1. x y ocean 2L L Island Recharge Problem ocean well R= 0.00305 ft/d T= 10,000 ft 2 /day

2. ocean R x = 0x = Lx = - L 1D approximation used by C.E. Jacob h(L) = 0 at x =0

3. h(x) = R (L 2 – x 2 ) / 2T Analytical solution for 1D “confined” version of the problem C.E. Jacob’s Model h(L) = 0 at x =0 Governing Eqn. Boundary conditions R = (2 T) h(x) / ( L 2 – x 2 ) Forward solution Inverse solution for R

4. R = (2 T) h(x) / ( L 2 – x 2 ) Inverse solution for R Solve for R with h(x) = h(0) = 20 ft. Re-arrange eqn to solve for T, given value for R and h(0) = 20 ft. Inverse solution for T L You will find that R  0.00305 ft/d

5. Head measured in an observation well is known as a target. Targets used in Model Calibration The simulated head at the node representing the observation well is compared with the measured head. During model calibration, parameter values (e.g., R and T) are adjusted until the simulated head matches the observed value. Model calibration solves the inverse problem.

6. x y ocean Solve the forward problem: R= 0.00305 ft/d T= 10,000 ft 2 /day 2L L = 12,000 ft Island Recharge Problem ocean well

7. Water Balance IN = Out IN = R x AREA Out = outflow to the ocean

8. Top 4 rows Red dots represent specified head cells, which are treated as inactive nodes. Note that the well node (not shown in this figure) is an active node. Head at a node is the average head in the area surrounding the node.

9. Top 4 rows IN =R x Area = R (L-  x/2) (2L -  y/2) 2L L Also: IN = R (2.5)(5.5)(a 2 )

10.  x/2 Top 4 rows OUT =  Qy +  Qx Qy = K (  x b) (  h/  y) or Qy = T  h Qx = K (  y b)(  h/  x) or Qx = T  h Qx Qy

11. Well  y/2 Bottom 4 rows

13. b h ocean groundwater divide R x = 0x = Lx = - L datum Unconfined version of the Island Recharge Problem Let b = 100 ft & K = 100 ft/d so that at x=L, Kb= 10,000 ft 2 /day

14. Let v = h 2