A free boundary problem arising in a kinematic wave model of channel flow with infiltration
Water flowing down a dry channel and infiltrating into the channel bed constitutes a free boundary problem. The free boundary is the time history of the water edge or front. In this paper we discuss a kinematic wave model of the problem. The problem is formulated in Sec. 1 and the results summarized in Sec. 2. In Sees. 3 and 4 the mathematical details are carried out, and in Sec. 5 a model using the continuity and momentum equations of hydraulics is discussed. 1. Formulation of the problem. The problem of irrigation has been studied for many years; we list, in the references, some of the more recent papers. We are dealing, essentially, with water flowing down a channel and infiltrating into the channel bed. The time history x = s(t) of the front of the water, i.e. the interface between the covered and uncovered part of the channel, is a free boundary which has to be determined along with the velocity u(x, t). This problem has been treated at various levels of mathematical complexity. We choose here a kinematic wave model that has been discussed in [1, 3, 7, 8, 9, 11]. Let x be the distance along a channel of uniform cross-section, u(x, t) the velocity, h(x, t) the depth, q(x, t) the lateral inflow rate, and f(x, t) the infiltration rate. The latter two are in volume per unit area per unit time. We can interpret q as rainfall. Further, let S be the slope of the channel, assumed constant, and Sf the friction slope. For Sf we take the Chezy formula Sf = vT/Ch, m and C positive constants. The continuity and momentum equations are [10, Chapter 11] h, + (uh)x = q — f u, + uux + ghx = g(S Sf) qu/h. (1) Conditions under which various terms in the momentum equation can be omitted have been discussed in the literature; we refer here to [4, 5, 6], The kinematic wave model is obtained by omitting all but the terms S — Ss. Thus we get u = ah1/m, v. = (SC)llm. (2) We assume now that q = 0 and that/is a function of t only. Then, writing n = 1 + m~x, we get, from (1) and (2), h, + (ah% = (3) * Received June 27, 1979; revised version received May 8, 1980. This research has been supported by the National Science Foundation through grant ENG78-25637.