik{G}_1-G_2$. Integrating the local relation with flux $-q_{xx}-ik{q}_x+k^{2}q$ proves the displayed global relation. The dispersion relation is invariant under multiplication of $k$ by a cube root of unity; the resulting three equations determine the three missing boundary transforms for data $q(0,t),q(L,t),q_x(L,t)$.

Here $\alpha =e^{2\pi i/3}$. The missing-trace matrix has determinant $-\sqrt{3}k\Delta (k)$. Its nonzero zeros are the eigenvalues in the parametrization $Av=i{k}^{3}v$ of $A=-d^3/d{x}^3$ with $v(0)=v(L)=v^{\prime }(L)=0$. Integration by parts gives $\mathrm{Re}⟨v,Av⟩=-|v^{\prime }(0){|}^2/2$. No nonzero zero lies in $\mathrm{Re}(-i{k}^3)\le 0$: on its boundary the determinant equation reduces, after rotation, to an equality between a unit-modulus exponential and $\cosh u-i\sqrt{3}\sinh u$, whose modulus exceeds one unless $u=0$. The zero at the origin is removable in the original boundary problem. There are nevertheless infinitely many genuine negative eigenvalues of $A$, obtained from $e^{-3s/2}=2\cos (\sqrt{3}s/2+\pi /3)$, $s>0$, with $k=-is/L$. Absence of contour poles in the Fokas method does not imply absence of discrete spectrum of the operator.

Let $q$ be a real harmonic function in a semistrip, $g_0(x)=q(x,0)$, $g_l(x)=q(x,l)$ and $h(y)=q(0,y)$. Let $G_j,N_j$ be the negative-exponential Half-range Fourier transforms of $g_j$ and $q_y(x,j)$, and put $H(k)={\int }_0^l{e}^{ky}h(y)dy$, $W(k)={\int }_0^l{e}^{ky}{q}_x(0,y)dy$. The closed differential form $e^{-ikz}{q}_{z}dz$ has zero boundary integral. Integration by parts of tangential derivatives cancels all corner values and gives the displayed global relation, for $\mathrm{Im}k\le 0$.

// Target: analysis.bigb

For $q_t=q_{xx}+\alpha q_x$ on $x>0$, write $Q(k)={\int }_0^{\infty }{e}^{-ikx}q(x,0)dx$ and $G_j(k,t)={\int }_0^t{e}^{\omega (k)s}{\partial }_x^{j}q(0,s)ds$. Twice integrating by parts gives $e^{\omega (k)t}\hat{q}(k,t)=Q(k)-G_1(k,t)-(ik+\alpha )G_0(k,t)$ in the lower transform half-plane. The dispersion relation is invariant under $k\mapsto i\alpha -k$. This global relation eliminates one unknown boundary trace after the symmetry maps the inversion contour into the transform's analytic domain.

For the linear dispersive Stokes equation $q_t+q_x+q_{xxx}=0$, let $\omega (k)=i(k-k^3)$. Its local relation is obtained with $X=q_{xx}+ik{q}_x+(1-k^2)q$. Integrating ${\partial }_t(e^{-ikx+\omega t}q)+{\partial }_x(e^{-ikx+\omega t}X)=0$ over the half-line gives the displayed global relation, with negative-exponential Half-range Fourier transforms and $G_j(k,t)={\int }_0^t{e}^{\omega (k)s}{\partial }_x^{j}q(0,s)ds$. The spatial transforms are holomorphic for $\mathrm{Im}k<0$, whereas the finite-time boundary transforms are entire functions.

Let $A_j$ be the weights of the cubic Stokes dispersion symmetry and set $F(k)={\sum }_j{A}_j(k){\hat{q}}_0({\nu }_j(k))$. Orient $\partial D_{+}$ with $D_{+}$ on its left. Evaluating the global relation at its two lower-half-plane roots and interpolating the linear expression $ik{G}_1+G_2$ removes both unknown boundary traces. The resulting Fokas method formula isThe discarded transform of the solution is holomorphic in $D_{+}$ and has zero integral after closing its contour for $x>0$. Initial compatibility and sufficient smoothness and spatial decay are required. At the boundary, use a fixed time horizon later than $t$ before Fourier inversion, to avoid an endpoint half-value.

Orient the upper-quadrant contour from $i\infty$ down to zero and then to $+\infty$. The reflected half-line transform has zero inverse contribution for positive $x$. Adding $c$ times this contribution, and deforming boundary integrands through the second quadrant where $|e^{ikx-i{k}^2(t-\tau )}|\le e^{-x\mathrm{Im}k}$, gives boundary coefficients $(1-c)k{\tilde{g}}_0-i(1+c){\tilde{g}}_1$. Thus $c=-1$ removes the unknown Neumann trace for a prescribed Dirichlet value, while $c=1$ removes the unknown Dirichlet trace for a prescribed Neumann value. The initial reflected term has coefficient $c$ and cannot be discarded after the time-dependent spectral decomposition.

A boundary trace transformed over a finite time interval using the dispersion relation of a linear partial differential equation. For integrable trace data and a polynomial dispersion function, it is an entire function of the complex spectral parameter. Its growth depends on $\mathrm{Re}\omega (k)$; multiplying by $e^{-\omega (k)t}$ changes the integrand to the causal factor $e^{-\omega (k)(t-s)}$.

In a Fokas method boundary integral over $\partial D_{+}$, where $\mathrm{Re}\omega <0$ inside, the contribution of replacing $G(k,t)$ by $G(k,T)$ is an integral of $e^{ikx}{\int }_t^T{e}^{\omega (k)(s-t)}g(s)ds$ times an holomorphic coefficient. For $x>0$ it vanishes by the Cauchy integral theorem and the exponential decay in the upper domain. The transformed amplitude is then independent of the observation time, which simplifies differentiation of the partial differential equation. Temporal inversion at $0<t<T$ recovers the full boundary value rather than the half-value of a truncated transform at its endpoint.

For $u_{z\bar{z}}={\beta }^{2}u$, pull back its spectral closed differential one-form along each straight side $z=m_j+s{h}_j$. Under counterclockwise traversal and outward normal derivatives $q_j$, the side density is $i{e}^{-i\beta (\lambda z-\bar{z}/\lambda )}[|h_j|q_j+\beta (\lambda h_j+{\bar{h}}_j/\lambda )g_j]$, where $g_j$ is the Dirichlet boundary data. The Generalized Stokes theorem gives the displayed global relation. Clockwise traversal reverses the normal-derivative coefficient; all orientations must be changed consistently.

For equal oriented Dirichlet boundary data on all three sides and a uniquely solvable elliptic problem, rotation invariance makes their outward normal derivatives equal. Pairing the global relation with its Schwarz conjugation of a spectral function at $\ell (k+\lambda /k)=2\pi in$ determines every Fourier coefficient of the common normal trace. The zero mode at $\lambda =0$ follows from zero total flux. At negative parameters, uniqueness and possible removable spectral denominators must be checked rather than assumed.

For the advection-diffusion equation $u_t=u_{xx}+\alpha u_x$ on $x>0$, the spectral divergence form uses $\omega =k^2-i\alpha k$ and factor $e^{-ikx+\omega t}$. With negative-exponential Half-range Fourier transforms, integration gives the displayed global relation for $\mathrm{Im}k\le 0$. The boundary transform $G_1$ refers to $u_x(0,t)$, the negative of the outward normal derivative.

For $q_t=q_{xx}+\alpha q_x$ with $\alpha >0$, the dispersion relation is $\omega =k^2-i\alpha k$ and its symmetry is $\nu =i\alpha -k$. On the upper domain $\mathrm{Re}\omega <0$, $\mathrm{Im}k>\alpha$ and therefore $\mathrm{Im}\nu <0$. After removing the unknown trace through the global relation, deform the boundary contour to the displayed two rays, with $0<\theta <\pi /4$, oriented from left infinity to $i\alpha$ to right infinity. The factor $e^{ikx}$ decays like $e^{-rx\sin \theta }$, while $e^{-\omega t}$ has Gaussian decay for large $r$. Apparent poles introduced by splitting a finite-time forcing transform must be treated as removable singularities of the unsplit expression.

These are chosen complex values of a spectral parameter for a linear boundary value problem at which a truncated global relation for a linear boundary value problem is enforced. Their directions and magnitudes determine which boundary side and which tangential mode are tested. Suitable row combinations can give much better conditioning than arbitrary raw samples.

The two modified Helmholtz adjoint plane waves $v_{\lambda }$ and ${\tilde{v}}_{\lambda }=\overline{v_{\overline{\lambda }}}$ give boundary identities $\mathcal{G}(\lambda )=0$ and $\tilde{\mathcal{G}}(\lambda )=0$. For real solution traces, $\tilde{\mathcal{G}}(\lambda )=\overline{\mathcal{G}(\overline{\lambda })}$. Reality is needed for this conjugation argument; the parametrization separately yields the reciprocal identity $\tilde{\mathcal{G}}(\lambda )=\mathcal{G}(1/\lambda )$.

For the modified Helmholtz equation $\Delta q-4\lambda q=0$ with identical oriented outward Neumann boundary data on all sides of an equilateral triangle, uniqueness forces rotation invariance. Put $\chi (k)=l(k+\lambda /k)/2$, $\Psi (k)={\int }_{-l/2}^{l/2}{e}^{(k+\lambda /k)s}f(s)ds$, and $a=e^{2\pi i/3}$. At $\chi (k_n)=i\pi n$, paired global relations giveFor $\lambda >0$ one may choose $k_n=i[\pi n/l+\sqrt{(\pi n/l{)}^2+\lambda }]$; both denominator factors are nonzero. The derivation retains the tangential-derivative term in each side spectrum. Its endpoint contribution vanishes at these samples because rotation invariance makes all vertex values equal. For $\lambda =0$ the data must have zero total flux and an additive constant remains; negative parameters at Neumann Laplacian eigenvalues require the Fredholm solvability condition and allow homogeneous additions.

On the square $[-1,1{]}^2$, write normal traces in ${\varphi }_n(s)=\sin [n\pi (s+1)/2]$. Set $p_n=n\pi /2$, ${\omega }_n=\sqrt{k^2+p_n^2}$, $r_n=({\omega }_n+p_n)/k$. Paired spectral samples $\{r_n,r_n^{-1}\}$, $\{-r_n,-r_n^{-1}\}$, $\{i{r}_n,i/r_n\}$, $\{-i{r}_n,-i/r_n\}$ give adjoint tests $e^{\mp {\omega }_{n}y}{\varphi }_n(x)$ and $e^{\mp {\omega }_{n}x}{\varphi }_n(y)$. Each test vanishes on the adjacent sides, removing their unknown normal derivatives.

Let $R_{jn}={\int }_{\partial \Omega }f{\partial }_n{v}_{jn}ds$ for a square sine adjoint test, and put ${\rho }_{jn}=e^{-{\omega }_n}{R}_{jn}$, ${\delta }_n=e^{-2{\omega }_n}$. The opposite-side normal-trace coefficients obeywith the same formulas for $L,R$. When the known Dirichlet boundary condition is integrated exactly, these are exact coefficients, since orthogonality eliminates all other unknown modes.

For sine collocation of square modified Helmholtz global relations, each opposite-side mode has a scaled matrixThe assembled system has strict diagonal dominance, and its spectral condition number of a positive-definite matrix is below $(1+e^{-\pi })/(1-e^{-\pi })<1.091$. The conclusion concerns the explicit paired sine rows, not arbitrary uncombined complex rows.