r1 = -b + b2 + c ,
r2 = -b - b2 + c
z′(0) - r2 z0
C2 = -C1 + z0
r1 - r2
It can be seen from Eqs. 14 and 15 that this solution is valid for relatively short times after
t = 0 because the term proportional to exp(r1t) diverges for long elapsed time (r1 > 0).
Substituting Eq. 14 into Eq. 9 and integrating gives
x = 2 qR ( exp(r1t ) -1) + ( exp(r2t ) -1)
Various simplified explicit expressions can be obtained for engineering quantities of
interest through limiting forms of the analytic solution for relatively short elapsed times
after dredging, as given next.
For small t, Eq. 14 can be expanded to give (retaining leading order in t, ad, and ab),
( ad + ab ) q2 t 2
indicating that the channel starts filling linearly with time. If ab = 0 (no bedload transport),
then x = 0 for all time, and Eq. 14 reduces to,
z = z0 1 - exp
which indicates exponential filling of the channel.
Similarly, for small t, Eq. 17 yields,
ab ad 2 2
x = qt +
showing that the channel fills linearly manner with time by intrusion of the updrift side into
the channel and that deposition by suspended material is governed by a lower order
quadratic dependence in time for relatively short elapsed time after dredging.
Kraus and Larson