Absorbing boundary conditions (ABC)

Part 4 — Wave equations in PINN form

Learning objectives

Recognise why open-domain seismic problems need absorbing boundaries
State the Clayton-Engquist 1st-order ABC: u_t + c u_x = 0 at the right edge
Train PINNs with reflecting vs absorbing BC and watch the difference in residual energy
Understand why Clayton-Engquist works only for near-normal incidence

A real seismic survey lives on an open Earth. The wave that propagates outward from the source eventually leaves the region we care about and never returns. In a numerical simulation we have to truncate the domain to a finite box. Reflective Dirichlet BCs at the truncation cause the wave to bounce back and contaminate everything inside. To avoid this we use absorbing boundary conditions (ABCs) that let the wave pass through the boundary cleanly.

The Clayton-Engquist 1st-order ABC

Clayton & Engquist (1977 BSSA) and Engquist & Majda (1977 PNAS) derived a hierarchy of absorbing BCs from the one-way wave equation. The simplest (1st-order) form, on the right boundary $x = L$ , is

\frac{\partial u}{\partial t} + c \frac{\partial u}{\partial x} = 0 \qquad \textrm{at } x = L .

This is the rightgoing-wave equation: any function $f(x - c t)$ satisfies it identically. So a rightgoing pulse exits cleanly; a leftgoing pulse satisfies $u_t = c u_x \ne -c u_x$ and reflects.

The catch: the 1st-order Clayton-Engquist BC is exact only for waves at normal incidence (perpendicular to the boundary). For oblique waves the absorption is partial; reflection coefficients grow as $\sin^2 \theta$ where $\theta$ is the incidence angle. Higher-order Clayton-Engquist BCs (Higdon, 2nd-order) reduce this but never fully eliminate it. PML (§4.7) does.

1D domain $x \in [0, 2]$ , $t \in [0, 2]$ . A rightgoing Gaussian pulse starts at $x = 0.4$ :

IC: $u(x, 0) = \exp(-50(x - 0.4)^2)$ .
IC: $u_t(x, 0) = 100 c (x - 0.4) \exp(-50(x - 0.4)^2)$ — the rightgoing-wave velocity field.
Left BC: $u(0, t) = 0$ (closed wall, doesn't matter much because the pulse moves right).
Right BC: REFLECTING ( $u(2, t) = 0$ ) vs ABSORBING ( $u_t + u_x = 0$ at $x = 2$ ).

Race the two PINNs and compare the $t = 1.4$ snapshot and the residual energy $E(t) = \int |u|^2 dx$ .

What you should observe

At $t = 1.4$ , the rightgoing pulse has just hit the right wall. In REFLECTING mode it has bounced and is now moving leftward, visible in the snapshot. In ABSORBING mode it has cleanly left the domain — the snapshot is essentially zero except for a tiny residual at the boundary.
The residual energy curve is flat for REFLECTING (energy is conserved in the closed system) and decays for ABSORBING (energy leaves with the pulse).

For 2D and 3D problems

The 1st-order Clayton-Engquist BC generalises straightforwardly: at each open boundary, write the one-way wave equation in the outward-normal direction. For 2D acoustic at the right edge: $u_t + c \partial_x u = 0$ unchanged. The oblique-incidence problem persists. For modern 2D and 3D seismic forward modelling, PML (§4.7) is the default; Clayton-Engquist is sometimes used as a cheap fallback.

References

Clayton, R., Engquist, B. (1977). Absorbing boundary conditions for acoustic and elastic wave equations. BSSA 67(6), 1529–1540.
Engquist, B., Majda, A. (1977). Absorbing boundary conditions for the numerical simulation of waves. PNAS.
Higdon, R.L. (1986). Absorbing boundary conditions for difference approximations to the multidimensional wave equation. Math. Comp. 47.