Sapping Channels of Permanent Form
March 5, 2002
Be careful!! Liable to have typos and real errors!!
MOTIVATION
Channels with rounded heads that migrate upstream due to sapping seem to be
of interest here at MIT (Rothman’s class, Whipple’s class). Some of these
channels look as though, to a first approximation, they extend themselves
upstream while preserving shape. Thus is suggested the possibility of a sapping
channel of permanent form, i.e. one that migrates upstream without ever
changing shape. Such a form will not occur in nature. Under the right
conditions, however, nature might get pretty close to it.
THE IDEA
Consider the illustration of Figure 1. Groundwater
flow is from bottom to top. The sapping channel is
the bold indentation. It attracts groundwater to it.
The groundwater removes erodible material in a
permeable layer, eventually causing the less
permeable layer from above to collapse. The
rubble is periodically cleaned out by the no-longer
metaphorical gully washer.
As a result, the
boundary of the sapping channel retreats
upstream.
DEFINITIONS
Ld
bc
Lc
J
Lh
The streamwise coordinate in Figure 1 is y and the
y
transverse coordinate is x.
The upstream
boundary of the aquifer is at y = 0, where
x
piezometric head is given. The distance to the
Lb
channel head in the streamwise direction is L h. Figure 1
The channel head has length Lc and half-width bc.
The head joins a body consisting of parallel walls at junction point J. The
streamwise distance from the junction point to the end of the channel is L d. The
half-width of the domain is Lb. The problem is assumed to be symmetrical in x.
From x = 0 to x = bc the shape of the upstream edge of the sapping channel is
given as
y  Lh  y s ( x, t )
(1)
where by definition ys(bc, t) = Lc. As noted in Figure 2, the angle  of the edge of
the sapping channel is given as
tan  
y s
x
(2)
and the unit normal vector to the channel n̂ is
given as
 y s

,  1

x

n̂  (sin , cos )  
 y 
1  s 
 x 

(3)
n̂
Figure 2
The aquifer itself, as well as the impermeable layer above it, is assumed to have
slope S in the y direction, and no slope in the x direction, as illustrated in the slice
along y = 0 of Figure 3.
GROUNDWATER
FLOW
Let  be the
elevation to the
bottom of the
permeable layer,
and p denote the
groundwater
pressure in that
layer.
The
piezometric head
 is then given as
    hp
impermeable
retreat
permeable
x
S
seepage
Lh
Figure 3
hp 
p

(4a,b)
where  denotes the specific weight of water. Noting that by definition
S

y
the groundwater velocity vector (u, v) is given as
(5)
h 
 h
   
(u, v )  k ,   k p ,  S  p 
y 
 x y 
 x
(6)
where k denotes a permeability. The equation governing groundwater flow is the
Laplace equation,
 2hp  2hp
 2 0
x 2
y
(7)
The boundary conditions (7) are as follows.
hp ( x, 0)  hpo , 0  x  Lb
hp
(0, y )  0 , 0  y  Lh
x
hp
(Lb , y )  0 , 0  y  Lh  L c  L d
x
hp x, Lh  y s ( x, t )  0 , 0  x  bc
(8a-f)
hp ( bb , y )  0 , Lh  L c  y  Lh  L c  L d
hp ( x, Lh  L c  L d )  0 , bc  x  Lb
The groundwater velocities in the ampitheater where the flow daylights are given
as (ud, vd) where
h

 h

ud  (ud, v d )  k p x, Lh  y s ( x, t ), S  p x, Lh  y s ( x, t )
y
 x

(9)
The component of flow velocity at the point of daylighting that is normal to the
boundary is given as udn, where

udn  ud  n̂  k
hp
h
x,Lh  y s ( x, t ) y s  S  p x,Lh  y s ( x, t )
x
x
y
 y 
1  s 
 x 
2
(10)
RELATION FOR EVOLUTION OF THE CHANNEL BOUNDARY
It is assumed here that the channel migrates in the direction normal to its own
boundary, i.e. in the direction n̂ of Figure 2. In addition, the magnitude of this
migration rate n is assumed (simply enough) to be linearly proportional to the
magnitude of the excess of the normal daylighting groundwater velocity above
some critical value ucr , so that,
  und  ucr  , und  ucr
n  
0 , und  ucr

(11)
where  is a dimensionless constant.
MIGRATING CHANNEL OF PERMANENT FORM
A solution of channel evolution of permanent form must have the characteristic
that the problem becomes independent of time in a coordinate system migrating
upstream at a constant speed c. If the sapping channel is to migrate upstream
with constant migration speed c while maintaining permanent form, it is seen
from the geometry of Figure 4 that the following condition must be satisfied;
c cos   n
(12)
Reducing, the above relation yields


 hp x, L  y ( x, t ) dy s  S  hp x, L  y ( x, t )

h
s
h
s
 x

c
dx
y
  k
 ucr 
2
2


 dy 
 dy 
1  s 
1  s 


 dx 
 dx 


(13)
uJ
J n
 t
ct

Figure 4
at the junction point, or thus
Figure 5
One more condition must
be satisfied for a channel
of permanent form; the
side
walls from the
junction point J to the end
of the domain y = Lh + Lc +
Ld must not erode. This
condition can be expected
to be satisfied if no erosion
occurs at the junction point
itself.
From Figure 5,
then,
the
necessary
condition is that und  ucr
k
hp
h
bc ,Lh  Lc  dy s  S  p x,Lh  Lc 
x
dx
y
 dy 
1  s 
 dx 
2
 ucr
(14)
If we assume a smooth match of the sapping boundary at point J, the following
condition must be satisfied there;
cos  J 
1
 dy

1   s (bc , Lh  L c )
 dx

2
0
(15a)
or thus
dy s
(bc , Lh  L c )  
dx
(15b)
Between (14) and (15b) it is seen that the junction condition reduces to
k
hp
(bb , Lh  Lc )  ucr
x
(16)
It should be noted that strictly speaking, the upstream and downstream
boundaries in the domain of Figure 1 must also be migrating upstream at the
same rate at the channel in order to yield a true solution of permanent form. In
practice, the solution (if there is one) is likely to be accurate over a wide range of
channel positions and time intervals as an approximation even when the domain
boundaries are not migrating.
ANALYTICAL VERSUS NUMERICAL SOLUTION
This section is not very well thought out.
It is possible that an approximate analytical solution to the problem is possible. A
necessary condition for this is that the streamline shown in Figure 5 should
deviate only slightly from the streamwise (y) direction at the point of daylighting.
If this is the case, then the solution for hp(x, y) may deviate only slightly from a
base solution hpb that would prevail in the absence of any transverse (x) effects;
 

y
 , 0  x  bc
 hpo 1 
  Lh  y s ( x ) 
hpb ( x, y )  

y
h 1 
 , b  x  Lb
po 
  Lh  L c  L d  c
If this is the case than it may be that a perturbation solution of the following form
exists;
y s ( x )   2~
ys
~
hp  hpb  2hp
where  = bb/Lh. The junction condition may, however, render a perturbation
solution difficult.
In the event that an analytical solution is not possible, it should still be possible to
execute a numerical solution. To study this we first make the problem
dimensionless.
The following nondimensionalizations are employed;
x  bc x
y  Lh y
h
S  po Ŝ
bc
hp  hpo h
c  k
hpo
c
bc
ucr  k
hpo
ucr
bc
(17a-g)
y s  Lc y s
In addition, the following definitions is made;

bc
Lh

Lc
bc
(18)
Here it is implicitly assumed that the area of interest is one for which  << 1 and 
~ o(1).
Equations (7) and (8) now become
2
 2 hp
2  hp

0
x 2
y 2
(19)
hp ( x, 0)  1 , 0  x  Lb
 hp
(0, y )  0 , 0  y  1
x
 hp
( Lb , y )  0 , 0  y  1    Ld
x
hp x,1  y s ( x )  0 , 0  x  1
(20a-f)
hp ( 1, y )  0 , 1    y  1    Ld
hp ( x,1    Ld )  0 , 1  x  Lb
where
Ld 
Ld
Lh
Lb 
Lb
bc
(21a-c)
Here the expected range of interest is one such Ld ~ o(1) and Lb >> 1. In
addition, (13) becomes

c
 dy 
1 2  s 
 dx 
2

 hp
h
x,1  y s  dy s  S   p x,1  y s 
x
dx
y
 dy 
1 2  s 
 dx 
2
 ucr
(22)
and (16) becomes
 hp
(1, 1  )  ucr
x
(23)
A possible numerical scheme is as follows. Impose values of Ld , Lb and c .
Assume some form for y s . Guess values of  and , and dink around with the
solution of (19) and (20) until a solution for hp that satisfied (23) is obtained. Use
this solution to compute a better estimate of y s from (22). Then with this new
solution, dink around with  and  until (23) is satisfied and repeat the process.