I am going to provide an excerpt from a textbook and then a project to complete similar to it Here is the excerpt: Asolution of the basic equations of groundwater flow, not so trivial as the previous
 one, in which the flow rate was constant, is the solution of the problem of flow in
 a very deep deposit, bounded by the horizontal surface z = 0, with a separation
 of two regions above that surface by a thin vertical wall at the location x = 0, see
 Fig.8.5. The water level at the right side of the wall is supposed to be at a height H
 above ground surface, and the water level at the left side of the wall is supposed to
 coincide with the ground surface. Under the influence of this water level difference
 groundwater will flow under the wall, from right to left.
 The solution of this problem can be obtained using the theory of functions of
 a complex variable (Strack 1989). The actual solution procedure is not considered
 here. It is assumed, without any derivation, that in this case the groundwater head h
 is
 Fig. 8.5 Flow under a wall
 z
 H
 x
74
 h = H
 8 Groundwater Flow
 π arctan(z/x).
 (8.9)
 In order to apply this solution, it should be verified first that it is indeed the cor
rect solution. For this purpose it is sufficient to check that the solution satisfies the
 differential equation, and that it is in agreement with the boundary conditions.
 Thatthesolution(8.9)satisfiesthedifferential equation (8.3) caneasilybeverified
 by substituting the solution into the differential equation. To verify the boundary
 conditions the behavior of the solution for z ↑ 0 must be investigated. The value of
 z/x then will approach 0 from below if x > 0, and it will approach 0 from above if
 x < 0. Let it now be assumed that the range of the function arctan(u) is from 0 to
 π/2 if the argument u goes from 0 to ∞, and from π/2toπ if the argument u goes
 from −∞ to 0, see Fig.8.6. In that case it indeed follows that h = H if x > 0 and
 z ↑ 0, and that h = 0ifx < 0 and z ↑ 0. All this means that Eq.(8.9) is indeed the
 correct solution of the problem, as it satisfies all necessary conditions.
 Theverticalcomponentofthespecificdischargecanbeobtainedbydifferentiation
 of the solution (8.9) with respect to z. This gives
 qz =−kH
 π
 x
 x2 +z2
 .
 In particular, it follows that along the horizontal axis, where z = 0,
 z =0 : qz =−kH
 πx .
 (8.10)
 (8.11)
 If x > 0this is negative, so that the water flows in downward direction. This means
 that to the right of the wall the water flows in vertical direction into the soil, as was
 to be expected. If x < 0, that means to the left of the wall, the specific discharge qz
 is positive, i.e. the water flows in upward direction, as also was to be expected. Very
 close to the wall, i.e. for small values of x, the velocity will be very large. Locally
 that might result in erosion of the soil.
 arctan(u)
 π
 π
 2
 −5 −4 −3 −2 −1 0 1 2 3 4 5
 Fig. 8.6 Function arctan(u)
 u
8.3 Flow Under a Wall
 Fig. 8.7 Flow under a sluice
 75
 2a
 d
 It also follows from the solution, because arctan(∞) = π/2, that on the vertical
 axis, i.e. for x = 0,thegroundwaterheadish = H/2.Thatcouldhavebeenexpected,
 noting the symmetry of the problem.
 The total discharge from the reservoir at the right side of the wall, between the
 two points x = a and x = b (with b > a) can be found by integration of Eq.(8.11)
 from x = a to x = b.Theresult is
 Q = kHB
 π ln(b/a),
 (8.12)
 in which B is the thickness of the plane of flow, perpendicular to the figure. This
 formula indicates that the total discharge is infinitely large if b →∞or if a → 0.
 In reality such situations do not occur, fortunately.
 Equation (8.12) can be used to obtain a first estimate for the discharge under a
 hydraulic structure, such as a sluice, see Fig.8.7. If the length of the sluice is denoted
 by 2a, and the thickness of the layer is d, it can be assumed that the water to the
 left and to the right of the sluice will mostly flow into the soil and out of it over a
 distance approximately equal to d. The flow then is somewhat similar to the flow in
 the problem of Fig.8.5 between x = a and x = b = a +d.InFig.8.7 it seems that
 the values of a and d are approximately equal, so that ln(b/a) = 0.693. This gives
 Q =0.22kHBasafirst estimate for the total discharge.
 Example 8.1 The thickness of a clay layer is 8m, and its volumetric weight is
 18kN/m3. It is covered by a layer of very permeable sand, having a thickness of
 4m, a saturated volumetric weight of 20kN/m3, and a dry volumetric weight of
 16kN/m3. The phreatic surface coincides with the soil surface. However, in the
 thick sand layer directly below the clay layer the groundwater head is at a level of
 4mabovethesoilsurface.Calculatetheeffective stress in the center of the clay layer.
 Solution
 The total stress at a depth of 8m below the surface is σzz = 20 × 4 + 18 × 4 =
 152 kPa. At the top of the clay layer the groundwater head is h = 0, and at the
 bottom of the clay layer the groundwater head is h = 4m. This means that there is
 a uniform upward flow through the clay layer, so that the groundwater head at the
 center of the clay layer is h = 2m. Because the depth below the surface at that level
76
 8 Groundwater Flow
 is 8 m, the pore pressure at that level is p = 100 kPa. The effective stress then is
 σzz = σzz − p = 52kPa.
 Example 8.2 Calculate the effective stress in the center of the clay layer if the
 groundwater level in the upper sand layer is lowered to 2m below the soil surface.
 Solution
 If the groundwater table in the upper sand layer is lowered by 2 m, this layer now
 consists of 2m dry soil and 2msaturated soil. The total stress at a depth of 8m below
 the surface now is σzz = 16 × 2 + 20 × 2 +18×4 = 144 kPa. The groundwater
 head at the top of the clay layer now is h =−2 kPa, and at the bottom of the clay
 layer the groundwater head again is h = 4 h. This means that at the center of the
 clay layer the groundwater head is h = 1 kPa. The pore pressure at that level now is
 90 kPa. The effective stress then is σzz = σzz − p = 54 kPa.
 Example 8.3 Next calculate the effective stress in the center of the clay layer if a
 layer of concrete is constructed on the soil surface, leading to a load of 40 kPa.
 Solution
 The only difference with the previous situation is the additional 40 kPa, which
 increases the total stress below it. At the center of the clay layer the total stress
 now is 184 kPa. Because the pore pressure remains at the value p = 90 kPa. The
 effective stress then is σzz = σzz − p = 94 kPa.
 Problem 8.1 A clay layer has a thickness of 3m, and a volumetric weight of
 18kN/m3. Above the clay layer the soil consists of a sand layer, of thickness 3 m, a
 saturated volumetric weight of 20kN/m3,andadryvolumetricweightof16kN/m3.
 The groundwater level in the sand is at 1m below the soil surface. Below the clay
 layer, in another sand layer, the groundwater head is variable, due to a connection
 with a tidal river. What is the maximum head (above the soil surface) that may occur
 before the clay layer will fail? End of excerpt. Here is the problem: The report should be typed (no more than 3 pages, including any graphs but not including the
appendix) and should have

Problem statement with equations
Description of the algorithm used
Calculation of the total flow rate Q (m^2 /s) and the shape factor alpha
Plot of the variation of the specific discharge q (m/s) along the downstream side of the flow
domain. This plot should also show the specific discharge q (m/s) given in Section 8.3 of the
textbook “Soil Mechanics” by A. Verruijt for the similar problem in a semi-infinite sand layer.
Consider the problem of flow under a wall in a 10 m thick sand layer underlain by an
impermeable clay deposit. The water level at the right side of the wall is supposed to be at a
height 2 m above ground surface, and the water level at the left side of the wall is supposed to
coincide with the ground surface. Under the influence of this water level difference groundwater
will flow under the wall, from right to left. Two sheet piles, located at a distance of 10 m from
the wall constrained the flow to a rectangular 10 m x 20 m region. The hydraulic conductivity is
k = 0.01 cm/s. The problem is similar to the flow under a wall, described in Section 8.3 of the
textbook “Soil Mechanics” by A. Verruijt, except for the finiteness of the flow domain
Using a finite difference scheme to solve Laplace equation, calculate
the total flow rate Q (m^2/s),
the shape factor alpha ,
the specific discharge q (m/s) along the downstream side of the flow domain.
Use symmetry considerations to reduce the flow region to a square 10 m × 10 m domain. Use a  grid of 11 × 11 nodes. Solve this problem by writing a MATLab code.