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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.

For $k>0$, a nonzero spectral parameter for a linear boundary value problem givesThe identity $A^2+B^2=k^2$ proves that this solves the modified Helmholtz equation. The reciprocal companion is $v_{1/\lambda }=\overline{v_{\overline{\lambda }}}$.

Use the linear dispersive Stokes equation with time variable $t$. Put $E_k=e^{-ikx+\omega (k)t}$. Expanding the required local relation givesChoose $X=q_{xx}+ik{q}_x+(1-k^2)q$. Direct differentiation then yieldsThe unwanted coefficient of $q$ cancels precisely when $\omega (k)=ik(1-k^2)$. HenceWith these choices, the local relation is exactly $E_k(q_t+q_x+q_{xxx})=0$ for every complex spectral parameter for a linear boundary value problem.

Use the spectral parameter for a linear boundary value problem $k\in \mathbb{C}$ and define the dispersion relation $\omega (k)=k^2-i\alpha k$. Direct differentiation gives the divergence formIndeed the coefficient of $u$ left over after expansion is $\omega -k^2+i\alpha k=0$, and the remaining factor is $u_t-u_{xx}-\alpha u_x$. Thus this is equivalent to the advection-diffusion equation for every $k$.

Introduce the Half-range Fourier transforms and finite-time spectral boundary transformswhere $g_1(t)=u_x(0,t)$ is the unknown normal derivative with the positive-$x$ convention. The outward normal at zero instead gives $-g_1$. The spatial transforms are analytic for $\mathrm{Im}k<0$ and continuous on the real axis under the stated decay assumptions. Integrating the divergence form on $(0,\infty )\times (0,t)$ gives the global relationThe sign follows from the lower spatial endpoint: the integrated spatial derivative is minus its value at zero. This sign will determine the boundary-forcing term in the solution.

Write $z=x+iy$ and $\overline{z}=x-iy$. A convenient spectral parameter for a linear boundary value problem is any $\lambda \in \mathbb{C}\setminus \{0\}$. Define the modified Helmholtz adjoint plane wavewhereIndeed $A^2+B^2=k^2$, so $(\Delta -k^2)v_{\lambda }=0$. Equivalently, $4{\partial }_z{\partial }_{\overline{z}}{v}_{\lambda }=k^2{v}_{\lambda }$. The exclusion of zero is only the coordinate singularity of this spectral parametrization; it still supplies a whole complex one-parameter family. The conjugate companion isIt also solves the adjoint equation.